FatMouse' Trade
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
题解：我建立了一个结构体，然后将食物价值用r表示，然后把结构体排序，依次从最大的开始递减，可以保证价值最大。
#include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> using namespace std; struct food { double j; double f; double r; }fo[1500]; bool compare(food a,food b) { return a.r>b.r; } int main() { int m,n; while(scanf("%d %d",&m,&n),m!=-1,n!=-1) { for(int i=0;i<n;i++) { scanf("%lf %lf",&fo[i].j,&fo[i].f); fo[i].r=fo[i].j/fo[i].f; } sort(fo,fo+n,compare); double ans=0; for(int i=0;i<n;i++) { if(m>=fo[i].f) { ans+=fo[i].j; m-=fo[i].f; } else { ans+=m*fo[i].r; break; } } printf("%.3lf ",ans); } return 0; }