1.题目：

Caesar密码作为一种最为古老的对称加密体制，它的基本思想是：通过把字母移动一定的位数来实现加密和解密。明文中的所有字母都在字母表上向后（或向前）按照一个固定数目进行偏移后被替换成密文。例如，当偏移量是3的时候，所有的字母A将被替换成D，B变成E，以此类推X将变成A，Y变成B，Z变成C。由此可见，位数就是Caesar密码加密和解密的密钥。

用Caesar密码对下列明文进行加密。

     “Public-key cryptography, also known as asymmetric cryptography, is a class of  cryptographic protocolsbased on algorithms that require two separatekeys, one of which         is secret (or private) and one of which is public.”加密密钥为k=11；

 对密文” Qbspbz Jhlzhy dhz ivyu pu Yvtl vu 89 vy 80 Qbsf 877 IJ puav aol wylzapnpvbz Qbsphu jshu. “进行解密，解密密钥为k=7

2.要求

大写字母加（解）密后仍为大写字母，小写字母（加）解密后仍为小写字母，数字解密后仍为数字，其他标点符号原样输出。

3.代码

#include<iostream>
#include<cmath>
#include<string>
#include<stdio.h>
using namespace std;

void MC(string &s1, int k1)         //加密
{
		for(unsigned int i=0; i<s1.length(); i++)
		{
			if(97 <= s1[i] && s1[i] <= 122)       //小写 
			{
				s1[i] = ((s1[i] + k1 - 97) % 26) + 97;
	            cout<<s1[i]; 
			}
	        else if(65 <= s1[i] && s1[i] <= 90)      //大写
			{	s1[i] = ((s1[i] + k1 - 65) % 26) + 65;
	            cout<<s1[i];
			}
			else if(48<= s1[i] && s1[i] <= 57)     //数字
			{
				s1[i] = ((s1[i] + k1 - 48) % 10) + 48;
				cout<<s1[i];
			}
			else
				cout<<s1[i];
		}
		cout<<endl;
}

void CM(string &s2, int k2)        //解密
{
		for(unsigned int i=0; i<s2.length(); i++)
		{
			if(97 <= s2[i] && s2[i] <= 122)      //小写 
			{
				s2[i] = ((s2[i] - k2 - 122) % 26) + 122;
	            cout<<s2[i]; 
			}
	        else if(65 <= s2[i] && s2[i] <= 90)         //大写
			{	s2[i] = ((s2[i] - k2 - 90) % 26) + 90;
	            cout<<s2[i];
			}
			else if(48 <= s2[i] && s2[i] <= 57)   //数字
			{
				s2[i] = ((s2[i] - k2 - 57) % 10) + 57;
				cout<<s2[i];
			}
			else
				cout<<s2[i];
		}
		cout<<endl;
}

int main()
{ 
	string s1 = "“Public-key cryptography, also known as asymmetric cryptography, is a class of cryptographic protocols based on algorithms that require two separatekeys, one of which is secret (or private) and one of which is public.”";
	string s2 = "“Qbspbz Jhlzhy dhz ivyu pu Yvtl vu 89 vy 80 Qbsf 877 IJ puav aol wylzapnpvbz Qbsphu jshu.”";
	int k1, k2;     //密钥
	cin>>k1>>k2;
	cout<<"加密后的密文为："<<endl;
	MC(s1, k1);
	cout<<endl;
	cout<<"解密后的明文为："<<endl;
	CM(s2, k2);
	return 0;
}

4.结果

5.总结

（1）不能用cin输入带有空格的字符串。

（2）’\n’和endl在C++里面具有同样的换行效果。

（3）也可以不把明文和密文提前输入，可先定义字符串s1,s2,再使用getline(cin,s1)输入字符串。也可使用循环语句，选择是加密还是解密。