嘟嘟嘟


这题跟上一道题有点像，但是我还是没推出来……菜啊

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

令\(f(n)\)表示小于等于\(n\)且与\(n\)互质的数的和，则

\[\begin{align*} ans &= n * \sum_{d | n} f(\frac{n}{d}) \\ &= n * \sum_{d | n} f(d) \end{align*}\]

如果\(i\)与\(n\)互质，那么\(n - i\)一定也和\(n\)互质，所以\(\varphi(n)\)个数两两配对等于\(n\)，得到\(f(n) = \frac{\varphi(n) * n}{2}\)。
但是这对\(1\)不成立，因此要特别处理\(f(1) = 1\)，于是

\[ans = n * (\sum_{d | n, d > 1} \frac{\varphi(d) * d}{2} + 1) \]

这个时候可以每一次\(O(\sqrt{n})\)枚举\(n\)的约数，总复杂度\(O(n + T *\sqrt{n} )\)，但是还可以再优化：我们像埃氏筛素数一样，\(O(n \log{n})\)预处理\(f(i)\)。然后\(O(1)\)询问。

#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 = 1e6 + 5;
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');
}

int n;
int prim[maxn], v[maxn], phi[maxn];
ll ans[maxn];
void init()
{
  phi[1] = 1;
  for(int i = 2; i < maxn; ++i)
    {
      if(!v[i]) v[i] = i, phi[i] = i - 1, prim[++prim[0]] = i;
      for(int j = 1; j <= prim[0] && i * prim[j] < maxn; ++j)
	{
	  v[i * prim[j]] = prim[j];
	  if(i % prim[j] == 0)
	    {
	      phi[i * prim[j]] = phi[i] * prim[j];
	      break;
	    }
	  else phi[i * prim[j]] = phi[i] * (prim[j] - 1);
	}
    }
  for(int i = 1; i < maxn; ++i)
    for(int j = 1; i * j < maxn; ++j)
      ans[i * j] += (ll)phi[i] * i;
  for(int i = 1; i < maxn; ++i) ans[i] = (ans[i] + 1) * i >> 1;
}
int main()
{
  init();
  int T = read();
  while(T--) n = read(), write(ans[n]), enter;
  return 0;
}