hdu1358(KMP+最小循环节)

Period

[b]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 2049    Accepted Submission(s): 1010
[/b]

[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

3
aaa
12
aabaabaabaab
0

 

[align=left]Sample Output[/align]

Test case #1
2 2
3 3

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

 

[align=left]Recommend[/align]
JGShining
 
本题要求所给字符串的长度大于等于2的可以由某一子串循环得到的前缀的长度和循环次数。
此题首先得求最小循环节，用KMP算法。
 

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

const int MAXN=1000000+100;
char str[MAXN];//W为模式串，T为主串
int next[MAXN];

//KMP算法中计算next[]数组
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\n",tag++);
for(i=2;i<=n;i++)
{
if(next[i]&&i%(i-next[i])==0)
printf("%d %d\n",i,i/(i-next[i]));
}
printf("\n");
}
return 0;
}