bzoj 2005: [Noi2010]能量采集【莫比乌斯反演】

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} ight floorleft lfloor frac{m}{dg} ight floor ]

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;
}