Added The ETF theory In Algorithms Section Of Cpp lang

CPP/algorithms/numbertheory/ETF1.cpp

@@ -0,0 +1,94 @@
+// It is function
+
+// denoted by ɸ(x)
+
+// ɸ(x) = count of number from [1,x] which are coprime with x
+
+//  if gcd(x,y) = 1 then x and y are co prime
+
+// ɸ(5) = 4
+
+// 5 means [1,5] = 1,2,3,4,5
+
+// gcd(1,5) = 1 , gcd(2,5) = 1  , gcd(3,5) = 1  , gcd(4,5) = 1 // count is 4 (means 4 numberes are there)
+
+// ɸ(10) = 4
+
+
+// Prime Number and ETF
+
+// ɸ(2) = 1
+// ɸ(3) = 2
+// ɸ(5) = 4
+// ɸ(7) = 6
+// ɸ(11) = 10
+
+// ɸ(PrimeNumber) = PrimeNumber-1  (it is very easy to understand because gcd with Prime will be 1 so from 1 to Prime-1 will be counted)
+
+
+// ɸ(Pˣ) = ?  , where is x is positive integer , x >= 1 && P = prime number
+
+// ɸ(2⁵) = 16
+
+// ɸ(3⁵) = 162
+
+/*
+*  ɸ(Pˣ) = Pˣ - number of integers that are not co prime with P 
+*  ɸ(Pˣ) = Pˣ - number of integers that are mutilple of P  
+*  ɸ(Pˣ) = Pˣ - number of inetgers that are mutilple of P from 1 to Pˣ (suppose 3 how many multiple of 3 from to 100 , ans = 100/3 )  
+*  ɸ(Pˣ) = Pˣ - Pˣ/P
+*  ɸ(Pˣ) = Pˣ - Pˣ⁻¹
+*  ɸ(Pˣ) = Pˣ⁻¹(P-1)
+
+
+*/
+
+
+// generalizing the formula
+
+
+// The one property about ETF
+
+// ETF is a multiplicative function 
+
+/*
+* multiplcative means 
+
+for a f(n) arithmetic function
+
+if f(n*m) = f(n) * f(m) , where gcd(n,m) = 1;
+*/
+
+/*
+
+using above property we can derive the generlize formula
+
+let ɸ(x) ETF is multiplicative
+using it to find ɸ(n)
+
+n =  p₁ˣ₁ X  p₂ˣ₂ ... pₖˣₖ () it is prime factorization
+
+
+ɸ(n) = ɸ(p₁ˣ₁ X  p₂ˣ₂ ... pₖˣₖ)
+
+since ETF is multiplicative
+
+ɸ(n) = ɸ(p₁ˣ₁) X  ɸ(p₂ˣ₂) X  ... ɸ(pₖˣₖ) 
+
+since pᵢ is prime and we know ɸ(Pˣ) = Pˣ⁻¹(P-1)
+
+ɸ(n) = P₁ˣ₁⁻¹(P₁-1) X P₂ˣ₂⁻¹(P₂-1) ... X Pₖˣₖ⁻¹(Pₖ-1) // generalized formula TC(O) = sqrt(n) * logn
+
+
+ɸ(n) = P₁ˣ₁((P₁-1)/P₁) X P₂ˣ₂((P₂-1)/P₂) ... X Pₖˣₖ((Pₖ-1)/Pₖ) 
+ɸ(n) = P₁ˣ₁ * P₂ˣ₂ ...* Pₖˣₖ  ((P₁-1)/P₁) X ((P₂-1)/P₂) ... X ((Pₖ-1)/Pₖ) 
+
+but P₁ˣ₁ * P₂ˣ₂ ...* Pₖˣₖ = n
+ɸ(n) = n  *  ((P₁-1)/P₁) X ((P₂-1)/P₂) ... X ((Pₖ-1)/Pₖ)  // easy to calculate TC(O) = sqrt(n)
+
+
+*/
+
+
+
+

