HDU 1358 Period (KMP的Next数组应用，循环节问题)

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#KMP #HDU #Next数组应用

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本文针对给定字符串的前缀，探讨如何判断这些前缀是否为循环字符串，并确定其循环次数及周期。采用KMP算法的Next数组进行高效计算。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1358

Problem 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 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 file 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
 

题目大意：求从位置2开始到字符串结尾，对于位置i，输出一行求其前缀的循环串次数，如果不是循环串则不输出

题目思路：KMP的Next数组应用，基本循环串问题

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const int N = 1e6+5;

int Next[N];
char mo[N];

void Get_next()
{
    int len = strlen(mo);
    int i,j;
    j = Next[0] = -1;
    i = 0;
    while(i<len)
    {
        while(j!=-1&&mo[i]!=mo[j])
            j = Next[j];
        Next[++i] = ++j;
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int cnt=0,n;
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        ++cnt;
        scanf(" %s",&mo);
        Get_next();
        printf("Test case #%d\n",cnt);
        //循环节问题
        //重要的性质len-next[i]为此字符串的最小循环节(i为字符串的结尾),即循环周期
        //另外如果len%(len-next[i])==0&&len!=(len-next[i]),此字符串的循环次数为len/(len-next[i])的循环串;

        for(int i=1;i<n;i++)
        {
            int len = i+1; //因为字符串下标从0开始，所以当前长度要加一
            int length = len - Next[len];  //循环节长度
            if(len%length==0&&len!=length)  //是一个循环节
                printf("%d %d\n",len,len/length);
        }
        puts("");
    }
    return 0;
}