POJ2406A- Power Strings

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.

思路：题意是让你求能构成循环的最多次数，关键是利用next数组的构建，其是KMP算法的精髓。

对于代码中i-next[i]代表了字符串最小前缀且满足能不但的复制得到

原字符串；len%(i-next[i])==0时代表字符串刚刚是子串的整数倍；

若len%(i-next[i])==0匹配时每一次移动的距离i-next[i]是相等的，若不等则只有最后一次不等；

#include <iostream>
#include<cstring>
#include<cstdio>
using namespace std;

char str[1000010],pat[1000010];//pat为模式串，str为主串
int Next[1000010]; //Next[x]下标x表示匹配失败处字符下标
//模式串pat的前缀与x位置的后缀的最大匹配字符个数-1
void GetNext(char *pat)
{
    int LenPat = strlen(pat);
    int i = 0,j = -1;
    Next[0] = -1;
    while(i < LenPat)
    {
        if(j == -1 || pat[i] == pat[j])
        {
            i++,j++;
            Next[i] = j;
        }
        else
            j = Next[j];
    }
   
}
/*      
int KMP()//返回模式串pat在str中第一次出现的位置
{
    int LenStr = strlen(str);
    int LenPat = strlen(pat);
    //GetNext(pat);
    int i = 0,j = 0;
    int ans = 0;//计算模式串在主串匹配次数
    while(i < LenStr)
    {
        if(j == -1 || str[i] == pat[j])
            i++,j++;
        else
            j = Next[j];
        if(j == LenPat)
        {
            //ans++; ans存放匹配次数，去掉return,最后返回ans
            return i - LenPat + 1;
        }
    }
    return -1;//没找到匹配位置
    //return ans;//返回匹配次数。
} */
int main()
{
    while(scanf("%s",str))
    {
        if(str[0]=='.')
            break;
        int len=strlen(str);
        GetNext(str);
        int k=len-Next[len];
        int ans;
        if(len%k==0)
            ans=len/k;
        else
            ans=1;
        printf("%d
",ans);
    }
    return 0;
}