hdu 1009 FatMouse' Trade（贪心）

struct数组用错，而且这个题不用考虑分母为0，还不报错，奇了怪了！！！

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

 

#include<iostream>#include<cstdio>#include<cstring>using namespace std;struct point{    double x,y;    double bl;}points[1001];int cp(const void *a,const void *b){    struct point *c=(point *)a;    struct point *d=(point *)b;    if(c->bl<=d->bl) return 1;    else return -1;//这儿是-1而不是0}int main(){    int n,m;    while(scanf("%d%d",&n,&m))    {        if(m==-1&&n==-1) break;        for(int i=0;i<m;i++)        {            scanf("%lf %lf",&points[i].x,&points[i].y);            points[i].bl=points[i].x/points[i].y;        }        qsort(points,m,sizeof(points[0]),cp);        double sum=0.0;        for(int j=0;j<m;j++)        {            if(n>=points[j].y)            {                sum+=points[j].x;                n-=points[j].y;            }            else            {                sum+=n*points[j].x/points[j].y;                break;            }        }        printf("%.3lf\n",sum);    }    return 0;}