Some hints please

#1

i have written the GP formula of sum of factors of a number x . but not able to process further .

#3

@JARVIS17 Its kind of brute force approach (using Sieve ) calculate sum of factors of, till 10000 and check for what number sum if equal to s, (time complexity will be nlogn ) just now ✓
#include <bits/stdc++.h>
using namespace std;
int a[100000000];
int b[100000000]={0};
void sumofFactors(int n)
{
int m=n;
int res = 1;
for (int i = 2; i <= sqrt(n); i++)
{
int curr_sum = 1;
int curr_term = 1;
while (n % i == 0) {
n = n / i;

        curr_term *= i; 
        curr_sum += curr_term; 
    } 

    res *= curr_sum; 
} 
if (n >= 2) 
    res *= (1 + n); 

b[res]=res;
a[res]= m;

}
int main() {
b[1]=1;a[1]=1;
for(int i=2;i<=10000;i++){
sumofFactors(i);

// b[sumfact+i]=sumfact+i;
// a[sumfact+i]=i;
}

int x;cin>>x;
while(x!=0){
    if(b[x]!=0){
        cout<<a[x]<<endl;
    }
    else{
        cout<<"-1"<<endl;
    }
    cin>>x;
}

} just now ✓
you can take reference from this code just now ✓
Feel free to ask any doubt in it

#4

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