seo整站优化吧,网站建设与开发试题与答案,福州网络推广平台,查域名是否注册老式的计算机只能按照固定次序进行运算#xff0c;华安大学就有这样一台老式计算机#xff0c;计算模式为AB#C#xff0c;和#为输入的运算符(可能是、-或*#xff0c;运算符优先级与C一致)#xff0c;现给出A#xff0c;B#xff0c;C的数值以及和#对应的运算符#xf…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AB#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ABC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
#includeiostream
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);}}}
}
