FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 66415    Accepted Submission(s): 22567

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
刚开始一直不懂a啥意思，最后我把a当成100，a%相当于1；这样就容易懂了。
<span style="font-family:Arial Black;font-size:18px;color:#ffffff;BACKGROUND-COLOR: #000000"><strong>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct bg{int a,b;double c;}s[1100];bool cmp(bg a,bg b){return a.c>b.c;}int main(){int m,n,i;double sum;while(scanf("%d%d",&m,&n)&&(m!=-1||n!=-1)){sum=0;for(i=0;i<n;i++){scanf("%d%d",&s[i].a,&s[i].b);s[i].c=s[i].a*1.0/s[i].b;}sort(s,s+n,cmp);for(i=0;i<n;i++){if(m>=s[i].b){sum+=s[i].a;m=m-s[i].b;}else{sum+=m*s[i].c;break;}}printf("%.3lf\n",sum);}return 0;}</strong></span>