POJ 2151.Check the difficulty of problems

384


题目



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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




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


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#include <cstdio>
#include <iomanip>
#include <iostream>
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;
}
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