10252  Common Permutation
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 New poster
 Posts: 30
 Joined: Wed Oct 23, 2002 6:53 am
10252  Common permutations
Why WA, so weird, please help me with this problem.
Here is my code:
If anyone knows why i got WA, please help me
Here is my code:
Code: Select all
/*
Common Permutation, Uva 10252
Nephtali Garrido
*/
#include <cstdio>
#include <iostream>
using namespace std;
char v1[1000],v2[1000],x;
int cuan1[29],cuan2[29],c,c2;
int main()
{ while(cin>>v1>>v2)
{ for(c=0;c<29;c++)
cuan1[c]=cuan2[c]=0;
for(c=0;v1[c]!='\0';c++)
cuan1[v1[c]'a']++;
for(c=0;v2[c]!='\0';c++)
cuan2[v2[c]'a']++;
for(c=0;c<29;c++)
{ x=c+'a';
if(cuan1[c]>=cuan2[c])
for(c2=0;c2<cuan2[c];c2++)
printf("%c",x);
else
for(c2=0;c2<cuan1[c];c2++)
printf("%c",x);
}
printf("\n");
}
return 0;
}
not sure if this will help, but i solved this problem a long time ago ... you can look at my source code if you want, it's at http://itnoise.com, you have to register though to see the solution repository. There are quite a few solutions there, and you can upload some of yours too if you like, it's a resource for students who are practising to participate in programming competitions.
cheers
cheers

 New poster
 Posts: 2
 Joined: Tue Aug 01, 2006 6:34 am
I have used "division by zero" method, and I can conclude that the strings a and b contain ONLY LOWERCASE character. Also, there can be blank line as a or b.
However, after making corresponding changes to my program, I am still getting WA. I use the method of counting frequency, stated by IIUC GOLD.
I just wonder is the algorithm wrong, or there is something really tricky.
However, after making corresponding changes to my program, I am still getting WA. I use the method of counting frequency, stated by IIUC GOLD.
I just wonder is the algorithm wrong, or there is something really tricky.

 New poster
 Posts: 23
 Joined: Tue Sep 12, 2006 9:46 pm
getting WA for 10252! PLZZZZZZZZ help
I have chked all the test cases from previous discussions
Also add the capital letter case & blank space
but still getting WA. plz HELP!
Thnx in Advance!
Also add the capital letter case & blank space
but still getting WA. plz HELP!
Code: Select all
removed after AC!
Last edited by dust_cover on Mon Oct 23, 2006 10:03 pm, edited 1 time in total.
i wanna give it a try....

 New poster
 Posts: 23
 Joined: Tue Sep 12, 2006 9:46 pm
hi corrected it. But still WA
input
output
can someone ac telle me whether my output is right?
input
Code: Select all
pretty woman
walking down
qwe
pretty
women
Code: Select all
anow
e
i wanna give it a try....

 New poster
 Posts: 23
 Joined: Tue Sep 12, 2006 9:46 pm
10252 (TLE) Help me!!
Who can give advice to me?

/*
No. 10252 Common Permutation
*/
#include <iostream>
#include <string>
using namespace std;
void quick_sort (char a[], int n)
{
char v, t;
int i, j;
if (n > 1)
{
v = a[n1];
i = 1;
j = n1;
while(1)
{
while(a[++i] < v);
while(a[j] > v);
if (i >= j) break;
t = a;
a = a[j];
a[j] = t;
}
t = a;
a = a[n1];
a[n1] = t;
quick_sort(a, i);
quick_sort(a+i+1, ni1);
}
}
int main()
{
char input1[1001], input2[1001], result[1001];
int inputLen1, inputLen2, resultLen, k, L, i, j;
while(!cin.eof())
{
k = 0;
// Array Clear
for (L = 0; L < 1001; L++)
input1[L] = 0;
for (L = 0; L < 1001; L++)
input2[L] = 0;
for (L = 0; L < 1001; L++)
result[L] = 0;
// String Input
cin.getline(input1, 1000);
cin.getline(input2, 1000);
// String Length Check
inputLen1 = strlen(input1);
inputLen2 = strlen(input2);
for (i = 0; i < inputLen1; i++)
{
for(j = 0; j < inputLen2; j++)
{
if (result[k] == 0 && input1 == input2[j])
{
result[k] = input1;
input2[j] = NULL;
j = inputLen2;
}
}
if (result[k] != 0)
k++;
}
quick_sort(result, k);
for(int r = 0; r < k; r++)
{
cout << result[r];
}
cout << endl;
}
return 0;
}

