Power Strings ( KMP )

本文介绍了一种用于解决字符串幂次分解问题的高效算法。通过分析输入字符串s,该算法可以找出最大的整数n,使得s可以表示为某个字符串a重复n次。文章提供了两种实现方式,一种使用了next数组,另一种使用了next数组的优化版本,同时给出了详细的代码实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

题目链接:https://cn.vjudge.net/contest/248066#problem/F

参考链接:https://blog.youkuaiyun.com/u013167299/article/details/46050405

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.

最长公共前后缀

 正版:

#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;
const int MAXN = 1e6+10;

char s1[MAXN];
char s2[MAXN];
int net[MAXN];
int lens1;
int lens2;
void getnet(){
      int i=0;
      int j=-1;
      net[0]=-1;
      while(i<lens1){

            if(j==-1 || s1[i]==s1[j]){
                i++;
                j++;
                net[i]=j;
            }else{
                j=net[j];
            }
      }
}

int main(){

     while(scanf("%s",s1)!= EOF){

      if(s1[0]=='.'){
            break;
      }

      lens1=strlen(s1);

      getnet();
    
      int k = lens1 - net[lens1];
      if(lens1%k==0){
           cout<<lens1/k<<endl;
      }
      else{
           cout<<1<<endl;
      }
   }
    return 0;
}

盗版: 

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 1e6+100;
int nxt[MAXN];

char  s1[MAXN];

int n,cut;
void getnxt(){
    int i=1;
    int j=0;
    nxt[0]=0;
    while(i<n){
        if(s1[i]==s1[j]){
            nxt[i]=j+1;
            i++;
            j++;
        }else if(!j){
            i++;
        }else{
            j=nxt[j-1];
        }
    }
}
int main(){
    
    while( scanf("%s",s1)!=EOF){

        if(s1[0]=='.'){
            break;
        }
        n=strlen(s1);
        getnxt();
        
        int k=n-nxt[n-1];
        if(n%k==0){
            cout<<n/k<<endl;
        }else{
            cout<<1<<endl;
        }
        memset(nxt,0,sizeof(nxt));
    }
    return 0;
}

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值