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poj2406 Power Strings
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
abcdaaaaababab.
Sample Output
143
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
题意是给一个串s，求有没有一个串能通过复制自己变成s。求出最大的长度
这题跟那个bzoj1355很像……
最长的串的长度应该是n-next[n]，然后判断n能不能整除n-next[n]，如果能输出商，不能输出1
#include<cstdio>#include<cstring>#define LL long longinline void write(LL a){ if (a<0){printf("-");a=-a;} if (a>=10)write(a/10); putchar(a%10+‘0‘);}inline void writeln(LL a){write(a);printf("\n");}int next[1000010];char s[1000010];int l,j,k;inline void pre(){ memset(next,0,sizeof(next)); j=0; for (int i=2;i<=l;i++) { while (j>0&&s[j+1]!=s[i])j=next[j]; if (s[j+1]==s[i])j++; next[i]=j; }}int main(){ while (scanf("%s",s+1)&&s[1]!=‘.‘) { l=strlen(s+1); pre(); int len=l-next[l]; if (l%len)writeln(1); else writeln(l/len); } return 0;}
poj2406 Power Strings
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