Robberies
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10526 Accepted Submission(s):
3868
movies, and knows that the bad guys usually gets caught in the end, often
because they become too greedy. He has decided to work in the lucrative business
of bank robbery only for a short while, before retiring to a comfortable job at
a university.
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.
For each scenario, the first line of input gives a floating point number P, the
probability Roy needs to be below, and an integer N, the number of banks he has
plans for. Then follow N lines, where line j gives an integer Mj and a floating
point number Pj .
Bank j contains Mj millions, and the probability of
getting caught from robbing it is Pj .
number of millions he can expect to get while the probability of getting caught
is less than the limit set.
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.
#include <iostream>
#include <cstdio>
using namespace std;
#define MAX 10003
double p[MAX],f[MAX];
int m[];
int main()
{
double P;
int T, N, i, j;
cin>>T;
while(T--)
{
int sum = ;
scanf("%lf %d",&P,&N);
P = -P; //不被抓的概率
for(i=; i<N; i++)
{
scanf("%d %lf",&m[i],&p[i]);
p[i] = -p[i]; //不被抓的概率
sum += m[i]; //可以抢到的最大金钱数目
} for(i=; i<=sum; i++)
f[i] = ;
f[] = ; //表示抢金钱为0的时候,不被抓的概率为1
for(i=; i<N; i++)
for(j=sum; j>=m[i]; j--)
f[j] = max(f[j],f[j-m[i]]*p[i]);
for(i=sum; i>=; i--) //从最大的金钱数目开始,依次查看不被抓概率是否和给定的相等
if(f[i]-P>0.000000001)
{
cout<<i<<endl;
break;
}
}
return ;
}