于神之怒加强版题解
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\sum_{i = 1}^{n} \sum_{j = 1}^{m} \gcd (i, j) ^ k \\ \sum_{i = 1}^{n} \sum_{j = 1}^{m} \sum_{g} g ^ k [\gcd (i, j) = g] \\ \sum_{g} g ^ k \sum_{i = 1}^{\lfloor \frac{n}{g} \rfloor} \sum_{j = 1}^{\lfloor \frac{m}{g} \rfloor} [\gcd (i, j) = 1] \\ \sum_{g} g ^ k \sum_{i = 1}^{\lfloor \frac{n}{g} \rfloor} \sum_{j = 1}^{\lfloor \frac{m}{g} \rfloor} \sum_{d \mid \gcd (i, j)} \mu (d) \\ \sum_{g}^{\min (n, m)} g ^ k \sum_{d}^{\min (\lfloor \frac{n}{g} \rfloor, \lfloor \frac{m}{g} \rfloor)} \mu (d) \lfloor \frac{n}{gd} \rfloor \lfloor \frac{m}{gd} \rfloor \\ \sum\limits_{T = 1}^{\min (n, m)} \sum\limits_{g \mid T} g^k \mu(\dfrac T g) \lfloor \dfrac n T \rfloor \lfloor \dfrac m T \rfloor \\ \sum\limits_{T = 1}^{\min (n, m)} \lfloor \dfrac n T \rfloor \lfloor \dfrac m T \rfloor \sum\limits_{g \mid T} g^k \mu(\dfrac T g) \\ h (T) = \sum\limits_{g \mid T} g^k \mu(\dfrac T g) \\ \sum\limits_{T = 1}^{\min (n, m)} \lfloor \dfrac n T \rfloor \lfloor \dfrac m T \rfloor h (T)
i=1∑nj=1∑mgcd(i,j)ki=1∑nj=1∑mg∑gk[gcd(i,j)=g]g∑gki=1∑⌊gn⌋j=1∑⌊gm⌋[gcd(i,j)=1]g∑gki=1∑⌊gn⌋j=1∑⌊gm⌋d∣gcd(i,j)∑μ(d)g∑min(n,m)gkd∑min(⌊gn⌋,⌊gm⌋)μ(d)⌊gdn⌋⌊gdm⌋T=1∑min(n,m)g∣T∑gkμ(gT)⌊Tn⌋⌊Tm⌋T=1∑min(n,m)⌊Tn⌋⌊Tm⌋g∣T∑gkμ(gT)h(T)=g∣T∑gkμ(gT)T=1∑min(n,m)⌊Tn⌋⌊Tm⌋h(T)
因为
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f (n) = \sum_{d \mid n} d^k, \mu (n)
f(n)=∑d∣ndk,μ(n) 都是积性函数,由迪利克雷卷积可知,
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(f * \mu)(n)
(f∗μ)(n) 也是积性函数,所以
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h (T)
h(T) 是一个积性函数。
参考代码
#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define int long long
#define PII pair <int, int>
#define ULL unsigned long long
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); i++)
#define per(i,j,k) for (int i = (j); i >= (k); i--)
template <typename T>
void read (T &x) {
x = 0; T f = 1;
char ch = getchar ();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar ();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + ch - '0';
ch = getchar ();
}
x *= f;
}
template <typename T, typename... Args>
void read (T &x, Args&... Arg) {
read (x), read (Arg...);
}
const int MaxPrint = 1000;
int Poi_For_Print, Tmp_For_Print[MaxPrint + 5];
template <typename T>
void write (T x) {
if (x == 0) {
putchar ('0');
return;
}
bool flag = (x < 0 ? 1 : 0);
x = (x < 0 ? -x : x);
while (x) Tmp_For_Print[++Poi_For_Print] = x % 10, x /= 10;
if (flag) putchar ('-');
while (Poi_For_Print) putchar (Tmp_For_Print[Poi_For_Print--] + '0');
}
template <typename T, typename... Args>
void write (T x, Args... Arg) {
write (x); putchar (' '); write (Arg...);
}
template <typename T, typename... Args>
void print (T x, char ch) {
