10298 - Power Strings
Moderator: Board moderators
10298
who can me why time limiit exceed!!!
#include<stdio.h>
#include<string.h>
void main()
{
char ch[1000001];
int in,len,i,j,sum,sum1,ans,count,h;
while(scanf("%s",ch)==1)
{
h=count=sum=sum1=0;
len=strlen(ch);
if(ch[0]=='.')
break;
for(i=1;i<=len;i++)
{
h=sum1=count=0;
sum+=ch[i-1]-'0';
j=1;
for(j;j<=len;)
{
while((count++)<i&&(j<=len))
{
if(ch[count-1]==ch[j-1])
sum1+=ch[j-1]-'0';
j++;
}
if(sum==sum1)
{
count=sum1=0;
ans=i;
h++;
}
else
break;
}
if((len-(h*i))==0)
{
printf("%d\n",h);
break;
}
}
}
}
#include<stdio.h>
#include<string.h>
void main()
{
char ch[1000001];
int in,len,i,j,sum,sum1,ans,count,h;
while(scanf("%s",ch)==1)
{
h=count=sum=sum1=0;
len=strlen(ch);
if(ch[0]=='.')
break;
for(i=1;i<=len;i++)
{
h=sum1=count=0;
sum+=ch[i-1]-'0';
j=1;
for(j;j<=len;)
{
while((count++)<i&&(j<=len))
{
if(ch[count-1]==ch[j-1])
sum1+=ch[j-1]-'0';
j++;
}
if(sum==sum1)
{
count=sum1=0;
ans=i;
h++;
}
else
break;
}
if((len-(h*i))==0)
{
printf("%d\n",h);
break;
}
}
}
}
my name is Lon
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- Learning poster
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- Joined: Sun Nov 30, 2008 5:00 pm
- Location: IUT-OIC, Dhaka, Bangladesh
Re: 10298 - Power Strings
Can anyone tell me that, is it possible to pass the time limit using brute force in this problem? I tried brute force but got TLE. Or is there any sort of algorithm for this type of problems? Please someone reply.
May be tomorrow is a better day............ ![:)](./images/smilies/icon_smile.gif)
![:)](./images/smilies/icon_smile.gif)
Re: 10298 - Power Strings
There's Knuth-Morris-Pratt algorithm. Find using it position of the first occurrence of s in string s*s.
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- Learning poster
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- Location: IUT-OIC, Dhaka, Bangladesh
Re: 10298 - Power Strings
mf, I have a silly question. What I knew was this algorithm can be used to find the occurence of a pattern in a string. But how can i use this to find a substring in a string when i dont know the substring? Can you explain it more please ![:-?](./images/smilies/icon_confused.gif)
![:-?](./images/smilies/icon_confused.gif)
May be tomorrow is a better day............ ![:)](./images/smilies/icon_smile.gif)
![:)](./images/smilies/icon_smile.gif)
Re: 10298 - Power Strings
The pattern is string s - the input string.
The string to search the pattern in is s*s, where * is concatenation (as in problem statement).
If s occurs at position k (1-based) in s*s, then the answer is length(s) / k.
The string to search the pattern in is s*s, where * is concatenation (as in problem statement).
If s occurs at position k (1-based) in s*s, then the answer is length(s) / k.
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- Learning poster
- Posts: 78
- Joined: Sun Nov 30, 2008 5:00 pm
- Location: IUT-OIC, Dhaka, Bangladesh
Re: 10298 - Power Strings
I have got it, everything is clear. Thanks a lot ![:D](./images/smilies/icon_biggrin.gif)
![:D](./images/smilies/icon_biggrin.gif)
May be tomorrow is a better day............ ![:)](./images/smilies/icon_smile.gif)
![:)](./images/smilies/icon_smile.gif)
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- Learning poster
- Posts: 78
- Joined: Sun Nov 30, 2008 5:00 pm
- Location: IUT-OIC, Dhaka, Bangladesh
Re: 10298 - Power Strings
Thanks mf, got AC. But one more thing. Sorry to bother you, but can you explain how and why this approach works?
May be tomorrow is a better day............ ![:)](./images/smilies/icon_smile.gif)
![:)](./images/smilies/icon_smile.gif)
Re: 10298 - Power Strings
Hmm, isn't it obvious?
Consider string t obtained from s by a cyclic shift of s by k characters right. How to determine for which values of k, the strings t and s are equal? By definition of equality, it's when s = t = s[(i+k)%n], for 0<=i<n. You can get rid of that modulus by simply concatenating s with itself, and it becomes a regular substring search problem, solved by KMP.
And the smallest such cyclic shift k is exactly equal to length of smallest periodic prefix of s. Number of repetitions of this prefix (which is what the problem asks for) is just length of s divided by length of this prefix.
Consider string t obtained from s by a cyclic shift of s by k characters right. How to determine for which values of k, the strings t and s are equal? By definition of equality, it's when s = t = s[(i+k)%n], for 0<=i<n. You can get rid of that modulus by simply concatenating s with itself, and it becomes a regular substring search problem, solved by KMP.
And the smallest such cyclic shift k is exactly equal to length of smallest periodic prefix of s. Number of repetitions of this prefix (which is what the problem asks for) is just length of s divided by length of this prefix.
Re: 10298 - Power Strings
mf wrote:Hmm, isn't it obvious?
