Balloon Comes!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21970 Accepted Submission(s): 8278
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
小模拟......
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <iostream>
using namespace std;
int t,a,b;
char s[5];
int main ( )
{
scanf ( "%d" , &t );
while ( t-- )
{
scanf ( "%s" , s );
scanf ( "%d%d" , &a , &b );
if ( s[0] == '+' )
printf ( "%d\n" , a+b );
if ( s[0] == '-' )
printf ( "%d\n" , a-b );
if ( s[0] == '*' )
printf ( "%d\n" , a*b );
if ( s[0] == '/' )
{
if ( a%b == 0 ) printf ( "%d\n" , a/b );
else printf ( "%.2lf\n" , a*1.0/b );
}
}
}
