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/* N = 100 Factor Power   2      2   5      2 N = 35 Factor  Power   5      1   7      1     nput: 2 100 35 Output: 2 2 5 2 5 1 7 1 we can see it as key value printing we can use map */ #include <bits/stdc++.h> using namespace std; void primeFactors(int n) { map<int,int> m; while(n%2==0) // runs two times for 100 making it 25, covered divisible by 4 { m[2]++; n=n/2; } for(int i=3;i<=n;i=i+2) // checking for all odd numbers note if 3 checked then it wont check for 9 only primes { while(n%i==0) { m[i]++; n=n/i; } } map<int,int> ::iterator itr; for(itr=m.begin();itr!=m.end();itr++) { cout<<itr->first<<" "<<itr->second<<" "; } cout<<endl; } int main() { int n,i,t; cin >> t; while(t--){     cin >> n; ...

Examples: LCS for input Sequences “ABCDGH” and “AEDFHR” is “ADH” of length 3. LCS for input Sequences “AGGTAB” and “GXTXAYB” is “GTAB” of length 4. #include<bits/stdc++.h>   int max(int a, int b);   int lcs( char *X, char *Y, int m, int n ) {    int L[m+1][n+1];    int i, j;  for (i=0; i<=m; i++)    {      for (j=0; j<=n; j++)      {        if (i == 0 || j == 0)          L[i][j] = 0;          else if (X[i-1] == Y[j-1])          L[i][j] = L[i-1][j-1] + 1;          else          L[i][j] = max(L[i-1][j], L[i][j-1]);      }    }       //L[m][n] contains length of LCS for X[0..n-1] and Y[0..m-1]    return L[m][n]; }   // Utility function to get max of 2 integers int max(int a, in...

Count all possible paths from top left to bottom right The task is to count all the possible paths from top left to bottom right of a mXn matrix with the constraints that from each cell you can either move only to right or down. Input:  First line consists of T test cases. First line of every test case consists of N and M, denoting the number of rows and number of column respectively. Output:  Single line output i.e count of all the possible paths from top left to bottom right of a mXn matrix. Since output can be very large number use %10^9+7. Constraints: 1<=T<=100 1<=N<=100 1<=M<=100 Example: Input: 1 3 3 Output: 6 #include <iostream> using namespace std; int path(int m , int n) { int a[m][n]; // now we want to find the total paths from left top to bottom right //we use the concept of longest common subsequence to do it //in l.c.s we check if a[i]==a[j] // if true then done add 1 to diagnol eleme...

// A Dynamic Programming based solution that uses  // table C[][] to calculate the Binomial Coefficient #include<bits/stdc++.h> using namespace std;   // Prototype of a utility function that // returns minimum of two integers int min(int a, int b);   // Returns value of Binomial Coefficient C(n, k) int binomialCoeff(int n, int k) {     int C[n + 1][k + 1];     int i, j;       // Caculate value of Binomial Coefficient     // in bottom up manner     for (i = 0; i <= n; i++)     {         for (j = 0; j <= min(i, k); j++)         {             // Base Cases             if (j == 0 || j == i)                 C[i][j] = 1;               // Calculate value using previously         ...

// sieve of Atkin is a popular way to find all the prime numbers in range using the dp approach and is //better than the erronthesis approach // C++ program for implementation of Sieve of Atkin #include <bits/stdc++.h> using namespace std;   int SieveOfAtkin(int limit) {     // 2 and 3 are known to be prime     if (limit > 2)         cout << 2 << " ";     if (limit > 3)         cout << 3 << " ";       // Initialise the sieve array with false values     bool sieve[limit];     for (int i = 0; i < limit; i++)         sieve[i] = false;     for (int x = 1; x * x < limit; x++) {         for (int y = 1; y * y < limit; y++) {                           // Main part of Sieve of Atkin       ...

/* Given two integers L and R find the difference of number of composites and primes between the range L and R (both inclusive). Input: First line of input contains of an integer 'T' denoting number of test cases . Then T test cases follow. Each test contains two integers L and R . Output: For each case print an integer corresponding to the answer . Constraints: 1<=T<=100 1<=L<=R<=10^7 Example: Input: 2 4 4 4 6 Output: 1 1 */ // given l , r print no of comp- no of prime  T.C->0.21 // normal approac i*i<=n gives TLE > 3.14 #include<bits/stdc++.h> using namespace std; bool prime[100001]; void SieveOfEratosthenes() {     memset(prime, true, sizeof(prime));     prime[0]=false;     prime[1]=false;     for (int p=2; p*p<=10000000; p++)     {             if (prime[p] == true) // we make all multiples of prime false       ...

/* Given an array of n integers where each value represents number of chocolates in a packet. Each packet can have variable number of chocolates. There are m students, the task is to distribute chocolate packets such that: Each student gets one packet. The difference between the number of chocolates in packet with maximum chocolates and packet with minimum chocolates given to the students is minimum. Examples: Input :  arr[] = {7, 3, 2, 4, 9, 12, 56} m = 3 Output:  Minimum Difference is 2 We have seven packets of chocolates and we need to pick three packets for 3 students If we pick 2, 3 and 4, we get the minimum difference between maximum and minimum packet sizes. Input :  arr[] = {3, 4, 1, 9, 56, 7, 9, 12} m = 5 Output:  Minimum Difference is 6 The set goes like 3,4,7,9,9 and the output is 9-3 = 6 Input :  arr[] = {12, 4, 7, 9, 2, 23, 25, 41, 30, 40, 28, 42, 30, 44, 48, 43, 50} m = 7 Output:  Minimum Difference is 10 We need to pick 7 pack...

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