神犇 YY 虐完数论后给傻× kAc 出了一题
给定 (N, M) 求 (1 leq x leq N)，(1 leq y leq M) 且 (gcd(x, y)) 为质数的 ((x, y)) 有多少对。
前置知识
[egin{aligned}
&sum_{pin prim}sum_{i=1}^{n}sum_{j=1}^{m}[gcd(i,j)=p] \
=& sum_{pin prim}sum_{i=1}^{lfloorfrac{n}{p}
floor}sum_{j=1}^{lfloorfrac{m}{p}
floor}[gcd(i,j)=1]\
=& sum_{pin prim}sum_{i=1}^{lfloorfrac{n}{p}
floor}sum_{j=1}^{lfloorfrac{m}{p}
floor}varepsilon(gcd(i,j)=1)\
=& sum_{pin prim}sum_{i=1}^{lfloorfrac{n}{p}
floor}sum_{j=1}^{lfloorfrac{m}{p}
floor}varepsilon(gcd(i,j))\
=& sum_{pin prim}sum_{i=1}^{lfloorfrac{n}{p}
floor}sum_{j=1}^{lfloorfrac{m}{p}
floor} sum_{dmid gcd(i,j)} mu(d)\
=&sum_{pin prim}sum_{pmid d} mu(d)sum_{i=1}^{lfloorfrac{n}{p}
floor}[dmid i]sum_{j=1}^{lfloorfrac{m}{p}
floor}[dmid j]\
=&sum_{pin prim}sum_{pmid d} mu(d) lfloor frac{n}{kd}
floorlfloor frac{m}{kd}
floor\
end{aligned}
]
设：
[f(p)=sumlimits_{pmid d} mu(d)
]
则原式化为：
[egin{aligned}
=&sum_{pin prim} f(p) imes lfloor frac{n}{kd}
floorlfloor frac{m}{kd}
floor\end{aligned}
]
我们先去预处理 (f) 函数
for(int i=1;i<=cnt;i++)
for(int j=1;j*prim[i]<N;j++)
f[j*prim[i]]+=mu[j];
就是这里的 (f) 数组。
之后我们就可以求解了。
我们可以使用数论分块来写这题。
代码：
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<typename T>inline void read(T &FF){
T RR=1;FF=0;char CH=getchar();
for(;!isdigit(CH);CH=getchar())if(CH=='-')RR=-1;
for(;isdigit(CH);CH=getchar())FF=(FF<<1)+(FF<<3)+(CH^48);
FF*=RR;
}
const int N=1e7+10;
int prim[N],mu[N],sum[N],cnt,k,T,f[N];
bool vis[N];
void init(){
mu[1]=1;
for(register int i=2;i<N;i++){
if(!vis[i]){
mu[i]=-1;
prim[++cnt]=i;
}
for(register int j=1;j<=cnt&&i*prim[j]<N;j++){
vis[i*prim[j]]=1;
if(i%prim[j]==0)break;
mu[i*prim[j]]=-mu[i];
}
}
for(int i=1;i<=cnt;i++)
for(int j=1;j*prim[i]<N;j++)
f[j*prim[i]]+=mu[j];
for(register int i=1;i<N;i++)sum[i]=sum[i-1]+f[i];
}//莫比乌斯反演的板子
ll calc(int a,int b){
ll ans=0;
for(register int l=1,r;l<=min(a,b);l=r+1){
r=min(a/(a/l),b/(b/l));
ans+=(1ll*a/l)*(1ll*b/l)*(sum[r]-sum[l-1]);
}
return ans;
}
int main(){
init();
for(read(T);T--;){
int x,y;
read(x);read(y);
printf("%lld
",calc(x,y));
}
return 0;
}