POJ：1961-Period（寻找字符串循环节）

Period

Time Limit: 3000MS Memory Limit: 30000K

Description

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 AK ,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

---

• 题意就是给你一个字符串，问分别以字符串中的每一个字符结尾是否可以形成循环节，可以形成多少个循环节。

• 其实就是考察的一个next数组。

---

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1e6+100;
char s[maxn];
int next[maxn];

void  cal_next()
{
    int k;
    next[0] = k = -1;
    int len = strlen(s);
    for(int i=1;i<len;i++)
    {
        while(k>-1 && s[i]!=s[k+1])
            k = next[k];
        if(s[i] == s[k+1])
            k++;
        next[i] = k;
    }
}

void solve()
{
    int len = strlen(s);
    for(int i=1;i<len;i++)
    {
        if(next[i] == -1)
            continue;
        if((i+1)%(i + 1 - next[i] - 1) == 0)//形成循环节是一定没有余数的情况下
            printf("%d %d\n",i+1,(i+1)/(i+1-next[i]-1));
    }
}

int main()
{
    int n,t = 1;
    while(scanf("%d",&n) && n)
    {
        scanf("%s",s);
        printf("Test case #%d\n",t++);
        cal_next();
        solve();
        printf("\n");
    }
    return 0;
}