FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 55094 Accepted Submission(s): 18478

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

解题思路:

这是一个简单的贪心的题目，我感觉做贪心的题的时候需要注意的情况就是排序，只要抓住按照什么排序的话，

贪心就迎刃而解了，那这个题的话，就是按照J[i] / F[i]从大到小排序，只要排完序了，在根据题意进行一下计算

那不就是ok了，

上代码:

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              using namespace std;#define MM(a) memset(a,0,sizeof(a))typedef long long LL;typedef unsigned long long ULL;const int maxn = 100000+5;const int mod = 1000000007;const double eps = 1e-7;struct Node{ double J, F, ave;}arr[maxn];double cmp(Node a, Node b){ return a.ave > b.ave;}int main(){ int m, n; while(cin>>m>>n) { if(m==-1 && n==-1) break; for(int i=0; i
              
               >arr[i].J>>arr[i].F; arr[i].ave = arr[i].J/arr[i].F; } sort(arr, arr+n, cmp); double ret = 0.0; for(int i=0; i
               
                = arr[i].F) { ret += arr[i].J; m -= arr[i].F; } else { ret += arr[i].J/arr[i].F*m; break;///注意要break哦 } } printf("%.3lf\n",ret); } return 0;}