【华为OD机试真题2023 C&D卷 Python&C++】符号运算

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符号运算

时间限制：1s?空间限制：32MB?限定语言：不限

题目描述：

给定一个表达式，求其分数计算结果?
?

表达式的限制如下：

1. 所有的输入数字皆为正整数(包括0)

2. 仅支持四则运算(+-*/)和括号

3. 结果为整数或分数, 分数必须化为最简格式(比如6, 3/4, 7/8, 90/7)

4. 除数可能为0，如果遇到这种情况，直接输出"ERROR"

5.?输入和最终计算结果中的数字都不会超出整型范围

用例的输入一定合法,?不会出现括号不匹配的情况

输入描述：

字符串格式的表达式，仅支持+-*/，数字可能超过两位，可能带有空格，没有负数

长度小于200个字符

输出描述：

表达式结果，以最简格式表达
如果结果为整数，那么直接输出整数

如果结果为分数，那么分子分母不可再约分，可以为假分数，不可表达为带分数

结果可能是负数,?负号放在最前面

示例1

输入：

1 + 5 * 7 / 8

输出：

43/8

示例2

输入：

1 / (0 - 5)

输出：

-1/5

说明：

负号需要提到最前面

示例3

输入：

1 * (3*4/(8-(7+0)))

输出：

12

说明：

注意括号可以多重嵌套

解题思路：

这道题主要还是在处理分数逻辑上有点乱。需要求最小公倍数和最大公约数来处理分子分母。我们的算法思路就是将所有的数字全部转化为分数，在分数的基础上进行加减乘除。

代码（Python）：

from collections import deque
def main():
    #str = input()
    str = '1 / 2 / 3 / 4 / (5 / 6) / 7 / 8 / 9 / 10'
    len_str = len(str)
    deque_str = deque()
    num = ''
    isError = False
    for i in range(len_str):
        c = str[i]
        # 如果是数字
        if c.isdigit():
            num += c
            if i == len_str - 1:
                # 判断符号
                fuhao = deque_str[-1] if deque_str else ''
                # 如果符号是乘除，则需要计算
                if fuhao == '*' or fuhao == '/':
                    if fuhao == '/' and num == '0':
                        isError = True
                        break
                    fuhao = deque_str.pop()
                    deque_str.append(jisuan(deque_str.pop(), fuhao, num))
                else:
                    deque_str.append(num)
        elif c == '(':
            deque_str.append(c)
        elif c == ')':
            if num == '':
                sum = deque_str.pop()
            else:
                sum = num
                num = ''
            while deque_str[-1] != '(':
                fuhao = deque_str.pop()
                temp = deque_str.pop()
                sum = jisuan(temp, fuhao, sum)
            # 左括号删除
            deque_str.pop()
            # 判读左括号前的符号
            fuhao = deque_str[-1] if deque_str else ''
            # 如果符号是乘除，则需要计算
            if fuhao and (fuhao == '*' or fuhao == '/'):
                if fuhao == '/' and sum == '0':
                    isError = True
                    break
                fuhao = deque_str.pop()
                deque_str.append(jisuan(deque_str.pop(), fuhao, sum))
            else:
                deque_str.append(sum)
        elif c == ' ':
            continue
        else:
            # 判断符号
            fuhao = deque_str[-1] if deque_str else ''
            # 如果符号是乘除，则需要计算
            if fuhao and (fuhao == '*' or fuhao == '/'):
                if fuhao == '/' and num == '0':
                    isError = True
                    break
                fuhao = deque_str.pop()
                deque_str.append(jisuan(deque_str.pop(), fuhao, num))
            elif num != '':
                deque_str.append(num)
            # 本次的符号也需要加入
            deque_str.append(c)
            # num置空
            num = ''

    if isError:
        print('ERROR')
    else:
        res = deque_str.popleft()
        while deque_str:
            res = jisuan(res, deque_str.popleft(), deque_str.popleft())

        print(res)

def jisuan(a, fuhao, b):
    as_ = zhuanhua(a)
    aFZ = int(as_[0])
    aFM = int(as_[1])

    bs = zhuanhua(b)
    if fuhao == '/':
        # 除号需要倒置分子分母
        bFZ = int(bs[1])
        bFM = int(bs[0])
    else:
        bFZ = int(bs[0])
        bFM = int(bs[1])

    # 最小公约数
    yueshu, fenzi, fenmu = 0, 0, 0
    if fuhao == '*' or fuhao == '/':
        fenzi = aFZ * bFZ
        fenmu = aFM * bFM
    else:
        gongbeishu_ = gongbeishu(aFM, bFM)
        fenmu = gongbeishu_
        aFZ = gongbeishu_ // aFM * aFZ
        bFZ = gongbeishu_ // bFM * bFZ
        if fuhao == '+':
            fenzi = aFZ + bFZ
        else:
            fenzi = aFZ - bFZ

    if fenzi == 0:
        return '0'

    max_ = max(abs(fenmu), abs(fenzi))
    min_ = min(abs(fenmu), abs(fenzi))
    yueshu = zuixiaogongyueshu(max_, min_)

    fenzi = fenzi // yueshu
    fenmu = fenmu // yueshu

    if fenzi % fenmu == 0:
        return str(fenzi // fenmu)
    else:
        if fenmu < 0:
            # 分母是负数，则需要将负号移到分子
            fenmu = -fenmu
            fenzi = -fenzi
        return str(fenzi) + '/' + str(fenmu)

