1410242114-hd-Balloon Comes!
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Balloon Comes!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20169 Accepted Submission(s): 7597
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
题目大意
给定一个符号和两个整数,根据符号对这两个数进行运算,如果存在小数的话就保留两位小数
代码
#include<stdio.h>int main(){int t;int a,b;char c;int i,j;scanf("%d",&t);getchar();while(t--){scanf("%c%d%d",&c,&a,&b);getchar();switch(c){case'+':printf("%d\n",a+b);break; case'-':printf("%d\n",a-b);break; case'*':printf("%d\n",a*b);break; case'/':{//如果得到的不是整数的话就需要保留2位小数。 if(a%b==0) printf("%d\n",a/b); else printf("%.2lf\n",a/(b*1.0)); }break;}}return 0;}
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