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  • hdu 1170 Balloon 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 -1 2 0.50
     
    #include <stdio.h>
    #include <string.h>
    const int maxn=500000;
    int a[maxn+5];
    int main()
    {
        int t,a,b;
        char ch;
        scanf("%d",&t);
        while(t--)
        {
            getchar();
            scanf("%c %d%d",&ch,&a,&b);
            if(ch=='+')
                printf("%d
    ",a+b);
            else if(ch=='-')
                printf("%d
    ",a-b);
            else if(ch=='*')
                printf("%d
    ",a*b);
            else if(ch=='/')
            {
                if(a/b*b==a)
                    printf("%d
    ",a/b);
                else
                    printf("%.2lf
    ",(double)a/b);
            }
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/cxbky/p/4907530.html
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