【POJ 2406】Power Strings
【題目】
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.
【分析】
大致意思:一個字串由它的一個子串不斷拼接而成,求最長拼接的長度
比如說樣例, 就是由 拼接 3 次形成的,最長的長度也就是 3
很容易想到用 KMP 中的 next 陣列求最小迴圈節,答案是
有一個要注意的地方是當 時,如 ,答案也只能為 1
【程式碼】
#include<cstdio> #include<cstring> #include<algorithm> #define L 1000005 using namespace std; char s[L]; int next[L]; int main() { int i,j,l,ans; scanf("%s",s+1); while(s[1]!='.') { l=strlen(s+1); j=0,next[1]=0; for(i=2;i<=l;++i) { while(j&&s[i]!=s[j+1]) j=next[j]; if(s[i]==s[j+1]) ++j; next[i]=j; } ans=1; if(l%(l-next[l])==0) ans=l/(l-next[l]); printf("%d\n",ans); scanf("%s",s+1); } return 0; }