Period_poj1961_kmp

Description

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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

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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

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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.

Translation

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对于给定长度为n的字符串s，求所有从第1位开始的字符串循环节，输出位置和循环节数量

Analysis

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之前做过类似的题，这次换成枚举第i位就可以
论熟练掌握kmp中next数组的用法的重要性

Code

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var
  s:ansistring;
  p:array[0..1000000]of longint;
  n,i,ans,count:longint;
procedure init(s:ansistring);
var
  i,j:longint;
begin
  p[1]:=0;
  j:=0;
  for i:=2 to length(s) do
  begin
    while (s[j+1]<>s[i])and(j>0) do j:=p[j];
    if (s[j+1]=s[i]) then
      inc(j);
    p[i]:=j;
  end;
end;
begin
  readln(n);
  while n>0 do
  begin
    readln(s);
    init(s);
    inc(count);
    writeln('Test case #',count);
    for i:=1 to length(s) do
    begin
      ans:=i div(i-p[i]);
      if (ans>1)and(i mod (i-p[i])=0) then
      writeln(i,' ',ans);
    end;
    writeln;
    readln(n);
  end;
end.