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
#include<iostream> #include<algorithm> #include<iomanip> using namespace std; struct cat_mouse { double mouse_food; double cat_food; double value; } ; bool cmp(cat_mouse a,cat_mouse b) { return a.value>b.value; } int main() { double m,s; int n; struct cat_mouse a[1000]; while((cin>>m>>n)&&!(m==-1&&n==-1)) { s=0; for(int i=0;i< n;i++) { cin>>a[i].mouse_food>>a[i].cat_food; a[i].value=1.0*a[i].mouse_food/a[i].cat_food; } sort(a ,a+n,cmp); for(int i=0;i<n;++i) { if(m>=a[i].cat_food) { m-=a[i].cat_food; s+=a[i].mouse_food; } else{ s+=1.0*m*a[i].mouse_food/a[i].cat_food; break; } } cout<<setprecision(3)<<fixed<<s<<endl; } return 0; }c++ 编译通过了 但是G++ 没通过； 这是一个简单的贪心算法；
