【POJ2406】【KMP】Power Strings

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01

【分析】

其实我是来贴诗句的。。

/*
唐代李白
《三五七言 / 秋风词》

秋风清，秋月明，
落叶聚还散，寒鸦栖复惊。
相思相见知何日？此时此夜难为情！
入我相思门，知我相思苦。
长相思兮长相忆，短相思兮无穷极。
早知如此绊人心，何如当初莫相识。
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <utility>
#include <iomanip>
#include <string>
#include <cmath>
#include <queue>
#include <assert.h>
#include <map>
#include <ctime>
#include <cstdlib>
#include <stack>
#define LOCAL
const int MAXN = 1000000 + 10;
const int INF = 100000000;
const int SIZE = 450;
const int MAXM = 1000000 + 10;
const int maxnode =  0x7fffffff + 10;
using namespace std;
int l1, l2;
char a[MAXN], b[MAXN];
int next[1000000];//不用开太大了.. 
void getNext(){
     //初始化next数组 
     next[1] = 0;
     int j = 0;
     for (int i = 2; i <= l1; i++){
         while (j > 0 && a[j + 1] != a[i]) j = next[j];
         if (a[j + 1] == a[i]) j++;
         next[i] = j;
     }
     return;
}
int kmp(){
    int j = 0, cnt = 0;
    for (int i = 1; i <= l2; i++){
        while (j > 0 && a[j + 1] != b[i]) j = next[j];
        if (a[j + 1] == b[i]) j++;
        if (j == l1){
           cnt++;
           j = next[j];//回到上一个匹配点 
        }
    }
    return cnt;
}

void init(){
     scanf("%s", a + 1);
     scanf("%s", b + 1);
     l1 = strlen(a + 1);
     l2 = strlen(b + 1);
}

int main(){
    int T;
    
    while (scanf("%s", a + 1)){
          if (a[1] == '.') break;
          l1 = strlen(a + 1);
          getNext();
          if (l1%(l1 - next[l1]) == 0) printf("%d
", l1/(l1 - next[l1]));
          else printf("1
");
    }
    return 0;
}