Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 32966    Accepted Submission(s): 12430 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   Author lcy

题目解析：

非常简单的题目，不多说，看代码

代码：

#include <bits/stdc++.h> using namespace std; int main() { //freopen("in.txt","r",stdin); int T; cin>>T; char c; int a,b; while(T--) { double ans=0; cin>>c>>a>>b; if(c=='+') ans=a+b; if(c=='-') ans=a-b; if(c=='*') ans=a*b; if(c=='/') ans=a*1.0/b; if(ans==(int)ans) cout<<(int)ans<<endl; else printf("%.2f\n",ans); } return 0; }