抢银行 扭向0 1

Problem V

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 24   Accepted Submission(s) : 8

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.<br><br><center><img src=../../../data/images/con211-1010-1.jpg></center> <br>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.<br><br><br>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 . <br>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.<br><br>Notes and Constraints<br> 0 < T <= 100<br> 0.0 <= P <= 1.0<br> 0 < N <= 100<br> 0 < Mj <= 100<br> 0.0 <= Pj <= 1.0<br> 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

30.04 31 0.022 0.033 0.050.06 32 0.032 0.033 0.050.10 31 0.032 0.023 0.05

 

Sample Output

246

 关于抢银行的问题

上来先把失败全部转化为成功的概率

然后逆向dp

对所有的 价钱

分别求出抢到这个价钱最大成功概率

抢到这个价钱的概率

最后 取出 钱数最大概率能成功的 案例

#include<iostream>#include<string.h>#include<set>#include<stdio.h>#include<vector>#include<algorithm>#include<numeric>#include<math.h>#include<string.h>#include<sstream>#include<stdio.h>#include<string>#include<cstdlib>#include<algorithm>#include<iostream>#include<map>#include<queue>#include<iomanip>#include<cstdio>using namespace std;double pao[10000];int a[10000];double b[10000];int main(){    int t;   scanf("%d",&t);    while(t--)    {        memset(pao,0,sizeof(pao));        pao[0]=1;        double t;        int x;        scanf("%lf %d",&t,&x);        t=1-t;        int i,j;        int sum=0;        for(i=1;i<=x;i++)        {             scanf("%d%lf",&a[i],&b[i]);            b[i]=1-b[i];            sum=sum+a[i];        }        for(i=1;i<=x;i++)        {            for(j=sum;j>=a[i];j--)            {                if(pao[j]<pao[j-a[i]]*b[i])                {                    pao[j]=pao[j-a[i]]*b[i];                }            }        }        for(i=sum;i>=0;i--)        {            if((pao[i]-t)>0.000000001)            {cout<<i<<endl;             break;            }        }    }    return 0;}