題意：

有一串字符串,問求出有多少個循環節連續重復組成，即可以用KMP直接求出循環節有多少個字符組成。答案就是l/next[l]（剛開始理解錯題意，認為是找出最多的重復子串）

題目

Given two strings a and b we define ab to be their concatenation. For example, if a = “abc” and b = “def” then ab = “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.

思路

1.即可以用KMP直接求出循環節有多少個字符組成。答案就是l/next[l]。
2.若不連續則循環節個數為一。

ac代碼

#include <cstdio> #include <cstring> #include <algorithm> const int M=1e6+10; int next[M] ; char str[M] ; void getnext(int l) {int j = 0 , k = -1 ;next[0] = -1 ;while(j < l){if( k == -1 || str[j] == str[k] )next[++j] =++k ;elsek = next[k] ;} } int main() {int l , m ;while(~scanf("%s", str)&&str[0]!= '.'){l = strlen(str);getnext(l) ;m = next[l];if( l % (l-next[l]) != 0 )printf("1\n");else{int k= l/( l-next[l]);printf("%d\n",k);}memset(str,0,sizeof(str));}return 0; }/**題目要求有循環節重復連續，即整個串是用循環節組成的*/

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