Balloon Comes! 1170
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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
#include <cstdio>#include <iostream>int main(int argc, const char *argv[]){ int n; while(scanf("%d", &n) == 1) { while(n --) { int a = 0, b = 0; char ch; std::cin >> ch >> a >> b; switch(ch) { 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); break; } else { printf("%.2f\n", static_cast<double>(a) / b); break; } } } } //system("pause"); return 0;}
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