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There are n balls with m colors. The possibility of that the color of the i-th ball is color j is ai,jai,1+ai,2+...+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.
Several test cases(about 5) For each cases, first come 2 integers, n,m(1≤n≤1000,1≤m≤1000) Then follows n lines with m numbers ai,j(1≤ai≤100)
For each cases, please output the answer with two decimal places.
2 2 1 1 3 5 2 2 4 5 4 2 2 2 2 4 1 4
3.00 2.96 3.20
附上中文题目：
附上官方题解：
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #include<math.h> 7 #include<algorithm> 8 #include<queue> 9 #include<set> 10 #include<bitset> 11 #include<map> 12 #include<vector> 13 #include<stdlib.h> 14 #include <stack> 15 using namespace std; 16 #define PI acos(-1.0) 17 #define max(a,b) (a) > (b) ? (a) : (b) 18 #define min(a,b) (a) < (b) ? (a) : (b) 19 #define ll long long 20 #define eps 1e-10 21 #define MOD 1000000007 22 #define N 1006 23 #define inf 1e12 24 int n,m; 25 double a[N][N]; 26 double p[N][N]; 27 int main() 28 { 29 while(scanf("%d%d",&n,&m)==2){ 30 for(int i=0;i<n;i++){ 31 double sum=0; 32 for(int j=0;j<m;j++){ 33 scanf("%lf",&a[i][j]); 34 sum+=a[i][j]; 35 } 36 for(int j=0;j<m;j++){ 37 p[i][j]=a[i][j]*1.0/sum; 38 } 39 } 40 double ans=0; 41 for(int j=0;j<m;j++){ 42 double sum=0; 43 for(int i=0;i<n;i++){ 44 ans+=p[i][j]*(1.0-p[i][j]); 45 } 46 for(int i=0;i<n;i++){ 47 sum+=p[i][j]; 48 } 49 ans+=sum*sum; 50 } 51 printf("%.2lf\n",ans); 52 } 53 return 0; 54 }
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原文地址：http://www.cnblogs.com/UniqueColor/p/4985873.html
