UVA 11728 - Alternate Task Solution

problem link-click here

This problem is easy to implement if you can understand it clearly. 1st of all pre-calculate the factors sum till 1000 and keep the sum in the corresponding index. for example, suppose the sum of the factors of 101 is 1+101 = 102. so keep array[101] = 102. similarly keep 1 to 1000 all the factors sum to its corresponding index. now for a number N just check for 1000 to 1 iteratively which index's sum is equal to N. print the index number and if not found just print -1.

an accepted code is as follows:

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

vector<int> v;


void pre()
{
    for(int i=1;i<=1000;i++)
    {
        int sum=0;

        for(int j=1;j<=i;j++)
        {
            if(i%j==0)
            {
                sum=sum+j;
            }
        }

        v.push_back(sum);
    }
}

int main()
{
    pre();

    int n,q=1;

    while(cin>>n && n)
    {
        int s=0,p;

        for(int i=v.size();i>=0;i--)
        {
            if(v[i]==n)
            {
                s=1;
                p=i+1;
                break;
            }
        }

        if(s==1)
        {
            printf("Case %d: %d\n",q,p);
        }
        else
        {
            printf("Case %d: %d\n",q,-1);
        }

        q++;

    }

    return 0;
}