@surabh sab jagah hath mar liya beywhy don't u help me out...vase bhi
aab toh tum free bhi ho gye ho ge..:P:P
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@payal why u need this..??...:P:P
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+1 @saurabh...:P
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+1
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I think logic is the difference is
2 2 4 2 2 2 8 so next will be 2 2 2 2 2 16
so ans will be 66 68 70
but first number 3 making some problem
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output depend upon complier how it store floating point number
actualy exactly 0/3 is not stored in memory some 0.299 like this is
stored so its totally implementation dependent..
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@atul by doing through link list it will give TLEu have to derive some
kind of formula .. and till now i am unable to find that
kartik.sac...@gmail.com
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if n is less than 10^6 hasing works fine ..and we count in O(1) time only
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its a compiler depentent...
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@amol I think it is not the behaviour of printf
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+1 amol sharma...what u want ??.output should be shown at every
step or u want final ans??
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@DAVE that was typing mistake...
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simply find max in o(n) by liner search then remove max then again find max
that will be second max:P :P
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@amol..we can simply break the loop where we find the slow and fast pointer
meeting,then its become a a problem of two link list having common ending
point and different staring point
and for detecting the extact point where loop began is two traverse both
the list and find its length lets say 'a'
yup using matrix method we can solve it in O(log(n)).
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yup.
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PLEASE ANY BODY CAN EXPLAIN ME ARTICULATION POINT ALGO IN DETAIL
PLZ TAKE SOME EXAMPLE TOO
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yup but didn't get much out of it...
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@rahul if u have to find first 100 even fibonacci term sum
the u have to simply find the t2n+1 term -1
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@AMOL i want array index i.e i to j that will be max subarry which has equal
no of zero and one's
but i think ur soln is not providing this
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ok amol i will do it..but i am unable to convience myself that
this algo will give the desire result
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@ratan suggest some efficient algo...
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@dheeraj i didn't get what u want to say explain me with the help of example
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i think i can solve this in O(n^2)
here is code http://ideone.com/Gk69A
# includestdio.h# includestring.hchar a[100][100];int findword(int
*b,int n,int m){
int i,j,flag=0;
char s[1];
for(i=0;in;i++)
for(j=0;jm;j++)
s[a[i][j]]++;
@sukran if it is undefined in ansi c so it can return any value???
at complilaton.but all time it is giving me same value
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http://ideone.com/xRnDo
# includestdio.h# includestring.hvoid findbig(char *a){
int i,j;
int flag=0;
for(i=strlen(a)-1;i0;i--)
if(a[i-1]a[i])
{
char temp=a[i];
a[i]=a[i-1];
a[i-1]=temp;
flag=1;
thanks sukran...
i have one more question evaluation of printf is also undefined in C??
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how no's are stored
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obivously it will be first largest
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@ravi i think ur concepts is correct the no is stored in 2's formif
negative
thanks ravi
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@arya ur code for 7695...giving ans 9765
but ans should be 7965
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@bharatkumar i think u r in thinking oder in lexicographic order .but in
question its not the same...
we have print the according to the sentence not according to dictonary
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+1 to yogesh
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given n nodes ,how many different possible tree can be formed??(not
necessary that all are binary treeit could be any tree...)
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@what's special in this ...
it is giving 0 in both cases.
basically a garbage value
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using gets we can get buffer over as gets doesn't check about the size input
so it mai go to that area where our stack base pointer(think in memory
layout section) is stored so our programmed doenn't return to orignal state
as address is stored there...so it is dangeroues to use
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@ prateek i am still not geeting why now out put is 4
but if we change second bit2:5 then out put again 8
plzz explain throughly
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becoz int is of 4btyes so 32 bits required
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@prateek check this i have complied on ideone
http://ideone.com/UV7MG
output is 4 only
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for 1st question ans will be a,b,e as (125)^1/3 is 5 sum is 8
for 2nd question we see pattern in 7 power i.e 7,9,3,1 and this pattern
reapeats for inerval of 4
and second term of of any power of 7=2 is 4
so diff is 5
ans will be 5
how to find the ans of third question??
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see the out put will the help of gcc -E.u will get ur ans
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THANKs amol piyush i got my mistake
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@PRATEEK if i make int bit1:2 then it will store 2 bytes
for example if bit1=4;then out will be 0 or 1
my question is it storing number from least significant bit or most
significant bit???
suppose for 4 0100 so out will be 00 or 01??
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it can be simply done in O(sqrt(n))
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koyi to test cases de jisme mera code fail kare?
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thanks piyush for replying but i don't have complier for java and i don't
know even java
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any body tell the test cases??
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thanks nitin got AC...:) silliy mistake
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For
]);
}
printf(\n);
}
return 0;
}
plzz help me..
