Re: [algogeeks] spoj problem
Any faster way to solve this problem?? On 3/12/11, Balaji Ramani rbalaji.psgt...@gmail.com wrote: Yes, the correct answer is 0. if(flag ==0 || ans ==0) i think can be changed to if(flag ==0) alone. Thanks, Balaji. On Sat, Mar 12, 2011 at 10:16 AM, Satyam Kapoor satyamkapoo...@gmail.comwrote: is case main answer 0 hona chahiye. -- Satyam Kapoor B.Tech 2nd year Computer Science And Engineering M.N.N.I.T ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] spoj problem
Find the gcd of n-1 differences ( a[1]..a[n-1] with a[0] ) ans = (a[n-1] - a[0])/gcd - n + 1 Thanks, Balaji. On Sat, Mar 12, 2011 at 2:00 PM, Akshata Sharma akshatasharm...@gmail.comwrote: Any faster way to solve this problem?? On 3/12/11, Balaji Ramani rbalaji.psgt...@gmail.com wrote: Yes, the correct answer is 0. if(flag ==0 || ans ==0) i think can be changed to if(flag ==0) alone. Thanks, Balaji. On Sat, Mar 12, 2011 at 10:16 AM, Satyam Kapoor satyamkapoo...@gmail.comwrote: is case main answer 0 hona chahiye. -- Satyam Kapoor B.Tech 2nd year Computer Science And Engineering M.N.N.I.T ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] spoj problem
this is gud but not the best soln. -- Satyam Kapoor B.Tech 2nd year Computer Science And Engineering M.N.N.I.T ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Interesting Toughest Problem On Telephone PC ..
Number 1, doesn't represent any letter.
pseudocode, (numbers is array of numbers, currentString is currently
composed string to show)
function printChar(numbers, currentString){
for (every letter coded as numbers[0]){
if (numbers.lenght1){
printChar(numbers[1:], currentString+letter)
} else {
print(currentString+letter)
}
}
}
On 11 Mar, 16:45, bittu shashank7andr...@gmail.com wrote:
Hy Guys I want to do . for any phone number .. the program should
print out all the possible strings it represents. For eg.) 2 in the
number can be replaced by 'a' or 'b' or 'c', 3 by 'd' 'e' 'f' etc. In
this way how many possible permutations can be formed from a given
phone number. I don't want anyone to write code for it ... a good
algorithm or psuedocode would be great.
I am talking about old small phone where a single digit corresponds
to 3charcter from digit 2 to 8 and digit 9 has 4 character w x y
z ..Hope everyone got my points what i wants well i tried my solution
but thats become so complex so now even i am facing to understand that
so will anybody try this interesting question .
more clearly for digit o/p is
AD
AE
AF
BD
BE
BF
CD
CE
CF
Now You Can Imegin what will be o/p for 9740852296 or any 10 digit
number
Think as much as you can but finally solution matters ..
Thanks Regards
Shashank Mani
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Re: [algogeeks] spoj problem
@Ankur: then what is the 'best' solution for this? :)
@Balaji: i tried implementing but I dont know which case it fails?? getting
WA now!!
Here is the code:
#includestdio.h
int main()
{
long n,gcd=1;
scanf(%d,n);
long long a[n],b[n],cnt=0,sum=0;
long long min=9;
scanf(%lld,a[0]);
for(long i=1;in;i++)
{
scanf(%lld,a[i]);
b[i-1]=a[i]-a[0];
if(minb[i-1])
min=b[i-1];
}
for(int k=min;k0;k--)
{
cnt=0;
for(int i=0;in-1;i++)
{
if(b[i]%k==0)
cnt++;
}
if(cnt==n-1)
{
gcd=k;
break;
}
}
sum=((a[n-1]-a[0])/gcd)-n+1;
printf(%lld\n,sum);
return 0;
}
On Sat, Mar 12, 2011 at 2:38 PM, Satyam Kapoor satyamkapoo...@gmail.comwrote:
this is gud but not the best soln.
--
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B.Tech 2nd year
Computer Science And Engineering
M.N.N.I.T ALLAHABAD
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Re: [algogeeks] spoj problem
sorry, @satyam: then what is the 'best' solution for this? :)
On Sat, Mar 12, 2011 at 7:06 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:
@Ankur: then what is the 'best' solution for this? :)
@Balaji: i tried implementing but I dont know which case it fails?? getting
WA now!!
