对一串数字加密和解密！（费了心血的，不顶勿看！）

using System;
using System.Threading;

class 加密
{
static void Main()
{
Console.WriteLine("请输入要加密的四位整数：");
int i=int.Parse(Console.ReadLine());
// 假设4395 /为除法，%求余！
int first = i/1000; // 整除得4 ，第一位4
int second = (i/100)%10; //43%10余3，第二位3
int third = (i/10)%10; //439%10余9，第3位9
int forth = i%10; //4395除10余5，第4位5

Console.WriteLine(first+" "+second+" "+third+" "+forth);

first = (first+7)%10; //加上7然后再%10求余
second =(second+7)%10;
third = (third+7)%10;
forth = (forth+7)%10;

int temp=third; //将第3位和第1位换位
third=first ;
first = temp;

temp= forth; //将第2位和第4位换位
forth= second;
second = temp;

int pass = first*1000+second*100+third*10+forth;

Console.WriteLine("加密后数变成："+pass); //得到的结果是6210

Thread.Sleep(5000);
}
}

using System;
using System.Threading;
class 破解
{
static void Main()
{
Console.WriteLine("请输入你加密过的四个数：");//输入上次加密后的结果6210
int i=int.Parse(Console.ReadLine());
int first = i/1000; // 整除得6 ，第一位6
int second = (i/100)%10;//62%10余2，第二位2
int third = (i/10)%10; //621%10余1，第3位1
int forth = i%10; //6210除10余0，第4位0

Console.WriteLine(first+" "+second+" "+third+" "+forth);

int temp=third; //将第3位和第1位换位
third=first ;
first = temp;

temp= forth; //将第2位和第4位换位
forth= second;
second = temp;

first = (first+10)-7; //此处不能直接加3，因为有可能会超过10！
second =(second+10)-7;
third = (third+10)-7;
forth = (forth+10)-7;

int pass = first*1000+second*100+third*10+forth;

Console.WriteLine("破解成功了！原数为："+pass);//得到原来的数4395

Thread.Sleep(5000);
}
}