bzoj #3853. GCD Array 题解

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Main idea: give a sequence $a$ , each position is initially$0$ , there are two operations: 1. Give$n,d,v$ , let all satisfy$\gcd (n,x)=d$ is$a\left[x\right]$ plus$v$ ; 2, request$A$ certain prefix sum.

answer

Maintain an auxiliary array $f$，满足 $a_i={\sum }_{d|i}{f}_d$。

Consider each $a_x$, Its increment in each operation is $v[(x,n)=d]$，推一推就是：
$$\begin{array}{rl}& v\left[\right(x,n)=d]\\ & =v\left[\right(\frac{x}{d},\frac{n}{d})=1]\\ & =\sum _{k|(\frac{x}{d},\frac{n}{d})}v\mu \left(k\right)\\ & =\sum _{dk|x,k|\frac{n}{d}}v\mu \left(k\right)\end{array}$$

It can be found that a nd \dfrac nd is enumerated$\frac{n}{d}$Factor $k$ , then$dk$ must meet the condition ofxxfactor of$x$ , so you can make$f_{dk}$Add $v\mu \left(k\right)$。

再考虑求解，${\sum }_{i=1}^x{a}_i={\sum }_{i=1}^x{\sum }_{j|i}{f}_j={\sum }_{j=1}^n{f}_j\lfloor \frac{x}{j}\rfloor$ , so just set a divisible block, hereffis requiredThe prefix sum of$f$ , so it needs to be maintained by a tree array, and the complexity is$O(n\sqrt{n}\log n)$。

code show as below:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 200010
#define ll long long

int n,m;ll tr[maxn];
void add(int x,int y){
    
    for(;x<=maxn-10;x+=(x&-x))tr[x]+=y;}
ll sum(int x){
    
    ll re=0;for(;x>=1;x-=(x&-x))re+=tr[x];return re;}
int mu[maxn],prime[maxn],t=0;
bool v[maxn];
struct edge{
    
    int y,next;}e[maxn<<4];
int first[maxn],len=0;
void buildroad(int x,int y){
    
    e[++len]=(edge){
    
    y,first[x]};first[x]=len;}
void work(){
    
    
	mu[1]=1;for(int i=2;i<=maxn-10;i++){
    
    
		if(!v[i])prime[++t]=i,mu[i]=-1;
		for(int j=1;j<=t&&i*prime[j]<=maxn-10;j++){
    
    
			v[i*prime[j]]=true;
			if(i%prime[j]==0)break;
			mu[i*prime[j]]=-mu[i];
		}
	}
	for(int i=1;i<=maxn-10;i++)
	for(int j=i;j<=maxn-10;j+=i)buildroad(j,i);
}

int main()
{
    
    
	work();int Case=0;
	while(scanf("%d %d",&n,&m),n!=0&&m!=0)
	{
    
    
		printf("Case #%d:\n",++Case);
		memset(tr,0,sizeof(tr));
		for(int i=1,id,N,D,V;i<=m;i++)
		{
    
    
			scanf("%d",&id);
			if(id==1){
    
    
				scanf("%d %d %d",&N,&D,&V);
				if(N%D==0)for(int j=first[N/D];j;j=e[j].next)add(e[j].y*D,V*mu[e[j].y]);
			}else{
    
    
				scanf("%d",&N);ll ans=0;
				for(int l=1,r;l<=N;l=r+1){
    
    
					r=N/(N/l);
					ans+=N/l*(sum(r)-sum(l-1));
				}
				printf("%lld\n",ans);
			}
		}
	}
}