Codeforces div2 E. Madoka and The Best University

题意： 对于任意三元组{(a,b,c)|a+b+c<n}，计算lcm(c,gcd(a,b))的累和

思路：首先想到的是枚举a,b，计算c，暴力出结果，但是n^2的时间复杂度让人望而却步，因此我们转换枚举对象，枚举gcd(a,b)，那么如何计算出c？加一层枚举（调和级数），枚举a与b的和，为gcd(a,b) 的倍数，那么问题就转化成了，在a与b的和一定的情况下，计算gcd(a,b)一定值的a,b个数，我们将a设为x,b设为k(a+b的和)-x

有如下转换gcd(x,k-x)==g(a,b的最大公约数)==>gcd(x/g,(k-x)/g)==1==>gcd(x/g,k/g)==1,因此,题目变成了求小于k/g的与其互质的数量，即phi[k/j]

欧拉函数求解

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<string>
#include<bitset>
#include<cmath>
#include<array>
#include<atomic>
#include<sstream>
#include<stack>
#include<iomanip>
//#include<bits/stdc++.h>
 
//#define int ll
#define pb push_back
#define endl '\n'
#define x first
#define y second
#define Endl endl
#define pre(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,b,a) for(int i=b;i>=a;i--)
#define si(x) scanf("%d", &x);
#define sl(x) scanf("%lld", &x);
#define ss(x) scanf("%s", x);
#define YES {puts("YES");return;}
#define NO {puts("NO"); return;}
#define all(x) x.begin(),x.end()
 
using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<int, PII> PIII;
typedef pair<char, int> PCI;
typedef pair<int, char> PIC;
typedef pair<double, double> PDD;
typedef pair<ll, ll> PLL;
const int N = 200010, M = 2 * N, B = N, MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;
 
int dx[4] = { -1,0,1,0 }, dy[4] = { 0,1,0,-1 };
int n, m, k;
int phi[N];
int primes[N], countNum;
bool st[N];
 
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lowbit(ll x) { return x & -x; }
ll qmi(ll a, ll b, ll mod) {
    ll res = 1;
    while (b) {
        if (b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}
 
inline void init() {}
 
void get_phi(int n)
{
    pre(i, 2, n)
    {
        if (!st[i]) { primes[countNum++] = i; phi[i] = i - 1; }
        for (int j = 0; primes[j]<=n/i; j++)
        {
            st[i * primes[j]] = true;
            if (i % primes[j] == 0) {
                phi[i * primes[j]] = phi[i] * primes[j];
                break;
            }
            phi[i * primes[j]] = phi[i] * (primes[j] - 1);
        }
    }
}
 
ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
 
void slove()
{
    cin >> n;
    ll ans = 0;
    get_phi(n);
 
    pre(g, 1, n)
    {
        for (int ab = g; ab < n; ab += g)
        {
            int c = n - ab;
            ans = (ans+(lcm(c, g) * phi[ab / g])%MOD)%MOD;
        }
    }
    cout << ans << Endl;
}
 
signed main()
{
    int _;
    //si(_);
    _ = 1;
    init();
    while (_--)
    {
        slove();
    }
    return 0;
}