CF1485C Floor and Mod

C. Floor and Mod

A pair of positive integers \( (a,b) \)  is called special if \(\lfloor \frac{a}{b} \rfloor = a \bmod b \). Here, \( \lfloor \frac{a}{b} \rfloor \)is the result of the integer division between \( a \) and \( b \), while \(a \bmod b\) is its remainder.You are given two integers \(x\) and \(y\). Find the number of special pairs \( (a,b) \) such that \(1\leq a \leq x\) and \(1 \leq b \leq y\).

Input

The first line contains a single integer \(t\) \(1 \le t \le 100\)— the number of test cases.The only line of the description of each test case contains two integers \(x\), \(y\) (\(1 \le x,y \le 10^9\)).

Output

For each test case print the answer on a single line.

题意

输入\(x\),\(y\)计算满足\(1\leq a \leq x\),\(1 \leq b \leq y\) 时，有几对 \( (a,b) \)满足\(\lfloor \frac{a}{b} \rfloor = a \bmod b \)

思路

当时试图模拟列举每一种情况，虽然也考虑到了\(a = k* b + k\)这一特征，但嵌套循环导致时间复杂度似乎还是太高了，看了看赛后提供的题解不禁再让人感受到数论的奇妙。

 

因为\(a = k* b + k\) 我们只需要针对\(1\leq k \leq \sqrt{x}\) (因为\(k^2 < kb+k = a \leq x\))

将\(1 \leq kb+k \leq x\) 变换形式，可以得到\(b \leq x/k-1\)

\(b = min(y,x/k-1)\) 因为 \(b > k\) 所以还需要再减去\(k\)

所以对于一个\(k\),有\(max(0, min(y,x/k-1) - k)\)对合法答案。

代码

 

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
using ll = long long;
inline int read()
{
    int X = 0; bool flag = 1; char ch = getchar();
    while (ch < '0' || ch>'9') { if (ch == '-') flag = 0; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { X = (X << 1) + (X << 3) + ch - '0'; ch = getchar(); }
    if (flag) return X;
    return ~(X - 1);
}
void solve()
{
    ll x, y;
    x = read();
    y = read();
    ll ans = 0;
    ll k;
    for (k = 1; k * k <= x; k++)
    {
        ans += max(0LL, min(y, x / k - 1) - k);
    }
    cout << ans << "\n";
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    t = read();
    while (t--)
    {
        solve();
    }
    return 0;
}

 

参考思路  

https://codeforces.com/blog/entry/87470

 

第一次在博客园上用数学公式，果然还是不熟练啊..折腾了好久。排版有点乱，见谅。