【题解】P2303 [SDOI2012] Longge 的问题

原题地址

一句话题意

求\(\begin{aligned}\sum\limits_{i=1}^n \gcd(i, n)\end{aligned}\)
\(=\begin{aligned}\sum\limits_{d|n}d*\sum\limits_{i=1}^n[gcd(i,n)=d]\end{aligned}\)
\(=\begin{aligned}\sum\limits_{d|n}d*\sum\limits_{i=1}^\frac nd [gcd(i,\frac nd)=1]\end{aligned}\)
\(=\begin{aligned}\sum\limits_{d|n}d*\varphi(\frac nd)\end{aligned}\)
对于数据范围\(1\leq n\leq 2^{32}\),考虑在\(\sqrt n\)时间内筛出\(d\)，用\(\varphi\)的定义求出\(\varphi\)，就可以过了，时间复杂度为O(K\(\sqrt n\))，\(k\)为所筛\(d\)的个数

\(Code\)

#include<bits/stdc++.h>
    
#define LL long long
    
#define int long long

using namespace std;
    
/*Grievous Lady*/
    
template <typename T> void read(T & t){
    t = 0;int f = 1;char ch = getchar();
    while(ch < '0' || ch > '9'){if(ch == '-')f =- 1;ch = getchar();}
    do{t = t * 10 + ch - '0';ch = getchar();}while(ch >= '0' && ch <= '9');t *= f;
}

int n , ans , i;

const int kato = 1e7 + 1;

int prime[kato] , cnt;

bool ispri[kato];

inline void pri(int atri){
    for(int i = 2;i <= atri;i ++){
        if(!ispri[i]){
            prime[++ cnt] = i;
        }
        for(int j = 1;j <= cnt && i * prime[j] <= atri;j ++){
            ispri[i * prime[j]] = 1;
            if(i % prime[j] == 0){
                break;
            }
        }
    }
}

inline int get_phi(int x){
    int ans = x;
    for(int i = 1;i <= cnt && prime[i] * prime[i] <= x;i ++){
        if(x % prime[i]) continue;
        ans = ans / prime[i] * (prime[i] - 1);
        while(x % prime[i] == 0){
            x /= prime[i];
        }
    }
    if(x > 1){
        ans = ans / x * (x - 1);
    }
    return ans;
}

inline int Ame_(){
    read(n);
    // cout << phi[1] << ' ' << phi[2];
    pri(sqrt(n));
    for(i = 1; i * i < n; i++){
        // cout << i << '\n';
        if(n % i == 0){
            ans += i * get_phi(n / i) + (n / i) * get_phi(i);
        }
    }
    if(i * i == n){
        ans += i * get_phi(n / i);
    }
    printf("%lld" , ans);
    return 0;
}
    
int Ame__ = Ame_();
    
signed main(){;}