标签：dp
一个背包问题将价值变成了概率，求最小的一个收不到的概率，然后用1减去就可以啦！#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <climits>
#include <cstdlib>
#include <cstring>
using namespace std;
int w[100005];
double p[10005];
double dp[2][10005];
int main(){
int n,t;
while(cin >> n >> t,n||t){
for(int i = 1;i <= t;i++){
scanf("%d%lf",&w[i],&p[i]);
p[i] = 1 - p[i];
}
for(int i = 0;i <= n;i++){
dp[0][i] = 1;
}
for(int i = 1;i <= t;i++){
for(int j = 0;j <= n;j++){
if(j >= w[i]){
dp[i%2][j] = min(dp[(i-1)%2][j],dp[(i-1)%2][j-w[i]] * p[i]);
}
else{
dp[i%2][j] = dp[(i-1)%2][j];
}
}
}
dp[t%2][n] = 1- dp[t%2][n];
dp[t%2][n] *= 100;
printf("%.1f%%\n",dp[t%2][n]);
}
return 0;
}
标签：dp
原文地址：http://blog.csdn.net/qq_24667639/article/details/46241991
