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 = ;
 const int N   = ;

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

 int main(){

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

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

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

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

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

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

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

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

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