最小公倍數計數

51nod 1222

求滿足a≤lcm(i,j)≤b$a\le lcm(i,j)\le b$ 的二元組[i,j]$[i,j]$ 的個數

a,b≤1011$a,b\le {10}^{11}$

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設

S(n)=∑i=1n∑j=1i[ijgcd(i,j)≤n]$$S(n)=\sum _{i=1}^n\sum _{j=1}^i[{\displaystyle \frac{ij}{gcd(i,j)}}\le n]$$

則

ans=S(b)−S(a−1)$$ans=S(b)-S(a-1)$$

S(n)=∑d=1n∑i=1n∑j=1n[ij≤dn][gcd(i,j)=d]$$S(n)=\sum _{d=1}^n\sum _{i=1}^n\sum _{j=1}^n[ij\le dn][gcd(i,j)=d]$$

i$i$ /=d$d$ , j$j$ /=d$d$

S(n)=∑d=1n∑i=1n/d∑j=1n/d[idjd≤dn][gcd(id,jd)=d]$$S(n)=\sum _{d=1}^n\sum _{i=1}^{n/d}\sum _{j=1}^{n/d}[idjd\le dn][gcd(id,jd)=d]$$

S(n)=∑d=1n∑i=1n/d∑j=1n/d[ij≤nd][gcd(i,j)=1]$$S(n)=\sum _{d=1}^n\sum _{i=1}^{n/d}\sum _{j=1}^{n/d}[ij\le {\displaystyle \frac{n}{d}}][gcd(i,j)=1]$$

S(n)=∑d=1n∑i=1n/d∑j=1n/d[ij≤nd]∑k|i且k|jμ(k)$$S(n)=\sum _{d=1}^n\sum _{i=1}^{n/d}\sum _{j=1}^{n/d}[ij\le {\displaystyle \frac{n}{d}}]\sum _{k|i且k|j}\mu (k)$$

枚舉一下k$k$

S(n)=∑d=1n∑k=1n/dμ(k)∑i=1n/dk∑j=1n/dk[ikjk≤nd]$$S(n)=\sum _{d=1}^n\sum _{k=1}^{n/d}\mu (k)\sum _{i=1}^{n/dk}\sum _{j=1}^{n/dk}[ikjk\le {\displaystyle \frac{n}{d}}]$$

S(n)=∑k=1nμ(k)∑d=1n/k∑i=1n/dk∑j=1n/dk[ij≤ndk2]$$S(n)=\sum _{k=1}^n\mu (k)\sum _{d=1}^{n/k}\sum _{i=1}^{n/dk}\sum _{j=1}^{n/dk}[ij\le {\displaystyle \frac{n}{d{k}^2}}]$$

我們發現如果k>n−−√,ij$k>\sqrt{n},ij$ 就必須小於0$0$ 了

S(n)=∑k=1n√μ(k)∑d=1n/k∑i=1n/dk∑j=1n/dk[ijd≤nk2]$$S(n)=\sum _{k=1}^{\sqrt{n}}\mu (k)\sum _{d=1}^{n/k}\sum _{i=1}^{n/dk}\sum _{j=1}^{n/dk}[ijd\le {\displaystyle \frac{n}{k^2}}]$$

此時的i,j$i,j$ 代表idk,jdk$idk,jdk$
我們發現後面的問題就是求出有多少滿足abc≤n$abc\le n$ 的三元組[a,b,c]$[a,b,c]$
我們設 a≤b≤c$a\le b\le c$ ，這樣就可以枚舉出有序的所有三元組。
if(a≠b≠c),ans+=3$if(a\ne b\ne c),ans+=3$
if(a=b≠c),ans+=2$if(a=b\ne c),ans+=2$
if(a=b=c),ans+=1$if(a=b=c),ans+=1$

code：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2000000;

bool is_prime[N]; 
int tot = 0, prime[N], mobius[N]; 
ll l, r, ans = 0;

inline void calc_mobius( int size )
{
    memset( is_prime, true, sizeof( is_prime ) );
    is_prime[1] = false; mobius[1] = 1;

    for( int i = 2; i <= size; i ++ )
    {
        if( is_prime[i] )
            prime[++tot] = i, mobius[i] = -1;

        for( int j = 1; j <= tot; j ++ )
        {
            int p = prime[j];
            if( i * p > size ) break;
            is_prime[i*p] = false;

            if( i % p == 0 )
            {
                mobius[i*p] = 0;
                break;
            }

            mobius[i*p] = -mobius[i];
        }
    }
}

ll calc( ll n )
{
    ll ret = 0;

    for( ll k = 1; k*k <= n; k ++ )
    {
        if( !mobius[k] ) continue;

        ll lim = n / ( k * k ), sum = 0;
        for( ll d = 1; d*d*d <= lim; d ++ )
        {
            for( ll i = d+1; d*i*i <= lim; i ++ )
            {
                ll j = lim / ( i * d ); 
                if( j < i ) continue;

                sum += 2; // i == j
                if( j < i + 1 ) continue;
                sum += 3 * ( j - i ); 
            }

            ll tmp = lim / ( d * d );
            if( tmp < d ) continue;

            sum ++; // d == i == j
            if( tmp - d < 1 ) continue;
            sum += ( tmp - d ) * 2; // d == i
        }

        ret += mobius[k] * sum;
    }

    return ret;
}

int main()
{
    calc_mobius( 1200000 );

    scanf( "%lld%lld", &l, &r );

    ans = calc( r ); 
    ans -= calc( l - 1 );

    printf( "%lld\n", ans );

    return 0;
}