The 2021 ICPC Asia Shanghai Regional Programming Contest D、E

Problem - D - Codeforces

解方程

正解设 $\frac{p}{q}=k$ , $\frac{a}{b}=x$ ,那么原方程就变成了 $x^{2}-k*x+1=0$ ,那么就成了二元一次方程组来判断有无实根的情况，判别式为 $\Delta =k^{2}-4$ ，幼儿园数学可知 $\Delta$ 大于等于0有实数根，小于0没有实数根，所以有解情况为 $p^{2}\geq 4*q^{2}$ ,然后再根据求根公式得到 $x=\frac{p\pm \sqrt{p^{2}-4*q^{2}}}{2*q}$ ,因此 $\sqrt{p^{2}-4*q^{2}}$ 一定是个整数，进而可知a、b的值，最后把a、b化简

AC代码：
#include <bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
using namespace std;
using LL = long long;int main() {ios::sync_with_stdio(false);cin.tie(nullptr);int T;cin >> T;rep (oo, 0, T) {LL p, q;cin >> p >> q;if (p * p < 4 * q * q) {cout << "0 0\n";continue;}LL tmp = p * p - 4 * q * q;LL z = sqrt(tmp);if (z * z != tmp) {cout << "0 0\n";continue;}function<LL(LL, LL)> gcd = [&](LL x, LL y) {return y == 0 ? x : gcd(y, x % y);};tmp = z;LL d = gcd(p + tmp, 2 * q);cout << (p + tmp) / d << " " << (2 * q) / d << '\n';}return 0;
}
E. Strange Integers

选择任意一些数，使得任意两个数之间的差值的绝对值大于等于k，求最多选多找个，直接贪心去选，排一下序从小到大能选即选

AC代码：
#include <bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
using namespace std;
using LL = long long;int main() {ios::sync_with_stdio(false);cin.tie(nullptr);int n, k;cin >> n >> k;vector<int> a(n);for (int i = 0; i < n; i++) {cin >> a[i];}sort(a.begin(), a.end());stack<int> st;st.push(a[0]);for (int i = 1; i < n; i++) {if (a[i] - st.top() >= k) {st.push(a[i]);}}cout << st.size() << '\n';return 0;
}

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The 2021 ICPC Asia Shanghai Regional Programming Contest D、E

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