P6271 [湖北省队互测2014]一个人的数论
$$
\begin {aligned}
\sum _ {i = 1} ^ n i ^ d [(i, n) = 1] & = \sum _ {i = 1} ^ n i ^ d \sum _ {k | (i, n)} \mu(k) \\
& = \sum _ {k | n} \mu(k) \sum _ {i = 1} ^ {\frac n k} (ki) ^ d \\
& = \sum _ {k | n} k ^ d \mu(k) \sum _ {i = 1} ^ {\frac n k} i ^ d
\end {aligned}
$$
后面的式子是一个自然数幂和，$\sum _ {i = 0} ^ {n - 1} i ^ d$ ，可以用伯努利数构建为一个 $d + 1$ 次的多项式。记 $i$ 次的系数为 $f _ i$ ，那么有：
$$
\begin {aligned}
\sum _ {i = 1} ^ n i ^ d [(i, n) = 1] & = \sum _ {k | n} k ^ d \mu(k) \sum _ {i = 1} ^ {\frac n k} i ^ d \\
& = \sum _ {k | n} k ^ d \mu(k) ((\frac n k) ^ d + \sum _ {i = 1} ^ {d + 1} f _ i (\frac n k) ^ i)\\
& = \sum _ {k | n} k ^ d \mu(k) \sum _ {i = 1} ^ {d + 1} f _ i (\frac n k) ^ i + \sum _ {k | n} k ^ d \mu(k) (\frac n k) ^ d\\
& = \sum _ {i = 1} ^ {d + 1} f _ i n _ i \sum _ {k | n}\mu(k) k ^ {d - i} + \sum _ {k | n} \mu(k) n ^ d
\end {aligned}
$$
对于 $\sum _ {k | n} \mu(k) n ^ d$ ，答案为 $[n = 1]$ ，在本题中恒为 $0$ 。
对于 $ \sum _ {i = 1} ^ {d + 1} f _ i n _ i \sum _ {k | n}\mu(k) k ^ {d - i}$ ，容易发现，后面是一个积性函数，可以将所有质数的值乘起来即可，又注意到，对于质数的大于 $1$ 的次幂，$\mu(k) = 1$ ，没有贡献。因此只有质数处有 $-p ^ {d - i}$ 的贡献。于是可以在 $O(dw)$ 的时间内完成。
查看代码
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| #include <cstdio>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 1) + (x << 3) + c - '0';
flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar('0' + x % 10);
}
typedef long long LL;
const int N = 110, mod = 1e9 + 7;
int d, w, p[N], c[N];
int inv[N], B[N], C[N][N], f[N];
int binpow (int b, int k)
{
if (k < 0) k += mod - 1;
int res = 1;
for (; k; k >>= 1, b = (LL)b * b % mod)
if (k & 1) res = (LL)res * b % mod;
return res;
}
void init()
{
for (int i = 0; i <= d + 1; ++i)
{
C[i][0] = 1;
for (int j = 1; j <= i; ++j)
C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
}
inv[1] = 1;
for (int i = 2; i <= d + 1; ++i)
inv[i] = -(LL)(mod / i) * inv[mod % i] % mod;
B[0] = 1;
for (int i = 1; i <= d; ++i)
{
int t = 0;
for (int j = 0; j < i; ++j)
t = (t + (LL)C[i + 1][j] * B[j]) % mod;
B[i] = -(LL)t * inv[i + 1] % mod;
}
for (int i = 0; i <= d; ++i)
f[d + 1 - i] = (LL)C[d + 1][i] * B[i] % mod * inv[d + 1] % mod;
}
int calc (int x)
{
int res = 1;
for (int i = 1; i <= w; ++i)
res = res * (1ll - binpow(p[i], x)) % mod;
return res;
}
int main ()
{
read(d, w);
init();
int n = 1;
for (int i = 1; i <= w; ++i)
{
read(p[i], c[i]);
n = (LL)n * binpow(p[i], c[i]) % mod;
}
int res = 0;
for (int i = 1, t = n; i <= d + 1; ++i, t = (LL)t * n % mod)
res = (res + (LL)f[i] * t % mod * calc(d - i)) % mod;
write((res + mod) % mod);
return 0;
}
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