CODETALK

Consider this small code sample in Javascript. var arrayOfNumbers=[1,2,3,4,5,6,78]; for(var index in arrayOfNumbers){     console.log(index+1); }; Output of this snippet will be 01 11 21 31 41 51 61 Now this is because Arrays are treated as Objects in Javascript, so index is iterating over keys in an object . In this case the object will be represented as follows arrayOfNumbers={ "0":1, "1":2, "2":3, "3":4, "4":5, "5":6, "6":78}

import java.util.*; public class anagrams {     public static void main(String[] args) {         int count=0,i;         Scanner sc= new Scanner(System.in);         System.out.println("enter first string");         String a=sc.nextLine();         System.out.println("enter second string");         String b=sc.nextLine();         a=a.toLowerCase();         b=b.toLowerCase();         char a1[]=a.toCharArray();         char a2[]=b.toCharArray();         Arrays.sort(a1);         Arrays.sort(a2);         if(a1.length!=a2.length)             count++;         if(count==0)         for(i=0;i<a1.length;i++)             if(a1[i]!=a2[i])                 count++;         if(count==0)             System.out.println("anagrams");         else             System.out.println("not anagrams");     } }

problem link-   http://www.spoj.com/problems/SAMER08F/ #include<stdio.h> #include<math.h> int main() { int n,i,sum=0; while(1) { scanf("%d",&n); if(n==0) break; for(i=1;i<=n;i++) sum=sum+ pow(i, 2); printf("%d\n",sum); sum=0; } return 0; } this problem can be done without  using loop. 

#include<stdio.h> main() { int n; scanf("%d",&n); printf("%d",( ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1) )); }

#include<stdio.h> main() { static int i=100; static int a=1; if(--i) main(); printf("%d\n",a++); }

#include<stdio.h> main() { int n,i; printf("\nenter the number\n"); scanf("%d",&n); i=n-1; if((n&i)==0)    printf("\ninteger is a power of 2\n"); else    printf("\ninteger is not a power of 2\n"); }