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

无聊刷的贪心水题，主要是研究c++的浮点型问题。

1.头文件里加#include<iomanip>。
2.cout<<setioflags(io::fixed)<<setprecision(位数)<<变量<<endl;

AC代码如下：

#include<iostream>#include<algorithm>#include<iomanip>using namespace std;struct H{    int x,y;    double z;}f[10005];bool cmp(H a,H b){    return a.z>b.z;}int main(){    int m,n;    int i,j;    while(cin>>m>>n,m!=-1&&n!=-1)    {        for(i=0;i<n;i++)        {            cin>>f[i].x>>f[i].y;            f[i].z=(double)f[i].x/f[i].y;        }        sort(f,f+n,cmp);        double sum=0;        for(i=0;i<n;i++)        {            m-=f[i].y;            if(m>0)            {                sum+=(double)f[i].x;            }            else            {                sum+=f[i].z*(m+f[i].y);                break;            }        }        cout<<setiosflags(ios::fixed);        cout<<setprecision(3)<<sum<<endl;    }return 0;}