题目链接：
题目大意：
快速求：
[sum_{pin Prime}sum_{i=1}^{n}sum_{j=1}^{m}left[operatorname{gcd}(i,j)==p
ight]
]
正文：
按照莫比乌斯反演的常规套路，将式子化简成我们能接受的时限：
[egin{aligned}sum_{pin Prime}sum_{i=1}^{n}sum_{j=1}^{m}left[operatorname{gcd}(i,j)==p
ight] &=sum_{pin Prime}sum_{i=1}^{lfloorfrac{n}{p}
floor}sum_{j=1}^{lfloorfrac{m}{p}
floor}left[operatorname{gcd}(i,j)==1
ight]\&=sum_{pin Prime}sum_{i=1}^{lfloorfrac{n}{p}
floor}sum_{j=1}^{lfloorfrac{m}{p}
floor}sum_{d|operatorname{gcd}(i,j)}mu(d)\&=sum_{pin Prime} sum_{d=1}mu(d)sum_{d|i}^{lfloorfrac{n}{p}
floor}sum_{k|j}^{lfloorfrac{m}{p}
floor}1\&=sum_{pin Prime}sum_{d=1}mu(d)leftlfloorfrac{n}{pd}
ight
floorleftlfloorfrac{m}{pd}
ight
floorend{aligned}
]
设 (k=pd) 枚举 (k)：
[egin{aligned}sum_{pin Prime}sum_{d=1}mu(d)leftlfloorfrac{n}{pd}
ight
floorleftlfloorfrac{m}{pd}
ight
floor&=sum_{k=1}leftlfloorfrac{n}{k}
ight
floorleftlfloorfrac{m}{k}
ight
floorsum_{p|k,pin Prime}mu(frac{k}{p})end{aligned}
]
预处理 (mu(frac{k}{p}))，就能 (O(tsqrt{n}))。
代码：
稍微卡卡常就能过了。
inline void prework()
{
miu[1] = 1;
for (int i = 2; i <= N - 10; i++)
{
if(!vis[i]) {pri[++cnt] = i, miu[i] = -1;}
for (int j = 1; j <= cnt && pri[j] * i <= N - 10; j++)
{
vis[pri[j] * i] = 1;
if (i % pri[j] == 0)
{
miu[i * pri[j]] = 0;
break;
}
else
miu[i * pri[j]] = -miu[i];
}
}
for (int i = 1; i <= cnt; i++)
for (int j = 1; j * pri[i] <= N - 10; j++)
sum[j * pri[i]] += miu[j];
for (int i = 1; i <= N - 10; i++)
sum[i] += sum[i - 1];
}
int main()
{
prework();
for (read(t); t--; )
{
ans = 0LL;
read(n);read(m);
if(n > m)
{
ll c = n; n = m; m = c;
}
for (register int l = 1, r; l <= n; l = r + 1)
{
r = min (n / (n / l), m / (m / l));
ans += (sum[r] - sum[l - 1]) * (n / l) * (m / l);
}
print(ans);
}
return 0;
}