luoguP2303 [SDOI2012]Longge的问题 化式子

求(sum limits_{i = 1}^n gcd(i, n))

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(sum limits_{i = 1}^n gcd(i, n))

(=sum limits_{i = 1}^n sumlimits_{d|i;and;d|n} varphi(d))

(=sum limits_{d |n} varphi(d) * frac{n}{d})

然后就可以以一个很低的复杂度过了

反正复杂度是不会超过(O(sqrt n * d(n)))的

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

#define ll long long

ll n, ans;

inline ll phi(ll m) {
    ll ret = m;
    for(ll t = 2; t * t <= m; t ++)
    if(m % t == 0) {
        while(m % t == 0) m /= t;
        ret /= t; ret *= (t - 1);
    }
    if(m > 1) ret /= m, ret *= (m - 1);
    return ret;
}

int main() {
    cin >> n;
    for(ll i = 1; i * i <= n; i ++) 
    if(n % i == 0) {
        ans += phi(i) * (n / i);
        if(n / i != i) ans += phi(n / i) * i;
    }
    printf("%lld
", ans);
    return 0;
}