AtCoder Regular Contest 064 F - Rotated Palindromes

Problem Statement

Takahashi and Aoki are going to together construct a sequence of integers.

First, Takahashi will provide a sequence of integers a, satisfying all of the following conditions:

• The length of a is N.
• Each element in a is an integer between 1 and K, inclusive.
• a is a palindrome, that is, reversing the order of elements in a will result in the same sequence as the original.

Then, Aoki will perform the following operation an arbitrary number of times:

• Move the first element in a to the end of a.

How many sequences a can be obtained after this procedure, modulo 109+7?

Constraints

• 1≤N≤109
• 1≤K≤109

---

Input

The input is given from Standard Input in the following format:

N K

Output

Print the number of the sequences a that can be obtained after the procedure, modulo 109+7.

---

Sample Input 1

Copy

4 2

Sample Output 1

Copy

6

The following six sequences can be obtained:

• (1,1,1,1)
• (1,1,2,2)
• (1,2,2,1)
• (2,2,1,1)
• (2,1,1,2)
• (2,2,2,2)

---

Sample Input 2

Copy

1 10

Sample Output 2

Copy

10

---

Sample Input 3

Copy

6 3

Sample Output 3

Copy

75

---

Sample Input 4

Copy

1000000000 1000000000

Sample Output 4

Copy

875699961

用1-k填序列，序列可以左移n次变为回文

不考虑移位总共有k^(n/2)个回文，所以就是一个容斥问题了

若回文串的最小循环节长度为c,经过c次操作后,得到c个序列
当c为偶数时,重复数为一半,奇数时,无重复

所以这个问题就解决了，其循环节只可能是其因子

#include <bits/stdc++.h>
using namespace std;
const int N=,MD=1e9+;
int v[N],f[N];
int Pow(int x,int y)
{
    int ans=;
    for(;y;x=x*1LL*x%MD,y>>=)if(y&)ans=1LL*ans*x%MD;
    return ans;
}
int main()
{
    int n,k,tot=,ans=,i;
    cin>>n>>k;
    for(i=; i*i<n; i++)
        if(n%i==)v[tot++]=i,v[tot++]=n/i;
    if(i*i==n)v[tot++]=i;
    sort(v,v+tot);
    for(int i=; i<tot; i++)
    {
        f[i]=Pow(k,(v[i]+)/);
        for(int j=; j<i; j++)if(v[i]%v[j]==) f[i]=(f[i]-f[j])%MD;
        if(v[i]&) ans=(ans+1LL*v[i]*f[i])%MD;
        else ans=(ans+v[i]*1LL*Pow(,MD-)%MD*f[i])%MD;
    }
    cout<<(ans+MD)%MD;
}