[解题报告]HDU 1170 Balloon Comes!

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16376    Accepted Submission(s): 5958

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

 

 

 

注意空格吸收，类型转换

 

#include<stdio.h>
int main()
{
    int t,a,b;
    char m[2];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%d%d",m,&a,&b);
        if(m[0]=='+') printf("%d\n",a+b);
        else if(m[0]=='-') printf("%d\n",a-b);
        else if(m[0]=='*') printf("%d\n",a*b);
        else if(a/b*b==a) printf("%d\n",a/b);
        else printf("%.2f\n",(double)a/b);
    }
    return 0;
}