Waterloo Cup-The realization of the Candy C language of the previous questions

All previous questions are divided into sweetsTime limit: 1.0s memory limit: 256.0MB

Problem description
There are N children sitting in a circle. The teacher randomly sends a couple of sweets to each child and then plays the following game:
Each child treats his or her sweets in half to the children on the left hand side.
After a round of sugar, the children with odd sugars were supplied 1 sweets by the teacher, which became even.
Repeat the game until all the children have the same number of sweets.
Your task is to predict how many sweets the teacher will need to reissue in the event of a known initial candy.
Input format
The program first reads an integer N (2<n<100), indicating the number of children.
followed by a row of n-even numbers separated by a space (every even number is no greater than 1000, not less than 2)
Output format
Requires the program to output an integer, indicating the number of sweets the teacher needs to reissue.
Sample input
3
2 2 4
Sample output
4

Analysis: This problem data is small (in fact Waterloo Cup test data is 2<n<1000, Waterloo Cup test data are unreliable), directly with the array and linked list are full marks, recommended by the linked list familiar.
Code here: (linked list)

#include <stdio.h> #include <stdlib.h> typedef struct node{int now;
struct Node*next;

}student;
	int main () {int n, x=0, y, K, s=0, o[100], m=0;
	int I, J;
	
	Student *head, *r, *p, *q;
	
	scanf ("%d", &n);
	p = (student*) malloc (sizeof (student));
	head = p;
	P->next = head->next;
	r = P;
		for (i = 0; i < n; i + +) {scanf ("%d", &k);
		p = (student*) malloc (sizeof (student));
		P->now = k;
		P->next = head->next;
		R->next = p;
	r = r->next;
		while (1) {p = head->next;
			for (i = 0; i < n; i + +, p = p->next) {q = p->next;
	            if (P->now = = Q->now) m + +;
				if (M = = N) {printf ("%d\n", s);
				return 0;
		} m = 0;
			for (i = 0;i < n; i + +) {y = p->now/2;
			P->now = y;
			P->now = P->now + x;
			x = y;
		p = p->next;
		} P->now = p->now+x;					x = 0;
				The number of global variables with loops does not affect the next round of loops for (i = 0; i < n; i + +) {if (p->now% 2!= 0) {(p->now) + +;
			s + +; }
			p = p->next; }
	}
}

code here (Array)

#include <stdio.h>  

#define SIZE 1000+10

int main () {  
    int a[size];  
    int n;  
    int s = 0;  
    int i;  
      
    scanf ("%d", &n);  
      
    for (i = 0; i < n; i + +)  
        scanf ("%d", &a[i]);  
  
    while (1) {  
        int k = 0;  
        for (i = 0; i < n; i + +) {  
            if (A[i]% 2 = 1) {  
                a[i] + +;  
                s + +;  
            }  
            if (a[i] = = A[0])  
                k + +;  
        }  
        if (k = = n) break  
            ;            
          
        int temp = a[n-1]/2;  
        int t;  
        for (i = 0; i < n; i + +) {  
            t = a[i]/2;  
            A[i] = t + temp;  
            temp = t;  
        }  
    }  
      
    printf ("%d", s);
    return 0;  
}