UVa 455 Periodic Strings

Description
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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

找字符串中的最长循环节，直接暴力。

AC代码：

////  Created by  CQUWEL//  Copyright (c) 2015年 CQUWEL. All rights reserved.//#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <string>#include <vector>#include <list>#include <map>#include <queue>#include <stack>#include <set>#include <algorithm>#include <numeric>#include <functional>#define cir(i,a,b)  for (int i=a;i<=b;i++)#define CIR(j,a,b)  for (int j=a;j>=b;j--)#define CLR(x) memset(x,0,sizeof(x))#define root 1,n,1#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define filetest freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin)#define BUG() printf("In here\n")#define seevalue(x) printf("x=%d\n",x)#define INF 0x3f3f3f3fconst double eps=1e-9;typedef long long  ll;using namespace std;int main(){#ifdef LOCAL      freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);#endif    int n;    char s[100];    cin >> n;    getchar();    getchar();    for (int t=1;t<=n;t++){        gets(s);    //  puts(s);        getchar();        int len;        int ls=strlen(s);        char tmp[100];        for (len=1;len<=ls;len++){            if (ls%len==0){                int flag=0,j;                for (j=0;j<len;j++) tmp[j]=s[j]; tmp[j]='\0';            //  puts(tmp);                int cnt=0;                for (j=0;s[j];j++){                    if (s[j]==tmp[cnt]){                        cnt++;                        if (cnt==len) cnt=0;                    }                    else {                        flag=1;                        break;                    }                }                if (flag) continue;                else break;            }               }        if (t==1) printf("%d\n",len);        else if (t==n) printf("\n%d\n",len);        else printf("\n%d\n",len);    }    return 0;}