📜  找到N%4(余数为4)以获得较大的N

📅  最后修改于: 2021-04-26 09:15:20             🧑  作者: Mango

给定一个表示大整数的字符串str ,任务是找到N%4的结果。
例子:

方法:除以4的余数仅取决于数字的最后两位数字,因此,除了对N进行除法运算外,我们仅对N的最后两位数进行除法并找到余数。
下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return s % n
int findMod4(string s, int n)
{
 
    // To store the number formed by
    // the last two digits
    int k;
 
    // If it contains a single digit
    if (n == 1)
        k = s[0] - '0';
 
    // Take last 2 digits
    else
        k = (s[n - 2] - '0') * 10
            + s[n - 1] - '0';
 
    return (k % 4);
}
 
// Driver code
int main()
{
    string s = "81";
    int n = s.length();
    cout << findMod4(s, n);
 
    return 0;
}


Java
// Java implementation of the approach
class GFG
{
     
// Function to return s % n
static int findMod4(String s, int n)
{
 
    // To store the number formed by
    // the last two digits
    int k;
 
    // If it contains a single digit
    if (n == 1)
        k = s.charAt(0) - '0';
 
    // Take last 2 digits
    else
        k = (s.charAt(n - 2) - '0') * 10
            + s.charAt(n - 1) - '0';
 
    return (k % 4);
}
 
// Driver code
public static void main(String[] args)
{
    String s = "81";
    int n = s.length();
    System.out.println(findMod4(s, n));
}
}
 
// This code is contributed by Code_Mech.


Python3
# Python 3 implementation of the approach
 
# Function to return s % n
def findMod4(s, n):
     
    # To store the number formed by
    # the last two digits
     
    # If it contains a single digit
    if (n == 1):
        k = ord(s[0]) - ord('0')
 
    # Take last 2 digits
    else:
        k = ((ord(s[n - 2]) - ord('0')) * 10 +
              ord(s[n - 1]) - ord('0'))
 
    return (k % 4)
 
# Driver code
if __name__ == '__main__':
    s = "81"
    n = len(s)
    print(findMod4(s, n))
 
# This code is contributed by
# Surendra_Gangwar


C#
// C# implementation of the approach
using System;
class GFG
{
     
// Function to return s % n
static int findMod4(string s, int n)
{
 
    // To store the number formed by
    // the last two digits
    int k;
 
    // If it contains a single digit
    if (n == 1)
        k = s[0] - '0';
 
    // Take last 2 digits
    else
        k = (s[n - 2]- '0') * 10
            + s[n - 1] - '0';
 
    return (k % 4);
}
 
// Driver code
public static void Main()
{
    string s = "81";
    int n = s.Length;
    Console.WriteLine(findMod4(s, n));
}
}
 
// This code is contributed by Code_Mech.


PHP


Javascript


输出:
1

时间复杂度: O(1)