| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 19570 | Accepted: 5260 |
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.
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
題意：給一個n元置換，給一個字符串（若小於n則用空格補齊），求k次置換後的字符串
輸入包含多個塊，輸出時每個塊後加個空行；
注意：不要直接進行k次置換，肯定超時；
先預先處理置換羣，找循環節，並儲存，然後，對於每個循環節內的字符串，只要進行k%（此循環節的長度）次即可；
代碼：
#include <iostream>
#include <vector>
#include <queue>
#include <math.h>
#include <set>
#include <map>
#include <stack>
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b)
#define N 210
using namespace std;
int a[N];
int m[N][N];///m[i][0]表示第i個循環節的元素的個數，m[i][j]表示第i個循環節內的第j-1個元素；
int nodecnt;
void translate(char ss[],int n,int k)
{
int i,j,k1;
char s[N];
strcpy(s,ss);
for(i=0; i<nodecnt; i++)
{
int d=k%(m[i][0]);
for(j=0; j<d; j++)
{
for(k1=1; k1<=m[i][0]; k1++)
{
ss[a[m[i][k1]]]=s[m[i][k1]];
}
strcpy(s,ss);
}
}
}
void find_node(int n)
{
bool visit[N];
int i,j;
nodecnt=0;
memset(visit,0,sizeof(visit));
for(i=0; i<n; i++)
{
if(visit[i])
continue;
visit[i]=1;
m[nodecnt][0]=1;
m[nodecnt][1]=i;
for(j=a[i]; j<n; j=a[j])
{
if(j==i)
break;
visit[j]=1;
m[nodecnt][m[nodecnt][0]+1]=j;
m[nodecnt][0]++;
}
nodecnt++;
}
}
int main()
{
char ss[N];
///freopen("output.txt","w",stdout);
int n,k,i,slen,cnt=0;
while(~scanf("%d",&n)&&n)
{
memset(ss,'\0',sizeof(ss));
memset(m,0,sizeof(m));
for(i=0; i<n; i++)
{
scanf("%d",&a[i]);
a[i]-=1;
}
find_node(n);///找尋環節
while(~scanf("%d",&k)&&k)
{
getchar();
gets(ss);
slen=strlen(ss);
while(slen<n)
{
ss[slen++]=' ';
}
translate(ss,slen,k);
printf("%s\n",ss);
memset(ss,'\0',sizeof(ss));
}
printf("\n");
}
return 0;
}