Farmer John commanded his cows to search for different sets of numbers that sum to a given number.
The cows use only numbers that are an integer power of 2. Here are the possible sets of numbers that
sum to 7:
1) 1+1+1+1+1+1+1
2) 1+1+1+1+1+2
3) 1+1+1+2+2
4) 1+1+1+4
5) 1+2+2+2
6) 1+2+4
Help FJ count all possible representations for a given integer N (1 <= N <= 1 ,000,000)

The first line of the input contains the number of test cases in the file. And t he first line of each case
contains one integer numbers n

For each test case, output a line with the ans % 1000000000.

``````1
7``````

``6``

``````#include<stdio.h>
int main()
{
int a[1000000];
int n;
scanf("%d",&n);
while(n--)
{
int s;
scanf("%d",&s);
a[1]=1;
int i;
a[2]=2;
a[3]=2;
for(i=4;i<=s;i++)
a[i]=(a[i-2]+a[i/2])%1000000000;
printf("%d\n",a[s]%1000000000);
}
return 0;
}
``````

``````#include <iostream>
using namespace std;
long long a[1000001];
int main()
{
long long N,T,i;
a[1]=1;
a[2]=2;
for(i=3; i<1000001; i++)
{
if(a[i]%2==1)
a[i]=a[i-1];
else
a[i]=a[i-2]+a[i/2];
a[i]%=1000000000;
}
cin>>N;
while(N--)
{
cin>>T;
cout<<a[T]<<endl;
}
return 0;
}
``````