Number Theory miscellaneous questions

①: given positive integer N, seeking to satisfy a + b≤N, a + b | a * b, the number of the number (a, b) is, N≤1e14.

We set d = gcd (a, b). Then a = a '* d, b = b' * d, readily available gcd (a ', b') == 1.

The a = a '* d and b = b' * d substituting a + b | a * b, readily available (a '+ b') | a'b'd.

Obviously we have (a '+ b') are not 'divisible, then (a' a'b + b ') | d.

We provided a '+ b' = k, k * d = n, t * k = d, then k ∈ [2, sqrt (N)].

That for a given k, the number of its legitimate it?

(A '+ b') * d = k * k * t ≤ n, then t ≤ N / (k * k), note that k is constant, i.e. t k valid range in the N-number range.

That for a given k, legal (a ', b') number for the number of it? Should k = a '+ b', gcd (a '+ b', a ') = 1, then obviously ans φ (k).

Yr = (n / k * k) * φ (k), k∈ [2.sqrt (N)], O (sqrt (N))

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define e exit(0)
#define re register
#define LL long long
const LL maxn = 1e7+1;
LL n,t,ans;
int cnt,prime[maxn],vis[maxn],Ins[maxn];
int main()
{
    scanf("%lld",&n);
    LL t = sqrt(n);
    Ins[1] = 1;
    for(re int i=2;i<=t;++i){    
        if(!vis[i]){
            prime[++cnt] = i;
            Ins[i] = i-1;
        }
        for(re int j=1;j<=cnt;++j){
            if(i*prime[j] > t)
                break;
            vis[i*prime[j]] = 1;
            if(i%prime[j] == 0){
                Ins[i*prime[j]] = Ins[i]*prime[j];
                break;
            }
            else Ins[i*prime[j]] = Ins[i]*Ins[prime[j]];
        }
    }
    for(re LL k=2;k<=t;++k)
        ans += (LL)(n/(k*k))*(LL)Ins[k];
    printf("%lld",ans);
    return 0;
}