[SDOi2012]Longge的问题

嘟嘟嘟


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\[\begin{align*} ans &= \sum_{i = 1} ^ {n} (i, n) \\ &= \sum _ {d | n} ^ {n} \sum _ {(i, n) = d} ^ {n} d \\ &= \sum _ {d | n} ^ {n} \sum _ {(i, \frac{n}{d}) = 1} ^ {\frac{n}{d}} d \\ &= \sum _ {d | n} ^ {n} d * \varphi(\frac{n}{d}) \end{align*}\]

然后\(O(\sqrt{n})\)枚举\(n\)的因数，\(O(\sqrt{n})\)求欧拉函数即可。时间复杂度为\(O(\)因数个数$ * \sqrt{n})$。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
//const int maxn = ;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

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

int main()
{
  ll n = read();
  write(solve(n)), enter;
  return 0;
}