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;
}