POJ 2406 Power Strings（KMP+最小循环节）-蒲公英云

POJ 2406 Power Strings（KMP+最小循环节）

电玩女神 2022-06-10 14:08 295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;
}