HDU1170四则运算

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30958 Accepted Submission(s): 11667

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

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思路

题目水是水，不过我也学到了一些！

1、题目要求若结果非整数四舍五入保留两位小数，结果我一直在搞保留两位小数的算法(小白~~)，现在才知道C是自动会四舍五入的

int main()
{
float a=1.346;
printf(“%.2f”,a);
return 0;
}
//输出1.5

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代码

#include <iostream>#include <stdio.h>using namespace std;int main(){    int n;    cin>>n;    while(n--)    {        char a;        int b,c;        cin>>a>>b>>c;        switch(a)        {             case '/':                {                   if(b%c==0)                   printf("%d\n",b/c);                   else                   printf("%.2f\n",(float)b/c);                   break;                }            case '+':            {                cout<<b+c<<endl;                break;            }            case '-':            {                cout<<b-c<<endl;                break;            }            case '*':            {                cout<<b*c<<endl;                break;            }        }    }    return 0;}