Reverse the part of the number which is present from
position i to j. Print the new number.[without using the array]
eg:
num=789876
i=2
j=5
778986
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#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
using namespace std;
int getdnum(int num)
{
int numd=0;
while(num!=0)
{
numd++;
num=num/10;
}
return numd;
}
int reversenum(int i,int j ,int d,int num)
{
int a=(num/(pow(10,d-i+1)));
int b=(num/(pow(10,d-j)));
int c=num%static_cast<int>(pow(10,d-j));
int n=0;
int k;
for(k=0;k<=(j-i);k++)
{
n+=(b%10)*(pow(10,j-i-k));
b=b/10;
}
n=a*pow(10,d-i+1)+c+n*pow(10,d-j);
return n;
}
int main()
{
int i,j,k,l,m;
cin>>i>>j>>k;
int d=getdnum(i);
m=reversenum(j,k,d,i);
cout<<m<<endl;
return 0;
}
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / abdur rab
#include <stdio.h>
int get_digits ( int number )
{
int n_digits = 0;
while ( number ) {
n_digits++;
number = number / 10;
}
return ( n_digits );
}
int main ( int argc, char* argv [] )
{
int number = 789876;
int digits = 0;
int temp_number = 0;
int final_number = 0;
int counter = 0;
int start_point = 2;
int end_point = 5;
digits = get_digits ( number );
if ( ( start_point < end_point ) && ( end_point <=
digits ) ) {
temp_number = number;
if ( start_point - 1 ) final_number =
number / pow ( 10, ( digits - ( start_point - 1 ) ) );
counter = digits;
while ( temp_number ) {
if ( ( counter <= ( end_point ) )
&& ( counter >= ( start_point ) ) ) {
final_number *= 10;
final_number += (
temp_number % 10 );
}
temp_number /= 10;
counter--;
}
if ( digits - end_point ) {
final_number *= pow ( 10, ( digits -
end_point ) );
final_number += number % (int) (
pow ( 10, ( digits - end_point ) ) );
}
}
printf ( "\n Number: %d", number );
printf ( "\nFinal Number: %d", final_number );
printf ( "\nS_Pos: %d, E_Pos: %d", start_point,
end_point );
return ( 0 );
}
| Is This Answer Correct ? | 0 Yes | 1 No |
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