codeforces 622F. The Sum of the k-th Powers 拉格朗日插值法

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const ll mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 1e6+5;
ll f[maxn], fac[maxn];
ll pow(ll a, ll b) {
    ll ret = 1;
    while(b) {
        if(b&1)
            ret = (ret*a)%mod;
        a = (a*a)%mod;
        b>>=1;
    }
    return ret;
}
int main()
{
    ll n, k;
    cin>>n>>k;
    for(int i = 1; i<=k+2; i++) {
        f[i] = (f[i-1]+pow(i*1LL, k))%mod;
    }
    fac[0] = 1;
    for(int i = 1; i<maxn; i++) {
        fac[i] = (fac[i-1]*i)%mod;
    }
    if(n<=k+2) {
        cout<<f[n]<<endl;
        return 0;
    }
    ll cur = 1, ans = 0;
    for(int i = 1; i<=k+2; i++) {
        cur = (cur*(n-i))%mod;
    }
    for(int i = 1; i<=k+2; i++) {
        ll tmp = pow(n-i, mod-2)%mod;
        ll tmp1 = pow(fac[i-1]%mod*fac[k+2-i]%mod, mod-2)%mod;
        int sign = (k+2-i)%2?-1:1;
        ans = (ans + sign*tmp*tmp1%mod*f[i]%mod*cur%mod)%mod;
    }
    ans = (ans+mod)%mod;
    cout<<ans<<endl;
    return 0;
}