Bzoj1101 Zap（莫比乌斯反演）

题面

Bzoj

题解

先化式子

$$ \sum_{x=1}^a\sum_{y=1}^b\mathbf f[gcd(x,y)==d] \\ = \sum_{x=1}^a\sum_{y=1}^b\sum_{d\mid x,d\mid y}\mathbf f[gcd(x,y)==1] \\ = \sum_{x=1}^{\lfloor \frac ad\rfloor}\sum_{y=1}^{\lfloor \frac bd\rfloor}\mathbf f[gcd(x,y)==1] $$

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\n", ans);
	}
    return 0;
}