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……
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 #define rep(i, a, b) for (int i(a); i <= (b); ++i) 6 #define dec(i, a, b) for (int i(a); i >= (b); --i) 7 8 const int mod = 1000000007; 9 const int N = 1000010; 10 11 int v[N], p[N], f[N], r[N], b[N], c[N], d[N]; 12 int n, m, ret, tmp, tot; 13 14 inline int Pow(int a, int b){ 15 int t = 1; 16 for (; b; b >>= 1, a = 1LL * a * a % mod) 17 if (b & 1) t = 1LL * t * a % mod; 18 return t; 19 } 20 21 int main(){ 22 23 scanf("%d%d", &n, &m); ++m; 24 25 f[1] = 1; 26 rep(i, 2, m){ 27 if (!v[i]){ 28 f[i] = Pow(i, m - 1); 29 p[++tot] = i; 30 } 31 32 rep(j, 1, tot){ 33 if (i * p[j] > m) break; 34 v[i * p[j]] = 1; 35 f[i * p[j]] = 1LL * f[i] * f[p[j]] % mod; 36 if (i % p[j] == 0) break; 37 } 38 } 39 40 rep(i, 2, m) (f[i] += f[i - 1]) %= mod; 41 42 if ((n %= mod) <= m) return 0 * printf("%d\n", f[n]); 43 44 r[0] = r[1] = 1; 45 rep(i, 2, m) r[i] = 1LL * (mod - r[mod % i]) * (mod / i) % mod; 46 rep(i, 2, m) r[i] = 1LL * r[i - 1] * r[i] % mod; 47 rep(i, 1, m + 1) b[i] = (n - i + 1 + mod) % mod; 48 49 c[0] = d[m + 2] = 1; 50 rep(i, 1, m + 1) c[i] = 1LL * c[i - 1] * b[i] % mod; 51 dec(i, m + 1, 1) d[i] = 1LL * d[i + 1] * b[i] % mod; 52 53 rep(i, 0, m){ 54 tmp = 1LL * f[i] * r[m - i] % mod * r[i] % mod * c[i] % mod * d[i + 2] % mod; 55 if ((m - i) & 1) (ret += mod - tmp) %= mod; 56 else (ret += tmp) %= mod; 57 } 58 59 return 0 * printf("%d\n", ret); 60 }
