Light OJ: 1276 – Very Lucky Numbers

0

Problem Link : http://lightoj.com:81/volume/problem/1276

Solution Idea:

Just pre-calculate the numbers and then run binary search.


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

vector<ll>v;
int cnt=0;
void func(ll n)
{
    if(n>1000000000000ULL) return;
    if(n!=0)
        v.pb(n);
    func(n*10+4);
    func(n*10+7);
}

int id;
ll maxx=1000000000000;

vector<ll>lucky;

void gen_num(int idx,ll num)
{
    if(idx>=SZ(v)) return ;
    for(int i=idx; i<SZ(v); i++)
    {
        ll temp=num*v[i];
        if(temp<=0 || temp>=maxx) return;
        lucky.pb(temp);
        gen_num(i,temp);
    }
}

int main()
{

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

    func(0);
    sort(all(v));

    gen_num(0,1);
    
    set<ll> st(all(lucky));
    lucky.assign(all(st));
    
    int t;
    sf(t);
    
    TEST_CASE(t)
    {
        ll a,b;
        sffl(a,b);
        ll r=upper_bound(all(lucky),b)-lucky.begin();
        ll l=lower_bound(all(lucky),a)-lucky.begin();
        int ans=(r-l);
        PRINT_CASE;
        pf("%d\n",ans);
    }

    return 0;
}

Light OJ: 1235 – Coin Change (IV)

0

Problem Link : http://lightoj.com:81/volume/problem/1235

Solution Idea:

You generate all possible combinations for the left half and right half, and then sort one side, and for each element on the other side, search its corresponding one in the sorted side using binary search.

However, you have 3 choices for each item, i.e. take 0, 1, or 2 of it. So, if we try to calculate exactly, generating all possible combination for each side should take 3^(n/2) operations which is 3^9 in the worst case. let m = 3^(n/2), then the rest of the thing can be done in O(m lg m)


#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 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 coins[30];
ll n,k;

vector<ll>set1,set2;

void gen_set1(int idx, ll val)
{
    set1.pb(val);
    if(idx==n/2) return;
    gen_set1(idx+1,val);
    gen_set1(idx+1,val+coins[idx]);
    gen_set1(idx+1,val+2*coins[idx]);
}

void gen_set2(int idx, ll val)
{
    set2.pb(val);
    if(idx==n) return;
    gen_set2(idx+1,val);
    gen_set2(idx+1,val+coins[idx]);
    gen_set2(idx+1,val+2*coins[idx]);
}

int main()
{

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

    int t;
    sf(t);
    TEST_CASE(t)
    {
       sffl(n,k);
       loop(i,n) sfl(coins[i]);
       set1.clear();
       set2.clear();

       gen_set1(0,0);
       gen_set2(n/2,0);

       sort(all(set1));

       PRINT_CASE;

       bool test=0;

       for(int i=0;i<SZ(set2);i++)
       {
           if(binary_search(all(set1),k-set2[i]))
           {pf("Yes\n");test=1;break;}
       }

       if(test==0)
            pf("No\n");

    }


    return 0;
}

Light OJ: 1170 – Counting Perfect BST

0

Problem Link : http://lightoj.com:81/volume/problem/1170


#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              100000007
#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 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 numbers[1000006];
int cnt=0;
void generate_num()
{
    for(ll i=2;i<=100000;i++)
    {
        ll num=i*i;
        while(num<=10000000000)
        {
          numbers[cnt++]=num;
          num*=i;
        }
    }

    sort(numbers,numbers+cnt);
    cnt=unique(numbers,numbers+cnt)-numbers;
    numbers[cnt++]=1000000000000000;

}

ll fact[1000006];

void gen_fact()
{
    fact[0]=1;
    for(int i=1;i<=1000005;i++)
    {
        fact[i]=(fact[i-1]*i)%MOD;
    }
}

ll mod_inv(ll n, ll pow)
{
    if(pow==0) return 1;
    if(pow%2==0)
    {
        ll ret=mod_inv(n,pow/2)%MOD;
        return (ret*ret)%MOD;
    }
    return ((n%MOD)*(mod_inv(n,pow-1)%MOD))%MOD;
}



int main()
{

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

    generate_num();
    gen_fact();

    int t;
    sf(t);
    TEST_CASE(t)
    {
        ll a,b;
        sffl(a,b);

        ll r=upper_bound(numbers,numbers+cnt,b)-numbers;
        ll l=lower_bound(numbers,numbers+cnt,a)-numbers;
        ll n=(r-l);
        PRINT_CASE;
        if(n==0)
            pf("0\n");
        else
        {
            ll ans=(fact[n+1]*fact[n])%MOD;
            ans=mod_inv(ans,MOD-2);
            ans=(fact[2*n]*ans)%MOD;
            pf("%lld\n",ans);
        }


