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xtu字符串 B. Power Strings

2024-07-14 01:31:24   215人阅读

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

abcdaaaaababab.

Sample Output

143

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。
 
 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 long10 #define INF 0x3f3f3f11 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 }
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