Here is the code:
#includestdio.h
int main()
{
long n,gcd=1;
scanf(%d,n);
long long a[n],b[n],cnt=0,sum=0;
long long min=9;
scanf(%lld,a[0]);
for(long i=1;in;i++)
{
scanf(%lld,a[i]);
b[i-1]=a[i]-a[0];
if(minb[i-1])
min=b[i-1];
}
for(int k=min;k0;k--)
{
cnt=0;
for(int i=0;in-1;i++)
{
if(b[i]%k==0)
cnt++;
}
if(cnt==n-1)
{
gcd=k;
break;
}
}
sum=((a[n-1]-a[0])/gcd)-n+1;
printf(%lld\n,sum);
return 0;
}
On Sat, Mar 12, 2011 at 2:38 PM, Satyam Kapoor
satyamkapoo...@gmail.comwrote:
this is gud but not the best soln.
--
Satyam Kapoor
B.Tech 2nd year
Computer Science And Engineering
M.N.N.I.T ALLAHABAD
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Re: [algogeeks] spoj problem
@akshata:i did this one by hcf method but it took too much time... -- Satyam Kapoor B.Tech 2nd year Computer Science And Engineering M.N.N.I.T ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] spoj problem
@Balaji:
that WA was due to using int in place of long long in loop. But still, this
is giving TLE.
On Sat, Mar 12, 2011 at 7:07 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:
sorry, @satyam: then what is the 'best' solution for this? :)
On Sat, Mar 12, 2011 at 7:06 PM, Akshata Sharma akshatasharm...@gmail.com
wrote:
@Ankur: then what is the 'best' solution for this? :)
@Balaji: i tried implementing but I dont know which case it fails??
getting WA now!!
Here is the code:
#includestdio.h
int main()
{
long n,gcd=1;
scanf(%d,n);
long long a[n],b[n],cnt=0,sum=0;
long long min=9;
scanf(%lld,a[0]);
for(long i=1;in;i++)
{
scanf(%lld,a[i]);
b[i-1]=a[i]-a[0];
if(minb[i-1])
min=b[i-1];
}
for(int k=min;k0;k--)
{
cnt=0;
for(int i=0;in-1;i++)
{
if(b[i]%k==0)
cnt++;
}
if(cnt==n-1)
{
gcd=k;
break;
}
}
sum=((a[n-1]-a[0])/gcd)-n+1;
printf(%lld\n,sum);
return 0;
}
On Sat, Mar 12, 2011 at 2:38 PM, Satyam Kapoor
satyamkapoo...@gmail.comwrote:
this is gud but not the best soln.
--
Satyam Kapoor
B.Tech 2nd year
Computer Science And Engineering
M.N.N.I.T ALLAHABAD
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Re: [algogeeks] spoj problem
@Akshata, Satyam: I think that might be because the way you are using to
calculate gcd is inefficient.
I got AC.
I used the following technique to calculate gcd:
long long int gcd(long long int a, long long int b){
long long int rem = a%b;
if(rem == 0) return b;
return gcd(b,rem);
}
long long int gcdvar = a[1]-a[0];
for(i=2;in;i++)
{
gcdvar = gcd(a[i]-a[0],gcdvar);
}
Thanks,
Balaji.
On Sat, Mar 12, 2011 at 7:16 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:
@Balaji:
that WA was due to using int in place of long long in loop. But still, this
is giving TLE.
On Sat, Mar 12, 2011 at 7:07 PM, Akshata Sharma akshatasharm...@gmail.com
wrote:
sorry, @satyam: then what is the 'best' solution for this? :)
On Sat, Mar 12, 2011 at 7:06 PM, Akshata Sharma
akshatasharm...@gmail.com wrote:
@Ankur: then what is the 'best' solution for this? :)
@Balaji: i tried implementing but I dont know which case it fails??
getting WA now!!
Here is the code:
#includestdio.h
int main()
{
long n,gcd=1;
scanf(%d,n);
long long a[n],b[n],cnt=0,sum=0;
long long min=9;
scanf(%lld,a[0]);
for(long i=1;in;i++)
{
scanf(%lld,a[i]);
b[i-1]=a[i]-a[0];
if(minb[i-1])
min=b[i-1];
}
for(int k=min;k0;k--)
{
cnt=0;
for(int i=0;in-1;i++)
{
if(b[i]%k==0)
cnt++;
}
if(cnt==n-1)
{
gcd=k;
break;
}
}
sum=((a[n-1]-a[0])/gcd)-n+1;
printf(%lld\n,sum);
return 0;
}
On Sat, Mar 12, 2011 at 2:38 PM, Satyam Kapoor satyamkapoo...@gmail.com
wrote:
this is gud but not the best soln.
