题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=2955
题目：
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
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：
给定一个概率，被抓的概率要小于它。然后给出若干个银行的金额和抢劫被抓的概率。求在满足被抓概率小于所给概率的情况下，能获得的最大金额数。
思路：
概率dp题。被抓的情况有很多种，不好一个个去算，所以我们换个角度来计算，即P(被抓)=1-P(逃脱)，最后用1-P（逃脱）< P(给定的)来判定。这道题显然是将金额当做背包容量，求得在获得金额相同时，逃脱的概率最大。
状态转移：dp[j]=max(dp[j], dp[j-bank[i].m]*(1-bank[i].p));我们让dp[0]=1,则第一次抢劫一个银行时，则逃脱的概率就等于1-P(被抓)。
代码：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 using namespace std;
5 int n;
6 double dp[10005];
7 double p;
8 struct node{
9 int m;
10 double p;
11 }bank[105];
12 int main(){
13 int t;
14 scanf("%d",&t);
15 while (t--) {
16 int total=0;
17 memset(dp, 0, sizeof(dp));
18 scanf("%lf%d",&p,&n);
19 for (int i=0; i<n; i++){
20 scanf("%d%lf",&bank[i].m,&bank[i].p);
21 total+=bank[i].m;//算出不计概率的情况下，金额总数
22 }
23 dp[0]=1;
24 for (int i=0; i<n; i++) {
25 for (int j=total; j>=bank[i].m; j--) {
26 dp[j]=max(dp[j], dp[j-bank[i].m]*(1-bank[i].p));
27 }
28 }
29 for (int i=total; i>=0; i--) {
30 if(1-dp[i]<p){//dp[i]是逃脱的概率，1-dp[i]是被抓的概率
31 printf("%d\n",i);
32 break;
33 }
34 }
35 }
36 return 0;
37 }