dwcs5做h5网站可以引流推广的app

CSP-J 2023 T3 一元二次方程 解题报告

Link

前言

今年$CSP$的原题, 回家$1h$内写$AC$, 但是考场上没有写出来 , 原因是脑子太不好了, 竟然调了两个小时没有调出来. 一等奖悬那…

正题

看完题目,第一眼就是大模拟, 并且$CCF$绝对不会让你好受,所以出了一个如此刁钻的题目, 并且要考虑到非常多的情况, 代码非常长…

最重要的一点: $\Delta$

$\Delta$是此题中最重要的分情况讨论的地方. 根据初三$22$章的所学知识 ,可知分三种情况:

1. $\Delta <0$

不用说了
直接输出NO

2. $\Delta =0$

同样的, 只有一种情况, 答案为$-\frac{b}{2a}$,但是, 需要严谨的判断.

if(delta == 0) {if(b == 0) {cout << 0;}else if(a * b > 0) {a = abs(a);b = abs(b);cout << "-";if(b % (2 * a) == 0) {cout << b / 2 / a;}else {cout << b / __gcd(2 * a, b) << "/" << (2 * a) / __gcd(2 * a, b);}}else {a = abs(a);b = abs(b);if(b % (2 * a) == 0) {cout << b / 2 / a;}else {cout << b / __gcd(2 * a, b) << "/" << (2 * a) / __gcd(2 * a, b);}}
}

3. $\Delta >0$

地狱.
我是分两种情况的, 一种是$a>0$, 一种是$a<0$. 这样可以分辨出是$+\sqrt{\Delta }$ 还是$-\sqrt{\Delta }$

如若$a<0$, 则可知答案为:
$$\frac{b+\sqrt{\Delta }}{-2a}$$

如若$a>0$, 则可知答案为:
$$\frac{\sqrt{\Delta }-b}{2a}$$

• 在这里有一个技巧, 就是不论怎样, 输出时, $\sqrt{\Delta }$永远是正的(符号为+)

可以分两种情况:
1.第一种: 不需要写sqrt, 也就是$\Delta$为完全平方数时,

比较好处理, 首先需要判断$b+\sqrt{\Delta }$是否为$0$. 如果是, 则直接输出$0$; 否则输出最简分数.

其中, 一定要记住如果$(b+\sqrt{\Delta })\%(2\ast a)=0$, 就直接输出一个整数.还要注意判断正负号.

2.第二种: 需要写sqrt, 很难.

首先, 先输出前面的内容, 也就是$-\frac{b}{2a}$, 判断同上.

然后, 输出+, 代表符号.

接着, 找出三个变量, 也就是: $\frac{x}{y}\sqrt{\frac{\Delta }{x^2}}$中的$x,y和\frac{\Delta }{x^2}$.其中,$\sqrt{\frac{\Delta }{x^2}}$为最简平方根数.

接下来是$4$种情况:

$x=y$, 只有$\sqrt{\frac{\Delta }{x^2}}$;

$x\%y=0$, 只有$\frac{x}{y}\sqrt{\frac{\Delta }{x^2}}$

$y\%x=0$, 只有$\frac{\sqrt{\frac{\Delta }{x^2}}}{y}$

其他情况, 输出$\frac{x\times \sqrt{\frac{\Delta }{x^2}}}{y}$

完结撒花!!

