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josephus.c
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josephus.c
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#include <stdio.h>
#include <stdlib.h>
// Josephus Problem Algorithm:
// 1. Create a circular linked list to represent the circle of people.
// 2. Initialize the linked list with people numbered from 1 to n.
// 3. Start with a pointer pointing to the first person.
// 4. Repeat the elimination process until only one person remains:
// a. Move the pointer k-1 steps in a circular manner.
// b. Remove the person at the current position.
// c. Update the pointer to the next position after the elimination, wrapping around if necessary.
// 5. Return the position of the last remaining person.
typedef struct Node {
int data;
struct Node* next;
} Node;
int josephus(int n, int k) {
Node* head = NULL;
Node* prev = NULL;
// Create the circular linked list with people numbered from 1 to n
for (int i = 1; i <= n; ++i) {
Node* newNode = (Node*)malloc(sizeof(Node));
newNode->data = i;
newNode->next = NULL;
if (head == NULL) {
head = newNode;
} else {
prev->next = newNode;
}
prev = newNode;
}
// Make the linked list circular
prev->next = head;
Node* curr = head;
// Elimination process
while (n > 1) {
// Move the pointer k-1 steps in a circular manner
for (int count = 1; count < k; ++count) {
prev = curr;
curr = curr->next;
}
// Remove the person at the current position
prev->next = curr->next;
Node* nextNode = curr->next;
free(curr);
curr = nextNode;
--n;
}
int lastPerson = curr->data;
// Free the remaining node
free(curr);
return lastPerson;
}
int main() {
int n, k;
// User input for n and k
printf("Enter the number of people in the circle: ");
scanf("%d", &n);
printf("Enter the elimination step size: ");
scanf("%d", &k);
int lastPerson = josephus(n, k); // Solve the Josephus Problem
printf("The last remaining person is at position: %d\n", lastPerson);
return 0;
}