传送门
题目所给的不合法的条件可以转化为
[exists p,p^2|gcd(a,b) Leftrightarrow mu(gcd(a,b))
e 0
]
那么
[ans=sum_{a=1}^{A}sum_{b=1}^{B}[mu(gcd(i,j))
e 0]frac{ab}{gcd(a,b)}
]
不妨假设 (Ale B)，枚举 (gcd) 之后经典莫比乌斯反演
设 (S(x)=sum_{i=1}^{x}i)
得到
[sum_{i=1}^{A}S(lfloorfrac{A}{i}
floor)S(lfloorfrac{B}{i}
floor)sum_{d|i}[mu(d)
e 0](frac{i}{d})^2mu(frac{i}{d})d
]
后面的狄利克雷卷积是一个积性函数，直接线性筛
# include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
const int maxn(4e6 + 5);
const int mod(1 << 30);
int test, pr[maxn], tot, n, m, cnt[maxn], pwd[maxn];
uint ans, s[maxn];
bitset <maxn> ispr;
inline uint S(uint x) {
return (x & 1) ? (x + 1) / 2 * x : x / 2 * (x + 1);
}
int main() {
register uint i, j;
ispr[1] = 1, s[1] = 1;
for (i = 2; i < maxn; ++i) {
if (!ispr[i]) pwd[i] = pr[++tot] = i, s[i] = i - i * i, cnt[i] = 1;
for (j = 1; j <= tot && i * pr[j] < maxn; ++j) {
ispr[i * pr[j]] = 1;
if (i % pr[j]) {
s[i * pr[j]] = s[i] * s[pr[j]];
pwd[i * pr[j]] = pr[j], cnt[i * pr[j]] = 1;
}
else {
cnt[i * pr[j]] = cnt[i] + 1, pwd[i * pr[j]] = pwd[i] * pr[j];
if (cnt[i] == 1) s[i * pr[j]] = -(uint)pr[j] * pr[j] * pr[j] * s[i / pwd[i]];
else s[i * pr[j]] = 0;
break;
}
}
}
for (i = 2; i < maxn; ++i) s[i] += s[i - 1];
scanf("%d", &test);
while (test) {
scanf("%d%d", &n, &m), test--;
if (n > m) swap(n, m);
for (ans = 0, i = 1; i <= n; i = j + 1) {
j = min(n / (n / i), m / (m / i));
ans += (s[j] - s[i - 1]) * S(n / i) * S(m / i);
}
printf("%u
", ans % mod);
}
return 0;
}