给定一个数字数组arr []和一个整数K ，任务是找到num(arr)+ K ，其中num(arr)是将数组的所有数字串联而成的数字。
例子：
Input: arr[] = {2, 7, 4}, K = 181
Output: 455
274 + 181 = 455
Input: arr[] = {6}, K = 815
Output: 821
6 + 815 = 821
方法：从末尾开始遍历数字数组，并开始将K的数字一一加到当前数字，如果产生了进位，则将其添加到下一个数字加法中。添加完所有K位数后，开始从左侧遍历digits数组，并开始打印元素，这些元素在级联时将给出结果数。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the vector containing the answer
vector addToArrayForm(vector& A, int K)
{
// Vector v is to store each digits sum
// and vector ans is to store the answer
vector v, ans;
// No carry in the beginning
int rem = 0;
int i = 0;
// Start loop from the end
// and take element one by one
for (i = A.size() - 1; i >= 0; i--) {
// Array index and last digit of number
int my = A[i] + K % 10 + rem;
if (my > 9) {
// Maintain carry of summation
rem = 1;
// Push the digit value into the array
v.push_back(my % 10);
}
else {
v.push_back(my);
rem = 0;
}
K = K / 10;
}
// K value is greater then 0
while (K > 0) {
// Push digits of K one by one in the array
int my = K % 10 + rem;
v.push_back(my % 10);
// Also maintain carry with summation
if (my / 10 > 0)
rem = 1;
else
rem = 0;
K = K / 10;
}
if (rem > 0)
v.push_back(rem);
// Reverse the elements of vector v
// and store it in vector ans
for (int i = v.size() - 1; i >= 0; i--)
ans.push_back(v[i]);
return ans;
}
// Driver code
int main()
{
vector A{ 2, 7, 4 };
int K = 181;
vector ans = addToArrayForm(A, K);
// Print the answer
for (int i = 0; i < ans.size(); i++)
cout << ans[i];
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the vector containing the answer
static ArrayList addToArrayForm(ArrayList A, int K)
{
// ArrayList v is to store each digits sum
// and ArrayList ans is to store the answer
ArrayList v = new ArrayList();
ArrayList ans = new ArrayList();
// No carry in the beginning
int rem = 0;
int i = 0;
// Start loop from the end
// and take element one by one
for (i = A.size() - 1; i >= 0; i--)
{
// Array index and last digit of number
int my = A.get(i) + K % 10 + rem;
if (my > 9)
{
// Maintain carry of summation
rem = 1;
// Push the digit value into the array
v.add(my % 10);
}
else
{
v.add(my);
rem = 0;
}
K = K / 10;
}
// K value is greater then 0
while (K > 0)
{
// Push digits of K one by one in the array
int my = K % 10 + rem;
v.add(my % 10);
// Also maintain carry with summation
if (my / 10 > 0)
rem = 1;
else
rem = 0;
K = K / 10;
}
if (rem > 0)
v.add(rem);
// Reverse the elements of vector v
// and store it in vector ans
for (int j = v.size() - 1; j >= 0; j--)
ans.add(v.get(j));
return ans;
}
// Driver code
public static void main (String[] args)
{
ArrayList A = new ArrayList();
A.add(2);
A.add(7);
A.add(4);
int K = 181;
ArrayList ans = addToArrayForm(A, K);
// Print the answer
for (int i = 0; i < ans.size(); i++)
System.out.print(ans.get(i));
}
}
// This code is contributed by ihritik Python3
# Python3 implementation of the approach
# Function to return the vector
# containing the answer
def addToArrayForm(A, K):
# Vector v is to store each digits sum
# and vector ans is to store the answer
v, ans = [], []
# No carry in the beginning
rem, i = 0, 0
# Start loop from the end
# and take element one by one
for i in range(len(A) - 1, -1, -1):
# Array index and last digit of number
my = A[i] + (K % 10) + rem
if my > 9:
# Maintain carry of summation
rem = 1
# Push the digit value into the array
v.append(my % 10)
else:
v.append(my)
rem = 0
K = K // 10
# K value is greater then 0
while K > 0:
# Push digits of K one by one in the array
my = (K % 10) + rem
v.append(my % 10)
# Also maintain carry with summation
if my // 10 > 0:
rem = 1
else:
rem = 0
K = K // 10
if rem > 0:
v.append(rem)
# Reverse the elements of vector v
# and store it in vector ans
for i in range(len(v) - 1, -1, -1):
ans.append(v[i])
return ans
# Driver code
if __name__ == "__main__":
A = [2, 7, 4]
K = 181
ans = addToArrayForm(A, K)
# Print the answer
for i in range(0, len(ans)):
print(ans[i], end = "")
# This code is contributed by Rituraj JainC#
// C# implementation of the approach
using System;
using System.Collections;
class GFG
{
// Function to return the vector containing the answer
static ArrayList addToArrayForm(ArrayList A, int K)
{
// ArrayList v is to store each digits sum
// and ArrayList ans is to store the answer
ArrayList v = new ArrayList();
ArrayList ans = new ArrayList();
// No carry in the beginning
int rem = 0;
int i = 0;
// Start loop from the end
// and take element one by one
for (i = A.Count - 1; i >= 0; i--)
{
// Array index and last digit of number
int my = (int)A[i] + K % 10 + rem;
if (my > 9)
{
// Maintain carry of summation
rem = 1;
// Push the digit value into the array
v.Add(my % 10);
}
else
{
v.Add(my);
rem = 0;
}
K = K / 10;
}
// K value is greater then 0
while (K > 0)
{
// Push digits of K one by one in the array
int my = K % 10 + rem;
v.Add(my % 10);
// Also maintain carry with summation
if (my / 10 > 0)
rem = 1;
else
rem = 0;
K = K / 10;
}
if (rem > 0)
v.Add(rem);
// Reverse the elements of vector v
// and store it in vector ans
for (int j = v.Count - 1; j >= 0; j--)
ans.Add((int)v[j]);
return ans;
}
// Driver code
static void Main()
{
ArrayList A = new ArrayList();
A.Add(2);
A.Add(7);
A.Add(4);
int K = 181;
ArrayList ans = addToArrayForm(A, K);
// Print the answer
for (int i = 0; i < ans.Count; i++)
Console.Write((int)ans[i]);
}
}
// This code is contributed by mitsPHP
= 0; $i--)
{
// Array index and last digit of number
$my = $A[$i] + $K % 10 + $rem;
if ($my > 9)
{
// Maintain carry of summation
$rem = 1;
// Push the digit value into the array
array_push($v,$my % 10);
}
else
{
array_push($v,$my);
$rem = 0;
}
$K = floor($K / 10);
}
// K value is greater then 0
while ($K > 0)
{
// Push digits of K one by one in the array
$my = $K % 10 + $rem;
array_push($v,$my % 10);
// Also maintain carry with summation
if ($my / 10 > 0)
$rem = 1;
else
$rem = 0;
$K = floor($K / 10);
}
if ($rem > 0)
array_push($v,$rem);
// Reverse the elements of vector v
// and store it in vector ans
for ($i = count($v) - 1; $i >= 0; $i--)
array_push($ans,$v[$i]);
return $ans;
}
// Driver code
$A = array( 2, 7, 4 );
$K = 181;
$ans = addToArrayForm($A, $K);
// Print the answer
for ($i = 0; $i < count($ans); $i++)
echo $ans[$i];
// This code is contributed by Ryuga
?>输出：
455