Method1 (Using recursion)
void printUp(int startNumber, int endNumber) { if (startNumber > endNumber) return; printf("[%d]\n", startNumber++); printUp(startNumber, endNumber); }
Method2 (Using goto)
`
void printUp(int startNumber, int endNumber)
{
start:
if (startNumber > endNumber)
{
goto end;
}
else
{
printf("[%d]\n”, startNumber++);
goto start;
}
end:
return;
}`
#include
int main()
{
decimal_to_anybase(10, 2);
decimal_to_anybase(255, 16);
getch();
}
decimal_to_anybase(int n, int base)
{
int i, m, digits[1000], flag;
i=0;
printf("\n\n[%d] converted to base [%d] : “, n, base);
while(n)
{
m=n%base;
digits[i]="0123456789abcdefghijklmnopqrstuvwxyz”[m];
n=n/base;
i++;
}
//Eliminate any leading zeroes
for(i–;i>=0;i–)
{
if(!flag && digits[i]!='0’)flag=1;
if(flag)printf("%c”,digits[i]);
}
}
n = sizeof(void *);
if n==8, than 64 bit
if n==4, than 32 bit.
The important realization for this problem is that the hour hand is always moving. In other words, at 1:30, the hour hand is halfway between 1 and 2. Once you remember that, this problem is fairly straightforward. Assuming you don’t care whether the function returns the shorter or larger angle.
int main()
{
int a\[10\]={1, 2, 4, 4, 7, 8, 9, 14, 14, 20};
int i;
for (i = 0;i<9;i++)
{
if (a\[i\] != a\[i+1\])
printf("%d\\n",a\[i\]);
}
return 0;
}
void putlong(unsigned long x)
{
// we know that 32 bits can have 10 digits. 2^32 = 4294967296
for (unsigned long y = 1000000000; y > 0; y /= 10) {
putchar( (x / y) + '0');
x = x % y;
}
}