P2303 [SDOI2012] Longge 的问题

思路

我们显然可以枚举每一对数的$gcd$进行求解，进而我们有如下推导：

$$=>\sum _{i=1}^{n}gcd(i,n)$$

$$=>\sum _{d\mid n}d\sum _{i=1}^n(gcd(i,d)==d)$$

$$=>\sum _{d\mid n}d\sum _{i=1}^{\frac{n}{d}}(gcd(i,d)==1)$$

$$=>\sum _{d\mid n}d\varphi (\frac{n}{d})$$

这就简单了，只要枚举$n$的所有因子就行。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
// #include <iostream>
// #include <algorithm>
// #include <stdlib.h>
// #include <cmath>
// #include <vector>
// #include <cstdio>
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define endl '\n'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 f = 1, x = 0;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 f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}ll eular(ll n) {ll ans = n;for(ll i = 2; i * i <= n; i++) {if(n % i == 0) {while(n % i == 0) n /= i;ans = ans / i * (i - 1);}}if(n != 1) ans = ans / n * (n - 1);return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);ll n = read(), Ans = 0;for(ll i = 1; i * i <= n; i++) {if(n % i == 0) {Ans += i * eular(n / i);if(i * i != n) {Ans += n / i * eular(i);}}}printf("%lld\n", Ans);return 0;
}