問題 E: FatMouse's Trade

題目描述

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.

 

輸入

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.

 

輸出

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.

 

樣例輸入

4 2
4 7
1 3
5 5
4 8
3 8
1 2
2 5
2 4
-1 -1

樣例輸出

2.286
2.500

思路：

貪心算法，具體思路可以看代碼註釋部分

#include<bits/stdc++.h>
using namespace std;

struct fat
{
    int j; //食物
    int f; // 貓糧
};

bool cmp(fat a, fat b)
{
    return ((double)a.f/a.j) < ((double)b.f/b.j); //數值越小，表示可以用更少的貓糧換取更多的食物
}
int main()
{
    int m, n; //分別表示 m 磅的貓糧 和 n行數據
    while(cin >>m >> n && m!=-1 && n!=-1)
    {
        struct fat s[n];
        for(int i=0;i<n;i++)
        {
            cin >> s[i].j >> s[i].f;
        }
        sort(s, s+n, cmp);
        double ans = 0.0;
        for(int i=0;i<n;i++)
        {
            if(m>=s[i].f) //當準備的貓糧可以換取一個房間的所有食物
            {
                ans += s[i].j;
                m -= s[i].f;
            }
            else //準備的貓糧可以換取一個房間的部分食物，按比例分配，此時，準備的貓糧完全換出了，可以直接退出循環
            {
                ans += (double)m / s[i].f * s[i].j;
                break;
            }
        }
        printf("%.3f\n", ans);

    }
    return 0;
}