HDu 1358(KMP+最小サイクル)

Period
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2049    Accepted Submission(s): 1010
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
   
   
   
   

 
Recommend
JGShining
 
この問題では、与えられた文字列の長さが2以上である、あるサブストリングループから得られるプレフィックスの長さとループ回数を要求する.
この問題はまず最小の循環節を求めなければならなくて、KMPアルゴリズムを使います.
 

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

const int MAXN=1000000+100;
char str[MAXN];//W    ，T   
int next[MAXN];

//KMP     next[]  
void getNext(char *p)
{
    int j,k,len=strlen(p);
    j=0;
    k=-1;
    next[0]=-1;
    while(j<len)
    {
        if(k==-1||p[j]==p[k])
        {
            next[++j]=++k;
        }
        else k=next[k];
    }
}

int main()
{
	int n,i,tag=1;
	while(~scanf("%d",&n),n)
	{
		scanf("%s",str);
		getNext(str);
		printf("Test case #%d
",tag++);
		for(i=2;i<=n;i++)
		{
			if(next[i]&&i%(i-next[i])==0)
				printf("%d %d
",i,i/(i-next[i]));
		}
		printf("
");
	}
	return 0;
}