$Code:$

• 心脏病患者请勿观看

#include <bits/stdc++.h>
using namespace std;int T, M;
int a, b, c;int pd(int x) {for(int i = sqrt(x) + 1; i >= 1; --i) {if(x % (i * i) == 0) {return i;}}
}int main() {cin >> T >> M;while(T--) {cin >> a >> b >> c;int delta;delta = b * b - 4 * a * c;if(delta < 0) {cout << "NO";}else if(delta == 0) {if(b == 0) {cout << 0;}else if(a * b > 0) {a = abs(a);b = abs(b);cout << "-";if(b % (2 * a) == 0) {cout << b / 2 / a;}else {cout << b / __gcd(2 * a, b) << "/" << (2 * a) / __gcd(2 * a, b);}}else {a = abs(a);b = abs(b);if(b % (2 * a) == 0) {cout << b / 2 / a;}else {cout << b / __gcd(2 * a, b) << "/" << (2 * a) / __gcd(2 * a, b);}}}else {if(a < 0) {int mother = - 2 * a;int x = pd(delta);int y = delta / x / x;if(b == 0) {mother = abs(mother);if(y == 1) {if(x == mother) {cout << "1";}else if(x % mother == 0) {cout << x / mother;}else {cout << x / __gcd(x, mother) << "/" << mother / __gcd(x, mother);}}else {if(x == mother) {cout << "sqrt(" << y << ")";}else if(mother % x == 0) {cout << "sqrt(" << y << ")";cout << "/" << mother / x;}else if(x % mother == 0) {cout << x / mother << "*sqrt(" << y << ")";}else {cout << x / __gcd(x, mother) << "*sqrt(" << y << ")" << "/" << mother / __gcd(x, mother);}}}else if(y == 1) { // 不需要sqrt// 说明可以合并为同一个式子int son = - b - x;if(son == 0) {cout << 0;}else if(son * mother < 0) { // 如果分子分母同号.son = abs(son);mother = abs(mother);if(son % mother == 0) {cout << son / mother;}else {cout << son / __gcd(son, mother) << "/" << mother / __gcd(son, mother);}}else { // 如果分子分母异号.son = abs(son);mother = abs(mother);cout << "-";if(son % mother == 0) {cout << son / mother;}else {cout << son / __gcd(son, mother) << "/" << mother / __gcd(son, mother);}}}else { // 需要sqrt.// 1. 先输出前面的if(b > 0) { // 不需要负号b = abs(b);mother = abs(mother);if(b % mother == 0) {cout << b / mother;}else {cout << b / __gcd(b, mother) << "/" << mother / __gcd(b, mother);}}else { // 需要负号b = abs(b);mother = abs(mother);cout << "-";if(b % mother == 0) {cout << b / mother;}else {cout << b / __gcd(b, mother) << "/" << mother / __gcd(b, mother);}}// 2. 输出sqrt部分(不管怎样都是+)cout << "+";if(x == 1) { // 不需要输出前缀.cout << "sqrt(" << y << ")";cout << "/" << - 2 * a;}else {if(x == mother) {cout << "sqrt(" << y << ")";}else if(x % mother == 0) {cout << x / mother << "*sqrt(" << y << ")";}else if(mother % x == 0) {cout << "sqrt(" << y << ")";cout << "/" << mother / x;}else {cout << x / __gcd(x, mother);cout << "*sqrt(" << y << ")";cout << "/" << mother / __gcd(x, mother);}}}}else {int mother = 2 * a;int x = pd(delta);int y = delta / x / x;if(b == 0) {mother = abs(mother);if(y == 1) {if(x == mother) {cout << "1";}else if(x % mother == 0) {cout << x / mother;}else {cout << x / __gcd(x, mother) << "/" << mother / __gcd(x, mother);}}else {if(x == mother) {cout << "sqrt(" << y << ")";}else if(mother % x == 0) {cout << "sqrt(" << y << ")";cout << "/" << mother / x;}else if(x % mother == 0) {cout << x / mother << "*sqrt(" << y << ")";}else {cout << x / __gcd(x, mother) << "*sqrt(" << y << ")" << "/" << mother / __gcd(x, mother);}}}else if(y == 1) { // 不需要sqrt// 说明可以合并为同一个式子int son = - b + x;if(son == 0) {cout << 0;}else if(son * mother > 0) { // 如果分子分母同号.son = abs(son);mother = abs(mother);if(son % mother == 0) {cout << son / mother;}else {cout << son / __gcd(son, mother) << "/" << mother / __gcd(son, mother);}}else { // 如果分子分母异号.son = abs(son);mother = abs(mother);cout << "-";if(son % mother == 0) {cout << son / mother;}else {cout << son / __gcd(son, mother) << "/" << mother / __gcd(son, mother);}}}else { // 需要sqrt.// 1. 先输出前面的if(b * mother < 0) { // 不需要负号b = abs(b);mother = abs(mother);if(b % mother == 0) {cout << b / mother;}else {cout << b / __gcd(b, mother) << "/" << mother / __gcd(b, mother);}}else { // 需要负号b = abs(b);mother = abs(mother);cout << "-";if(b % mother == 0) {cout << b / mother;}else {cout << b / __gcd(b, mother) << "/" << mother / __gcd(b, mother);}}// 2. 输出sqrt部分(不管怎样都是+)cout << "+";if(x == 1) { // 不需要输出前缀.cout << "sqrt(" << y << ")";cout << "/" << 2 * a;}else {mother = 2 * a;if(x == mother) {cout << "sqrt(" << y << ")";}else if(x % mother == 0) {cout << x / mother << "*sqrt(" << y << ")";}else if(mother % x == 0) {cout << "sqrt(" << y << ")";cout << "/" << mother / x;}else {cout << x / __gcd(x, mother);cout << "*sqrt(" << y << ")";cout << "/" << mother / __gcd(x, mother);}}}}}cout << endl;}return 0;
}