zoukankan      html  css  js  c++  java
  • ACM HDU 1170 Balloon Comes! (完全的水题,加减乘除)

    Balloon Comes!

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 10546    Accepted Submission(s): 3623


    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
     
    水题,注意
    The result should be rounded to 2 decimal places If and only if it is not an integer.就是做除法如果整除要输出整数,因为这个WR了
    #include<stdio.h>
    #include
    <iostream>
    using namespace std;
    int main()
    {
    int T,a,b;
    char ch;
    scanf(
    "%d",&T);
    while(T--)
    {
    cin
    >>ch>>a>>b;
    if(ch=='+') printf("%d\n",a+b);
    else if(ch=='-')
    printf(
    "%d\n",a-b);
    else if(ch=='*')
    printf(
    "%d\n",a*b);
    else
    {
    if(a%b==0)printf("%d\n",a/b);//这里注意,看清题目意思
    else
    printf(
    "%.2f\n",(float)a/b);
    }
    }
    return 0;
    }

  • 相关阅读:
    Entity Framework Code First 数据迁移
    Tekla Structures 使用类库概览
    从IT的角度思考BIM(三):敏捷开发
    在 IIS MIME 类型中添加 md 扩展名
    使用 windows 计划任务播放音乐文件
    Win10 IIS以及ASP.NET 4.0配置问题日志
    从IT的角度思考BIM(二):模式与框架
    最小生成树算法总结(Kruskal,Prim)
    最短路径算法总结(floyd,dijkstra,bellman-ford)
    大整数运算模板总结
  • 原文地址:https://www.cnblogs.com/kuangbin/p/2124185.html
Copyright © 2011-2022 走看看