Dados dos enteros N y K , la tarea es encontrar la suma del módulo K de los primeros N números naturales, es decir, 1%K + 2%K + ….. + N%K.
Ejemplos:
Input : N = 10 and K = 2. Output : 5 Sum = 1%2 + 2%2 + 3%2 + 4%2 + 5%2 + 6%2 + 7%2 + 8%2 + 9%2 + 10%2 = 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 = 5.
Método 1:
Iterar una variable i de 1 a N, evaluar y sumar i%K.
A continuación se muestra la implementación de este enfoque:
C++
// C++ program to find sum of // modulo K of first N natural numbers. #include <bits/stdc++.h> using namespace std; // Return sum of modulo K of // first N natural numbers. int findSum(int N, int K) { int ans = 0; // Iterate from 1 to N && // evaluating and adding i % K. for (int i = 1; i <= N; i++) ans += (i % K); return ans; } // Driver Program int main() { int N = 10, K = 2; cout << findSum(N, K) << endl; return 0; }
Java
// Java program to find sum of modulo // K of first N natural numbers. import java.io.*; class GFG { // Return sum of modulo K of // first N natural numbers. static int findSum(int N, int K) { int ans = 0; // Iterate from 1 to N && evaluating // and adding i % K. for (int i = 1; i <= N; i++) ans += (i % K); return ans; } // Driver program static public void main(String[] args) { int N = 10, K = 2; System.out.println(findSum(N, K)); } } // This code is contributed by vt_m.
Python3
# Python3 program to find sum # of modulo K of first N # natural numbers. # Return sum of modulo K of # first N natural numbers. def findSum(N, K): ans = 0; # Iterate from 1 to N && # evaluating and adding i % K. for i in range(1, N + 1): ans += (i % K); return ans; # Driver Code N = 10; K = 2; print(findSum(N, K)); # This code is contributed by mits
C#
// C# program to find sum of modulo // K of first N natural numbers. using System; class GFG { // Return sum of modulo K of // first N natural numbers. static int findSum(int N, int K) { int ans = 0; // Iterate from 1 to N && evaluating // and adding i % K. for (int i = 1; i <= N; i++) ans += (i % K); return ans; } // Driver program static public void Main() { int N = 10, K = 2; Console.WriteLine(findSum(N, K)); } } // This code is contributed by vt_m.
PHP
<?php // PHP program to find sum // of modulo K of first N // natural numbers. // Return sum of modulo K of // first N natural numbers. function findSum($N, $K) { $ans = 0; // Iterate from 1 to N && // evaluating and adding i % K. for ($i = 1; $i <= $N; $i++) $ans += ($i % $K); return $ans; } // Driver Code $N = 10; $K = 2; echo findSum($N, $K), "\n"; // This code is contributed by ajit ?>
Javascript
<script> // JavaScript program to find sum // of modulo K of first N natural // numbers. // Return sum of modulo K of // first N natural numbers. function findSum(N, K) { let ans = 0; // Iterate from 1 to N && evaluating // and adding i % K. for(let i = 1; i <= N; i++) ans += (i % K); return ans; } // Driver Code let N = 10, K = 2; document.write(findSum(N, K)); // This code is contributed by code_hunt </script>
Producción :
5
Complejidad temporal: O(N).
Método 2:
Dos casos surgen en este método.
Caso 1: Cuando N < K , para cada número i, N >= i >= 1, dará como resultado i al operar con módulo K. Entonces, la suma requerida será la suma del primer N número natural, N* (N+1)/2.
Caso 2: Cuando N >= K , entonces los números enteros de 1 a K en secuencia de números naturales producirán, 1, 2, 3, ….., K – 1, 0 como resultado cuando se opera con módulo K. Similarmente, de K + 1 a 2K, producirá el mismo resultado. Entonces, la idea es contar cuántas veces aparece esta secuencia y multiplicarla por la suma de los primeros K – 1 números naturales.
