hdu-Balloon Comes!

hdu-Balloon Comes!一道水题的报告

今天闲来无事，想刷几道水题来增加自信心。但是在刷这道水题时，我发现了自己的一个知识漏洞！！！！

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   这个题目是个十足的水题，但是我今天却在这个题目上卡壳了。不过通过这道题，我对于 %c这个符号有了新的认识。 先附上我最后ac的代码吧 #include<stdio.h> int main() { int t,a,b,i; char c; scanf("%d",&t); while(t--) { getchar();//注意这个getchar(),这是千万少不得的，我就是栽在这个上面！！！！！ scanf("%c %d %d",&c,&a,&b);//%c放在一个位置时，前面必须要有getchar()，不然他会吃掉一个回车符。会导致程序不输出结果。但是如果%c在第二或者后面的位置，那就不需要getchar了 if(c=='+') printf("%d\n",a+b); if(c=='-') printf("%d\n",a-b); if(c=='*') printf("%d\n",a*b); if(c=='/') { if(a%b==0)printf("%d\n",a/b); else printf("%.2f\n",(float)a/b); } } return 0; }