[国家集训队] 拉拉队排练 - Manacher

用 Manacher 跑出回文串长，注意这里不需要偶数长度所以不需要对串做一些奇怪的处理

然后用前缀和搞一下，计算答案时跑快速幂即可

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
namespace man {
const int N = 1100005;
char str[N], s[N<<1];
int a[N<<1];

int manacher(int len){
    a[0] = 0;
    int ans = 0, j;
    for(int i = 0; i < len; ){
        while(i-a[i]>0 && s[i+a[i]+1]==s[i-a[i]-1]) a[i]++;
        if(ans < a[i])ans = a[i];
        j = i+1;
        while(j<=i+a[i] && i-a[i]!=i+i-j-a[i+i-j]){
            a[j] = min(a[i+i-j], i+a[i]-j);
            j++;
        }
        a[j] = max(i+a[i]-j, 0ll);
        i = j;
    }
    return ans;
}

int solve(){
    int len;
    len = strlen(str);
    for(int i=0;i<len;i++) s[i]=str[i];
    return manacher(len);
}
}
const int N = 1000005;
const int modulo = 19930726;
int qpow(int p,int q) {
    ll r = 1;
    for(; q; p*=p, p%=modulo, q>>=1) if(q&1) r*=p, r%=modulo;
    return r;
}
int n,k,a[N];
signed main() {
    scanf("%lld%lld%s",&n,&k,man::str);
    man::solve();
    for(int i=0;i<n;i++) {
        a[man::a[i]]++;
    }
    for(int i=N-2;i>=0;--i) a[i]+=a[i+1];
    int ans=1;
    for(int i=N-2;i>=0;--i) {
        ans*=qpow(i*2+1,min(a[i],k));
        ans%=modulo;
        k-=min(a[i],k);
    }
    cout<<ans;
}