Robberies
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13554 Accepted Submission(s): 4989
For a few months now, Roy has been assessing the security of various banks and the amount of cash they hold. He wants to make a calculated risk, and grab as much money as possible.
His mother, Ola, has decided upon a tolerable probability of getting caught. She feels that he is safe enough if the banks he robs together give a probability less than this.
Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .
Notes and Constraints
0 < T <= 100
0.0 <= P <= 1.0
0 < N <= 100
0 < Mj <= 100
0.0 <= Pj <= 1.0
A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.
//dp[i]表示偷i錢不被抓到的最大概率dp[i]=max{dp[j],dp[j-c[i]]*(1-w[i])}
#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
int main()
{
int T,N,i,m,c[105],sum,j;
scanf("%d",&T);
double P,p,w[105],dp[10050];
while(T--)
{
memset(dp,0,sizeof(dp));
sum=0;
scanf("%lf%d",&P,&N); //表示小偷被抓到的概率應小於P,而不被抓到的概率應大於1-P
for(i=0; i<N; i++)
{
scanf("%d%lf",&c[i],&w[i]); //w[i]爲小偷被抓到的概率
sum+=c[i];
}
dp[0]=1;
for(i=0;i<N;i++)
{
for(j=sum;j>=c[i];j--)
{
dp[j]=dp[j]>dp[j-c[i]]*(1-w[i])?dp[j]:dp[j-c[i]]*(1-w[i]);
}
}
for(i=sum;i>=0;i--)
{
if(dp[i]>=1-P)
break;
}
printf("%d\n",i);
}
return 0;
}