杭电ACM 1170解题报告
Balloon Comes!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14636 Accepted Submission(s): 5270
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
代码:
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
int n,a,b;
char s;
cin>>n;
while(n--)
{
double sum;
cin>>s>>a>>b;
switch(s)
{
case '+':sum=a+b;
break;
case '-':sum=a-b;
break;
case '*':sum=a*b;
break;
case '/':sum=(float)a/(float)b;
break;
default:
break;
}
if(sum!=(int) sum)
printf("%.2lf\n",sum);
else
printf("%.0lf\n",sum);
}
return 0;
}