POJ2151Check the difficulty of problems
作者:互联网
题目: 概率dp||组合数学;
http://poj.org/problem?id=2151
一套题,有T个题,M个人应考,已知每个人做来某题的概率。问X的概率。X满足,每个考生至少做来一道题。至少有一人做的题不少于N道
思路:
p[i][j][k] 表示第i个考试前j个题会做k道的概率。再根据题意进行DP。
p[i][j][k] = p[i][j-1][k]*(1-a[i][j]) + p[i][j-1][k-1]*a[i][j];
然后用 每个人至少做出一道题的概率,减去 每个人至少做一道,至多做n-1道的概率; just answer。
#include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<iostream> #define LL long long #define ULL unsigned long long #define rep(i,j,k) for(int i=j;i<=k;i++) #define dep(i,j,k) for(int i=k;i>=j;i--) #define INF 0x3f3f3f3f #define mem(i,j) memset(i,j,sizeof(i)) #define make(i,j) make_pair(i,j) #define pb push_back using namespace std; double p[1010][32][32],P1,P2,a[1010][32],tmp,res; int main() { int i,j,k,m,n,t; while(~scanf("%d%d%d",&m,&t,&n)){//t人 m题 if(m==0&&t==0&&n==0) return 0; memset(p,0,sizeof(p)); for(i=1;i<=t;i++) for(j=1;j<=m;j++) scanf("%lf",&a[i][j]); for(i=1;i<=t;i++) for(j=0;j<=m;j++) for(k=0;k<=j;k++){ if(j==0&&k==0) p[i][j][k]=1; else if(j==k) p[i][j][k]=p[i][j-1][k-1]*a[i][j]; else if(k==0) p[i][j][k]=p[i][j-1][k]*(1-a[i][j]); else p[i][j][k]=p[i][j-1][k-1]*a[i][j]+p[i][j-1][k]*(1-a[i][j]); } res=1;P1=1; for(i=1;i<=t;i++){ P1*=(1-p[i][m][0]); tmp=0; for(j=1;j<n;j++) tmp+=p[i][m][j]; res*=tmp; } res=P1-res; printf("%.3f\n",res); } return 0; }View Code
参考:https://www.cnblogs.com/hua-dong/p/8044969.html
标签:概率,POJ2151Check,int,32,problems,long,difficulty,include,define 来源: https://www.cnblogs.com/Willems/p/10946176.html