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
2
2 1
output
1
input
3
1 3 2
output
-1
input
2
1 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]

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#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]



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