hdoj1009FatMouse' Trade（贪心）

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

 代码：

#include<algorithm>#include<iostream>#include<string.h>using namespace std;struct stu{double j,f;double sum;}a[1100];bool cmp(stu a,stu b){return a.sum>b.sum;}int main(){int n,m,i;double p;while(scanf("%d%d",&n,&m)!=EOF){p=0;if(n==-1&&m==-1)break;for(i=0;i<m;i++){scanf("%lf%lf",&a[i].j,&a[i].f);a[i].sum=a[i].j/a[i].f;}sort(a,a+m,cmp);for(i=0;i<m;i++){if(a[i].f<=n){n=n-a[i].f;p+=a[i].j;}else{p+=(n/a[i].f)*a[i].j;break;}}printf("%.3lf\n",p);}return 0;}

思路：简单的贪心题，比较比值大小然后加起来就行了。