What is the necessary and sufficient condition?

The necessary and sufficient condition for a beautiful sequence is that there exist one $$$i$$$, such that $$$a_{i} \le i$$$. Just check the sequence for the condition.

#include<bits/stdc++.h>
using namespace std;
int a[100005];
void solve()
{
    int n;
    scanf("%d",&n);
    for(int i  =1;i <= n;i++) scanf("%d",&a[i]);
    for(int i = 1;i <= n;i++) {
        if(a[i] <= i) {
            puts("YES");
            return;
        }
    }
    puts("NO");
}
int main()
{
    int t;scanf("%d",&t);
    while(t--) solve();
}

How the binary representation changes after an operation?

First, we notice that after each operation, the number of candies is always a odd number. So even numbers can not be achieved.

Then consider how the binary representation changes for a odd number $$$x$$$, after turn it into $$$2x+1$$$ or $$$2x-1$$$.

\begin{itemize}

\item For the $$$2x + 1$$$ operation: $$$\overline{\dots 1}$$$ turns into $$$\overline{\dots 11}$$$.

\item For the $$$2x - 1$$$ operation: $$$\overline{\dots 1}$$$ turns into $$$\overline{\dots 01}$$$.

\end{itemize}

So, the operation is just insert a $$$0/1$$$ before the last digit. And the answer for an odd $$$n$$$ is just the binary representation of $$$n$$$, after removing the last digit.

#include<bits/stdc++.h>
using namespace std;
void solve()
{
    int n;scanf("%d",&n);
    if(n % 2 == 0) {
        puts("-1");return;
    }
    vector<int> v;
    int f = 0;
    for(int i = 29;i >= 1;i--) {
        if((n >> i) & 1) {
            f = 1;
            v.push_back(2);
        }
        else if(f) {
            v.push_back(1);
        }
    }
    printf("%d\n",v.size());
    for(auto x : v) {
        printf("%d ",x);
    }
    printf("\n");
}
int main()
{
    int t;scanf("%d",&t);
    while(t--) solve();
    return 0;
}