Cipher
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 20821 | Accepted: 5708 |
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
题解：
给你一个位置，再给你字符串，问转换k次后的字串，暴力了下，超时了。。。然后就是用置换群，找出循环节，主要是ans[b[(j+k)%num]]=s[b[j]];
每一个位置变换（k%num）次
大神的解释拿来：
对于置换的数组（我们用next数组代替），里面的每一个元素都会存在于（有且只有）一个置换子群里面。
例如 题中数据: （4 -> 7 -> 1 -> 4）（5 -> 2 -> 3 -> 3） （8 -> 6 -> 8）（10 -> 9 -> 10）。 可找到4个子群。
思路：每个子群都有一定数目的各不相同的元素，用变量t记录子群元素个数，用son[]存储子群元素，就有son[0],son[1]...son[t-1]个元素。
那么我们可以得到如下变换规则-> 新字符数组gain[son[(k + j) % t]] = 原字符数组str[son[j]] (0 <= j <= t-1)。
公式推导过程：一个子群 有t个元素说明它的循环节为t， 经过k次置换 那么肯定就有son[(k + j) % t] = son[j]；
例如 题目数据：4-1 5-1 3-1 7-1 2-1 8-1 1-1 6-1 10-1 9-1 （为了对应字符串的下标自减1） Hello+Bob+ 空格用+表示
拿子群（3 -> 6 -> 0 -> 3【已经减一】）来说，其中t = 3,son[0] = 3, son[1] = 6, son[2] = 0。
当k = 1时， { j = 0,son[(k + j) % t] = 6，gain[6] = str[3] = l
j = 1,son[(k + j) % t] = 0，gain[0] = str[6] = B
j = 2,son[(k + j) % t] = 3，gain[3] = str[0] = H
}
暴力超时：
#include<cstdio> #include<iostream> #include<cmath> #include<algorithm> #include<cstring> #include<queue> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) typedef long long LL; const int MAXN=210; int pos[MAXN]; int main(){ int N; while(~scanf("%d",&N),N){ int k; char s[MAXN],ans[MAXN]; for(int i=0;i<N;i++)SI(pos[i]); while(scanf("%d%*c",&k),k){ mem(s,0); gets(s); while(k--){ for(int i=0;i<N;i++){ if(i>=strlen(s))ans[pos[i]-1]=' '; else ans[pos[i]-1]=s[i]; } ans[N]=' '; strcpy(s,ans); } puts(ans); } } return 0; }
置换群：
#include<cstdio> #include<iostream> #include<cmath> #include<algorithm> #include<cstring> #include<queue> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) typedef long long LL; const int MAXN=210; int next[MAXN]; int vis[MAXN]; char ans[MAXN]; int b[MAXN]; int main(){ int N; while(~scanf("%d",&N),N){ int k; char s[MAXN]; for(int i=0;i<N;i++)SI(next[i]),next[i]--; while(scanf("%d%*c",&k),k){ gets(s); mem(vis,0); for(int i=strlen(s);i<N;i++)s[i]=' '; for(int i=0;i<N;i++){ if(!vis[i]){ int j=i,num=0; while(!vis[j]){ vis[j]=1; b[num++]=j; j=next[j]; } for(j=0;j<num;j++){ ans[b[(j+k)%num]]=s[b[j]]; } } } ans[N]=' '; puts(ans); } puts(""); } return 0; }