破译密码

破译密码

对于给定的整数 a,b, 和 d，有多少正整数对 x, y，满足 x≤a，y≤b, 并且 gcd(x,y)=d。

莫比乌斯函数 :

对于 \(x=p_1^{a_1}p_2^{a_2}p_3^{a_3}...p_k^{a_k}\).

\[u(x)= \begin{cases} 1,&\text{if } k \text{ is even} \\ -1,&\text{if } k \text{ is odd} \\ 1,&n=1 \\ 0, &\exist \ a_i>1\end{cases} \]

\(gcd(a,b)=d \Longleftrightarrow gcd(a',b') =1\\ \text{其中,} \ a'=[\frac{a}{d}], b'=[\frac{b}{d}]\).

设 \(g(x)\) 为满足 \([\frac{a}{x}]\) 不变的最大值, 使得 \([\frac{a}{x}]=[\frac{a}{g(x)}]\) 且 \([\frac{a}{x}]>[\frac{a}{g(x)+1}]\).

\(g(x)\) 怎么求 ? \(g(x)=[\frac{a}{[\frac{a}{x}]}]\).

\(ans=a'b'-[\frac{a}{2}][\frac{b}{2}]-[\frac{a}{3}][\frac{b}{3}]-...+[\frac{a}{6}][\frac{b}{6}]+...\)

后面那一坨就是容斥原理, 但是如果暴力枚举因子时间是 \(O(n)\) 的, 这时就用到了 \(g(x)\), 可以把枚举降到 \(O(\sqrt n)\).

\(ans = a'b'-\sum_{i=1}^{min(a',b')}[\frac{a'}{i}][\frac{b'}{i}]\ * mobius[i]\).

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define ull unsigned long long
#define pb push_back
#define PII pair<int, int>
#define VIT vector<int>
#define x first
#define y second
#define inf 0x3f3f3f3f
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
int pri[N], cnt;
bool st[N];
int mobius[N], sum[N];

void init() {
    mobius[1] = 1;
    for (int i = 2; i < N; ++i) {
        if (!st[i]) {
            pri[cnt++] = i;
            mobius[i] = -1;
        }
        for (int j = 0; pri[j] * i < N; ++j) {
            st[pri[j] * i] = true;
            if (i % pri[j] == 0) {
                mobius[pri[j] * i] = 0;
                break;
            }
            mobius[pri[j] * i] = mobius[i] * -1;
        }
    }
    for (int i = 1; i < N; ++i) sum[i] = sum[i - 1] + mobius[i];
}

int main() {
    IO;
    init();
    int T;
    cin >> T;
    while (T--) {
        int a, b, d;
        cin >> a >> b >> d;
        a = a / d, b = b / d;
        int n = min(a, b);
        ll ans = 0;
        for (int l = 1, r; l <= n; l = r + 1) {
            r = min(n, min(a / (a / l), b / (b / l)));
            ans += (sum[r] - sum[l - 1]) * (ll)(a / l) * (b / l);
        }
        cout << ans << '\n';
    }
    return 0;
}