Count a * b

推式子

$$\begin{gathered} f(n)=\sum _{i=0}^{n-1}\sum _{j=0}^{n-1}n\nmid ij \\ =n^2-\sum _{i=1}^n\sum _{j=1}^{n}n\mid ij \\ =n^2-\sum _{i=1}^n\sum _{j=1}^n\frac{n}{gcd(i,n)}\mid \frac{i}{gcd(i,n)}j \\ =n^2-\sum _{i=1}^n\frac{n}{\frac{n}{gcd(i,n)}} \\ =n^2-\sum _{i=1}^{n}gcd(i,n) \\ =n^2-\sum _{d\mid n}d\sum _{i=1}^n(gcd(i,n)==d) \\ =n^2-\sum _{d\mid n}d\sum _{i=1}^{\frac{n}{d}}(gcd(i,\frac{n}{d})==1) \\ =n^2-\sum _{d\mid n}d\varphi (\frac{n}{d}) \end{gathered}$$

$$\begin{gathered} g(n)=\sum _{m\mid n}f(m) \\ =\sum _{m\mid n}{m}^2-\sum _{m\mid n}\sum _{d\mid m}d\varphi (\frac{m}{d}) \\ =\sum _{m\mid n}{m}^2-\sum _{d\mid n}d\sum _{d\mid m,m\mid n}\varphi (\frac{m}{d}) \\ =\sum _{m\mid n}{m}^2-\sum _{d\mid n}d\sum _{k\mid \frac{n}{d}}\varphi (k) \\ =\sum _{m\mid n}{m}^2-\sum _{d\mid n}n \end{gathered}$$

有约数个数等于${\prod }_{i=1}^{sum}(nu{m}_i+1)$

约数之和等于${\prod }_{i=1}^{sum}(1+p_i^1+p_i^2+\dots \dots +p_i^{nu{m}_i})$

约数平方之和等于${\prod }_{i=1}^{sum}(1+p_i^2+p_i^4+\dots \dots +p_i^{2nu{m}_i})$

可以推导，也就是多个等比数列求前n项和，然后累乘。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}const int N = 1e5 + 10;ull prime[N];
int cnt;bool st[N];void init() {for(int i = 2; i < N; i++) {if(!st[i]) prime[cnt++] = i;for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;if(i % prime[j] == 0) break;}}
}ull quick_pow(ull a, int n) {ull ans = 1;while(n) {if(n & 1) ans = ans * a;a = a * a;n >>= 1;}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T = read();while(T--) {ull n = read(), m = n;ull ans = 1, sum = 1;for(int i = 0; i < cnt && prime[i] * prime[i] <= n; i++) {if(n % prime[i]) continue;int num = 0;while(n % prime[i] == 0) {n /= prime[i];num++;}ull res = 1 + prime[i] * prime[i] * ((quick_pow(prime[i], 2 * num) - 1) / (prime[i] * prime[i] - 1));ans *= res;sum *= 1ull * (num + 1);}if(n != 1) {ans *= 1 + 1ull * n * n; sum *= 2ull;}printf("%llu\n", ans - sum * m);}return 0;
}