计算机挑战赛3

老式的计算机只能按照固定次序进行运算，华安大学就有这样一台老式计算机，计算模式为A@B#C，@和#为输入的运算符(可能是+、-或*，运算符优先级与C++一致)，现给出A，B，C的数值以及@和#对应的运算符，请你编写程序，验证老式计算机的运行结果。输入说明:第一行是一个整数N(IN≤10000)，表示有N组计算数据需要处理，接下来N行，每行是相应的数据，包括三个整数和两个运算符，分别对应A、@、B、#和C。输出说明:对每行输入的数据，输出计算结果。输入样例:

3

1 - 12 + 12

12 * 1 - 8

13 + 13 * 2

输出样例：

23

4

39

代码：

C++：

#include<iostream>
using namespace std;
int main() {int n;cin >> n;for (int i = 0; i < n; i++) {int a, b, c;char d, e;cin >> a >> d >> b >> e >> c;if (d == '+' && e == '+') {cout << a + b + c << endl;}else if (d == '+' && e == '-') {cout << a + b - c << endl;}else if (d == '+' && e == '*') {cout << a + b * c << endl;}else if (d == '-' && e == '-') {cout << a - b - c << endl;}else if (d == '-' && e == '+') {cout << a - b + c << endl;}else if (d == '-' && e == '*') {cout << a - b * c << endl;}else if (d == '*' && e == '+') {cout << a * b + c << endl;}else if (d == '*' && e == '-') {cout << a * b - c << endl;}else if (d == '*' && e == '*') {cout << a * b * c << endl;}}return 0;
}

Python：

n = int(input())
for i in range(0, n, 1):a, d, b, e, c = map(str, input().split())if d == '+' and e == '+':print(int(a) + int(b) + int(c))elif d == '+' and e == '-':print(int(a) + int(b) - int(c))elif d == '+' and e == '*':print(int(a) + int(b) * int(c))elif d == '-' and e == '+':print(int(a) - int(b) + int(c))elif d == '-' and e == '-':print(int(a) - int(b) - int(c))elif d == '-' and e == '*':print(int(a) - int(b) * int(c))elif d == '*' and e == '+':print(int(a) * int(b) + int(c))elif d == '*' and e == '-':print(int(a) * int(b) - int(c))elif d == '*' and e == '*':print(int(a) * int(b) * int(c))

Java：

package com.my.gududu;import java.util.*;public class Main {public static void main(String[] args) {Scanner input = new Scanner(System.in);int n = input.nextInt();for (int i = 0; i < n; i++) {int a, b, c;char d, e;a = input.nextInt();d = input.next().charAt(0);b = input.nextInt();e = input.next().charAt(0);c = input.nextInt();if (d == '+' && e == '+') {System.out.println(a + b + c);}else if (d == '+' && e == '-') {System.out.println(a + b - c);}else if (d == '+' && e == '*') {System.out.println(a + b * c);}else if (d == '-' && e == '+') {System.out.println(a - b + c);}else if (d == '-' && e == '-') {System.out.println(a - b - c);}else if (d == '-' && e == '*') {System.out.println(a - b * c);}else if (d == '*' && e == '+') {System.out.println(a * b + c);}else if (d == '*' && e == '-') {System.out.println(a * b - c);}else if (d == '*' && e == '*') {System.out.println(a * b * c);}}}
}