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

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.

Sample test(s)

input

22 1

output

1

input

31 3 2

output

-1

input

21 2

output

0

        题意：对数组排序，你只能将最后一个元素放到第一个位置。问要操作几次。如果不能这样实现排序，输出-1。

        思路：统计第n个元素比第n-1个元素小的个数。如果有两个以上，直接输出-1。如果只有一个，比较最后一个元素和第一个元素，如果最后一个元素小于第一个元素，那么可以排序，操作次数由出现a[n]<a[n-1]的位置得出。

#include <iostream>  #include <stdio.h>  #include <cmath>  #include <algorithm>  #include <iomanip>  #include <cstdlib>  #include <string>  #include <memory.h>  #include <vector>  #include <queue>  #include <stack>  #include <map>#include <set>#define ll long long#define INF 1000000using namespace std;int num[100010];int main(){int n;while(cin>>n){for(int i=1;i<=n;i++){scanf("%d",&num[i]);}int cnt=0;int k;for(int i=2;i<=n;i++){if(num[i]<num[i-1]){cnt++;k=i;}}if(cnt==0){cout<<"0"<<endl;}else if(cnt==1){if(num[n]<=num[1]){cout<<n-k+1<<endl;}else{cout<<"-1"<<endl;}}else{cout<<"-1"<<endl;}}return 0;}