标签：des style blog http io color ar os java
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20356 Accepted Submission(s): 7673
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
解题思路：水。直接的就是简单的四则运算。不过要注意，最大的坑在于：当两数相除时，若能整除，则要输出整数，否则输出两位小数。
AC代码：
#include <iostream>
#include <cstdio>
using namespace std;
int main(){
// freopen("in.txt", "r", stdin);
int n, a, b;
char p[2];
scanf("%d", &n);
for(int i=0; i<n; i++){
scanf("%s", p);
scanf("%d%d", &a, &b);
if(p[0] == '-') printf("%d\n", a-b);
else if(p[0] == '+') printf("%d\n", a+b);
else if(p[0] == '*') printf("%d\n", a*b);
else{
if(a%b) printf("%.2f\n", a*1.0/b);
else printf("%d\n", a/b);
}
}
return 0;
}
HDU 1170 Balloon Comes!
标签：des style blog http io color ar os java
原文地址：http://blog.csdn.net/u013446688/article/details/40746225
