减去给定基数的两个整数的程序
给定三个正整数X 、 Y和B ，其中X和Y是Base-B整数，任务是找到X - Y的值，使得X >= Y 。
例子：
Input: X = 1212, Y = 256, B = 8
Output: 0734
Explanation: The value of 1212 – 256 in base 8 is 734.
Input: X = 546, Y = 248, B = 9
Output: 287
方法：给定的问题可以通过使用基本的数学减法来解决。请按照以下步骤解决给定的问题：
- 初始化两个变量，说power = 1 ，进位 = 0 ，以分别跟踪当前的功率和相减时产生的进位。
- 初始化一个变量，比如finalVal = 0 ，以存储X – Y的结果值。
- 迭代一个循环直到X > 0并执行以下步骤：
- 将X和Y的当前值的最后一位存储在两个变量中，分别表示n1 = X % 10和n2 = Y % 10 。
- 通过更新X = X / 10和Y = Y / 10从X和Y中删除最后一位数字。
- 初始化temp = n1 – n2 + 进位。
- 如果temp < 0 ，则将基数B添加到N ，即N = N + B并设置进位 = -1 ，这将充当借位。否则，设置进位 = 0 。
- 将当前temp * power添加到 finalVal，即finalVal = finalVal + temp * power并设置power = power * 10 。
- 完成上述步骤后，打印finalVal的值作为结果。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find X - Y in base B
int getDifference(int B, int X, int Y)
{
// To store final answer
int finalVal = 0;
// To store carry generated
int carry = 0;
// To keep track of power
int power = 1;
while (X > 0) {
// Store last digits of current
// value of X and Y in n1 and
// n2 respectively
int n1 = X % 10;
int n2 = Y % 10;
// Remove last digits from
// X and Y
X = X / 10;
Y = Y / 10;
int temp = n1 - n2 + carry;
if (temp < 0) {
// Carry = -1 will act
// as borrow
carry = -1;
temp += B;
}
else {
carry = 0;
}
// Add in final result
finalVal += temp * power;
power = power * 10;
}
// Return final result
return finalVal;
}
// Driver Code
int main()
{
int X = 1212;
int Y = 256;
int B = 8;
cout << (getDifference(B, X, Y));
return 0;
}
// This code is contributed by rakeshsahni Java
// Java program for the above approach
import java.io.*;
class GFG {
// Function to find X - Y in base B
public static int getDifference(
int B, int X, int Y)
{
// To store final answer
int finalVal = 0;
// To store carry generated
int carry = 0;
// To keep track of power
int power = 1;
while (X > 0) {
// Store last digits of current
// value of X and Y in n1 and
// n2 respectively
int n1 = X % 10;
int n2 = Y % 10;
// Remove last digits from
// X and Y
X = X / 10;
Y = Y / 10;
int temp = n1 - n2 + carry;
if (temp < 0) {
// Carry = -1 will act
// as borrow
carry = -1;
temp += B;
}
else {
carry = 0;
}
// Add in final result
finalVal += temp * power;
power = power * 10;
}
// Return final result
return finalVal;
}
// Driver Code
public static void main(String[] args)
{
int X = 1212;
int Y = 256;
int B = 8;
System.out.println(
getDifference(B, X, Y));
}
}Python3
# Python3 program for the above approach
# Function to find X - Y in base B
def getDifference(B, X, Y) :
# To store final answer
finalVal = 0;
# To store carry generated
carry = 0;
# To keep track of power
power = 1;
while (X > 0) :
# Store last digits of current
# value of X and Y in n1 and
# n2 respectively
n1 = X % 10;
n2 = Y % 10;
# Remove last digits from
# X and Y
X = X // 10;
Y = Y // 10;
temp = n1 - n2 + carry;
if (temp < 0) :
# Carry = -1 will act
# as borrow
carry = -1;
temp += B;
else :
carry = 0;
# Add in final result
finalVal += temp * power;
power = power * 10;
# Return final result
return finalVal;
# Driver Code
if __name__ == "__main__" :
X = 1212;
Y = 256;
B = 8;
print(getDifference(B, X, Y));
# This code is contributed by AnkThonC#
// C# program for the above approach
using System;
public class GFG
{
// Function to find X - Y in base B
public static int getDifference(int B, int X, int Y)
{
// To store final answer
int finalVal = 0;
// To store carry generated
int carry = 0;
// To keep track of power
int power = 1;
while (X > 0) {
// Store last digits of current
// value of X and Y in n1 and
// n2 respectively
int n1 = X % 10;
int n2 = Y % 10;
// Remove last digits from
// X and Y
X = X / 10;
Y = Y / 10;
int temp = n1 - n2 + carry;
if (temp < 0) {
// Carry = -1 will act
// as borrow
carry = -1;
temp += B;
}
else {
carry = 0;
}
// Add in final result
finalVal += temp * power;
power = power * 10;
}
// Return final result
return finalVal;
}
// Driver Code
public static void Main(string[] args)
{
int X = 1212;
int Y = 256;
int B = 8;
Console.WriteLine(getDifference(B, X, Y));
}
}
// This code is contributed by AnkThonJavascript
输出：
734时间复杂度： O(log 10 N)
辅助空间： O(1)