hdu sort排序

最新推荐文章于 2022-12-03 19:48:18 发布

smilelingling

最新推荐文章于 2022-12-03 19:48:18 发布

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Problem F

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 21 Accepted Submission(s) : 19

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 3 7 2 4 3 5 2 20 3 25 18 24 15 15 10 -1 -1

Sample Output

13.333 31.500

这道题很简单，只是英语太长，没有看，说一下题意。。。。。。。。

第一行代表老鼠有限定的猫粮和有几种兑换的方式，接着n行就是老鼠用猫粮去换自己的粮食；

那么要想在有限的猫粮中换更多自己的粮食，就要选出最佳的兑换方式，用sort排序就可以，然后需要减去用过的猫粮，继续兑换，剩下不足的，按照不足的兑换。。。。。

#include<bits/stdc++.h>
using namespace std;
struct node
{
    double a,b,k;
}m[1010];
int cmp(node x,node y)
{
    return x.k>y.k;
}
int main()
{
    int t,n;
    double sum;
    while(~scanf("%d%d",&t,&n))
    {
        if(t==-1&&n==-1) break;
        sum=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&m[i].a,&m[i].b);
            m[i].k=m[i].a/m[i].b;
        }
        sort(m,m+n,cmp);
        for(int i=0;i<n;i++)
        {
            if(t>m[i].b)
            {
                sum=sum+m[i].a;
                t=t-m[i].b;
            }
            else
            {
                sum=sum+t*m[i].k;
                break;
            }
        }
        printf("%.3lf\n",sum);
    }
}