SPOJ: LPS – Longest Palindromic Substring

0

Problem Link : http://www.spoj.com/problems/LPS/


Solution Idea:
——————
This one is the straight forward implementation of manachar’s algorithm. You can learn about manachar algorithm form the following links-
1. Hackerearth’s Note
2. Hackerank Topic
3. Youtube


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define LINE_PRINT_CASE  printf("Case %d:\n",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

#define mx 600005

char s1[mx];
int p[mx];
int n;

string process(string &str)
{
    string ret;
    ret+='$';
    ret+='#';
    for(int i=0;i<SZ(str);i++)
    {
       ret+=str[i];
       ret+='#';
    }
    ret+='@';
    return ret;
//    cout<<str<<endl;
}

int manacher(string &str)
{
    int c=0,r=0;
    for(int i=1;i<SZ(str)-1;i++)
    {
        int miror=c+c-i;
        if(i<r)
            p[i]=min(p[miror],r-i);
        while(str[i+1+p[i]]==str[i-1-p[i]])
            p[i]++;
        if(i+p[i]>r)
        {
            c=i;
            r=i+p[i];
        }
    }
    int ret=0;
    for(int i=0;i<SZ(str);i++)
        ret=max(ret,p[i]);
    return ret;
}

int main()
{

//    freopen("in.txt","r",stdin);
//	  freopen("out.txt","w",stdout);

      string str;

      scanf("%d",&n);
      scanf("%s",s1);

      str=s1;
      str=process(str);
      int ans=manacher(str);
      printf("%d\n",ans);

    return 0;
}



SPOJ: LCS – Longest Common Substring

0

Problem Link : http://www.spoj.com/problems/LCS/


Solution Idea:

This is the problem no 6 on Suffix array paper .
The question need to find the maximum value of LCP[] array between two consecutive suffixes where this two consecutive suffixes are from different input string.


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define LINE_PRINT_CASE  printf("Case %d:\n",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

#define mx 500005

int SA[mx],tempSA[mx];
int RA[mx],tempRA[mx];
int c[mx];
int Phi[mx],PLCP[mx],LCP[mx];
char str[mx],str1[mx];
int n;

void counting_sort(int k)
{
    int maxi=max(n,300);
    ms(c,0);
    for(int i=0; i<n; i++)
    {
        int a=i+k<n?RA[i+k]:0;
        c[a]++;
    }
    for(int i=0,sum=0; i<maxi; i++)
    {
        int x=c[i];
        c[i]=sum;
        sum+=x;
    }

    for(int i=0; i<n; i++)
    {
        int a=SA[i]+k<n?RA[SA[i]+k]:0;
        int b=c[a];
        c[a]++;
        tempSA[b]=SA[i];
    }

    for(int i=0; i<n; i++)
        SA[i]=tempSA[i];
}

void build_SA()
{
    for(int i=0; i<n; i++)
    {
        RA[i]=str[i];
        SA[i]=i;
    }

    for(int k=1; k<n; k*=2)
    {
        counting_sort(k);
        counting_sort(0);

        int r=0;
        tempRA[SA[0]]=r=0;

        for(int i=1; i<n; i++)
        {
            if(RA[SA[i]]==RA[SA[i-1]] && RA[SA[i]+k]==RA[SA[i-1]+k])
                tempRA[SA[i]]=r;
            else
                tempRA[SA[i]]=++r;
        }

        for(int i=0; i<n; i++)
            RA[i]=tempRA[i];

        if(RA[SA[n-1]]==n-1) break;
    }
}

void build_LCP()
{
    Phi[SA[0]]=-1;
    for(int i=1; i<n; i++)
        Phi[SA[i]]=SA[i-1];
    for(int i=0,L=0; i<n; i++)
    {
        if(Phi[i]==-1)
        {
            PLCP[i]=0;
            continue;
        }
        while(str[i+L]==str[Phi[i]+L]) L++;
        PLCP[i]=L;
        L=max(L-1,0);
    }

    for(int i=0; i<n; i++)
        LCP[i]=PLCP[SA[i]];

