HDOJ 题目1439Cipher(置换群)
2014-09-06 20:32
393 查看
Total Submission(s): 386 Accepted Submission(s): 171
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
10
4 5 3 7 2 8 1 6 10 9
1 Hello Bob
1995 CERC
0
0
[align=left]Sample Output[/align]
BolHeol b
C RCE
[align=left]Source[/align]
ACM暑期集训队练习赛(5)
[align=left]Recommend[/align]
lcy | We have carefully selected several similar problems for you: 1415 1443 1427 1423 1455
给一个序列a,a[ i ]表示当前在i位置,经过一次encode变到a[ i ]的位置。再给一个字符串,s0, s1, s2,...,sn-1(长度不够的用空格
补上)。s[ i ]在a[ i ]位置输出得到一个新的串,重复这个过程k次,求最后的串。
ac代码
#include<stdio.h>
#include<string.h>
int a[300],c[300];
char s[300],ans[300];
int main()
{
int n;
while(scanf("%d",&n)!=EOF,n)
{
int i,temp,k;
memset(c,0,sizeof(c));
//memset(ans,0,sizeof(ans));
memset(a,0,sizeof(a));
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
a[i]--;
}
for(i=0;i<n;i++)//求个个点的循环节
{
if(!c[i])
{
temp=a[i];
c[i]++;
while(temp!=i)
{
c[i]++;
temp=a[temp];
}
}
}
while(scanf("%d",&k)!=EOF,k)
{
int pos,len;
getchar();
gets(s);
memset(ans,0,sizeof(ans));
len=strlen(s);
for(i=len;i<n;i++)
s[i]=' ';
s
='\0';
//k--;
for(i=0;i<n;i++)
{
temp=k%c[i];
pos=i;
while(temp--)
{
pos=a[pos];
}
ans[pos]=s[i];
}
ans
='\0';
//printf("%s\n",ans);
for(i=0;i<n;i++)
printf("%c",ans[i]);
printf("\n");
}
printf("\n");
}
}
Cipher
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 386 Accepted Submission(s): 171
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
10
4 5 3 7 2 8 1 6 10 9
1 Hello Bob
1995 CERC
0
0
[align=left]Sample Output[/align]
BolHeol b
C RCE
[align=left]Source[/align]
ACM暑期集训队练习赛(5)
[align=left]Recommend[/align]
lcy | We have carefully selected several similar problems for you: 1415 1443 1427 1423 1455
给一个序列a,a[ i ]表示当前在i位置,经过一次encode变到a[ i ]的位置。再给一个字符串,s0, s1, s2,...,sn-1(长度不够的用空格
补上)。s[ i ]在a[ i ]位置输出得到一个新的串,重复这个过程k次,求最后的串。
ac代码
#include<stdio.h>
#include<string.h>
int a[300],c[300];
char s[300],ans[300];
int main()
{
int n;
while(scanf("%d",&n)!=EOF,n)
{
int i,temp,k;
memset(c,0,sizeof(c));
//memset(ans,0,sizeof(ans));
memset(a,0,sizeof(a));
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
a[i]--;
}
for(i=0;i<n;i++)//求个个点的循环节
{
if(!c[i])
{
temp=a[i];
c[i]++;
while(temp!=i)
{
c[i]++;
temp=a[temp];
}
}
}
while(scanf("%d",&k)!=EOF,k)
{
int pos,len;
getchar();
gets(s);
memset(ans,0,sizeof(ans));
len=strlen(s);
for(i=len;i<n;i++)
s[i]=' ';
s
='\0';
//k--;
for(i=0;i<n;i++)
{
temp=k%c[i];
pos=i;
while(temp--)
{
pos=a[pos];
}
ans[pos]=s[i];
}
ans
='\0';
//printf("%s\n",ans);
for(i=0;i<n;i++)
printf("%c",ans[i]);
printf("\n");
}
printf("\n");
}
}
相关文章推荐
- 【置换群】 HDOJ 1439 && POJ 1026 Cipher
- hdu 1439 Cipher(置换群)
- 【HDU 1439】Cipher(置换群)
- HDOJ 1439 Cipher
- Num 4: HDOJ: 题目1106 : 排序(以5为分隔符进行分割)
- HDOJ 题目4500 小Q系列故事——屌丝的逆袭(模拟)
- HDOJ 题目1258 Sum It Up(DFS)
- HDOJ 题目1509 Windows Message Queue(priority_queue)
- HDOJ题目3440 House Man(差分约束)
- HDOJ 题目4512 吉哥系列故事——完美队形I(LICS变形)
- HDOJ 题目1798 Tell me the area
- HDOJ 题目1465 不容易系列一 (全错排)
- HDOJ 题目2603 Wiskey's Power(数学)
- HDOJ 题目4339 Query(线段树,单点更新)
- HDOJ题目2516 取石子小游戏(博弈)
- HDOJ题目分类
- HDOJ 题目1466 计算直线的交点数(动态规划)
- HDOJ 题目1527 取石子游戏(威佐夫博奕)
- HDOJ 题目1532 Drainage Ditches(最大流)
- HDOJ 题目2191 悼念512汶川大地震遇难同胞——珍惜现在,感恩生活(多重背包问题)