Balloon Comes!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22017 Accepted Submission(s): 8291
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
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 這個題 真他媽的傻逼 氣死我啦 靠 今天本來就不爽
#include
int main()
{int a,b;
int T;
scanf("%d",&T);
char ch[2];
while(T--)
{
scanf("%s%d%d",ch,&a,&b);
switch(ch[0])
{
case '+':
printf("%d\n",a+b);
break;
case'-':
printf("%d\n",a-b);
break;
case'*':
printf("%d\n",a*b);
break;
case'/':
if(a%b==0)
printf("%d\n",a/b);
else
printf("%.2f\n",a/(b*1.0));
break;
}
}
return 0;
}