题意
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。
思路
f[i][j][k]表示 第i个队伍在前i个题目里过了k道题
dp[i][j] 表示第i个队伍过了1道到 j 道题
ans1 表示1～T个队伍过题数都在 1～M之间的概率
ans2 表示1～T个队伍过题数都在 1～N-1之间概率
ans1 - ans2 就是所求答案
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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
#include <vector>
using namespace std;
const int INF = ~0u>>1;
typedef pair <int,int> P;
#define MID(x,y) ((x+y)>>1)
#define iabs(x) ((x)>0?(x):-(x))
#define REP(i,a,b) for(int i=(a);i<(b);i++)
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define pb push_back
#define mp make_pair
#define print() cout<<"--------"<<endl
#define maxn 1010
double p[maxn][40],f[maxn][40][40],dp[maxn][40];
int main(){
int m,n,t;
while (~scanf("%d%d%d", &m, &t, &n) && (m + t + n)){
for (int i = 1; i <= t; i ++){
for (int j = 1; j <= m; j ++){
scanf("%lf", &p[i][j]);
}
}
memset(f,0,sizeof(f));
memset(dp,0,sizeof(dp));
for (int i = 1; i <= t; i ++){
f[i][1][0] = 1 - p[i][1];
f[i][1][1] = p[i][1];
for (int j = 2; j <= m; j ++){
for (int k = 0; k <= j; k ++){
if (k)
f[i][j][k] = f[i][j - 1][k] * (1 - p[i][j])
+ f[i][j - 1][k - 1] * p[i][j];
else
f[i][j][k] = f[i][j - 1][k] * (1 - p[i][j]);
}
}
}
for (int i = 1; i <= t; i ++){
for (int j = 1; j <= m; j ++){
for (int k = 1; k <= j; k ++){
dp[i][j] += f[i][m][k];
}
}
}
double ans1 = 1;
double ans2 = 1;
for (int i = 1; i <= t; i ++){
ans1 *= dp[i][m];
ans2 *= dp[i][n - 1];
}
printf("%.3fn", ans1 - ans2);
}
return 0;
}
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