FatMouse' Trade（贪心）

A - FatMouse’ Trade
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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Status
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

题意：多组输入，每组数据第一行给出M和N，接着N行，每行两个正整数J[i],F[i]。代表有M磅猫食，N中兑换JavaBeans的汇率，F[i]磅的猫食可以换J[i]的JavaBeans，如果剩下的M不足F[i]，则按照比例部分兑换，输出保留三位小数

思路：对J[i]/F[i]的比值进行降序排列，然后逐个取值即可，注意数据类型最好全用double

代码

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<math.h>#include<queue>using namespace std;struct node{    double J;    double F;    double num;//J/F} num[1000];bool cmp(node x,node y){    return x.num>y.num;}int main(){    int M,N;    while(~scanf("%d%d",&M,&N)&&M!=-1&&N!=-1)    {        for(int i=0; i<N; i++)        {            scanf("%lf%lf",&num[i].J,&num[i].F);            num[i].num=num[i].J/num[i].F;            //printf("%.3lf\n",num[i].num);        }        sort(num,num+N,cmp);        double sum=0.0;        for(int i=0; i<N; i++)        {            if(M>=num[i].F)            {                sum+=num[i].J;                M=M-num[i].F;            }            else            {                sum=sum+M/num[i].F*num[i].J;                break;            }        }        printf("%.3lf\n",sum);    }    return 0;}