Loading [MathJax]/jax/element/mml/optable/BasicLatin.js

bzoj 3853 : GCD Array

搬运题解
Claris:
1 n d v相当于给a[x]+=v[gcd

\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*}
a[i]=\sum_{j|i}f[j]
则每次修改相当于枚举k|\frac{n}{d},然后给f[dk]+=v\mu(k)
查询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}
可以分块统计,用树状数组维护f[]的前缀和

大概维护一个数列
支持
1.对所有x的倍数的位置加上v
2.查询前缀和
可以用分块的方法把复杂度降为n\sqrt{n}logn

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#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<vector>
#define ll long long
#define pb(x) push_back(x)
#define N 200005
using namespace std;
int n,q;
int su[N],tot,pr[N],miu[N];
const int inf = 200000;
vector<int>v[N];
void shai()
{
    miu[1]=1;
    for(int i=1;i<=inf;i++)v[i].pb(1);
    for(int i=2;i<=inf;i++)
    {
        if(!pr[i])
        {
            pr[i]=i;
            su[++tot]=i;
            miu[i]=-1;
        }
        for(int j=1;j<=tot&&su[j]*i<=inf;j++)
        {
            pr[su[j]*i]=su[j];
            if(su[j]==pr[i])
            {
                break;
            }
            else miu[su[j]*i]=-miu[i];
        }
        for(int j=i;j<=inf;j+=i)v[j].pb(i);
    }
    return ;
}
ll c[N];
void add(int x,int z)
{
    for(int i=x;i<=n;i+=(i&(-i)))
    {
        c[i]+=z;
    }
    return ;
}
ll qur(int x)
{
    ll ans=0;
    for(int i=x;i;i-=(i&(-i)))
    {
        ans+=c[i];
    }
    return ans;
}
int main()
{
    shai();int cnt=0;
    while(~scanf("%d%d",&n,&q))
    {
        if(!n&&!q)break;
        printf("Case #%d:\n",++cnt);
        for(int i=1;i<=n;i++)c[i]=0;
        int t1,t2,t3,t4;
        for(int i=1;i<=q;i++)
        {
            scanf("%d",&t1);
            if(t1==1)
            {
                scanf("%d%d%d",&t2,&t3,&t4);
                if(t2%t3!=0)continue;
                int num=t2/t3;
                for(int j=0;j<v[num].size();j++)
                {
                    int k=v[num][j];
                    add(k*t3,miu[k]*t4);
                }
            }
            else
            {
                scanf("%d",&t2);
                ll ans=0;int r;
                for(int l=1;l<=t2;l=r+1)
                {
                    r=t2/(t2/l);
                    ans+=1LL*(t2/l)*(qur(r)-qur(l-1));
                }
                printf("%lld\n",ans);
            }
             
        }
         
    }
    return 0;
}

  

 

posted @   SD_le  阅读(254)  评论(0编辑  收藏  举报
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