[BZOJ2820]YY的GCD
题目大意
给你 \(n, m\) 求 $\displaystyle \sum_{i = 1}^n \sum_{j = 1}^m [ gcd(x, y) == p ] $
解析
这道题实际上是上一道解决的HAOI2011 ProblemB的弱化版
同样的我们的我们可以设出 \(F(x)\) , \(f(x)\),然后用莫比乌斯反演推出 \(f(x)\)
然后进行数论分块就行了
#include <bits/stdc++.h>
using namespace std;
int T;
const int maxn = 1e7 + 10;
int prime[maxn], mu[maxn], n, m, tot;
long long f[maxn];
inline void sieve() {
mu[1] = 1; fill(prime, prime + maxn, 1); tot = 0;
for (int i = 2; i < maxn; ++ i) {
if (prime[i]) prime[++tot] = i, mu[i] = -1;
for (int j = 1; j <= tot && i * prime[j] < maxn; ++ j) {
prime[i * prime[j]] = 0;
if (i % prime[j] == 0) {
mu[i * prime[j]] = 0;
break;
} else {
mu[i * prime[j]] = -mu[i];
}
}
}
for (int i = 1; i <= tot; ++ i) {
for (int j = 1; j * prime[i] < maxn; ++ j) {
f[j * prime[i]] += mu[j];
}
}
for (int i = 1; i < maxn; ++ i) f[i] += f[i - 1];
}
int main() {
sieve();
scanf("%d", &T);
while (T --) {
scanf("%d%d", &n, &m);
if (n > m) swap(n, m);
long long ans = 0;
for (int i = 1, nxt; i <= n; i = nxt + 1) {
nxt = min(n / (n / i), m / (m / i));
ans += (f[nxt] - f[i - 1]) * (n / i) * (m / i);
}
printf("%lld\n", ans);
}
}