[POI 2006]OKR-Periods of Words

Description

题库链接

定义 $A$ 串为 $B$ 串的循环串，当且仅当 $A$ 是 $B$ 的前缀（不包括 $B$ 本身），且 $B$ 为连续的 $A$ 串拼接的字符串的前缀。给出一个长度为 $n$ 字符串，求它的所有前缀（包括本身）的最长循环串的长度之和。

$1\leq n\leq 1000000$

Solution

容易发现最长循环串长度=总长度-最短相同前后缀长度。

$KMP$ 递推时同时处理出一个 $low$ 数组即可，表示最短相同前后缀长度。

Code

//It is made by Awson on 2018.3.17
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 1e6;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void print(unsigned LL x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(unsigned LL x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, nxt[N+5], low[N+5]; unsigned LL ans;
char ch[N+5];

void work() {
    read(n); scanf("%s", ch+1);
    for (int i = 2; i <= n; i++) {
    int j = nxt[i-1];
    while (j && ch[j+1] != ch[i]) j = nxt[j];
    if (ch[j+1] == ch[i]) nxt[i] = j+1, low[i] = (low[j+1] == 0 ? j+1 : low[j+1]);
    ans += low[i] ? i-low[i] : low[i];
    }
    writeln(ans);
}
int main() {
    work(); return 0;
}