CodeForces 454B Little Pony and Sort by Shift

D - Little Pony and Sort by Shift

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 454B

Appoint description: System Crawler  (Jul 20, 2016 9:07:41 PM)

Description

One day, Twilight Sparkle is interested in how to sort a sequence of integers a₁, a₂, ..., a_n 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, ..., a_n → a_n, a₁, a₂, ..., a_n - 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, ..., a_n(1 ≤ a_i ≤ 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

22 1

Output

1

Input

31 3 2

Output

-1

Input

21 2

Output

0

一定是不小心犯迷糊了，明明很水的题，却被自己搞的那么复杂，还一直没过，我也是无语了。。

题意：给出一个序列，问最少经过几步可以使它变为一个非递减序列，每步只能把最后一个放到第一个。

思路：判断一下这个序列式由几个递增序列组成，如果是由大于两个递增序列组成，则不行，如果是1个，

直接输出0，如果是两个再比较一下最后一个和第一个。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){int n,i,j,k,l,a[110000];while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++)scanf("%d",&a[i]);int flag=0;for(i=2;i<=n;i++){if(a[i]<a[i-1]){flag++;k=i;} }if(flag==0)printf("0\n");else if(flag>1)printf("-1\n");else if(flag==1&&a[n]<=a[1]){printf("%d\n",n-k+1);}else printf("-1\n");}return 0; }