poj 2406 Power Strings-优快云博客

本文链接：https://blog.youkuaiyun.com/zhang__liuchen/article/details/78493701

本文介绍如何使用KMP算法中的next数组来找出一个字符串的最大重复子串数量，通过实例详细解析了算法的工作原理，并提供了完整的C++代码实现。

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 52540		Accepted: 21871

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.

大意：给出一个字符串问它最多由多少相同的字串组成

如 abababab由4个ab组成

分析：

kmp中的next数组求最小循环节的应用

例如

ababab next[6] = 4; 即

ababab

1~4位与2~6位是相同的

那么前两位

就等于3、4位

3、4位就等于5、6位

……

所以如果能整除也就循环到最后了

如果不能整除

就最后余下的几位不在循环内

例如

1212121

最后剩余1不能等于循环节

#include<iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1000005;
char str[N];
int Next[N];
void getNext()
{
	int len=strlen(str);
    Next[0] = -1;
    int i = 0, j = -1;
    while(i < len)//注意
    {
        if(j == -1 || str[i] == str[j])
        {
            i++;
            j++;
            Next[i] = j;
        }
        else
            j = Next[j];
    }
}
int main()
{
    while(scanf("%s", str) && str[0] != '.')
    {
        int len=strlen(str);
        getNext();
        int ans = 1;
        if(len%(len-Next[len])==0)
            ans=len/(len-Next[len]);
        printf("%d\n", ans);
    }
    return 0;
}

参考网址

http://www.cnblogs.com/zhanzhao/p/4761477.html