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
贪心法水题，
个人认为这句话难理解：he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food
这样表达能够购买几分之几的。
#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
struct twoInts
{
int j, f;
bool operator<(const twoInts two) const
{
double a = (double)j / (double)f;
double b = (double)two.j / (double)two.f;
return a > b;
}
};
int main()
{
int M, N;
while (scanf("%d %d", &M, &N) && -1 != M)
{
vector<twoInts> vt(N);
for (int i = 0; i < N; i++)
{
scanf("%d", &vt[i].j);
scanf("%d", &vt[i].f);
}
sort(vt.begin(), vt.end());
double maxBean = 0.0;
for (int i = 0; i < N; i++)
{
if (M >= vt[i].f)
{
maxBean += vt[i].j;
M -= vt[i].f;
}
else
{
maxBean += (double)vt[i].j * M / (double)vt[i].f;
break;
}
}
printf("%.3lf\n", maxBean);
}
return 0;
}