nt

Number Theory/binaryExp.cpp

@@ -0,0 +1,67 @@
+/*
+*Binary Exponentiation*
+(a^b)%m sometimes becomes out of range
+
+properties of modulo
+(a+b)%m = (a%m) + (b%m)
+(a*b)%m = (a%m) * (b%m)
+(a-b)%m = (a%m) - (b%m)
+(a/b)%m = (a%m) * (b^-1%m)
+
+recursive case
+a^n
+when n is odd
+a^n = a^n/2 * a^n/2 * a
+
+when a is even
+a^n = a^n/2 * a^n/2
+
+base case 
+n=0 then return 1
+
+
+iterative
+a^45
+45 = 1x2^5 + 0x2^4 + 1x2^3 + 1x2^2 + 0x2^1 + 1x2^0
+45 = (101101)2
+7^45 can be calculated easily
+2^3 = 2^2 + 2^2
+
+start with var x = 1
+keep multiplying it with itself
+when there is 1 in binary rep
+multiply it with ans
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+#define int long long
+const int N = 1e5 + 2, MOD = 1e9 + 7;
+
+int power(int a, int n)
+{
+
+    if (n == 0)
+    {
+        return 1;
+    }
+    int p = (power(a, n / 2) % MOD);
+
+    if (n & 1) //if n is odd
+    {
+        return (((p * p) % MOD) * a) % MOD;
+    }
+    else
+    {
+        return (p * p) % MOD;
+    }
+}
+
+signed main()
+{
+    int a, n;
+    cin >> a >> n;
+    a %= MOD;
+    cout << power(a, n) << endl;
+    return 0;
+}

Number Theory/pigeon.cpp

@@ -0,0 +1,82 @@
+/*
+Divisible Sub-arrays
+*Pigeonhole Principle*
+
+array containing N elements
+we have to find number of good sub arrays
+sub array whose sum is divisible by N
+
+n no of holes 4
+and n+1 pigeons 3
+allocate pigeons in holes
+1 hole will have 2 pigeons
+
+cumulative sum array -> will contain sum upto particular index
+1, (3, 2,) 6, 4
+   a    b  
+   total sum upto this point is b
+   and total sum upto this point is a
+   sum of sub array will be b-a
+   (b-a)%N = 0
+   b%N = a%N
+
+   1, 3, 2, 6, 4 -> original array
+0, 1, 4, 6,12,16 -> sum upto a point
+
+0, 1, 4, 1, 2, 1 -> sum%n (cumulative sum array) 6 diff values after taking mod 5
+x % 5 -> max 5 unique values
+atleast 2 boxes will have same value
+
+buckets 
+0->1
+1->3 if we take any two pos we will get the same sum  -> mC2 sub arrays
+2->1
+3->0
+4->1
+we will add whereever we have atleast 2 freq
+and our ans will be 3c2 as m =3
+
+ans+= fc2
+where f>=2
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+#define ll long //constraints are long multiplying 2 numbers which are 10^9
+
+ll a[1000005];   //array
+ll pre[1000005]; //frequency array
+
+int main()
+{
+    int t;
+    cin >> t;
+    while (t--)
+    {
+        int n;
+        cin >> n;
+        memset(pre, 0, sizeof(pre)); //initializing every bucket with 0
+
+        pre[0] = 1; //freq of 0 will be 1 because of null sub array
+
+        //read input
+        int sum = 0;
+        for (int i = 0; i < n; i++)
+        {
+            cin >> a[i];         //input array element
+            sum += a[i];         //checking sum till that point
+            sum %= n;            //mod with n is cumulative sum
+            sum = (sum + n) % n; //for negative case
+            pre[sum]++;          //increasing frequency of that number (creating freq array)
+        }
+
+        ll ans = 0;
+        for (int i = 0; i < n; i++) //going to all positions
+        {
+            ll m = pre[i];
+            ans += (m) * (m - 1) / 2; //nC2
+        }
+        cout << ans << endl;
+    }
+    return 0;
+}

Number Theory/topic.txt

@@ -3,8 +3,8 @@
 2. Scaline Algorithm
 3. Sliding Window + Two Pointer Technique
 4. Modular Arithmetic
-5. Binary and Modular Exponentiation
+5. Binary and Modular Exponentiation         -> done
 6. Euclid GCD                                -> done
 7. Extended Euclid's Theorem                 -> done
-8. Pigeonhole Principle
+8. Pigeonhole Principle                      -> done
 9. Prime sieve                               -> done