FatMouse' Trade

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

 

简单贪心，思路是每次取javabean和catfood比值最大的。。。

AC代码:

<span style="font-size:14px;">#include<stdio.h>#include<algorithm>using namespace std;typedef struct{    int javab,catf;}node;node a[1010];int cmp(node a,node b){    return (a.javab*1.0/a.catf)>(b.javab*1.0/b.catf);}int main(){    //freopen("in.txt","r",stdin);    int n,m;    while(scanf("%d %d",&n,&m)!=EOF)    {        if(n==-1&&m==-1)            break;        for(int i=0;i<m;i++)            scanf("%d %d",&a[i].javab,&a[i].catf);        sort(a,a+m,cmp);        int tot=n;        double ans=0;        for(int i=0;i<m&&tot>=0;i++)        {            if(tot>=a[i].catf)            {                ans+=a[i].javab;                tot-=a[i].catf;            }            else            {                ans+=a[i].javab*1.0/a[i].catf*tot;                tot=0;            }        }        printf("%.3lf\n",ans);    }    return 0;}</span>