FatMouse's Trade

题目描述
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.

输入描述
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.

输出描述
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.

输入样例
4 2
4 7
1 3
5 5
4 8
3 8
1 2
2 5
2 4
-1 -1

输出样例
2.286
2.500

贪心题

#include"iostream"#include"algorithm"using namespace std; struct node{    double x,y,bi;}; bool compare(node a,node b){    return a.bi>b.bi;}int main(){    int m,n;         while(cin>>m>>n)    {        if(m==-1&&n==-1)        break;        node ai[n];        for(int i=0;i<n;i++)        {            cin>>ai[i].x>>ai[i].y;            ai[i].bi=ai[i].x/ai[i].y;        }        sort(ai,ai+n,compare);        double sum=0;        for(int i=0;i<n;i++)        {            if(m<=ai[i].y)            {                sum+=m*ai[i].bi;                break;            }            sum+=ai[i].x;            m-=ai[i].y;        }        cout.precision(3);        cout.setf(ios::fixed);        cout<<sum<<endl;    }    return 0;}