51nod 1188 最大公约数之和 V2(欧拉函数)

1188 最大公约数之和 V2

思路

用欧拉函数可以化简式子如下

$\sum_{i = 1} ^{n} \sum _{j = 1} ^{i - 1} gcd(i, j)$

$= \sum_{i = 1} ^{n} \sum_{j = 1} ^{i} \gcd(i, j) - \frac{(n + 1)(n)}{2}$

$= \sum_{i = 1} ^{n} \sum_{d \mid i} d\sum_{j = 1}^{i}(gcd(i, d) == d) - \frac{(n + 1)(n)}{2}$

$= \sum_{i = 1} ^{n} \sum_{d\mid i}d\phi(\frac{i}{d}) - \frac{(n + 1)(n)}{2}$

我们再通过类似于埃筛来求得 $\sum_{i = 1} ^{n} \sum_{d\mid i}d\phi(\frac{i}{d})$，接下来就可以直接输出答案了。

代码

/*
  Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>

#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1

using namespace std;


typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;

inline ll read() {
    ll x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return x * f;
}

const int N = 5e6 + 10;

int phi[N], n;

bool st[N];

vector<int> prime;

ll ans[N];

void init() {
    st[0] = st[1] = 1;
    phi[1] = 1;
    for(int i = 2; i < N; i++) {
        if(!st[i]) {
            prime.pb(i);
            phi[i] = i - 1;
        }
        for(int j = 0; j < prime.size() && i * prime[j] < N; j++) {
            st[i * prime[j]] = 1;
            if(i % prime[j]) {
                phi[i * prime[j]] = phi[i] * (prime[j] - 1);
            }
            else {
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
        }
    }
    for(int i = 1; i < N; i++) {
        for(int j = i; j < N; j += i) {
            ans[j] += 1ll * i * phi[j / i];
        }
    }
    for(int i = 1; i < N; i++) ans[i] += ans[i - 1];
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    init();
    int T;
    cin >> T;
    while(T--) {
        int n;
        cin >> n;
        cout << ans[n] - 1ll * (n + 1) * n / 2 << endl;
    }
	return 0;
}