题目链接：
题目大意：
快速求：
[sum_{i=1}^{n}sum_{j=1}^{m}left[operatorname{gcd}(i,j)==d
ight]
]
正文：
将式子化简：
[egin{aligned}sum_{i=1}^{n}sum_{j=1}^{m}left[operatorname{gcd}(i,j)==d
ight] &= sum_{i=1}^{lfloorfrac{n}{d}
floor}sum_{j=1}^{lfloorfrac{m}{d}
floor}left[operatorname{gcd}(i,j)==1
ight]\ &=sum_{i=1}^{lfloorfrac{n}{d}
floor}sum_{j=1}^{lfloorfrac{m}{d}
floor}sum_{k|operatorname{gcd}(i,j)}mu(k)end{aligned}
]
把 (mu) 那儿移到前面：
[egin{aligned}sum_{i=1}^{lfloorfrac{n}{d}
floor}sum_{j=1}^{lfloorfrac{m}{d}
floor}sum_{k|operatorname{gcd}(i,j)}mu(k) &= sum_{k=1}mu(k)sum_{k|i}^{lfloorfrac{n}{d}
floor}sum_{k|j}^{lfloorfrac{m}{d}
floor}1\ &=sum_{k=1}mu(k)leftlfloorfrac{n}{kd}
ight
floorleftlfloorfrac{m}{kd}
ight
floorend{aligned}
]
接着预处理一下 (mu) 函数再求个关于 (leftlfloorfrac{n}{kd} ight floorleftlfloorfrac{m}{kd} ight floor) 的 整除分块 就行了。
代码：
ll n, m, d, t;
ll ans;
ll pri[N], miu[N], cnt, sum[N];
bool vis[N];
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 <= N - 10; i++)
sum[i] = sum[i - 1] + miu[i];
}
int main()
{
prework();
for (scanf ("%lld", &t); t--; )
{
ans = 0LL;
scanf("%lld%lld%lld", &n, &m, &d);
if(n > m)
{
int c = n; n = m; m = c;
}
n /= d, m /= d;
for (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);
}
printf ("%lld
", ans);
}
return 0;
}