# 杭电 acm 1170 Balloon Comes!

xiaoxiao2021-02-28  6

# Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 32728    Accepted Submission(s): 12329 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  基本加减乘除，输入必为整数只考虑除法中的精度问题，若余数为0则直接int型输出，若余数不为零则转化为double型并保留两位小数输出。 #include<iostream> #include<stdio.h> using namespace std; int main() { int n; char a; int b, c; cin >> n; for (int i = 0; i < n;i++) { cin >> a >> b >> c; if (a == '+') cout << b + c << endl; if (a == '-') cout << b - c<<endl; if (a == '*') cout << b*c << endl; if (a == '/'&&(b%c)==0) cout<<b/c<<endl; if (a == '/' && (b%c) != 0) { double d = b, e = c; printf("%.2f\n", d / e); } } return 0; }