hdu_1009_FatMouse' Trade

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Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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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 
简单的贪心问题，按照j/f比值由大到小排序，而且要用double型数据。
#include <stdio.h>#include <algorithm>using namespace std;struct node{    double j,f,ave;};node a[1002];double m;int n;bool cmp(node a,node b){    return a.ave>b.ave;}int main(){    while(scanf("%lf%d",&m,&n)!=EOF&&m!=-1&&n!=-1)    {        double sum=0;        for(int i=1;i<=n;i++) {            scanf("%lf%lf",&a[i].j,&a[i].f);            a[i].ave=(a[i].j/a[i].f);        }        sort(a+1,a+n+1,cmp);        for(int i=1;i<=n;i++)        {            if(m>=a[i].f)            {                sum+=a[i].j;                m-=a[i].f;}            else {sum+=a[i].ave*m;break;}        }        printf("%.3lf\n",sum);    }    return 0;}

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