# 题目

{% fold 点击显/隐题目 %}

Now the organizer has studied out the contest problems, and through the result of preliminary contest, the organizer can estimate the probability that a certain team can successfully solve a certain problem.

Given the number of contest problems M, the number of teams T, and the number of problems N that the organizer expect the champion solve at least. We also assume that team i solves problem j with the probability Pij (1 <= i <= T, 1<= j <= M). Well, can you calculate the probability that all of the teams solve at least one problem, and at the same time the champion team solves at least N problems?

# 题解

`dp[i][j][k]`

表示第i个队伍在前j道题中做出k题

有递推式

`dp[i][j][k] += dp[i][j - 1][k - 1] * Prob[i][j] + dp[i][j - 1][k] * (1 - Prob[i][j])`

计算出来后,需要用概率的知识计算满足**每一队都至少过一题,并且至少有一队过N题**的概率

首先计算出所有队伍至少过一题的数量

# 代码

{% fold 点击显/隐代码 %}```cpp Check the difficulty of problems https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份

#include

#include

#include

using namespace std;

const int maxT = 1005;

const int maxM = 35;

double Prob[maxT][maxM];

double dp[maxT][maxM][maxM];

int main() {

int M, T, N;

while (scanf("%d%d%d", &M, &T, &N) != EOF) {

if (!(M | T | N))

break;

```
for (int i = 1; i <= T; ++i)
for (int j = 1; j <= M; ++j)
scanf("%lf", &Prob[i][j]);
for (int i = 1; i <= T; ++i)
dp[i][0][0] = 1.0;
// dp[i][j][k] 第i个队伍在前j道题中做出k题
for (int i = 1; i <= T; ++i)
for (int j = 1; j <= M; ++j)
for (int k = 0; k <= j; ++k) {
dp[i][j][k] = 0.0;
if (k != 0)
dp[i][j][k] += dp[i][j - 1][k - 1] * Prob[i][j];
// if (k != j)
dp[i][j][k] += dp[i][j - 1][k] * (1 - Prob[i][j]);
}
double P1 = 1.0; //有队伍达到N题
double P2 = 1.0; //有队伍未达到1题
for (int i = 1; i <= T; ++i) {
double sum = 0.0;
for (int j = 1; j < N; ++j)
sum += dp[i][M][j];
P1 *= sum;
P2 *= 1 - dp[i][M][0];
}
// P1 = 1 - P1;
// P2 = 1 - P2;
// printf("%.3f\n", P1 - P2);
cout << fixed << setprecision(3) << P2 - P1 << endl;
}
return 0;
```

}

{% endfold %}