joj1011

 1011: If only I had a Venn diagram

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	524288K	3990	1595	Standard

The symmetric difference of two sets is the set of elements belonging to one but not both of the two sets. For example, if we have two sets A = {1,2,3,4,5} and B = {3,4,5,6,7,8}, then the symmetric difference of A and B is the set {1,2,6,7,8}.

Given two sets of positive integers, display their symmetric difference.

Input

 

A positive integer will denote the number of cases. Both sets will be input from a single line. A zero (0) marks the end of each set. There will be no more than 20 numbers in each set, and no number within a set will be repeated. Each number in a set is a positive integer less than 65535.

Output

 

The set of integers making up the symmetric difference of the two sets. The numbers within a set may be sorted from smaller to bigger.

Sample Input

3
1 2 3 4 5 0 3 4 5 6 7 8 0
1 2 3 0 1 2 3 0
129 34 5 6 7 0 129 0

Sample Output

{1,2,6,7,8}
{}
{5,6,7,34}

 

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This problem is used for contest: 50 

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将全部数据存入一个数组，对数组进行快排。用一个类似于索引数组的东西将不重复的数输出。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int cmp(const void *a, const void *b)
{
    return *(int *)a - *(int *)b;
}

void output(int a[], int t)
{
    int b[50];
    memset(b, 0, sizeof(b));
    int i;

    for (i=0; i<t-1; i++)
    {
        if (a[i] == a[i+1])
        {
            b[i] = b[i+1] = 1;
        }
    }
    int flag = 1;
    printf("{");
    for (i=0; i<t; i++)
    {
        if (flag && 0 == b[i])
        {
            flag = 0;
            printf("%d", a[i]);
        }
        else
        {
            if (0 == b[i])
            {
                printf(",%d", a[i]);
            }
        }
    }
    printf("}\n");

}

int main(void)
{
    int sample, val;
    int a[50];

    freopen("in.txt", "r", stdin);

    while (scanf("%d", &sample) == 1)
    {
        for (int i=0; i<sample; i++)
        {
            int j = 0;
            while(scanf("%d", &val), val)
            {
                a[j++] = val;
            }

            while (scanf("%d", &val), val)
            {
                a[j++] = val;
            }

            qsort(a, j, sizeof(int), cmp);

            output(a, j);
        }
    }

    return 0;
}