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

题目链接：http://codeforces.com/contest/454/problem/B

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

思路：

判断有存在几个“下坡”，也就是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；

代码如下：

#include <cstdio>#include <cstring>int main(){int n;int i, j;int a[100017];while(~scanf("%d",&n)){for(i = 1; i <= n; i++){scanf("%d",&a[i]);}int cont = 0;int t;for(i = 2; i <= n; i++){if(a[i] < a[i-1]){cont++;t = i;}}if(cont == 0)printf("0\n");else if(cont > 1)printf("-1\n");else if(cont == 1 && a[n] <= a[1])printf("%d\n",n-t+1);elseprintf("-1\n");}return 0;}