杭电ACM 1170解题报告

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14636    Accepted Submission(s): 5270

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

-1

2

0.50

 

代码：
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
 　　int n,a,b;
 　　char s;
 　　cin>>n;
 　　while(n--)
 　　{
  　　　　double sum;
  　　　　cin>>s>>a>>b;
  　　　　switch(s)
  　　　　{
  　　　　　　case '+':sum=a+b;
   　　　　　　　　break;
  　　　　　　case '-':sum=a-b;
   　　　　　　　　break;
  　　　　　　case '*':sum=a*b;
   　　　　　　　　break;
  　　　　　　case '/':sum=(float)a/(float)b;
   　　　　　　　　break;
  　　　　　　default:
   　　　　　　　　break;
  　　}
  　　if(sum!=(int) sum)
  　　　　printf("%.2lf\n",sum);
  　　else
   　　　　printf("%.0lf\n",sum);
 　　}
 　　return 0;
}