九、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 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;