Consider string t obtained from s by a cyclic shift of s by k characters right. How to determine for which values of k, the strings t and s are equal? By definition of equality, it's when s = t = s[(i+k)%n], for 0<=i<n. You can get rid of that modulus by simply concatenating s with itself, and it becomes a regular substring search problem, solved by KMP.
And the smallest such cyclic shift k is exactly equal to length of smallest periodic prefix of s. Number of repetitions of this prefix (which is what the problem asks for) is just length of s divided by length of this prefix.
You help me a lot! Thank you very much.
Re: 10298 - Power Strings
I am trying a variation with kmp. I calculate Prefix table and tries to infer length from table directly.
My logic is that eventually prefix table should have a sequence of x1,x2,x3.....xN,1,2,3,4,5,6,7,8.....Y
I tried to detect xN and conclude power of string from that. I am not able to understand what is wrong in this approach.
I have tried many test cases and they seems to work.
abcd
aaaa
ababab
abcabababcababdf
abcabababcabab
.
Can someone give me a critical test case for my approach
My logic is that eventually prefix table should have a sequence of x1,x2,x3.....xN,1,2,3,4,5,6,7,8.....Y
I tried to detect xN and conclude power of string from that. I am not able to understand what is wrong in this approach.
Code: Select all
#include<set>
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstdio>
#include<cstring>
#include<cctype>
#include<string>
#include<map>
#include<sstream>
#include<functional>
using namespace std;
char data[1000005];
int table[1000001];
void myfillTables(int num)
{
memset(table,0,sizeof(table));
table[0] = 0;
table[1] = 0;
int i;
for(int j=2;j<=num; j++)
{
i = table[j-1];
while(1)
{
//cout<<i<<endl;
if(data[i] == data[j-1])
{
table[j] = i+1;
break;
}
else if(i>0)
{
i = table[i];
}
else
{
table[j] = 0;
break;
}
}
}
}
int main()
{
while(gets(data) && !(data[0]=='.' && data[1]=='\0'))
{
int len = strlen(data);
myfillTables(len);
int i, first = len;
for(i=0; i<=len; i++)
{
if(i<len && table[i] == table[i+1]-1 && table[i]>0)
{
first = i-1;
i++;
while(i<=len && table[i]==table[i-1]+1)
{
i++;
}
if(i<=len)
first = len;
}
}
if(len%first==0)
printf("%d\n",len/first);
else
puts("1");
}
return 0;
}
abcd
aaaa
ababab
abcabababcababdf
abcabababcabab
.
Can someone give me a critical test case for my approach
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- Posts: 5947
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- Location: San Jose, CA, USA
Re: 10298 - Power Strings
input:AC output:
Code: Select all
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bhcdarzbhcdarzbhcdarz
kkyhidd
cdx
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xxxxxxx
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Check input and AC output for thousands of problems on uDebug!
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- Location: Santa Clara / Mountain View, CA, USA
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Re: 10298 - Power Strings
I used the same strategy, but still got TLE, I believe my algorithm is O(n):mf wrote:There's Knuth-Morris-Pratt algorithm. Find using it position of the first occurrence of s in string s*s.
Code: Select all
inline int GetLargestN(const char* cs)
{
string s(cs);
string ss(s, 1); ss += s; ss += s[0];
int len = s.length();
int pos = ss.find(s);
return len/(pos+1);
}
Please help.
I Believe I Can - leestime.com
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- New poster
- Posts: 33
- Joined: Tue Apr 27, 2004 7:41 pm
- Location: Santa Clara / Mountain View, CA, USA
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Re: 10298 - Power Strings
OK. I switched to use the native C function strstr, and it got accepted, which does NOT make too much sense!
http://stackoverflow.com/questions/19506571
http://stackoverflow.com/questions/19506571
I Believe I Can - leestime.com
Re: 10298 - Power Strings
I have got a lot of TLE while trying to solve this problem. After trying for a certain period I came up with a HIGHLY modified brute force algorithm to solve this problem. I saw this thread's discussion about the KMP algorithm & tried to learn it. The problem is I don't know how to use this KMP algorithm to solve this problem! Then I went to http://e-maxx.ru/algo/prefix_function
website as I heard it's a good source. The problem is the portion where the article discussed the process called "Compression Line". As the original site is russian & I used google translator to translate it, some portion wasn't finely translated. So, I didn't understand that portion. Could some one please explain how this portion>>//Consider it the last value, ie \ Pi [n-1], and introduce notation k = n - \ pi [n-1]. We show that if ndivisible by k, then it kwill be a long answer, otherwise effective compression does not exist, and the answer is n.//>>from the website is working(i.e. what does that actually mean?!)?
and also I want to know if this problem can be solved using Z-function.
website as I heard it's a good source. The problem is the portion where the article discussed the process called "Compression Line". As the original site is russian & I used google translator to translate it, some portion wasn't finely translated. So, I didn't understand that portion. Could some one please explain how this portion>>//Consider it the last value, ie \ Pi [n-1], and introduce notation k = n - \ pi [n-1]. We show that if ndivisible by k, then it kwill be a long answer, otherwise effective compression does not exist, and the answer is n.//>>from the website is working(i.e. what does that actually mean?!)?
and also I want to know if this problem can be solved using Z-function.