查找给定n的(n ^ 1 + n ^ 2 + n ^ 3 + n ^ 4)mod 5的值

给定一个函数f(n)=(n ¹ + n ² + n ³ + n ⁴ )，对于任何给定的正整数n值，必须找到f(n)mod 5的值。
注意：n可能足够大，使得f(n)> 10 ¹⁸ 。
例子 ：

Input : n = 4
Output : 0
Explanation : f(4) = 4 + 16 + 64 + 256 = 330, 
f(4) mod 5 = 330 mod 5 = 0.

Input : n = 1 
Output : 4
Explanation : f(1) = 1 + 1 + 1 + 1 = 4, 
f(1) mod 5 = 4.

首先，为解决该问题，您可以直接借助任何幂函数和模运算符找到(n ¹ + n ² + n ³ + n ⁴ )mod 5的值。
但是对于较大的n值，您的结果将是错误的，因为对于较大的n值，f(n)可能超出long long int的范围，在这种情况下，您必须选择其他有效的方法。
为了解决这个问题，让我们对f(n)做一些小的数学推导。

f(n) = (n1 + n2 + n3 + n4)
     = (n) (n+1) (n2+1)
Now, for finding f(n) mod 5 we must take care of unit digit of f(n) only,
also as f(n) mod 5 is dependent on n%5, (n+1)%5 & (n2+1)%5,
if any of these three result in zero then our whole result is 0.
So, if n = 5, 10, .. 5k then n mod 5 = 0 hence f(n) mod 5 = 0.
if n = 4, 9, .., (5k-1) then (n+1) mod 5 = 0 hence f(n) mod 5 = 0.
if n = 3, 8, 13..., (5k-2) f(n) mod 5 = (3 * 4 * 10) mod 5 = 0
if n = 2, 7, 12..., (5k-3) f(n) mod 5 = (2 * 3 * 5) mod 5 = 0.
if n = 1, 6, 11..., (5k-4) f(n) mod 5 = (1 * 2 * 2) mod 5 = 4.

经过以上分析，我们可以看到，如果n的形式为5k + 1或5k-4，则f(n)mod 5 = 4，否则f(n)= 0。
IE浏览器if(n％5 == 1)结果= 4，
否则结果= 0。

C++

// finding the value of f(n) mod 5 for given n.
#include 
using namespace std;
 
// function for f(n) mod 5
int fnMod(int n)
{
    // if n % 5 == 1 return 4
    if (n % 5 == 1)
        return 4;
 
    // else return 0
    else
        return 0;
}
 
// driver program
int main()
{
    int n = 10;
    cout << fnMod(n) << endl;
    n = 11;
    cout << fnMod(n) << endl;
    return 0;
}

Java

// Java code to finding the value
// of f(n) mod 5 for given n.
import java.io.*;
 
class GFG
{
    // function for f(n) mod 5
    static int fnMod(int n)
    {
        // if n % 5 == 1 return 4
        if (n % 5 == 1)
            return 4;
     
        // else return 0
        else
            return 0;
    }
     
    // Driver program
    public static void main (String[] args)
    {
        int n = 10;
        System.out.println(fnMod(n));
        n = 11;
        System.out.println(fnMod(n));
 
    }
}
 
// This code is contributed by vt_m.

Python3

# Python3 program to find the value
# of f(n) mod 5 for given n.
 
# Function for f(n) mod 5
def fnMod(n):
 
    # if n % 5 == 1 return 4
    if (n % 5 == 1):
        return 4
 
    # else return 0
    else:
        return 0
 
# Driver Code
n = 10
print(fnMod(n))
 
n = 11
print(fnMod(n))
 
# This code is contributed by Smitha Dinesh Semwal

C#

// Code for finding the value
// of f(n) mod 5 for given n.
using System;
 
class GFG {
    // function for f(n) mod 5
    static int fnMod(int n)
    {
        // if n % 5 == 1 return 4
        if (n % 5 == 1)
            return 4;
 
        // else return 0
        else
            return 0;
    }
 
    // Driver program
    public static void Main()
    {
        int n = 10;
        Console.WriteLine(fnMod(n));
        n = 11;
        Console.WriteLine(fnMod(n));
    }
}
 
// This code is contributed by vt_m.

Javascript

// JavaScript program to finding the value
// of f(n) mod 5 for given n.
 
// function for f(n) mod 5
    function fnMod(n)
    {
     
        // if n % 5 == 1 return 4
        if (n % 5 == 1)
            return 4;
       
        // else return 0
        else
            return 0;
    }
  
// Driver Code
        let n = 10;
        document.write(fnMod(n) + "<br/>");
        n = 11;
        document.write(fnMod(n) + "<br/>");
                
        // This code is contributed by chinmoy1997pal.      

输出 ：

0
4