拉拉队排练

传送门

这道题好像只要跑一个Manacher就能过……

要求的女生和谐小群体就是求一个回文串，然后同时要计算这个回文串的子串。这个很简单，我们跑一遍Manacher算出每个点的最长回文半径，然后长度为偶数的忽略（题目要求），奇数的计算一下前缀和，然后对于每一个长度用快速幂计算一下同时k减去回文串个数就行，直到k=0。

注意此题要开longlong而且别把模数打错了……

看一下代码。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<set>
#include<vector>
#include<queue>
#define pb push_back
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define enter putchar('\n')

using namespace std;
typedef long long ll;
const int M = 40005;
const int N = 1000005;
const ll mod = 19930726;

ll read()
{
    ll ans = 0,op = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
    if(ch == '-') op = -1;
    ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
    ans *= 10;
    ans += ch - '0';
    ch = getchar();
    }
    return ans * op;
}

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

int p[N<<1],mx,mid,len;
char s[N<<1],c[N];
ll n,k,sum[N],ans = 1;

int change()
{
    int j = 2,l = strlen(c);
    s[0] = '!',s[1] = '#';
    rep(i,0,l-1) s[j++] = c[i],s[j++] = '#';
    s[j] = '@';
    return j;
}

void manacher()
{
    len = change(),mx = mid = 1;
    rep(i,1,len-1)
    {
    if(i < mx) p[i] = min(mx-i,p[(mid<<1)-i]);
    else p[i] = 1;
    while(s[i-p[i]] == s[i+p[i]]) p[i]++;
    if(mx < i + p[i]) mid = i,mx = i + p[i];
    sum[p[i]-1]++;
    }
}

int main()
{
    n = read(),k = read();
    scanf("%s",c);
    manacher();
    //rep(i,1,n) printf("%d ",p[i]);enter;
    //per(i,n,1) printf("%lld ",sum[i]);enter;
    per(i,n,1)
    {
    if(!(i&1)) continue;
    sum[i] += sum[i+2];
    //printf("%d %lld %lld\n",i,k,sum[i]);
    if(k >= sum[i]) ans *= qpow(i,sum[i]),ans %= mod,k -= sum[i];
    else
    {
        ans *= qpow(i,k),ans %= mod;
        k = 0;
        break;
    }
    }
    if(k) printf("-1\n");
    else printf("%lld\n",ans);
    return 0;
}