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 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1
Sample Output 13.333 31.500
Author CHEN, Yue
Source ZJCPC2004 这应该是我第一道写出来的贪心，贪心的题目一向比较玄，但是总归还是有一点套路的，做多了就能发现了 这道题我先是按J降序排列的，发现只有第一个例子可以， 然后我接着再按F升序排列，发现也只有第一个例子可以， 我慢慢的观察，发现只要按照J/F降序排序就可以了。 然后在每个例子中，m的值都不停的发生变化，直到为0为止
#include <stdio.h> #include <algorithm> using namespace std; struct node{ int J;//JavaBeans int F;//cat food }a[9000005]; bool cmp(struct node a,struct node b) { double x = a.J*1.0/a.F; double y = b.J*1.0/b.F; return x > y; } int main() { int m,n; while(scanf("%d %d",&m,&n) && m != -1 && n != -1){ for(int i = 0;i < n;i++){ scanf("%d %d",&a[i].J,&a[i].F); } sort(a,a+n,cmp); double sum = 0.0; for(int i = 0;i < n;i++){ if(a[i].F <= m){ sum += a[i].J; m -= a[i].F; }else { sum += m*1.0*a[i].J/a[i].F; m = 0; } if(m == 0) break; } printf("%.3lf\n",sum); } return 0; }