Problem 10

March 2, 2021

15 comments on “Problem 10”


  • Pallipattu

    n=input()
    k=n[::-1]
    dec=0
    for i in range(len(k)):
    if k[i]==’1′:
    dec=dec+2**i
    x=str(oct(dec))
    print(x[2:])

    0

  • Daipayan

    public static void main(String[] args) {
    ArrayList arr = new ArrayList();
    int binary = Integer.parseInt(args[0]);
    int dec = 0, count = 0;
    while (binary > 0) {
    if (binary % 10 > 0) {
    dec += Math.pow(2, count);
    }

    binary = binary / 10;
    count++;
    }
    while (dec > 0) {
    arr.add(dec % 8);
    dec = dec / 8;
    }
    for (int i = arr.size() – 1; i >= 0; i–) {
    System.out.print(arr.get(i));
    }

    }

    0

  • amit

    int main()
    {
    int num,call;
    int arr[100];
    int count=0;
    int i=0;
    scanf(“%d”,&num);
    num=b_to_d(num);
    while(num!=0)
    {
    arr[i]=num%8;
    num=num/8;
    i++;
    }
    for(int j=i-1;j>=0;j–)
    {
    printf(“%d”,arr[j]);
    }

    return 0;
    }
    int b_to_d(int input)
    {
    int i=0;
    int num=0;
    while(input!=0)
    {
    num=num+(input%10)*pow(2,i);
    input=input/10;
    i++;
    }
    return num;
    }

    0

  • Raman

    // Online C compiler to run C program online
    #include

    int main() {
    int x,n,a[20],rem,p;
    int sum=0;
    int count=0,i=0;
    printf(“enter the number :”);
    scanf(“%d”,&x);
    //binary to deciamal
    p=1;
    while(x>0){
    rem=x%10;
    sum=sum+(rem*p);
    p*=2;
    x=x/10;
    i++;
    }
    n=sum;

    //decimal to octal
    i=0;
    count=0;
    while(n>0){
    rem=n%8;
    a[i]=rem;
    n=n/8;
    i++;
    count++;
    }
    for(i=count-1;i>=0;i–){
    printf(“%d”,a[i]);
    }

    return 0;
    }

    0

  • swapnil

    public static void main(String[] args) {
    // TODO Auto-generated method stub

    Scanner s=new Scanner(System.in);
    String s1=s.nextLine();
    String[] s2=s1.split(“”);
    ArrayList arrList=new ArrayList();
    ArrayList res=new ArrayList();
    for(String i:s2) {
    arrList.add(Integer.parseInt(i));
    }
    System.out.println(Math.pow(2,2));
    Math.pow(4, 3);
    if(arrList.size()%3!=0) {
    int m = arrList.size()%3;
    int k=3-m;
    for(int i=0;i<k;i++) {
    arrList.add(i,0);
    }
    }
    int size=arrList.size()/3;
    for(int i=0;i=0) {
    if(arrList.get(0)==1) {
    sum+= Math.pow(2,p);
    }
    arrList.remove(0);
    p-=1;
    }

    res.add(sum);
    } System.out.println(res);
    int index=0;
    for(float i:res) {
    arrList.add(0, (int) i);

    index+=1;
    }
    Collections.reverse(arrList);
    for(int i:arrList) {
    System.out.print(i);
    }
    }

    0

  • swapnil

    arr=[int(i) for i in input()]
    if len(arr)%3!=0:
    extraZeroToAdd=len(arr)%3
    for i in range(3-extraZeroToAdd):
    arr.insert(0,0)
    size=len(arr)//3
    res=[]
    for i in range(int(size)):
    c=2
    sum=0
    while c>=0:
    if(arr[0]==1):
    sum=sum+2**c
    c=c-1
    arr.pop(0)
    res.append(sum)

    print(”.join(str(i) for i in res))

    0

  • The Bong

    public class BinaryToOctal {
    public static void main(String[] args) {
    System.out.println(binToOct(“101001”));
    }
    static int binToOct(String number) {
    int dec = Integer.parseInt(number , 2);
    int oct = Integer.parseInt(Integer.toOctalString(dec));
    return oct;
    }
    }

    1

  • Febin

    c=0
    s=0
    n=int(input(“Enter number:”))
    while n > 0:
    r=int(n%10)
    p=pow(8,c)
    s=s+(r*p)
    n=int(n/10)
    c=c+1
    print(s)

    0

  • Bathula

    #python
    binary = input(“Enter any number in Binary Format: “);
    if binary == ‘x’:
    exit();
    else:
    temp = int(binary, 2);
    p=oct(temp)
    a=p.replace(“0o”,””)
    print(a)

    1

  • Bathula

    #python
    binary = input(“Enter any number in Binary Format: “);
    if binary == ‘x’:
    exit();
    else:
    temp = int(binary, 2);
    print(temp)
    p=oct(temp)
    a=p.replace(“0o”,””)
    print(a)

    0