G - Power Strings
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.
题意:
求最小循环节个数
注:
1)当输入" . " (最后一个样例),结束。
2)数据范围1~1000000
3)当S长度不能除尽循环节时,输出1;
代码:
#include<map>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define Mlm 1000005 //数据范围1~1000000
int ne[Mlm];
char mo[Mlm];
int lm,lt;
void nex()
{
int i=0,j=-1;
ne[0]=-1;
while(i<lm)
{
while(j!=-1&&mo[j]!=mo[i])
j=ne[j];
ne[++i]=++j;
}
}
int main()
{
while(~scanf("%s",mo))
{
if(mo[0]=='.')
break;
memset(ne,0,sizeof(ne));
lm=strlen(mo);
nex();
int Xun=lm-ne[lm]; //求最小循环节
if(lm%(Xun)!=0) //当S长度不能除尽循环节时,输出1;
printf("1\n");
else
printf("%d\n",lm/Xun);
}
return 0;
}