Q 11.)
Is there a way to multiply two matrix in less than O(n^3) time complexity..
For this Q algorithm was asked, if you guys don't know the answer then make a comment below and I will tell the answer or just take some efforts to search it.
Q 12.)
#include
using namespace std;
int main()
{
int a,c,foo;
char *p;
cin>>a>>c;// give two numbers..
p=(char *)a;
foo= (int)&p[c];
cout< return 0;
}
what this code snippet does...
even if u get the answer explain the logic.
without executing
Q 13.)
whats wrong with this program..
#include
using namespace std;
class foo
{
const int a;
public:
foo():a(10)
{}
void show()
{
cout<}
};
int main()
{
const foo obj;
obj.show();
}
Q 14.)
#include
using namespace std;
int main()
{
int *p,*c;
p=(int*)10;
c=(int*)20;
cout<<(int)p<<(int)c;
}
is this a valid program..
if no why..
if yes why..
Q 15.) #include
using namespace std;
int main()
{
void *vptr = malloc(sizeof(int)*2);
int p;
p=(int)vptr;
*((int*)p +0)=9;
*((int*)p +1)=10;
cout<<1[(int*)vptr]; }
comment on this program and write the output if any
Is there a way to multiply two matrix in less than O(n^3) time complexity..
For this Q algorithm was asked, if you guys don't know the answer then make a comment below and I will tell the answer or just take some efforts to search it.
#include
using namespace std;
int main()
{
int a,c,foo;
char *p;
cin>>a>>c;// give two numbers..
p=(char *)a;
foo= (int)&p[c];
cout<
}
what this code snippet does...
even if u get the answer explain the logic.
without executing
Q 13.)
whats wrong with this program..
#include
using namespace std;
class foo
{
const int a;
public:
foo():a(10)
{}
void show()
{
cout<}
};
int main()
{
const foo obj;
obj.show();
}
Q 14.)
#include
using namespace std;
int main()
{
int *p,*c;
p=(int*)10;
c=(int*)20;
cout<<(int)p<<(int)c;
}
is this a valid program..
if no why..
if yes why..
Q 15.) #include
using namespace std;
int main()
{
void *vptr = malloc(sizeof(int)*2);
int p;
p=(int)vptr;
*((int*)p +0)=9;
*((int*)p +1)=10;
cout<<1[(int*)vptr]; }
comment on this program and write the output if any
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