CodeForces 34A Reconnaissance 2

 Reconnaissance 2

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

n soldiers stand in a circle. For each soldier his height a_i is known. A reconnaissance unit can be made of such two neighbouringsoldiers, whose heights difference is minimal, i.e. |a_i - a_j|

Input

The first line contains integer n (2 ≤ n ≤ 100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — n space-separated integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 1000). The soldier heights are given in clockwise or counterclockwise direction.

Output

Output two integers — indexes of neighbouring

Sample Input

Input

5 10 12 13 15 10

Output

5 1

Input

4 10 20 30 40

Output

1 2

大体题意：n个人围成一个圈，下面分别是n个人的身高，求这个圈里面的相邻的两个人身高差最小的编号是那两个？其中编号按输入数据的顺序排列，第一个是1，最后一个是n

AC代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int a[2101];
        int ii,jj;
        int cha;
        int min = 100000;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            if(i!=1)
            {
                cha = fabs(a[i] - a[i-1]);
                if(min > cha)
                {
                    min  = cha;
                    ii = i-1;
                    jj = i;
                }
            }

        }
        cha = fabs(a[1] - a[n]);
        if(min > cha)
        {
            min  = cha;
            ii = n;
            jj = 1;
        }
        printf("%d %d\n",ii,jj);
    }
    return 0;
}