HDU 5570 balls 期望 数学

balls

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5570

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,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.

Input

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)

Output

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

Sample Input

2 2
1 1
3 5

2 2
4 5
4 2

2 2
2 4
1 4

Sample Output

3.00
2.96
3.20

HINT

题意

题解：

贡献为x^2，其实就是问你期望的对数有多少个

假设p[i][j]表示第一个位置出现颜色为j的的概率是多少，那么我们可以推出来该颜色期望的对数的公式是

ans = p[i][1] + p[i][2]+ ... + p[i][m] + p[i][1]*p[i][2] + p[i][1]*p[i][3] + .... + p[i][2]*p[i][1]+ ... + p[i][m]*p[i][m-1]

然后化简，就可以得到ans = sigma(p[i][k]) + sigma(p[i][k]) * sigma(p[i][k]) - sigma(p[i][k]*p[i][k])

就可以On来处理答案啦~

代码：

#include<iostream>
#include<stdio.h>
using namespace std;

int n,m;
double a[][];
double s[];
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=;i<=n;i++)
        {
            s[i]=;
            for(int j=;j<=m;j++)
            {
                scanf("%lf",&a[i][j]);
                s[i]+=a[i][j];
            }
        }
        double ans = ;
        for(int j=;j<=m;j++)
        {
            double s1 = ;
            double s2 = ;
            for(int i=;i<=n;i++)
            {
                double p = a[i][j]/s[i];
                s1+=p;
                s2+=p*p;
            }
            ans += s1 + s1*s1 - s2;
        }
        printf("%.2f\n",ans);
    }
}