题解 mate

传送门

尝试手推exLucas无果，果断数据结构暴力
将式子写出来发现每个值的每次变化差值都只有1
所以线性筛分解质因数扔到线段树上暴力维护即可
复杂度略大于 \(O(nlogn)\)

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 1000010
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, x, y;
ll mod;
inline ll qpow(ll a, int b) {ll ans=1; for (; b; a=a*a%mod,b>>=1) if (b&1) ans=ans*a%mod; return ans;}

namespace task1{
	ll fac[N], inv[N], ans;
	void solve() {
		x=abs(read()); y=abs(read());
		fac[0]=fac[1]=1; inv[0]=inv[1]=1;
		for (int i=2; i<=n; ++i) fac[i]=fac[i-1]*i%mod;
		for (int i=2; i<=n; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
		for (int i=2; i<=n; ++i) inv[i]=inv[i-1]*inv[i]%mod;
		n-=x+y;
		if (n<0 || n&1) {puts("0"); exit(0);}
		for (int i=0; i<=n; i+=2) {
			ans=(ans+fac[x+y+n]*inv[x+i/2]%mod*inv[i/2]%mod*inv[y+(n-i)/2]%mod*inv[(n-i)/2]%mod)%mod;
		}
		cout<<ans<<endl;
	}
}

namespace task2{
	bool npri[N], vis[N];
	int pri[N], pcnt, low[N], lowp[N], lowc[N], id[N], sta[N], top;
	int tl[N<<2], tr[N<<2], c[N], tem[N];
	ll prod[N<<2], val[N], ans;
	#define tl(p) tl[p]
	#define tr(p) tr[p]
	#define val(p) val[p]
	#define prod(p) prod[p]
	#define pushup(p) prod(p)=prod(p<<1)*prod(p<<1|1)%mod
	void build(int p, int l, int r) {
		tl(p)=l; tr(p)=r;
		if (l==r) {val(p)=pri[l]; c[p]=tem[l]; prod(p)=qpow(val(p), c[p]); return ;}
		int mid=(l+r)>>1;
		build(p<<1, l, mid);
		build(p<<1|1, mid+1, r);
		pushup(p);
	}
	void upd(int p, int pos, int dat) {
		if (tl(p)==tr(p)) {
			c[p]+=dat;
			if (dat==1) prod(p)=prod(p)*val(p)%mod;
			else if (dat==2) prod(p)=prod(p)*val(p)%mod*val(p)%mod;
			else prod(p)=qpow(val(p), c[p]);
			return ;
		}
		int mid=(tl(p)+tr(p))>>1;
		if (pos<=mid) upd(p<<1, pos, dat);
		else upd(p<<1|1, pos, dat);
		pushup(p);
	}
	void prep_upd(int i, int k) {
		for (int t=i; t>1; t/=lowp[t])
			tem[id[low[t]]]+=lowc[t]*k;
	}
	void upd(int i, int k) {
		for (int t=i; t>1; t/=lowp[t]) {
			// upd(1, id[low[t]], lowc[t]*k);
			if (!vis[low[t]]) sta[++top]=low[t], vis[low[t]]=1;
			tem[low[t]]+=lowc[t]*k;
		}
	}
	void update() {
		while (top) {
			if (tem[sta[top]]) upd(1, id[sta[top]], tem[sta[top]]);
			tem[sta[top]]=0; vis[sta[top]]=0;
			--top;
		}
	}
	void init() {
		for (int i=2; i<N; ++i) {
			if (!npri[i]) pri[++pcnt]=low[i]=lowp[i]=i, lowc[i]=1, id[i]=pcnt;
			for (int j=1,x; j<=pcnt&&i*pri[j]<N; ++j) {
				npri[x=i*pri[j]]=1;
				if (!(i%pri[j])) {
					low[x]=pri[j];
					lowp[x]=lowp[i]*pri[j];
					lowc[x]=lowc[i]+1;
					break;
				}
				else low[x]=lowp[x]=pri[j], lowc[x]=1;
			}
		}
	}
	void solve() {
		x=abs(read()); y=abs(read()); n-=x+y;
		init();
		if (n<0 || n&1) {puts("0"); exit(0);}
		for (int i=1; i<=x+y+n; ++i) prep_upd(i, 1);
		for (int i=1; i<=y+n/2; ++i) prep_upd(i, -1);
		for (int i=1; i<=n/2; ++i) prep_upd(i, -1);
		for (int i=1; i<=x; ++i) prep_upd(i, -1);
		build(1, 1, pcnt);
		ans=prod(1);
		memset(tem, 0, sizeof(tem));
		for (int i=2; i<=n; i+=2) {
			upd(x+i/2, -1); upd(i/2, -1); upd(y+(n-i+2)/2, 1); upd((n-i+2)/2, 1);
			update();
			ans=(ans+prod(1))%mod;
		}
		printf("%lld\n", ans);
		exit(0);
	}
}

signed main()
{
	n=read(); mod=read();
	// task1::solve();
	task2::solve();
	
	return 0;
}