Check the difficulty of problems（概率+DP）

http://poj.org/problem?id=2151

看的题解。。表示没看懂状态转移方程。。

#include<stdio.h>
#include<string.h>
int m,t,n;
double dp[][][],p[][],tt[][];
int main()
{
    int i,j,k;
    while(~scanf("%d%d%d",&m,&t,&n))
    {
        if (!m && !t && !n)
            break;
        memset(dp,0.0,sizeof(dp));
        memset(tt,0.0,sizeof(tt));
        for (i = ; i <= t; i ++)
            for (j = ; j <= m; j ++)
            {
                scanf("%lf",&p[i][j]);
            }

        for (i = ; i <= t; i ++)
        {
            dp[i][][] = 1.0;
            for (j = ; j <= m; j ++)
            {
                dp[i][j][] = dp[i][j-][]*(-p[i][j]);
            }
            for (j = ; j <= m; j ++)
            {
                for (k = ; k <= j; k ++)
                {
                    dp[i][j][k] = dp[i][j-][k-]*p[i][j] + dp[i][j-][k]*(-p[i][j]);
                }
            }
            tt[i][] = dp[i][m][];
            for (k = ; k <= m; k ++)
            {
                tt[i][k] = tt[i][k-] + dp[i][m][k];
            }
        }
        double p1 = 1.0,p2=1.0;
        for (i = ; i <= t; i ++)
        {
            p1 *= (tt[i][m]-tt[i][]);
            p2 *= (tt[i][n-]-tt[i][]);
        }
        printf("%.3f\n",p1-p2);
    }
    return  ;
}