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 这个题 真他妈的傻逼 气死我啦 靠 今天本来就不爽
#includeint 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; }