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Page 1257 of 2109
Geometry using Complex Numbers in C++
In this section, we will see how to make point class using complex class from STL in C++. And apply them on some geometry related problems. The complex number is present inside the complex class from STL (#include )Defining Point ClassTo make complex to point, we will change the name of the complex as point, then change x to real() of complex class and y to imag() of complex class. Thus, we can simulate the point class.# include typedef complex point; # define x real() # define y imag()We have to keep in mind that the x and y ...
Read MoreUser Defined Literals in C++
Here we will see the concept of the user-defined literals in C++. From C++ version 11, the User Defined Literals (UDL) are added in C++. C++ also provides literals for a variety of built-in types but these are limited.Built-in Literals −31 (Integer)3.5 (Double)4.2F (Float)'p' (Character)31ULL (Unsigned Long Long)0xD0 (Unsigned Hexadecimal Integer)"pq" (String)Apart from the built-in literals, sometimes we need user defined literals. There are few reasons behind that. Let us see with few examples −Suppose we want to define one weight variable, but we cannot specify the units, like if we define as follows −long double Weight = 3.5;We have ...
Read MoreCount number of primes in an array in C++
We are given with an array of numbers. The goal is to find the count of prime numbers in that array.A prime number is the one which is divisible by 1 and the number itself. It has only two factors. We will check if the number is prime starting from the first element till the last and increase the count of prime numbers found so far.To check if the number N is prime, check if numbers between the range [2 to N/2], fully divides N. If yes then it is non-prime. Else it is prime.Let’s understand with examples.Input − arr[]= ...
Read MoreCount number of solutions of x^2 = 1 (mod p) in given range in C++
We are given with integers x and p. The goal is to find the number of solutions of the equation −x2=1 ( mod p ) such that x lies in range [1, N].We will do this by traversing from 1 to N and take each number as x check if (x*x)%p==1. If yes then increment the count.Let’s understand with examples.Input − n=5, p=2Output − Number of Solutions − 3Explanation − Between the range 1 to 5.12=1%2=1, count=1 22=4%2=0, count=1 32=9%2=1, count=2 42=16%2=0, count=2 52=25%2=1, count=3 Total number of solutions=3.Input − n=3, p=4Output − Number of Solutions − 2Explanation − Between ...
Read MoreCount the triplets such that A[i] < B[j] < C[k] in C++
We are given with three arrays A[], B[] and C[]. The goal is to find all triplets of elements of these arrays such that A[i]
Read MoreCount total divisors of A or B in a given range in C++
We are given four integers L, R, A and B. The goal is to find the count of numbers in range [L, R] that fully divide either A or B or both.We will do this by traversing from L to R and for each number if number%A==0 or number%B==0 then increment count of divisors.Let’s understand with examples.Input − L=10, R=15, A=4, B=3Output − Count of divisors of A or B − 2Explanation −Number 12 is fully divisible by 3 and 4. Number 15 is fully divisible by 3 only. Total divisors=2Input − L=20, R=30, A=17, B=19Output − Count of divisors ...
Read MoreCount the numbers divisible by 'M' in a given range in C++
We are given three numbers A,B and M. A and B define the range [A,B] of numbers.The goal is to count numbers between A and B that are divisible by M.We will start from i=A till first multiple of M. Increment count if i%M=0. Now increment i till i
Read MoreCount the number of possible triangles in C++
We are given an array which contains the length of sides of triangles. The goal is to find the number of possible triangles that can be made by taking any three sides from that array.We will do this by checking if the sum of any two is always > third side. If yes these three sides can make a triangle. Increment count of possible triangles that can be made.Let’s understand with examples.Input − arr[]= {1, 2, 4, 5}Output − Count of possible triangles − 1Explanation − Sides (2, 4, 5) can only make a triangle as 2+4>5 & 4+5>2 & ...
Read MoreCount the number of pairs (i, j) such that either arr[i] is divisible by arr[j] or arr[j] is divisible by arr[i] in C++
We are given with an array arr[] of N elements. The goal is to find the count of all valid pairs of indexes (i, j) such that either arr[i] is divisible by arr[j] or arr[j] is divisible by arr[i] and i!=j.We will do this by traversing the array arr[] using two for loops for each number of pair and check if arr[i]%arr[j]==0 or arr[j]%arr[i]==0 when i!=j. If true increment count of pairs.Let’s understand with examples.Input − Arr[]= { 2, 4, 3, 6 } N=4Output − Count of valid pairs − 3Explanation − Valid pairs are −Arr[0] & Arr[1] → (2, ...
Read MoreCount numbers with same first and last digits in C++
We are given an interval [first, last]. The goal is to find the count of numbers that have the same first and last digit within this interval. For example, 232 has the same first and last digit as 2.We will do this by traversing from i=first to i=last. For each number I compare its first digit with the last digit, if they are the same increment the count.Let’s understand with examples.Input − first=8 last=40Output − Count of numbers with same first and last digits − 5Explanation − Numbers between 8 and 40 with same first and last digit −8, 9, ...
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