POJ 2406 Power Strings (KMP)

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 44165		Accepted: 18427

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.

分析：KMP，next表示模式串如果第i位(设str[0]为第0位)与文本串第j位不匹配则要回到第next[i]位继续与文本串第j位匹配。则模式串第1位到next[n]与模式串第n-next[n]位到n位是匹配的。所以思路和上面一样，如果n%（n-next[n]）==0,则存在重复连续子串，长度为n-next[n]。

例如：a b a b a b

next:-1 0 0 1 2 3 4

next[n]==4,代表着，前缀abab与后缀abab相等的最长长度，这说明，ab这两个字母为一个循环节，长度=n-next[n]。

#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1000000 + 10;
char str[maxn];
int next[maxn];
int len;

void getNext()
{
    int i = 0, j = -1;
    next[0] = -1;
    while (i != len){
        if (j == -1 || str[i] == str[j]){
            next[++i] = ++j;
        }
        else
            j = next[j];
    }
}

int main()
{
    while (scanf("%s", str) != EOF){
        if (str[0] == '.')
            break;
        len = strlen(str);
        getNext();
        if (len % (len - next[len]) == 0)
            printf("%d\n", len / (len - next[len]));
        else
            printf("1\n");
    }
    return 0;
}