Sum of GCD of all possible sequences
Given two numbers N and K. A sequence A1, A2, ….AN of length N can be created by placing numbers from 1 to K at each position, making a total of KN sequences. The task is to find the sum of GCD of all the sequences formed.
Note: The answer can be very large, so take modulo of 109 + 7.
Examples:
Input: N = 3, K = 2
Output: 9
Explanation:
The gcd of all the subsequences are:
gcd(1, 1, 1) = 1
gcd(1, 1, 2) = 1
gcd(1, 2, 1) = 1
gcd(1, 2, 2) = 1
gcd(2, 1, 1) = 1
gcd(2, 1, 2) = 1
gcd(2, 2, 1) = 1
gcd(2, 2, 2) = 2
The sum of GCD is 1 + 1 + 1 + 1 + 1 + 1 + 1 + 2 = 9.
Input: N = 3, K = 200
Output: 10813692
Naive Approach: The idea is to generate all the possible subsequences of length N recursively. The summation of GCD of all the sequences formed is the required result.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
const int MOD = ( int )1e9 + 7;
int calculate( int pos, int g, int n, int k)
{
if (pos == n) {
return g;
}
int answer = 0;
for ( int i = 1; i <= k; i++) {
answer = (answer % MOD
+ calculate(pos + 1, __gcd(g, i), n, k) % MOD);
answer %= MOD;
}
return answer;
}
int sumofGCD( int n, int k)
{
return calculate(0, 0, n, k);
}
int main()
{
int N = 3, K = 2;
cout << sumofGCD(N, K);
return 0;
}
|
Java
class GFG{
static int MOD = ( int )1e9 + 7 ;
static int calculate( int pos, int g, int n, int k)
{
if (pos == n) {
return g;
}
int answer = 0 ;
for ( int i = 1 ; i <= k; i++) {
answer = (answer % MOD
+ calculate(pos + 1 , __gcd(g, i), n, k) % MOD);
answer %= MOD;
}
return answer;
}
static int sumofGCD( int n, int k)
{
return calculate( 0 , 0 , n, k);
}
static int __gcd( int a, int b)
{
return b == 0 ? a:__gcd(b, a % b);
}
public static void main(String[] args)
{
int N = 3 , K = 2 ;
System.out.print(sumofGCD(N, K));
}
}
|
Python3
MOD = 1e9 + 7
def gcd(a, b):
if (b = = 0 ):
return a
else :
return gcd(b, a % b)
def calculate(pos, g, n, k):
if (pos = = n):
return g
answer = 0
for i in range ( 1 , k + 1 ):
answer = (answer % MOD +
calculate(pos + 1 ,
gcd(g, i), n, k) % MOD)
answer % = MOD
return answer
def sumofGCD(n, k):
return calculate( 0 , 0 , n, k)
if __name__ = = "__main__" :
N = 3
K = 2
print (sumofGCD(N, K))
|
C#
using System;
public class GFG{
static int MOD = ( int )1e9 + 7;
static int calculate( int pos, int g, int n, int k)
{
if (pos == n) {
return g;
}
int answer = 0;
for ( int i = 1; i <= k; i++) {
answer = (answer % MOD
+ calculate(pos + 1, __gcd(g, i), n, k) % MOD);
answer %= MOD;
}
return answer;
}
static int sumofGCD( int n, int k)
{
return calculate(0, 0, n, k);
}
static int __gcd( int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
public static void Main(String[] args)
{
int N = 3, K = 2;
Console.Write(sumofGCD(N, K));
}
}
|
Javascript
<script>
var MOD = 1000000007;
function calculate(pos, g, n, k)
{
if (pos == n) {
return g;
}
var answer = 0;
for ( var i = 1; i <= k; i++) {
answer = (answer % MOD
+ calculate(pos + 1, __gcd(g, i), n, k) % MOD);
answer %= MOD;
}
return answer;
}
function __gcd(a, b)
{
return b == 0? a:__gcd(b, a % b);
}
function sumofGCD(n, k)
{
return calculate(0, 0, n, k);
}
var N = 3, K = 2;
document.write( sumofGCD(N, K));
</script>
|
Time Complexity: O(NK)
Auxiliary Space: O(k * log N)
Efficient Approach:
- Since the numbers of the sequence can be from 1 to K, the gcd value of the sequence will be in the range 1 to K.
- Let count[i] represent the number of sequences with gcd = i. For i = 1, we have no constraints on which elements can belong to the sequence, so at each of the N places, we have K possibilities to place elements making the total sequences be KN. But the resulting sequences may have higher GCD, So subtract the over counted values:
count[1] = KN - count[2] - count[3] - count[4] - .... count[K]
- Similarly, for i = 2, since every number must be a multiple of 2, we have K/2 possibilities at each place, making the total be (K/2)N. And Subtract all the over counted values by subtracting the sequence count with the GCD of all multiples of 2.
count[2] = (K/2)N - count[4] - count[6] - count[8] - ... all multiples of 2
- Similarly, follow the above steps for each gcd value to K.
