Power Strings
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 56055 | Accepted: 23310 |
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
这类题遇到蛮多次了 要提醒自己 如果它不是正好一个周期串 要直接输出1
这需要特判 其他就是用循环长度利用n2-Next[n2]实现
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
const int MAX_N = 1000024;
int Next[MAX_N];
char mo[MAX_N];
int n2;
void getnext(){
int i=0,j=-1;
while(i<n2){
if(j==-1||mo[i]==mo[j]){++i,++j,Next[i] = j;}
else j=Next[j];
}
return;
}
int main(){
while(~scanf("%s",mo)){
if(mo[0]=='.') break;
memset(Next,0,sizeof(Next));
n2 = strlen(mo);
Next[0] = -1;
getnext();
if(n2%(n2-Next[n2])==0)
printf("%d\n",n2/(n2-Next[n2]));
else printf("1\n");
}
return 0;
}