【POI2007】ZAP-Queries

题面

题解

$$ \sum_{i=1}^a\sum_{j=1}^b[gcd(i,\;j)=d] \\ =\sum_{i=1}^{\left\lfloor\frac ad\right\rfloor}\sum_{j=1}^{\left\lfloor\frac bd\right\rfloor}[gcd(i,\;j)=1] \\ =\sum_{i=1}^p\mu(i)\lfloor\frac a{id}\rfloor\lfloor\frac b{id}\rfloor $$

筛一下$\mu$即可

代码

#include<bits/stdc++.h>
#define RG register
#define clear(x, y) memset(x, y, sizeof(x));
using namespace std;

inline int read()
{
	int data=0, w=1;
	char ch=getchar();
	while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
	if(ch=='-') w=-1, ch=getchar();
	while(ch>='0'&&ch<='9') data=(data<<3)+(data<<1)+(ch^48), ch=getchar();
	return data*w;
}

const int maxn(100010);
int mu[maxn], prime[maxn], cnt, sum[maxn], n=100000, T, a, b, d;
bool not_prime[maxn];

inline void getMu()
{
	not_prime[1]=true; mu[1]=1;
	for(RG int i=2;i<=n;i++)
	{
		if(!not_prime[i]) prime[++cnt]=i, mu[i]=-1;
		for(RG int j=1;j<=cnt && i*prime[j] <= n;j++)
		{
			not_prime[i*prime[j]]=true;
			if(i%prime[j]) mu[i*prime[j]]=-mu[i];
			else { mu[i*prime[j]]=0; break; }
		}
	}
	for(RG int i=1;i<=n;i++) sum[i]=sum[i-1]+mu[i];
}

inline long long solve(int a, int b, int d)
{
	a/=d; b/=d;
	if(a > b) swap(a, b);
	long long ans=0;
	RG int i=1, j, k, l;
	while(i<=a)
	{
		k=a/i; l=b/i;
		j=min(a/k, b/l);
		ans+=1ll*(sum[j]-sum[i-1])*k*l;
		i=j+1;
	}
	return ans;
}

int main()
{
	getMu();
	T=read();
	while(T--) a=read(), b=read(), d=read(), printf("%lld\n", solve(a, b, d));
	return 0;
}