Balloon Comes!【水题中的水题】

最新推荐文章于 2020-07-04 17:00:16 发布

原创最新推荐文章于 2020-07-04 17:00:16 发布 · 474 阅读

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本博客提供了一个简单的程序示例，用于解决加、减、乘、除四种基本算术运算。通过输入操作符和两个整数，程序将输出相应的结果。特别地，对于除法，如果结果为非整数，则会四舍五入到小数点后两位。快来挑战你的编程技能吧！

Balloon Comes!

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 102 Solved: 44
[ Submit][ Status][ Web Board]

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

HINT

Source

#include <stdio.h>
#include <stdlib.h>

int main()
{

	int n;

	
	while (~scanf("%d", &n))
	{
	  int a, b;
  	char x;
		while ( n --)
		{
			getchar();
			scanf("%c%d%d",&x,&a,&b);
			if ( x == '+')
			{
				printf("%d\n", a + b);
			
			}
			if ( x == '-')
			{
				printf("%d\n", a - b);
			
			}
			if ( x == '*')
			{
				printf("%d\n", a * b);
			
			}
			if ( x == '/')
			{
				double y = double (a) / double (b);
				if ( y == a / b)
						printf("%d\n", a / b);
				else 
						printf("%.2f\n", y);
			
			}
		
		}
	}

}

水题中的战斗机~~~