00062.cpp

通过一些基础的程序

//将A-Z和a-z分别连成两个圆环，将原字符后移四位得到加密算法
//动态创建内存和加密完美结合
#include <iostream>
using namespace std;
void KaiSa(char Point_S[], int k,int size);
void main()
{   int num1,num2;
	char *p;
	cout<<"加密字符的个数:"<<endl;
    cin>>num1;
	p= new char[num1]; 
    cout<<"加密字符:"<<endl;
	for(int j=0;j<num1;j++)
	{ 
 			cin>>p[j];	
	} 
	char S[100];
    for(int k=0;k<num1;k++)
	{
		S[k]=p[k];	
	} 
	cout<<"原字符为:"<<endl; 
    for(int m=0;m<=(num1-1);m++)
		 {
			 cout<<S[m];	
		 }
	cout<<endl; 
	cout<<"字符后移位数:"<<endl;
	cin>>num2;
	KaiSa(S,num2,num1);
	delete [] p;
	}


void KaiSa(char Point_S[], int k,int size)
	{int i;
	 for(i=0;i<=size;i++)
	 { 
		 if((Point_S[i]>='a'&&Point_S[i]<='z')||(Point_S[i]>='A'&&Point_S[i]<='Z'))
			Point_S[i]=(Point_S[i]+k);  //从 97开始为A-Z，共26个字母
		 if((Point_S[i]>='Z'&&Point_S[i]<='Z'+6)||Point_S[i]>'z')
			Point_S[i]=Point_S[i]-26;
	 }
	 cout<<"加密后字符为:"<<endl; 
	 for(i=0;i<=(size-1);i++)
	 {   
		 cout<<Point_S[i];	
	 }
     cout<<endl; 
     
}
/*
测试结果
加密字符的个数:
8
加密字符:
z l j m r q b o
原字符为:
zljmrqbo
字符后移位数:
3
加密后字符为:
computer
Press any key to continue*/