题意分析

就是让你求

\[\sum_{i=1}^{|S|}val[i][gcd(a[i],x)=y]\]

那么接下来就是化简式子

\[\sum_{i=1}^{|S|}val[i][gcd(\frac{a[i]}{y},\frac{x}{y})=1]\]

\[\sum_{i=1}^{|S|}val[i]\sum_{d|\frac{a[i]}{y}d|\frac{x}{y}}μ(d)\]

\[\sum_{i=1}^{|S|}val[i]\sum_{dy|a[i]dy|x}μ(d)\]

我们考虑枚举\(x\)的因子\(k=dy\)

那么贡献就是\(μ(\frac{k}{y})*s[i]\)

其中\(s[i]=\sum_{i|d}val[d]\)

这里直接让\(val[a[i]]=val[i]\)了

然后查询修改我们都可以\(\sqrt n\)了

CODE:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<string>
#include<queue>
#include<map>
#include<stack>
#include<list>
#include<set>
#include<deque>
#include<vector>
#include<ctime>
#define ll long long
#define inf 0x7fffffff
#define N 10000008
#define IL inline
#define M 1008611
#define D double
#define maxn 10000000
#define mod 20020303
#define R register
using namespace std;
template<typename T>IL void read(T &_)
{
    T __=0,___=1;char ____=getchar();
    while(!isdigit(____)) {if(____=='-') ___=0;____=getchar();}
    while(isdigit(____)) {__=(__<<1)+(__<<3)+____-'0';____=getchar();}
    _=___ ? __:-__;
}
/*-------------OI使我快乐-------------*/
int n,m,tot;
struct Node
{
    int xx,yy;
}e[M];
int prime[N],mul[N],val[N];
ll sum[N];
bool mark[N];
IL void work()
{
    mul[1]=1;
    for(R int i=2;i<=maxn;++i)
    {
        if(!mark[i]) {prime[++tot]=i;mul[i]=-1;}
        for(R int j=1;j<=tot&&prime[j]*i<=maxn;++j)
        {
            mark[prime[j]*i]=1;
            if(i%prime[j]==0) 
            {
                mul[prime[j]*i]=0;
                break;
            }
            else mul[prime[j]*i]=-mul[i];
        }
    }
    
}
int main()
{
    freopen("t1.in","r",stdin);
    freopen("t1.out","w",stdout);
    read(n);read(m);work();
    for(R int i=1,x,y;i<=n;++i)
    {
        read(x);read(y);
        val[x]=y;
    }
    for(R int i=1;i<=maxn;++i)
     for(R int j=i;j<=maxn;j+=i)
      sum[i]+=val[j];
    while(m--)
    {
        int knd,x,y;ll ans=0;
        read(knd);read(x);read(y);
        if(knd==1)
        {
            for(R int i=1;i*i<=x;++i)
            {
                if(x%i) continue;
                int cdy=i,wzy=x/i;
                if(cdy%y==0)
                {
                    int now=cdy/y;
                    ans=(ans+mul[now]*sum[cdy]%mod+mod)%mod;
                }
                if(cdy==wzy) continue;
                if(wzy%y==0)
                {
                    int now=wzy/y;
                    ans=(ans+mul[now]*sum[wzy]%mod+mod)%mod;
                }
            }
            printf("%lld\n",ans);
        }
        else
        {
            if(val[x])
            {
                for(R int i=1;i*i<=x;++i)
                {
                    if(x%i) continue;
                    int cdy=i,wzy=x/i;
                    sum[cdy]-=val[x];
                    if(cdy==wzy) continue;
                    sum[wzy]-=val[x];
                }
                val[x]=0;
            }
            else
            {
                val[x]=y;
                for(R int i=1;i*i<=x;++i)
                {
                    if(x%i) continue;
                    int cdy=i,wzy=x/i;
                    sum[cdy]+=val[x];
                    if(cdy==wzy) continue;
                    sum[wzy]+=val[x];
                }
            }
        }
    }
    fclose(stdin);
    fclose(stdout);
    return 0;
}

HEOI 2019 RP++

考试题 T1

原文：https://www.cnblogs.com/LovToLZX/p/10638180.html