簡單的數學題

求

∑i=1n∑j=1nij gcd(i,j)$$\sum _{i=1}^n\sum _{j=1}^{n}ijgcd(i,j)$$

n≤1010$n\le {10}^{10}$

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ans=∑d=1nd∑i=1n∑j=1nij [gcd(i,j)=d]$$ans=\sum _{d=1}^{n}d\sum _{i=1}^n\sum _{j=1}^{n}ij[gcd(i,j)=d]$$

=∑d=1nd3∑i=1n/d∑j=1n/dij [gcd(i,j)=1]$$=\sum _{d=1}^n{d}^3\sum _{i=1}^{n/d}\sum _{j=1}^{n/d}ij[gcd(i,j)=1]$$

=∑d=1nd3∑i=1n/d∑j=1n/dij∑e|gcd(i,j)μ(e)$$=\sum _{d=1}^n{d}^3\sum _{i=1}^{n/d}\sum _{j=1}^{n/d}ij\sum _{e|gcd(i,j)}\mu (e)$$

=∑d=1nd3∑i=1n/d∑j=1n/dij∑e|i且e|jμ(e)$$=\sum _{d=1}^n{d}^3\sum _{i=1}^{n/d}\sum _{j=1}^{n/d}ij\sum _{e|i且e|j}\mu (e)$$

=∑d=1nd3∑e=1n/dμ(e)e2∑i=1n/de∑j=1n/deij$$=\sum _{d=1}^n{d}^3\sum _{e=1}^{n/d}\mu (e)e^2\sum _{i=1}^{n/de}\sum _{j=1}^{n/de}ij$$

設f(x)=∑xi=1i$f(x)=\sum _{i=1}^{x}i$

ans=∑d=1nd3∑e=1n/dμ(e)e2f2(nde)$$ans=\sum _{d=1}^n{d}^3\sum _{e=1}^{n/d}\mu (e)e^2{f}^2({\displaystyle \frac{n}{de}})$$

我們知道de≤n$de\le n$
設T=de$T=de$

ans=∑T=1n∑d|Td3μ(Td)(Td)2f2(nT)$$ans=\sum _{T=1}^n\sum _{d|T}{d}^3\mu ({\displaystyle \frac{T}{d}})({\displaystyle \frac{T}{d}}{)}^2{f}^2({\displaystyle \frac{n}{T}})$$

=∑T=1nf2(nT)∑d|Td3μ(Td)(Td)2$$=\sum _{T=1}^n{f}^2({\displaystyle \frac{n}{T}})\sum _{d|T}{d}^3\mu ({\displaystyle \frac{T}{d}})({\displaystyle \frac{T}{d}}{)}^2$$

=∑T=1nT2f2(nT)∑d|Td μ(Td)$$=\sum _{T=1}^n{T}^2{f}^2({\displaystyle \frac{n}{T}})\sum _{d|T}d\mu ({\displaystyle \frac{T}{d}})$$

我們知道

∑d|nμ(d)d=ϕ(n)n$$\sum _{d|n}{\displaystyle \frac{\mu (d)}{d}}={\displaystyle \frac{\varphi (n)}{n}}$$

∑d|nμ(d)nd=ϕ(n)$$\sum _{d|n}\mu (d){\displaystyle \frac{n}{d}}=\varphi (n)$$

∴ans=∑T=1nT2f2(nT)∑d|TTd μ(d)$$\therefore ans=\sum _{T=1}^n{T}^2{f}^2({\displaystyle \frac{n}{T}})\sum _{d|T}{\displaystyle \frac{T}{d}}\mu (d)$$

=∑T=1nT2f2(nT)ϕ(T)$$=\sum _{T=1}^n{T}^2{f}^2({\displaystyle \frac{n}{T}})\varphi (T)$$

我們知道

∑i=1ni2=n(n+1)(2n+1)6$$\sum _{i=1}^n{i}^2={\displaystyle \frac{n(n+1)(2n+1)}{6}}$$

f(x)=∑i=1x=x(x+1)2$$f(x)=\sum _{i=1}^x={\displaystyle \frac{x(x+1)}{2}}$$

所以我們只需杜教篩出

S(n)=∑i=1ni2ϕ(i)$$S(n)=\sum _{i=1}^n{i}^2\varphi (i)$$

即可在O(n23)$O(n^{\frac{2}{3}})$ 內解決此題

code：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 5000000;
const int LIM = 4000000;

ll MOD, INV_6, INV_2;

map <ll, ll> mp; 

bool is_prime[N]; 
int tot = 0, prime[N]; 
ll n, s[N], euler[N];

inline ll ksm( ll a, ll b )
{
    ll ret = 1; a %= MOD;
    for( ; b; b >>= 1, a = a * a %MOD )
        if( b & 1 ) ret = ret * a %MOD;
    return ret;
}

inline ll calc_sum( ll x ) {
    x %= MOD; return x * ( x + 1 ) %MOD * INV_2 %MOD;
}

inline ll calc_poi( ll x ) {
    x %= MOD; return x * ( x + 1 ) %MOD * ( 2*x + 1 ) %MOD * INV_6 %MOD;
}

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

    for( int i = 2; i <= size; i ++ )
    {
        if( is_prime[i] )
            prime[++tot] = i, euler[i] = 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 )
            {
                euler[i*p] = euler[i] * p;
                break;
            }

            euler[i*p] = euler[i] * euler[p];
        }
    }

    for( int i = 1; i <= size; i ++ )
        s[i] = ( s[i-1] + (ll)euler[i] * i %MOD * i %MOD ) %MOD;
}

inline ll calc_euler( ll x )
{
    if( x <= LIM ) return s[x];
    if( mp[x] ) return mp[x];

    ll next = 0, ret = calc_sum( x ) * calc_sum( x ) %MOD;
    for( ll i = 2; i <= x; i = next + 1 )
    {
        next = x / ( x / i );
        ll tmp = ( calc_poi( next ) - calc_poi( i - 1 ) ) %MOD;
        if( tmp < 0 ) tmp += MOD;

        ret -= calc_euler( x / i ) * tmp %MOD;
        while( ret < 0 ) ret += MOD;
    }

    mp[x] = ret; 
    return ret;
}

int main()
{ 
    scanf( "%lld%lld", &MOD, &n );

    INV_2 = ksm( 2, MOD-2 );
    INV_6 = ksm( 6, MOD-2 );
    init( LIM );

    ll next = 0, ans = 0;
    for( ll i = 1; i <= n; i = next + 1 )
    {
        next = n / ( n / i );
        ll tmp = calc_euler( next ) - calc_euler( i - 1 );
        while( tmp < 0 ) tmp += MOD;
        ll _tmp = calc_sum( n / i ) * calc_sum( n / i ) %MOD; 

        ans += _tmp * tmp %MOD;
        while( ans >= MOD ) ans -= MOD;
    }

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

    return 0;
}