湖北民族学院oj 1775（贪心）（杭电1009） 之 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

题意：

FatMouse有m镑cat food，他想用cat food去换JavaBeans，有n个房间可以换，交换规则为a%*f[i]磅的cat food可以换a%*j[i]磅的

JavaBeans，问他最多可以换取多少JavaBeans？

贪心，按比例j[i]/f[i]优先换取即可

AC代码如下：

#include 
#include 
#include 
using namespace std;
const int maxn=1000+10;

struct point
{
    int a,b;
    double value;
}s[maxn];

bool cmp(point x,point y)
{
    return x.value>=y.value;
}


int main()
{
    int n,m;
    double ans,cur;
    while(cin>>n>>m && n!=-1 && m!=-1)
    {
        ans=0;
        for(int i=0;i>s[i].a>>s[i].b;
            s[i].value=(double)s[i].a/s[i].b;
        }
        sort(s,s+m,cmp);
        cur=(double)n;
        for(int i=0;is[i].b)
            {
                cur-=s[i].b;
                ans+=s[i].a;
            }
            else
            {
                ans+=(s[i].value*cur);
                break;
            }
        }
        printf("%.3lf\n",ans);
    }
    return 0;
}