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10718 - Bit Mask

Posted: Wed Sep 22, 2004 10:14 am
by liulike
How to solve it?
Plz give me some hints,thanks!

Posted: Wed Sep 22, 2004 10:35 am
by abishek
I solved it in the contest using the following idea:

1. i decide upon the bits of M from the left most one by one.
2. when deciding upon i ensure that value of M is inbetween low and high.
3. and finally with this constraint, i will turn on a bit in M if the same bit is off in N. (this is to minimise M but to maximise N).
bye
abi

Posted: Wed Sep 22, 2004 10:55 am
by liulike
Is it a brute force solution ?
The time complexity is O(2 ^ 32)?

Posted: Wed Sep 22, 2004 11:02 am
by abishek
no it is not a brute force solution at all
the complexity is O(n) where n is number of bits of the given numbers (32 steps in this case)

we need to build this number bit by bit starting from the Most significant bit.

Posted: Wed Sep 22, 2004 11:37 am
by liulike
Thank you, I got AC now :D

Posted: Wed Sep 22, 2004 6:54 pm
by windows2k
liulike wrote:Thank you, I got AC now :D
Hello, I got WA all the time.
Could you give me tricky input/output? Thx :D

Posted: Wed Sep 22, 2004 11:27 pm
by anupam
abishek, I used as follows: (but got WA)
1. convert N to binary
2. for up to low of binary N
{
tmp=prev. val+2^i;
iif(tmp<l) {val=tmp;continue;}
if(tmp>u) continue;
if(a==0) val=tmp;
}
3. before step 2 I used the folowing special cases:
let n = no. of bit in N and m=no. of bit in M
if(n==m) and all 1 bits in both then output lowerlimit
if m>n then output the value of (bits from n+1 pos to m of M concatenated by n zeroes.

What am I doing wrong??

Posted: Thu Sep 23, 2004 1:04 am
by Adrian Kuegel
I don't understand all what you are doing. For example, what does:
"if m>n then output the value of (bits from n+1 pos to m of M concatenated by n zeroes. " mean?
It doesn't look correct to me, especially that you concatenate these n zeros. what if N is 101 binary? you could need the bit of 2^1.

And I think this step is wrong:
iif(tmp<l) {val=tmp;continue;}
think of N = 6, l = 0, u = 7
I think you would answer 5, but it should be 1.

Posted: Thu Sep 23, 2004 9:14 pm
by anupam
So, what may be the algorithm that abishek described. I think I didn't understand it properly. Can you help how to proceed?

Posted: Fri Sep 24, 2004 3:56 pm
by shamim
Let us clarify with an example.

100 50 60


100 in binary mode is: 0001100100 ( Considering 10 bits, although in general it should be 32 bits.)

Now, start from the leftmost bit to decide, whether this bit should be set or not. The decision is based on the current status of the bit in (0001100100). Here the first three bit is 0, so if they are set, then the smallest value obtained would be more than the upper limit. So they can not be set.

Now, come to the fourth, it is already set, and if we don't set it we could still obtain a number larger than or equal to lower limit.
Now, the fifth bit is already set, but if we don't set it we could never obtain a numer more than or eqaul to 50. So it has to be set.
If this process is continued, we would get a binary pattern as follows:

0 0 0 0 1 1 1 0 1 1

Which is that for 59, hence 59 is the answer.

Good Luck. :wink:

10718

Posted: Fri Sep 24, 2004 4:00 pm
by zzzzzddddd
I used brute froce to check my program, my code can pass all test data I made. Can anyone told me why I got WA?
Thanks a lot!
[cpp]#include <stdio.h>
#include <iostream.h>

unsigned int n,lo,hi;

int main()
{
unsigned int alr;
int i;
while(cin>>n>>lo>>hi)
{
alr=0;
for(i=31;i>=0;i--)
if((n&(1<<i))==(1<<i))
{
if(!(alr+(1<<i)-1>=lo&&alr<=hi))
alr+=(1<<i);
}
else
if(alr+(1<<(i+1))-1>=lo&&alr+(1<<i)<=hi)
alr+=(1<<i);
cout<<alr<<endl;
}
return(0);
}

[/cpp]

Posted: Fri Sep 24, 2004 6:08 pm
by liulike
If anyone want, I would sent him my ac code, Good Luck!

to liulike

Posted: Sat Sep 25, 2004 5:58 am
by zzzzzddddd
can u tell me why i got wa?
thanks a lot.
see my program in message i have post.

Posted: Sat Sep 25, 2004 11:51 am
by Krzysztof Duleba
Some random input:
22 1 21
17 3 5
12 9 17
622 435 516
774 887 905
398 119 981
550 99 427
823 684 966
22398 14719 27341
29787 21141 30633
3113 24242 29497
5021 11052 19571
7359 6669 19790
993331 367623 921769
810986 227838 492138
987964 208251 1034287
1002648 217556 236589
680047 402532 767548
27719912 10747535 22001285
16862072 13339946 28844391
20421198 11724734 31928748
1541522 10226990 20764468
22651935 2478699 31095674
1995360670 908222743 1868651617
305542918 594331857 1426036714
1265639777 1580376576 1885248297
1442823820 658800174 1919310996
604563406 1050668699 2128532112
Output:
9
4
17
435
905
625
409
712
14721
23460
29462
19554
17216
382924
237589
257219
234343
499600
14223127
16692359
13133233
19429997
10902496
957429345
1305069817
1880088222
704659827
1542920241

Here are my output.

Posted: Sat Sep 25, 2004 12:46 pm
by zzzzzddddd
9
4
17
435
905
625
409
712
14721
23460
29462
19554
17216
382924
237589
257219
234343
499600
14223127
16692359
13133233
19429997
10902496
957429345
1305069817
1880088222
704659827
1542920241