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

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

SubmitStatusPracticeZOJ 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;}