Balloon Comes!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30053 Accepted Submission(s): 11318
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
#include<stdio.h>int main()
{
int t,a,b;
char m[2];
scanf("%d",&t);
while(t--)
{
scanf("%s%d%d",m,&a,&b);
if(m[0]=='+') printf("%d
",a+b);
else if(m[0]=='-') printf("%d
",a-b);
else if(m[0]=='*') printf("%d
",a*b);
else
{
if(a%b==0) printf("%d
",a/b);
else printf("%.2f
",(double)a/b);
}
}
return 0;
}