F - Balloon Comes!
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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
题意分析：
这道题就是有T组输入，每组输入包括一个操作符和两个操作数。根据操作符对两个操作数做相应的加减乘除运算。
解题思路：
这道题本身就判断符号，进行相应的运算就可以了，需要注意的一点就是对于如果结果是整数的情况，要按int类型输出，如果结果是小数，要控制小数点后两位。因为在加减乘中因为输入类型为int，所以直接进行整数的运算就好了，对于除法先要进行强制数据类型转换，然后在运算。
编码：
#include <bits/stdc++.h> using namespace std; int main() { int n; cin >> n; while (n--) { char op; int a, b; cin >> op >> a >> b; if (op == '+') cout << a + b << endl; else if (op == '-') cout << a - b << endl; else if (op == '*') cout << a * b << endl; else if (op == '/') { double ans = (double)a / (double)b; if (a % b == 0) cout << a / b << endl; else { printf("%.2f\n", (float) a / b); } } } return 0; }最后：
通过这道题的做题过程，最大的问题就在于最后这条语句：printf("%.2f\n", (float) a / b);上，之前一直用cout，再加控制格式，但是交了两发都是Wrong Answer。另外，这道题控制了操作数的大小范围，所以在运算本身上没有难度。有关这个地方printf和cout的区别还没有搞清楚，另外还有一个疑问，举例： / 1555 1000 应该输出1.56，但是结果是1.55 但是仍能AC，虽然可以知道1.555本身可能在计算机中存储的是1.5549999......
