***Little Pony and Sort by Shift***
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.
Examples
Input
2
2 1
Output
1
Input
3
1 3 2
Output
-1
Input
2
1 2
Output
0
题意：将一串数字按照非递减的顺序排列并输出，但是排列的规则是只能将最后一个数据挪到第一个数据，而后倒数第二个变成最后一个，然后接着进行刚才的操作。问能不能按照递增顺序排列出来。
思路：整串数据中只能有一个下坡（也就是整串数据中只能有一个后一位的数据比前一位的数据小），并且，原始的第一个数据必须比最后一个大。这样才能实现数据的按照递增顺序的排列。
在这里插入代码片
#include"stdio.h"
int main()
{
int n,a[100005];
while(~scanf("%d",&n))
{
int t=0;
int q;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
for(int i=2;i<=n;i++)
{
if(a[i]<a[i-1])
{
t++;
q=i;
}
}
if(t0)
{
printf(“0\n”);
}
else if(t1&&a[1]>=a[n])
{
printf("%d\n",n-q+1);
}
else if(t>1)
{
printf("-1\n");
}
else
printf("-1\n");
}
return 0;
}
来源：https://blog.csdn.net/weixin_44824383/article/details/99059899