Operation of sorted set对集合的操作

Description:
There are two groups of some numbers（0~50). each group should be input in a set(remove the duplicate numbers).
And then output the two set, intersection and union of the two set.

Input format:
first line: a couple of numbers and end of -1.
second line: a couple of numbers and end of -1.
Output format:
first line：Output the first set.
second line: Output the second set.
third line: Output the intersection of the two set.
fourth line：Output the union of the two set.

All the numbers in the set should be output by ascending oder.
Each line behind the numbers, there is a space ’ ‘.

For example:
[Input]
1 2 3 6 5 4 1 2 3 -1
3 2 3 2 1 0 -1
[Output]
1 2 3 4 5 6
0 1 2 3
1 2 3
0 1 2 3 4 5 6

其实就是对集合排序，然后求两个集合的交集和并集●ω●

#include<stdio.h>int set(int *a) {    int i, j, temp, count = 1;    scanf("%d", &temp);    a[0] = temp;    while (1) {        scanf("%d", &temp);        if (temp == -1) break;        for (i = 0; i < count; i++) {            if (a[i] == temp) break;  //判断该数是否已存在        }        if (i == count) {            a[count] = temp;            count++;        }    }  //完成集合的构造    for (i = 0; i < count; i++) {        for (j = 0; j < count - i - 1; j++) {            if (a[j] > a[j+1]) {                temp = a[j];                a[j] = a[j+1];                a[j+1] = temp;            }        }    }  //用冒泡排序对集合元素排序    return count;}int main() {    int a[50], b[50], count1, count2;    count1 = set(a);    count2 = set(b);    int i, j, k = 0;    for (i = 0; i < count1; i++) {        printf("%d ", a[i]);    }  //输出集合A    printf("\n");    for (i = 0; i < count2; i++) {        printf("%d ", b[i]);    }  //输出集合B    printf("\n");    for (i = 0; i < count1; i++) {        for (j = k; j < count2; j++) {            if (a[i] == b[j]) {                printf("%d ", b[j]);  //输出相同元素                k = j + 1;            }        }    }  //输出A∩B    printf("\n");    i = 0;    j = 0;    //由于set是排好序的，所以可以将两个set的元素逐个进行比较    while (1) {        if (a[i] < b[j]) {        //如果，A的第一个比B的第一个小，输出A的第一个数，A的下标加1            printf("%d ", a[i]);            i++;        } else if (a[i] > b[j]) {          //如果，A的第一个比B的第一个大，输出B的第一个数，B的下标加1            printf("%d ", b[j]);            j++;        } else {          //如果，相等，输出任何一个，AB下标加1            printf("%d ", a[i]);            i++;            j++;        }        if (i >= count1 || j >= count2) break;  //继续比较，直到一个下标等于其长度为止，输出另外一个set的剩余元素    }    if (i >= count1 && j < count2) {        for (j; j < count2; j++)            printf("%d ", b[j]);  //若B还有未输出的元素继续输出    } else if (i < count1 && j >= count2) {        for (i; i < count1; i++)            printf("%d ", a[i]);  //若A还有未输出的元素继续输出    }    printf("\n");    return 0;}