杭电 acm 1170 Balloon Comes!

最新推荐文章于 2020-06-17 21:16:30 发布

原创最新推荐文章于 2020-06-17 21:16:30 发布 · 414 阅读

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本文介绍了一个简单的程序设计问题——基本的四则运算处理方法。文章详细解释了如何通过编程来实现加、减、乘、除操作，并特别关注了除法中的精度处理问题。对于整数运算结果为整数的情况，程序将直接输出；而对于结果带有小数的情况，则会保留两位小数进行输出。

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32728 Accepted Submission(s): 12329

Problem Description

The contest starts now! How excited it is to see balloons floating around. You, one of the best programmers in HDU, can get a very beautiful balloon if only you have solved the very very very... easy problem.
Give you an operator (+,-,*, / --denoting addition, subtraction, multiplication, division respectively) and two positive integers, your task is to output the result.
Is it very easy?
Come on, guy! PLMM will send you a beautiful Balloon right now!
Good Luck!

Input

Input contains multiple test cases. The first line of the input is a single integer T (0<T<1000) which is the number of test cases. T test cases follow. Each test case contains a char C (+,-,*, /) and two integers A and B(0<A,B<10000).Of course, we all know that A and B are operands and C is an operator.

Output

For each case, print the operation result. The result should be rounded to 2 decimal places If and only if it is not an integer.

Sample Input

Sample Output

基本加减乘除，输入必为整数只考虑除法中的精度问题，若余数为0则直接int型输出，若余数不为零则转化为double型并保留两位小数输出。

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
	int n;
	char a;
	int b, c;
	cin >> n;
	for (int i = 0; i < n;i++)
	{
		cin >> a >> b >> c;
		if (a == '+')
			cout << b + c << endl;
		if (a == '-')
			cout << b - c<<endl;
		if (a == '*')
			cout << b*c << endl;
		if (a == '/'&&(b%c)==0)
			cout<<b/c<<endl;
		if (a == '/' && (b%c) != 0)
		{
			double d = b, e = c;
			printf("%.2f\n", d / e);
		}
	}
	return 0;
}