BZOJ3853 : GCD Array

1 n d v相當於給$a[x]+=v[\gcd(x,n)=d]$

\[\begin{eqnarray*}&&v[\gcd(x,n)=d]\\&=&v[\gcd(\frac{x}{d},\frac{n}{d})=1]\\&=&v\sum_{k|\gcd(\frac{x}{d},\frac{n}{d})}\mu(k)\\&=&\sum_{k|\frac{n}{d},dk|x}v\mu(k)\end{eqnarray*}\]

設$a[i]=\sum_{j|i}f[j]$

則每次修改相當於列舉$k|\frac{n}{d}$，然後給$f[dk]+=v\mu(k)$

查詢$x=\sum_{i=1}^x a[i]=\sum_{i=1}^x\sum_{d|i}f[d]=\sum_{d=1}^x f[d]\frac{x}{d}$

可以分塊統計，用樹狀陣列維護f[]的字首和

 

#include<cstdio>
typedef long long ll;
const int N=200001,M=2480000;
int T,n,m,i,j,x,y,z,p[N],tot,mu[N],g[N],nxt[M],v[M],ed,op;ll b[N],t;bool vis[N];
inline void addedge(int x,int y){v[++ed]=y;nxt[ed]=g[x];g[x]=ed;}
inline void add(int x,int y){for(;x<=n;x+=x&-x)b[x]+=y;}
inline ll sum(int x){ll t=0;for(;x;x-=x&-x)t+=b[x];return t;}
int main(){
  for(mu[1]=1,i=2;i<N;i++){
    if(!vis[i])p[++tot]=i,mu[i]=-1;
    for(j=1;j<=tot;j++){
      if(i*p[j]>=N)break;
      vis[i*p[j]]=1;
      if(i%p[j])mu[i*p[j]]=-mu[i];else{mu[i*p[j]]=0;break;};
    }
  }
  for(i=1;i<N;i++)for(j=i;j<N;j+=i)addedge(j,i);
  while(1){
    scanf("%d%d",&n,&m);
    if(!n)return 0;
    for(i=1;i<=n;i++)b[i]=0;
    printf("Case #%d:\n",++T);
    while(m--){
      scanf("%d%d",&op,&x);
      if(op==1){
        scanf("%d%d",&y,&z);
        if(x%y==0)for(i=g[x/y];i;i=nxt[i])add(v[i]*y,z*mu[v[i]]);
      }else{
        for(t=0,i=1;i<=x;i=j+1)j=x/(x/i),t+=(sum(j)-sum(i-1))*(x/i);
        printf("%lld\n",t);
      }
    }
  }
}