G - Power Strings POJ 2406 （字符串的周期）

G - Power Strings POJ 2406 （字符串的周期）

G - Power Strings
Time Limit:3000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 2406

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.

1.直接暴力枚舉

 

#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn=1000005;
typedef long long LL;
char s[maxn];
int main(){
    int T;
    //freopen("Text//in.txt","r",stdin);
    while(~scanf("%s",s)&&s[0]!='.'){
        int len=strlen(s);
        for(int i=1;i<=len;i++)if(len%i==0){
            int ok=1;
            for(int j=i;j
 
2.kmp算法 i-next[i]:表示已i結尾的前綴的循環節的長度
 
#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn=1000005;
typedef long long LL;
char s[maxn];
int next[maxn];
void getnext(char*s){
    int len=strlen(s);
    int i=0,j=-1;
    next[0]=-1;
    while(i