BZOJ3512：DZY Loves Math IV

传送门

Sol

好神仙的题目。。

一开始就直接莫比乌斯反演然后就 \(GG\) 了

orz 题解

permui

枚举 \(n\)，就是求 \(\sum_{i=1}^{n}S(i,m)\)

其中\(S(n,m)=\sum _{i=1}^m\varphi (ni)\)

设 \(n=\prod_{i}p_i^{c_i}\)

设 \(y=\prod _{i=1} p_i^{c_i-1}\)，\(w=\prod _{i=1}p_j\)

那么

\[\begin{aligned}
S(n,m)&=y\sum _{i=1}^m\varphi (wi) \\
&=y\sum _{i=1}^m\varphi (\frac{w}{gcd(i,w)})\varphi (i)gcd(i,w) \\
&=y\sum _{i=1}^m\varphi (\frac{w}{gcd(i,w)})\varphi (i)\sum _{e|gcd(i,w)}\varphi (e) \\
&=y\sum _{i=1}^m\varphi (i)\sum _{e|gcd(i,w)}\varphi (\frac{w}{e}) \\
&=y\sum _{i=1}^m\varphi (i)\sum _{e|i,e|w}\varphi (\frac{w}{e}) \\
&=y\sum _{e|w} \varphi (\frac{w}{e})S(e,\lfloor\frac{m}{e}\rfloor) \\
\end{aligned}
\]

运用了 \(n=\sum _{d|n}\varphi (d)\) 去掉了 \(gcd\)

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int mod(1e9 + 7);
const int maxn(1e6 + 5);
const int blk(80);

inline void Inc(int &x, int y) {
	if ((x += y) >= mod) x -= mod;
}

int pr[maxn], phi[maxn], tot, d, id1[maxn], id2[maxn], idx, m, ans, sphi[maxn], low[maxn];
bitset <maxn> ispr;
map <int, int> s[maxn];

# define ID(x) ((x) <= d ? id1[x] : id2[m / (x)])

int Sumphi(int x) {
    if (x < maxn) return phi[x];
    if (sphi[ID(x)]) return sphi[ID(x)];
    register int ans = (ll)(x + 1) * x / 2 % mod, i, j;
    for (i = 2; i <= x; i = j + 1) j = x / (x / i), Inc(ans, mod - (ll)Sumphi(x / i) * (j - i + 1) % mod);
    return sphi[ID(x)] = ans;
}

int Calc(int n, int v) {
	if (!v || !n) return 0;
	if (n == 1) return Sumphi(v);
	if (v == 1) return (phi[n] - phi[n - 1] + mod) % mod;
	if (s[n].count(v)) return s[n][v];
	register int i, j, ret, y = 1, w = 1, x, cnt, e, dv[30];
	for (cnt = 0, x = n; x > 1; ) {
		w *= low[x], dv[++cnt] = low[x], x /= low[x];
		while (low[x] == dv[cnt]) y *= low[x], x /= low[x];
	}
	for (ret = 0, i = (1 << cnt) - 1; ~i; --i) {
		for (e = 1, j = 0; j < cnt; ++j) if (i >> j & 1) e *= dv[j + 1];
		Inc(ret, (ll)(phi[w / e] - phi[w / e - 1] + mod) * Calc(e, v / e) % mod);
	}
	return s[n][v] = (ll)ret * y % mod;
}

int main() {
	register int i, j, n;
	phi[1] = 1, ispr[1] = 1;
	for (i = 2; i < maxn; ++i) {
		if (!ispr[i]) pr[++tot] = i, phi[i] = i - 1, low[i] = i;
		for (j = 1; j <= tot && i * pr[j] < maxn; ++j) {
			ispr[i * pr[j]] = 1;
			if (i % pr[j]) phi[i * pr[j]] = phi[i] * (pr[j] - 1), low[i * pr[j]] = pr[j];
			else {
				phi[i * pr[j]] = phi[i] * pr[j], low[i * pr[j]] = low[i];
				break;
			}
		}
	}
	for (i = 1; i < maxn; ++i) Inc(phi[i], phi[i - 1]);
	scanf("%d%d", &n, &m);
	for (d = sqrt(m), i = 1; i <= m; i = j + 1) {
		j = m / (m / i);
		(m / i <= d) ? id1[m / i] = ++idx : id2[j] = ++idx;
	}
	for (i = 1; i <= n; ++i) Inc(ans, Calc(i, m));
	printf("%d\n", ans);
    return 0;
}