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Power Strings（KMP）
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
abcdaaaaababab.
Sample Output
143
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source
Waterloo local 2002.07.01
#include<iostream>#include<stdio.h>#include<string>#include<string.h>#define N 1000010using 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\n",m/(m-next[m])); else puts("1"); } return 0;}
Power Strings（KMP）
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