Codeforces 622F The Sum of the k-th Powers（数论）

题目链接 The Sum of the k-th Powers

其实我也不懂为什么这么做的……看了无数题解觉得好厉害哇……

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)    for (int i(a); i <= (b); ++i)
#define dec(i, a, b)    for (int i(a); i >= (b); --i)

const int mod = 1000000007;
const int N   = 1000010;

int v[N], p[N], f[N], r[N], b[N], c[N], d[N];
int n, m, ret, tmp, tot;

inline int Pow(int a, int b){
    int t = 1;
    for (; b; b >>= 1, a = 1LL * a * a % mod)
        if (b & 1) t = 1LL * t * a % mod;
    return t;
}

int main(){

    scanf("%d%d", &n, &m); ++m;

    f[1] = 1; 
    rep(i, 2, m){
        if (!v[i]){
            f[i] = Pow(i, m - 1);
            p[++tot] = i;
        }

        rep(j, 1, tot){
            if (i * p[j] > m) break;
            v[i * p[j]] = 1;
            f[i * p[j]] = 1LL * f[i] * f[p[j]] % mod;
            if (i % p[j] == 0) break;
        }
    }

    rep(i, 2, m) (f[i] += f[i - 1]) %= mod;

    if ((n %= mod) <= m) return 0 * printf("%d
", f[n]);

    r[0] = r[1] = 1;
    rep(i, 2, m) r[i] = 1LL * (mod - r[mod % i]) * (mod / i) % mod;
    rep(i, 2, m) r[i] = 1LL * r[i - 1] * r[i] % mod;
    rep(i, 1, m + 1) b[i] = (n - i + 1 + mod) % mod;
    
    c[0] = d[m + 2] = 1;
    rep(i, 1, m + 1) c[i] = 1LL * c[i - 1] * b[i] % mod;
    dec(i, m + 1, 1) d[i] = 1LL * d[i + 1] * b[i] % mod;

    rep(i, 0, m){
        tmp = 1LL * f[i] * r[m - i] % mod * r[i] % mod * c[i] % mod * d[i + 2] % mod;
        if ((m - i) & 1) (ret += mod - tmp) %= mod;
        else (ret += tmp) %= mod;
    }

    return 0 * printf("%d
", ret);
}