P2568 GCD

合集 - 数论(18)

1.D. Birthday Gift2024-03-272.P2613 【模板】有理数取余2024-04-043.P2657 [SCOI2009] windy 数2024-04-064.P8754 [蓝桥杯 2021 省 AB2] 完全平方数2024-05-085.C. Nikita and LCM2024-05-286.G. D-Function2024-06-137.A. Bitwise Operation Wizard2024-06-178.M. 渚千夏的串2024-06-189.E. Look Back2024-06-2910.数论里的欧拉定理，简单证明，非常简单2024-07-0811.D. Same GCDs2024-07-1012.B. Missing Subsequence Sum2024-07-10

13.P2568 GCD2024-07-11

14.D. Small GCD2024-07-1115.C. Jellyfish and Green Apple2024-07-1816.C. Even Subarrays2024-07-3117.F - x = a^b2024-08-0818.D - Square Pair2024-08-11

收起

原题链接

题解

令 $g=gcd(i,j)$ 则 $i=t_{1}g,j=t_{2}g$
所以原题等价于求 $\sum _{i\in prime}\sum gcd(x,y)==1,x\in [1,n/i],y\in [1,n/i]$

也就是对于每个素数 $i$，求 $[1,n/i]$ 内有几对数互质，我们可以求欧拉函数前缀和得出

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;

vector<ll> prime;
bool mark[10000005]={0};
ll phi[10000005]={0};
ll sumgcd1[10000005]={0};

void solve()
{
    ll n;
    cin>>n;
    ll ans=0;
    for(auto it:prime)
    {
        if(it>n) break;
        ans+=sumgcd1[n/it];
    }
    cout<<ans<<'\n';

}
int main()
{

    sumgcd1[1]=1;
    for(ll i=2;i<=1e7;i++)
    {
        if(!mark[i])
        {
            prime.push_back(i);
            phi[i]=i-1;
        }

        for(auto it:prime)
        {
            if(it*i>1e7) break;
            mark[it*i]=1;

            if(i%it==0)
            {
                phi[it*i]=it*phi[i];
                break;
            }
            else
            {
                phi[it*i]=phi[it]*phi[i];
            }
        }

        sumgcd1[i]=sumgcd1[i-1]+2LL*phi[i];
    }

    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--) solve();
    return 0;
}