zoj 2109 FatMouse' Trade简单的贪心（注意double数组的排序问题）

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本文详细阐述了FatMouse在利用猫粮与JavaBean进行交易时的优化算法，通过输入猫粮数量与房间内JavaBean与所需猫粮比例，计算出最大可获取的JavaBean数量，实现资源的有效利用。

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2109

1、问题描述

FatMouse' Trade

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 2109

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

2、代码：

#include<stdio.h>
#include<stdlib.h>
struct node
{
    double j,f;
    double p;
}a[1010];
int cmp(const void *a,const void *b)
{
    struct node *c=(node *)a;
    struct node *d=(node *)b;
    if(c->p > d->p) return -1;
    else return 1;
}
int main()
{
    int N;
    double M;
    double ans;
    while(scanf("%lf%d",&M,&N))
    {
        if(M==-1&&N==-1) break;
        for(int i=0;i<N;i++)
        {
           scanf("%lf%lf",&a[i].j,&a[i].f);
           a[i].p=a[i].j/a[i].f;
        }
        qsort(a,N,sizeof(a[0]),cmp);
        ans=0;
        for(int i=0;i<N;i++)
        {
            if(M>=a[i].f)
            {
                ans+=a[i].j;
                M-=a[i].f;
            }
            else
            {
                ans+=(a[i].j/a[i].f)*M;
                break;
            }
        }
        printf("%.3lf\n",ans);
    }
    return 0;
}