UVa 455 - Periodic Strings解题报告

UVa OJ 455

Periodic Strings

A character string is said to have period k if it can be formed by concatenating one or more repetitions of another string of length k. For example, the string "abcabcabcabc" has period 3, since it is formed by 4 repetitions of the string "abc". It also has periods 6 (two repetitions of "abcabc") and 12 (one repetition of "abcabcabcabc").

Write a program to read a character string and determine its smallest period.

Input

The first line oif the input file will contain a single integer N indicating how many test case that your program will test followed by a blank line. Each test case will contain a single character string of up to 80 non-blank characters. Two consecutive input will separated by a blank line.

Output

An integer denoting the smallest period of the input string for each input. Two consecutive output are separated by a blank line.

Sample Input

1

HoHoHo

Sample Output

2

寻找周期字符串的最小周期。从小到大枚举各个周期，一旦符合条件就输出。注意每一个测试用例的输出结果之间空一行，最后一个用例的输出后面无空行。

此题个人认为最关键在于对题意的把握，即例如ABCDEF这样的串最后的结果是6而不是0，把握了这一点，多考虑一下即可AC

/**
 * UVa 455 Periodic Strings
 * Author: Sleigh
 * Last Modified: 2016.03.23
 */
#include <stdio.h>
#include <string.h>
int main()
{
    int T,len,i,temp,first=1;//T代表测试块的个数,len是字符串长度,temp存储可能的周期值
    char str[90]={0};//存储字符串
    scanf("%d",&T);
    while(T--){
        scanf("%s",str);
        temp=len=strlen(str);//针对ABC这样的情况
        for(i=1;i<len;i++){
            if(str[i]==str[0]){
                temp=i;
                for(int k=0;k<temp;k++,i++){
                    if(str[k]!=str[i]){//反向思维，针对ABABACC的情况
                        temp=len;
                        i--;
                        break;
                    }
                    if(k!=temp-1&&i==len-1){//主要针对ABCDAB的情况
                        temp=len;
                        break;
                    }
                    if((k==temp-1)&&(i!=len-1))
                        k=-1;//始终注意k的自加
                }
            }
        }
        if(first){
            printf("%d\n",temp);
            first=0;
        }
        else
            printf("\n%d\n",temp);
    }
    return 0;
}