hdu 1170 Balloon Comes!
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!
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> #include <string.h> const int maxn=500000; int a[maxn+5]; int main() { int t,a,b; char ch; scanf("%d",&t); while(t--) { getchar(); scanf("%c %d%d",&ch,&a,&b); if(ch=='+') printf("%d\n",a+b); else if(ch=='-') printf("%d\n",a-b); else if(ch=='*') printf("%d\n",a*b); else if(ch=='/') { if(a/b*b==a) printf("%d\n",a/b); else printf("%.2lf\n",(double)a/b); } } return 0; }