Power Strings POJ - 2406

Power Strings

POJ - 2406

时限: 3000MS

内存: 65536KB

64位IO格式: %I64d & %I64u

提交 状态

已开启划词翻译

问题描述

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).

输入

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.

输出

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

样例输入

abcd
aaaa
ababab
.

样例输出

1
4
3

提示

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

来源

Waterloo local 2002.07.01

题意：一个字符串可以由其循环节循环n次得到，问n最大可以是多少……

kmp练习。

 #include <iostream>
 #include <cstdio>
 #include <cstring>

 using namespace std;

 #define maxn 10000008

 char s[maxn];
 int next[maxn];

 void getNext()
 {
     int j, k, i;
     i = strlen(s);

     j = ;
     k = -;
     next[] = -;
     while(j < i)
     {
         if(k == - || s[j] == s[k])
         {
             j++;
             k++;
             next[j] = k;
         }
         else
             k = next[k];
     }
 }

 int main()
 {
     while()
     {
         scanf("%s", s);
         if(s[] == '.')
             break;

         int q = strlen(s);

         memset(next, , sizeof(next));
         getNext();

         int k = next[q];   // k表示s串 q位置的最长的前缀和后缀相等的长度。q-k代表最小循环节的长度

         if(q % (q - k) == )  // q-k就是最小循环节的长度
             printf("%d\n", q / (q - k));
         else
             printf("1\n");
     }
     return ;
 }