11713 - Abstract Names

[Mohsin Reza Razib]

I am continuously getting WA for this. Here is my code:

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Accepted removed

Can anyone give any critical output or hint on this. And what should be output for this:
5
pele
pele
jkl
jkl


dfg
dag
bad
bcd
My output are:
No
No
No
No
No
And can anybody explain the problem description:"a name which can be obtained by replacing zero or more vowels by other vowels to obtain a new name are considered same, provided they have same length". Considering this at first my program gave output "Yes" for 4th input and "No" for 5th input. Looking forward to ur reply.

[r2ro]

My Accepted code outputs the following:

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Yes
Yes
Yes
No
No

[Mohsin Reza Razib]

Thanks Mr. r2ro. I got it accepted. That was so nice of you. I still have complaints about problem specification "a name which can be obtained by replacing zero or more vowels by other vowels to obtain a new name are considered same"
It should be like : "a name which can be obtained by replacing ONE or more vowels by other vowels to obtain a new name are considered same".
Because for input:
dfg
dag
output should be "Yes" if some one follow the problem specification strictly. I got confused there.
Thanks & Regards
Mohsin

[Taman]

I still have complaints about problem specification "a name which can be obtained by replacing zero or more vowels by other vowels to obtain a new name are considered same"
It should be like : "a name which can be obtained by replacing ONE or more vowels by other vowels to obtain a new name are considered same".
Because for input:
dfg
dag
output should be "Yes" if some one follow the problem specification strictly. I got confused there.

@Razib: I do not think so, if anyone follows the problem he will find that 'f' is not a vowel. So, no question arise to replace it with anything!
If the input were:
dug
dfg
then the same thing occurs. 'u' is a vowel, so it can be replaced but 'f' is not a vowel so 'u' can't be replaced with 'f'!so, if you follow the instruction, the output should be "No" in both cases. obviously your statement is a bit more clear .

[Mohsin Reza Razib]

I still have complaints about problem specification "a name which can be obtained by replacing zero or more vowels by other vowels to obtain a new name are considered same"
It should be like : "a name which can be obtained by replacing ONE or more vowels by other vowels to obtain a new name are considered same".
Because for input:
dfg
dag
output should be "Yes" if some one follow the problem specification strictly. I got confused there.

@Razib: I do not think so, if anyone follows the problem he will find that 'f' is not a vowel. So, no question arise to replace it with anything!
If the input were:
dug
dfg
then the same thing occurs. 'u' is a vowel, so it can be replaced but 'f' is not a vowel so 'u' can't be replaced with 'f'!so, if you follow the instruction, the output should be "No" in both cases. obviously your statement is a bit more clear .

But consider the case:
dfg
dug
Here 'f' is not a vowel, so that zero vowel at position 2. As we recall from problem specification "a name which can be obtained by replacing zero or more vowels by other vowels", it should be replaced by a vowel like 'u'. So as i get output should be "Yes". Similarly,
jkl
aeu
output should be: "Yes". But for
jkl
aeb
should be: "No". AS 'l' is not replaced by a vowel. Anyway it over now . But problem specification should be clear.

[sadia_atique]

I'm continusly getting WA in this problem,so frustrating can anyone please help me,where's my fault?or give me some critical input?

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

int vow(char p)
{
    if(p^'a' && p^'e' && p^'i' && p^'o' && p^'u') return 0;
    else return 1;
}
int main()
{
    int t,i,f;
    char s1[22],s2[22];
    scanf("%d",&t);
    getchar();
    while(t--)
    {
              gets(s1);
              gets(s2);
              f=1;
              if(strlen(s1)==0 && strlen(s2)==0)
              {
                            printf("Yes\n");
                            continue;
                            }
              if(!strcmp(s1,s2))
              {
                                printf("Yes\n");
                                continue;
                                }
              if(strlen(s1)!=strlen(s2))
              {
                                printf("No\n");
                                continue;
                                }      
              for(i=0; i<strlen(s1); i++)
              {
                       if(vow(s1[i]) && !vow(s2[i]))
                       {
                                    printf("No\n");
                                    f=0;
                                    break;
                                    }
                       else if(vow(s2[i]) && !vow(s1[i]))
                       {
                            printf("No\n");
                            f=0;
                            break;
                            }
                       }
              if(f) printf("Yes\n");
              }
              
    return 0;
}

[uvasarker]

Apo, Kemon achen? I hear your name from my university's elder brother. Actually I am a new coder I joined Uva at 06-06-2011. So, I can not understand well of your code. So, I give my simple AC code of 11713 and my rank is 170 and Time is 0.004. I will delete my code after your response.......................Shahadat

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I removed my AC code. I understand your code. So, I have given some test case below. Try and get AC, Apo.

