（周日赛）Little Pony and Sort by Shift

题意及题解：判断有存在几个后面的数比前面的小；
如果没有则输出0；如果有不止一个，则输出-1；如果恰有一个， 那么就需要判断a[1]和a[n]的大小关系。

 

Description
One day, Twilight Sparkle is interested in how to sort a sequence of integers a1, a2, ..., an in non-decreasing order. Being a young unicorn, the only operation she can perform is a unit shift. That is, she can move the last element of the sequence to its beginning:

a1, a2, ..., an → an, a1, a2, ..., an - 1.
Help Twilight Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?

Input
The first line contains an integer n(2 ≤ n ≤ 105). The second line contains n integer numbers a1, a2, ..., an(1 ≤ ai ≤ 105).

Output
If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.

Sample Input
Input
2
2 1
Output
1

Input
3
1 3 2
Output
-1

Input
2
1 2
Output
0

 

#include<stdio.h>
int main()
{
    int n,i,j,t;
    int a[100001];
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        int ans=0;
        for(i=1;i<n;i++)
        {
            if(a[i]<a[i-1])
            {
                t=i+1;
                ans++;
            }
        }
        if(ans==0)
            puts("0");
        else if(ans==1&&a[n-1]<=a[0])
        {
            printf("%d\n",n-t+1);
        }
        else
        puts("-1");
    }
    return 0;
}