嘟嘟嘟
这题跟上一道题有点像，但是我还是没推出来……菜啊
[egin{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
[egin{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;
}