/*
No. 10252 Common Permutation
*/
#include <iostream>
#include <string>
using namespace std;
void quick_sort (char a[], int n)
{
char v, t;
int i, j;
if (n > 1)
{
v = a[n1];
i = 1;
j = n1;
while(1)
{
while(a[++i] < v);
while(a[j] > v);
if (i >= j) break;
t = a;
a = a[j];
a[j] = t;
}
t = a;
a = a[n1];
a[n1] = t;
quick_sort(a, i);
quick_sort(a+i+1, ni1);
}
}
int main()
{
char input1[1001], input2[1001], result[1001];
int inputLen1, inputLen2, resultLen, k, L, i, j;
while(!cin.eof())
{
k = 0;
// Array Clear
for (L = 0; L < 1001; L++)
input1[L] = 0;
for (L = 0; L < 1001; L++)
input2[L] = 0;
for (L = 0; L < 1001; L++)
result[L] = 0;
// String Input
cin.getline(input1, 1000);
cin.getline(input2, 1000);
// String Length Check
inputLen1 = strlen(input1);
inputLen2 = strlen(input2);
for (i = 0; i < inputLen1; i++)
{
for(j = 0; j < inputLen2; j++)
{
if (result[k] == 0 && input1 == input2[j])
{
result[k] = input1;
input2[j] = NULL;
j = inputLen2;
}
}
if (result[k] != 0)
k++;
}
quick_sort(result, k);
for(int r = 0; r < k; r++)
{
cout << result[r];
}
cout << endl;
}
return 0;
}
Dont open a new thread if there is one already. Check the following thread...
http://onlinejudge.uva.es/board/viewto ... ight=10252
http://onlinejudge.uva.es/board/viewto ... ight=10252
Ami ekhono shopno dekhi...
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this problem can't be that difficult
I've tried it hundreds of times but always get WA. I'm getting tired of this silly problem, it's too straightforward to worth doing it. My approach is based on counting the frequencies of characters in the two input strings, handling empty lines too. Here is my short code for this problem:
Thanks in advance!
Code: Select all
#include <stdio.h>
#include <string.h>
#define MAXLEN 1000
char cadena1[MAXLEN+2], cadena2[MAXLEN+2], permutacion[MAXLEN+1];
char letras1[256], letras2[256];
int len1, len2;
int main()
{
int i, j, k;
int min;
while (gets(cadena1) && gets(cadena2)) {
len1 = strlen(cadena1);
len2 = strlen(cadena2);
for (i = 0; i < len1; i++) letras1[cadena1[i]]++;
for (i = 0; i < len2; i++) letras2[cadena2[i]]++;
for (i = 'a', k = 0; i <= 'z'; i++) {
min = (letras1[i] < letras2[i]) ? letras1[i] : letras2[i];
for (j = 0; j < min; j++) {
permutacion[k++] = i;
}
letras1[i] = letras2[i] = 0;
}
permutacion[k] = '\0';
printf("%s\n", permutacion);
}
return 0;
}
Oh my gO_od!
OK, i've just found the subtle mistake that was spoiling my code and hence my results, notice the difference:
Before:
After:
Sometimes i feel a bit embarrassed
Before:
Code: Select all
for (j = 0; j < min; j++)
permutacion[k++] = i;
Code: Select all
for (j = 0; j < min; j++)
permutacion[k++] = (char)i;
i'm getting PE!!! can anyone help?
Code: Select all
Last edited by kana on Wed Sep 12, 2007 9:31 pm, edited 1 time in total.
Read the description again.
So, you dont need this line.
Hope it helps.
So, there can be blank strings.Each string is on a separate line and consists of at most 1000 lowercase letters
Code: Select all
if(l1 == 0)
continue;
Hope it helps.
Ami ekhono shopno dekhi...
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Can it have two answers
Can this problem have two answers or that is just to confuse the extra cautions.If several x satisfy the criteria above, choose the first one in alphabetical order.
Thank You,
HG