}

int main()
{

//    freopen("in.txt","r",stdin);
//	  freopen("out.txt","w",stdout);

    int pos;

    scanf("%s %s",str,str1);
    strcat(str,"$");
    pos=strlen(str)-1;
    strcat(str,str1);
    strcat(str,"#");
    n=strlen(str);
    build_SA();
    build_LCP();
//    string ss=str;
//    for(int i=0;i<n;i++)
//        cout<<SA[i]<<" ";
//    cout<<endl;
//    for(int i=0;i<n;i++)
//        cout<<LCP[i]<<" ";
//    cout<<endl;
//    for(int i=0;i<n;i++)
//        cout<<ss.substr(SA[i])<<endl;

    int ans=0;

    for(int i=1; i<n; i++)
    {
        if((SA[i]<pos && (SA[i-1]>pos)) || ((SA[i]>pos && SA[i-1]<pos)))
        {
            ans=max(ans,LCP[i]);
        }
    }

    pf("%d\n",ans);

    return 0;
}


SPOJ: SUBLEX – Lexicographical Substring Search

0

Problem Link : http://www.spoj.com/problems/SUBLEX/


Solution Idea:

My solution for this problem is using suffix array and LCP array.
At first Build Suffix array and LCP array. Then find from which position of the suffix array you can get the K’th substring. Determine the position and determine the length for which it is the K’th smallest substring. Then print it.

We can get number of distinct substring using Suffix Array and LCP array. If you don’t know it how to do this job then solve this problem first.
SPOJ: DISUBSTR – Distinct Substrings


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define LINE_PRINT_CASE  printf("Case %d:\n",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

#define mx 100005

int SA[mx],tempSA[mx];
int RA[mx],tempRA[mx];
int c[mx];
int LCP[mx],PLCP[mx],Phi[mx];
char str[mx];
int n;

void counting_sort(int k)
{
    int maxi=max(300,n);
    ms(c,0);
    for(int i=0; i<n; i++)
    {
        int a=i+k<n?RA[i+k]:0;
        c[a]++;
    }
    for(int i=0,sum=0; i<maxi; i++)
    {
        int x=c[i];
        c[i]=sum;
        sum+=x;
    }

    for(int i=0; i<n; i++)
    {
        int a=SA[i]+k<n?RA[SA[i]+k]:0;
        int b=c[a];
        c[a]++;
        tempSA[b]=SA[i];
    }

    for(int i=0; i<n; i++)
        SA[i]=tempSA[i];
}

void build_Suffix_Array()
{
    for(int i=0; i<n; i++)
    {
        RA[i]=str[i];
        SA[i]=i;
    }

    for(int k=1; k<n; k*=2)
    {
        counting_sort(k);
        counting_sort(0);
        int r=0;
        tempRA[SA[0]]=r=0;
        for(int i=1; i<n; i++)
        {
            if(RA[SA[i]]==RA[SA[i-1]] && RA[SA[i]+k]==RA[SA[i-1]+k])
                tempRA[SA[i]]=r;
            else
                tempRA[SA[i]]=++r;
        }
        for(int i=0; i<n; i++)
        {
            RA[i]=tempRA[i];
        }
        if(RA[SA[n-1]]==n-1) break;
    }
}

void build_LCP()
{
    Phi[SA[0]]=-1;
    for(int i=1; i<n; i++)
        Phi[SA[i]]=SA[i-1];
    for(int i=0,L=0; i<n; i++)
    {
//        D(i);
        if(Phi[i]==-1)
        {
            PLCP[i]=0;
            continue;
        }
        while(str[i+L]==str[Phi[i]+L]) L++;
        PLCP[i]=L;
        L=max(L-1,0);
    }

    for(int i=0; i<n; i++)
        LCP[i]=PLCP[SA[i]];
}



int main()
{

//    freopen("in.txt","r",stdin);
//	  freopen("out.txt","w",stdout);

//    int t;
//    sf(t);
//    TEST_CASE(t)
    {
        scanf("%s",str);
        string str1=str;
        strcat(str,"$");
        n=strlen(str);
        build_Suffix_Array();
        build_LCP();

//        for(int i=0; i<n; i++)
//            cout<<SA[i]<<" ";
//        cout<<endl;
//        for(int i=0; i<n; i++)
//            cout<<LCP[i]<<" ";
//        cout<<endl;


        int string_len=SZ(str1);

        int q;
        sf(q);
        while(q--)
        {
            ll k;
            sfl(k);
            ll ans=0;
            int pos=0,len=0;
            for(int i=1;i<n;i++)
            {
                int temp=string_len-SA[i]-LCP[i];
                if(ans+temp<k)
                    ans+=temp;
                else
                {
                    pos=SA[i];
                    len=ans+temp-k;
                    len=LCP[i]+temp-len;
                    break;
                }
            }
            pf("%s\n",str1.substr(pos,len).c_str());
        }

    }

    return 0;
}

SPOJ: DISUBSTR – Distinct Substrings

1

Problem Link : http://www.spoj.com/problems/DISUBSTR/en/


Solution Idea: Build Suffix Array and LCP array. Then the answer will be subtraction of length of the ith suffix and LCP[i].

