Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18054 Accepted Submission(s): 6668

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int t,n;
float p,dp[10005];
int wight[105];
float cost[105];
int main()
{
scanf("%d",&t);
for(int i=1;i<=t;i++)
{
memset(dp,0,sizeof(dp));
memset(wight,0,sizeof(wight));
memset(cost,0,sizeof(cost));
scanf("%f%d",&p,&n);
int sum=0;
for(int j=1;j<=n;j++)
{
scanf("%d %f",&wight[j],&cost[j]);
sum+=wight[j];
cost[j]=(1-cost[j]);
}
dp[0]=1;
for(int j=1;j<=n;j++)
{
for(int k=sum;k>=wight[j];k--)
dp[k]=max(dp[k],dp[k-wight[j]]*cost[j]);
}
int gg;
for(gg=sum;gg>=1;gg--)
{
if((1-dp[gg])<=p)
{
break;
}
}
printf("%d\n",gg);
}
return 0;
}
原文:http://www.cnblogs.com/hsd-/p/5185800.html