题目1.2.5 Ballon Comes
2015-11-10 23:16
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| [align=left]Problem Description[/align] 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! |
| [align=left]Input[/align] 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. |
| [align=left]Output[/align] 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. |
[align=left]Sample Input[/align]4 + 1 2 - 1 2 * 1 2 / 1 2 |
[align=left]Sample Output[/align]3 -1 2 0.50 |
除法运算时，输出结果为小数时，对输出结果的处理。
二. 总结知识点：
1.对cout输出结果小数点位数的控制。
先声明头文件#include<iomanip>
cout.precision(n);//n为输出结果小数点右边保留的位数
cout.setf(ios::fixed);
cout<<Result;
2.掌握两个运算符
%-->取余运算，可用来判断一个数是否可以整除另一个数。
／-->取模运算，如果结果需要保留小数，除数和被除数需要有一个为float或者double类型。
3.实现N个案例的输入输出。
先输入一个正整数N，表示测试案例的个数
然后通过判断语句去判断输入的测试案例数目是否达到N，若未达到，继续输入测试案例，否则退出程序。
三. 题目代码：
#include <iostream>
#include <iomanip>
#include "math.h"
using namespace std;
int main()
{
int T,A,B;
char C;
int i=0;
cin>>T;
while(i != T) {
cin>>C;
cin>>A>>B;
switch (C) {
case '+':
cout<<A+B;
break;
case '-':
cout<<A-B;
break;
case '*':
cout<<A*B;
break;
case '/':
if(A%B!=0)
{
cout.precision(2);
cout.setf(ios::fixed);
cout<<float(A)/B;
}
else
cout<<A/B;
break;
default:
break;
}
i++;
cout<<"\n";
}
}
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