    }

    return 0;
}


Light OJ: 1137 – Expanding Rods

0

Problem Link : http://lightoj.com:81/volume/problem/1137

Solution Idea: The problem become easy for binary search if you find any way to calculate the radius of the circle which contains that arc. If you can do this then you can calculate the arc length using the formula S=R*theta. where S=arc length and R= calculated radius.

Now we can guess the ans H through binary search and adjust on the basis of calculated arc length of that taken H and original final arc length.

Here is a link which contains the formula and formula proof for calculate Radius R .

Formula and Proof Link : http://www.mathopenref.com/arcradiusderive.html


#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 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 main()
{

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

    int t;
    sf(t);
    TEST_CASE(t)
    {
        double l,n,c;
        sc("%lf %lf %lf",&l,&n,&c);
        double final_length=(1.0+n*c)*l;

        double lo=0.0, hi=l;

        for(int i=0;i<60;i++)
        {
            double mid=(lo+hi)/2.0;
            double radius=(mid/2.0)+(sqr(l)/(8.0*mid));
            double theta=2.0*asin((l/2.0)/radius);
            double temp_length=theta*radius;
            if(temp_length>=final_length)
                hi=mid;
            else
                lo=mid;
        }

        PRINT_CASE;
        pf("%.10lf\n",lo);

    }

    return 0;
}


Light OJ: 1088 – Points in Segments

0

Problem Link : http://lightoj.com:81/volume/problem/1088

Solution Idea: Simple binary search. Just search in the given range and output the ans.


#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:\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 main()
{

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

    int t;
    sf(t);
    TEST_CASE(t)
    {
        int n,q;
        sff(n,q);
        vector<int>v;
        loop(i,n)
        {
            int a;
            sf(a);
            v.pb(a);
        }

        PRINT_CASE;
        while(q--)
        {
            int a,b;
            sff(a,b);
            int r=upper_bound(all(v),b)-v.begin();
            int l=lower_bound(all(v),a)-v.begin();
            printf("%d\n",r-l);

        }
    }

    return 0;
}

Light OJ: 1076 – Get the Containers

0

Problem Link : http://lightoj.com:81/volume/problem/1076

Solution Idea: Guess the capacity through binary search and check that capacity is valid or not and thus find the optimal one.


#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 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 ammount[1005];
int n,m;

bool check(ll capacity)
{
    int i=0;
    ll current=0;
    int cnt=1;
    while(i<n)
    {
        if(ammount[i]>capacity) return 0;
        if(current+ammount[i]<=capacity)
            current+=ammount[i],i++;
        else
        {
            if(cnt==m && i<n) return 0;
            current=0,cnt++;
        }

    }

    if(i<n) return 0;
    return 1;
}


int main()
{

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

    int t;
    sf(t);
    TEST_CASE(t)
    {
        sff(n,m);
        loop(i,n) sfl(ammount[i]);
        ll lo=0,hi=10000000000000000;
        ll ans=0;
        while(lo<=hi)
        {
            ll mid=(lo+hi)>>1;
            if(check(mid))
            {
                ans=mid;
                hi=mid-1;
            }
            else
                lo=mid+1;
        }

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

    }

    return 0;
}

Light OJ: 1062 – Crossed Ladders

0

Problem Link : http://lightoj.com:81/volume/problem/1062

Solution Idea: http://ojsolver.blogspot.com/2016/04/uva-10566-lightoj-1062-crossed-ladders.html


#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 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 main()
{

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

    int t;
    sf(t);
    TEST_CASE(t)
    {
        double x,y,c;
        sc("%lf %lf %lf",&x,&y,&c);

        double lo=0.0,hi=max(x,y);

        for(int i=0;i<50;i++)
        {
            double mid=(lo+hi)/2;
            double h2=sqrt(sqr(y)-sqr(mid));
            double h1=sqrt(sqr(x)-sqr(mid));
            double temp_ans=(1.0/h1)+(1.0/h2);
            temp_ans=1.0/temp_ans;
            if(temp_ans>=c)
                lo=mid;
            else
                hi=mid;
        }
        PRINT_CASE;
        pf("%.10lf\n",lo);

    }

    return 0;
}