Cipher
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 12776 Accepted: 3194
Description
Bob and Alice started to use a brand-new encoding scheme. Surprisingly it is not a Public Key Cryptosystem, but their encoding and decoding is based on secret keys. They chose the secret key at their last meeting in Philadelphia on February 16th, 1996. They chose as a secret key a sequence of n distinct integers, a1 ; . . .; an, greater than zero and less or equal to n. The encoding is based on the following principle. The message is written down below the key, so that characters in the message and numbers in the key are correspondingly aligned. Character in the message at the position i is written in the encoded message at the position ai, where ai is the corresponding number in the key. And then the encoded message is encoded in the same way. This process is repeated k times. After kth encoding they exchange their message.
The length of the message is always less or equal than n. If the message is shorter than n, then spaces are added to the end of the message to get the message with the length n.
Help Alice and Bob and write program which reads the key and then a sequence of pairs consisting of k and message to be encoded k times and produces a list of encoded messages.
Input
The input file consists of several blocks. Each block has a number 0 < n <= 200 in the first line. The next line contains a sequence of n numbers pairwise distinct and each greater than zero and less or equal than n. Next lines contain integer number k and one message of ascii characters separated by one space. The lines are ended with eol, this eol does not belong to the message. The block ends with the separate line with the number 0. After the last block there is in separate line the number 0.
Output
Output is divided into blocks corresponding to the input blocks. Each block contains the encoded input messages in the same order as in input file. Each encoded message in the output file has the lenght n. After each block there is one empty line.
Sample Input
10
4 5 3 7 2 8 1 6 10 9
1 Hello Bob
1995 CERC
0
0
Sample Output
BolHeol b
C RCE
Source
Central Europe 1995
給定1~n的置換F,求其變換m次的變換F^m.
先找到循環節,再用m對循環節的長度取模即可.
#include <iostream>
using namespace std;
int main()
{
const int MAX=300;//最大長度
char str[MAX];//讀入串
int n;//變換的長度
int data[MAX]={0};//存放原始變換
int used[MAX]={0};//標志數組
int cir[MAX][MAX]={0};//每個循環節的成員
int num[MAX]={0};//循環節對應長度
int cnt=0;//循環節的個數
int time=0;//變換次數
int change[MAX]={0};//原始循環變換time次之后的變換
char res[MAX]={0};//變換之后的字符串
int i,j;
while(cin>>n && n)
{
memset(used,0,sizeof(used));
memset(num,0,sizeof(num));
for(i=1;i<=n;i++)
cin>>data[i];
cnt=0;//計數循環節個數
for(i=1;i<=n;i++)
{
if(used[i]==0)
{
used[i]=1;
int temp=data[i];
cir[cnt][num[cnt]]=temp;
num[cnt]=1;
while(used[temp]==0)//獲得循環節
{
used[temp]=1;
temp=data[temp];
cir[cnt][num[cnt]++]=temp;
}
cnt++;
}
}
while(cin>>time && time)//讀入變換次數
{
memset(res,0,sizeof(res));
memset(str,0,sizeof(str));
gets(str);
int len=strlen(str);
for(i=len;i<=n;i++)//位數不足n,補空格
str[i]=' ';
//獲得變換
for(i=0;i<cnt;i++)
{
for(j=0;j<num[i];j++)
{
change[cir[i][j]]=cir[i][(j+time)%num[i]];
}
}
//對讀入數據變換,獲得結果
for(i=1;i<=n;i++)
{
res[change[i]]=str[i];
}
cout<<res+1<<endl;
}
cout<<endl;
}
return 0;
}