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📜  不带进位的两个数字相加

📅  最后修改于: 2021-05-25 09:10:38             🧑  作者: Mango

给出两个正数n和m。您只需要简单地找到两个数字的加法,但在给定条件下,该加法中没有任何进位系统。在更高的MSB上不添加进位。
例子 :

Input : m = 456, n = 854
Output : 200

Input : m = 456, n = 4
Output : 450

算法 :

Input n, m while(n||m)
    {
        // Add each bits 
        bit_sum = (n%10) + (m%10);

        // Neglect carry
        bit_sum %= 10;

        // Update result
        // multiplier to maintain place value
        res = (bit_sum * multiplier) + res;
        n /= 10;
        m /= 10;

        // Update multiplier
        multiplier *=10;
    } print res

方法 :
为了解决这个问题,我们将需要对数字进行逐位加法,即从最右边的位(LSB)开始加两个数,并从两个相同位置的nuber中加整数。同样,我们将忽略每个位置的进位,以使进位不会影响更高的位。
开始将两个数字逐位相加,然后对每个位取整数和,然后通过对bit_sum取模乘以10来忽略它们的进位,通过将bit_sum与指定位值的乘数相乘,再将bit_sum加到res上。 (乘数在每次迭代中增加10倍。)
下面是上述方法的实现:

C++
// CPP program for special
// addition of two number
#include 
using namespace std;
 
int xSum(int n, int m)
{
    // variable to store result  
    int res = 0;
 
    // variable to maintain
    // place value
    int multiplier = 1;
 
    // variable to maintain
    // each digit sum
    int bit_sum;
 
    // Add numbers till each
    // number become zero
    while (n || m) {
 
        // Add each bits
        bit_sum = (n % 10) + (m % 10);
         
        // Neglect carry
        bit_sum %= 10;
         
        // Update result
        res = (bit_sum * multiplier) + res;
        n /= 10;
        m /= 10;
         
        // Update multiplier
        multiplier *= 10;
    }
    return res;
}
 
// Driver program
int main()
{
    int n = 8458;
    int m = 8732;
    cout << xSum(n, m);
    return 0;
}


Java
// Java program for special
// addition of two number
import java.util.*;
import java.lang.*;
 
public class GfG {
 
    public static int xSum(int n, int m)
    {
        int res = 0;
        int multiplier = 1;
        int bit_sum;
 
        // Add numbers till each
        // number become zero
        while (true) {
                 
            if(n==0 && m==0)
            break;
 
            // Add each bits
            bit_sum = (n % 10) + (m % 10);
 
            // Neglect carry
            bit_sum %= 10;
 
            // Update result
            res = (bit_sum * multiplier) + res;
            n /= 10;
            m /= 10;
 
            // Update multiplier
            multiplier *= 10;
           
        }
        return res;
    }
 
    // Driver function
    public static void main(String args[])
    {
        int n = 8458;
        int m = 8732;
        System.out.println(xSum(n, m));
    }
}
/* This code is contributed by Sagar Shukla */


Python3
# Python3 program for special
# addition of two number
import math
 
def xSum(n, m) :
 
    # variable to
    # store result
    res = 0
 
    # variable to maintain
    # place value
    multiplier = 1
 
    # variable to maintain
    # each digit sum
    bit_sum = 0
 
    # Add numbers till each
    # number become zero
    while (n or m) :
 
        # Add each bits
        bit_sum = ((n % 10) +
                   (m % 10))
         
        # Neglect carry
        bit_sum = bit_sum % 10
         
        # Update result
        res = (bit_sum *
               multiplier) + res
        n = math.floor(n / 10)
        m = math.floor(m / 10)
         
        # Update multiplier
        multiplier = multiplier * 10
     
    return res
 
# Driver code
n = 8458
m = 8732
print (xSum(n, m))
 
# This code is contributed by
# Manish Shaw(manishshaw1)


C#
// C# program for special
// addition of two number
using System;
 
public class GfG {
 
    public static int xSum(int n, int m)
    {
        int res = 0;
        int multiplier = 1;
        int bit_sum;
 
        // Add numbers till each
        // number become zero
        while (true) {
 
            // Add each bits
            bit_sum = (n % 10) + (m % 10);
 
            // Neglect carry
            bit_sum %= 10;
 
            // Update result
            res = (bit_sum * multiplier) + res;
            n /= 10;
            m /= 10;
 
            // Update multiplier
            multiplier *= 10;
            if (n == 0)
                break;
            if (m == 0)
                break;
        }
        return res;
    }
 
    // Driver function
    public static void Main()
    {
        int n = 8458;
        int m = 8732;
        Console.WriteLine(xSum(n, m));
    }
}
 
/* This code is contributed by Vt_m */


PHP


Javascript


输出 :

6180