CPP/algorithms/numbertheory/ETF2.cpp

@@ -0,0 +1,55 @@
+// ɸ(n) = n  *  ((P₁-1)/P₁) X ((P₂-1)/P₂) ... X ((Pₖ-1)/Pₖ)
+
+#include<bits/stdc++.h>
+#define raj_01 ios::sync_with_stdio(false);cin.tie(0);
+#define speed1 cin.tie(0);
+#define speed2 cout.tie(0);
+#define print(val) cout << val << endl;
+#define for1(i,n) for(int i = 0; i < n ; i++)
+#define fork(i,n,k) for(int i = 0; i < n ; i+=k)
+#define printcontainer(i,container) for1(i,container.size()) print(container[i])
+#define INF 1e18
+#define INIT_MATRIX(rows, cols, value) \
+int matrix[rows][cols];            \
+for (int i = 0; i < rows; ++i)     \
+for (int j = 0; j < cols; ++j) \
+cin >> matrix[i][j]
+#define MOD 1000000007
+#define ll long long
+#define lli long long int
+#define vi vector<int>
+#define pii pair<int, int>
+#define vvi vector<vi>
+#define all(x) (x).begin(), (x).end() 
+#define endl "\n"
+#define debug(x) cout << #x << " = " << x << endl
+#define debug2(x, y) cout << #x << " = " << x << #y << " = " << y << endl
+#define debug3(x, y, z) cout << #x << " = " << x << #y << " = " << y << int << #z << " = " << z << endl
+using namespace std;
+
+int phi(int n) {
+    int res = n;
+    for(int i = 2 ; i * i <= n ; i++) {
+        if(n%i==0) {
+            res /= i;
+            res *= (i-1);
+            while(n%i==0) n/=i;
+        }
+    }
+    if(n>1) {
+        res /= n;
+        res *= (n-1);
+    }
+    return res;
+}
+
+int main(){
+    raj_01
+    int test,n;
+    cin >> test;
+    while(test--) {
+        cin >> n;
+        cout << phi(n) << endl;
+    }
+    return 0;
+}

CPP/algorithms/numbertheory/ETF3.cpp

@@ -0,0 +1,55 @@
+// efficient way to calculate all etf values from 1 to 10⁶
+
+
+#include<bits/stdc++.h>
+#define raj_01 ios::sync_with_stdio(false);cin.tie(0);
+#define speed1 cin.tie(0);
+#define speed2 cout.tie(0);
+#define print(val) cout << val << endl;
+#define for1(i,n) for(int i = 0; i < n ; i++)
+#define fork(i,n,k) for(int i = 0; i < n ; i+=k)
+#define printcontainer(i,container) for1(i,container.size()) print(container[i])
+#define INF 1e18
+#define INIT_MATRIX(rows, cols, value) \
+int matrix[rows][cols];            \
+for (int i = 0; i < rows; ++i)     \
+for (int j = 0; j < cols; ++j) \
+cin >> matrix[i][j]
+#define MOD 1000000007
+#define ll long long
+#define lli long long int
+#define vi vector<int>
+#define pii pair<int, int>
+#define vvi vector<vi>
+#define all(x) (x).begin(), (x).end() 
+#define endl "\n"
+#define debug(x) cout << #x << " = " << x << endl
+#define debug2(x, y) cout << #x << " = " << x << #y << " = " << y << endl
+#define debug3(x, y, z) cout << #x << " = " << x << #y << " = " << y << int << #z << " = " << z << endl
+using namespace std;
+
+int ETF[1000001];
+
+void init() {
+    ETF[0] = -1; // not defined for 0
+    for(int i = 1 ; i < 1000001 ; i++) ETF[i] = i;
+
+    for(int i = 2 ; i  < 1000001 ; i++) {
+        if(ETF[i] == i) {
+            for(int j = i ; j < 1000001 ;j += i) {
+                ETF[j] /= i;
+                ETF[j] *= (i-1);
+            }
+        } 
+    }
+}
+int main(){
+    raj_01
+    init();
+    int t,n;
+    cin >> t;
+    while(t--) {
+        cin >> n , cout << ETF[n] << endl;
+    }
+    return 0;
+}

CPP/algorithms/numbertheory/ETF4.cpp

@@ -0,0 +1,30 @@
+#include<bits/stdc++.h>
+#define raj_01 ios::sync_with_stdio(false);cin.tie(0);
+#define speed1 cin.tie(0);
+#define speed2 cout.tie(0);
+#define print(val) cout << val << endl;
+#define for1(i,n) for(int i = 0; i < n ; i++)
+#define fork(i,n,k) for(int i = 0; i < n ; i+=k)
+#define printcontainer(i,container) for1(i,container.size()) print(container[i])
+#define INF 1e18
+#define INIT_MATRIX(rows, cols,value) \
+int matrix[rows][cols];            \
+for (int i = 0; i < rows; ++i)     \
+for (int j = 0; j < cols; ++j) \
+matrix[i][j] = value;
+#define MOD 1000000007
+#define ll long long
+#define lli long long int
+#define vi vector<int>
+#define pii pair<int, int>
+#define vvi vector<vi>
+#define all(x) (x).begin(), (x).end() 
+#define endl "\n"
+#define debug(x) cout << #x << " = " << x << endl
+#define debug2(x, y) cout << #x << " = " << x << #y << " = " << y << endl
+#define debug3(x, y, z) cout << #x << " = " << x << #y << " = " << y << int << #z << " = " << z << endl
+using namespace std;
+int main(){
+    raj_01
+    return 0;
+}

