Atcoder Regular Contest 064 F Rotated Palindromes
2018-02-25 18:04
489 查看
Rotated Palindromes
Problem Statement
给出n,k。问有多少个长度为n的序列a,对于任意1≤i≤n满足1≤ai≤k,并且序列a经过若干次循环左移能变成回文序列.
For example:循环左移一次1,1,2,3 -> 1,2,3,1
1≤n,k≤109
Solution
给出一个显然的结论,一个回文串的最小循环节一定回文。先找到n的所有约数,设第i个约数为Xi。
设fi表示序列的最小循环节长度为Xi,最小循环节的个数。
考虑fi对答案的贡献。
循环节对应的序列左移操作Xi次,就可以得到Xi种不同的序列。
但可以发现,若Xi为偶数时会算重一半,乘上12。
fi的求法也不难,容斥即可。即fi=k⌊Xi+12⌋-∑Xd|Xifd
时间复杂度O(σ20(n))
Code
#include<bits/stdc++.h> #define fo(i,j,l) for(int i=j;i<=l;++i) #define fd(i,j,l) for(int i=j;i>=l;--i) using namespace std; typedef long long ll; const ll N=2e5,mo=1e9+7; int n,o,p; ll f ,a ,k; ll ksm(ll o,ll t) { ll y=1; for(;t;t>>=1,o=o*o%mo) if(t&1)y=y*o%mo; return y; } int main() { cin>>n>>k; for(int i=1;i*i<=n;++i) if(n%i==0){ a[++o]=i; if(i*i%n)a[++o]=n/i; } sort(a+1,a+o+1); ll ans=0; fo(i,1,o){ f[i]=ksm(k,(a[i]+1)>>1); fo(l,1,i-1)if(a[i]%a[l]==0)f[i]=(f[i]-f[l]+mo)%mo; (a[i]&1)?ans=(ans+f[i]*a[i])%mo:ans=(ans+(f[i]*a[i]>>1))%mo; } cout<<ans; }
相关文章推荐
- AtCoder Regular Contest 064 F - Rotated Palindromes 乱搞
- [AtCoder Regular Contest 064] F: Rotated Palindrome (arc064F)
- AtCoder Regular Contest D - Remainder Reminder 取余问题
- AtCoder Regular Contest 092 C - 2D Plane 2N Points 贪心 匈牙利算法模板
- AtCoder Regular Contest 093 D - Grid Components
- AtCoder Regular Contest 059 F Unhappy Hacking
- AtCoder Regular Contest 078
- AtCoder Regular Contest 079 D - Decrease (Contestant ver.)
- AtCoder Regular Contest 079 C D E
- AtCoder Regular Contest 061 DSnuke's Coloring
- AtCoder Regular Contest 082-F-Sandglass
- 【AtCoder Regular Contest 082 F】Sandglass
- AtCoder Regular Contest 088
- 【AtCoder Regular Contest 092】C.2D Plane 2N Points(匈牙利算法/tuple+set 贪心)
- AtCoder Regular Contest 092 C - 2D Plane 2N Points 贪心 匈牙利算法模板
- AtCoder Regular Contest 093 D - Grid Components
- 从AtCoder Regular Contest 077D: 11 中学习逆元的线性求法
- AtCoder Regular Contest 082
- 【AtCoder Regular Contest 082 A】Together
- AtCoder Regular Contest 067 F - Yakiniku Restaurants