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.
InputThe 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 .OutputFor 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:其实这道题还是很坑的，就是有个人去抢银行了，每个银行可以抢的钱是Mj，抢钱被抓住的概率是Pj，给你一个最大被抓概率，问这时最多抢多少钱。
- 思路：这道题的关键在于一般情况下，我们用背包容量来表示dp的数位，这个题目中背包大小变成了浮点数，就只能反过来存，dp[获得钱数]=概率；最后从大到小输出第一个概率大于等于Pj的dp的标点
代码：
import java.util.Arrays;
import java.util.Scanner;
public class E {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
double v = sc.nextDouble();
int n = sc.nextInt();
double vo[] = new double[1050];
int val[] = new int[1050];
int sum = 0;
for(int i=0;i<n;i++){
val[i] = sc.nextInt();
vo[i] = sc.nextDouble();
sum+=val[i];
}
double[] dp = new double[10500];
Arrays.fill(dp, 0);
dp [0] = 1;
for(int i=0;i<n;i++){
for(int j=sum;j>=val[i];j--){
dp[j] = Math.max(dp[j], dp[j-val[i]]*(1-vo[i]));
// System.out.println(j+" :"+dp[j]);
}
}
for(int i=sum;i>=0;i--){
if(dp[i]>=(1-v)){
System.out.println(i);
break;
}
}
}
}
}