51nod1847 奇怪的数学题 (Min_25筛+第二类斯特林数)

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$\sum_{i=1}^n\sum_{j=1}^n\mathrm{sgcd}(i,j)^k=\sum_{p=1}^ns(p)^k\sum_{i=1}^n\sum_{j=1}^n[\gcd(i,j)=p]=\sum_{p=1}^ns(p)^k(-1+2\sum_{i=1}^{n/p}\varphi(i))$

由于 $n$ 的范围是 $10^9$ ，对于后面的我们最多只有根号种取值，根据套路，可以杜教筛/Min_25筛一波。

至于前面的东西，我们可以考虑Min_25筛的过程：

Min_25筛我们设 $g(n,j)$ 为2~n内素数或最小质因子$\ge p_j$的数的$k$次方的和

考虑转移：

$g(n,j)=g(n,j-1)-p_j^k(g(a/p_j,j-1)-g(p_{j-1},j-1)) (n\ge p_j^2)$

我们发现后面减去的 $p_j^k(g(a/p_j,j-1)-g(p_{j-1},j-1))$ 就是最小质因子恰为 $p_j$ 的合数的k次方的和

那么 $g(a/p_j,j-1)-g(p_{j-1},j-1)$ 就是[1, n]内这最小质因子为 $p_j$ 的合数的 $s(x)^k$ 的和。

我们对于每个n/x向下取整开一个数，记录每个 j 的这个值，然后差分一下就是区间的答案了。（注意加上质数的贡献，也就是区间质数个数）

预处理Min_25筛的数组时，由于模数是合数，所以不能拉格朗日插值，要用第二类斯特林数：

$\sum_{i=1}^ni^k=\sum_{i=1}^n\sum_{j=0}^k\{^k_j\}{i\choose j}j!=\sum_{j=0}^k\{^k_j\}j!\sum_{i=1}^n{i\choose j}=\sum_{j=0}^k\{^k_j\}j!{n+1\choose j+1}=\sum_{j=0}^k\{^k_j\}\frac{n+1^{\underline{j+1}}}{j+1}$

phi用Min_25筛的代码：(观察txc巨佬的)

#include <cstdio>
#include <cmath>
#define int unsigned
using namespace std;

int n, m, k, sqt, tot;
int a[233333], g0[233333], g1[233333], gk[233333], gphi[233333], ans[233333], prime[233333], s[233][233];
int getid(int x) { return x <= sqt ? x : m - n / x + 1; }

int qsum(int n)
{
	int ans = 0;
	for (int i = 0; i <= k; i++)
	{
		int tmp = 1;
		for (int j = 0; j <= i; j++) //(n - j + 1)
			if ((n - j + 1) % (i + 1) == 0) tmp *= ((n - j + 1) / (i + 1));
			else tmp *= (n - j + 1);
		ans += tmp * s[k][i];
	}
	return ans;
}
signed main()
{
	scanf("%u%u", &n, &k), sqt = sqrt(n);
	s[0][0] = 1;
	for (int i = 1; i <= k; i++) for (int j = 1; j <= i; j++) s[i][j] = s[i - 1][j] * j + s[i - 1][j - 1];
	for (int i = 1; i <= n; i = a[m] + 1)
	{
		a[++m] = n / (n / i);
		gk[m] = qsum(a[m]) - 1;
		g1[m] = a[m] * (long long)(a[m] + 1) / 2 - 1;
		g0[m] = a[m] - 1;
	}
	for (int i = 1; i <= sqt; i++) if (g0[i] != g0[i - 1])
	{
		prime[++tot] = i;
		int tmp = 1;
		for (int t = 1; t <= k; t++) tmp *= i;
		for (int j = m; a[j] >= i * i; j--)
		{
			int id = getid(a[j] / i);
			gk[j] -= tmp * (gk[id] - gk[i - 1]);
			ans[j] += gk[id] - gk[i - 1];
			g1[j] -= i * (g1[id] - g1[i - 1]);
			g0[j] -= g0[id] - g0[i - 1];
		}
	}
	for (int i = 1; i <= m; i++) gphi[i] = (g1[i] -= g0[i]);
	for (int i = tot; i >= 1; i--)
		for (int j = m; a[j] >= prime[i] * prime[i]; j--)
			for (int x = prime[i], f = x - 1; x * (long long)prime[i] <= a[j]; x *= prime[i], f *= prime[i])
				gphi[j] += f * (gphi[getid(a[j] / x)] - g1[prime[i]] + prime[i]);
	int res = 0;
	for (int i = 2; i <= m; i++) res += (ans[i] - ans[i - 1] + g0[i] - g0[i - 1]) * (2 * gphi[m - i + 1] + 1);
	printf("%u\n", res);
	return 0;
}

phi用杜教筛写的代码：

#include <cmath>
#include <cstdio>
#include <cstring>
using namespace std;

#define int unsigned

int n, k, sqt;
int a[233333], g0[233333], gk[233333], prime[233333], ans[233333], tot, m;
int s[60][60];
int phi[233333], mem[233333];
bool vis[233333];

int qid(int x) { return x <= sqt ? x : m - n / x + 1; }

int qsum(int x, int k)
{
	int ans = 0;
	for (int i = 0; i <= k; i++)
	{
		int tmp = 1;
		for (int j = 0; j <= i; j++)
		{
			if ((x + 1 - j) % (i + 1) == 0) tmp *= (x + 1 - j) / (i + 1);
			else tmp *= x + 1 - j;
		}
		ans += tmp * s[k][i];
	}
	return ans;
}

int chuphi(int id)
{
	if (a[id] <= sqt) { return phi[a[id]]; }
	if (mem[id] != 0xffffffff) return mem[id];
	int n = a[id], ans = n * (n + 1) / 2;
	for (int i = 2; i <= n; i = n / (n / i) + 1)
		ans -= chuphi(qid(n / i)) * (n / (n / i) - i + 1);
	return mem[id] = ans;
}

signed main()
{
	memset(mem, 0xff, sizeof(mem)); phi[1] = 1;
	scanf("%u%u", &n, &k), sqt = sqrt(n);
	s[0][0] = 1;
	for (int i = 1; i <= k; i++) for (int j = 1; j <= k; j++) s[i][j] = s[i - 1][j - 1] + j * s[i - 1][j];
	for (int i = 1; i <= n; i = a[m] + 1)
	{
		a[++m] = n / (n  / i);
		g0[m] = a[m] - 1;
		gk[m] = qsum(a[m], k) - 1;
	}
	for (int i = 1; i <= sqt; i++)
	{
		if (g0[i] != g0[i - 1])
		{
			prime[++tot] = i, phi[i] = i - 1;
			int tmp = 1; for (int j = 1; j <= k; j++) tmp *= i;
			for (int j = m; a[j] >= i * i; j--)
			{
				int id = qid(a[j] / i);
				g0[j] -= g0[id] - g0[i - 1];
				gk[j] -= tmp * (gk[id] - gk[i - 1]);
				ans[j] += gk[id] - gk[i - 1];
			}
		}
		for (int j = 1; j <= tot && i * prime[j] <= sqt; j++)
		{
			if (i % prime[j] == 0) { phi[i * prime[j]] = phi[i] * prime[j]; break; }
			phi[i * prime[j]] = phi[i] * (prime[j] - 1);
		}
		phi[i] += phi[i - 1];
	}
	int res = 0;
	for (int i = 1; i <= m; i++)
	{
		res += (ans[i] - ans[i - 1] + g0[i] - g0[i - 1]) * (2 * chuphi(m - i + 1) - 1);
	}
	printf("%u\n", res);
	return 0;
}