Balloon Comes!
来源:互联网 发布:大连 张丕林 知乎 编辑:程序博客网 时间:2024/05/16 19:43
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23182 Accepted Submission(s): 8774
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!
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
3-120.50
#include<stdio.h>int main(){ int n,a,b; char k; scanf("%d",&n); while(n--) { getchar(); scanf("%c %d%d",&k,&a,&b); if(k=='+') printf("%d\n",a+b); if(k=='-') printf("%d\n",a-b); if(k=='*') printf("%d\n",a*b); if(k=='/') { if(a%b==0) printf("%d\n",a/b); else printf("%.2f\n",a*1.0/b); } } return 0;}
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