D2. Reverse Card (Hard Version)

D2. Reverse Card (Hard Version)

The two versions are different problems. You may want to read both versions. You can make hacks only if both versions are solved.

You are given two positive integers $n$, $m$.

Calculate the number of ordered pairs $(a, b)$ satisfying the following conditions:

• $1\le a\le n$, $1\le b\le m$;
• $b \cdot \gcd(a,b)$ is a multiple of $a+b$.

Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1\le t\le 10^4$). The description of the test cases follows.

The first line of each test case contains two integers $n$, $m$ ($1\le n,m\le 2 \cdot 10^6$).

It is guaranteed that neither the sum of $n$ nor the sum of $m$ over all test cases exceeds $2 \cdot 10^6$.

Output

For each test case, print a single integer: the number of valid pairs.

Example

input

6
1 1
2 3
3 5
10 8
100 1233
1000000 1145141

output

0
1
1
6
423
5933961

Note

In the first test case, no pair satisfies the conditions.

In the fourth test case, $(2,2),(3,6),(4,4),(6,3),(6,6),(8,8)$ satisfy the conditions.

 

解题思路

  令 $d = \gcd(a, b)$，则有 $a = p \cdot d$，$b = q \cdot d$，且 $\gcd(p,q) = 1$。对于 $(a+b) \mid b \cdot d$，等价于 $(p+q) \mid q \cdot d$。又因为 $\gcd(q, p + q) = \gcd(q, p) = 1$，因此有 $(p+q) \mid d$。

  因此 $(a+b) \mid b \cdot \gcd(a, b)$ 等价于 $(p+q) \mid q \cdot d$。

  此时考虑能否枚举数对 $(p,q)$。事实上 $p < d = \frac{n}{p} \Longrightarrow p^2 < n$，同理 $q^2 < m$，因此这样的数对最多有 $O(\sqrt{n} \sqrt{m})$。所以我们直接暴力枚举找到满足 $\gcd(p,q) = 1$ 的 $p$ 和 $q$，由于此时 $d$ 最大能取到 $\min\left\{ \left\lfloor\frac{n}{p}\right\rfloor, \left\lfloor\frac{m}{q}\right\rfloor \right\}$，因此满足 $(p+q) \mid q \cdot d$ 对应的 $(a,b)$ 的数量就是 $\left\lfloor\frac{\min\left\{ \left\lfloor\frac{n}{p}\right\rfloor, \left\lfloor\frac{m}{q}\right\rfloor \right\}}{p+q}\right\rfloor$。

  AC 代码如下，时间复杂度为 $O\left(\sqrt{n} \sqrt{m} \, \log{\max\{n,m\}}\right)$：

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

void solve() {
    int n, m;
    scanf("%d %d", &n, &m);
    LL ret = 0;
    for (int i = 1; i * i < n; i++) {
        for (int j = 1; j * j < m; j++) {
            if (__gcd(i, j) == 1) ret += min(n / i, m / j) / (i + j);
        }
    }
    printf("%lld\n", ret);
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

 

参考资料

  Codeforces Round 942 (Div. 1, Div. 2) Editorial：https://codeforces.com/blog/entry/129027