杭电-1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 66325    Accepted Submission(s): 22518

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

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

Recommend

JGShining

 

AC代码：

<pre name="code" class="cpp">#include <stdio.h>#include <algorithm>using namespace std;struct Node{    double j,f,p;} node[10000];int cmp(Node x,Node y){    return x.p>y.p;}int main(){    int m,n;    while(~scanf("%d%d",&n,&m) && (m!=-1 || n!=-1))    {        double sum = 0,max = 0;        int i,j;        for(i = 0; i<m; i++)        {            scanf("%lf%lf",&node[i].j,&node[i].f);            node[i].p = node[i].j/node[i].f;        }        sort(node,node+m,cmp);        for(i = 0; i<m; i++)        {            if(n>=node[i].f)            {                sum+=node[i].j;                n-=node[i].f;            }            else            {                sum+=node[i].p*n;                break;            }        }        printf("%.3lf\n",sum);    }    return 0;}