POJ2151-Check the difficulty of problems

/*首先算出所有队伍至少都有A 1道的概率sum.再算出只A 1~n-1的概率t2答案就是sum-t2咯*/#include<iostream>#include<cstdio>#include<cstring>#define M 31#define T 1001using namespace std;double dp[T][M][M];//dp[i][j][k] 表示第i个队伍前k道题A了j道..double map[M],sum,t2,tt;int m,t,n;int main(){    freopen("test.txt","r",stdin);    while(scanf("%d%d%d",&m,&t,&n)&&m!=0&&t!=0&&n!=0)    {        memset(dp,0,sizeof(dp));        sum=1;        t2=1;        for(int i=1;i<=t;i++)        {            dp[i][0][0]=1;//明显前0道题A零道德概率为0            for(int j=1;j<=m;j++)            {                scanf("%lf",&map[j]);            }            for(int j=1;j<=m;j++)            {                for(int k=j;k>=1;k--)                {                    dp[i][k][j]=dp[i][k-1][j-1]*map[j]+dp[i][k][j-1]*(1-map[j]);                }                dp[i][0][j]=dp[i][0][j-1]*(1-map[j]);            }        }        for(int i=1;i<=t;i++)        {            sum*=1-dp[i][0][m];        }        for(int i=1;i<=t;i++)        {            tt=0;            for(int j=1;j<n;j++)            {                tt+=dp[i][j][m];            }            t2*=tt;        }        printf("%.3f\n",sum-t2);    }}