给定三个整数X ， Y和B ，其中X和Y是Base-B整数。任务是找到整数X和Y的总和。
例子：
Input: X = 123, Y = 234, B = 6
Output: 401
Explanation:
Sum of two integers in base 6 -
1 1
1 2 3
+ 2 3 4
-------------
4 0 1
Input: X = 546, Y = 248 B = 9
Output: 805
Explanation:
Sum of two integers in base 9 -
1 1
5 4 6
+ 2 4 8
-------------
8 0 5
方法：想法是利用这样的事实：每当数字的两位数字相加时，则位值将是数字总和乘以底数的模数，而进位将是数字总和除以底数的整数除法。 IE
Let two digits of the number be D1 and D2 -
Place Value = (D1 + D2) % B
Carry = (D1 + D2) / B
同样，将最后一位加上每个数字以得到所需的结果。
下面是上述方法的实现：
C++
// C++ implementation to find the
// sum of two integers of base B
#include
using namespace std;
// Function to find the sum of
// two integers of base B
string sumBaseB(string a,
string b, int base)
{
int len_a, len_b;
len_a = a.size();
len_b = b.size();
string sum, s;
s = "";
sum = "";
int diff;
diff = abs(len_a - len_b);
// Padding 0 in front of the
// number to make both numbers equal
for (int i = 1; i <= diff; i++)
s += "0";
// Condition to check if the strings
// have lengths mis-match
if (len_a < len_b)
a = s + a;
else
b = s + b;
int curr, carry = 0;
// Loop to find the find the sum
// of two integers of base B
for (int i = max(len_a, len_b) - 1;
i > -1; i--) {
// Current Place value for
// the resultant sum
curr = carry + (a[i] - '0') +
(b[i] - '0');
// Update carry
carry = curr / base;
// Find current digit
curr = curr % base;
// Update sum result
sum = (char)(curr + '0') + sum;
}
if (carry > 0)
sum = (char)(carry + '0') + sum;
return sum;
}
// Driver Code
int main()
{
string a, b, sum;
int base;
a = "123";
b = "234";
base = 6;
// Function Call
sum = sumBaseB(a, b, base);
cout << sum << endl;
return 0;
} Java
// Java implementation to find the
// sum of two integers of base B
class GFG {
// Function to find the sum of
// two integers of base B
static String sumBaseB(String a, String b, int base)
{
int len_a, len_b;
len_a = a.length();
len_b = b.length();
String sum, s;
s = "";
sum = "";
int diff;
diff = Math.abs(len_a - len_b);
// Padding 0 in front of the
// number to make both numbers equal
for (int i = 1; i <= diff; i++)
s += "0";
// Condition to check if the strings
// have lengths mis-match
if (len_a < len_b)
a = s + a;
else
b = s + b;
int curr, carry = 0;
// Loop to find the find the sum
// of two integers of base B
for (int i = Math.max(len_a, len_b) - 1;
i > -1; i--) {
// Current Place value for
// the resultant sum
curr = carry + (a.charAt(i) - '0') +
(b.charAt(i) - '0');
// Update carry
carry = curr / base;
// Find current digit
curr = curr % base;
// Update sum result
sum = (char)(curr + '0') + sum;
}
if (carry > 0)
sum = (char)(carry + '0') + sum;
return sum;
}
// Driver Code
public static void main (String[] args)
{
String a, b, sum;
int base;
a = "123";
b = "234";
base = 6;
// Function Call
sum = sumBaseB(a, b, base);
System.out.println(sum);
}
}
// This code is contributed by AnkitRai01Python 3
# Python 3 implementation to find the
# sum of two integers of base B
# Function to find the sum of
# two integers of base B
def sumBaseB(a,b,base):
len_a = len(a)
len_b = len(b)
s = "";
sum = "";
diff = abs(len_a - len_b);
# Padding 0 in front of the
# number to make both numbers equal
for i in range(1,diff+1):
s += "0"
# Condition to check if the strings
# have lengths mis-match
if (len_a < len_b):
a = s + a
else:
b = s + b;
carry = 0;
# Loop to find the find the sum
# of two integers of base B
for i in range(max(len_a, len_b) - 1,-1,-1):
# Current Place value for
# the resultant sum
curr = carry + (ord(a[i]) -ord('0')) +( ord(b[i]) - ord('0'));
# Update carry
carry = curr // base
# Find current digit
curr = curr % base;
# Update sum result
sum = chr(curr + ord('0')) + sum
if (carry > 0):
sum = chr(carry + ord('0')) + sum;
return sum
# Driver Code
a = "123"
b = "234"
base = 6
# Function Call
sum = sumBaseB(a, b, base);
print(sum)
# This code is contributed by atul_kumar_shrivastavaC#
// C# implementation to find the
// sum of two integers of base B
using System;
class GFG {
// Function to find the sum of
// two integers of base B
static string sumBaseB(string a, string b, int base_var)
{
int len_a, len_b;
len_a = a.Length;
len_b = b.Length;
string sum, s;
s = "";
sum = "";
int diff;
diff = Math.Abs(len_a - len_b);
// Padding 0 in front of the
// number to make both numbers equal
for (int i = 1; i <= diff; i++)
s += "0";
// Condition to check if the strings
// have lengths mis-match
if (len_a < len_b)
a = s + a;
else
b = s + b;
int curr, carry = 0;
// Loop to find the find the sum
// of two integers of base B
for (int i = Math.Max(len_a, len_b) - 1;
i > -1; i--) {
// Current Place value for
// the resultant sum
curr = carry + (a[i] - '0') +
(b[i] - '0');
// Update carry
carry = curr / base_var;
// Find current digit
curr = curr % base_var;
// Update sum result
sum = (char)(curr + '0') + sum;
}
if (carry > 0)
sum = (char)(carry + '0') + sum;
return sum;
}
// Driver Code
public static void Main (string[] args)
{
string a, b, sum;
int base_var;
a = "123";
b = "234";
base_var = 6;
// Function Call
sum = sumBaseB(a, b, base_var);
Console.WriteLine(sum);
}
}
// This code is contributed by AnkitRai01输出：
401