网站设计属于什么分类号/说到很多seo人员都转行了

/*===============================================*\
**	设计目的：简单计算器，计算形如10*(20.2-30.6)+5.0/2的表达式值
**	简要原理：中缀式转逆波兰式（后缀式）
**	IDE：Dev-Cpp 4.9.9.2
**	注意事项：括号必须是英文状态的 
**	时间： 2014-6-17
\*===============================================*/#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define maxn 1000
char buf[maxn], str[maxn], signStack[maxn], ch[2];
int len, id, idSign, idAns, i, n;
double ans[maxn];void checkSign(char sign){if(sign == '(') signStack[idSign++] = sign;else if(sign == '*' || sign == '/'){while(idSign && (signStack[idSign-1] == '*' || signStack[idSign-1] == '/'))str[id++] = signStack[--idSign];signStack[idSign++] = sign;}else if(sign == ')'){while(signStack[idSign-1] != '(') str[id++] = signStack[--idSign];--idSign;}else{while(idSign && signStack[idSign-1] != '(') str[id++] = signStack[--idSign];signStack[idSign++] = sign;}str[id++] = ' ';
}int check(double a, char sign){if(a == 0 && sign == '/'){printf("除数不能为0!，程序结束。\n");system("pause");exit(EXIT_FAILURE);}return 1;
}double cal(double a, double b, char sign){switch(sign){case '+': return a + b;case '-': return a - b;case '*': return a * b;case '/': return a / b;}
}int main(){	printf("\t\t\t\t简单计算器\n\n");printf("\t\t   ======================================\n");printf("\t\t  ||  支持+ - * / ( ) 等符号和负数输入  ||\n");printf("\t\t   ======================================\n");printf("\t\t\t\t\t\t原创作者：邱正钢\n");printf("\t\t\t\t\t\t完毕时间：2014年6月17日\n");printf("\t\t\t\t\t\tEmail：2276479303@qq.com\n");printf("请输入一个表达式，如 -10*(20.2-30.6)+5.0/2\n以回车结束:\n");	do{gets(buf); id = idSign = idAns = 0;if((len = strlen(buf)) == 0) continue;int flag = 0; //这个是用来推断'-'字符是负号还是减号的。1表示数字，0表示 +*/(/*推断原理：假设第一次读取就碰到-，那么一定是负号，若近期一次读取是+*-/(那么也一定是负号，假设是右括号或数字那么是减号*/for(i = 0; i < len; ++i){if(buf[i] == ' ') continue;if(buf[i] >= '0' && buf[i] <= '9' || buf[i] == '.' || buf[i] == '-' && !flag){str[id++] = buf[i]; flag = 1;}else { str[id++] = ' '; checkSign(buf[i]); if(buf[i] != ')') flag = 0;else flag = 1;}}while(idSign) str[id++] = signStack[--idSign];for(i = 0, n = 0; i < id; ++i){if(str[i] == ' ') continue;/*用添加空格的方式推断‘-’字符是负号还是减号，若‘-’后面挨着数字或小数点，那么一定是负号*/if(str[i] >= '0' && str[i] <= '9' || str[i] == '.' || str[i] == '-' && (str[i+1] >= '0' && str[i+1] <= '9' || str[i+1] == '.')){sscanf(str + i, "%lf%n", &ans[idAns++], &n);i += n - 1;  continue;}else if(check(ans[idAns-1], str[i])){ans[idAns-2] = cal(ans[idAns-2], ans[idAns-1], str[i]);--idAns; }}printf("结果是 %.2lf\n", ans[0]);printf("输入Y 继续, N 退出: ");scanf("%s", ch);}while(ch[0] == 'Y' || ch[0] == 'y');printf("感谢您的使用！再见\n");system("pause");return 0;
}