A continuación se muestra la implementación de este enfoque:
C++
// C++ program to find sum of modulo // K of first N natural numbers. #include <bits/stdc++.h> using namespace std; // Return sum of modulo K of // first N natural numbers. int findSum(int N, int K) { int ans = 0; // Counting the number of times 1, 2, .., // K-1, 0 sequence occurs. int y = N / K; // Finding the number of elements left which // are incomplete of sequence Leads to Case 1 type. int x = N % K; // adding multiplication of number of // times 1, 2, .., K-1, 0 sequence occurs // and sum of first k natural number and sequence // from case 1. ans = (K * (K - 1) / 2) * y + (x * (x + 1)) / 2; return ans; } // Driver program int main() { int N = 10, K = 2; cout << findSum(N, K) << endl; return 0; }
Java
// Java program to find sum of modulo // K of first N natural numbers. import java.io.*; class GFG { // Return sum of modulo K of // first N natural numbers. static int findSum(int N, int K) { int ans = 0; // Counting the number of times 1, 2, .., // K-1, 0 sequence occurs. int y = N / K; // Finding the number of elements left which // are incomplete of sequence Leads to Case 1 type. int x = N % K; // adding multiplication of number of times // 1, 2, .., K-1, 0 sequence occurs and sum // of first k natural number and sequence // from case 1. ans = (K * (K - 1) / 2) * y + (x * (x + 1)) / 2; return ans; } // Driver program static public void main(String[] args) { int N = 10, K = 2; System.out.println(findSum(N, K)); } } // This Code is contributed by vt_m.
Python3
# Python3 program to find sum of modulo # K of first N natural numbers. # Return sum of modulo K of # first N natural numbers. def findSum(N, K): ans = 0; # Counting the number of times # 1, 2, .., K-1, 0 sequence occurs. y = N / K; # Finding the number of elements # left which are incomplete of # sequence Leads to Case 1 type. x = N % K; # adding multiplication of number # of times 1, 2, .., K-1, 0 # sequence occurs and sum of # first k natural number and # sequence from case 1. ans = ((K * (K - 1) / 2) * y + (x * (x + 1)) / 2); return int(ans); # Driver Code N = 10; K = 2; print(findSum(N, K)); # This code is contributed by mits
C#
// C# program to find sum of modulo // K of first N natural numbers. using System; class GFG { // Return sum of modulo K of // first N natural numbers. static int findSum(int N, int K) { int ans = 0; // Counting the number of times 1, 2, .., // K-1, 0 sequence occurs. int y = N / K; // Finding the number of elements left which // are incomplete of sequence Leads to Case 1 type. int x = N % K; // adding multiplication of number of times // 1, 2, .., K-1, 0 sequence occurs and sum // of first k natural number and sequence // from case 1. ans = (K * (K - 1) / 2) * y + (x * (x + 1)) / 2; return ans; } // Driver program static public void Main() { int N = 10, K = 2; Console.WriteLine(findSum(N, K)); } } // This code is contributed by vt_m.
PHP
<?php // PHP program to find sum of modulo // K of first N natural numbers. // Return sum of modulo K of // first N natural numbers. function findSum($N, $K) { $ans = 0; // Counting the number of times // 1, 2, .., K-1, 0 sequence occurs. $y = $N / $K; // Finding the number of elements // left which are incomplete of // sequence Leads to Case 1 type. $x = $N % $K; // adding multiplication of number // of times 1, 2, .., K-1, 0 // sequence occurs and sum of // first k natural number and // sequence from case 1. $ans = ($K * ($K - 1) / 2) * $y + ($x * ($x + 1)) / 2; return $ans; } // Driver program $N = 10; $K = 2; echo findSum($N, $K) ; // This code is contributed by anuj_67. ?>
Javascript
<script> // Javascript program to find sum of modulo // K of first N natural numbers. // Return sum of modulo K of // first N natural numbers. function findSum(N, K) { let ans = 0; // Counting the number of times // 1, 2, .., K-1, 0 sequence occurs. let y = N / K; // Finding the number of elements // left which are incomplete of // sequence Leads to Case 1 type. let x = N % K; // adding multiplication of number // of times 1, 2, .., K-1, 0 // sequence occurs and sum of // first k natural number and // sequence from case 1. ans = (K * (K - 1) / 2) * y + (x * (x + 1)) / 2; return ans; } // Driver code let N = 10; let K = 2; document.write(findSum(N, K)); // This code is contributed by _saurabh_jaiswal </script>
Producción :
5
Complejidad temporal: O(1).
Este artículo es una contribución de Anuj Chauhan . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA