AtCoder Regular Contest 064 F - Rotated Palindromes

最新推荐文章于 2024-03-25 21:55:19 发布

转载最新推荐文章于 2024-03-25 21:55:19 发布 · 154 阅读

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原文链接：http://www.cnblogs.com/BobHuang/p/9607224.html

本文探讨了构造特定条件下的回文序列问题，通过算法分析，详细解释了如何计算在给定长度N和元素范围K的情况下，能通过旋转操作获得的不同回文序列数量。文章深入讨论了容斥原理的应用，并提供了完整的C++代码实现。

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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

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

Sample Input 3

Copy

6 3

Sample Output 3

Copy

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=2010,MD=1e9+7;
int v[N],f[N];
int Pow(int x,int y)
{
    int ans=1;
    for(;y;x=x*1LL*x%MD,y>>=1)if(y&1)ans=1LL*ans*x%MD;
    return ans;
}
int main()
{
    int n,k,tot=0,ans=0,i;
    cin>>n>>k;
    for(i=1; i*i<n; i++)
        if(n%i==0)v[tot++]=i,v[tot++]=n/i;
    if(i*i==n)v[tot++]=i;
    sort(v,v+tot);
    for(int i=0; i<tot; i++)
    {
        f[i]=Pow(k,(v[i]+1)/2);
        for(int j=0; j<i; j++)if(v[i]%v[j]==0) f[i]=(f[i]-f[j])%MD;
        if(v[i]&1) ans=(ans+1LL*v[i]*f[i])%MD;
        else ans=(ans+v[i]*1LL*Pow(2,MD-2)%MD*f[i])%MD;
    }
    cout<<(ans+MD)%MD;
}

转载于:https://www.cnblogs.com/BobHuang/p/9607224.html