Ternary Calculation

最新推荐文章于 2021-03-22 22:56:14 发布

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最新推荐文章于 2021-03-22 22:56:14 发布

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分类专栏：算法文章标签：数论

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本文指导您如何解析并计算包含多种运算符的复杂数学表达式，涵盖了加、减、乘、除、取余等基本运算。

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Description

Complete the ternary calculation.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

There is a string in the form of "number₁operator_anumber₂operator_bnumber₃". Each operator will be one of {'+', '-' , '*', '/', '%'}, and each number will be an integer in [1, 1000].

Output

For each test case, output the answer.

Sample Input

Sample Output

7
-1
0
11
3

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main(){
	int a,b,c,T,ans;
	char e,f;
	cin >> T;
	while(T--){
		cin >> a >> e >> b >> f >> c;
		if((e=='+' || e=='-')&&(f=='*' || f=='/'|| f=='%')){
			switch(f){
				case '*':ans=b*c;break;
				case '/':ans=b/c;break;
				case '%':ans=b%c;break;
			}
			switch(e){
				case '+':ans=a+ans;break;
				case '-':ans=a-ans;break;
			}
		}
		else{
			switch(e){
				case '*':ans=a*b;break;
				case '/':ans=a/b;break;
				case '%':ans=a%b;break;
				case '+':ans=a+b;break;
				case '-':ans=a-b;break;
			}
			switch(f){
				case '*':ans=ans*c;break;
				case '/':ans=ans/c;break;
				case '%':ans=ans%c;break;
				case '+':ans=ans+c;break;
				case '-':ans=ans-c;break;
			}
		}
		cout << ans << endl;
	}
	return 0;
}