[uvasarker]

Sample Input

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6
zxv
zxv
pala
palah
palah
pala
pala
pale
nala
pala
nala
pala

Your program

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Yes
No
No
Yes
Yes
Yes

Apo, last two is incorrect because the first character is 'n' and 'p'.
So Correct Output Is

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Yes
No
No
Yes
No
No

Remove This part of your program
if(strlen(s1)==0 && strlen(s2)==0)
{
printf("Yes\n"); // don't print Yes
continue; // don't continue
}
that means if null string or character is given it's nothing that means no '\n' . But, in your program when I input 2 (number of test cases) and press Enter four times your program is terminated, but, never terminated in that case. When strlen is zero it's also nothing don't print 'Yes' .

Good Luck....................Shahadat

[sadia_atique]

Thanks a lotttt ..n sorry I coudn't browse earlier..stupid mistake that forgot to check..

[Evan72]

getting WA..plz help figure out the problem..

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

int main()
{
char real[21], game[21];
int length, i, tst;
scanf("%d", &tst);
while(tst--)
{
scanf("%s", real);
scanf("%s", game);
if(strlen(game) != strlen(real))
{
printf("No\n");
continue;
}
else
{
length = strlen(game);
for(i = 0; i < length; i++)
{
if(game == real)
continue;
else if(game == 'a' || game == 'e' || game == 'i' || game == 'o' || game == 'u')
continue;
else
break;
}
}
if(i == length)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}

[brianfry713]

Check if both game and real are vowels if they differ. For input

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2
a
b
b
a

AC output is:

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No
No

[Evan72]

Thanx bro..got AC

[hello]

Clearly the data set have a great lacking.
I solve this problem in this way

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#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<algorithm>
#include<set>
#include<queue>
#include<stack>
#include<list>
#include<iostream>
#include<fstream>
#include<numeric>
#include<string>
#include<vector>
#include<cstring>
#include<map>
#include<iterator>
#define LLU long long unsigned int
#define LLD long long double
#define FOR(i,N) for(int i=0;i<(N);i++)
#define Vec vector<int>
#define Vit vector<int>::iterator
#define PI acos(-1)
//define ULLD unsigned __int64 u64
using namespace std;

int main()
{

    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    string s;
    string a;
    int cases;
    cin>>cases;
    while(cases--)
    {

        cin>>s;
        cin>>a;
        int flag=1;
        int l1=s.size();
        int l2=a.size();
        if(l1!=l2)cout<<"No"<<endl;
        else
        {
            for(int i=0; s[i]; i++)
            {
                if(a[i]==s[i])flag++;
                else if((s[i]=='a' || s[i]=='e' || s[i]=='i' || s[i]=='o' || s[i]=='u') && (a[i]=='a' || a[i]=='e' || a[i]=='i' || a[i]=='o' || a[i]=='u')){
                        flag++;
                }
                else {
                    flag=0;
                    cout<<"No"<<endl;
                    break;
                }
            }
            if(flag!=0)cout<<"Yes"<<endl;
        }
    }


    return 0;
}

later on viewing some thread in this forum I take one of my friend's code & post that. Which is in different logic.

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#include <stdio.h>
#include <string.h>
char str[30],str1[30];
int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%s",str);
        scanf("%s",str1);
        int i=strlen(str);
        int j=strlen(str1);
        if(i!=j)
            printf("No\n");
        else
        {
            int c=0;
            for(i=0; i<j; i++)
            {
                if(str[i]==str1[i])
                {
                    c++;
                }
                else if(str[i]=='a'|| str[i]=='e'||str[i]=='i'||str[i]=='o'||str[i]=='u')
                {
                    c++;
                }
                else
                {
                    c=100;
                }
            }
            if(c>=100)
                printf("No\n");
            else
                printf("Yes\n");
        }
    }
    return 0;
}

The second code not even match the i/o given by brianfry713. Just look for the comment from the GURU @ brianfry713

[uDebug]

brianfry713, uvasarker,

Thanks very much for the test input / output.