HDOJ 1237 简单计算器（堆栈）

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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 + 24 + 2 * 5 - 7 / 110

Sample Output

3.0013.36

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;}

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