按照對映關係進行加密,解密
阿新 • • 發佈:2018-11-09
應該還有其他彩蛋
1程式碼
/* 字串加密 加密 由S1向S2對映,這個關係由P1儲存,P1(S1[i])-> S2[i], 解密 反向 P2(S2[i])-> S1[i], 利用ASCII碼值,表明對映關係,先進行加密,加密後字串解密 應該還有其他好玩的彩蛋 */ #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> using namespace std; char s1[30]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'}; char s2[30]={'n','g','z','q','t','c','o','b','m','u','h','e','l','k','p','d','a','w','x','f','y','i','v','r','s','j'}; char p1[26],p2[26]; int main(){ char s[1000]; int s_len,i; for(i=0;i<26;i++){ p1[s1[i]-'a']=s2[i]; p2[s2[i]-'a']=s1[i]; } cout<<"請輸入您想加密的字串:"<<endl; cin>>s; s_len=strlen(s); for(i=0;i<s_len;i++){ s[i]=p1[s[i]-'a']; }cout<<"加密後字串:"<<endl<<s<<endl; for(i=0;i<s_len;i++){ s[i]=p2[s[i]-'a']; }cout<<"原字串:"<<endl<<s; return 0; }