传送门
文章目录
- 题意：
- 思路：
题意：
思路：
转换一下题意，就是求一个最小公共前后缀，显然可以暴跳nenene数组，复杂度O(n2)O(n^2)O(n2)，注意到我们每次都跳的话会跳到很多重复的位置，假设当前跳到了xxx，我们每次跳完都将当前ne[i]=xne[i]=xne[i]=x，这样就会减少很多不必要的跳，复杂度O(n)O(n)O(n)。
// Problem: P3435 [POI2006]OKR-Periods of Words
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3435
// Memory Limit: 128 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
char s[N];
int ne[N];
int ans[N];int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);scanf("%d%s",&n,s+1);for(int i=2,j=0;i<=n;i++) {while(j&&s[i]!=s[j+1]) j=ne[j];if(s[i]==s[j+1]) j++;ne[i]=j;}LL res=0;for(int i=2;i<=n;i++) {int x=ne[i];while(ne[x]>0) x=ne[x];ne[i]=x;res+=(i-x)==i? 0:(i-x);}printf("%lld\n",res);return 0;
}
/**/
