C 解法, 执行用时: 1ms, 内存消耗: 368KB, 提交时间: 2019-05-26

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>
 
int main()
{
    char s[100000];
    char s1[100000];
    int i,j;
    int flag=0;
    int len1;
    scanf("%s",s);
    int len=strlen(s);
    int k;
     
    for(i=2;i<=len;i++)
    {
        if(len%i==0)
        {
             
            len1=len/i;//printf("________%d\n",len1);
            for(j=0;j<len1;j++)
            {
 
                k=j+len1;
                while(k<len)
                {
                    if(s[j]==s[k])
                        k+=len1;
                    else
                    {
                        flag=2;
                        break;
                    }
                 
                }
                if(flag==2)
                {
                    break;
                }  
            }
            //printf("________%d\n",k);
 
            if(j==len1||k>=len)
            {
                 
                strncpy(s1,s,len1);
                s1[len1]='\0';
                printf("%s\n",s1);
                return 0;  
 
            }
             
 
 
        }
         
 
    }
    if(i==len+1)
    {
        printf("false\n"); 
        return 0;  
    }
    return 0;  
 
 
 
     
}

C 解法, 执行用时: 2ms, 内存消耗: 232KB, 提交时间: 2019-07-23

#include <stdio.h>
#include <string.h>
int main()
{
    char a[10000];
    scanf("%s",a);
    char b[10000];
    int len = strlen(a);
    int len1 ;
    int i , k , j;
    int flag = 0;
    for (i = 2; i <= len; i++)
    {
        flag=0;
        if (len%i == 0)
        {
            len1 = len / i;
            for ( j = 0; j < len1; j++)
            {
                 k = j + len1;
                while (k < len)
                {
                    if (a[j] == a[k])
                        k += len1;
                    else
                    {
                        flag = 2;
                        break;
                    }
                }
                if (flag == 2)
                {
                    break;
                }
            }
          
            if (j == len1 || k >= len)
            {
            strncpy(b, a, len1);
            b[len1] = '\0';
            printf("%s", b);
            return 0;
            }
        }
    }
    if (i == len + 1)
    {
        printf("false");
        return 0;
    }
}

C 解法, 执行用时: 2ms, 内存消耗: 232KB, 提交时间: 2018-11-15

#include <stdio.h>
#include <string.h>

int check(char *str, int len, int shortlen){
    int n = len/shortlen;
    int i, j;
    for(i = 0; i < shortlen; i ++){
        for(j = 1; j < n; j ++){
            if(str[j] != str[i*shortlen + j])
                return -1;
        }
    }
    return 0;
}

int main(void){
    char buf[512];
    scanf("%s", buf);
    if( buf[0] == '\0' )
        goto NO;
    int times = 0;
    int len = strlen(buf);
    int i, j;
    for(i = 0; i < len; i ++)
        if( buf[i] == buf[0] )
            times ++;
    if( times == 1 || 0 != len % times )
        goto NO;
    
    if( 0 == times % 2 && 0 == check(buf, len, i) ){
        i = 2;
        goto YES;
    }
    for(i = 3; i <= times; ){
        if( 0 == check(buf, len, i) )
            goto YES;
        i ++;
        for(j = 2; j <= i/2 && i <= times; j += 2 ){
            if( 0 == i % j ){
                j = 2;
                i ++;
            }
        }
    }
YES:
    buf[len/i] = '\0';
    printf("%s", buf);
    return 0;
NO:
    printf("false");
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 328KB, 提交时间: 2019-04-07

#include <stdio.h>
#include <string.h>
 
int check(char *str, int len, int shortlen){
    int n = len/shortlen;
    int i, j;
    for(i = 0; i < shortlen; i ++){
        for(j = 1; j < n; j ++){
            if(str[j] != str[i*shortlen + j])
                return -1;
        }
    }
    return 0;
}
 
int main(void){
    char buf[512];
    scanf("%s", buf);
    if( buf[0] == '\0' )
        goto NO;
    int times = 0;
    int len = strlen(buf);
    int i, j;
    for(i = 0; i < len; i ++)
        if( buf[i] == buf[0] )
            times ++;
    if( times == 1 || 0 != len % times )
        goto NO;
     
    if( 0 == times % 2 && 0 == check(buf, len, i) ){
        i = 2;
        goto YES;
    }
    for(i = 3; i <= times; ){
        if( 0 == check(buf, len, i) )
            goto YES;
        i ++;
        for(j = 2; j <= i/2 && i <= times; j += 2 ){
            if( 0 == i % j ){
                j = 2;
                i ++;
            }
        }
    }
YES:
    buf[len/i] = '\0';
    printf("%s", buf);
    return 0;
NO:
    printf("false");
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 348KB, 提交时间: 2018-09-08

# include <stdio.h>
# include <stdlib.h>
# include <string.h>
int main(){
    char str[100];
    char *p=str;
    int n,i,j,m,z,c;
    gets(str);
    n=strlen(str);
    for(i=0;i<n;i++){
        for(j=1;j<n;j++){
            if(*(p+i)==*(p+j))
                for(m=0;m<n;m=m+j){
                    for(z=0;z<j;z++)
                        if(*(p+z)==*(p+m+z))
                            c=j;
                        else c=0;
                    }
            else continue;
            break;
        }
        break;
    }
    i=0;
    if(c)
        while(i<j){
        printf("%c",*(p+i));
        i++;
    }
    else printf("false");
    return 0;
}