拉格朗日插值练习1

题目链接：https://codeforces.com/problemset/problem/622/F
拉格朗日插值入门题，请熟悉一下要怎么传参就行
update:
注意虽然原理上是说要k次多项式的前k+1个值，但用模板的时候是a[i]的前k+1个值

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9+7;

ll powmod(ll a, ll b){
    ll ans = 1;
    while(b > 0){
        if(b & 1){
            ans = ans * a % mod;
        }
        a = a * a % mod;
        b >>= 1; 
    } 
    return ans;
}

namespace polysum {
    #define rep(i,a,n) for (int i=a;i<n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    const int D=101000;
    ll a[D],f[D],g[D],p[D],p1[D],p2[D],b[D],h[D][2],C[D];
    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+1) {
            ll t=(n-i+mod)%mod;
            p1[i+1]=p1[i]*t%mod;
        }
        rep(i,0,d+1) {
            ll t=(n-d+i+mod)%mod;
            p2[i+1]=p2[i]*t%mod;
        }
        ll ans=0;
        rep(i,0,d+1) {
            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 M) {//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,M+5) f[i]=f[i-1]*i%mod;
        g[M+4]=powmod(f[M+4],mod-2);
        per(i,1,M+4) 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+2) 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=powmod(R,mod-2),p3=0,p4=0,c,ans;
        h[0][0]=0;h[0][1]=1;
        rep(i,1,m+2) {
            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+2) {
            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=powmod(p4,mod-2)*(mod-p3)%mod;
        rep(i,0,m+2) h[i][0]=(h[i][0]+h[i][1]*c)%mod;
        rep(i,0,m+2) C[i]=h[i][0];
        ans=(calcn(m,C,n)*powmod(R,n)-c)%mod;
        if (ans<0) ans+=mod;
        return ans;
    }
}

const int maxn = 1e6+100;
ll a[maxn], b[maxn];

int main()
{
    ll n, k;
    cin >> n >> k;
    if(k == 0) {
        printf("%lld",n);
        return 0;
    }
    polysum::init(k+100);
    b[0] = 0;
    for(int i = 1; i <= k+1; i++)
        b[i] = powmod(i, k);
    ll ans = polysum::polysum(n+1, b, k+1); 
    cout << ans << endl;
    return 0;
}