--
Satyam Kapoor
B.Tech 2nd year
Computer Science And Engineering
M.N.N.I.T ALLAHABAD
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Re: [algogeeks] spoj problem
@balaji:i hve got ac with gcd method but the time is 0.32 sec best soln is 0.03 how is that achievable? -- Satyam Kapoor B.Tech 2nd year Computer Science And Engineering M.N.N.I.T ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] spoj problem
Yeah, I too am wondering how to implement more efficiently. On Sat, Mar 12, 2011 at 7:36 PM, Satyam Kapoor satyamkapoo...@gmail.comwrote: @balaji:i hve got ac with gcd method but the time is 0.32 sec best soln is 0.03 how is that achievable? -- Satyam Kapoor B.Tech 2nd year Computer Science And Engineering M.N.N.I.T ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] spoj problem
Here is my solution.i got it AC
#includestdio.h
int gcd(int a, int b)
{ int temp;
while(b)
{
temp = a % b;
a = b;
b = temp;
}
return(a);
}
int main()
{
int n,i,min,ans=0;
scanf(%d,n);
int arr[n],ard[n];
ard[0]=1;
if(n0)
scanf(%d,arr[0]);
for(i=1;in;i++)
{
scanf(%d,arr[i]);
ard[i]=arr[i]-arr[i-1];
if(i==1)
min=ard[i];
else
min=gcd(min,ard[i]);
}
for(i=1;in;i++)
{
ans+=((ard[i]/min)-1);
}
printf(%d,ans);
return 0;
}
On Sat, Mar 12, 2011 at 7:37 PM, Balaji Ramani rbalaji.psgt...@gmail.comwrote:
Yeah, I too am wondering how to implement more efficiently.
On Sat, Mar 12, 2011 at 7:36 PM, Satyam Kapoor
satyamkapoo...@gmail.comwrote:
@balaji:i hve got ac with gcd method but the time is 0.32 sec
best soln is 0.03
how is that achievable?
--
Satyam Kapoor
B.Tech 2nd year
Computer Science And Engineering
M.N.N.I.T ALLAHABAD
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Re: [algogeeks] spoj problem
yeah I too got AC. The way I was calculating gcd was not efficient. :)
On 3/12/11, Logic King crazy.logic.k...@gmail.com wrote:
Here is my solution.i got it AC
#includestdio.h
int gcd(int a, int b)
{ int temp;
while(b)
{
temp = a % b;
a = b;
b = temp;
}
return(a);
}
int main()
{
int n,i,min,ans=0;
scanf(%d,n);
int arr[n],ard[n];
ard[0]=1;
if(n0)
scanf(%d,arr[0]);
for(i=1;in;i++)
{
scanf(%d,arr[i]);
ard[i]=arr[i]-arr[i-1];
if(i==1)
min=ard[i];
else
min=gcd(min,ard[i]);
}
for(i=1;in;i++)
{
ans+=((ard[i]/min)-1);
}
printf(%d,ans);
return 0;
}
On Sat, Mar 12, 2011 at 7:37 PM, Balaji Ramani
rbalaji.psgt...@gmail.comwrote:
Yeah, I too am wondering how to implement more efficiently.
On Sat, Mar 12, 2011 at 7:36 PM, Satyam Kapoor
satyamkapoo...@gmail.comwrote:
@balaji:i hve got ac with gcd method but the time is 0.32 sec
best soln is 0.03
how is that achievable?
--
Satyam Kapoor
B.Tech 2nd year
Computer Science And Engineering
M.N.N.I.T ALLAHABAD
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Re: [algogeeks] spoj problem
which probem ? pls gve the link
On Sat, Mar 12, 2011 at 8:27 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:
yeah I too got AC. The way I was calculating gcd was not efficient. :)
On 3/12/11, Logic King crazy.logic.k...@gmail.com wrote:
Here is my solution.i got it AC
#includestdio.h
int gcd(int a, int b)
{ int temp;
while(b)
{
temp = a % b;
a = b;
b = temp;
}
return(a);
}
int main()
{
int n,i,min,ans=0;
scanf(%d,n);
int arr[n],ard[n];
ard[0]=1;
if(n0)
scanf(%d,arr[0]);
for(i=1;in;i++)
{
scanf(%d,arr[i]);
ard[i]=arr[i]-arr[i-1];
if(i==1)
min=ard[i];
else
min=gcd(min,ard[i]);
}
for(i=1;in;i++)
{
ans+=((ard[i]/min)-1);
}
printf(%d,ans);
return 0;
}
On Sat, Mar 12, 2011 at 7:37 PM, Balaji Ramani
rbalaji.psgt...@gmail.comwrote:
Yeah, I too am wondering how to implement more efficiently.
On Sat, Mar 12, 2011 at 7:36 PM, Satyam Kapoor
satyamkapoo...@gmail.comwrote:
@balaji:i hve got ac with gcd method but the time is 0.32 sec
best soln is 0.03
how is that achievable?
