问题描述
题解
求 \(\sum\limits_{i=1}^{n}{\sum\limits_{j=1}^{m}{[gcd(i,j)==p]}}\) ，其中 \(p\)为质数，\(n<m\) 。
考虑不要求 \(gcd(i,j)\) 为质数时的做法：
可以转化为
\[\sum\limits_{g=1}^{n}{g \times \sum\limits_{i=1}^{\lfloor \frac{n}{g} \rfloor}{\sum\limits_{j=1}^{\lfloor \frac{m}{g} \rfloor}{[gcd(i,j)==1]}}}\]
转化为枚举倍数，得
\[\sum\limits_{g=1}^{n}{g \times \sum\limits_{i=1}^{\lfloor \frac{n}{g} \rfloor}{\sum\limits_{j=1}^{\lfloor \frac{m}{g} \rfloor}{\sum\limits_{x|gcd(i,j)}{\mu(x)}}}}\]
\(\mathrm{Code}\)
#include<bits/stdc++.h>
using namespace std;
const int maxn=10000000;
int p[maxn+7],pr[maxn+7],miu[maxn+7];
int h[maxn+7],T,tot;
long long s[maxn+7];
void preprocess(void){
miu[1]=1;
for(int i=2;i<=maxn;i++){
if(!p[i]) p[i]=i,pr[++tot]=i,miu[i]=-1;
for(int j=1;j<=tot;j++){
if((long long)i*pr[j]>maxn||pr[j]>p[i]) break;
p[i*pr[j]]=pr[j];
if(i%pr[j]) miu[i*pr[j]]=-miu[i];
else miu[i*pr[j]]=0;
}
}
for(int i=1;i<=tot;i++){
for(int j=1;(long long)pr[i]*j<=maxn;j++){
h[pr[i]*j]+=miu[j];
}
}
for(int i=1;i<=maxn;i++) s[i]=s[i-1]+(long long)h[i];
}
void Init(void){
scanf("%d",&T);
}
long long calc(int x,int y){
if(x>y) swap(x,y);
long long res(0);
for( int l=1,r;l<=x;l=r+1){
r=min(x/(x/l),y/(y/l));
res+=(long long)(s[r]-s[l-1])*(long long)(x/l)*(y/l);
}
return res;
}
void Work(void){
preprocess();
while(T--){
int x,y;
scanf("%d%d",&x,&y);
printf("%lld\n",calc(x,y));
}
}
signed main(){
Init();
Work();
}