write (x); putchar (ch);
}
template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
const int Mod = 1e9 + 7;
const int Maxn = 5 * 1e6;
int t, k, n, m;
int quick_pow (int x, int y) {
int res = 1;
while (y) {
if (y & 1) res = (res * x) % Mod;
x = (x * x) % Mod; y >>= 1;
}
return res;
}
int cnt, primes[Maxn + 5];
int mu[Maxn + 5], h[Maxn + 5], pre[Maxn + 5], px[Maxn + 5];
bool vis[Maxn + 5];
void Euler () {
mu[1] = 1;
h[1] = 1;
rep (i, 2, Maxn) {
if (!vis[i]) {
primes[++cnt] = i;
mu[i] = -1;
h[i] = quick_pow (i, k) * mu[1] % Mod + quick_pow (1, k) * mu[i] % Mod;
h[i] %= Mod;
px[i] = i;
}
rep (j, 1, cnt) {
if (primes[j] > Maxn / i) break;
vis[primes[j] * i] = 1;
if (i % primes[j] == 0) {
mu[i * primes[j]] = 0;
if (i == px[i]) h[i * primes[j]] = quick_pow (i * primes[j], k) + mu[primes[j]] * quick_pow (i, k) % Mod;
else h[i * primes[j]] = h[i / px[i]] * h[primes[j] * px[i]] % Mod; // i * primes[j] = primes[j] ^ q 要特判
px[i * primes[j]] = px[i] * primes[j] % Mod;
break;
}
mu[i * primes[j]] = (mu[i] * mu[primes[j]]) % Mod;
h[i * primes[j]] = (h[i] * h[primes[j]]) % Mod;
px[i * primes[j]] = primes[j];
}
}
rep (i, 1, Maxn) pre[i] = (pre[i - 1] + h[i]) % Mod;
}
signed main () {
// freopen ("C:\\Users\\Administrator\\Desktop\\lihan\\1.in", "r", stdin);
// freopen ("C:\\Users\\Administrator\\Desktop\\lihan\\1.out", "w", stdout);
read (t, k);
Euler ();
while (t--) {
read (n, m);
int l = 1, r, res = 0;
while (l <= Min (n, m)) {
r = Min (n / (n / l), m / (m / l));
res += (n / l) * (m / l) % Mod * (pre[r] - pre[l - 1]) % Mod;
res = (res % Mod + Mod) % Mod;
l = r + 1;
}
print (res, '\n');
}
return 0;
}
又是推错的废式子
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\begin{aligned} \sum_{i = 1}^{n} \sum_{j = 1}^{m} \gcd (i, j) ^ k \\ \sum_{i = 1}^{n} \sum_{j = 1}^{m} \sum_{g} g ^ k [\gcd (i, j) = g] \\ \sum_{g} g ^ k \sum_{i = 1}^{\lfloor \frac{n}{g} \rfloor} \sum_{j = 1}^{\lfloor \frac{m}{g} \rfloor} [\gcd (i, j) = 1] \\ \sum_{g} g ^ k \sum_{i = 1}^{\lfloor \frac{n}{g} \rfloor} \sum_{j = 1}^{\lfloor \frac{m}{g} \rfloor} \sum_{d \mid \gcd (i, j)} \mu (d) \\ \sum_{g}^{\min (n, m)} g ^ k \sum_{d}^{\min (\lfloor \frac{n}{g} \rfloor, \lfloor \frac{m}{g} \rfloor)} \mu (d) \lfloor \frac{n}{gd} \rfloor \lfloor \frac{m}{gd} \rfloor \\ h (g) = \sum_{d}^{\min (\lfloor \frac{n}{g} \rfloor, \lfloor \frac{m}{g} \rfloor)} \mu (d) \lfloor \frac{\lfloor \frac{n}{g} \rfloor}{d} \rfloor \lfloor \frac{\lfloor \frac{m}{g} \rfloor}{d} \rfloor \\ \sum_{g}^{\min (n, m)} g ^ k h (g) \end{aligned}
i=1∑nj=1∑mgcd(i,j)ki=1∑nj=1∑mg∑gk[gcd(i,j)=g]g∑gki=1∑⌊gn⌋j=1∑⌊gm⌋[gcd(i,j)=1]g∑gki=1∑⌊gn⌋j=1∑⌊gm⌋d∣gcd(i,j)∑μ(d)g∑min(n,m)gkd∑min(⌊gn⌋,⌊gm⌋)μ(d)⌊gdn⌋⌊gdm⌋h(g)=d∑min(⌊gn⌋,⌊gm⌋)μ(d)⌊d⌊gn⌋⌋⌊d⌊gm⌋⌋g∑min(n,m)gkh(g)