Average duration (kmp)

Average duration (kmp)

Question: Give a string with a length of n and ask whether the prefix of the string is a periodic string. If it is a periodic string, the position of the last letter of the output prefix and the shortest cycle

Idea: the use of the Nature of kmp string matching.

For the first I character, if f [I] is not equal to zero, it indicates that part of the prefix of this string is [0, f [I] is the same as this part of [I-f [I], I. If the I character is a periodic string, the staggered part [f [I], I] is exactly a cyclic section. (In other words, if f [I] is not equal to zero & [f [I], and I] is the two points in the cyclic section, it can be said that the first I characters constitute a periodic string)

 

Code:

 

#include
 
  #include
  
   #includeusing namespace std;const int maxn = 1000005;int f[maxn];char s[maxn];int n;void kmp(char* p,int* f){    f[0] = 0; f[1] = 0;    for(int i = 1; i < n; i++){        int j = f[i];        while(j && p[j]!=p[i]) j = f[j];        f[i+1] = p[i] == p[j] ? j+1 : 0;    }}int main(){    int cas = 0;    while(scanf("%d",&n)){        if(n == 0) break;        scanf("%s",s);        kmp(s,f);        printf("Test case #%d\n",++cas);        for(int i = 2; i <= n; i++){            if(f[i] != 0 && i%(i-f[i])==0){                printf("%d %d\n",i,i/(i-f[i]));            }        }        printf("\n");    }    return 0;}
  
 

Understanding kmp can improve the accuracy rate