Power Strings POJ

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 = 0;
    k = -1;
    next[0] = -1;
    while(j < i)
    {
        if(k == -1 || s[j] == s[k])
        {
            j++;
            k++;
            next[j] = k;
        }
        else
            k = next[k];
    }
}

int main()
{
    while(1)
    {
        scanf("%s", s);
        if(s[0] == '.')
            break;

        int q = strlen(s);

        memset(next, 0, 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) == 0)  // q-k就是最小循环节的长度
            printf("%d
", q / (q - k));
        else
            printf("1
");
    }
    return 0;
}