Codeforces 622F. The Sum of the k-th Powers Lagrange interpolation method

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Ask Sigma (I:1 to N) i^k.

In order to do this question these two days really fill up a lot of number theory, before even multiplication inverse yuan do not know ...

About Lagrange interpolation, I'm looking at the http://www.guokr.com/post/456777/here, and it's kinda fun ...

According to the example given in the topic, we can find that the highest order of the K-th formula is k+1 times, and according to the Lagrange interpolation method, it is necessary to construct a k+1 equation k+2 term.

Then the formula is, for this problem, P[i] is i^k+ (i-1) ^k+ (i-2) ^k+.....+1^k, this part can be preprocessed out. I will not make the formula, stolen from the http://www.cnblogs.com/qscqesze/p/5207132.html here (fog

We find that above is (n-1) * (n-2) ... * (n-k-2)/(N-i), the upper part of the preprocessing, except (n-i) equivalent multiplication (n-i) inverse.

The following part is (i-1) * (i-2) * ... (i-i+1) * (i-i-1) * (i-i-2) * ... (I-k-2), which is equivalent to two factorial multiplication, factorial preprocessing, and then pay attention to the following sign on it.

#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;
}

Codeforces 622F. The Sum of the k-th Powers Lagrange interpolation method