一道高一的集合类的题，求解答。
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已知集合A=x|x平方-ax+a平方-19=0,B=x|x平方-5x+6=0，Cx|x平方+2x-8=0，若A交B不等于空集，A交C等于空集，求实数a的值。
我算出来a等于2或-5，但觉着好像要舍去其中一个，或还差几个。
x²-5x+6=0，（x-2)(x-3)=0，解得x=2或x=3
所以B=｛2，3｝
x²+2x-8=0，(x-2)(x+4)=0，解得x=2或x=-4
所以C=｛2，-4｝
因A交B不等于空集，所以A中含有2和3中至少一个
因A交C等于空集，所以A中不含2和-4
所以A含有3，不含2和-4
把x=3代入方程x²-ax+a²-19=0
得9-3a+a²-19=0，即a²-3a-10=0
(a-5)(a+2)=0，解得a=5或a=-2
当a=5时，方程为x²-5x+6=0，解得x=2或x=3 不合
当a=-2时，方程为x²+2a-15=0，解得a=-5或a=2
综上可得a值为-2
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手机提问的朋友在客户端右上角评价点【满意】即可。 参考技术A 因为5∈A
所以a²+2a-3=5
解得a=2或-4
当a=2时，|a+3|=5不合题意
当a=-4时，|a+3|=1
所以a=-4
本人碰见一道C语言难题，寻大神帮助，利用C语言实现：求任意两个集合的交集、并集、差集，
以前写过一个纯C的， 用的是数组，模拟C++ STL里面的set_intersection，set_union和set_difference的实现。 稍作了修改，添加了些注释，希望能帮到你。注意：必须先对输入集合排序；输出结果和C++ STL的测试结果吻合。
#include <stdio.h>#include <stdlib.h>
#include <string.h>
int set_intersection (const int Set1[], const unsigned int SizeofSet1,
const int Set2[], const unsigned int SizeofSet2,
int Res[], unsigned int* pSizeofRes);
int set_union (const int Set1[], const unsigned int SizeofSet1,
const int Set2[], const unsigned int SizeofSet2,
int Res[], unsigned int* pSizeofRes);
int set_difference (const int Set1[], const unsigned int SizeofSet1,
const int Set2[], const unsigned int SizeofSet2,
int Res[], unsigned int* pSizeofRes);
int compare (const void * a, const void * b);
void print_array(const int arr[], const size_t len);
int main(int argc, char** argv)
int first[] = 5,10,15,20,25;
int second[] = 50,40,30,20,10;
unsigned int size1, size2, size_intxn, size_union, size_diff, retcode;
int *pr_intxn, *pr_union, *pr_diff;
/* Pre-requirement, set MUST be sorted. */
size1 = sizeof(first) / sizeof(first[0]);
size2 = sizeof(second) / sizeof(second[0]);
qsort(first, size1, sizeof(int), compare);
qsort(second, size2, sizeof(int), compare);
/* Intersection */
size_intxn = (size1 > size2) ? size1 : size2; /* estimate size of result */
pr_intxn = (int *)malloc(sizeof(int) * size_intxn); /* pre-allocate result */
if (NULL == pr_intxn)
printf("Intersection memory error.\\n");
return -1;
printf("1) Set of Intersection:\\n");
retcode = set_intersection(first, size1, second, size2, pr_intxn, &size_intxn);
if (retcode == 0)
print_array(pr_intxn, size_intxn);
else
printf("Error in set_intersection, code %d\\n", retcode);
free(pr_intxn);
/* Union */
size_union = size1 + size2; /* estimate size of result */
pr_union = (int *)malloc(sizeof(int) * size_union); /* pre-allocate result */
if (NULL == pr_union)
printf("Union memory error.\\n");
return -1;
printf("2) Set of Union:\\n");
retcode = set_union(first, size1, second, size2, pr_union, &size_union);
if (retcode == 0)
print_array(pr_union, size_union);
else
printf("Error in set_union, code %d\\n", retcode);
free(pr_union);
/* Difference */
size_diff = size1 + size2; /* estimate size of result */
pr_diff = (int *)malloc(sizeof(int) * size_diff); /* pre-allocate result */
if (NULL == pr_diff)
printf("Difference memory error.\\n");
return -1;
printf("3) Set of Difference:\\n");
retcode = set_difference(first, size1, second, size2, pr_diff, &size_diff);
if (retcode == 0)
print_array(pr_diff, size_diff);
else
printf("Error in set_difference, code %d\\n", retcode);
free(pr_diff);
return 0;
/*
Input:
Set1 - First set.
