首页 > ACM题库 > HDU-杭电 > HDU 1170 Balloon Comes!-模拟-[解题报告] C++
2013
12-03

HDU 1170 Balloon Comes!-模拟-[解题报告] C++

Balloon Comes!

问题描述 :

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!

输入:

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.

输出:

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.

样例输入:

4
+ 1 2
- 1 2
* 1 2
/ 1 2

样例输出:

3
-1
2
0.50

题目大意:简单的四则运算

解题思路:四则运算

代码如下:

/*
 * 1170_1.cpp
 *
 *  Created on: 2013年8月10日
 *      Author: Administrator
 */


#include <stdio.h>

int main(){
	int t;
	scanf("%d",&t);

	while(t--){
		int a , b;
		char c;
		//在这里%c前面的空白符一定要有。他表示忽略%c前面的所有空白符
		scanf(" %c%d%d",&c,&a,&b);

		if(c == '+'){
			printf("%d\n",a+b);
		}else if( c == '-'){
			printf("%d\n",a-b);
		}else if(c == '*'){
			printf("%d\n",a*b);
		}else{
			if(a%b == 0){
				printf("%d\n",a/b);
			}else{
				//.2lf表示显示2两位小数
				printf("%.2lf\n",(float)a/b);
			}
		}
	}
}