[CSP-J 2023] 一元二次方程 题解

大模拟！

解题思路

题目比较长，但中心就是求一元二次方程的最大实数解，否则输出 NO，读题就可以知道:

$\Delta \ = \ b^{2} \ - \ 4ac$

如果 $\Delta \ < \ 0$，就没有实数解

如果 $\Delta \ \ge 0$，就依照题意来写:

两个实数解 $x_{1,2} \ = \ \frac{-b\pm \sqrt{\Delta} }{2a}$

但 $2a \ < \ 0$ 时，应当用 $\frac{-b\pm \sqrt{\Delta} }{2a}$

AC Code

# include <bits/stdc++.h>
using namespace std;
bool kaifang(int x) {
	int k = sqrt(x);
	return k * k == x;
}
void huajie(int fenzi, int fenmu) {
	int fu = 1;
	if (fenzi  < 0) {
		fu = -1;
		fenzi = -fenzi;
	}
	int k = __gcd(fenzi, fenmu);
	fenzi /= k;
	fenmu /= k;
	if (fenzi % fenmu == 0) {
		cout << fu * fenzi / fenmu;
	} else {
		cout << fenzi * fu << "/" << fenmu;
	}
}
int check(int x) {
	int ret = 1;
	for (int i = 2; i * i <= x; i++) {
		if (x % (i * i) == 0) {
			ret = i;
		}
	}
	return ret;
}
void huajie2(int fenzi, int fenmu, int delta) {
	if (fenzi % fenmu == 0) {
		if (fenzi / fenmu == 1) {
			cout << "sqrt(" << delta << ")" << "\n";
		} else {
			cout << fenzi / fenmu << "*sqrt(" << delta << ")" << "\n";
		}
		return;
	}
	int k = __gcd(fenzi, fenmu);
	fenzi /= k;
	fenmu /= k;
	if (fenzi == 1) {
		cout << "sqrt(" << delta << ")/" << fenmu << "\n";
	} else {
		cout << fenzi << "*sqrt(" << delta << ")/" << fenmu << "\n";
	}
}
int main() {
    freopen("uqe.in", "r", stdin);
    freopen("uqe.out", "w", stdout);
	int T, M;
	scanf("%d%d", &T, &M);
	while (T--) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		if (a < 0) {
			a *= -1;
			b *= -1;
			c *= -1;
		}
		int delta = b * b - 4 * a * c;
		if (delta < 0) {
			printf("NO\n");
			continue;
		}
		if (kaifang(delta)) {
			int ret;
			int flag = 0;
			int k = sqrt(delta);
			ret = (-b + k) / (2 * a);
			if ((-b + k) % (2 * a) == 0) {
				printf("%d\n", ret);
				continue;
			}
			huajie(-b + sqrt(delta), 2 * a);
			printf("\n");
			continue;
		}
		if (b != 0) {
			huajie(-b, 2 * a);
			printf("+");
		}
		int yinzi = check(delta);
		delta /= yinzi * yinzi;
		huajie2(yinzi, 2 * a, delta);
	}
    return 0;
}