POJ 2406 Power Strings KMP循环节+升级版 POJ 1961 Period

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.

remark ：字符串的循环节：if（len%(len-next[len])） ans=len/(len-next[len])

#include<iostream>  
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef long long ll;
int n,lb,la,crt;
char a[1000005];
int nextt[1000005];
void kmp()
{
    int k = -1, j = 0;
    nextt[0] = -1;
    while (j < la)
    {
        if (k == -1 || a[j] == a[k])
        {
            j++; k++;
            nextt[j] = k;
        }
        else k = nextt[k];
    }
}
void solve()
{
    if(la% (la - nextt[la]) == 0)
        printf("%d
", la / (la - nextt[la]));
    else puts("1");
}
int main()
{
    while (scanf("%s", a)!=EOF)
    {
        if (a[0] == '.') break;
        la = strlen(a);
        kmp();
        solve();
    }
    return 0;
}

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A ^K ,that is A concatenated K times, for some string A. Of course, we also want to know the period K.Input

The input consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S.The second line contains the string S. The input file ends with a line, having the
number zero on it.

Output

For each test case, output "Test case #" and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

Sample Input

3
aaa
12
aabaabaabaab
0

Sample Output

Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

emmmmmmm……没什么好说的，跟上边代码一样一样的……

#include<iostream>  
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef long long ll;
int n,la,crt=0;
char a[1000005];
int nextt[1000005];
void kmp()
{
    int k = -1, j = 0;
    nextt[0] = -1;
    while (j < n)
    {
        if (k == -1 || a[j] == a[k])
        {
            j++; k++;
            nextt[j] = k;
        }
        else k = nextt[k];
    }
}
void solve()
{
    printf("Test case #%d
", crt);
    for(int i=1;i<=n;i++)
    if(i%(i - nextt[i]) == 0&& i!=(i - nextt[i]))
        printf("%d %d
",i, i / (i - nextt[i]));
    puts("");
}
int main()
{
    while (~scanf("%d", &n)&&n)
    {
        crt++;
        scanf("%s", a);
        kmp();
        solve();
    }
    return 0;
}