注意到k=gcd(x,y)-1，所以答案是
\[2*(\sum_{i=1}^{n}\sum_{i=1}^{m}gcd(i,j))-n*m
\]
去掉前面的乘和后面的减，用莫比乌斯反演来推，设n<m：
\[\sum_{i=1}^{n}\sum_{i=1}^{m}gcd(i,j)
\]
\[\sum_{d=1}^{n}d*\sum_{i=1}^{n}\sum_{i=1}^{m}[gcd(i,j)==d]
\]
\[\sum_{d=1}^{n}d*\sum_{i=1}^{\frac{n}{d}}\sum_{i=1}^{\frac{m}{d}}[gcd(i,j)==1]
\]
\[\sum_{d=1}^{n}d*\sum_{g=1}^{\frac{n}{d}}\mu(g)\left \lfloor \frac{n}{dg} \right \rfloor\left \lfloor \frac{m}{dg} \right \rfloor
\]
分块求即可
#include<iostream>
#include<cstdio>
using namespace std;
const int N=100005;
long long n,m,mb[N],s[N],q[N],tot,ans;
bool v[N];
long long mobi(long long n,long long m)
{
long long r=0ll;
for(long long i=1,la;i<=n;i=la+1)
{
long long ni=n/i,mi=m/i;
la=min(n/ni,m/mi);
r+=(s[la]-s[i-1])*ni*mi;
}
return r;
}
int main()
{
scanf("%lld%lld",&n,&m);
if(n>m)
swap(n,m);
mb[1]=1;
for(long long i=2;i<=n;i++)
{
if(!v[i])
{
mb[i]=-1;
q[++tot]=i;
}
for(long long j=1;j<=tot&&q[j]*i<=n;j++)
{
long long k=q[j]*i;
v[k]=1;
if(i%q[j]==0)
{
mb[k]=0;
break;
}
mb[k]=-mb[i];
}
}
for(long long i=1;i<=n;i++)
s[i]=s[i-1]+mb[i];
for(long long i=1,la;i<=n;i=la+1)
{
long long ni=n/i,mi=m/i;
la=min(m/mi,n/ni);
ans+=(i+la)*(la-i+1)/2ll*mobi(ni,mi);
}
printf("%lld",2*ans-n*m);
return 0;
}