个人练习-PAT甲级-1088 Rational Arithmetic

题目链接https://pintia.cn/problem-sets/994805342720868352/problems/994805378443755520

不难，但是细节有些多要注意。

用到的分数类，分为整数部分、分子、分母

class Node {
    
public:long int part1, a, b;
};

首先写好输出函数，输入一个假分数，输出带分数，负数时要带括号。
（1）为零的话只需输出零，我们统一在带分数部分输出这个零，即整数部分为零的话不输出；
（2）仅有整数部分或仅有分数部分时空格不输出；
（3）符号在一开始就判断好

void Output(Node x) {
    bool isNeg = false;if (x.a < 0) {
    isNeg = true;x.a = -x.a;printf("(-");}Simplify(x);x.part1 = x.a / x.b;x.a -= x.b * x.part1;if (x.part1 != 0)printf("%ld", x.part1);if (x.part1 != 0 && x.a != 0)printf(" ");if (x.a == 0) {
    if (x.part1 == 0)printf("0");}elseprintf("%ld/%ld", x.a, x.b);if (isNeg)printf(")");
}

其中Simplify()函数是将分子分母约到最简，注意gcd()函数输入要都为正

void Simplify(Node& x) {
    long int tmp = gcd(abs(x.a), abs(x.b));x.a /= tmp;x.b /= tmp;
}

加减乘都不难，除法时要注意，要保持负号始终在分子的位置。

Node Divide(Node x, Node y) {
    Node ret;ret.a = x.a * y.b;ret.b = x.b * y.a;if (ret.a > 0 && ret.b < 0) {
    ret.b = -ret.b;ret.a = -ret.a;}else if (ret.a < 0 && ret.b < 0) {
    ret.b = -ret.b;ret.a = -ret.a;}else;return ret;
}

完整代码

#include <iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<map>
#include<string.h>
#include<string>using namespace std;class Node {
    
public:long int part1, a, b;
};long int gcd(long int a, long int b) {
    int tmp;if (a < b) {
    tmp = b;b = a;a = tmp;}while (b) {
    tmp = b;b = a % b;a = tmp;}return a;
}void Change(Node& x, long int d) {
    x.a *= d;x.b *= d;
}void Simplify(Node& x) {
    long int tmp = gcd(abs(x.a), abs(x.b));x.a /= tmp;x.b /= tmp;
}void Output(Node x) {
    bool isNeg = false;if (x.a < 0) {
    isNeg = true;x.a = -x.a;printf("(-");}Simplify(x);x.part1 = x.a / x.b;x.a -= x.b * x.part1;if (x.part1 != 0)printf("%ld", x.part1);if (x.part1 != 0 && x.a != 0)printf(" ");if (x.a == 0) {
    if (x.part1 == 0)printf("0");}elseprintf("%ld/%ld", x.a, x.b);if (isNeg)printf(")");
}Node Add(Node x, Node y) {
    long int new_de = x.b * y.b / gcd(x.b, y.b);Change(x, new_de / x.b);Change(y, new_de / y.b);Node ret;ret.b = new_de;ret.a = x.a + y.a;Simplify(ret);return ret;
}Node Minus(Node x, Node y) {
    y.a = -y.a;return Add(x, y);
}Node Product(Node x, Node y) {
    Node ret;ret.a = x.a * y.a;ret.b = x.b * y.b;return ret;
}Node Divide(Node x, Node y) {
    Node ret;ret.a = x.a * y.b;ret.b = x.b * y.a;if (ret.a > 0 && ret.b < 0) {
    ret.b = -ret.b;ret.a = -ret.a;}else if (ret.a < 0 && ret.b < 0) {
    ret.b = -ret.b;ret.a = -ret.a;}else;return ret;
}int main() {
    Node x, y, ret;scanf("%ld/%ld %ld/%ld", &x.a, &x.b, &y.a, &y.b);// Addret = Add(x, y);Output(x);printf(" + ");Output(y);printf(" = ");Output(ret);printf("\n");// Minusret = Minus(x, y);Output(x);printf(" - ");Output(y);printf(" = ");Output(ret);printf("\n");// Productret = Product(x, y);Output(x);printf(" * ");Output(y);printf(" = ");Output(ret);printf("\n");// Divideret = Divide(x, y);Output(x);printf(" / ");Output(y);printf(" = ");if (ret.b == 0)printf("Inf");elseOutput(ret);printf("\n");return 0;
}