CPP/algorithms/numbertheory/ETF5.cpp

@@ -0,0 +1,115 @@
+// problem to be solved is 
+
+// ans  = Σ (1 to N) gcd(i , N)   // for q queries
+
+
+// observation one = GCD(i,N) = one of the divisors of N
+
+// Assume N = 12
+
+// gcd(1,12) = 1
+// gcd(2,12) =  2
+// gcd(3,12) = 3
+// gcd(4,12) = 4
+// gcd(5,12) = 1
+// gcd(6,12) = 6
+// gcd(7,12) = 1
+// gcd(8,12) = 4
+// gcd(9,12) = 3
+// gcd(10,12) = 2
+// gcd(11,12) = 1
+// gcd(12,12)  12
+
+// sum of gcds =   1 + 2 + 3 + 4 + 1 + 6 +1 + 4 +3 +2 + 1 +12
+
+// sum of gcds  = 1*4 + 2*2 + 3*2 + 4*2 + 12*1
+
+// we can say that 
+
+// sum of gcd = (for divisor 1 * how many nymber are from 1 to 12 have gcd equal divisor(1))
+
+
+// like for divisor 2 , how many numbers from 1 to 12 have gcd 2 that will be the contibution  of gcd 2 with that count of numbers
+
+// this is the way to calculate gcd sum
+
+// for x1 , x2 , ...xm for a divisor d such that gcd(xi , N) = d
+
+// have to find m (count)
+
+// gcd(xi/d , N/d) = 1;
+
+// means how many xi/d are there in 1 to N such that their gcd with N/D is 1
+
+// 1 means co prime ==> co prime count ==> ETF(N/d) that is the answer
+
+#include<bits/stdc++.h>
+#define raj_01 ios::sync_with_stdio(false);cin.tie(0);
+#define speed1 cin.tie(0);
+#define speed2 cout.tie(0);
+#define print(val) cout << val << endl;
+#define for1(i,n) for(int i = 0; i < n ; i++)
+#define fork(i,n,k) for(int i = 0; i < n ; i+=k)
+#define printcontainer(i,container) for1(i,container.size()) print(container[i])
+#define INF 1e18
+#define INIT_MATRIX(rows, cols, value) \
+int matrix[rows][cols];            \
+for (int i = 0; i < rows; ++i)     \
+for (int j = 0; j < cols; ++j) \
+cin >> matrix[i][j]
+#define MOD 1000000007
+#define ll long long
+#define lli long long int
+#define vi vector<int>
+#define pii pair<int, int>
+#define vvi vector<vi>
+#define all(x) (x).begin(), (x).end() 
+#define endl "\n"
+#define debug(x) cout << #x << " = " << x << endl
+#define debug2(x, y) cout << #x << " = " << x << #y << " = " << y << endl
+#define debug3(x, y, z) cout << #x << " = " << x << #y << " = " << y << int << #z << " = " << z << endl
+using namespace std;
+
+
+int ETF[1000001];
+
+void init() {
+    ETF[0] = -1; // not defined for 0
+    for(int i = 1 ; i < 1000001 ; i++) ETF[i] = i;
+
+    for(int i = 2 ; i  < 1000001 ; i++) {
+        if(ETF[i] == i) {
+            for(int j = i ; j < 1000001 ;j += i) {
+                ETF[j] /= i;
+                ETF[j] *= (i-1);
+            }
+        } 
+    }
+}
+
+int getCount(int d , int N) {
+    return ETF[N/d];
+}
+
+
+int main(){
+    raj_01
+    init();
+    int t , N;
+    cin >> t;
+    while(t--){
+        cin >> N;
+        int res = 0;
+        for(int i = 1 ; i*i<=N ; i++) {
+            if(N%i==0) {
+                int d1 = i , d2 = N/i;
+                res += (d1*getCount(d1,N));
+                if(d1 != d2) res += (d2*getCount(d2,N));
+            }
+        } // sqrt(n) = TC(O )
+        cout << res << endl;
+    }
+    return 0;
+}
+
+