For each $i$ from $1$ to $N,$ try setting $a_1=i.$ Then we can determine the rest of the elements of $a$ by setting $a_i=b_{i-1}-a_{i-1}$ for each $2\le i\le N.$ If this indeed produces a valid permutation (all elements of $a$ are in $[1,N]$ and none repeat), then return the result. This runs in $O(N^2)$ time.

Dhruv Rohatgi's code:

#include <iostream>
#include <algorithm>
using namespace std;
 
int N;
int b[100000], d[100000], ans[100000];
bool used[100000];
 
int main() {
	freopen("photo.in","r",stdin);
	freopen("photo.out","w",stdout);
	cin >> N;
	for(int i=0;i<N-1;i++)
		cin >> b[i];
	for(int i=2;i<N;i++)
		d[i] = b[i-1]-b[i-2];
	for(int a=1;a<=N;a++)
	{
		ans[0] = a, ans[1] = b[0] - a;
		for(int i=2;i<N;i++)
			ans[i] = ans[i-2] + d[i];
		for(int i=1;i<=N;i++)
			used[i] = 0;
		bool bad = 0;
		for(int i=0;i<N;i++)
		{
			if(used[ans[i]] || ans[i] <= 0 || ans[i] > N)
			{
				bad = 1;
				break;
			}
			used[ans[i]] = 1;
		}
		if(!bad)
		{
			for(int i=0;i<N;i++)
			{
				cout << ans[i];
				if(i<N-1) cout << ' ';
			}
			cout << '\n';
			return 0;
		}
	}
}

Bonus: Solve the problem in $O(N).$