POJ2406 Power Strings

嘟嘟嘟

用kmp求最短循环节出现的次数，失配数组的妙用。

不想画图了～～

举个栗子，s = "ababab"，则f[6] = 4,即s[1~4] == s[3~6]，全写出来就是s[1] = s[3]. s[2] = s[4], s[3] = s[5], s[4] = s[6]，然后就发现这些等式有关联的地方：化简一下，得到了：s[1] = s[3] = s[5], s[2] = s[4] = s[6]，就是s[1~2] = s[3~4] = s[5~6]，所以刚好把整个串都包含了，所以循环节就是2.

那也就是说，只要看(n - f[n])能否整除n，如果能的话，那这个循环节出现的次数就是n / (n - f[n])，否则循环节是s本身，输出1.

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<stack>
#include<queue>
#include<vector>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e6 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), las = ' ';
  while(!isdigit(ch)) las = ch, ch = getchar();
  while(isdigit(ch)) ans = ans * 10 + ch - '0', ch = getchar();
  if(las == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) putchar('-'), x = -x;
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

char s[maxn];
int f[maxn], cnt = 0;

void init()
{
  int m = strlen(s + 1);
  for(int i = 2, j = 0; i <= m; ++i)
    {
      while(j && s[j + 1] != s[i]) j = f[j];
      if(s[j + 1] == s[i]) j++;
      f[i] = j;
    }
}

int main()
{
  while(1)
    {
      scanf("%s", s + 1);
      if(s[1] == '.') break;
      init();
      int n = strlen(s + 1);
      if(!(n % (n - f[n]))) write(n / (n - f[n]));
      else write(1);
      enter;
    }
  return 0;
}