Answer = Li-LCP[i]

This works because LCP[i] holds the length of the longest common prefix of Suffix SA[i] and SA[i-1] where SA[] is suffix array.

Don’t forget to add an smaller extra or invalid symbol at the end of the string after taking input. Here I add a ‘$’.


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define LINE_PRINT_CASE  printf("Case %d:\n",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

#define mx 1005

int SA[mx],tempSA[mx];
int RA[mx],tempRA[mx];
int c[mx];
int LCP[mx],PLCP[mx],Phi[mx];
char str[mx];
int n;

void counting_sort(int k)
{
    int maxi=max(300,n);
    ms(c,0);
    for(int i=0; i<n; i++)
    {
        int a=i+k<n?RA[i+k]:0;
        c[a]++;
    }
    for(int i=0,sum=0; i<maxi; i++)
    {
        int x=c[i];
        c[i]=sum;
        sum+=x;
    }

    for(int i=0; i<n; i++)
    {
        int a=SA[i]+k<n?RA[SA[i]+k]:0;
        int b=c[a];
        c[a]++;
        tempSA[b]=SA[i];
    }

    for(int i=0; i<n; i++)
        SA[i]=tempSA[i];
}

void build_Suffix_Array()
{
    for(int i=0; i<n; i++)
    {
        RA[i]=str[i];
        SA[i]=i;
    }

    for(int k=1; k<n; k*=2)
    {
        counting_sort(k);
        counting_sort(0);
        int r=0;
        tempRA[SA[0]]=r=0;
        for(int i=1; i<n; i++)
        {
            if(RA[SA[i]]==RA[SA[i-1]] && RA[SA[i]+k]==RA[SA[i-1]+k])
                tempRA[SA[i]]=r;
            else
                tempRA[SA[i]]=++r;
        }
        for(int i=0; i<n; i++)
        {
            RA[i]=tempRA[i];
        }
        if(RA[SA[n-1]]==n-1) break;
    }
}

void build_LCP()
{
    Phi[SA[0]]=-1;
    for(int i=1; i<n; i++)
        Phi[SA[i]]=SA[i-1];
    for(int i=0,L=0; i<n; i++)
    {
//        D(i);
        if(Phi[i]==-1)
        {
            PLCP[i]=0;
            continue;
        }
        while(str[i+L]==str[Phi[i]+L]) L++;
        PLCP[i]=L;
        L=max(L-1,0);
    }

    for(int i=0; i<n; i++)
        LCP[i]=PLCP[SA[i]];
}



int main()
{

//    freopen("in.txt","r",stdin);
//	  freopen("out.txt","w",stdout);

    int t;
    sf(t);
    TEST_CASE(t)
    {
        scanf("%s",str);
        strcat(str,"$");
        n=strlen(str);
        build_Suffix_Array();
        build_LCP();

//        for(int i=0; i<n; i++)
//            cout<<SA[i]<<" ";
//        cout<<endl;
//        for(int i=0; i<n; i++)
//            cout<<LCP[i]<<" ";
//        cout<<endl;

        int ans=0;

        int string_len=n-1;

        for(int i=1; i<n; i++)
        {
            ans+=(string_len-SA[i]-LCP[i]);
        }

        pf("%d\n",ans);

    }

    return 0;
}


SPOJ: BEADS – Glass Beads

0

Problem Link : http://www.spoj.com/problems/BEADS/


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define LINE_PRINT_CASE  printf("Case %d:\n",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

