FatMouse' Trade（杭电ACM---1009）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 103515    Accepted Submission(s): 36159

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

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

题意：

这一道题意思就是老鼠用猫食物换取自己最喜爱的食物javaBean的过程，当然换取的最终结果是保证最后的JavaBean是最多的，

而且是当自己手中的猫食物小于每个仓库所需交换的猫食物时候，可以手中有多少就交换多少。

所以在解这道题时候要想到按照每个仓库javaBean最大的比率排序才能保证最后的交换的javaBean是最大的。

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 using namespace std;
 const int maxn = 1e3+;
 int m,n;
 struct node{
     int j,f;
     double r;
 } food[maxn];
 class cmp{
 public:
     bool operator()(node a,node b)const{
         return a.r>b.r;
     }
 };
 int main(){
     while(~scanf("%d%d",&m,&n)){
         if(m==-) break;
         for(int i=; i<n; i++ ){
             cin>>food[i].j>>food[i].f;
             food[i].r=1.0*food[i].j/food[i].f;
         }
         sort(food,food+n,cmp());
         double ans=;
         for( int i=; i<n; i++ ){
             if(m>=food[i].f){
                 ans+=food[i].j;
                 m-=food[i].f;
             }
             else{
                 if(m==) break;
                 else{
                     ans+=m*food[i].r;
                     break;
                 }
             }
         }
         printf("%.3f\n",ans);
     }
     return ;
 }