FatMouse' Trade

前言：水题，感觉一道很经典的贪心的算法，关于题目的理解，有句话很重要“he may get J[i]* a% pounds of JavaBeans if he pays F[i]* a% pounds of cat food”。

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

C语言代码：

#include<stdio.h>#include<stdlib.h>struct node{    double j;    double f;}s[1001];     //用结构体感觉会方便很多int cmp(const void *a,const void *b){//这里传入的参数是const void类型的指针    struct node *J=(struct node *)a;  //这里一定不要忘了把指针a转化为struct node类型的    struct node *F=(struct node *)b;    return J->f*F->j-J->j*F->f>0?1:-1;}int main(){    int m,n;    double fsum;    while(scanf("%d %d",&m,&n)!=EOF&&(m!=-1&&n!=-1)){    //这里一开始写的时候掉了掉了EOF，我也是日了狗了    for(int i=0;i<n;i++){        scanf("%lf %lf",&s[i].j,&s[i].f);    }    qsort(s,n,sizeof(s[0]),cmp);    fsum=0.000;    for(int i=0;i<n;i++){        if(m>0){            if(m>s[i].f){                fsum+=s[i].j;                m-=s[i].f;            }            else{                fsum+=m*s[i].j/s[i].f;                m=0;            }        //这里的写法是参照别人的，这种写法真的很好        }        else        break;    }    printf("%.3f\n",fsum);    }    return 0;}