Balloon Comes!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16586 Accepted Submission(s): 6047
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
Author
lcy
1 #include<cstdio> 2 #include<cmath> 3 #include<iostream> 4 #include<algorithm> 5 #include<cstring> 6 using namespace std; 7 8 9 char c; 10 int a,b; 11 12 int main() 13 { 14 int n; 15 while(scanf("%d",&n)!=EOF) 16 { 17 int i=0; 18 for(i=0;i<n;i++) 19 { 20 getchar(); 21 scanf("%c",&c); 22 scanf("%d%d",&a,&b); 23 if(c=='+')printf("%d ",a+b); 24 if(c=='-')printf("%d ",a-b); 25 if(c=='*')printf("%d ",a*b); 26 if(c=='/') 27 { 28 if(a%b==0) 29 { 30 printf("%.0lf ",a*1.0/b); 31 } 32 else 33 printf("%0.2lf ",a*1.0/b); 34 } 35 36 } 37 } 38 return 0; 39 }