杭电 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40975    Accepted Submission(s): 13563

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
简单背包     J[i]与F[i]比例大的先拿  注意J F数组要随着比例一起排序。。。。。

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;struct food{    double j;    double f;    double p;}a[1005];int cmp(food b,food c){    return b.p>c.p;}int main(){    int m,n,i,k;    double sum,l,s;    while(~scanf("%d%d",&n,&m))    {        sum=0;        s=0;        l=0;        if((n==-1)&&(m==-1))            break;        for(i=0;i<m;i++)            scanf("%lf%lf",&a[i].f,&a[i].j);        for(i=0;i<m;i++)            {                a[i].p=a[i].f/a[i].j;            }       sort(a,a+m,cmp);        for(i=0;i<m;i++)            {                    {s+=a[i].j;k=i;}                if(s>n)                    {l=n-(s-a[i].j);break;}            }            for(i=0;i<k;i++)                    {sum+=a[i].f;}                    if((int)l)                        sum+=l*a[k].p;                printf("%.3lf\n",sum);    }return 0;}