Set2 - Second set.
SizeofSet1 - Set length of First set.
SizeofSet2 - Set length of Second set.
Input/Output:
Res - Set for storing results.
pSizeofRes - Point to SizeofRes, length of result. If SizeofRes is less than
expected, a proper size will be returned to caller.
Return:
0 - If successfully.
1 - SizeofRes is smaller than expected.
-1 - Internal memory error.
*/
int set_difference (const int Set1[], const unsigned int SizeofSet1,
const int Set2[], const unsigned int SizeofSet2,
int Res[], unsigned int* pSizeofRes)
int i, j, k;
unsigned int size_pre;
int *pr = 0;
size_pre = SizeofSet1 + SizeofSet2;
if ( *pSizeofRes < size_pre)
*pSizeofRes = size_pre;
return 1;
pr = (int *)malloc(size_pre * sizeof(int));
if ( pr == NULL )
return -1;
i = 0; j = 0; k = 0;
while ( i < SizeofSet1 && j < SizeofSet2 )
if (Set1[i] < Set2[j]) pr[k++] = Set1[i++];
else if (Set2[j] < Set1[i]) ++j;
else
i++; j++;
memcpy(pr+k, Set1+i-1, sizeof(int)*(SizeofSet1-i+1));
k += SizeofSet1-i;
memcpy(Res, pr, k*sizeof(int));
*pSizeofRes = k;
free(pr);
return 0;
int set_union (const int Set1[], const unsigned int SizeofSet1,
const int Set2[], const unsigned int SizeofSet2,
int Res[], unsigned int* pSizeofRes)
int i, j, k;
unsigned int size_pre;
int *pr = 0;
size_pre = SizeofSet1 + SizeofSet2;
if ( *pSizeofRes < size_pre)
*pSizeofRes = size_pre;
return 1;
pr = (int *)malloc(size_pre * sizeof(int));
if ( pr == NULL )
return -1;
i = 0; j = 0; k = 0;
while ( 1 )
if (i > SizeofSet1 - 1)
memcpy(pr+k, Set2+j-1, sizeof(int)*(SizeofSet2-j+1));
k += SizeofSet2 - j;
break;
if (j > SizeofSet2 - 1)
memcpy(pr+k, Set1+i-1, sizeof(int)*(SizeofSet1-i+1));
k += SizeofSet1 - i;
break;
if (Set1[i] < Set2[j]) pr[k] = Set1[i++];
else if (Set2[j] < Set1[i]) pr[k] = Set2[j++];
else pr[k] = Set1[i]; ++i; ++j;
++k;
memcpy(Res, pr, k*sizeof(int));
*pSizeofRes = k;
free(pr);
return 0;
int set_intersection (const int Set1[], const unsigned int SizeofSet1,
const int Set2[], const unsigned int SizeofSet2,
int Res[], unsigned int* pSizeofRes)
int i, j, k;
unsigned int size_pre;
int *pr = 0;
size_pre = (SizeofSet1 > SizeofSet2) ? SizeofSet1 : SizeofSet2;
if ( *pSizeofRes < size_pre)
*pSizeofRes = size_pre;
return 1;
pr = (int *)malloc(size_pre * sizeof(int));
if ( pr == NULL )
return -1;
i = 0; j = 0; k = 0;
while ( i < SizeofSet1 && j < SizeofSet2 )
if (Set1[i] < Set2[j]) ++i;
else if (Set2[j] < Set1[i]) ++j;
else
pr[k++] = Set1[i];
i++; j++;
memcpy(Res, pr, k*sizeof(int));
*pSizeofRes = k;
free(pr);
return 0;
void print_array(const int arr[], const size_t len)
int i;
for (i = 0; i < len; i++)
printf("%d ", arr[i]);
printf("\\n");
int compare (const void * a, const void * b)
return ( *(int*)a - *(int*)b );
参考技术A 具体忘了，利用链表可以求解，参考严蔚敏 数据结构（C语言版）
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