Problem V

Problem Description

The aspiring
Roy the Robber has seen a lot of American 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.



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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.

Input

The first line
of input gives T, the number of cases. 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 .

Output

For each test
case, output a line with the maximum 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.

Sample Input

3

0.04
3

1
0.02

2
0.03

3
0.05

0.06
3

2
0.03

2
0.03

3
0.05

0.10
3

1
0.03

2
0.02

3
0.05

Sample Output

2

4

6

题意：ROY想测试银行的安全系数；给你I个银行的钱数和被抓的概率；让你求被抓的条件下，能抢到的最多钱数；

解题思路：一开始想的是就是正着求，但是太麻烦了，还有交集，看了论坛上的留言才知道翻着求很简单；

以前做的DP都是累加，这个是累乘，剩下的就是01背包问题了；

感悟：冷静一下可能还有思路；

代码：

#include

#include

#include

#define maxn 105

using namespace std;

double dp[maxn*maxn],hold[maxn*maxn];

int cost[maxn*maxn];

int main()

{

   
//freopen("in.txt", "r", stdin);

    int
t,n,total_cost;

    double
p;

   
scanf("%d",&t);

   
while(t--)

    {

       
scanf("%lf %d",&p,&n);

       
p=1-p;//安全逃走的概率

       
total_cost=0;

       
for(int i=0;i

       
{

           
scanf("%d %lf",&cost[i],&hold[i]);

           
hold[i]=1-hold[i];

           
total_cost+=cost[i];

       
}

       
for(int i=1;i<=total_cost;i++)

           
dp[i]=0;

       
dp[0]=1;//什么也没偷安全的概率就是1；

       
for(int i=0;i

           
for(int j=total_cost;j>=cost[i];j--)

           
{

               
dp[j]=max(dp[j],dp[j-cost[i]]*hold[i]);

           
}

       
for(int i=total_cost;i>=0;i--)

           
if(dp[i]-p>0.00000000001)

           
{

               
printf("%d\n",i);

               
break;

           
}

    }

    return
0;

}