HDU1170：Balloon Comes!

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

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

 

Sample Output

3-120.50

 

 

#include <stdio.h>int main(){    char x;    int a,b,n;    double ans;    scanf("%d%*c",&n);    while(n--)    {        scanf("%c%d%d%*c",&x,&a,&b);        if(x == '/')        {            ans = (double)a/b;            if(ans == (int)ans)            printf("%.0lf\n",ans);            else            printf("%.02lf\n",ans);            continue;        }        if(x == '+')        ans = a+b;        else if(x == '-')        ans = a-b;        else if(x == '*')        ans = a*b;        printf("%.0lf\n",ans);    }    return 0;}