UVA 1583 - Digit Generator Solution

Problem link-Click here

it's an easy implement based problem.just follow the statement carefully. you need to process 100 number back to the given number to get desired result. for example if n=2005 then you have to check 2005-100=1905 to 2004 to get the generator of 2005. there's a constraint and which is if n is less than 100 then you have to go to n/2 to n-1 to get the generator and if can't be found just print 0.

an accepted code is as follows:

#include<bits/stdc++.h>
using namespace std;

int process(int n)
{
    int sum=n;

    while(n!=0)
    {
        sum=sum+(n%10);
        n=n/10;
    }
    return sum;
}

int main()
{
    int t,n;

    scanf("%d",&t);
    while(t--)
    {
        bool flag=0;
        int s;

        scanf("%d",&n);

        if(n>100)
        {
            for(int i=n-100;i<n;i++)
            {
                if(process(i)==n)
                {
                    flag=1;
                    s=i;
                    break;
                }
            }
        }
        else
        {
            for(int i=n/2;i<n;i++)
            {
                if(process(i)==n)
                {
                    flag=1;
                    s=i;
                    break;
                }
            }
        }

        if(flag==1)
        {
            printf("%d\n",s);
        }
        else
        {
            printf("%d\n",0);
        }
    }
    return 0;
}