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 <= 10100) 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; } |
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