#define mx 101005

int SA[mx],tempSA[mx];
int RA[mx],tempRA[mx];
int cnt_sort[mx];
char T[mx];
string str;
int n;

void counting_sort(int k)
{
    ms(cnt_sort,0);
    int maxi=max(300,n);
    for(int i=0;i<n;i++)
    {
        int a=(i+k<n)?RA[i+k]:maxi-1;
        cnt_sort[a]++;
    }

    for(int i=0,sum=0;i<maxi;i++)
    {
        int x=cnt_sort[i];
        cnt_sort[i]=sum;
        sum+=x;
    }

    for(int i=0;i<n;i++)
    {
        int a=SA[i]+k<n?RA[SA[i]+k]:maxi-1;
        int x=cnt_sort[a];
        cnt_sort[a]++;
        tempSA[x]=SA[i];
    }

    for(int i=0;i<n;i++)
        SA[i]=tempSA[i];
}

void build_Suffix_Array()
{
    for(int i=0;i<n;i++)
    {
        RA[i]=str[i];
        SA[i]=i;
    }
    for(int k=1;k<n;k*=2)
    {
        counting_sort(k);
        counting_sort(0);
        tempRA[SA[0]]=0;
        int r=0;

        for(int i=1;i<n;i++)
        {
            if(RA[SA[i]]==RA[SA[i-1]] && RA[SA[i]+k]==RA[SA[i-1]+k])
                tempRA[SA[i]]=r;
            else
                tempRA[SA[i]]=++r;
        }
        for(int i=0;i<n;i++)
            RA[i]=tempRA[i];
        if(RA[SA[n-1]]==n-1)break;
    }

//    for(int i=0;i<n;i++)
//        pf("%s\n",str.substr(SA[i]).c_str());
}

int main()
{

//    freopen("in.txt","r",stdin);
//	  freopen("out.txt","w",stdout);

    int t;
    sf(t);
    TEST_CASE(t)
    {
        scanf("%s",T);
        str.clear();
        str=T;
        str+=str;
        n=SZ(str);
//        cout<<str<<endl;
        build_Suffix_Array();
        for(int i=0;i<n;i++)
        {
            if(SA[i]<n/2)
            {
                pf("%d\n",SA[i]+1);
                break;
            }
        }
    }

    return 0;
}


SPOJ KALTSUM – k Alternating Sum

0

Problem Link : http://www.spoj.com/problems/KALTSUM/

Solution Idea:

Break the problem into two cases: When k is greater than sqrt(n) and when k is less tahn or equal sqrt(n). Deal with them in different ways.

Case 1: For any k>sqrt(N) there will be less than sqrt(N) alternating segment. If you have an array, having cumulative sum, you can get the sum of a contiguous segment in O(1). So you can just loop over the alternating parts and get the sum in O(sqrt(N)) for a single query.

Case 2: Now when k <= sqrt(n), you need to do some preprocessing.

let arr[k][i] be the "k alternating sum" of a subarray which starts at position i and keeps alternating untill it reaches a position x such that if you add another segment of size k then it will go out of the array. [ This array can be computed in O(N) , and as you're doing it for every k < sqrt(n), the total complexity is O(N * sqrt(N))] With the help of this array, you can answer every query having k < sqrt(N) in O(1).

This solution idea is from this link.


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define LINE_PRINT_CASE  printf("Case %d:\n",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

ll dp[318][100005];
ll csum[100005];
ll ara[100005];

int main()
{

//    freopen("in.txt","r",stdin);
//	  freopen("out.txt","w",stdout);

    int n,q;

    sff(n,q);
    for(int i=1;i<=n;i++)
    {
        sfl(ara[i]);
        csum[i]=csum[i-1]+ara[i];
    }

    int root=ceil(sqrt(n));

    for(int i=1;i<=root;i++)
    {
        ll temp=0,sub=0;
        for(int j=n;j>=1;j--)
        {
            temp+=ara[j];
            if(j+i>n)
                sub=0;
            else
                sub=ara[j+i];
            temp-=sub;

            if(j+i-1>n)
                dp[i][j]=0;
            else
                dp[i][j]=temp-dp[i][j+i];
        }
    }
//
//    for(int i=1;i<=root;i++,cout<<endl)
//        for(int j=1;j<=n;j++)
//        cout<<dp[i][j]<<" ";

    while(q--)
    {
        int a,b,k;
        sfff(a,b,k);
        ll ans=0;
        if(k>root)
        {
            for(int i=a,j=0;i<=b;i+=k,j++)
            {
                if(j%2==0)
                    ans+=csum[i+k-1]-csum[i-1];
                else
                    ans-=csum[i+k-1]-csum[i-1];
            }
            pf("%lld\n",ans);
        }
        else
        {
            int turn=(b-a+1)/k;
            if(turn%2==1)
                pf("%lld\n",dp[k][a]+dp[k][b+1]);
            else
                pf("%lld\n",dp[k][a]-dp[k][b+1]);
        }
    }


    return 0;
}


SPOJ: AMR10E – Stocks Prediction

0

Problem Link : http://www.spoj.com/problems/AMR10E/


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
//#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define LINE_PRINT_CASE  printf("Case %d:\n",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/



