poj2406 Power Strings-------KMP
Power Strings
Time Limit:3000MS |
Memory Limit:65536K |
Total Submissions:23083 |
Accepted:9679 |
Description
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.
題意:就是問一個字串寫成(a)^n的形式,求最大的n.
根據KMP的next函式的性質,已知字串t第K個字元的next[k],那麼d=k-next[k],如果k%d==0,那麼t[1……k]最多可均勻的分成k/d份。也就是可以生成一個長度為d的重複度為k/d的字串。
#include<iostream> #include<cstdlib> #include<stdio.h> #include<string.h> using namespace std; const int maxm = 1000010; // 模式串的最大長度 char p[maxm]; int m, next[maxm]; void getNext() { int i = 0, j = next[0] = -1; while(i < m) { if (j == -1 || p[i] == p[j]) { ++i; ++j; next[i] = p[i] != p[j] ? j:next[j]; } else j = next[j]; } } int main() { while(scanf("%s",p)!=EOF) { if(p[0]=='.') break; m=strlen(p); getNext(); int cc=1; if(m%(m-next[m])==0) cc=m/(m-next[m]); cout<<cc<<endl; } }