Here you will find solutions of many problems on spoj. If you want solution of some problem which is not listed in blog or have doubt regarding any spoj problem (which i have solved) or any programming concept (data structure) you can mail me @ raj.nishant360@gmail.com
And my humble request to you all that don't copy the code only try to understand the logic and algorithm behind the code. I have started this because if you tried as hard as you can and still can't find any solution to the problem then you can refer to this.
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Wednesday, April 2, 2014

INS14E-Glorious Gamblers

Glorious Gamblers

below given code is for ins14e spoj or Glorious Gamblers spoj. This is simple problem of DP(as of LCS)
#include <bits/stdc++.h>
using namespace std;
#define MAX 510
#define gc getchar_unlocked    //scan function is for fast input;
inline void scan(int &x)
{
    register int c = gc();
    x = 0;
    int neg = 0;
    for(;((c<48 || c>57) && c != '-');c = gc());
    if(c=='-') {neg=1;c=gc();}
    for(;c>47 && c<58;c = gc()) {x = (x<<1) + (x<<3) + c - 48;}
    if(neg) x=-x;
}
template <class T>
T mi(T a, T b)
{
 return a<b?a:b;
}
template <class T>
inline T min_o(T a,T b,T c)
{
 return mi(a,mi(b,c));
}
template <class T>
T m(T a, T b)
{
 return a<b?b:a;
}
template <class T>
inline T max_o(T a,T b,T c)
{
 return m(a,m(b,c));
}
int main()
{
 int t;
 scan(t);
 while(t--)
 {
  double a[MAX][MAX];
  int n,m,temp,i,j;
  scan(m);scan(n);
  for(i=1;i<=m;i++)
   for(j=1;j<=n;j++){
    scan(temp);
    a[i][j] = temp;
   }
  for(i=m-1;i>=1;i--)
   a[i][n] += a[i+1][n];
  for(j=n-1;j>=1;j--)
   a[m][j] += a[m][j+1];
  for(i=m-1;i>=1;i--)
  {
   for(j=n-1;j>=1;j--)
    a[i][j] += 0.5*(min_o(a[i+1][j],a[i][j+1],a[i+1][j+1]) + max_o(a[i+1][j],a[i][j+1],a[i+1][j+1]));
  }
  printf("%0.6lf\n",a[1][1]);
 }
 return 0;
}