306 - Cipher

[johnnydog33]

program p306;
type arr=array[1..500] of longint;
var n,i,k,r,j,y:longint;
a:arr;
w,get:array[1..500] of char;
ch:char;
function lcm(a,b:longint):longint;
var t:longint;
begin
if b=0 then lcm:=a else lcm:=lcm(b,a mod b);
end;
function circle(w:arr):longint;
var flag:array[1..500] of boolean;
i,k1,ans,p,e:longint;
begin
fillchar(flag,sizeof(flag),false);
ans:=1;
repeat
k1:=0;
for i:=1 to n do
if not flag then begin
k1:=i;
break;
end;
if k1<>0 then begin
p:=k1;e:=0;flag[p]:=true;
while a[p]<>k1 do begin
inc(e);
p:=a[p];
flag[p]:=true;
end;
e:=e+1;
if ans>e then ans:=ans*e div lcm(ans,e) else ans:=ans*e div lcm(e,ans);
end;
until k1=0;
circle:=ans;
end;
begin
repeat
readln(n);
if n<>0 then begin
fillchar(a,sizeof(a),0);
for i:=1 to n do read(a);
y:=circle(a);
readln;
repeat
read(k);
if k<>0 then begin
read(ch);r:=0;
fillchar(w,sizeof(w),' ');
fillchar(get,sizeof(get),' ');
while not eoln do begin
read(ch);
inc(r);
w[r]:=ch;
end;
readln;
for i:=1 to k mod y do begin
for j:=1 to n do get[j]:=w[j];
for j:=1 to n do w[a[j]]:=get[j];
end;
for i:=1 to n do write(w);
writeln;
end;
until k=0;
readln;
end;
writeln;
until n=0;
end.

[StatujaLeha]

Hi all!
I have got many times WA in this problem, but i don't know where is the mistake Could you give me the test cases please and check up my results.

10
4 5 3 7 2 8 1 6 10 9
1 Hello Bob
2 Hello Bob
3 Hello Bob
4 Hello Bob
5 Hello Bob
6 Hello Bob
0
10
4 3 7 5 6 1 8 9 10 2
1 acm.uva.es
2 acm.uva.es
3 acm.uva.es
4 acm.uva.es
5 acm.uva.es
6 acm.uva.es
7 acm.uva.es
8 acm.uva.es
9 acm.uva.es
10 acm.uva.es
11 acm.uva.es
12 acm.uva.es
0
0

Result:

BolHeol b
lelBo Hob
HolleoB b
BelHo lob
lolBeoH b
Hello Bob

vsca.uma.e
uesva.cma.
..euvascma
aa..uvescm
vmaa.u.esc
ucmva.a.es
.scuvama.e
aes.uvcma.
v.ea.uscma
ua.va.escm
.mauva.esc
acm.uva.es

[StatujaLeha]

I solved my problem and got Accepted! Thanks!

[jaracz]

Hi!

Anyone knows how to speed-up this algo??

Here's my TLE code:

Code: Select all

#include <stdio.h>
#include <string.h>

class asd{
    public:
        char sign;
        int value;
    };    

int main()
{
    int n,k;
    while(scanf("%d",&n)==1 && n)
    {
        int key[n];
        char line[n];
        asd szyfr[n];
        for(int i = 0; i < n; i++)
        scanf("%d",&key[i]);
        while(scanf("%d",&k)==1 && k)
        {
            gets(line);int len = strlen(line);
            if(strlen(line)<n)
            {
                for(int i = strlen(line); i < n; i++)
                line[i] = ' ';
                line[n] = '\0';                
            }
            k %= len;
            for(long x = 0; x < k; x++){
            for(int i = 0; i < n; i++)
            {
                szyfr[i].sign = line[i];
                szyfr[i].value = key[i];
            }
            for(int i = 1; i <= n; i++)
            {
                for(int j = 0; j < n; j++)
                if(szyfr[j].value == i)
                {
                    line[i-1] = szyfr[j].sign;
                    break;
                }    
            }}
            printf("%s\n",line);
            line[0] = '\0';
        }    
        printf("\n");
    }    
    return 0;
}   

Regards!

[shamim]

I am getting WA, could some one post some test cases.
The previous cases could be wrong as it was asked when the author was getting WA.

[mf]

Input:

Code: Select all

10
4 5 3 7 2 8 1 6 10 9
65540 CERC
1 ABCDEFGHIJ
2 ABCDEFGHIJ
3 ABCDEFGHIJ
4 ABCDEFGHIJ
5 ABCDEFGHIJ
6 ABCDEFGHIJ
7 ABCDEFGHIJ
8 ABCDEFGHIJ
9 ABCDEFGHIJ
10 ABCDEFGHIJ
11 ABCDEFGHIJ
12 ABCDEFGHIJ
13 ABCDEFGHIJ
14 ABCDEFGHIJ
15 ABCDEFGHIJ
0
0

My output:

Code: Select all

CER   C   
GECABHDFJI
DBCGEFAHIJ
AECDBHGFJI
GBCAEFDHIJ
DECGBHAFJI
ABCDEFGHIJ
GECABHDFJI
DBCGEFAHIJ
AECDBHGFJI
GBCAEFDHIJ
DECGBHAFJI
ABCDEFGHIJ
GECABHDFJI
DBCGEFAHIJ
AECDBHGFJI

[Jan]

Find the k-th permutation of the values first. The rest is simple.

