when user give a number it multiply with 9 without useing '+' and &

when user give a number it multiply with 9 without
useing '+' and '*' oprator

Question Posted / yogendra

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when user give a number it multiply with 9 without useing '+' and '*' oprator..

Answer / guest

#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{clrscr();
int a,i,r=0;
scanf("%d",&a);

for(i=a;i>0;i--)
r=r-9;
r=abs(r);
printf("%d",r);
getch();
}

Is This Answer Correct ?

4 Yes

2 No

when user give a number it multiply with 9 without useing '+' and '*' oprator..

Answer / tknowledge05

The above guest is me .. okk....
by the time i solved and executing i was logged out and when i posted the answer it considered guest.......

Is This Answer Correct ?

0 Yes

0 No

when user give a number it multiply with 9 without useing '+' and '*' oprator..

Answer / tknowledge05

The above guest is me .. okkkk....
by the time i solved and executing i was logged out and when i posted the answer it considered guestt.........

Is This Answer Correct ?

0 Yes

0 No

when user give a number it multiply with 9 without useing '+' and '*' oprator..

Answer / vaibhav nigam

/*
your answer is quite correct but its better to use
r=-r; instead of r=abs(r);

well i would have done it like :

*/

#include <iostream>
using namespace std;
int main()
{
int k;
cin >> k;
k=(k<<4)-(k<<2)-(k<<1)-k;
cout << k << endl;
return 0;
}

//here no loop is used..

Is This Answer Correct ?

0 Yes

1 No

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