poj 2480 Longge's problem（欧拉函数）

Description

Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now a problem comes: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i, N) 1<=i <=N. 
"Oh, I know, I know!" Longge shouts! But do you know? Please solve it. 

Input

Input contain several test case. 
A number N per line. 

Output

For each N, output ,∑gcd(i, N) 1<=i <=N, a line

Sample Input

2
6

Sample Output

3
15
解题思路：给出一个数n，求1-n这n个数与n的最大公约数之和。举个栗子：当n=4时，1，2，3，4与4的最大公约数分别为1，2，1，4，累加和为8。正解：1-n中每个数与n的最大公约数肯定是n的一个因子，所以我们只需要枚举n的每一个因子x∈[1,√n]，然后看有多少个满足gcd(k,n)==x，即求满足gcd(k/x,n/x)==1中k的个数（用欧拉函数求解），则公式为：∑x*[gcd(k/x,n/x)==1]。
AC代码(204ms)：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <vector>
#include <set>
using namespace std;
typedef long long LL;
const int maxn = 1e6+5;
LL n, ans;
LL get_Euler(LL x){
    LL res = x;
    for(LL i = 2LL; i * i <= x; ++i) {
        if(x % i == 0) {
            res = res / i * (i - 1);
            while(x % i == 0) x /= i;
        }
    }
    if(x > 1LL) res = res / x * (x - 1);
    return res;
}

int main(){
    while(cin >> n) {
        ans = 0LL;
        for (LL i = 1LL; i * i <= n; ++i) {
            if(n % i == 0) {
                ans += i * get_Euler(n / i);
                if(i * i != n) ans += n / i * get_Euler(i); ///避免重复计数
            }
         }
         cout << ans << endl;
    }
    return 0;
}

AC代码二(32ms)：思路和上面相同，只是将问题求解转换一下gcd(i, n) == (p_i)^j，即求Σ(p_i)^j [gcd(i/((p_i)^j)), n/((p_i)^j)==1]，化简公式得 (k+1)* p^k - k*p^(k-1)，再根据积性函数的性质得n的欧拉函数值为每种素因子对应的欧拉函数值φ((p_i)^a_i)相乘即可。时间复杂度是O(sqrt(n))。具体推导过程：传送门

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
LL n;
LL solve(LL x) {
    LL p_i, k, ans = 1LL;
    for(LL i = 2LL; i * i <= x; ++i) {
        if(x % i == 0) {
            p_i = 1LL, k = 0; 
            while(x % i == 0) {k++, p_i *= i, x /= i;}
            ans *= (k + 1) * p_i - k * p_i / i; ///(k+1)*p^k - k*p^(k-1)
        }
    }
    if(x > 1LL) ans *= 2 * x - 1LL; 
    return ans;
}
int main() {
    while(cin >> n) {
        cout << solve(n) << endl;
    }
    return 0;
}