HDU - 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

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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 

 

#include<iostream>  //贪心#include<algorithm>#include<iomanip>using namespace std;struct a{double f;double j;double w;};bool cmp(a b1, a b2){if (b1.w>b2.w) return true;else return false;}int main(){int n;double m;while (cin >> m >> n){if (n == -1 && m == -1) break;a aa[1050];for (int i = 0; i<n; i++){cin >> aa[i].j >> aa[i].f;aa[i].w = aa[i].j / aa[i].f;}sort(aa, aa + n, cmp);double sum = 0;for (int i = 0; i<n; i++){if (m - aa[i].f >= 0) { m = m - aa[i].f; sum = sum + aa[i].j; }else { sum = sum + m*aa[i].w; break; }}cout << fixed << setprecision(3) << sum << endl;}}