La and LB are complementary angles, and half of LB is 30 degrees smaller than LA. How many degrees are La and LB?

La and LB are complementary angles, and half of LB is 30 degrees smaller than LA. How many degrees are La and LB?

Let LB be degree X
X＋（X/2＋30）＝180
3X/2＝150
X＝100
So Lb is 100 degrees and La is 80 degrees

Data structure algorithm implementation: two linear tables La and LB are used to represent two sets a and B respectively. Now a new set a = A and B is required
Using two linear tables La and lb to represent two sets a and B respectively, a new set a = A and B is required
The algorithm is
void union(List &La,List Lb){
La_ len=ListLength(La);
Lb_ len=ListLength(Lb);
for(i=1;i

I wrote, you compile it, change it
Void Union (LinkList * L1, linklist * L2, linklist * & L3) / / intersection
{
LinkList *p=L1->next,*q=L2->next,*s,*c;
L3=(LinkList *)malloc(sizeof(LinkList));
L3->next=NULL;
c=L3;
while(p!=NULL&&q!=NULL)
{ if(p->datadata)
{s = (LinkList *) malloc (sizeof (LinkList)); / / copy node
s->data=p->data;
c->next=s;c=s;
p=p->next;
}
else if(p->data>q->data)
{ s=(LinkList *)malloc(sizeof(LinkList));
s->data=q->data;
c->next=s;c=s;
q=q->next;
}
else
{
s=(LinkList *)malloc(sizeof(LinkList));
s->data=p->data;
c->next=s;c=s;
p=p->next;
q=q->next;
}
}
while(q!=NULL)
{
s=(LinkList *)malloc(sizeof(LinkList));
s->data=q->data;
c->next=s;c=s;
q=q->next;
}
c->next=NULL;
while(p!=NULL)
{
s=(LinkList *)malloc(sizeof(LinkList));
s->data=p->data;
c->next=s;c=s;
p=p->next;
}
c->next=NULL;
}

Data structure (C language version) has linear table La (3,5,8110) and lb (2,6,8,9,11,15,20)?
1: If LA and LB represent two sets a and B respectively, the new set a = a ∪ B (the same element is not reserved) prediction output La = (3,5,8,11,2,6,9,15,20)
2: If LA and LB represent two sets a and B respectively, the new set a = a ∪ B (the same element is reserved) prediction output La = (2,3,5,6,8,8,9,11,11,15,20)
The data structure (C language version) is provided with linear tables La (3,5,8,11) and lb (2,6,8,9,11,15,20)

#include
#include
#define list_ init_ size 100
#define listincrement 10
typedef struct
{ int *elem;
int length;
int listsize;
} sqlist;
Int initsqlist (SqList * l) / / initialization
{
l->elem=(int *)malloc(list_ init_ size*sizeof(int));
\x09if(!l->elem)
\x09\x09exit(0);
\x09l->length=0;
\x09l->listsize=list_ init_ size;
\x09return 0;
}
int listinsert_ SQ (SqList * l, int i, int E) / / insert an element
{
\x09int *p,*q;
\x09if(il->length+1)
\x09\x09exit(0);
\x09q=&(l->elem[i-1]);
\x09for(p=&(l->elem[l->length-1]);p>=q;--p)
\x09\x09*(p+1)=*p;
\x09*q=e;
\x09++l->length;
\x09return 0;
}
Void add (SqList * l, int E) / / add to last
{
listinsert_ sq(l,l->length+1,e);
}
void disp(sqlist *l)
{
int i;
for(i=0;ilength;i++)
\x09 printf("%d ",l->elem[i]);
printf("\n");
}
Int find (SqList * l, int E) / / find whether the element exists
{
int i,t=-1;
for(i=0;ilength;i++)
\x09if(l->elem[i]==e)
\x09{t=i;break;}
return t;
}
void opt_ 1 (SqList * La, SqList * lb) / / (the same element is not reserved)
{
int i,j;
for(i=0;ilength;i++)
{
\x09j = find(la,lb->elem[i]);
if(j==-1)
\x09\x09listinsert_ sq(la,la->length+1,lb->elem[i]);
}
}
Void sort (SqList * LA) / / sort
{
\x09int i,j,k;
\x09for(i=0;ilength;i++)
\x09\x09for(j=i+1;jlength;j++)
\x09\x09{
\x09\x09 if(la->elem[i]>la->elem[j])
\x09\x09 {
\x09\x09\x09 k = la->elem[i];
\x09\x09\x09 la->elem[i] = la->elem[j];
\x09\x09\x09 la->elem[j] = k;
\x09\x09 }
\x09\x09}
}
void opt_ 2 (SqList * La, SqList * lb) / / (same element reserved)
{
int i,j;
for(i=0;ilength;i++)
{
\x09j = find(la,lb->elem[i]);
if(j!=-1)
\x09\x09listinsert_ sq(la,j+1,lb->elem[i]);
\x09else
\x09\x09listinsert_ sq(la,la->length+1,lb->elem[i]);
}
sort(la);
}
int main()
{
\x09sqlist La,Lb;
\x09initsqlist(&La);
add(&La,3);
add(&La,5);
add(&La,8);
add(&La,11);
\x09initsqlist(&Lb);
add(&Lb,2);
add(&Lb,6);
add(&Lb,8);
add(&Lb,9);
add(&Lb,11);
add(&Lb,15);
add(&Lb,20);
\x09disp(&La);
\x09disp(&Lb);
\x09opt_ 1 (& LA, & lb); / / operation (the same element is not reserved)
\x09disp(&La);
\x09 La.length =0; / / reinitialize La
add(&La,3);
add(&La,5);
add(&La,8);
add(&La,11);
\x09opt_ 2 (& LA, & lb); / / operation (same element reserved)
\x09disp(&La);
\x09
\x09return 0;
}
Output:
3 5 8 11
2 6 8 9 11 15 20
3 5 8 11 2 6 9 15 20
2 3 5 6 8 8 9 11 11 15 20