- The summation of each GCD value(say g) with count[g] is the sum of GCD of all the sequences formed.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
const int MOD = ( int )1e9 + 7;
int fastexpo( int a, int b)
{
int res = 1;
a %= MOD;
while (b) {
if (b & 1)
res = (res * a) % MOD;
a *= a;
a %= MOD;
b >>= 1;
}
return res;
}
int sumofGCD( int n, int k)
{
int count[k + 1] = { 0 };
for ( int g = k; g >= 1; g--) {
int count_multiples = k / g;
int temp;
temp = fastexpo(count_multiples, n);
temp %= MOD;
int extra = 0;
for ( int j = g * 2; j <= k; j += g) {
extra = (extra + count[j]);
extra %= MOD;
}
count[g] = (temp - extra + MOD);
count[g] %= MOD;
}
int sum = 0;
int add;
for ( int i = 1; i <= k; ++i) {
add = (count[i] % MOD * i % MOD);
add %= MOD;
sum += add;
sum %= MOD;
}
return sum;
}
int main()
{
int N = 3, K = 2;
cout << sumofGCD(N, K);
return 0;
}
|
Java
class GFG{
static int MOD = ( int )1e9 + 7 ;
static int fastexpo( int a, int b)
{
int res = 1 ;
a %= MOD;
while (b > 0 ) {
if (b % 2 == 1 )
res = (res * a) % MOD;
a *= a;
a %= MOD;
b >>= 1 ;
}
return res;
}
static int sumofGCD( int n, int k)
{
int []count = new int [k + 1 ];
for ( int g = k; g >= 1 ; g--) {
int count_multiples = k / g;
int temp;
temp = fastexpo(count_multiples, n);
temp %= MOD;
int extra = 0 ;
for ( int j = g * 2 ; j <= k; j += g) {
extra = (extra + count[j]);
extra %= MOD;
}
count[g] = (temp - extra + MOD);
count[g] %= MOD;
}
int sum = 0 ;
int add;
for ( int i = 1 ; i <= k; ++i) {
add = (count[i] % MOD * i % MOD);
add %= MOD;
sum += add;
sum %= MOD;
}
return sum;
}
public static void main(String[] args)
{
int N = 3 , K = 2 ;
System.out.print(sumofGCD(N, K));
}
}
|
Python3
MOD = ( int )( 1e9 + 7 )
def fastexpo(a, b) :
res = 1
a = a % MOD
while (b > 0 ) :
if ((b & 1 ) ! = 0 ) :
res = (res * a) % MOD
a = a * a
a = a % MOD
b = b >> 1
return res
def sumofGCD(n, k) :
count = [ 0 ] * (k + 1 )
for g in range (k, 0 , - 1 ) :
count_multiples = k / / g
temp = fastexpo(count_multiples, n)
temp = temp % MOD
extra = 0
for j in range (g * 2 , k + 1 , g) :
extra = extra + count[j]
extra = extra % MOD
count[g] = temp - extra + MOD
count[g] = count[g] % MOD
Sum = 0
for i in range ( 1 , k + 1 ) :
add = count[i] % MOD * i % MOD
add = add % MOD
Sum = Sum + add
Sum = Sum % MOD
return Sum
N, K = 3 , 2
print (sumofGCD(N, K))
|
C#
using System;
class GFG{
static int MOD = ( int )1e9 + 7;
static int fastexpo( int a, int b)
{
int res = 1;
a %= MOD;
while (b > 0) {
if (b % 2 == 1)
res = (res * a) % MOD;
a *= a;
a %= MOD;
b >>= 1;
}
return res;
}
static int sumofGCD( int n, int k)
{
int []count = new int [k + 1];
for ( int g = k; g >= 1; g--) {
int count_multiples = k / g;
int temp;
temp = fastexpo(count_multiples, n);
temp %= MOD;
int extra = 0;
for ( int j = g * 2; j <= k; j += g) {
extra = (extra + count[j]);
extra %= MOD;
}
count[g] = (temp - extra + MOD);
count[g] %= MOD;
}
int sum = 0;
int add;
for ( int i = 1; i <= k; ++i) {
add = (count[i] % MOD * i % MOD);
add %= MOD;
sum += add;
sum %= MOD;
}
return sum;
}
public static void Main(String[] args)
{
int N = 3, K = 2;
Console.Write(sumofGCD(N, K));
}
}
|
Javascript
<script>
var MOD = 1000000007;
function fastexpo(a, b)
{
var res = 1;
a %= MOD;
while (b) {
if (b & 1)
res = (res * a) % MOD;
a *= a;
a %= MOD;
b >>= 1;
}
return res;
}
function sumofGCD( n, k)
{
var count = Array(k+1).fill(0);
for ( var g = k; g >= 1; g--) {
var count_multiples = k / g;
var temp;
temp = fastexpo(count_multiples, n);
temp %= MOD;
var extra = 0;
for ( var j = g * 2; j <= k; j += g) {
extra = (extra + count[j]);
extra %= MOD;
}
count[g] = (temp - extra + MOD);
count[g] %= MOD;
}
var sum = 0;
var add;
for ( var i = 1; i <= k; ++i) {
add = (count[i] % MOD * i % MOD);
add %= MOD;
sum += add;
sum %= MOD;
}
return sum;
}
var N = 3, K = 2;
document.write( sumofGCD(N, K));
</script>
|
Time Complexity: O( K*log(N) + K*log(log(K)) )
Auxiliary Space: O(K)
Last Updated :
24 Mar, 2022
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