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the output will be hi
even if u have putted equal to sign there this is becoz 0.7 do not exactly
stored like 0.7 in a but some thing lesser as describe in ASCII format
that's why it is awlays lesser than 0.7
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I think in this case count sort type thing would work better way
just take a array of max 26(as 26 CHARACTER ARE THERE)
and using index as count['M']=store how many times M has comes
similary store other character
now print array whose value 0
but IN this case u might lost the
@priya its some thing like this
for(i=0;a[i]!='\0';i++)
count['a[i]']++;
so its O(n) only
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@amit it's given that array is increasing then decreasing..so where
there is change from incre to drece that value of i in loop will be k
in this we can find out k if not given
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@ankit i am saying at the time of insertion of first element to last we
have to maintain a min element which store the min number
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i think this will be the only solution
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For
I think then output should be 17 not 13??
correct me if i am wrong and explain too
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oh..got it i didn't read that statement:(
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similar type algo is there in geeksforgeeks.org in array section from there
u can understand it better way
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I think its stack where deletion insertion and finding min (as we will cal
min at the time of insertion only by taking min val ) will take O(1) time
correct me if i am worng...!
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#include stdio.h
void foo(const char **p) { }
int main(int argc, char **argv)
{
foo(argv);
return 0;
}
why the above code giving me warning can body explain me??
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*@AMOL *sending a normal pointer to a const pointer doesn't give any warning
why this is giving??
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@thanks ceg sirfinally got acceptedur test
case worked..silly mistake...= error
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@sunny atof(ab) is giving me as zero.so it will not affect
the calculation i think so...
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GOT AC IN .02 SEC .:)))
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@VAIBHAV.ur logic fails in many
cases...like G
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anonymous u have to play with the index no of either G or
B...and see how many shift it will require to reach it original
postion then max will be the ans..
well GBGBBB
if we start from right to left then G=3 and G=5
first G have reach to 0 pos and second G have to reach
@ vaibhav i have submitted my concept in spoj and it got AC in .02 sec
i am saying for u have to play with index of EITHER of G or B
now u have to think logic by taking two or more exmaple and shift it
in...do it in paperu will see some pattern
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this is the link
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if u want code then mail me.
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HEY DUDE I AM NOT GETTING UR LOGIC AT ALL I THINK HOW U WILL SATISFY THIS
CASE GBGBBB
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@jitendra that's what i am asking forwhat algo i should
implement to get process in 1 sec?
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@ anonymous here u see the formula where it is clearly mention for single
step the value of k is not changing
so we can solve this question by taking single dimensional array instead of
two dimensional array
and in 1D array we have large size to store value for DP as compare to 2D
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@saurabh circular link list implementation i think tough as camparision to
single array
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@anoyomous take array size as a[106]
and dude to intilize array to zero u si,ply write a[106]={0};
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;
flag=count;
}
printf(%d\n,count);
}
return 0;
}
PLZZ HELP ME OUT OR SUGGEST ANY OTHER LOGIC IF IT HAS SOME MISTAKES
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but for big numbers this method is very expensiveand hope
give TLE in 1 sec question where n=1000
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@ sunny i am not getting ur apporach
but i am thinking like this..
taking an array from and intilize it to n to n-r
a[1000];
int k=0;
for(int i=n;i=n-r;i--)
a[k++]=i;
for(int y=n-r;y1;y--)
for(j=0;jk;j++)
if(a[j]%y==0)
{a[j]=a[j]/y;break;}
for example we take 7c3
so first array is
hey sunny suppose u have to calculate 100c50 then there is a lot of chances
of over flow becoz u have to multiply 100 to 50...i don't think
it will work..
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hey anika here n=1 and n=0 so its 1 only
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sir i think the same ans i gave the above..
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without priority queue its .08 and with priority queue its .0.00
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ACTUALLY FIBBONACCI SERIERS REAPEAT AFTER 60 TERM SO 246%60 WILLBE 6 SO WE
ONLY HAVE TO FIND M+6 TERM
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@vipul what algo u have applied for o(1)??
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most simple with bash..using formula 11*f(n+6)+11*f(n+6)%10
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hey is itoa() is supported by g++ compliers???
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ya dude finally i applied that algo only
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it is correct ...in c++ 4.3.2 compiler
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got AC silliy mistake i am sorting array only one time..
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case like
5
1 2 3 45
got worng ans
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amit it is working fine in gcc complier and giving out too i.e
10
what's error in it??
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hey dude nominate me as a group moderator kartik.sac...@gmail.com
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@shady sorry dude.i am nominating pramendra prathi...@gmail.com
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what dignity u are going to maintain.apne desh k liye kuch to
karo mere dost.apne desh ki dignity nhi bache gi to is group ki
bacha k kya karo ge??and by the i didn't put any thing wrong
in the group.many educated indian are there in the group i
;
}
return 0;
}
why i am getting TLE...can anyone explain
this...plzz
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*WITH REGARDS,*
*
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*KARTIK SACHAN*
*B.TECH 2ND YEAR*
*COMPUTER SCIENCE AND ENGINEERING*
*NIT ALLAHABAD*
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what i understand from the problem was that.
n no of higests are given
and a and b and city no i.e from city a a ball is thrown to city no b
so i have done
from array a[] i.e hieght { n...no of times}
find all the element between b[i].city A and
how should we get TLE loop total running less than 10^6??
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