标签:style class code ext width color string int html set htm
主要的子问题是每一个队伍有一个做出题目的概率,求做出k个题目的概率。简单的简单的组合数DP。想清楚即可。
1: #include <iostream>
2: #include <cstdio>
3: #include <cstring>
4: using namespace std;
5:
6: double dp[35][35];
7: double p[1005][35];
8: int main()
9: {
10: // freopen("1.txt","r",stdin);
11: int M,T,N;
12: while(cin>>M>>T>>N && M!=0)
13: {
14: memset(p, 0, sizeof(p));
15: memset(dp, 0, sizeof(dp));
16: for(int i=0; i<T; i++)
17: {
18: for(int j=0; j<M; j++)
19: cin>>p[i][j];
20: }
21: double ans = 1.0;
22: for(int i=0; i<T; i++)
23: {
24:
25: double ret = 1.0f;
26: for(int j=0; j<M; j++)
27: ret *= (1-p[i][j]);
28: ret = 1 - ret;
29: ans *= ret;
30: }
31: double ant = 1.0f;
32: for(int i=0; i<T; i++)
33: {
34: // first i solved num is j
35: memset(dp, 0, sizeof(dp));
36: dp[1][0] = 1- p[i][0];
37: dp[1][1] = p[i][0];
38: for(int j=2; j<=M; j++) dp[j][0] = dp[j-1][0] * (1 - p[i][j-1]);
39: for(int j=2; j<= M; j++)
40: {
41: for(int k = 1; k<=j; k++)
42: {
43: dp[j][k] = dp[j-1][k-1]*(p[i][j-1])+ dp[j-1][k]*(1 - p[i][j-1]);
44: }
45: }
46: double sum = 0.0f;
47: for(int i=1; i<N; i++)
48: sum += dp[M][i];
49: ant *= sum;
50: }
51: printf("%.3f\n", ans - ant);
52: }
53: return 0;
54: }
标签:style class code ext width color string int html set htm
原文地址:http://www.cnblogs.com/sosi/p/3701609.html