Language:
Check the difficulty of problems
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 Source
POJ Monthly,鲁小石
|
题意: n个人,m道题,求每个人做出题,并且至少一个做出k道题的概率
思路:等于每个人做出题的概率--每个人做出题并且没有一个人做出k到题的概率
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
#define eps 1e-8
typedef __int64 ll;
#define fre(i,a,b) for(i = a; i < b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define ssf(n) scanf("%s", n)
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define bug pf("Hi\n")
using namespace std;
#define INF 0x3f3f3f3f
#define N 1005
double dp[N][35][35]; // dp[i][j][k] 第 i 个人 前 j 道题 做 k 道概率
double p[N][35];
int n,m,k;
void solve()
{
int i,j,t;
mem(dp,0);
fre(i,1,n+1) //初始每一个人第一道题 对 与不对
{
dp[i][1][0]=1-p[i][1];
dp[i][1][1]=p[i][1];
}
fre(i,1,n+1) //每一个人m到题都没有作对
{
fre(j,2,m+1)
dp[i][j][0]=dp[i][j-1][0]*(1-p[i][j]);
}
fre(i,1,n+1) //每一个人j道题 做t道概率
fre(j,2,m+1)
fre(t,1,j+1)
{
if(t<j)
dp[i][j][t]=dp[i][j-1][t]*(1-p[i][j])+dp[i][j-1][t-1]*p[i][j];
else
dp[i][j][t]=dp[i][j-1][t-1]*p[i][j];
}
double p1=1,p2;
fre(i,1,n+1)
p1*=(1-dp[i][m][0]); //每个人都做出提的概率
p2=1;
fre(i,1,n+1)
{
double te=0;
fre(j,1,k)
te+=dp[i][m][j];
p2*=te; //每个人做出题但是没有一个做了k道题的概率
}
pf("%.3lf\n",p1-p2);
}
int main()
{
int i,j;
while(~sfff(m,n,k),n+m+k)
{
fre(i,1,n+1)
fre(j,1,m+1)
scanf("%lf",&p[i][j]);
solve();
}
return 0;
}