poj 1026 Cipher（置换群循环节）

Cipher
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Time Limit: 1000MS Memory Limit: 10000KTotal 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.

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

104 5 3 7 2 8 1 6 10 91 Hello Bob1995 CERC00

Sample Output

BolHeol  bC RCE

Source

题意：给一个n元置换，给一个字符串（若小于n则用空格补齐），求k次置换后的字符串

输入包含多个块，输出时每个块后加个空行；

注意：不要直接进行k次置换，肯定超时；

先预先处理置换群，找循环节，并储存，然后，对于每个循环节内的字符串，只要进行k%（此循环节的长度）次即可；

代码：

#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 210using namespace std;int  a[N];int m[N][N];///m[i][0]表示第i个循环节的元素的个数，m[i][j]表示第i个循环节内的第j-1个元素； 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);///找寻环节        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;}