Power Strings （ KMP ）

最新推荐文章于 2023-06-30 19:30:28 发布

guozuofeng

最新推荐文章于 2023-06-30 19:30:28 发布

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分类专栏： KMP

本文链接：https://blog.youkuaiyun.com/guozuofeng/article/details/81810228

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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。

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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;
}