POJ 2930 加减乘除

时间限制: 

1000ms

内存限制:  

65536kB

描述

根据输入的运算符对输入的整数进行简单的整数运算。运算符只会是加+、减-、乘*、除/、求余%、阶乘！六个运算符之一。输出运算的结果，如果出现除数为零，则输出“error”,如果求余运算的第二个运算数为0，也输出“error”。

输入

输入为一行。先输入第一个整数，空格输入运算符，然后再空格输入第二个整数，回车结束本次输入。如果运算符为阶乘！符号，则不输入第二个整数，直接回车结束本次输入。

输出

输出为一行。输出对输入的两个（或一个）数，根据输入的运算符计算的结果，或者“error”。

样例输入

12 + 3454 – 253 * 645 / 05 !34 % 0

样例输出

462918error120error

 

（1）、源代码：

#include  

using   namespace   std;

 

int   main(){

                 int   i, n, a, b, sum;

                 char   c;

                

                 while (cin >> a >> c){

                                 switch (c){

                                 case   '*' :cin >> b;

                                                cout << a * b << endl;

                                                 break ;

                                 case   '/' :cin >> b;

                                                 if (0 == b)

                                                                cout <<   "error\n" ;

                                                 else

                                                                cout << a/b << endl;

                                                 break ;

                                 case   '%' :cin >> b;

                                                 if (0 == b)

                                                                cout <<   "error\n" ;

                                                 else

                                                                cout << a%b << endl;

                                                 break ;

                                 case   '+' :cin >> b;

                                                cout << a + b << endl;

                                                 break ;

                                 case   '-' :cin >> b;

                                                cout << a - b << endl;

                                                 break ;

                                 case   '!' :sum = 1;

                                                 for (i = 2; i <= a; i++)

                                                                                 sum *= i;

                                                cout << sum << endl;

                                                 break ;

                                }

                }

      return   0;

}

  

（2）、解题思路：略

（3）、可能出错：略

 

 

转载于:https://www.cnblogs.com/lydf-2012/archive/2012/05/11/2496622.html