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http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2109
1、问题描述
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
2、代码：
#include<stdio.h>
#include<stdlib.h>
struct node
{double j,f;double p;
}a[1010];
int cmp(const void *a,const void *b)
{struct node *c=(node *)a;struct node *d=(node *)b;if(c->p > d->p) return -1;else return 1;
}
int main()
{int N;double M;double ans;while(scanf("%lf%d",&M,&N)){if(M==-1&&N==-1) break;for(int i=0;i<N;i++){scanf("%lf%lf",&a[i].j,&a[i].f);a[i].p=a[i].j/a[i].f;}qsort(a,N,sizeof(a[0]),cmp);ans=0;for(int i=0;i<N;i++){if(M>=a[i].f){ans+=a[i].j;M-=a[i].f;}else{ans+=(a[i].j/a[i].f)*M;break;}}printf("%.3lf\n",ans);}return 0;
}
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