HDU1009--FatMouse' Trade

FatMouse' Trade
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33038 Accepted Submission(s): 10693


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


Author
CHEN, Yue


Source
ZJCPC2004


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JGShining

#include<cstdio> 
#include<cmath> 
#include<iostream> 
#include<algorithm> 
#include<cstring> 
using namespace std; 
    
struct node 
{ 
    int ja,va; 
    double sor; 
}edge[10001]; 
    
bool cmp(node a,node b) 
{ 
    return a.sor<b.sor; 
} 
    
int main() 
{ 
    int m,n; 
    while(scanf("%d%d",&m,&n)!=EOF) 
    { 
        if(m==-1&&n==-1)break; 
        int i; 
        for(i=0;i<n;i++) 
        { 
            scanf("%d%d",&edge[i].ja,&edge[i].va); 
            edge[i].sor=edge[i].va*1.0/edge[i].ja; 
        } 
    
        sort(edge,edge+n,cmp); 
    
        double sum=0; 
        for(i=0;i<n;i++) 
        { 
            if(m>edge[i].va) 
            { 
                sum+=edge[i].ja; 
                m-=edge[i].va; 
            } 
            else
            { 
                sum+=m*1.0/edge[i].va*edge[i].ja; 
                break; 
            } 
        } 
        printf("%0.3lf
",sum); 
    } 
    return 0; 
}