Poj 2159 / OpenJudge 2159 Ancient Cipher

时间:2022-04-24 07:19:38

1.链接地址:

http://poj.org/problem?id=2159

http://bailian.openjudge.cn/practice/2159

2.题目:

Ancient Cipher
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 28064   Accepted: 9195

Description

Ancient Roman empire had a strong government system with various departments, including a secret service department. Important documents were sent between provinces and the capital in encrypted form to prevent eavesdropping. The most popular ciphers in those times were so called substitution cipher and permutation cipher.
Substitution cipher changes all occurrences of each letter to some
other letter. Substitutes for all letters must be different. For some
letters substitute letter may coincide with the original letter. For
example, applying substitution cipher that changes all letters from 'A'
to 'Y' to the next ones in the alphabet, and changes 'Z' to 'A', to the
message "VICTORIOUS" one gets the message "WJDUPSJPVT".

Permutation cipher applies some permutation to the letters of the
message. For example, applying the permutation <2, 1, 5, 4, 3, 7, 6,
10, 9, 8> to the message "VICTORIOUS" one gets the message
"IVOTCIRSUO".

It was quickly noticed that being applied separately, both
substitution cipher and permutation cipher were rather weak. But when
being combined, they were strong enough for those times. Thus, the most
important messages were first encrypted using substitution cipher, and
then the result was encrypted using permutation cipher. Encrypting the
message "VICTORIOUS" with the combination of the ciphers described above
one gets the message "JWPUDJSTVP".

Archeologists have recently found the message engraved on a stone
plate. At the first glance it seemed completely meaningless, so it was
suggested that the message was encrypted with some substitution and
permutation ciphers. They have conjectured the possible text of the
original message that was encrypted, and now they want to check their
conjecture. They need a computer program to do it, so you have to write
one.

Input

Input
contains two lines. The first line contains the message engraved on the
plate. Before encrypting, all spaces and punctuation marks were removed,
so the encrypted message contains only capital letters of the English
alphabet. The second line contains the original message that is
conjectured to be encrypted in the message on the first line. It also
contains only capital letters of the English alphabet.

The lengths of both lines of the input are equal and do not exceed 100.

Output

Output
"YES" if the message on the first line of the input file could be the
result of encrypting the message on the second line, or "NO" in the
other case.

Sample Input

JWPUDJSTVP
VICTORIOUS

Sample Output

YES

Source

3.思路:

4.代码:

 #include "stdio.h"
//#include "stdlib.h"
#define N 26
int b[][N];
int Partition(int a[],int p,int r)
{
int tmp = a[p];
while(p<r)
{
while(p<r && a[r]>=tmp) r--;
a[p]=a[r];
while(p<r && a[p]<=tmp) p++;
a[r]=a[p];
}
a[p]=tmp;
return p;
}
void qsort(int a[],int p,int r)
{
int q=;
if(p<r)
{
q=Partition(a,p,r);
qsort(a,p,q-);
qsort(a,q+,r);
}
}
int main()
{
int i,j;
char ch;
for(i=;i<;i++)
{
while((ch=getchar())!='\n') b[i][ch-'A']++;
qsort(b[i],,N-);
}
//for(i=0;i<N;i++) printf("%d ",b[0][i]);
//for(i=0;i<N;i++) printf("%d ",b[1][i]);
for(i=;i<N;i++) if(b[][i]!=b[][i]) break;
if(i>=N) printf("YES\n");
else printf("NO\n");
//system("pause");
return ;
}