[国家集训队]拉拉队排练

嘟嘟嘟

看到回文串，就想到了manacher。而且这道题只用求长度为奇数的回文串，所以不用添加字符，在原回文串上跑一边即可。

需要注意的是，如果一个回文串的半径是r，那么这个回文串同时也包含了r - 1, r - 2,……１的回文串，因为r表示的是以 i 为回文中心的最长的回文串。于是我们开一个桶，记录回文半径长度为 i 的回文串有多少个，用前缀和的思想，差分维护区间加１。最后倒着扫一边，快速幂加速乘法。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
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;
const ll mod = 19930726;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) {last = ch; ch = getchar();}
  while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n;
ll k;
char s[maxn];
int r[maxn];

void manacher()
{
  s[0] = '@'; s[n + 1] = '$';
  int mx = 0, id;
  for(int i = 1; i <= n; ++i)
    {
      if(i < mx) r[i] = min(r[(id << 1) - i], mx - i);
      else r[i] = 1;
      while(s[i - r[i]] == s[i + r[i]]) r[i]++;
      if(i + r[i] > mx) mx = i + r[i], id = i;
    }
}

ll dif[maxn];
ll ans = 1;

ll quickpow(ll a, ll b)
{
  a %= mod;
  ll ret = 1;
  while(b)
    {
      if(b & 1) ret = ret * a % mod;
      a = a * a % mod; b >>= 1;
    }
  return ret;
}

int main()
{
  n = read(); k = read(); scanf("%s", s + 1);
  manacher();
  for(int i = 1; i <= n; ++i) dif[1]++, dif[r[i] + 1]--;
  for(int i = 1; i <= n; ++i) dif[i] = dif[i - 1] + dif[i];
  for(int i = (n + 1) >> 1; i; --i)
    {
      if(k - dif[i] > 0)
    {
      k -= dif[i];
      ans *= quickpow((i << 1) - 1, dif[i]);
      ans %= mod;
    }
      else
    {
      ans *= quickpow((i << 1) - 1, k);
      ans %= mod;
      k = 0; break;
    }
    }
  write(k ? -1 : ans); enter;
  return 0;
}