[jaracz]

you're right:)
wasn't necessery to change string each time
thanks anyway
<peace>

[jaracz]

Actually, your suggestion gives TLE too
there is 2kn operations to get the final sequence and than print them
maybe you did it in better way??
I tried to speed up it several times but it must be k-permutaion, otherwise i get WA;(
anyway here is my code

Code: Select all

#include <stdio.h>
#include <string.h>

int main()
{
    int n,k,counter = 0;char line[201];
    while(scanf("%d",&n)==1 && n)
    {
        if(counter)printf("\n");
        else counter = 1;
        int key[n],seq[n],pom[n];
        for(int i = 0; i < n; i++)scanf("%d",&key[i]);
        while(scanf("%d ",&k)==1 && k)
        {
            gets(line);
            if(strlen(line)<n)
            {
                for(int i = strlen(line); i < n; i++)
                line[i] = ' ';
                line[n] = '\0';                
            }    
            for(int i = 0; i < n; i++)pom[i] = i+1;
            for(int c = 0; c < k; c++)
            {
                for(int i = 0; i < n; i++)seq[key[i]-1] = pom[i];
                for(int i = 0; i < n; i++)pom[i] = seq[i];
            }
            for(int i = 0; i < n; i++)
            printf("%c",line[seq[i]-1]);
            printf("\n");
        }        
    }    
    return 0;
}    

Any other suggestions??

[trulo17]

Code: Select all

#include<cstdio>
#include<cstring>
int main()
{
    int n,i,j,ciclo,t,k;
    int T[210][210];
    int A[210];
    char S[210],S2[210];
    char c;
    for( i = 0; i < 210; ++i )
    {
        T[0][i] = i;
    }
    for( ; ; )
    {
        scanf( "%d", &n );
        if( n == 0 )
        {
            break;
        }
        for( i = 0; i < n; ++i )
        {
            scanf( "%d", &A[i] );
        }
        /*problematic part
        for( i = 1; ; ++i )
        {
            for( j = 0; j < n; ++j )
            {
                T[i][ A[j] - 1 ] = T[ i - 1 ][j];
            }
            for( j = 0; j < n; ++j )
            {
                if( T[i][j] != j )
                {
                    break;
                }
            }
            if( j == n )
            {
                break;
            }
        }*/
        ciclo = i;
        for( ; ; )
        {
            scanf( "%d%c", &k,&c );
            if( k == 0 && c == '\n' )
            {
                break;
            }
            gets( S );
            for( i = strlen( S ); i < n; ++i )
            {
                S[i] = ' ';
            }
            S[i] = '\0';
            t = k % ciclo;
            for( i = 0; i < n; ++i )
            {
                S2[i] = S[ T[t][i] ];
            }
            S2[i] = '\0';
            puts( S2 );
        }
        putchar( '\n' );
    }
}

I can`t see why i am getting tle once and again( because it's supossed cycle <= 200. any suggestions?

[mamun]

Oh! This is irritating! RTE.
From other posts, I think I can take it for sure that the cycle is less than 200. But I'm getting RTE.
I use gets() to read input. So I consider the input is as exactly as in the description, ie. n is given in one line, the n integers in next line, k and a message in each line and a 0 in last line of each block. And there is no blank line. I think I should better post my code. Any help is greatly appreciated.

Code: Select all

Wrong method!

[mamun]

Would anybody give a look at my code above plz?

[mf]

Input:

Code: Select all

28
2 1 4 5 3 7 8 9 10 6 12 13 14 15 16 17 11 19 20 21 22 23 24 25 26 27 28 18
2006 microsoft.com
0
0

Output:

Code: Select all

miroc.soft    com

[mamun]

Thank you mf.

Hmm... Segmentation fault. So it is not that cycle is less than 200 or is it that my coding is faulty? If the first case is true then what should be the idea for solving?

[mf]

I guess, by cycle of a permutation pi, you meant the smallest integer c>0, such that for all x, pi^c (x) = x. This number is usually called order of permutation, and in this problem it could be as large as 24067258815600.

There are several correct ways to solve this problem.

1. You can use the fact, that composition of permutations is associative, and binary exponentiation to get O(n log k) algorithm.

2. Write permutation in cyclic notation (if you don't know what it is, look in google.) If an element x is in a cycle of length c, then pi^k (x) = pi^{k mod c} (x). You can use this to compute pi^k in O(n) time.

For example, in the sample input, you are given a permutation pi=(1 4 7)(2 5)(3)(6 8)(9 10). How do you compute, say pi^1995 (5) from it? Element 5 is in a cycle of length 2, so after every 2 applications of pi it returns to its original place, so pi^1995 (5) = pi^(1995 mod 2) (5) = pi^1 (5) = 2. Hence, after 1995 encodings of CERC, the 5th character (a space) is put to position 2.