两个十六进制数的模
给定两个十六进制数 N 和 K ，任务是找到NT8】模 T10】K .
示例:
输入: N = 3E8，K = 13 输出: C 说明: N(= 3e 8)的十进制表示为 1000 K(= 13)的十进制表示为 19 的十进制表示为(N % K) = 1000 % 19 = 12 ( = C)。 因此，要求的输出是 c。
输入: N = 2A3，K = 1A T3】输出: 19
方法:按照以下步骤解决问题:
- 将十六进制数 N 和 K 分别转换成它们等价的十进制数比如说 X 和 Y 。
- 将十进制数(X % Y)转换成其等价的十六进制数比如说 res
- 最后，打印 res 的值。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate modulus of
// two Hexadecimal numbers
void hexaModK(string s, string k)
{
// Store all possible
// hexadecimal digits
map<char, int> mp;
// Iterate over the range ['0', '9']
for(char i = 1; i <= 9; i++)
{
mp[i + '0'] = i;
}
mp['A'] = 10;
mp['B'] = 11;
mp['C'] = 12;
mp['D'] = 13;
mp['E'] = 14;
mp['F'] = 15;
// Convert given string to long
long m = stoi(k, 0, 16);
// Base to get 16 power
long base = 1;
// Store N % K
long ans = 0;
// Iterate over the digits of N
for(int i = s.length() - 1;
i >= 0; i--)
{
// Stores i-th digit of N
long n = mp[s[i]] % m;
// Update ans
ans = (ans + (base % m * n
% m) % m) % m;
// Update base
base = (base % m * 16 % m) % m;
}
// Print the answer converting
// into hexadecimal
stringstream ss;
ss << hex << ans;
string su = ss.str();
transform(su.begin(), su.end(),
su.begin(), ::toupper);
cout << (su);
}
// Driver Code
int main()
{
// Given string N and K
string n = "3E8";
string k = "13";
// Function Call
hexaModK(n, k);
return 0;
}
// This code is contributed by sallagondaavinashreddy7
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
import java.util.*;
public class Main {
// Function to calculate modulus of
// two Hexadecimal numbers
static void hexaModK(String N, String k)
{
// Store all possible
// hexadecimal digits
HashMap<Character, Integer> map
= new HashMap<>();
// Iterate over the range ['0', '9']
for (char i = '0'; i <= '9'; i++) {
map.put(i, i - '0');
}
map.put('A', 10);
map.put('B', 11);
map.put('C', 12);
map.put('D', 13);
map.put('E', 14);
map.put('F', 15);
// Convert given string to long
long m = Long.parseLong(k, 16);
// Base to get 16 power
long base = 1;
// Store N % K
long ans = 0;
// Iterate over the digits of N
for (int i = N.length() - 1;
i >= 0; i--) {
// Stores i-th digit of N
long n
= map.get(N.charAt(i)) % m;
// Update ans
ans = (ans + (base % m * n % m) % m) % m;
// Update base
base = (base % m * 16 % m) % m;
}
// Print the answer converting
// into hexadecimal
System.out.println(
Long.toHexString(ans).toUpperCase());
}
// Driver Code
public static void main(String args[])
{
// Given string N and K
String n = "3E8";
String k = "13";
// Function Call
hexaModK(n, k);
}
}
Python 3
# Python3 program to implement
# the above approach
# Function to calculate modulus of
# two Hexadecimal numbers
def hexaModK(s, k) :
# Store all possible
# hexadecimal digits
mp = {};
# Iterate over the range ['0', '9']
for i in range(1, 10) :
mp[chr(i + ord('0'))] = i;
mp['A'] = 10;
mp['B'] = 11;
mp['C'] = 12;
mp['D'] = 13;
mp['E'] = 14;
mp['F'] = 15;
# Convert given string to long
m = int(k);
# Base to get 16 power
base = 1;
# Store N % K
ans = 0;
# Iterate over the digits of N
for i in range(len(s) - 1, -1, -1) :
# Stores i-th digit of N
n = mp[s[i]] % m;
# Update ans
ans = (ans + (base % m * n
% m) % m) % m;
# Update base
base = (base % m * 16 % m) % m;
# Print the answer converting
# into hexadecimal
ans = hex(int(ans))[-1].upper()
print(ans)
# Driver Code
if __name__ == "__main__" :
# Given string N and K
n = "3E8";
k = "13";
# Function Call
hexaModK(n, k);
# This code is contributed by AnkThon
C
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to calculate modulus of
// two Hexadecimal numbers
static void hexaModK(String N, String k)
{
// Store all possible
// hexadecimal digits
Dictionary<char,
int> map = new Dictionary<char,
int>();
// Iterate over the range ['0', '9']
for(char i = '0'; i <= '9'; i++)
{
map.Add(i, i - '0');
}
map.Add('A', 10);
map.Add('B', 11);
map.Add('C', 12);
map.Add('D', 13);
map.Add('E', 14);
map.Add('F', 15);
// Convert given string to long
long m = long.Parse(k);
// Base to get 16 power
long Base = 1;
// Store N % K
long ans = 0;
// Iterate over the digits of N
for(int i = N.Length - 1; i >= 0; i--)
{
// Stores i-th digit of N
long n = map[N[i]] % m;
// Update ans
ans = (ans + (Base % m * n % m) % m) % m;
// Update base
Base = (Base % m * 16 % m) % m;
}
// Print the answer converting
// into hexadecimal
Console.WriteLine(ans.ToString("X"));
}
// Driver Code
public static void Main(String []args)
{
// Given string N and K
String n = "3E8";
String k = "13";
// Function Call
hexaModK(n, k);
}
}
// This code is contributed by Princi Singh
java 描述语言
<script>
// Javascript program to implement
// the above approach
// Function to calculate modulus of
// two Hexadecimal numbers
function hexaModK(s, k)
{
// Store all possible
// hexadecimal digits
var mp = new Map();
// Iterate over the range ['0', '9']
for(var i = 1; i <= 9; i++)
{
mp.set(String.fromCharCode(
i + '0'.charCodeAt(0)), i);
}
mp.set('A', 10);
mp.set('B', 11);
mp.set('C', 12);
mp.set('D', 13);
mp.set('E', 14);
mp.set('F', 15);
// Convert given string to long
var m = parseInt(k, 16);
// Base to get 16 power
var base = 1;
// Store N % K
var ans = 0;
// Iterate over the digits of N
for(var i = s.length - 1;
i >= 0; i--)
{
// Stores i-th digit of N
var n = mp.get(s[i]) % m;
// Update ans
ans = (ans + (base % m * n % m) % m) % m;
// Update base
base = (base % m * 16 % m) % m;
}
document.write(ans.toString(16).toUpperCase());
}
// Driver Code
// Given string N and K
var n = "3E8";
var k = "13";
// Function Call
hexaModK(n, k);
// This code is contributed by famously
</script>
Output:
C
时间复杂度:O(N) T5辅助空间:** O(1)
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