HDU-1009（简单贪心）

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 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

Sample Output

13.333 31.500

分析：按J[i]/F[i]的价值从大到小贪。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int maxn=;
 struct javabean{
     int j,f;
     double v;
 }value[maxn];
 bool cmp(javabean a,javabean b){
     return a.v>b.v;
 }
 int main()
 {
     int m,n;
     while(scanf("%d%d",&m,&n)){
         if(m==-&&n==-) break;
         for(int i=;i<n;i++){
             scanf("%d%d",&value[i].j,&value[i].f);
             value[i].v=(double)value[i].j/(value[i].f*1.0);
         }
         sort(value,value+n,cmp);
         double sum=0.0;
         for(int i=;i<n;i++){
             if(m>=value[i].f){
                 sum+=value[i].j;
                 m-=value[i].f;
             }
             else{
                 sum+=value[i].v*m;
                 break;
             }
         }
         printf("%.3f\n",sum);
     }
     return ;
 }