HDU-1170-Ballon Comes！（细节题）

最新推荐文章于 2020-07-21 12:05:42 发布

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本文介绍了一道简单的算术运算题目的解决方法，包括加、减、乘、除四种基本运算，并提供了完整的C++代码实现，特别关注了除法运算中结果的精确表示。

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题目链接：点击打开链接

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 26449 Accepted Submission(s): 10051

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

还是一道水题，但是有几点注意的

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#define LL long long
using namespace std;
char str;
int n,m;
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		//scanf("%s %lf %lf",str,&n,&m);
		cin>>str>>n>>m;
		if(str=='+')	printf("%d\n",n+m);
		if(str=='-')	printf("%d\n",n-m);
		if(str=='*')	printf("%d\n",n*m);
		if(str=='/')
		{   // double 型不能用 % 取余 
			if(n%m==0)	printf("%d\n",n/m); // 能整除就不输出浮点型，这里一直 WA  
			else printf("%.2lf\n",(double)(n+0.0)/m);
		}
	}
	return 0;
}