hdu 5570 balls（期望好题）

Problem Description

There are n balls with m colors. The possibility of that the color of the i-th ball is color j is ai,jai,+ai,+...+ai,m. If the number of balls with the j-th is x, then you should pay x2 as the cost. Please calculate the expectation of the cost.

Input

Several test cases(about )

For each cases, first come  integers, n,m(≤n≤,≤m≤)

Then follows n lines with m numbers ai,j(≤ai≤)

 

Output

For each cases, please output the answer with two decimal places.

 

Sample Input

Sample Output

3.00
2.96
3.20

Source

BestCoder Round #63 (div.2)

附上中文题目：

附上官方题解：

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 #include <stack>
 using namespace std;
 #define PI acos(-1.0)
 #define max(a,b) (a) > (b) ? (a) : (b)
 #define min(a,b) (a) < (b) ? (a) : (b)
 #define ll long long
 #define eps 1e-10
 #define MOD 1000000007
 #define N 1006
 #define inf 1e12
 int n,m;
 double a[N][N];
 double p[N][N];
 int main()
 {
    while(scanf("%d%d",&n,&m)==){
       for(int i=;i<n;i++){
          double sum=;
          for(int j=;j<m;j++){
             scanf("%lf",&a[i][j]);
             sum+=a[i][j];
          }
          for(int j=;j<m;j++){
             p[i][j]=a[i][j]*1.0/sum;
          }
       }
       double ans=;
       for(int j=;j<m;j++){
          double sum=;
          for(int i=;i<n;i++){
             ans+=p[i][j]*(1.0-p[i][j]);
          }
          for(int i=;i<n;i++){
             sum+=p[i][j];
          }
          ans+=sum*sum;
       }
       printf("%.2lf\n",ans);
    }
     return ;
 }