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.
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 .
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.
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
一个人去抢银行，已知在每个银行里能够抢到的钱数和被抓住的概率，又给出了一个被抓住的概率的最大值，问不超过最大被抓概率的前提下最多能够抢多少钱。
思路：
以总钱数为背包容量，求不被抓到的最大概率，注意求此最大概率时dp[0]初始化为1，其他的初始化为0，最后找出得到最大钱数并且概率合法的情况。
代码：
1 #include<iostream>
2 #include<string>
3 #include<cstdio>
4 #include<cmath>
5 #include<cstring>
6 #include<algorithm>
7 #include<vector>
8 #include<iomanip>
9 #include<queue>
10 using namespace std;
11 double dp[10002];
12 double min(double x,double y)
13 {
14 return x>y?x:y;
15 }
16 void zeroonepack(int v,double w,int n)
17 {
18 for(int j=n;j>=v;j--)
19 {
20 dp[j]=max(dp[j],dp[j-v]*w);
21 }
22 }
23 int main()
24 {
25 int t,n,mn[102];
26 double p,pn[102];
27 scanf("%d",&t);
28 while(t--)
29 {
30 int sum=0;
31 cin>>p>>n;
32 for(int i=0;i<n;i++)
33 {
34 cin>>mn[i]>>pn[i];
35 pn[i]=1-pn[i];
36 sum+=mn[i];
37 }
38 for(int i=0;i<=10000;i++)
39 dp[i]=0;
40 dp[0]=1;
41 for(int i=0;i<n;i++)
42 {
43 zeroonepack(mn[i],pn[i],sum);
44 }
45 for(int i=sum;i>=0;i--)
46 {
47 if(1-dp[i]<=p)
48 {
49 printf("%d\n",i);
50 break;
51 }
52 }
53 }
54 return 0;
55 }
