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  • 杭电1170 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
    -1
    2
    0.50
     
    解决方案:
     
    #include <cstdlib>
    #include<iomanip>
    #include <iostream>

    using namespace std;
    void f(char optr,int a,int b)
    {
         switch(optr)
         {
             case '+':cout<<a+b<<endl;break;
             case '-':cout<<a-b<<endl;break;
             case '*':cout<<a*b<<endl;break;
             case '/':if(a%b==0) cout<<a/b<<endl;//能整除
                      else cout<<fixed<<setprecision(2)<<1.0*a/b<<endl;//不能整除
                     break;
         }
    }
    int main(int argc, char *argv[])
    {
        char optr;
        int a,b,n;
        cin>>n;
        while(n--)
        {
            cin>>optr>>a>>b;
            f(optr,a,b);
        }
        
        system("PAUSE");
        return EXIT_SUCCESS;
    }

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  • 原文地址:https://www.cnblogs.com/wft1990/p/4298185.html
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