# Divide a big number into two parts that differ by k

Given a big positive number **N**. The task is divide N into two number **‘A’** and **‘B’** such that difference between them is **K** (1 <= K <= 10^{100}) i.e A – B = K.

**Examples:**

Input : N = 10, K = 2 Output : A = 6 B = 4 Input : N = 20, K = 4 Output : A = 12 B = 8

Let the two required number be ‘A’ and ‘B’. So, we know sum of ‘A’ and ‘B’ will end upto N.

So, one equation became,

A + B = N

And also, we want the difference between ‘A’ and ‘B’ equals to ‘K’.

So, another equation becomes,

A – B = K

On adding both the equation, we get

2*A = N + K

So, A = (N + K)/2

Then we can find B by,** B = (N – A)**

Now, to handle the Big Integer, we have store Integers in character array and define a function to perform the operation on them.

Below is C implementation of this approach:

`// C program to Divide a Big ` `// Number into two parts ` `#include <stdio.h> ` `#include <string.h> ` `#define MAX 100 ` ` ` `// Function to adds two Numbers ` `// represnted as array of character. ` `void` `add(` `char` `v1[], ` `char` `v2[]) ` `{ ` ` ` `int` `i, d, c = 0; ` ` ` ` ` `// length of string ` ` ` `int` `l1 = ` `strlen` `(v1); ` ` ` `int` `l2 = ` `strlen` `(v2); ` ` ` ` ` `// initializing extra character ` ` ` `// position to 0 ` ` ` `for` `(i = l1; i < l2; i++) ` ` ` `v1[i] = ` `'0'` `; ` ` ` ` ` `for` `(i = l2; i < l1; i++) ` ` ` `v2[i] = ` `'0'` `; ` ` ` ` ` `// Adding each element of character ` ` ` `// and storing the carry. ` ` ` `for` `(i = 0; i < l1 || i < l2; i++) { ` ` ` `d = (v1[i] - ` `'0'` `) + (v2[i] - ` `'0'` `) + c; ` ` ` `c = d / 10; ` ` ` `d %= 10; ` ` ` `v1[i] = ` `'0'` `+ d; ` ` ` `} ` ` ` ` ` `// If remainder remains. ` ` ` `while` `(c) { ` ` ` `v1[i] = ` `'0'` `+ (c % 10); ` ` ` `c /= 10; ` ` ` `i++; ` ` ` `} ` ` ` ` ` `v1[i] = ` `'\0'` `; ` ` ` `v2[l2] = ` `'\0'` `; ` `} ` ` ` `// Function to subtracts two numbers ` `// represnted by string. ` `void` `subs(` `char` `v1[], ` `char` `v2[]) ` `{ ` ` ` `int` `i, d, c = 0; ` ` ` ` ` `// Finding the length of the string. ` ` ` `int` `l1 = ` `strlen` `(v1); ` ` ` `int` `l2 = ` `strlen` `(v2); ` ` ` ` ` `// initializing extra character position to 0. ` ` ` `for` `(i = l2; i < l1; i++) ` ` ` `v2[i] = ` `'0'` `; ` ` ` ` ` `// Substrating each element of character. ` ` ` `for` `(i = 0; i < l1; i++) { ` ` ` `d = (v1[i] - ` `'0'` `- c) - (v2[i] - ` `'0'` `); ` ` ` ` ` `if` `(d < 0) { ` ` ` `d += 10; ` ` ` `c = 1; ` ` ` `} ` ` ` `else` ` ` `c = 0; ` ` ` ` ` `v1[i] = ` `'0'` `+ d; ` ` ` `} ` ` ` ` ` `v2[l2] = ` `'\0'` `; ` ` ` `i = l1 - 1; ` ` ` ` ` `while` `(i > 0 && v1[i] == ` `'0'` `) ` ` ` `i--; ` ` ` ` ` `v1[i + 1] = ` `'\0'` `; ` `} ` ` ` `// Function divides a number represented by ` `// character array a constant. ` `int` `divi(` `char` `v[], ` `int` `q) ` `{ ` ` ` `int` `i, l = ` `strlen` `(v); ` ` ` ` ` `int` `c = 0, d; ` ` ` ` ` `// Dividing each character element by constant. ` ` ` `for` `(i = l - 1; i >= 0; i--) { ` ` ` `d = c * 10 + (v[i] - ` `'0'` `); ` ` ` `c = d % q; ` ` ` `d /= q; ` ` ` `v[i] = ` `'0'` `+ d; ` ` ` `} ` ` ` ` ` `i = l - 1; ` ` ` ` ` `while` `(i > 0 && v[i] == ` `'0'` `) ` ` ` `i--; ` ` ` ` ` `v[i + 1] = ` `'\0'` `; ` ` ` `return` `c; ` `} ` ` ` `// Function to reverses the character array. ` `void` `rev(` `char` `v[]) ` `{ ` ` ` `int` `l = ` `strlen` `(v); ` ` ` `int` `i; ` ` ` `char` `cc; ` ` ` ` ` `// Reversing the array. ` ` ` `for` `(i = 0; i < l - 1 - i; i++) { ` ` ` `cc = v[i]; ` ` ` `v[i] = v[l - 1 - i]; ` ` ` `v[l - i - 1] = cc; ` ` ` `} ` `} ` ` ` `// Wrapper Function ` `void` `divideWithDiffK(` `char` `a[], ` `char` `k[]) ` `{ ` ` ` ` ` `// Reversing the character array. ` ` ` `rev(a); ` ` ` `rev(k); ` ` ` ` ` `// Adding the each element of both array ` ` ` `// and storing the sum in array a[]. ` ` ` `add(a, k); ` ` ` ` ` `// Dividing the array a[] by 2. ` ` ` `divi(a, 2); ` ` ` ` ` `// Reversing the character array to get output. ` ` ` `rev(a); ` ` ` `printf` `(` `"%s "` `, a); ` ` ` ` ` `// Substracting each element of array ` ` ` `// i.e calculating a = a - b ` ` ` `rev(a); ` ` ` `subs(a, k); ` ` ` ` ` `// Reversing the character array to get output. ` ` ` `rev(a); ` ` ` `printf` `(` `"%s"` `, a); ` `} ` ` ` `// Driven Program ` `int` `main() ` `{ ` ` ` `char` `a[MAX] = ` `"100"` `, k[MAX] = ` `"20"` `; ` ` ` `divideWithDiffK(a, k); ` ` ` `return` `0; ` `} ` |

*chevron_right*

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**Output:**

60 40

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