听小乐乐的去做NOI了~\(≧▽≦)/~啦啦啦

一开始sha bi了，没考虑2,3->4,6。。。样例骗人QAQ

容易得到答案就是∑i(1<=i<=n)∑j(1<=j<=m)gcd(i,j)*2-1

但暴力是nm的，GG

我们转换一下思路，设f[i]为gcd为i的点对个数，怎么求f[i]呢？直接算好像并不行，考虑容斥，显然(n/i)*(m/i)就是有公因数为i的点对个数，然后减去f[2*i],f[3*i]...就好了

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<string>
#include<ctime>
#include<queue>
#include<map>
#include<set>
#include<vector>
typedef long long LL;
using namespace std;
const int N=100010;
LL n,m,k,ans,f[N];
LL read() {LL d=0,f=1; char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();} while (c>='0'&&c<='9') d=(d<<3)+(d<<1)+c-48,c=getchar(); return d*f;}
void judge(){freopen(".in","r",stdin); freopen(".out","w",stdout);}
int main()
{
    //judge();
    n=read(); m=read(); int ljj=max(n,m),pjy=min(n,m);
    for (int i=ljj;i>=1;i--)
    {
        f[i]=(n/i)*(m/i);
        for (int j=2*i;j<=pjy;j+=i)
            f[i]-=f[j];
    }
    for (LL i=1;i<=ljj;i++)
        ans+=f[i]*(2*i-1);
    printf("%lld",ans);
    return 0;
}
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