Power Strings
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 45008 | Accepted: 18794 |
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.
Source
#include<iostream> #include<stdio.h> #include<string> #include<string.h> #define N 1000010 using namespace std; int next[N]; char str[N]; void getnext(char str[],int next []) { next[0]=next[1]=0; int len=strlen(str); for(int i=1;i<len;i++) { // cout<<len<<endl; int k=next[i]; while(k&&str[i]!=str[k]) { k=next[k]; //cout<<k<<endl; } next[i+1]=(str[i]==str[k])?k+1:0; } } int main() { //freopen("C:\Users\acer\Desktop\in.txt","r",stdin); while(scanf("%s",&str)!=EOF) { //cout<<str<<endl; int m=strlen(str); if(m==1) break; getnext(str,next); // for(int i=0;i<m;i++) // cout<<next[i]<<" "; // cout<<endl; if(m%(m-next[m])==0) printf("%d ",m/(m-next[m])); else puts("1"); } return 0; }