HDU

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25266    Accepted Submission(s): 9310

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.


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

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

题意：小偷偷银行，给定一个最大失败几率，给每个银行的钱数和失败概率。失败几率内偷得的最大钱数。

解决：01背包问题入门。

细节：概率是小数，无法当做背包重量。学习了网上的代码，可以将钱数当做重量。最后得到的数组是，获得这么多钱的最大成功概率。最后从大往小读，知道概率允许就输出钱数。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <set>#include <stack>#include <queue>#define fin freopen("in.txt", "r", stdin)using namespace std;const int INF = 0x3f3f3f;const int maxn = 10000;int t, n, m[maxn+5], sum;double P, p[maxn+5], ans[maxn+5];int main(){    cin >> t;    while(t--){        sum = 0;        memset(ans,0,sizeof(ans));        cin >> P >> n;        for(int i = 0; i < n; i++){            cin >> m[i] >> p[i];            sum += m[i];        }        ans[0] = 1;        for(int i = 0; i < n; i++){            for(int j = sum; j >= m[i]; j--)                ans[j] = max(ans[j], ans[j - m[i]]*(1-p[i]));        }        for(int i = sum; i >= 0; i--){            if(ans[i] > 1-P){                cout << i << endl;                break;            }        }    }}