POJ 2151 Check the difficulty of problems
以前做过的题目了。。。。补集+DP
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 4091 | Accepted: 1811 |
Description
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
Source
POJ Monthly,鲁小石
#include <iostream>
#include <cstdio>
#include <cstring> using namespace std; int M,T,N;
double a[][][],s[][],p1,pn,solve[][]; int main()
{
while(~scanf("%d%d%d",&M,&T,&N))
{
if((M||T||N)==) break;
for(int i=;i<=T;i++) for(int j=;j<=M;j++) scanf("%lf",&solve[i][j]);
memset(a,,sizeof(a)); memset(s,,sizeof(s));
for(int i=;i<=T;i++)
{
a[i][][]=;
for(int j=;j<=M;j++)
{
a[i][j][]=a[i][j-][]*(-solve[i][j]);
}
}
for(int i=;i<=T;i++)
{
for(int j=;j<=M;j++)
{
for(int k=;k<=j;k++)
{
a[i][j][k]=a[i][j-][k-]*solve[i][j]+a[i][j-][k]*(-solve[i][j]);
}
}
}
for(int i=;i<=T;i++)
{
s[i][]=a[i][M][];
for(int j=;j<=M;j++)
{
s[i][j]=s[i][j-]+a[i][M][j];
}
}
p1=pn=.;
for(int i=;i<=T;i++)
{
p1*=s[i][M]-s[i][];
pn*=s[i][N-]-s[i][];
}
printf("%.3lf\n",p1-pn);
}
return ;
}
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