Timeout error code problem in one C++ question

Quote:

Someone can please help me

Help what ?

Quote:

I am getting Timeout Error Code in this code.

This error message is typical from challenges sites, it simply means that you code takes too much time to execute.
You simply forgot to tell which site and which problem you try to solve.
Without those, it is impossible to give any help.

The only thing is that your code is too simplistic and they want you to give a better solution, faster.
[Update]
even if not the solution, I would start by simplifying the code:

int plenti(int n) {
    int sum=0,a;
    for(int i=1;i<=n;i++) {
        int sum1=0;
        if(i==1) {
            continue;
        }
        for(int j=1;j<i;j++) {
            a=0;
            if(i%j==0) {
                sum1=sum1+j;
            }
            if(sum1>i) {
                a=i;
                sum=sum+a+i;
                break;
            }
            else {
                a=0;
            }
        }
    }
    return sum;
}

If ever you decide to post code again, please have the courtesy to comment it.

A comment looks like this:

//  The following code does blah blah blah...

I glanced at the code but don't have the patience, after dinner with wine, to try to figure out what it's trying to do, which sadly is a prerequisite to being able to help. Sure, I could spend many minutes studying it, but frankly I couldn't be bothered when it's not even commented.

Edit1: Thanks for describing what your code is trying to do. Here are some suggestions:

1. How could you eliminate the if(i == 1) check?
2. If you removed a entirely, how could you write sum = sum + a?
3. If 1 is allowed as a divisor, how can you streamline the initialization of sum1 and the for(int j = 1…) loop?
4. In for(int j = 1, j < i…), how could you reduce the maximum value of j?

Edit2: If you knew the result of i/j, what would be the stopping point of the for(int j…) loop?

There are probably other optimizations that rely on analysis, such as numbers that can be immediately excluded if not divisible by both 2 and 3, but this starts to drag in as much math as software, which may not be the intention.