HDU 1170Balloon Comes!

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

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

 

Sample Output

3-120.50
注意输入字符的时候之前加getchar().
#include<stdio.h>int main(){int t,i,a,b;char c;double n;scanf("%d",&t);while(t--){getchar();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);elseprintf("%.2lf\n",(double)a/b);}}return 0;}