CSP-J倒计时1天。

最近教练给我们做了10+套模拟赛，疯狂整理后，每套模拟都写了1~2篇题解。

广刷题，确实有用，今天早上做出来一道第二题题目传送门

顺便在这放个题解

化简

decode

\(n = p \times q\)

\(e \times d = (p - 1)(q - 1) + 1\)

令\(x = e \times d - 2\)

则\(x = p \times q - (p + q)\)

因为，\(p \times q\)，且\(x = n - (n - x)\)

所以，\(x = n - n - e \times d + 2\)

所以，\(x = n - m\)

所以，\(p + q = m\)

\(\left\{\begin{aligned}p \times q & = & n \\p + q & = &n-e \times d+2 \\\end{aligned}\right.\)

\(\alpha=n-ed+2=p+q\)

\((p-q)^2=p^2-2qp+q^2 \dots \boxed1\)

\((p+q)^2=p^2 + 2qp+q^2 \dots \boxed2\)

\(qp=n\)

\(\boxed 2-4n=\boxed1\)

\(\sqrt{\boxed1}=p-q\)

\(p-q = \sqrt{\alpha ^ 2 - 4n} = \beta\)

\(q = \frac{\alpha + \beta}{2}\)

\(p = \frac{\alpha - \beta}{2}\)

求值

#include<bits/stdc++.h>
#define ll long long
ll k;
using namespace std;
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> k;
    while(k--){
        ll n, e, d, p = 0, q = 0, a, b;
        cin >> n >> e >> d;
        a = n - e * d + 2;
        b = a * a - 4 * n;
        ll c = sqrt(b);
        if(c * c != b){
            cout << "NO\n";
            continue;
        }
        b = c;
        q = (a + b) / 2;
        p = (a - b) / 2;
        if(p * q != n || (p - 1) * (q - 1) + 1 != e * d){
            cout << "NO\n";
            continue;
        }
        cout << p << ' ' << q << '\n';
    }
    return 0;
}