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

只用堆疊即可，c程式碼如下：

#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;
}