hdu FatMouse' Trade

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FatMouse' Trade

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) : 1

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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

贪心，能买性价比高的，就买高的。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct node{    int p,q;    double v;} s[1005];int cmp(node x,node y){    return x.v>y.v;}int main(){    int m,n;    while(scanf("%d%d",&m,&n))    {        if(m==-1&&n==-1)            break;        for(int i=0; i<n; i++)        {            scanf("%d%d",&s[i].p,&s[i].q);            s[i].v=s[i].p*1.0/s[i].q;        }        sort(s,s+n,cmp);        double ans=0;        for(int i=0;i<n;i++)        {            if(m>=s[i].q)            {                ans+=s[i].p;                m-=s[i].q;            }            else            {                ans+=s[i].p*1.0/s[i].q*m;                break;            }        }        printf("%.3lf\n",ans);    }    return 0;}