洛谷 P4948 拉格朗日多项式插值(杜老师板子)

https://www.luogu.org/problemnew/show/P4948

这篇博客主要目的是存一下的dls的神奇板子,本来应该是推公式或者二分做的

但是dls的插值板子直接写好了这个特殊式子的算法......

#include <bits/stdc++.h>
#define endl '\n'
#define ll long long
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<int,int>
#define ull unsigned long long
#define all(x) x.begin(),x.end()
#pragma GCC optimize("unroll-loops")
#define inline inline __attribute__(  \
(always_inline, __gnu_inline__, __artificial__)) \
__attribute__((optimize("Ofast"))) __attribute__((target("sse"))) \
__attribute__((target("sse2"))) __attribute__((target("mmx")))
#define IO ios::sync_with_stdio(false);
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define for_node(x,i) for(int i=head[x];i;i=e[i].next)
#define show(x) cout<<#x<<"="<<x<<endl
#define showa(a,b) cout<<#a<<'['<<b<<"]="baidu<a[b]<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
using namespace std;
const int maxn=1e6+10,maxm=2e6+10;
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;
const double PI=acos(-1.0);
//head
int casn,n,m,k;
int num[maxn];
ll a[maxn];
ll pow_mod(ll a,ll b,ll c=mod,ll ans=1){while(b){if(b&1) ans=(a*ans)%c;a=(a*a)%c,b>>=1;}return ans;}

namespace polysum {
	const int maxn=101000;
	const ll mod=1e9+7;
	ll a[maxn],f[maxn],g[maxn],p[maxn],p1[maxn],p2[maxn],b[maxn],h[maxn][2],C[maxn];
	ll calcn(int d,ll *a,ll n) {//d次多项式(a[0-d])求第n项
		if (n<=d) return a[n];
		p1[0]=p2[0]=1;
		rep(i,0,d) {
			ll t=(n-i+mod)%mod;
			p1[i+1]=p1[i]*t%mod;
		}
		rep(i,0,d) {
			ll t=(n-d+i+mod)%mod;
			p2[i+1]=p2[i]*t%mod;
		}
		ll ans=0;
		rep(i,0,d) {
			ll t=g[i]*g[d-i]%mod*p1[i]%mod*p2[d-i]%mod*a[i]%mod;
			if ((d-i)&1) ans=(ans-t+mod)%mod;
			else ans=(ans+t)%mod;
		}
		return ans;
	}
	void init(int maxm) {//初始化预处理阶乘和逆元(取模乘法)
		f[0]=f[1]=g[0]=g[1]=1;
		rep(i,2,maxm+4) f[i]=f[i-1]*i%mod;
		g[maxm+4]=pow_mod(f[maxm+4],mod-2);
		per(i,1,maxm+3) g[i]=g[i+1]*(i+1)%mod;
	}
	ll polysum(ll n,ll *a,ll m) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]
		// m次多项式求第n项前缀和
		a[m+1]=calcn(m,a,m+1);
		rep(i,1,m+1) a[i]=(a[i-1]+a[i])%mod;
		return calcn(m+1,a,n-1);
	}
	ll qpolysum(ll R,ll n,ll *a,ll m) { // a[0].. a[m] \sum_{i=0}^{n-1} a[i]*R^i
		if (R==1) return polysum(n,a,m);
		a[m+1]=calcn(m,a,m+1);
		ll r=pow_mod(R,mod-2),p3=0,p4=0,c,ans;
		h[0][0]=0;
		h[0][1]=1;
		rep(i,1,m+1) {
			h[i][0]=(h[i-1][0]+a[i-1])*r%mod;
			h[i][1]=h[i-1][1]*r%mod;
		}
		rep(i,0,m+1) {
			ll t=g[i]*g[m+1-i]%mod;
			if (i&1) p3=((p3-h[i][0]*t)%mod+mod)%mod,p4=((p4-h[i][1]*t)%mod+mod)%mod;
			else p3=(p3+h[i][0]*t)%mod,p4=(p4+h[i][1]*t)%mod;
		}
		c=pow_mod(p4,mod-2)*(mod-p3)%mod;
		rep(i,0,m+1) h[i][0]=(h[i][0]+h[i][1]*c)%mod;
		rep(i,0,m+1) C[i]=h[i][0];
		ans=(calcn(m,C,n)*pow_mod(R,n)-c)%mod;
		if (ans<0) ans+=mod;
		return ans;
	}
}

int main() {
//#define test
#ifdef test
	auto _start = chrono::high_resolution_clock::now();
	freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#endif
	IO;
	ll n,r,k;
	cin>>n>>r>>k;
	polysum::init(k+5);
	rep(i,0,2010) a[i]=pow_mod(i,k);
	ll ans=polysum::qpolysum(r,n+1,a,k+1);
	if(k==0) ans=(ans-1+mod)%mod;
	cout<<ans<<endl;
#ifdef test
	auto _end = chrono::high_resolution_clock::now();
  cerr << "elapsed time: " << chrono::duration<double, milli>(_end - _start).count() << " ms\n";
	fclose(stdin);fclose(stdout);system("out.txt");
#endif
	return 0;
}