Esempio n. 1
0
int prin(int g[N][M],int i, int w)
{
	if (i==0||w==0)
	{
		printf("%d",i);
		return 0;
	}
	if (g[i][w]==1)
		prin(g[N][M],i-1,w-w[i]);
	else
		prin(g[N][M],i-1,w);
}
Esempio n. 2
0
void prin(int i,int j)
{
if(i==j)
printf("a%d",i);
else
{
printf("(");
prin(i,s[i][j]);
prin((s[i][j]+1),j);
printf(")");
}
}
Esempio n. 3
0
int main()
{
    struct node* a=NULL;
    struct node* b=NULL;
    struct node* c=NULL;
    push(&a,50);push(&a,45);push(&a,40);push(&a,35);push(&a,30);push(&a,25);push(&a,20);push(&a,15);
    prin(a);
    push(&b,49);push(&b,46);push(&b,38);push(&b,25);push(&b,23);push(&b,13);
    prin(b);
    c=merg(a,b);
    prin(c);
    return 0;
}
Esempio n. 4
0
read_and_parse()
{
#define BUFLEN 512
   char buf[BUFLEN];	/* this will have to be dynamically expanded */
   int bufpos = 0;
   int ret;
   Value *v = NULL;
   Value *match_data;
   Value *pattern = vmake_cons(vmake_symbol_c("integer"),
			       vmake_var(integer, (void **) &match_data));

   while (1) {
      ret = read(0, buf + bufpos, BUFLEN - bufpos);
      if (ret < 0) {
	 perror("read");
	 exit(1);
      }
      else {
	 bufpos += ret;

	 do {
	    if (v != NULL) {
	       free_value(v);
	       v = NULL;
	    }
	    ret = parse(bufpos, buf, &v);
	    if (ret > 0) {
	       bcopy(buf + ret, buf, bufpos - ret);
	       bufpos -= ret;
	       printf("parsed: ");
	       prin(stdout, v);
	       fputc('\n', stdout);

	       if (destructure(pattern, v)) {
		  printf("match_data = ");
		  prin(stdout, match_data);
		  fputc('\n', stdout);
	       }
	       else {
		  printf("destructure failed\n");
	       }

	       free_value(v);
	    }
	    else
	       printf("EOF\n");
	 } while (ret > 0);
      }
   }
}
Esempio n. 5
0
main(int argc, char *argv[])
{
   read_and_parse();
#if 0
      Value *v;
      v = ALLOC_VALUE();

      v->tag = cons;
      v->value.cons.car = ALLOC_VALUE();
      v->value.cons.car->tag = symbol;
      v->value.cons.car->value.s.length = 6;
      v->value.cons.car->value.s.string = "symbol";

      v->value.cons.cdr = ALLOC_VALUE();
      v->value.cons.cdr->tag = cons;

      v->value.cons.cdr->value.cons.car = ALLOC_VALUE();
      v->value.cons.cdr->value.cons.car->tag = string;
      v->value.cons.cdr->value.cons.car->value.s.length = 6;
      v->value.cons.cdr->value.cons.car->value.s.string = "string";

      v->value.cons.cdr->value.cons.cdr = ALLOC_VALUE();
      v->value.cons.cdr->value.cons.cdr->tag = integer;
      v->value.cons.cdr->value.cons.cdr->value.integer.i = 23;
      prin(stdout, v);
      fputc('\n', stdout);
#endif
}
Esempio n. 6
0
int main(int argc,char **argv)
{
	const	int	a=100;
	int zz[a];
	prin();
	return 0;
}
Esempio n. 7
0
CELLP princ_f(CELLP args)
{
	CELLP result;
	FILE* nfp = cur_fpo;

	result = prin(args, ESCOFF);
	cur_fpo = nfp;
	return result;
}
Esempio n. 8
0
CELLP print_f(CELLP args)
{
	CELLP result;
	FILE* nfp = cur_fpo;

	result = prin(args, ESCON);	ec;
	fputc('\n', cur_fpo);
	cur_fpo = nfp;
	return result;
}
Esempio n. 9
0
void main()
{
int n,p[100],l,q,temp,k;
clrscr();
printf("enter the no of matrices\n");
scanf("%ld",&n);
printf("enter the orders of matrices\n");
for(i=0;i<(n+1);i++)
scanf("%d",&p[i]);
for(i=1;i<=n;i++)
{
m[i][i]=0;
}
for(l=2;l<=n;l++)
{
for(i=1;i<=(n-l+1);i++)
{
j=i+l-1;
m[i][j]=32767;
for(k=i;k<=(j-1);k++)
{
q=m[i][k]+m[k+1][j]+p[i-1]*p[k]*p[j];
if(q<m[i][j])
{
temp=m[i][j];
m[i][j]=q;
q=temp;

s[i][j]=k;
}
}
}
}
printf("Cost matrix::\n");
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
printf("%d\t",m[i][j]);
}
printf("\n");
}
printf("\nSequence matrix::\n");
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
printf("%d\t",s[i][j]);
printf("\n");
}

prin(1,n);
printf("cost of matrix multi is =%d",m[1][n]);
getch();
}
Esempio n. 10
0
void
prinlis(FILE *f, Value *v, int first)
{
   switch(VTAG(v)) {
    case cons:				/* continue printing list */
      if (! first)
	 putc(' ', f);
      prin(f, v->value.cons.car);
      prinlis(f, v->value.cons.cdr, 0);
      break;
    case nil:				/* last elt in list */
      putc(')', f);
      break;
    default:				/* dotted pair */
      putc(' ', f);
      putc('.', f);
      putc(' ', f);
      prin(f, v);
      putc(')', f);
      break;
   }
}
Esempio n. 11
0
int main()
{	
	int n,W,i;
	//int ans[1005];//要保存的游戏
	scanf("%d%d",&n,&W);//输入 数目n和容量W
	for ( i = 1; i <= n; i++)
		scanf("%d",&w[i]);
	for ( i = 1; i <= n; i++)
		scanf("%d",&v[i]);
	DPKnapsack(w, v,n,W);
	prin(g[N][M],n,W);
	return 0;
}
int main()
{
node *head,**s;
head=addnode(5);
head->next=addnode(15);
head->next->next=addnode(20);
head->next->next->next=addnode(38);
head->next->next->next->next=addnode(67);
s=&head;
printf("address of s is %lld and s it's self %lld\n",&s,s);
prin(head);
printf("double pointer address is %lld \n",&head);
insert_sort_way(&head,26);
printf("double pointer address is %lld \n",&head);
prin(head);
insert_sort_way(&head,2);
printf("double pointer address is %lld \n",&head);
prin(head);

return 0;

}
Esempio n. 13
0
int main(void)
{
  int argn;
  int ** arg, **** exg, * grp;
  rel ** relh;
  argn = argnum(M);
  grp = grpnum(M, N, G);
  arg = estarg(argn, M);
  exg = estexg(N, grp, G);
  relh = estrelh(argn);
  relargexg(arg, exg, M, N, argn, grp, G, relh);
  prin(arg, argn, relh, exg, G, grp, N, M, C);
  frerel(rel, argn);
  freexg(exg, N, grp, G);
  frearg(arg, argn, M);
  fregrp(grp, N);
  return 0;
}
Esempio n. 14
0
void main()
{ int x;
	for(x=97;x<=122;x++)
	prin("%s",&x);
	getch();
}