杭电(1009)FatMouse' Trade (贪心)

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FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 62841 Accepted Submission(s): 21210

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 37 24 35 220 325 1824 1515 10-1 -1

Sample Output

13.33331.500
题意（一直肥老鼠用f[i]的东西可以换取j[i]的爱吃的食物）
相当于部分背包问题（可以分割）
按照j[i]/f[i]的价值比从大往小排序。
#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <string>using namespace std;struct food{    int f;    int j;}s[1005];bool cmp(struct food a,struct food b){    return ((a.j*1.0)/a.f*1.0)>((b.j*1.0)/b.f*1.0);}int  main(){int n,m;while(cin>>n>>m&&(n!=-1,m!=-1)){    int i;    for(i=0;i<m;i++)    {        cin>>s[i].j>>s[i].f;    }    double sum=0;    sort(s,s+m,cmp);    for(i=0;i<m;i++)    {        if(n>=s[i].f)        {
        n-=s[i].f;        sum+=s[i].j;        }        else        {            sum+=n*((s[i].j*1.0)/s[i].f*1.0);            break;        }    }    printf("%.3lf\n",sum);}    return 0;}

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