Q1009 FatMouse' Trade C++

问题描述：

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 <vector>
#include <iomanip>

using namespace std;

struct food
{
double j, f, percent;
};

bool cmp(food a, food b)
{
return a.percent > b.percent;
}

int main()
{
double M, N;

while(cin >> M >> N && M != -1 || N != -1)
{
double final = 0;
food temp;
vector<food> v;
for (int i = 0; i < N; i++)
{
cin >> temp.j >> temp.f;
temp.percent = temp.j / temp.f;
v.push_back(temp);
}
sort(v.begin(),v.end(),cmp);
int count = 0;
while(M>0 && count < N)
{
if(v[count].f <= M)
{
M -= v[count].f;
final += v[count].j;
count++;
}
else
{
final += M * v[count].percent;
count++;
M = 0;
}
}
cout << setprecision(3) << fixed << final << endl;
}
return 0;
}