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HDU 1170

2015-01-02 21:15 316 查看

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4951 Accepted Submission(s): 1589

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
3-120.50

Author
lcy

水题是水题，还是要注意细节，注意读题。坑点，a/b是整数直接输出整数。

The result should be rounded to 2 decimal places If and only if it is not an integer.

代码。。

#include <stdio.h>
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
char s[5];
int a,b;
double sum;
scanf("%s%d%d",s,&a,&b);
if(s[0]=='/')
{
if(!(a%b))
{
printf("%d\n",a/b);  //如果a除以b是整数，输出整数，
}
else
{
sum=double(a)*1.0/b;
printf("%.2lf\n",sum);  //否则保留两位。
}
}
else
{
if(s[0]=='+')
a+=b;
if(s[0]=='-')
a-=b;
if(s[0]=='*')
a*=b;
printf("%d\n",a);
}
}
return 0;
}

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标签： ACM算法 amp c math.h printf

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