暑假培训题

寻找排队时间最长者的位置

最新推荐文章于 2021-07-21 03:02:21 发布

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本文探讨了一种算法，用于解决在餐厅排队时找到等待时间最长者的最优位置问题。通过输入人员的A、B值，算法能计算出每个人实际等待时间，并找出等待时间最长的人所在位置。

Description

In HUST,there are always many students go to the mell hall at the same time as soon as the bell rings.

Students have to queue up for a meal ,and the queue is awalys long,So it takes much time.Suposse there

are N people in a queue,each person has two characteristic value A and B(both of them are integers,read

input for more details),the i-th person in the queue have to spend m(i) =

A1A2

:::Ai 1

minutes,Where

,Bi

is the i-th person.s value A,B.Notice that if the order of the queue changes,the waiting time one

spend(that is,the value of m(i))may changes too. Of course,every student want to reduce the time he

spend.

Apparently,it is impossible to make everyone satis ed,in this problem,we only need to minimize the

waiting time of one who spend the longest time in the queue,that is,minimize Max m(1),m(2),,,m(n).You

can change the order of the queue in anyway in order to complete this problem.

You are asked to output the original location in the queue of the person who will cost the longest time

under optimal solution. Uniquity is insured by the given data.

Input

There are multiple test cases.

For each case, the rst line contains one integerN(1N1000).

The second line containsNintegers Ai

.(0 < Ai <100000)

The third line containsNintergers Bi

(0< Bi <10).

(Bi <10< Ai*bi)

Output

You are asked to output the original location in the queue of the person who will cost the longest time

under optimal solution. Uniquity is insured by the given data.

Sample Input

3
15 20 25
1 3 2

Sample Output

这道题刚看的时候真的被那个求时间的公式坑了，其实这道题就是求排队人中的A*B最大人的编号

#include<stdio.h>
int n;
struct node{
      int x;//记录权值A
      int y;//记录权值B
      int num;//记录编号
      int f;//记录A和B的乘积
}s[1005];
int main()
{
     int i,max=0,j;
     while(scanf("%d",&n)!=EOF)
     {
        for(i=0;i<n;i++)
             scanf("%d",&s[i].x);
         for(i=0;i<n;i++)
         {
            scanf("%d",&s[i].y);
             s[i].f=s[i].x*s[i].y;
             s[i].num=i+1;
         }
        for(i=0;i<n;i++)
        {
        if(max<s[i].f)
        {max=s[i].f;
        j=i;}
        }
        printf("%d\n",s[j].num);
        max=0;
     }
     return 0;
     }