HDU--杭电--1358--Period--KMP--next值的应用
2013-07-29 10:28
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Period
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1959 Accepted Submission(s): 961
[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
#include <iostream>
#include <cstring>
using namespace std;
int next[1111111];
char s[1111111];
void getnext(int l) //求next数组
{
int i=0,j=-1;next[i]=j;
while(i<l)
{
if(j==-1||s[i]==s[j])
{
next[++i]=++j;
if(i%(i-next[i])==0&&i/(i-next[i])>1)
//对每次求出来的next进行判别,符合输出就输出
cout<<i<<" "<<i/(i-next[i])<<endl;
}
else j=next[j];
}
}
int main (void)
{
int i,k=1,l;
while(cin>>l&&l)
{
cin>>s;
cout<<"Test case #"<<k++<<endl;
getnext(l);
cout<<endl;
}
return 0;
}
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