POJ 2406 Power Strings（KMP+最小循环节）

电玩女神 2022-06-10 14:08 181阅读 0赞

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.

题解：

给一个串，问是由最多几个循环节组成的，直接用最小循环节长度len-next\[len\]判断一下就好了，如果能整除len就输出len/最小循环节长度，否则输出1

代码：

#include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<stdio.h>
    #include<math.h>
    #include<string>
    #include<stdio.h>
    #include<queue>
    #include<stack>
    #include<map>
    #include<vector>
    #include<deque>
    using namespace std;
    #define lson k*2
    #define rson k*2+1
    #define M (t[k].l+t[k].r)/2
    #define INF 1008611111
    #define ll long long
    #define eps 1e-15
    int Next[1000005];
    char T[1000005];
    void init()
    {
        int i=0,j=-1;
        Next[0]=-1;
        while(T[i])
        {
            if(j==-1||T[i]==T[j])
            {
                i++;
                j++;
                Next[i]=j;
            }
            else
                j=Next[j];
        }
    }
    int main()
    {
        int i,j,n,len;
        while(gets(T)!=NULL)
        {
            if(T[0]=='.')
                break;
            init();
            len=strlen(T);
            n=len-Next[len];
            if(len%n==0)
                printf("%d\n",len/n);
            else
                printf("1\n");
        }
        return 0;
    }