鏈接

對於gcd（i，j）的位置來說，對答案的貢獻是2*（gcd（i，j）-1）+1，所以答案ans

ans=Σ(1<=i<=n)(1<=j<=m)2*(gcd(i,j)-1)+1

ans=2*Σ(1<=i<=n)(1<=j<=m)gcd(i,j)-n*m

前者可以通過莫比烏斯反演來計算，便很容易得出答案

//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define mod 1000000007
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pil pair<int,ll>
#define
 
 pii pair<int,int>
#define ull unsigned long long
#define base 1000000000000000000
#define fio ios::sync_with_stdio(false);cin.tie(0)

using namespace std;

const double g=10.0,eps=1e-12;
const int N=100000+10,maxn=400000+10,inf=0x3f3f3f3f;

int mu[N],prime[N],sum[N];
bool mark[N];
int num[N];
void init()
{
    mu[
 
1]=1;
    int cnt=0;
    for(int i=2;i<N;i++)
    {
        if(!mark[i])prime[++cnt]=i,mu[i]=-1,num[i]=1;
        for(int j=1;j<=cnt;j++)
        {
            int t=i*prime[j];
            if(t>N)break;
            mark[t]=1;
            num[t]=num[i]+1;
            if(i%prime[j]==0){mu[t]=0;break;}
            else mu[t]=-mu[i];
        }
    }
    for(int i=1;i<N;i++)sum[i]=sum[i-1]+mu[i];
}
int main()
{
    init();
    int n,m;
    scanf("%d%d",&n,&m);
    ll ans=0;
    for(int j=1;j<=max(n,m);j++)
    {
        ll te=0;
        int ten=n/j,tem=m/j;
        for(int i=1,last;i<=min(ten,tem);i=last+1)
        {
            last=min(ten/(ten/i),tem/(tem/i));
            te+=(ll)(sum[last]-sum[i-1])*(ten/i)*(tem/i);
        }
//        printf("%lld\n",te);
        ans+=te*j;
    }
    printf("%lld\n",2*ans-(ll)n*m);
    return 0;
}
/********************

********************/

HYSBZ - 2005 莫比烏斯反演