查找(1 ^ n + 2 ^ n + 3 ^ n + 4 ^ n)mod 5 |套装2

给定非常大的N。任务是找到(1 ⁿ + 2 ⁿ + 3 ⁿ + 4 ⁿ )mod 5 。
例子：

Input: N = 4 
Output: 4 
(1 + 16 + 81 + 256) % 5 = 354 % 5 = 4
Input: N = 7823462937826332873467731 
Output: 0 
 

方法： (1 ⁿ + 2 ⁿ + 3 ⁿ + 4 ⁿ )mod 5 =(1 ^{n mod?(5)} + 2 ^{n mod?(5)} + 3 ^{n mod?(5)} + 4 ^{n mod?(5)} )mod 5。
该公式是正确的，因为5是质数，并且与1，2，3，4互质。
知道?(n)和大数的模
α(5)= 4 ，因此(1 ⁿ + 2 ⁿ + 3 ⁿ + 4 ⁿ )mod 5 =(1 ^{n mod 4} + 2 ^{n mod 4} + 3 ^{n mod 4} + 4 ^{n mod 4} )mod 5
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return A mod B
int A_mod_B(string N, int a)
{
    // length of the string
    int len = N.size();
 
    // to store requried answer
    int ans = 0;
    for (int i = 0; i < len; i++)
        ans = (ans * 10 + (int)N[i] - '0') % a;
 
    return ans % a;
}
 
// Function to return (1^n + 2^n + 3^n + 4^n) % 5
int findMod(string N)
{
    // ?(5) = 4
    int mod = A_mod_B(N, 4);
 
    int ans = (1 + pow(2, mod) + pow(3, mod)
               + pow(4, mod));
 
    return (ans % 5);
}
 
// Driver code
int main()
{
    string N = "4";
    cout << findMod(N);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
// Function to return A mod B
static int A_mod_B(String N, int a)
{
    // length of the string
    int len = N.length();
 
    // to store requried answer
    int ans = 0;
    for (int i = 0; i < len; i++)
        ans = (ans * 10 + (int)N.charAt(i) - '0') % a;
 
    return ans % a;
}
 
// Function to return (1^n + 2^n + 3^n + 4^n) % 5
static int findMod(String N)
{
    // ?(5) = 4
    int mod = A_mod_B(N, 4);
 
    int ans = (1 + (int)Math.pow(2, mod) +
                (int)Math.pow(3, mod) +
                (int)Math.pow(4, mod));
 
    return (ans % 5);
}
 
// Driver code
public static void main(String args[])
{
    String N = "4";
    System.out.println(findMod(N));
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach
 
# Function to return A mod B
def A_mod_B(N, a):
     
    # length of the string
    Len = len(N)
 
    # to store requried answer
    ans = 0
    for i in range(Len):
        ans = (ans * 10 + int(N[i])) % a
 
    return ans % a
 
# Function to return (1^n + 2^n + 3^n + 4^n) % 5
def findMod(N):
 
    # ?(5) = 4
    mod = A_mod_B(N, 4)
 
    ans = (1 + pow(2, mod) +
               pow(3, mod) + pow(4, mod))
 
    return ans % 5
 
# Driver code
N = "4"
print(findMod(N))
 
# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return A mod B
static int A_mod_B(string N, int a)
{
    // length of the string
    int len = N.Length;
 
    // to store requried answer
    int ans = 0;
    for (int i = 0; i < len; i++)
        ans = (ans * 10 + (int)N[i] - '0') % a;
 
    return ans % a;
}
 
// Function to return (1^n + 2^n + 3^n + 4^n) % 5
static int findMod(string N)
{
    // ?(5) = 4
    int mod = A_mod_B(N, 4);
 
    int ans = (1 + (int)Math.Pow(2, mod) +
                (int)Math.Pow(3, mod) +
                (int)Math.Pow(4, mod));
 
    return (ans % 5);
}
 
// Driver code
public static void Main()
{
    string N = "4";
    Console.WriteLine(findMod(N));
}
}
 
// This code is contributed by Code_Mech.

Javascript

 
// Javascript implementation of the approach
 
// Function to return A mod B
function A_mod_B(N, a)
{
    // length of the string
    var len = N.length;
 
    // to store requried answer
    var ans = 0;
    for (var i = 0; i < len; i++)
        ans = (ans * 10 + parseInt(N.charAt(i) - '0')) % a;
 
    return ans % a;
}
 
// Function to return (1^n + 2^n + 3^n + 4^n) % 5
function findMod(N)
{
    // ?(5) = 4
    var mod = A_mod_B(N, 4);
 
    var ans = (1 + parseInt(Math.pow(2, mod) +
                Math.pow(3, mod) +
                Math.pow(4, mod)));
 
    return (ans % 5);
}
 
// Driver Code
var N = "4";
     
// Function call
document.write(findMod(N));
 
// This code is contributed by Kirti
 

输出：

4