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hdu 1170 Balloon Comes!

时间:2016-08-16 21:55:06      阅读:183      评论:0      收藏:0      [点我收藏+]

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26455    Accepted Submission(s): 10055


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
 
分析:简单题,只要注意除法的时候,如果能整除则直接输出就行,否则保留两位小数
 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 
 5 int main(){
 6     char c;
 7     int t, a, b;
 8     cin >> t;
 9     while(t--){
10         cin >> c >> a >> b;
11         if(c == +)
12             cout << (a + b) << endl;
13         else if(c == -)
14             cout << (a - b) << endl;
15         else if(c == *)
16             cout << (a * b) << endl;
17         else if(c == /){
18             if(a % b == 0)
19                 cout << (a / b) << endl;
20             else
21                 printf("%.2f\n", (float)a / b);
22         }        
23     }
24     return 0;
25 }

 

hdu 1170 Balloon Comes!

原文:http://www.cnblogs.com/qinduanyinghua/p/5777834.html

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