2013
12-03

Balloon Comes!

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 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.

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.

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

3
-1
2
0.50

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

#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);
}
}
}
}

1. 要灯光、服装配合才好，如果灯光闪闪的，衣服也是有反光膜，你会感觉跳舞的人在漂移，一下在这一下又在那边，配合肢体的动作，感觉很酷。