/*----------------------Matrix-----------------------*/

// int MOD=
ll MOD= 1000000007;

struct matrix
{
    ll mat[11][11];
    int row,col;

    matrix()
    {
        row=col=0;
        memset(mat,0,sizeof mat);
    }
    matrix(int a, int b)
    {
        row=a,col=b;
        memset(mat,0,sizeof mat);
    }

    matrix operator*(const matrix &p) const
    {
        assert(col == p.row);
        matrix temp;
        temp.row = row;
        temp.col = p.col;
        for (int i = 0; i < temp.row; i++)
        {
            for (int j = 0; j < temp.col; j++)
            {
                ll sum = 0;
                for (int k = 0; k <col;  k++)
                {
                    sum += ((mat[i][k]%MOD) * (p.mat[k][j]%MOD))%MOD;
                    sum%=MOD;
                }
                temp.mat[i][j] = sum;
            }
        }
        return temp;
    }
    matrix operator+ (const matrix &p) const
    {
        assert(row==p.row && col==p.col);
        matrix temp;
        temp.row=row;
        temp.col=col;
        for(int i=0; i<temp.row; i++)
        {
            for(int j=0; j<temp.col; j++)
                temp.mat[i][j]=((mat[i][j]%MOD)+(p.mat[i][j]%MOD))%MOD;;
        }
        return temp;
    }

    matrix identity()
    {
        matrix temp;
        temp.row=row;
        temp.col=col;
        for(int i=0; i<row; i++)
            temp.mat[i][i]=1;
        return temp;
    }

    matrix pow(ll pow)
    {
        matrix temp=(*this);
        matrix ret=(*this).identity();
        while(pow)
        {
            if(pow % 2==1)
                ret=ret*temp;
            temp=temp*temp;
            pow/=2;
        }
        return ret;
    }

    void show()
    {
        printf("-----------------------------\n");
        for(int i=0; i<row; i++)
        {
            for(int j=0; j<col; j++)
                printf("%lld ",mat[i][j]);
            printf("\n");
        }
        printf("-----------------------------\n");
    }

};

/*--------------------------Matrix End---------------------*/

matrix identity;

matrix func(matrix m, ll p)
{
//    if(p==0) return identity;
    if(p==1) return identity;
//
//    matrix ret=func(m,p/2);
//
//    matrix temp=m.pow(p/2);

    if(p%2==0)
    {
        return func(m,p/2)*(identity+m.pow(p/2));
    }
    else
    {
        return func(m,p-1)+m.pow(p-1);
    }
}

ll s[15],ara[15];

int main()
{

//    freopen("in.txt","r",stdin);
//	  freopen("out.txt","w",stdout);

    int t;
    sf(t);
    TEST_CASE(t)
    {
        ll n,r,k;

        sfffl(n,r,k);

        matrix base(r,1);

        ms(s,0);
        ms(ara,0);

        for(int i=1; i<=r; i++)
            sfl(s[i]);

        matrix a(r,r);

        for(int i=0; i<r; i++)
        {
            sfl(a.mat[0][i]);
            ara[i+1]=a.mat[0][i];
            if(i+1!=r)
                a.mat[i+1][i]=1;
        }

//        base.show();

//        a.show();

        a=a.pow(k);

//        a.show();

//        base.show();

        identity=a.identity();

        a=func(a,n);

//        a.show();

        ll ans=0;

        if(k>r)
        {
            for(int i=r+1; i<=k; i++)
            {
                for(int j=1; j<=r; j++)
                {
                    s[i]+=s[i-j]*ara[j];
                    s[i]%=MOD;
                }
            }

            for(int i=0; i<r; i++)
            {
                ans=(ans+(s[k-i]*a.mat[0][i])%MOD);
                ans%=MOD;

            }

        }
        else
        {
            for(int i=0; i<r; i++)
            {
                ans=(ans+(a.mat[r-k][r-i-1]*s[i+1])%MOD);
                ans%=MOD;
            }

        }
        pf("%lld\n",ans);

    }

    return 0;
}