[HAOI2011] Problem B

Statement

$Ans={\sum }_{i=a}^b{\sum }_{j=c}^d[\gcd (i,j)=k]$

Solution

设: $s(a,b,c,d,k)={\sum }_{i=a}^b{\sum }_{j=b}^c[\gcd (i,j)=k]$

考虑容斥可得$Ans=s(a,b,c,d,k)=s(1,b,1,d)-s(1,b,1,c-1,k)-s(1,a-1,1,d,k)+s(1,a-1,1,c-1,k)$.

我们考虑一个子问题$s(1,i,1,j,k)={\sum }_{i=1}^N{\sum }_{j=1}^M[\gcd (i,j)=k]$.

式子化简后和YY的GCD式子是一致的.

设: $f(d)={\sum }_{i=1}^N{\sum }_{j=1}^M[\gcd (i,j)=d]$.

设: $g(d)={\sum }_{d|d^{\prime }}f(d^{\prime })={\sum }_{i=1}^N{\sum }_{j=1}^M[d|i][d|j]={\sum }_{i=1}^N[d|i]{\sum }_{i=1}^M[d|j]={\sum }_{i=1}^N[d|i]\lfloor \frac{M}{d}\rfloor =\lfloor \frac{N}{d}\rfloor \cdot \lfloor \frac{M}{d}\rfloor$.

则: $f(d)={\sum }_{d|d^{\prime }}\mu (\frac{d^{\prime }}{d})g(d^{\prime })$.

由 $Ans={\sum }_{p\in prime}{\sum }_{i=1}^N{\sum }_{j=1}^N[\gcd (i,j)=p]$.

可得: $Ans={\sum }_{p\in prime}f(p)$.
A n s = ∑ p ∈ p r i m e ∑ i = 1 N ∑ j = 1 N [ gcd ⁡ ( i , j ) = p ] = ∑ p ∈ p r i m e f ( p ) = ∑ p ∈ p r i m e ∑ p ∣ d min ⁡ { N , M } μ ( d p ) ⌊ N d ⌋ ⌊ M d ⌋ = ∑ d = 1 min ⁡ { N , M } ∑ p ∣ d 且 p ∈ p r i m e μ ( d p ) ⌊ N d ⌋ ⌊ M d ⌋ = ∑ d = 1 min ⁡ { N , M } ⌊ N d ⌋ ⌊ M d ⌋ ∑ p ∣ d 且 p ∈ p r i e m μ ( d p ) \begin{aligned}Ans&=\sum_{p\in prime}\sum_{i=1}^N\sum_{j=1}^N[\gcd(i,j)=p]\\ &=\sum_{p\in prime}f(p)\\ &=\sum_{p\in prime}\sum_{p|d}^{\min\{N,M\}}\mu(\frac{d}{p})\lfloor\frac{N}{d}\rfloor\lfloor\frac{M}{d}\rfloor\\ &=\sum_{d=1}^{\min\{N,M\}}\sum_{p|d且p\in prime}\mu(\frac{d}{p})\lfloor\frac{N}{d}\rfloor\lfloor\frac{M}{d}\rfloor\\ &=\sum_{d=1}^{\min\{N,M\}}\lfloor\frac{N}{d}\rfloor\lfloor\frac{M}{d}\rfloor\sum_{p|d且p\in priem}\mu(\frac{d}{p}) \end{aligned} Ans​=p∈prime∑​i=1∑N​j=1∑N​[gcd(i,j)=p]=p∈prime∑​f(p)=p∈prime∑​p∣d∑min{N,M}​μ(pd​)⌊dN​⌋⌊dM​⌋=d=1∑min{N,M}​p∣d且p∈prime∑​μ(pd​)⌊dN​⌋⌊dM​⌋=d=1∑min{N,M}​⌊dN​⌋⌊dM​⌋p∣d且p∈priem∑​μ(pd​)​
我们发现$\sum _{p|d且p\in prime}\mu (\frac{d}{p})$是可以预处理的, 该部分我们枚举$p\in prime$即可, 时间复杂度是$O(n\log \log n)$的和$\text{Eratosthenes}$筛法(埃拉托特尼筛法, 简称埃氏筛法)该部分时间复杂度和写法相似.

然后一次询问我们只需要用数论分块就可以将复杂度做到$O(N\log \log N+T\sqrt{N})$.

Code

# include "iostream"
# include "cstdio"

using namespace std;

const int maxm=5e4+10;

int Mu[maxm],Prime[maxm],Cnt;
bool Visit[maxm];

void Init(int N){
	Mu[1]=1;
	register int i,j;
	for(i=2;i<=N;i++){
		if(!Visit[i]){
			Prime[++Cnt]=i;
			Mu[i]=-1;
		}
		for(j=1;j<=Cnt && Prime[j]*i<=N;j++){
			Visit[i*Prime[j]]=1;
			if(i%Prime[j]==0){Mu[Prime[j]*i]=0;break;}
			Mu[i*Prime[j]]=-Mu[i];
		}
	}
	for(i=2;i<=N;i++) Mu[i]+=Mu[i-1];
	return;
}

int Solve(int N,int M,int K){
	N/=K;M/=K;
	register int i,j,Ans=0;
	if(N>M) swap(N,M);
	for(i=1;i<=N;i=j+1){
		j=min(N/(N/i),M/(M/i));
		Ans+=(N/i)*(M/i)*(Mu[j]-Mu[i-1]);
	}
	return Ans;
}

int main(){
	register int T;
	int A,B,C,D,K;
	scanf("%d",&T);
	Init(5e4);
	while(T--){
		scanf("%d%d%d%d%d",&A,&B,&C,&D,&K);
		printf("%d\n",Solve(B,D,K)-Solve(B,C-1,K)-Solve(A-1,D,K)+Solve(A-1,C-1,K));
	}
	return 0;
}

Link

[1] Luogu P2522 [HAOI2021]Problem B