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。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 using namespace std;
 5 const int maxn = 1e3+10;
 6 int m,n;
 7 struct node{
 8     int j,f;
 9     double r;
10 } food[maxn];
11 class cmp{
12 public:
13     bool operator()(node a,node b)const{
14         return a.r>b.r;
15     }
16 };
17 int main(){
18     while(~scanf("%d%d",&m,&n)){
19         if(m==-1) break;
20         for(int i=0; i<n; i++ ){
21             cin>>food[i].j>>food[i].f;
22             food[i].r=1.0*food[i].j/food[i].f;
23         }
24         sort(food,food+n,cmp());
25         double ans=0;
26         for( int i=0; i<n; i++ ){
27             if(m>=food[i].f){
28                 ans+=food[i].j;
29                 m-=food[i].f;
30             }
31             else{
32                 if(m==0) break;
33                 else{
34                     ans+=m*food[i].r;
35                     break;
36                 }
37             }
38         }
39         printf("%.3f\n",ans);
40     }
41     return 0;
42 }