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Why do we use const qualifier in C++?

Srinivas Gorla
Srinivas Gorla
Updated on 10-Feb-2020 2K+ Views

We use the const qualifier to declare a variable as constant. That means that we cannot change the value once the variable has been initialized. Using const has a very big benefit. For example, if you have a constant value of the value of PI, you wouldn't like any part of the program to modify that value. So you should declare that as a const.Objects declared with const-qualified types may be placed in read-only memory by the compiler, and if the address of a const object is never taken in a program, it may not be stored at all.  For ...

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Count all possible position that can be reached by Modified Knight in C++

Ayush Gupta
Ayush Gupta
Updated on 10-Feb-2020 187 Views

In this tutorial, we will be discussing a program to find the number of possible positions that can be reached by Modified Knight.For this we will be provided with a 8*8 chessboard. Our task is to find the number of positions Modified Knight can capture with the given number of steps.Example#include using namespace std; //finding the positions void findSteps(int current_row, int current_column, int curr, int board_size, int steps, int* visited){    //bound checking    if (current_row >= board_size || current_row < 0       || current_column >= board_size || current_column < 0       || curr > ...

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Count all possible paths between two vertices in C++

Ayush Gupta
Ayush Gupta
Updated on 10-Feb-2020 492 Views

In this tutorial, we will be discussing a program to find the number of paths between two vertices.For this we will be provided with a directed graph. Our task is to find the number of paths possible between two given vertices.Example#include using namespace std; //constructing a directed graph class Graph{    int V;    list *adj;    void countPathsUtil(int, int, bool [], int &);    public:       //constructor       Graph(int V);       void addEdge(int u, int v);       int countPaths(int s, int d); }; Graph::Graph(int V){    this->V = V;    adj ...

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Count all possible N digit numbers that satisfy the given condition in C++

Ayush Gupta
Ayush Gupta
Updated on 10-Feb-2020 239 Views

In this tutorial, we will be discussing a program to find the number of possible N digit numbers that satisfy the given condition.For this we will be provided with an integer. Our task is to check which one of number having N digits followNumber + Reverse(Number) = 10N -1Example#include using namespace std; //returning the count of numbers string count_num(int N){    if (N % 2 == 1)       return 0;    string result = "9";    for (int i = 1; i

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Count all perfect divisors of a number in C++

Ayush Gupta
Ayush Gupta
Updated on 10-Feb-2020 369 Views

In this tutorial, we will be discussing a program to find the number of all perfect divisors of a number.For this we will be provided with a number. Our task is to count all the perfect divisors of that given number.Example#include using namespace std; //checking perfect square bool if_psquare(int n){    int sq = (int) sqrt(n);    return (n == sq * sq); } //returning count of perfect divisors int count_pdivisors(int n){    int count = 0;    for (int i=1; i*i

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Count all Palindromic Subsequence in a given String in C++

Ayush Gupta
Ayush Gupta
Updated on 10-Feb-2020 302 Views

In this tutorial, we will be discussing a program to find the number of all palindromic subsequences in a given string.For this we will be provided with a string. Our task is to find the number of palindromic subsequences that can be made in that given string.Example#include #include using namespace std; //returning total palindromic sequence int count_palin(string str){    int N = str.length();    //creating a 2D array    int cps[N+1][N+1];    memset(cps, 0 ,sizeof(cps));    for (int i=0; i

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Count all palindrome which is square of a palindrome in C++

Ayush Gupta
Ayush Gupta
Updated on 10-Feb-2020 232 Views

In this tutorial, we will be discussing a program to find the number of palindromes which are squares of a palindrome.For this we will be provided with two values L and R. Our task is to find the number of super palindromes in the given range. A super palindrome is the one in which the number and its square both are palindromes.Example#include using namespace std; //checking if the number is a palindrome bool if_palin(int x){    int ans = 0;    int temp = x;    while (temp > 0){       ans = 10 * ans + ...

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Count all Palindrome Sub-Strings in a String in C++

Ayush Gupta
Ayush Gupta
Updated on 10-Feb-2020 282 Views

In this tutorial, we will be discussing a program to find the number of palindrome sub strings in a string.For this we will be given a string. Our task is to count the number of palindrome sub strings in the given string with length greater than 3.Example#include using namespace std; //counting palindrome strings int count_pstr(char str[], int n){    int dp[n][n];    memset(dp, 0, sizeof(dp));    bool P[n][n];    memset(P, false , sizeof(P));    for (int i= 0; i< n; i++)       P[i][i] = true;    for (int i=0; i

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Count all pairs with given XOR in C++

Ayush Gupta
Ayush Gupta
Updated on 10-Feb-2020 300 Views

In this tutorial, we will be discussing a program to find the number of pairs with the given XOR.For this we will be provided with an array and a value. Our task is to find the number of pairs whose XOR is equal to the given value.Example#include using namespace std; //returning the number of pairs //having XOR equal to given value int count_pair(int arr[], int n, int x){    int result = 0;    //managing with duplicate values    unordered_map m;    for (int i=0; i

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Count all pairs of an array which differ in K bits in C++

Ayush Gupta
Ayush Gupta
Updated on 10-Feb-2020 239 Views

In this tutorial, we will be discussing a program to find the number of pairs of an array which differ in K bits.For this we will be provided with an array and an integer K. Our task is to find the number of pairs who differ by K bits in their binary representation.Example#include using namespace std; //counting number of bits in //binary representation int count_bit(int n){    int count = 0;    while (n) {       if (n & 1)          ++count;       n >>= 1;    }    return count; } ...

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