题目大意
记 (d(x)) 为 (x) 的约数个数，求 [
sum{i = 1}^{n} sum{j = 1}^{m} d(ij)
] 多组询问。
(1 leqslant n, ; m, ; T leqslant 500,000)
题目链接
BZOJ 3994
题解
首先有 [
d(nm) = sum{i | n} sum{j | m} [gcd(i, ; j) = 1]
] 那么就可以开始推式子了（默认 (n) 为更小的）： [
begin{align}
&sum{i = 1}^{n} sum{j = 1}^{m} d(ij) \
= &sum{i = 1}^{n} sum{j = 1}^{m} sum{a | i} sum{b | j} [gcd(a, ; b) = 1] \
= &sum{i = 1}^{n} sum{j = 1}^{m} sum{a | i} sum{b | j} sum{d | a, ; d | b} mu(d) qquad (mu times 1 = e) \
= &sum{i = 1}^{n} sum{j = 1}^{m} sum{d | i, ; d | j} mu(d) d(lfloor frac{i}{d} rfloor) d(lfloor frac{j}{d} rfloor) \
= &sum{d = 1}^{n} mu(d) sum{x | i} d(lfloor frac{i}{d} rfloor) sum{d | j} d(lfloor frac{j}{d} rfloor) \
= &sum{d = 1}^{n} mu(d) sum{i = 1}^{lfloor frac{n}{d} rfloor} d(i) sum{j = 1}^{lfloor frac{m}{d} rfloor} d(j)
end{align}
] 可以用线性筛在 (O(n)) 求出 (d(x)) 函数、莫比乌斯函数及其前缀和，每次询问 (O(sqrt{n})) 分块求出答案。
代码
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#include <algorithm>
const int MAXN = 50005;
long long mu[MAXN], d[MAXN];
int prime[MAXN], primeCnt;
bool notPrime[MAXN];
void () {
static int minPrimeCnt[MAXN];
notPrime[0] = notPrime[1] = true;
mu[1] = d[1] = 1;
for (int i = 2; i < MAXN; i++) {
if (!notPrime[i]) {
prime[++primeCnt] = i;
mu[i] = -1;
d[i] = 2;
minPrimeCnt[i] = 1;
}
for (int j = 1; j <= primeCnt && i * prime[j] < MAXN; j++) {
notPrime[i * prime[j]] = true;
if (i % prime[j] == 0) {
mu[i * prime[j]] = 0;
minPrimeCnt[i * prime[j]] = minPrimeCnt[i] + 1;
d[i * prime[j]] = d[i] / (minPrimeCnt[i] + 1) * (minPrimeCnt[i] + 2);
break;
} else {
mu[i * prime[j]] = -mu[i];
d[i * prime[j]] = d[i] * 2;
minPrimeCnt[i * prime[j]] = 1;
}
}
}
for (int i = 2; i < MAXN; i++) {
mu[i] += mu[i - 1];
d[i] += d[i - 1];
}
}
long long calc(int n, int m) {
if (n > m) std::swap(n, m);
long long res = 0;
for (int i = 1, last; i <= n; i = last + 1) {
last = std::min(n / (n / i), m / (m / i));
res += (mu[last] - mu[i - 1]) * d[n / i] * d[m / i];
}
return res;
}
int main() {
linearShaker();
int T;
scanf("%d", &T);
while (T--) {
int n, m;
scanf("%d %d", &n, &m);
printf("%lldn", calc(n, m));
}
return 0;
}
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