xiaoxiao2021-03-01  21


int a[100]; for(int i=1;i<=10;i++) { scanf("%d",&a[i]); } sort(a,a+10);//OK,排完了




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.


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.


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




#include<iostream> #include<algorithm> #include<stdio.h> using namespace std; struct mao { int f; int j; double v; } c[1001]; bool cmp(mao a,mao b)//根据结构体中的数据v,对结构体进行排序,从大到小排 { return a.v>b.v; } int main() { int m,n,i; while(scanf("%d %d",&m,&n)!=EOF,m!=-1&&n!=-1) { for(i=1; i<=n; i++) { scanf("%d %d",&c[i].j,&c[i].f); c[i].v=1.0*c[i].j/c[i].f; } sort(c+1,c+n+1,cmp); double sum=0; for(i=1; i<=n; i++) { if(m>=c[i].f) { m=m-c[i].f; sum=sum+c[i].j; } else { sum=sum+m*c[i].v; break; } } printf("%.3lf\n",sum); } return 0; }