簡單計算器

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14833 Accepted Submission(s): 5050

 

Problem Description 讀入一個只包含 +, -, *, / 的非負整數計算表達式，計算該表達式的值。

Input 測試輸入包含若干測試用例，每個測試用例占一行，每行不超過200個字符，整數和運算符之間用一個空格分隔。沒有非法表達式。當一行中只有0時輸入結束，相應的結果不要輸出。

Output 對每個測試用例輸出1行，即該表達式的值，精確到小數點後2位。

Sample Input

1 + 2
4 + 2 * 5 - 7 / 11
0

Sample Output

3.00
13.36

 

只用堆棧即可，c代碼如下：

 

#include
#include
double a[210];
int main()
{
	int i,sign,count,j;
	double sum;
	char c;
	while(1)
	{
		sign=1;i=0;
		memset(a,0,sizeof(a));
		scanf(%lf,&a[0]);
		while(getchar()!='
')
		{
			sign=0;
			scanf(%c %d,&c,&count);
			if(c=='+')
			  a[++i]=count;
			else if(c=='-')
			  a[++i]=-count;
			else if(c=='*')
			  a[i]=a[i]*count;
			else if(c=='/')
			  a[i]=a[i]/(count*(1.0));
		}
		if(sign)
		  break;
		sum=0;
		for(j=0;j<=i;j++)
		  sum+=a[j];
		printf(%0.2lf
,sum);
	}
	return 0;
}

 

c++ STL——stack解法：

 

 

#include
#include
using namespace std;
int main()
{
	int i,sign,j;
	double sum,num;
	char c;
	while(1)
	{
		sign=1;
		stackcount;
		scanf(%lf,&num);
		count.push(num);
		while(getchar()!='
')
		{
			sign=0;
			scanf(%c %lf,&c,&num);
			if(c=='+')
			   count.push(num);
			else if(c=='-')
			   count.push(-num);
			else if(c=='*')
			{
				double x=count.top();
				x=x*num;
				count.pop();
				count.push(x);
			}
			else
			{
				double x=count.top();
				x=x/num;
				count.pop();
				count.push(x);
			}
		}
		if(sign)
		   break;
		sum=0;
		while(!count.empty())
		{
			sum+=count.top();
			count.pop();
		}
		printf(%.2lf
,sum);
	}
	return 0;
}