Playfair is a kind of substitution cipher.And the encryption role is simple.In general,there are three steps in Playfair algorithm.
Step 1: Creat a 5*5 secret key table.
Step 2:Tidy plaintext : only leave letters and omit others And if there are capital letters in the plaintext,change it into lower letters.
Step 3: Work out the ciphertext.
Now, let’s see an example.
Firstly,we should creat a 5*5 secret key table.Let’s assume that “ulysses” is the key.If there is a letter appearing more than once we should only retain the first one and delete the others.So we can get the real key “ulyse”.Then we will fill the secret key table with the key “ulyse” and with the other letters which haven’t appeared in the key “ulyse” in order(assume j is equal to i,so every element in the table is unique and j won’t be in this table).Finally, we will get a secret key table as follows:
Secondly,assuming that we will send a important message “balloon”.We should divide ballon into a few groups with exactly two letters.But if the two letters are the same,a single x will be inserted between the same two letters.So “balloon” will be divided into “ba” “lx” “lo” “on”.
Finally,we will encrypt the message.The substitution rules are followed:
1.If the group’s two letters are in the same row in the key table,each letter will be changed into the letter which is just on the right of it.(the first column is seen as on the right of the fifth column).
2.If the group’s two letters are in the same column in the key table,each letter will be changed into the letter which is adjacently below it.(the first row is seen as adjacent below fifth column).
3.If the group’s two letters are neither in the same row nor in the same column in the key table,each letter will be changed into the letter whose position is the same row as itself and the same column as another letter.
So “balloon” will be changed into “cbsvbvpo”.
Ps: j in the plaintext is equal to i and all the letter both in plaintext and ciphertext should be lower letter.And if there are capital letters in the plaintext,change it into lower letters.
Input
The first line will contain the secret key(no more than 25 letters). Then you will get a article(including capital letter and the lower letter).You should omit all the spacing , the punctuation and the numbers(maybe has two or more rows).If there are odd letters after tidying plaintext,delete the last letter.The article would end with a ‘*’.
Ps:You can assume that the article is no more than 10000 letters.And “xx” won’t exist. And the secret key only contains lower letters.
Output
The ciphertext which only includes lower letters.
Sample input and output
| Sample Input | Sample Output |
|---|---|
cipher balloon * |
dbspgsug |
cipher fill book* |
aespslsvqg |
Source
#include <iostream> #include<cstdio> #include<cstring> using namespace std; int len1,i,j,k,x1,x2,y1,y2; int f[10005],mp[6][6],a[10005]; char ch1[10005]; int xt(char s) { if ((int)s>=97) return (int)s-96; return (int)s-64; } void solve() { int i,j,f[30]; len1=strlen(ch1); j=0; i=1; memset(f,0,sizeof(f)); for(int ii=0; ii<len1; ii++) { int k=xt(ch1[ii]); if (!f[k]) { if (k==10 || k==9) if (f[9]) continue; else k=9; j++; f[k]=1; mp[i][j]=k; if (j==5) { j=0; i++; } } } for(int ii=1; ii<=26; ii++) if (!f[ii]) { if (ii==10) continue; j++; mp[i][j]=ii; if (j==5) { j=0; i++; } } return; } void findxy(int *x,int *y,int z) { int i,j; for(i=1; i<=5; i++) for(j=1; j<=5; j++) if (mp[i][j]!=z) continue; else { *x=i; *y=j; return; } } int main() { freopen("in.txt", "r", stdin); while(~scanf("%s",&ch1)) { solve(); memset(a,0,sizeof(a)); while (~scanf("%s",&ch1)) { len1=strlen(ch1); for(i=0; i<len1; i++) { if (ch1[i]=='*') goto mm; if(!(ch1[i]>='a' && ch1[i]<='z' || ch1[i]>='A' && ch1[i]<='Z')) continue; if (ch1[i]=='j' || ch1[i]=='J') ch1[i]='i'; a[++a[0]]=xt(ch1[i]); } } mm: k=1; while(k<a[0]) { findxy(&x1,&y1,a[k]); if (a[k+1]==a[k]) findxy(&x2,&y2,24); else { k++; findxy(&x2,&y2,a[k]); } if (x1==x2 && y1!=y2) printf("%c%c",(char)(96+mp[x1][y1%5+1]),(char)(96+mp[x2][y2%5+1]) ); if (x1!=x2 && y1==y2) printf("%c%c",(char)(96+mp[x1%5+1][y1]),(char)(96+mp[x2%5+1][y2]) ); if (x1!=x2 && y1!=y2) printf("%c%c",(char)(96+mp[x1][y2]),(char)(96+mp[x2][y1]) ); k++; } printf(" "); } return 0; }
#include <cstdio> #include <iostream> #include <string> #include <sstream> #include <cstring> #include <stack> #include <queue> #include <algorithm> #include <cmath> #include <map> #define PI 3.1415926 #define ms(a) memset(a,0,sizeof(a)) #define msp memset(mp,0,sizeof(mp)) #define msv memset(vis,0,sizeof(vis)) using namespace std; #define LOCAL char mp[10][10]; int cha[30]; string key,line,art; struct Node { char a,b; }; void mpfill(int idx,char c)//填密码表 { int i=idx/5,j=idx%5; mp[i][j]=c; return; } void mpmake()//生成密码表 { ms(cha),msp,cha[9]=1; int idx=0; string::iterator it; for(it=key.begin(); it!=key.end(); it++) { if(isalpha(*it)==0)continue; *it=tolower(*it); if(cha[*it-'a']==0) { mpfill(idx++,*it); cha[*it-'a']=1; } } for(int i=0; i<26; i++) { if(cha[i]==0) { mpfill(idx++,i+'a'); cha[i]=1; } } } void scanart() { line.clear(),art.clear(); while(cin>>line) { string::iterator it; for(it=line.begin(); it!=line.end(); it++) { if(*it=='*')return; if(isalpha(*it)) { if(tolower(*it)=='j') art.push_back('i'); else art.push_back(tolower(*it)); } } } } void mpfind(int i1,int j1,int i2,int j2) { if(i1==i2)//同行 { j1=(j1+1)%5,j2=(j2+1)%5; printf("%c%c",mp[i1][j1],mp[i2][j2]); return; } if(j1==j2)//同列 { i1=(i1+1)%5,i2=(i2+1)%5; printf("%c%c",mp[i1][j1],mp[i2][j2]); return; } //不同行不同列 printf("%c%c",mp[i1][j2],mp[i2][j1]); } void scanmp(char c1,char c2) { int i1,i2,j1,j2; for(int i=0;i<25;i++) { int x=i/5,y=i%5; if(mp[x][y]==c1)i1=x,j1=y; if(mp[x][y]==c2)i2=x,j2=y; } mpfind(i1,j1,i2,j2); } void change() { string::iterator it; for(it=art.begin(); it!=art.end(); it++) { Node t; if(it+1==art.end())break; if(*it!=*(it+1))t.a=*it,t.b=*(it+1),it++; else t.a=*it,t.b='x'; scanmp(t.a,t.b); } printf(" "); return; } int main() { #ifdef LOCAL freopen("in.txt", "r", stdin); #endif // LOCAL ios::sync_with_stdio(false); while(cin>>key) { mpmake();//密码表 scanart();//读取明文 change();//转换密文 } return 0; }