题解 稀疏阶乘问题

传送门

即为统计 \(m\mid F(x)\) 的个数
有性质 \(m\mid F(x)\to m\mid F(x+m)\)，所以按 \(x \bmod m\) 的余数分类处理
考虑 \(m\) 的每个质因子，限制变为 \(\prod p_i^{c_i}\mid F(x)\)
那么对每个 \(r\) 和每个 \(p_i^{c_i}\) 求出符合上述限制的最小 \(x\)，再整体取 max 即可得到 \(m\mid F(x), x\equiv r\pmod m\) 的最小 \(x\)
然后问题即变为区间内 \(x\equiv r\pmod m\) 的 \(x\) 的个数
康康怎么对每个 \(r\) 和每个 \(p_i^{c_i}\) 求符合限制的最小 \(x\)
转为求 \(\prod\limits_{k=0}^{lim}(r-k^2)\equiv 0\pmod{p_i^{c_i}}\) 的最小 \(lim\)
由同余性质知 \(lim<p_i\times c_i\) 否则无解
于是枚举 \(k\in[0, p_i\times c_i)\)，它能影响到 \(r\equiv k^2\pmod{p_i}\) 的 \(r\)，更新其 \(lim\)
然后对每个 \(r\) 根据 \(lim\) 的定义（\(\lfloor\sqrt{x-1}\rfloor\)）不难反推出最小的 \(x\)
复杂度 \(O(\sum \lfloor\frac{m}{p_i}\rfloor p_ic_i)=O(m\log m)\)

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f3f3f3f3f
#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 ll read() {
	ll 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;
}

ll l, r, m;

namespace force{
	ll ans;
	void solve() {
		for (int i=l; i<=r; ++i) {
			int lim=floor(sqrt(i-1));
			ll tem=1;
			for (ll k=0; k<=lim&&tem; ++k)
				tem=tem*(i-k*k)%m;
			if (!tem) ++ans;
		}
		cout<<ans<<endl;
	}
}

namespace task{
	int p[N], c[N], ans[N], rest[N], lim[N], top, sum;
	inline ll qpow(ll a, ll b) {ll ans=1; for (; b; a=a*a,b>>=1) if (b&1) ans=ans*a; return ans;}
	int F(int x) {
		int lim=floor(sqrt(x-1));
		ll tem=1;
		for (ll k=0; k<=lim&&tem; ++k)
			tem=tem*(x-k*k)%m;
		return tem;
	}
	inline int qcnt(int t, int r) {return r/m+(r%m>=t%m)-(t!=0);}
	void solve() {
		int tem=m;
		for (int i=2; i<=m; ++i) if (tem%i==0) {
			p[++top]=i;
			do {tem/=i; ++c[top];} while (tem%i==0);
		}
		// cout<<"p: "; for (int i=1; i<=top; ++i) cout<<p[i]<<' '; cout<<endl;
		// cout<<"c: "; for (int i=1; i<=top; ++i) cout<<c[i]<<' '; cout<<endl;
		for (int i=1; i<=top; ++i) {
			// cout<<"i: "<<i<<endl;
			int mod=qpow(p[i], c[i]);
			for (int j=0; j<m; ++j) lim[j]=INF, rest[j]=1;
			for (int k=0; k<p[i]*c[i]; ++k) {
				int r=k*k%p[i];
				for (int t=r; t<m; t+=p[i]) {
					rest[t]=rest[t]*(t-k*k)%mod;
					if (!rest[t]) lim[t]=min(lim[t], k);
				}
			}
			// cout<<"lim: "; for (int j=0; j<m; ++j) cout<<lim[j]<<' '; cout<<endl;
			for (int t=0; t<m; ++t) {
				int x=lim[t]*lim[t]+1;
				ans[t]=max(ans[t], x);
			}
			// cout<<"ans: "; for (int j=0; j<m; ++j) cout<<ans[j]<<' '; cout<<endl;
		}
		for (int t=0; t<m; ++t) if (ans[t]<=r)
			sum+=qcnt(t, r)-qcnt(t, max(l, ans[t])-1);
		printf("%lld\n", sum);
	}
}

signed main()
{
	freopen("sfac.in", "r", stdin);
	freopen("sfac.out", "w", stdout);

	l=read(); r=read(); m=read();
	// force::solve();
	task::solve();
	
	return 0;
}