Poj1961 Poj2406 shortest cycle Section

POJ1961:

First, use KMP "preprocessing" to get the pre (I am used to using the p array in the program ).

Then the key lies in the judgment that (I + 1) % (I-pre [I]) = 0 has a loop. How can this problem be solved? This is to divide the array str into (I + 1)/(I-pre [I]) parts. From the pre nature, we can see that all these parts are equal, that is, the minimum cyclic section, it is not clear that you can draw a few strokes on the number axis. This is the key to this question.

#include "stdio.h"int p[1000010],N;char str[1000010];void get_p(int n){    int i,j=-1;    p[0]=-1;    for(i=1;i<n;i++)    {        while(j>-1 && str[i]!=str[j+1]) j=p[j];        if(str[i] == str[j+1]) j++;        p[i]=j;    }}int main(){    int i,j,cas=1;    while(scanf("%d",&N),N)    {        scanf("%s",str);        get_p(N);        printf("Test case #%d\n",cas++);        for(i=1;i<N;i++)        {            if(p[i]!=-1 && (i+1)%(i-p[i])==0)               printf("%d %d\n",i+1,(i+1)/(i-p[i]));        }        printf("\n");    }}

Poj2406

#include "stdio.h"#include "string.h"int ans;int p[1000010];char str[1000010];void get_p(int n){    int i,j=-1;    p[0]=-1;    for(i=1;i<n;i++)    {        while(j>-1 && str[i]!=str[j+1]) j=p[j];        if(str[i] == str[j+1]) j++;        p[i]=j;    }}int main(){    int i,j,k,N;    while(1)    {        scanf("%s",str);        N=strlen(str);        if(N==1 && str[0]=='.') break;        get_p(N);        ans=1;        if(N==1) printf("1\n");        else{            if(N%(N-1-p[N-1])==0) printf("%d\n",N/(N-1-p[N-1]));            else printf("1\n");        }    }}