Check the difficulty of problems
| Time Limit: 2000MS | Memory Limit: 65536K | |
|---|---|---|
| Total Submissions: 5779 | Accepted: 2518 |
Description
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?\
Input
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.
Output
For each test case, please output the answer in a separate line. The result should be rounded to three digits after the decimal point.
Sample Input
2 2 2
0.9 0.9
1 0.9
0 0 0
Sample Output
0.972
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看完题目后如果知道概率DP的话，应该都能想到。关键是怎么用。
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题意：
ACM比赛中，共M道题，T个队，pij表示第i队解出第j题的概率
问 每队至少解出一题且冠军队至少解出N道题的概率。
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思路：
用dp[i][j][k]表示，第i个队，在前j题，解出k道题的概率。
s[i][k]代表第i个队解出题<=k个的概率
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#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
double p[1100][40],dp[1100][40][40],pro[1100][40];
int n,m,t;
int main()
{
int i,j,k;
while(scanf("%d%d%d",&m,&t,&n)!=EOF&&n&&m&&t)
{
memset(dp,0,sizeof(dp));
memset(pro,0,sizeof(pro));
for(i=1;i<=t;i++)
{
for(j=1;j<=m;j++)
{
scanf("%lf",&p[i][j]);//第i个人解出第j个题的概率，（输入最好，以1为下标开始，以防数组跃界，并容易选择基）
}
}
for(i=1;i<=t;i++)
{
dp[i][0][0]=1;//题目数为1-m，所以解出第零个题的概率一定1,（以此作为加的基数）
for(j=1;j<=m;j++)
dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]); //先推i个队，在前j道题中做出0题的概率，以此做基推后边
for(j=1;j<=m;j++)
{
for(k=1;k<=j;k++)
{
dp[i][j][k]=dp[i][j-1][k-1]*(p[i][j])+dp[i][j-1][k]*(1-p[i][j]);//当前的状态和前面的状态及现在能做出题目的概率相关
}
}
pro[i][0]=dp[i][m][0];//第i队做出0<=题的概率，和在m个题中没做出题目的概率相同。
for(k=1;k<=m;k++)
pro[i][k]=pro[i][k-1]+dp[i][m][k];//当前做出题目<=k道的概率一定比在m道中做出k道题的概率大，肯定是由上一中情况推来的加上
}
double an1=1,an2=1;
for(i=1;i<=t;i++)
{
an1*=(1-pro[i][0]);//此是各队做出题数>=1到的总概率
an2*=(pro[i][n-1]-pro[i][0]);//每个队做出的题目个数在1~n-1的总概率<span style="white-space:pre"> </span>
}
printf("%.3f\n",an1-an2);
}
return 0;
}
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