Dados dois inteiros N anos K a tarefa é encontrar a soma do módulo K dos primeiros N números naturais, ou seja, 1%K + 2%K + ..... + N%K.
Exemplos:
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.Recommended Practice Codificação reversa Experimente!
Método 1:
Itere uma variável i de 1 a N, avalie e adicione i%K.
Abaixo está a implementação desta abordagem:
C++// C++ program to find sum of // modulo K of first N natural numbers. #include
// 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 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# 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 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 ?> <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>
Saída :
5
Complexidade de tempo: SOBRE).
Espaço Auxiliar: O(1)
Método 2:
Dois casos surgem neste método.
Caso 1: Quando N< K para cada número i N >= i >= 1 dará i como resultado quando operar com o módulo K. Portanto, a soma necessária será a soma do primeiro N número natural N*(N+1)/2.
Caso 2: Quando N >= K então os inteiros de 1 a K na sequência numérica natural produzirão 1 2 3 ..... K - 1 0 como resultado quando operar com o módulo K. Da mesma forma, de K + 1 a 2K, produzirá o mesmo resultado. Então a ideia é contar quantas vezes essa sequência aparece e multiplicá-la pela soma dos primeiros K - 1 números naturais.
Abaixo está a implementação desta abordagem:
C++// C++ program to find sum of modulo // K of first N natural numbers. #include
// 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 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# 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 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. ?> <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>
Saída :
5
Complexidade de tempo: O(1).
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Espaço Auxiliar: O(1)