Future Failure CodeForces

大意: 两人轮流操作一个长$n$, 只含前$k$种小写字母的串, 每次操作删除一个字符或者将整个串重排, 每次操作后得到的串不能和之前出现过的串相同, 求多少种串能使先手必胜.

找下规律发现$n$为奇数必胜, 否则假设$a_i$为字符$i$出现次数, 如果$frac{n!}{a_1!a_2!...a_k!}$为奇数则必败

$n!$中$2$的幂次为n-__builtin_popcount(n)

所以必败就等价于$a_1+...+a_n=a_1|...|a_n$

设$f_{i,j}$表示前$i$个字符, 状态为$j$的方案数除以总字符数的阶乘

可以得到转移为$f_{i,S}=sum frac{1}{x!} f_{i-1,Soplus x}$

做$O(log k)$次子集卷积即可, 复杂度是$O(nlog ^2nlog k)$

我写的好像常数太大的没卡过去, 先这样吧

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
const int N = 1e6+10;
int n,k,P,fac[N],ifac[N],cnt[N];
int dp[N],f[20][N],g[20][N],h[20][N];
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}


void FMT(int *a, int n, int tp) {
	int mx = (1<<n)-1;
	REP(i,0,n-1) REP(j,0,mx) {
		if (j>>i&1) a[j]=(a[j]+tp*a[j^1<<i])%P;
	}
}

void mul(int *a, int *b, int *c, int n) {
	int mx = (1<<n)-1;
	REP(i,0,n) REP(j,0,mx) f[i][j]=g[i][j]=h[i][j]=0;
	REP(i,0,mx) {
		f[cnt[i]][i] = a[i];
		g[cnt[i]][i] = b[i];
	}
	REP(i,0,n) FMT(f[i],n,1),FMT(g[i],n,1);
	REP(i,0,n) {
		REP(j,0,i) REP(k,0,mx) {
			h[i][k] = (h[i][k]+(ll)f[j][k]*g[i-j][k])%P;
		}
		FMT(h[i],n,-1);
		REP(k,0,mx) if (cnt[k]==i) c[k] = h[i][k];
	}
}

int main() {
	REP(i,0,N-1) cnt[i] = __builtin_popcount(i);
	scanf("%d%d%d",&n,&k,&P);
	fac[0] = 1;
	REP(i,1,N-1) fac[i]=(ll)fac[i-1]*i%P;
	ifac[N-1] = inv(fac[N-1]);
	PER(i,0,N-2) ifac[i]=(ll)ifac[i+1]*(i+1)%P;
	int tot = qpow(k,n);
	if (n&1) return printf("%d
",tot),0;
	int len = 1;
	while ((1<<len)<=n) ++len;
	dp[0] = 1;
	for (; k; mul(ifac,ifac,ifac,len),k>>=1) {
		if (k&1) mul(dp,ifac,dp,len);
	}
	int ans = (tot-(ll)dp[n]*fac[n])%P;
	if (ans<0) ans += P;
	printf("%d
", ans);
}