通过连接前 n 个自然数的二进制表示形成的数字的十进制值

Question

Given a number n , find the decimal value of the number formed by concatenating the binary representations of first n natural numbers.
Print answer modulo 10^9+7.

Also, n can be as big as 10^9 and hence logarithmic time approach is needed.

Eg: n = 4 , Answer = 220

Explanation : Number formed= 11011100 ( 1=1 , 2=10 , 3=11 , 4=100 ). Decimal value of 11011100 = "220" .

The code I am using below only works for first integers N<=15

    String input = "";
    for(int i = 1;i<=n;i++) {
        input += (Integer.toBinaryString(i));
    }
    return Integer.parseInt(input,2);

Answer 1

This solution to this question requires O(N) time. Luckily this can be solved in O(logN) time. Also, this is the A047778 sequence:

1,6,27,220,1765,14126,113015,1808248,28931977, 462911642,7406586283,118505380540,1896086088653, 30337377418462,485398038695407,15532737238253040, 497047591624097297,15905522931971113522

The sequence follows this recurrence relation: where ⌊.⌋ is floor function

a(n) can also be expressed as sum of multiple arithmetico–geometric series .

If we are interested in a(14) , here's how it can be calculated.

Multiplying with powers of two on both sides of the above equations gives equations like the following:

If we add all the above equations, a(14) can be expressed as sum of four arithmetico–geometric series.

It's important to note that in all sequences except the first one, the first term of the arithmetic progression is of form and the last term

The sum of n terms of arithmetico–geometric sequence can be calculated using this formula :

a(First term of AP), n(Number of terms), d(Common Difference of AP), b(First term of GP), r(Common ratio of GP). 

Since we're interested in a(n) mod 1000000007 and not the actual term a(n) , these modulo arithmetics may come in handy.

This is a good starting point for implementing division modulo which requires some number theory basics.

Once we figure out the number of sequences required and the five variables a, n, d, b, r for each sequence, a(n) modulo 1000000007 can be calculated in O(logn) time.

Here's a working C++ code:

#include <numeric>
#include <iostream>
#define mod long(1e9+7)

long multiply(long a,long b){
    a%= mod;b%= mod;
    return (a*b)%mod;
}

void inverseModulo(long a,long m,long *x,long *y){ //ax congruent to 1 mod m

    if(!a){
        *x=0;
        *y=1;
        return ;
    }
    long x1,y1;
    inverseModulo(m%a,a,&x1,&y1);
    *x=y1-(m/a)*x1;
    *y=x1;
    return;
}

long moduloDivision(long a,long b,long m){  // (a*(returnValue))mod m congruent to b mod m
    //https://www.geeksforgeeks.org/modular-division/ and https://www.geeksforgeeks.org/multiplicative-inverse-under-modulo-m/
    long x,y;
    inverseModulo(b, m, &x, &y);
    x%=m;
    return (x*a)%m;
}

long power(long n,long r){ //calculates (n^r)%mod in logarithmic time
    if(r==0) return 1;
    if(r==1) return n;
    if(r%2){
        auto tmp=power(n, (r-1)/2);
        return multiply(multiply(n,tmp),tmp);
    }
    auto tmp=power(n, r/2);
    return multiply(tmp, tmp);
}
long sumOfAGPSeries(long a,long d,long b,long r,long n){
    if(r==1) return multiply(n, multiply(a, 2)+multiply(n-1, d))/2;
    long left=multiply(multiply(d, r), power(r,n-1)-1);
    left=a+moduloDivision(left,r-1,mod);
    left=multiply(left, b);
    left%=mod;
    long right=multiply(multiply(b, power(r, n)), a+multiply(n-1, d));
    long ans=right-left;
    ans=(ans%mod + mod) % mod;
    return moduloDivision(ans,r-1,mod);

}
signed main(){
    long N=1000;
    long ans = 0;
    long bitCountOfN = log2(N) + 1;
    long nearestPowerOfTwo = pow(2, bitCountOfN - 1);
    long startOfGP = 0;
    while (nearestPowerOfTwo) { // iterating over each arithmetico–geometric sequence
        long a, d, b, r, n;
        a = N;
        d = -1;
        b = power(2, startOfGP);
        r = power(2, bitCountOfN);
        n = N - nearestPowerOfTwo + 1;
        ans += sumOfAGPSeries(a, d, b, r, n);
        ans %= mod;
        startOfGP += n * bitCountOfN;
        N = nearestPowerOfTwo - 1;
        nearestPowerOfTwo >>= 1;
        bitCountOfN--;
    }
    std::cout << ans << std::endl;
    return 0;
}

The validity of the above C++ code can be verified using this trivial python code:

def a(n): 
  return int("".join([(bin(i))[2:] for i in range(1, n+1)]), 2)
for n in range(1,100):
  print (a(n)%1000000007)

Answer 2

Note that working with string representation is not necessary (moreover, is not useful after task changing). Look at approach with bitwise arithmetics (Python, but principle is the same)

With new condition concerning modulo 1000000007 we have just add modulo operation to result calculation line at every step, because shift left and or-ing is equivalent to multiplication by power of two and adding, these operations are obeyed to equivalence relations for modulo properties. Note that intermediate results don't exceed 1000000007*n , so long type is suitable here for reasonable n values.

n = 100  
size = 0   #bit length of addends
result = 0   # long accumulator
for i in range(1, n + 1):    
    if i & (i - 1) == 0:    #for powers of two we increase bit length
        size += 1
    result = ((result << size) | i) % 1000000007   #shift accumulator left and fill low bits with new addend
print(result)

variant without bitwise operations:

pow2 = 1
nextpow = 2
result = 0   # long accumulator
for i in range(1, n + 1):
    if i == nextpow:    #for powers of two we increase bit length
        pow2 = nextpow
        nextpow = nextpow * 2
    result = (result * pow2  + i) % 1000000007  #shift accumulator left and fill low bits with new addend

Answer 3

    cin>>n;
    ll ans=1;
    ll one=1;
    for(int i=2;i<=n;i++)
    {
        ll digit=log2(i)+1;
        ans=(((ans%N*(one<<digit)%N)%N+i%N)%N);
    }
    cout<<ans<<Ed;