Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 41982    Accepted Submission(s): 13962

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

  

题意就是老鼠用猫粮换鼠粮。

。

。（Orz）。。

求它最多能换多少。。一道贪心水题。

。

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #define f1(i, n) for(int i=0; i
             
               b.c;}int main(){ while(scanf("%d%d", &n, &m)!=EOF) { ans = 0; t = (double) n; memset(Q, 0, sizeof(Q)); if(n==-1 && m==-1) break; f1(i, m) { scanf("%lf%lf", &Q[i].J, &Q[i].F); Q[i].c = Q[i].J / Q[i].F; } sort(Q, Q+m, cmp); f1(i, m) { if( Q[i].F<=t ) { ans += Q[i].J; t -= Q[i].F; } else { ans += t * Q[i].c; break; } } printf("%.3lf\n", ans); } return 0;}