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 <iostream>
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
 
 
posted on 2012-05-11 20:11  谷堆旁边  阅读(418)  评论(0编辑  收藏  举报