Count a * b
推式子
f ( n ) = ∑ i = 0 n − 1 ∑ j = 0 n − 1 n ∤ i j = n 2 − ∑ i = 1 n ∑ j = 1 n n ∣ i j = n 2 − ∑ i = 1 n ∑ j = 1 n n g c d ( i , n ) ∣ i g c d ( i , n ) j = n 2 − ∑ i = 1 n n n g c d ( i , n ) = n 2 − ∑ i = 1 n g c d ( i , n ) = n 2 − ∑ d ∣ n d ∑ i = 1 n ( g c d ( i , n ) = = d ) = n 2 − ∑ d ∣ n d ∑ i = 1 n d ( g c d ( i , n d ) = = 1 ) = n 2 − ∑ d ∣ n d ϕ ( n d ) 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 \phi(\frac{n}{d})\\ f(n)=i=0∑n−1j=0∑n−1n∤ij=n2−i=1∑nj=1∑nn∣ij=n2−i=1∑nj=1∑ngcd(i,n)n∣gcd(i,n)ij=n2−i=1∑ngcd(i,n)nn=n2−i=1∑ngcd(i,n)=n2−d∣n∑di=1∑n(gcd(i,n)==d)=n2−d∣n∑di=1∑dn(gcd(i,dn)==1)=n2−d∣n∑dϕ(dn)
g ( n ) = ∑ m ∣ n f ( m ) = ∑ m ∣ n m 2 − ∑ m ∣ n ∑ d ∣ m d ϕ ( m d ) = ∑ m ∣ n m 2 − ∑ d ∣ n d ∑ d ∣ m , m ∣ n ϕ ( m d ) = ∑ m ∣ n m 2 − ∑ d ∣ n d ∑ k ∣ n d ϕ ( k ) = ∑ m ∣ n m 2 − ∑ d ∣ n n g(n) = \sum_{m \mid n} f(m)\\ = \sum_{m \mid n} m ^ 2 - \sum_{m \mid n} \sum_{d \mid m} d \phi(\frac{m}{d})\\ = \sum_{m \mid n} m ^ 2 - \sum_{d \mid n} d \sum_{d \mid m, m \mid n} \phi(\frac{m}{d})\\ = \sum_{m \mid n} m ^ 2 - \sum_{d \mid n} d \sum_{k \mid \frac{n}{d}} \phi(k)\\ = \sum_{m \mid n} m ^ 2 - \sum_{d \mid n}n\\ g(n)=m∣n∑f(m)=m∣n∑m2−m∣n∑d∣m∑dϕ(dm)=m∣n∑m2−d∣n∑dd∣m,m∣n∑ϕ(dm)=m∣n∑m2−d∣n∑dk∣dn∑ϕ(k)=m∣n∑m2−d∣n∑n
有约数个数等于 ∏ i = 1 s u m ( n u m i + 1 ) \prod _{i = 1} ^{sum}(num_i + 1) ∏i=1sum(numi+1)
约数之和等于 ∏ i = 1 s u m ( 1 + p i 1 + p i 2 + … … + p i n u m i ) \prod _{i = 1} ^{sum}(1 + p_i ^ {1} + p_i ^{2} + …… + p_i ^{num_i}) ∏i=1sum(1+pi1+pi2+……+pinumi)
约数平方之和等于 ∏ i = 1 s u m ( 1 + p i 2 + p i 4 + … … + p i 2 n u m i ) \prod _{i = 1} ^{sum}(1 + p_i ^ {2} + p_i ^{4} + …… + p_i ^{2num_i}) ∏i=1sum(1+pi2+pi4+……+pi2numi)
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 | 1
using 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;
}
