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?
Input
Output
Sample Input
2 2 2
0.9 0.9
1 0.9
0 0 0
Sample Output
0.972
題目大意:
有
解題思路:
首先我們將問題轉化一下:
設所有隊伍至少做出一道題的概率爲
設所有隊伍做出題數都在區間
那麼可以推出我們要求的
考慮 DP,設
dp初始化:
設有
那麼就有:
至此,此題結束。
代碼:
#include <iostream>
#include <string.h>
#include <string>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <map>
using namespace std;
typedef long long LL;
const int MAXN = 1e3+5;
const double PI = acos(-1);
const double eps = 1e-8;
const LL MOD = 1e9+7;
double dp[MAXN][35][35], s[MAXN][35], p[MAXN][35];
int main(){
//freopen("C:/Users/yaonie/Desktop/in.txt", "r", stdin);
//freopen("C:/Users/yaonie/Desktop/out.txt", "w", stdout);
int T, M, N;
while(~scanf("%d%d%d",&M, &T, &N)){
if(M==0 && T==0 && N==0) break;
for(int i=1; i<=T; i++) for(int j=1; j<=M; j++) scanf("%lf",&p[i][j]);
//初始化
memset(dp, 0, sizeof(dp));
for(int i=1; i<=T; i++){
dp[i][0][0] = 1;
for(int j=1; j<=M; j++) dp[i][j][0] += (1-p[i][j])*dp[i][j-1][0];
}
for(int i=1; i<=T; i++)
for(int j=1; j<=M; j++)
for(int k=1; k<=j; k++)
dp[i][j][k] = p[i][j]*dp[i][j-1][k-1]+(1-p[i][j])*dp[i][j-1][k];
for(int i=1; i<=T; i++){
for(int j=0; j<=M; j++){
s[i][j] = 0;
for(int k=0; k<=j; k++) s[i][j] += dp[i][M][k];
}
}
double p1 = 1, p2 = 1;
for(int i=1; i<=T; i++){
p1 *= (s[i][M] - s[i][0]);
p2 *= (s[i][N-1] - s[i][0]);
}
printf("%.3f\n",p1-p2);
}
return 0;
}