hdu 1170
2012-12-04 08:59
127 查看
Balloon Comes!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14959 Accepted Submission(s): 5382
[align=left]Problem Description[/align]
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!
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
4 + 1 2 - 1 2 * 1 2 / 1 2
[align=left]Sample Output[/align]
3 -1 2 0.50
[align=left]Author[/align]
lcy
#include <iostream> using namespace std; int main() { int n, a, b; char c; cin>>n; while (n--) { cin>>c>>a>>b; if (c == '+') { cout<<a+b<<endl; } else if (c == '-') { cout<<a-b<<endl; } else if (c == '*') { cout<<a*b<<endl; } else if (c == '/') { if (a%b == 0)//attention: it is integer cout<<a/b<<endl; else { cout.setf(ios::fixed); cout.precision(2); cout<<1.0*a/b<<endl; } } } return 0; }
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