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xtu字符串 B. Power Strings

时间:2014-07-26 01:25:26      阅读:300      评论:0      收藏:0      [点我收藏+]

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B. Power Strings

Time Limit: 3000ms
Memory Limit: 65536KB
64-bit integer IO format: %lld      Java class name: Main
 
Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).
 

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.
 

Output

For each s you should print the largest n such that s = a^n for some string a.
 

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.
 
解题:求字符串的循环节长度。利用KMP的适配数组。如果字符长度可以被(字符长度-fail[字符长度])整除,循环节这是这个商,否则循环节长度为1,即就是这个字符本身。
 
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 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <vector>
 6 #include <climits>
 7 #include <algorithm>
 8 #include <cmath>
 9 #define LL long long
10 #define INF 0x3f3f3f
11 using namespace std;
12 const int maxn = 1000100;
13 char str[maxn];
14 int fail[maxn];
15 void getFail(int &len) {
16     int i,j;
17     len = strlen(str);
18     fail[0] = fail[1];
19     for(i = 1; i < len; i++) {
20         j = fail[i];
21         while(j && str[j] != str[i]) j = fail[j];
22         fail[i+1] = str[j] == str[i] ? j+1:0;
23     }
24 }
25 int main() {
26     int len;
27     while(gets(str) && str[0] != .) {
28         getFail(len);
29         if(len%(len-fail[len])) puts("1");
30         else printf("%d\n",len/(len-fail[len]));
31     }
32     return 0;
33 }
View Code

 

xtu字符串 B. Power Strings,布布扣,bubuko.com

xtu字符串 B. Power Strings

标签:style   blog   http   java   color   os   io   for   

原文地址:http://www.cnblogs.com/crackpotisback/p/3868915.html

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