HDU 5528 Count a * b

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 

#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;
}