hdu4947 GCD Array

Problem Description
Teacher Mai finds that many problems about arithmetic function can be reduced to the following problem:

Maintain an array a with index from 1 to l. There are two kinds of operations:

Input
There are multiple test cases, terminated by a line “0 0”.

For each test case, the first line contains two integers l,Q(1<=l,Q<=5*10^4), indicating the length of the array and the number of the operations.

In following Q lines, each line indicates an operation, and the format is “1 n d v” or “2 x” (1<=n,d,v<=2*10^5,1<=x<=l).

Output
For each case, output “Case #k:” first, where k is the case number counting from 1.

Then output the answer to each query.

Sample Input
6 4 1 4 1 2 2 5 1 3 3 3 2 3 0 0

Sample Output
Case #1: 6 7

Author
xudyh

Source
2014 Multi-University Training Contest 8

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考虑定义$a_i=\sum\limits_{d|i}^{n} f(d)$ 那么考虑题意中一个v需要加到哪几个$a_i$上 在原式中：$a_x+=v\times[gcd(x,n)=d]=v\times[gcd(x/d,n/d)=1]$ 然后可以套路反演一下$a_x+=v\times \sum\limits_{p|\frac{n}{d},p|\frac{x}{d}}\mu(p)$ 那么我们发现枚举$\frac{n}{d}$的所有约数的d倍就是我这次添加操作会影响的关于$a_x$的那些f数组的位置 所以添加$f(pd)+=v*\mu(p)$ 我们询问一个点$a[x]=\sum\limits_{d|x}f(d)$ $\sum\limits_{i=1}^{x}\sum\limits_{d|i}f(d)$可以枚举d $\sum\limits_{d=1}^{x}\frac{x}{d}f(d)$针对$\frac{x}{d}$数论分块即可 然后树状数组求前缀和 查询的时候树状数组求和即可

#include<bits/stdc++.h>
#define ll long long
using namespace std;
inline char gc(){
    static char now[1<<16],*S,*T;
    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(!isdigit(ch)) {if (ch=='-') f=-1;ch=gc();}
    while(isdigit(ch)) x=x*10+ch-'0',ch=gc();
    return x*f;
}
const int N=2e5+10;
int n,q,prime[N],tot,mu[N];ll s[N];
vector<int> fac[N];bool not_prime[N];
inline void init(){
    mu[1]=1;
    for (int i=2;i<=2e5;++i){
        if(!not_prime[i]) prime[++tot]=i,mu[i]=-1;
        for (int j=1;prime[j]*i<=2e5;++j){
            not_prime[prime[j]*i]=1;
            if (i%prime[j]==0){
                mu[prime[j]*i]=0;break;
            }else mu[prime[j]*i]=-mu[i];
        }
    }
    for (int i=1;i<=2e5;++i)
        for (int j=i;j<=2e5;j+=i) fac[j].push_back(i);
}
inline void add(int x,int v){
    while(x<=2e5) s[x]+=v,x+=x&-x;
}
inline ll query(int x){
    static ll tmp;tmp=0;
    while(x) tmp+=s[x],x-=x&-x;return tmp;
}
int main(){
    freopen("hdu4947.in","r",stdin);
    int cnt=0;init();
    while(1){
        n=read();q=read();if(!n&&!q) break;
        printf("Case #%d:\n",++cnt);
        memset(s,0,sizeof(s));
        for (int owo=1;owo<=q;++owo){
            int op=read();
            if(op==1){
                int x=read(),d=read(),v=read();
                if (x%d) continue;x/=d;
                for (int i=0;i<fac[x].size();++i) add(fac[x][i]*d,mu[fac[x][i]]*v);
            }else{
                ll pre=0,now=0,ans=0;int x=read();
                for (int i=1,nxt;i<=x;i=nxt+1){
                    nxt=x/(x/i);now=query(nxt);
                    ans+=(now-pre)*(x/i);pre=now;
                }
                printf("%lld\n",ans);
            }
        }
    }
    return 0;
}