求 N 的大值 N % 4(余数为 4)
给定一个代表大整数的字符串 str ，任务是找到 N % 4 的结果。 例:
输入: N = 81 输出: 1 输入:N = 46234624362346435768440 输出: 0
方法:除以 4 的余数只依赖于一个数的最后 2 位数字，所以我们不除 N，而是只除 N 的最后两位数字，求余数。 以下是上述办法的实施情况:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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.
Python 3
# 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.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return s % n
function findMod4($s, $n)
{
// To store the number formed by
// the last two digits
$k;
// If it contains a single digit
if ($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
{
$s = "81";
$n = strlen($s);
echo(findMod4($s, $n));
}
// This code is contributed by Code_Mech.
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return s % n
function findMod4(s, n)
{
// To store the number formed by
// the last two digits
var k=0;
// 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
var s = "81";
var n = s.length;
document.write(findMod4(s, n));
// This code is contributed by nood2000.
</script>
Output:
1
时间复杂度: O(1)
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