P2303 [SDOI2012] Longge 的问题

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：

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

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

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

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

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;
}