题面
题解
先化式子
$$ sum_{x=1}^asum_{y=1}^bmathbf f[gcd(x,y)==d] \ = sum_{x=1}^asum_{y=1}^bsum_{dmid x,dmid y}mathbf f[gcd(x,y)==1] \ = sum_{x=1}^{lfloor frac ad floor}sum_{y=1}^{lfloor frac bd floor}mathbf f[gcd(x,y)==1] $$
然后套路就类似于$Bzoj2818 Gcd$了，只不过直接$mathbf f$的逆直接变成了$mu$。（我直接在那题基础上改的）
#include <cstdio>
#include <cstring>
#include <algorithm>
using std::min; using std::max;
using std::swap; using std::sort;
typedef long long ll;
template<typename T>
void read(T &x) {
int flag = 1; x = 0; char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') flag = -flag; ch = getchar(); }
while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); x *= flag;
}
const int N = 5e4 + 10;
int t, n, m, d, mu[N], g[N], prime[N], cnt;
long long sum[N]; bool notprime[N];
void getmu(int k) {
mu[1] = 1;
for(int i = 2; i <= k; ++i) {
if(!notprime[i]) prime[++cnt] = i, mu[i] = -1;
for(int j = 1; j <= cnt && prime[j] * i <= k; ++j) {
notprime[prime[j] * i] = true;
if(!(i % prime[j])) break;
mu[prime[j] * i] = -mu[i];
}
}
for(int i = 1; i <= k; ++i)
sum[i] = sum[i - 1] + 1ll * mu[i];
}
int main () {
read(t); getmu(50000);
while(t--) {
read(n), read(m), read(d); ll ans = 0;
n /= d, m /= d;
if(n > m) swap(n, m);
for(int l = 1, r; l <= n; l = r + 1) {
r = min(n / (n / l), m / (m / l));
ans += (sum[r] - sum[l - 1]) * (m / l) * (n / l);
} printf("%lld
", ans);
}
return 0;
}