Codeforces Round #259 (Div. 2) B. Little Pony and Sort by Shift

B. Little Pony and Sort by Shift

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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：给你一个序列，问你只能把最后一个数移到第一个位置，问最少需要几次可将序列变成递增序列

思路：直接扫一遍，记录断点，查看分块是否是递增的即可，注意到最后检查是否能组成一个序列，若两部分都是递增，那么查看第二部分的最后一个数和第一部分第一个数的大小

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
int a[MAXN];
int main()
{
int n,i;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
scanf("%d",&a[i]);
int k;
int cnt=0;
for(i=0;i<n;i++)
{
if(i==0)
k=a[i];
else
{
if(a[i]>=k)
k=a[i];
else
{
cnt=i;
break;
}
}
}
int bz=0;
for(int j=i;j<n;j++)
{
if(j==i)
k=a[j];
else
{
if(a[j]>=k)
k=a[j];
else
{
bz=1;
break;
}
}
}
if(bz==0)
{
if(i!=n&&a[n-1]>a[0])
bz=1;
}
if(bz)
printf("-1\n");
else
printf("%d\n",(n-cnt)%n);
}
return 0;
}