109A - Lucky Sum of Digits - CodeForces Solution

brute force implementation *1000

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Python Code: 

n = int(input())

for i in range(n // 7 + 1):
    for j in range((n - 7 * (n // 7 - i)) // 4 + 1):
        if (n // 7 - i) * 7 + j * 4 == n:
            print("4" * j + "7" * (n // 7 - i))
            quit()
        elif (n // 7 - i) * 7 + j * 4 > n:
            break
print(-1)

C++ Code: 

#include <bits/stdc++.h>
 
using namespace std;
 
int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
        int n;
    cin>>n;
    string ans = "";
    while(n >= 7 && n%4 != 0){
        n-=7;
        ans+='7';
    }
    if(n%4 != 0){
        cout<<-1<<'\n';
        exit(0);
    }
    while(n >= 28){
        n-=28;
        ans+="7777";
    }
    while(n != 0){
        n-=4;
        ans+='4';
    }
    sort(ans.begin(), ans.end());
    cout<<ans<<'\n';
}


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