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

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:

Help Twiligh   t Sparkle to calculate: what is the minimum number of operations that she needs to sort the sequence?

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).

If it's impossible to sort the sequence output -1. Otherwise output the minimum number of operations Twilight Sparkle needs to sort it.

2
2 1

1

3
1 3 2

-1

2
1 2

0

思路：剛開始思路就錯了，只考慮相鄰的三位數- -！，試數據的時候可以試出來錯。

看有多少個a[i-1]>a[i]，用S記下，然後用now分別記下當前數的下標。要移動（shift）的數就是n-now-1，然後看S的值，如果S==1，表示可能會有移動成為

#include
#include
#include
#include
#include
#include
#define LL  int
#define inf 0x3f3f3f3f
using namespace std;
int  a[1000001];
int main()
{
   int n,m,i,j;int k,x,y;
    while(~scanf(%d,&n))
    {
        for(i=0;ia[i+1])
            {
                now=i;
                s++;
            }
        }
        if(s==0)
        {
            printf(0
);
            continue;
        }
        else if(s==1&&a[0]>=a[n-1])
        {
            printf(%d
,n-now-1);
        }
        else
        {
            printf(-1
);
        }
    }
    return 0;
}