考试题 T1

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++