[LA3026]Period

試題描述

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 ≤ i ≤ N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A^K, that is A concatenated K times, for some string A. Of course, we also want to know the period K.

輸入

The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 ≤ N ≤ 1000000) the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.

輸出

For each test case, output ‘Test case #’ and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

輸入示例

3 aaa 12 aabaabaabaab 0

輸出示例

Test case #1 2 2 3 3Test case #2 2 2 6 2 9 3 12 4

數據規模及約定

見“輸入”

題解

KMP 裸題，對于位置 i，它指向的失配的位置為 f[i+1]，那么當 f[i+1] > 1 且 (i - f[i+1] + 1) | i 時答案為 i / (i - f[i+1] + 1). 我 KMP 從 1 開始做的所以前面的式子可能會奇怪一些。

#include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <stack> #include <vector> #include <queue> #include <cstring> #include <string> #include <map> #include <set> using namespace std;int read() {int x = 0, f = 1; char c = getchar();while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }return x * f; }#define maxn 1000010 int n, f[maxn]; char S[maxn];int main() {n = read();int kase = 0; // bool fl = 1;while(n) {scanf("%s", S + 1); // if(!fl) putchar('\n'); fl = 0;printf("Test case #%d\n", ++kase);f[1] = f[2] = 1;for(int i = 2; i <= n; i++) {int u = f[i];while(u > 1 && S[u] != S[i]) u = f[u];f[i+1] = S[u] == S[i] ? u + 1 : u;if(f[i+1] > 1 && i % (i + 1 - f[i+1]) == 0) printf("%d %d\n", i, i / (i + 1 - f[i+1]));}putchar('\n');n = read();}return 0; }

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轉載于:https://www.cnblogs.com/xiao-ju-ruo-xjr/p/5753884.html

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