Light OJ: 1347 – Aladdin and the Magical Lamp

0

Problem Link : http://www.lightoj.com/volume_showproblem.php?problem=1347


Solution Idea:
###############
Solution idea for this problem is following-
1. Concatenate 3 input string one after another and make one single string.
2. Build Suffix array and store every step rank.
3. Run sliding window or two pointer on sorted suffix array and find the range in which 3 string presents. and determine the LCP of beginning and ending string of that range segment.
4. Largest LCP value find during step 3 is the answer.


#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 30005

int SA[mx],tempSA[mx];
int RA[mx],tempRA[mx];
int c[mx];
//int LCP[mx],PLCP[mx],Phi[mx];
int P[18][mx],stp;
string str;
char s1[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()
{
    stp=0;
    for(int i=0; i<n; i++)
    {
        RA[i]=str[i];
        SA[i]=i;
        P[stp][i]=RA[i];
    }

    stp++;

    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];
            P[stp][i]=RA[i];
        }
        stp++;
        if(RA[SA[n-1]]==n-1) break;
    }
    stp--;
}


int find_lcp(int x, int y)
{
    int ret=0;
    if(x==y) return n-x;
    for(int k=stp; k>=0 && x<n && y<n; k--)
    {
        if(P[k][x]==P[k][y])
        {
            x+=(1<<k);
            y+=(1<<k);
            ret+=(1<<k);
        }
    }
    return ret;
}


int vis[5];

int getRange(int idx, int pos1, int pos2)
{
    if(idx<pos1) return 1;
    if(idx<pos2) return 2;
    return 3;
}



int main()
{

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

    int t;
    sf(t);
    TEST_CASE(t)
    {
        str.clear();
        scanf("%s",s1);
        str+=string(s1);
        str+="@";
        int pos1=SZ(str)-1;
        scanf("%s",s1);
        str+=string(s1);
        str+="$";
        int pos2=SZ(str)-1;
        scanf("%s",s1);
        str+=string(s1);
        str+="#";
        n=SZ(str);
//        cout<<str<<endl;
        build_Suffix_Array();

//        for(int i=0; i<n; i++)
//            cout<<str.substr(SA[i])<<endl;

        int ans=0;

        int i=3,j=3;
        ms(vis,0);
        int cnt=0;
        while(j<n)
        {
            int a=getRange(SA[j],pos1,pos2);
            if(vis[a]==0)
            {
                cnt++;
            }
            vis[a]++;
            if(cnt==3)
            {
                int temp=find_lcp(SA[i],SA[j]);
                ans=max(ans,temp);
            }
            while(cnt>=3)
            {
                a=getRange(SA[i],pos1,pos2);
                vis[a]--;
                i++;
                if(vis[a]==0)
                    cnt--;
                if(cnt==3)
                {
                    int temp=find_lcp(SA[i],SA[j]);
                    ans=max(ans,temp);
                }
            }
            j++;
        }
        PRINT_CASE;
        pf("%d\n",ans);
    }


    return 0;
}


Light OJ: 1314 – Names for Babies

0

Problem Link : http://www.lightoj.com/volume_showproblem.php?problem=1314


Solution Idea:

Build Suffix array and LCP array. Then find the distinct substring using LCP arrray by the formula (Length of the suffix – LCP of this and the previous suffix). But in this problem we need to calculate the length which lie in the given range. Think about it.


#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];
//int P[18][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++)
    {
        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();


//        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 a,b;
        sff(a,b);

        int ans=0;
        for(int i=1;i<n;i++)
        {
            int len=n-SA[i]-1;
            int temp=len-LCP[i];
            if(len>b)
                temp-=(len-b);
            if(LCP[i]<a)
                temp-=(a-LCP[i]-1);
            temp=max(temp,0);
            ans+=temp;
        }

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

    }

    return 0;
}


Light OJ: 1424 – New Land

0

Problem Link : http://www.lightoj.com/volume_showproblem.php?problem=1424

Solution Idea: I think this problem can be solved with a 2D Segment Tree. But here I use the histogram technique to solve this problem. You can try Spoj – HISTOGRA – Largest Rectangle in a Histogram and Light OJ – 1083 – Histogram . Both are same problem on different website. This problems can be using a niche technique called histogram which can be implemented using a stack. the algorithm determine the index of the bar before the current bar which is smaller than the current bar also determine the index of the bar after the current bar which is smaller than the current bar. Here the function func is the answer of the above two problem.

In this problem for each row we consider a histogram from current row and the rows which are above the current row with continuous 0. Then determine the maximum answer.


#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 2002

char grid[mx][mx];

int dp[mx][mx];

int func(int ara[], int n)
{
    int left[mx];

    stack<int>st;

    for(int i=0; i<n; i++)
    {
        while(!st.empty() && ara[st.top()]>=ara[i]) st.pop();
        left[i]=(st.empty()?-1:st.top());
        st.push(i);
    }

    while(!st.empty()) st.pop();

    int ret=0;
    int right;
    for(int i=n-1; i>=0; i--)
    {
        while(!st.empty() && ara[st.top()]>=ara[i]) st.pop();
        right=(st.empty()?n:st.top());
        ret=max((right-left[i]-1)*ara[i],ret);
        st.push(i);
    }
    return ret;
}

int main()
{

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

    int t;
    sf(t);
    TEST_CASE(t)
    {
        int n,m;
        sff(n,m);
        getchar();
        for(int i=0; i<n; i++)
            sc("%s",grid[i]);

        for(int j=0; j<m; j++)
        {
            dp[0][j]=(grid[0][j]=='0')?1:0;
        }

        int ans=func(dp[0],m);

        for(int i=1; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                dp[i][j]=dp[i-1][j]+((grid[i][j]=='0')?1:-dp[i-1][j]);
            }
            ans=max(ans,func(dp[i],m));
        }
        PRINT_CASE;
        pf("%d\n",ans);

