题目

{% fold 点击显/隐题目 %}

Organizing a programming contest is not an easy job. To avoid making the problems too difficult, the organizer usually expect the contest result satisfy the following two terms: 1. All of the teams solve at least one problem. 2. The champion (One of those teams that solve the most problems) solves at least a certain number of problems.

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?

The input consists of several test cases. The first line of each test case contains three integers M (0 < M <= 30), T (1 < T <= 1000) and N (0 < N <= M). Each of the following T lines contains M floating-point numbers in the range of [0,1]. In these T lines, the j-th number in the i-th line is just Pij. A test case of M = T = N = 0 indicates the end of input, and should not be processed.
For each test case, please output the answer in a separate line. The result should be rounded to three digits after the decimal point.
2 2 2 0.9 0.9 1 0.9 0 0 0
0.972
{% endfold %}

题解

dp[i][j][k] 表示第i个队伍在前j道题中做出k题

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

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 %}`