--
Satyam Kapoor
B.Tech 2nd year
Computer Science And Engineering
M.N.N.I.T ALLAHABAD
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Re: [algogeeks] spoj problem
https://www.spoj.pl/problems/STREETR/
On 3/12/11, kumar anurag anurag.it.jo...@gmail.com wrote:
which probem ? pls gve the link
On Sat, Mar 12, 2011 at 8:27 PM, Akshata Sharma
akshatasharm...@gmail.comwrote:
yeah I too got AC. The way I was calculating gcd was not efficient. :)
On 3/12/11, Logic King crazy.logic.k...@gmail.com wrote:
Here is my solution.i got it AC
#includestdio.h
int gcd(int a, int b)
{ int temp;
while(b)
{
temp = a % b;
a = b;
b = temp;
}
return(a);
}
int main()
{
int n,i,min,ans=0;
scanf(%d,n);
int arr[n],ard[n];
ard[0]=1;
if(n0)
scanf(%d,arr[0]);
for(i=1;in;i++)
{
scanf(%d,arr[i]);
ard[i]=arr[i]-arr[i-1];
if(i==1)
min=ard[i];
else
min=gcd(min,ard[i]);
}
for(i=1;in;i++)
{
ans+=((ard[i]/min)-1);
}
printf(%d,ans);
return 0;
}
On Sat, Mar 12, 2011 at 7:37 PM, Balaji Ramani
rbalaji.psgt...@gmail.comwrote:
Yeah, I too am wondering how to implement more efficiently.
On Sat, Mar 12, 2011 at 7:36 PM, Satyam Kapoor
satyamkapoo...@gmail.comwrote:
@balaji:i hve got ac with gcd method but the time is 0.32 sec
best soln is 0.03
how is that achievable?
--
Satyam Kapoor
B.Tech 2nd year
Computer Science And Engineering
M.N.N.I.T ALLAHABAD
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[algogeeks] Re: Interesting Toughest Problem On Telephone PC ..
You cn also use modified BFS with mappings from numbers to its
corresponding paths and print the paths from the root.
On Mar 12, 7:45 am, xyz69 pawelszc...@gmail.com wrote:
Number 1, doesn't represent any letter.
pseudocode, (numbers is array of numbers, currentString is currently
composed string to show)
function printChar(numbers, currentString){
for (every letter coded as numbers[0]){
if (numbers.lenght1){
printChar(numbers[1:], currentString+letter)
} else {
print(currentString+letter)
}
}
}
On 11 Mar, 16:45, bittu shashank7andr...@gmail.com wrote:
Hy Guys I want to do . for any phone number .. the program should
print out all the possible strings it represents. For eg.) 2 in the
number can be replaced by 'a' or 'b' or 'c', 3 by 'd' 'e' 'f' etc. In
this way how many possible permutations can be formed from a given
phone number. I don't want anyone to write code for it ... a good
algorithm or psuedocode would be great.
I am talking about old small phone where a single digit corresponds
to 3charcter from digit 2 to 8 and digit 9 has 4 character w x y
z ..Hope everyone got my points what i wants well i tried my solution
but thats become so complex so now even i am facing to understand that
so will anybody try this interesting question .
more clearly for digit o/p is
AD
AE
AF
BD
BE
BF
CD
CE
CF
Now You Can Imegin what will be o/p for 9740852296 or any 10 digit
number
Think as much as you can but finally solution matters ..
Thanks Regards
Shashank Mani
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[algogeeks] Bhavesh agrawal wants to chat
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[algogeeks] SPoj maximum sum subseuence
https://www.spoj.pl/problems/MAXSUMSQ/
Hi in above problem , i am getting TLE but according to given contraints ,
i think my code is good enough to run. Can any body help me here
#include vector
#include map
#include algorithm
#include cstring
#include iostream
#include cstdio
#include cmath
#include cstdlib
#include climits
#define VI vector int
typedef long long int LL;
using namespace std;
VI v;
LL x,nos;
//finding maximum sum subsequence
LL findx()
{
int sz=v.size();
LL maxsum=INT_MIN;
int maxstart=0,maxend=0;
LL currentsum=0;
int currentstart=0,currentend=0;
for(currentend=0;currentendsz;currentend++)
{
currentsum=currentsum+v[currentend];
if(currentsummaxsum)
{
maxsum=currentsum;
maxstart=currentstart;
maxend=currentend;
}
if(currentsum0)
{
currentstart=currentend+1;
currentsum=0;
}
}
return maxsum;
}
// main Program
int main()
{
// freopen(input.txt,r,stdin);
int test;
int num;
map LL , LL m;
cintest;
LL sum=0;
int n;
int i;
while(test--)
{
sum=0;
nos=0;
m.clear();
m[0]=1;
scanf(%d,n);
v.resize(n);
for(i=0;in;i++)
{
scanf(%d,v[i]);
}
x=findx();
nos=0;
for(i=0;in;i++)
{
sum=sum+v[i];
nos=nos+m[sum-x];
m[sum]++;
}
coutx nosendl;
}
return 0;
}
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