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Author : Ankit Sagwekar
Tester : Vikesh Tiwari
Editorialist : Ankit Sagwekar

DIFFICULTY

Easy

PREREQUISITES

basics,array,implementation

PROBLEM

Given two integers 'm' and 'n'. Find the number of combinations of bit-size 'n' in which there are no 'm' consecutive '1'.

EXPLANATION

We are given with m and n. There are 3 cases for any given value of n and m

• Case 1: when n is less than m for this the ans will be equal to ans=pow(2,n).
  eg: 3 2 all the 2 bit words i.e 00 01 10 11 will not have 3 consecutive 1 ,thus all of them are selected.

• Case 2: when n is equal to m for this the ans will be equal to ans=pow(2,n)-1.
  eg: 3 3 all the 3 bit words i.e 000 001 010 011 100 101 110 111 out of which only one word will have 3 consecutive '1'.

• Case 3: when n is more than m This test case needs some computation which we will see through eg.
  eg: 3 4
  Here we first calculate for
  m n
  3 3 = 7
  3 2 = 4
  3 1 = 2
  then adding this ,we get for 3 4 as 7+4+2=13.
  here m=3 thus we calculate for 3 values and add them .
  eg:4 6
  here we first calculate for
  m n
  4 5 = 29
  4 4 = 15
  4 3 = 8
  4 2 = 4
  then adding this ,we get for 4 6 as 29+15+8+4=56.
  here m=4 thus we calculated for 4 values and add them .
  

Note: for faster computation we can precompute all the values and can store them in 2-D matrix.

Complexity:

(O(N^2.logN))

C++ Code

#include <stdio.h>
#include<math.h>
int main(void) {
    // your code goes here
    long long int t,n,m,i,j,k,a[15][105],l;
    scanf("%lld",&t);
    for(i=2;i<=10;i++)
    for(j=1;j<=50;j++)
    {
        if(j<i)
        a[i][j]=pow(2,j);
        else if(j==i)
        a[i][j]=pow(2,j)-1;
        else
        {
            k=i;
            l=j-1;
        while(k>0)
        {   k--;
            a[i][j]+=a[i][l--];
        }
        }
    }
    /*

         for(i=2;i<=10;i++)
    {
    for(j=1;j<=50;j++)
    {
        printf("%lld\t",a[i][j]);
    }
    printf("\n");
    }
         */
         // you can uncomment the above section to see all the precomputed values 
    while(t--)
    {
        scanf("%lld%lld",&m,&n);
        printf("%lld\n",a[m][n]);
    }
    return 0;
}