ZOJ3782-Ternary Calculation

最新推荐文章于 2020-05-18 12:06:49 发布

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最新推荐文章于 2020-05-18 12:06:49 发布

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本文介绍了一种解决三元运算问题的方法，通过解析输入的运算表达式并计算其结果。该算法可以处理加、减、乘、除及取余运算，并提供多个测试案例的解答。

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Ternary Calculation

Time Limit: 2 Seconds Memory Limit: 65536 KB

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 "number₁ operator_a number₂ operator_b number₃". 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

Note

The calculation "A % B" means taking the remainder of A divided by B, and "A / B" means taking the quotient.

Author: ZHOU, Yuchen
Source: The 11th Zhejiang Provincial Collegiate Programming Contest

题意：给你一串算式，计算出答案

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <map>
#include <vector>
#include <queue>
#include <bitset>
#include <stack>

using namespace std;

int main()
{
    char ch1,ch2;
    int a,b,c;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %c %d %c %d",&a,&ch1,&b,&ch2,&c);
        if(ch1=='+'||ch1=='-')
        {
            int ans;
            if(ch2=='+') ans=b+c;
            else if(ch2=='-') ans=b-c;
            else if(ch2=='*') ans=b*c;
            else if(ch2=='/') ans=b/c;
            else ans=b%c;
            if(ch1=='+') printf("%d\n",ans+a);
            else
            {
                if(ch2=='-') printf("%d\n",a-b-c);
                else if(ch2=='+') printf("%d\n",a-b+c);
                else printf("%d\n",a-ans);
            }
        }
        else
        {
            int ans;
            if(ch1=='*') ans=a*b;
            else if(ch1=='/') ans=a/b;
            else if(ch1=='%') ans=a%b;
            if(ch2=='+') printf("%d\n",ans+c);
            else if(ch2=='-') printf("%d\n",ans-c);
            else if(ch2=='*') printf("%d\n",ans*c);
            else if(ch2=='/') printf("%d\n",ans/c);
            else printf("%d\n",ans%c);
        }
    }
    return 0;
}