FatMouse' Trade

Problem Description

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.
The warehouse has N rooms. The i-th room contains J[i] pounds of JavaBeans and requires F[i] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

 

Input

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J[i] and F[i] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.

 

Output

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

 

Sample Input

5 37 24 35 220 325 1824 1515 10-1 -1

 

Sample Output

13.33331.500

 

将兑换比例从大到小排序

#include<iostream>#include<string.h>#include<algorithm>#include<cmath>#include<cstdio>using namespace std;struct node{int x, y;double z;};node s[1005];bool cmp(node &a, node &b){return a.z > b.z;}int main(){int M, N;double ans;while(cin>>M>>N && M >= 0 && N >= 0){ans = 0.0;for(int i = 0; i < N; i++){cin>>s[i].x>>s[i].y;s[i].z = s[i].x / (s[i].y * 1.0);}sort(s, s + N, cmp);for(int i = 0; i < N; i++){if(M <= 0)break;else if(M >= s[i].y){M -= s[i].y;ans += s[i].x;}else if(M < s[i].y){ans += (s[i].x * (1.0 / s[i].y) * M);M = 0;}}printf("%.3lf\n", ans);}return 0;}