# 将数字转化为分子分母格式
def zhuanhua(shuzi):
    shuzis = shuzi.split('/')
    szFenzi = shuzis[0]
    szFenmu = shuzis[1] if len(shuzis) == 2 else '1'
    return [szFenzi, szFenmu]

def zuixiaogongyueshu(a, b):
    while a % b != 0:
        b = a % b
    return b

def gongbeishu(a, b):
    gongyueshu = zuixiaogongyueshu(a, b)
    return a // gongyueshu * b // gongyueshu * gongyueshu

if __name__ == '__main__':
    main()

代码（C++）：

#include <iostream>
#include <deque>
#include <sstream>

using namespace std;

string jisuan(string a, string fuhao, string b);

string* zhuanhua(string shuzi);

int zuixiaogongyueshu(int a, int b);

int gongbeishu(int a, int b);

int main() {
    string str;
    getline(cin, str);
    //string str = "1 / 2 / 3 / 4 / (5 / 6) / 7 / 8 / 9 / 10";
    int len = str.length();
    deque<string> deque;
    string num = "";
    bool isError = false;
    for (int i = 0; i < len; i++) {
        char c = str[i];
        //如果是数字
        if (isdigit(c)) {
            num += c;
            if (i == len - 1) {
                //判断符号
                string fuhao = deque.back();
                //如果符号是乘除，则需要计算
                if (fuhao == "*" || fuhao == "/") {
                    if (fuhao == "/" && num == "0") {
                        isError = true;
                        break;
                    }
                    fuhao = deque.back();
                    deque.pop_back();
                    deque.push_back(jisuan(deque.back(), fuhao, num));
                } else {
                    deque.push_back(num);
                }
            }
        } else if (c == '(') {
            deque.push_back(string(1, c));
        } else if (c == ')') {
            string sum;
            if (num == "") {
                sum = deque.back();
                deque.pop_back();
            } else {
                sum = num;
                num = "";
            }
            while (deque.back() != "(") {
                string fuhao = deque.back();
                deque.pop_back();
                string temp = deque.back();
                deque.pop_back();
                sum = jisuan(temp, fuhao, sum);
            }
            //左括号删除
            deque.pop_back();
            //判读左括号前的符号
            string fuhao = deque.back();
            //如果符号是乘除，则需要计算
            if (fuhao != "" && (fuhao == "*" || fuhao == "/")) {
                if (fuhao == "/" && sum == "0") {
                    isError = true;
                    break;
                }
                fuhao = deque.back();
                deque.pop_back();
                deque.push_back(jisuan(deque.back(), fuhao, sum));
            } else {
                deque.push_back(sum);
            }
        } else if (c == ' ') {
            continue;
        } else {
            //判断符号
            string fuhao = deque.back();
            //如果符号是乘除，则需要计算
            if (fuhao != "" && (fuhao == "*" || fuhao == "/")) {
                if (fuhao == "/" && num == "0") {
                    isError = true;
                    break;
                }
                fuhao = deque.back();
                deque.pop_back();
                deque.push_back(jisuan(deque.back(), fuhao, num));
            } else if (num != "") {
                deque.push_back(num);
            }
            //本次的符号也需要加入
            deque.push_back(string(1, c));
            //num置空
            num = "";
        }
    }
    if (isError) {
        cout << "ERROR" << endl;
    } else {
        string res = deque.front();
        deque.pop_front();
        while (!deque.empty()) {
            res = jisuan(res, deque.front(), deque[1]);
            deque.pop_front();
            deque.pop_front();
        }
        cout << res << endl;
    }
    return 0;
}

string jisuan(string a, string fuhao, string b) {
    string* as = zhuanhua(a);
    int aFZ = stoi(as[0]);
    int aFM = stoi(as[1]);
    string* bs = zhuanhua(b);
    int bFZ, bFM;
    if (fuhao == "/") {
        //除号需要倒置分子分母
        bFZ = stoi(bs[1]);
        bFM = stoi(bs[0]);
    } else {
        bFZ = stoi(bs[0]);
        bFM = stoi(bs[1]);
    }
    //最小公约数
    int yueshu, fenzi, fenmu;
    if (fuhao == "*" || fuhao == "/") {
        fenzi = aFZ * bFZ;
        fenmu = aFM * bFM;
    } else {
        int gongbeishu = gongbeishu(aFM, bFM);
        fenmu = gongbeishu;
        aFZ = gongbeishu / aFM * aFZ;
        bFZ = gongbeishu / bFM * bFZ;
        if (fuhao == "+") {
            fenzi = aFZ + bFZ;
        } else {
            fenzi = aFZ - bFZ;
        }
    }
    if (fenzi == 0) {
        return "0";
    }
    int max = max(abs(fenmu), abs(fenzi));
    int min = min(abs(fenmu), abs(fenzi));
    yueshu = zuixiaogongyueshu(max, min);
    fenzi = fenzi / yueshu;
    fenmu = fenmu / yueshu;
    if (fenzi % fenmu == 0) {
        return to_string(fenzi / fenmu);
    } else {
        if (fenmu < 0) {
            //分母是负数，则需要将负号移到分子
            fenmu = -fenmu;
            fenzi = -fenzi;
        }
        return to_string(fenzi) + '/' + to_string(fenmu);
    }
}

string* zhuanhua(string shuzi) {
    stringstream ss(shuzi);
    string* shuzis = new string[2];
    getline(ss, shuzis[0], '/');
    if (ss.peek() != EOF) {
        getline(ss, shuzis[1], '/');
    } else {
        shuzis[1] = "1";
    }
    return shuzis;
}

int zuixiaogongyueshu(int a, int b) {
    while (a % b != 0) {
        b = a % b;
    }
    return b;
}

int gongbeishu(int a, int b) {
    int gongyueshu = zuixiaogongyueshu(a, b);
    return a / gongyueshu * b / gongyueshu * gongyueshu;
}