FatMouse' Trade

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

代码如下

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#define max(a,b)   (a>b?a:b)
#define min(a,b)   (a<b?a:b)
#define swap(a,b)  (a=a+b,b=a-b,a=a-b)
#define maxn 320007
#define N 100000000
#define INF 0x3f3f3f3f
#define mod 1000000009
#define e  2.718281828459045
#define eps 1.0e18
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define ll long long
using namespace std;

struct food
{
    int x;
    int y;
    double z;
} a[1111];
bool cmp(struct food a,struct food b)
{
    return a.z>b.z;
}
int main()
{
    int m,n;
    while(cin>>m>>n)
    {
        if(m==-1&&n==-1)
            break;
        for(int i=0; i<n; i++)
        {
            cin>>a[i].x>>a[i].y;
            a[i].z=a[i].x*1.0/a[i].y;
        }
        sort(a,a+n,cmp);
        double sum=0;
        for(int i=0; i<n; i++)
            if(m>=a[i].y)
            {
                m-=a[i].y;
                sum+=a[i].x;
            }
            else
            {
                sum+=m*a[i].z;
                break;
            }
        printf("%.3lf\n",sum);
    }
    return 0;
}