求第N个自然数的模K的和

给定两个整数N ans K ，任务是找到前N个自然数的模K的和，即1％K + 2％K +….. + N％K。

例子 ：

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.

方法1：

将变量i从1迭代到N，求值并加上i％K。

以下是此方法的实现：

C++

// C++ program to find sum of
// modulo K of first N natural numbers.
#include 
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.

Javascript

 
// 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
 

输出 ：

5

时间复杂度： O(N)。

方法2：
此方法出现两种情况。
情况1：当N ，对于每个数字i，N> = i> = 1时，以模K进行运算时将得到i。因此，所需的总和将是第一个N个自然数之和N * (N + 1)/ 2。
情况2：当N> = K时，自然数序列中从1到K的整数将对模K进行运算，结果将产生1，2，3，….. K – 1，0。 1至2K，将产生相同的结果。因此，想法是计算该序列出现的次数，然后将其乘以第一个K – 1个自然数的总和。

以下是此方法的实现：

C++

// C++ program to find sum of modulo
// K of first N natural numbers.
#include 
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.

Java脚本

 
// 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
 

输出 ：

5

时间复杂度： O(1)。