HDU 1170 Balloon Comes!（计算表达式）

HDU 1170 Balloon Comes!（计算表达式）

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25560    Accepted Submission(s): 9664

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

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

题意：没什么好说的。。。。

AC代码：

#include#include#include#include#include#include#include#includetypedef long long LL;using namespace std;int main(){ int n; scanf("%d",&n); getchar(); while (n--) { char c; int a,b; scanf("%c%d%d",&c,&a,&b); getchar(); //记得把回车消除，不然会WA switch (c) { case '+': printf("%d\n",a+b);break; case '-': printf("%d\n",a-b);break; case '*': printf("%d\n",a*b);break; case '/': a%b==0?printf("%d\n",a/b):printf("%.2lf\n",(double)a/(double)b);break; } } return 0;}

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