A mod B in O(number_of_digit_in(A))

Lets say you have a Big Integer A and an integer B . You want to calculate A mod B in O(length_of_(A)) time complexity . How to do that?

Lets say : A = 1234 (Not that big :-P ) and B = 6

I want to calculate 1234 % 6 . Now ,

s1 = 1234 % 6  =  ((123%6)*10 + 4)%6
                 =   (s2*10 + 4) %6  [If I know s2 I can calculate 1234 % 6 in O(1) now]

s2 = (123%6)  =  ((12%6)*10 + 3)%6
                        =  (s3*10 + 3)%6   [If I know s3 I can calculate 1234 % 6 in O(1) now]

s3 = (12 % 6 ) =  ( (1%6)*10 + 2)%6
                        =      ( s4*10 + 2) % 6 [If I know s4 I can calculate 1234 % 6 in O(1)  and s4 can be calculated in O(1)]


Now, We can easily implement this with a single loop which runs in O(n) . where,length(A) = n

So,

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//A is taken as string //mod = B int AmodB = 0; for(int a=0;a<s.length();a++) { AmodB = (AmodB*10 + (int)(s[a] - '0'))%mod; }

AmodB is initially (0*10 + 1)%6 = 1 (s4)
then,                       (1*10 + 2)%6 = 0 (s3)
then,                       (0*10 + 3)%6 = 3 (s2)
and finally,             (3*10 + 4)%6 = 4  (s1)

Theory : For any X , (X*k + m)%d = ((k%d)*X + m)%d

as, k can be written as a (multiple of d + remainder) this way  k = a*d + k%d ,for some integer a.
now, (X*k + m)%d

   =   (X*(a*d + k%d) + m)%d
   =   (X*a*d + X*k%d + m)%d
   =   (k%d*X + m)%d  [as,X*a*d % d = 0]

Now,put X = 10



Another approach : (1234)%6 = (4 + 3*10 + 2*100 + 1*1000)%6


so, ans =  (sigma(i = 0 to n-1)  (10^i*digit[i])%mod
The problem is how can I calculate (10^i%mod) efficiently and avoiding overflow?

Look, 10^5%mod = ((10^4)%mod*(10%mod))%mod
So,we can just save 10^(i-1)%mod in the loop and in the i'th iteration I'll find 10^i%mod by multiplying 10^(i-1)%mod [which I have already saved] and 10%mod [I can also save this into a variable to avoid repeated calculation] and "mod"ing the multiplied result.

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int mod_loop = 1; int tmp = 0; int ten_mod = 10%mod; //saving this for(int a=s.length()-1;a>=0;a--) { tmp+=(mod_loop*(int)(s[a] - '0' ) )%mod; mod_loop =((mod_loop)*ten_mod)%mod; } int ans = tmp%mod

Problems : Type 1) You'd need to mod a Big Integer very fast / You'd need to check divisibility (if B divides A) avoiding overflow and in O(length(A))

Type 2) You'll be given B and you have to find A such that B divides A and digits of A has a certain pattern such as A = 1111111........  , 12121212121212.................

LOJ 1010 Knights

Problem id :    LOJ 1010


To maximize the number of knights we have to place knights in a particular pattern.

Code :

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#include<stdio.h> int main() { int tc,cas = 0 ; int ans = 0; scanf("%d",&tc); while(tc--) { int m,n; scanf("%d %d",&m,&n); //ans = solve(m,n); //ROW / COL = 1 if(m==1) ans = n; else if(n==1) ans = m; //ROW / COL = 2 else if(m==2) { ans = 0; for(int i=1,cn=1; i<=n; i+=4,cn++) { // if(cn&1) { if(i==n) ans +=2; else ans += 4; } } } else if(n==2) { ans = 0; for(int i=1,cn=1; i<=m; i+=4,cn++) { // if(cn&1) { if(i==m) ans +=2; else ans += 4; } } } //ROW / COL > 2 else if(m&1)//odd { if(n&1)//odd ans= (m/2 + 1)*(n/2 + 1) + (m/2)*(n/2); else //m odd n even ans= (m/2 + 1)*(n/2) + (m/2)*(n/2); } else { if(n&1)//odd ans= (m/2)*(n/2 + 1) + (m/2)*(n/2); else ans= (m/2)*(n/2) + (m/2)*(n/2); } printf("Case %d: %d\n",++cas,ans); } return 0; }