    }


    return 0;
}

1217 – Neighbor House (II)

0

Problem Link : http://lightoj.com:81/volume/problem/1217


#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));}
/*------------------------------------------------*/

int n;
int ara[1001];

int    dp[1001][1001][3];
bool xxxxx[1001][1001][3];
int yy=1;

int func(int idx, int last, int state)
{

    if(idx>n)
    {
        if(state==0 && last!=n)
            return ara[1];
        return 0;
    }

    int &ret=dp[idx][last][state];
    bool &cas=xxxxx[idx][last][state];
    if(cas) return ret;
    cas=1;

////    if(ret!=-1) return ret;

    int p=0,q=0;

    if(idx==1)
    {
        p=ara[1]+func(idx+2,idx,1);
        q=func(idx+1,last,state);
    }
    else if(idx==2)
    {
        p=ara[2]+func(idx+2,idx,2);
        q=func(idx+1,0,state);
    }
    else if(idx==n)
    {
        if(last!=n-1 && state!=1)
            p=ara[n]+func(idx+2,idx,state);
        q=func(idx+1,last,state);
    }
    else
    {
        p=ara[idx]+func(idx+2,idx,state);
        q=func(idx+1,last,state);
    }

    return ret=max(p,q);


}

int main()
{

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

    int t;
    sf(t);
    TEST_CASE(t)
    {
        sf(n);
        for(int i=1; i<=n; i++) sf(ara[i]);
//        ms(dp,-1);
//        yy++;
        ms(xxxxx,0);
        PRINT_CASE;
        printf("%d\n",func(1,0,0));

    }

    return 0;
}

Light OJ: 1326 – Race

0

Problem Link : http://lightoj.com:81/volume/problem/1326

Solution Idea:

Let’s define dp(i) as the number of ways in which a race with “i” horses can finish.

Suppose we have “i” horses at the race and we will choose k of them for the first place, leaving “i – k” for the second (or higher) place. The process to count the number of ways to choose horses for the second place is equals to the original problem. This is the reason why we use dp.

(This Solution idea is from Manuel Pinda)


#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              10056
#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[1003][1003];
ll ans[1003];

ll nCr(int n, int r)
{
    if(r==1) return n;
    if(n==r) return 1;

    ll &ret=dp[n][r];

    if(ret!=-1) return ret;

    return ret=(nCr(n-1,r-1)%MOD+nCr(n-1,r)%MOD)%MOD;
}

ll func(int n)
{
    if(n==0) return 1;
    if(ans[n]!=-1) return ans[n];

    ll ret=0;

    for(int i=1;i<=n;i++)
    {
        ret+=(nCr(n,i)*func(n-i))%MOD;
        ret%=MOD;
    }

    return ans[n]=ret;
}

int main()
{

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

    int t;
    sf(t);

    ms(dp,-1);
    ms(ans,-1);

    TEST_CASE(t)
    {
        int n;
        sf(n);
        PRINT_CASE;

        printf("%lld\n",func(n));

    }

    return 0;
}


Light OJ: 1102 – Problem Makes Problem

0

Problem Link : http://lightoj.com:81/volume/problem/1102


#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 fact[2000006];

ll bigmod(ll n, ll pow)
{
    ll ret=1;
    while(pow)
    {
        if(pow%2==1)
            {ret*=n; ret%=MOD;}
        n*=n;
        n%=MOD;
        pow/=2;

    }
    return ret;
}

int main()
{

    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    fact[0]=1;
    for(ll i=1;i<2000005;i++)
    {
        fact[i]=(fact[i-1]*i)%MOD;
    }

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

        sff(n,k);

        ll down=(fact[k-1]*fact[n])%MOD;

        down=bigmod(down,MOD-2);

        down=(fact[n+k-1]*down)%MOD;

        PRINT_CASE;

        printf("%lld\n",down);



    }

    return 0;
}

Light OJ: 1095 – Arrange the Numbers

0

Problem Link : http://lightoj.com:81/volume/problem/1095


#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 fact[1005];

ll dp[1005][1005];

ll nCk(int n, int k)
{
    if(k==1) return n;
    if(n==k) return 1;

    if(dp[n][k]!=-1) return dp[n][k];

    return dp[n][k]= (nCk(n-1,k-1)+nCk(n-1,k))%MOD;
}


int main()
{

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

    fact[0]=1;

    for(ll i=1; i<1002; i++) fact[i]=(fact[i-1]*i)%MOD;

    int t;
    sf(t);
    ms(dp,-1);

    TEST_CASE(t)
    {
        ll n,m,k;
        sfffl(n,m,k);
        ll ans=nCk(m,k);

        int nn=n-k;

        ll ans1=fact[n-k];

        for(int i=1; i<=(m-k); i++)
        {
            if(i%2==1)
                ans1-= (nCk(m-k,i)*fact[nn-i])%MOD;
            else
                ans1+= (nCk(m-k,i)*fact[nn-i])%MOD;
            ans1=(ans1+MOD)%MOD;
        }

        PRINT_CASE;

        printf("%lld\n",(ans*ans1)%MOD);


    }


    return 0;
}