【SDOI2012】 Longgue的问题

【题目链接】

点击打开链接

【算法】

gcd(i,n)是n的约数

不妨设gcd(i,n) = d

考虑枚举d和gcd(i,n) = d有多少个

gcd(i,n) = d

gcd(i/d,n/d) = 1

因为i<=n,所以i/d<=n/d

因此满足gcd(i,n) = d一共有phi(n/d)个

【代码】

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll i,n,ans;

template <typename T> inline void read(T &x) {
        ll f = ; x = ;
        char c = getchar();
        for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
        for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
        x *= f;
}
template <typename T> inline void write(T x) {
    if (x < ) { putchar('-'); x = -x; }
    if (x > ) write(x/);
    putchar(x%+'');
}
template <typename T> inline void writeln(T x) {
    write(x);
    puts("");
}

ll phi(ll x) {
        ll i,ret=x;
        for (i = ; i <= sqrt(x); i++) {
                if (x % i == ) {
                        while (x % i == ) x /= i;
                        ret = ret / i * (i - );
                }
        }
        if (x > ) ret = ret / x * (x - );
        return ret;
} 

int main() {

        read(n);
        for (i = ; i <= sqrt(n); i++) {
                    if (n % i == ) {
                            ans += phi(n/i) * i;
                            if (i * i != n) ans += phi(i) * n / i;
                    }
        }

        writeln(ans);

        return ;

}