Poj 2151 check the difficulty of problems (DP, probability)

Link: poj 2151

Q: There are m questions in the competition, and t teams. PIJ indicates the probability that the I team will solve the J question,

Calculate the probability that each team should solve at least one question, and the champion army should solve at least N questions.

Analysis: the probability that each team should solve at least one question, and the champion army should solve at least N questions

Then the probability of the original request can be converted:

Probability of each team having at least one question P1 minus the probability of each team having a question between 1 and N-1 p2

Set DP [I] [J] [k] to indicate the probability that team I has solved K questions in the previous J questions.

DP [I] [J] [k] = DP [I] [J-1] [k-1] * PIJ + dp [I] [J-1] [k] * (1-pij)

Set sum [I] [J] to the probability that team I can solve the question J at most.

Sum [I] [J] = DP [I] [m] [0] + dp [I] [m] [1] +... + dp [I] [m] [J]

Sum [I] [m]-sum [I] [0] indicates the probability that team I should solve at least one question.

Sum [I] [N-1]-sum [I] [0] the probability of Team I solving the question [1, N-1]

#include<stdio.h>#include<string.h>double p[1005][32],dp[32][32],sum[32],p1,p2;int main(){    int t,m,n,i,j,k;    while(scanf("%d%d%d",&m,&t,&n)!=EOF){        if(t==0&&m==0&&n==0)            break;        for(i=1;i<=t;i++)            for(j=1;j<=m;j++)                scanf("%lf",&p[i][j]);        memset(dp,0,sizeof(dp));        memset(sum,0,sizeof(sum));        p1=p2=1.0;        for(i=1;i<=t;i++){            dp[0][0]=1.0;            for(j=1;j<=m;j++){                dp[j][0]=dp[j-1][0]*(1-p[i][j]);                for(k=1;k<=j;k++)                    dp[j][k]=dp[j-1][k]*(1-p[i][j])+dp[j-1][k-1]*p[i][j];            }            sum[0]=dp[m][0];            for(j=1;j<=m;j++)                sum[j]=sum[j-1]+dp[m][j];            p1*=(sum[m]-sum[0]);            p2*=(sum[n-1]-sum[0]);        }        printf("%.3lf\n",p1-p2);    }    return 0;}

Poj 2151 check the difficulty of problems (DP, probability)