HDOJ 1237 简单计算器（堆栈）

简单计算器

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

Total Submission(s): 14833 Accepted Submission(s): 5050



Problem Description
读入一个只包含 +, -, *, / 的非负整数计算表达式，计算该表达式的值。



Input
测试输入包含若干测试用例，每个测试用例占一行，每行不超过200个字符，整数和运算符之间用一个空格分隔。没有非法表达式。当一行中只有0时输入结束，相应的结果不要输出。



Output
对每个测试用例输出1行，即该表达式的值，精确到小数点后2位。



Sample Input

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

Sample Output

3.00
13.36

只用堆栈即可，c代码如下：

#include<cstdio>
#include<cstring>
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()!='\n')
		{
			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\n",sum);
	}
	return 0;
}

c++ STL——stack解法：

#include<cstdio>
#include<stack>
using namespace std;
int main()
{
	int i,sign,j;
	double sum,num;
	char c;
	while(1)
	{
		sign=1;
		stack<double>count;
		scanf("%lf",&num);
		count.push(num);
		while(getchar()!='\n')
		{
			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\n",sum);
	}
	return 0;
}