hd 1009 FatMouse' Trade (贪心)

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 66411    Accepted Submission(s): 22564

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 37 24 35 220 325 1824 1515 10-1 -1

 

Sample Output

13.33331.500

 

#include<cstdio>#include<algorithm>using namespace std;struct note{int J;int F;double ave;}q[10010];bool cmp(note a ,note b){return a.ave > b.ave ;}int main(){int n , m ;while(~scanf("%d %d",&n,&m)){if(n == -1 && m == -1)   break;for(int i = 0 ; i < m ; i++){scanf("%d %d",&q[i].J,&q[i].F);q[i].ave = (q[i].J * 1.0) / (q[i].F * 1.0) ;}sort(q,q+m,cmp);double sum = 0 ;for(int i = 0 ;i < m ; i++){if(n == 0) break;if(n >= q[i].F){n-=q[i].F ;sum+=q[i].J * 1.0 ;}else{sum+=n * 1.0 * q[i].ave ;break;}}printf("%.3lf\n",sum);}return 0;}