Poj 2151 check the difficulty of problems (probability DP)

/*一次比赛中，共M道题，T个队，p[i][j]表示队i解出题j的概率；问每队至少解出一题且冠军队至少解出N道题的概率。dp[i][j][k]表示第i支队伍在前j道题中解出k道的概率问题的解可以转化为：每队均至少做一题的概率（用P1表示）减去每队做题数均在1到N-1之间的概率（用P2表示）。*/# include <stdio.h># include <algorithm># include <string.h># include <iostream>using namespace std;double dp[1010][35][35];double p[1010][35];int main(){    int i,j,k,n,t,m;    while(~scanf("%d%d%d",&m,&t,&n),m+t+n)    {        for(i=1; i<=t; i++)        {            for(j=1; j<=m; j++)                scanf("%lf",&p[i][j]);        }        memset(dp,0,sizeof(dp));        for(i=1; i<=t; i++)        {            dp[i][0][0]=1;            for(j=1; j<=m; j++)            {                dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]);                for(k=1; k<=m; k++)                {                    dp[i][j][k]=dp[i][j-1][k-1]*p[i][j]+dp[i][j-1][k]*(1-p[i][j]);                }            }        }        double p1=1,p2=1;        for(i=1;i<=t;i++)///每队至少做一题的概率            p1*=1-dp[i][m][0];        double sum=0;        for(i=1;i<=t;i++)///每队做题数均在1到N-1之间的概率        {            sum=0;            for(j=1;j<=n-1;j++)///单独一支队伍,同支队伍是用+               sum+=dp[i][m][j];            p2*=sum;        }        printf("%.3lf\n",p1-p2);    }    return 0;}

Poj 2151 check the difficulty of problems (probability DP)