An unexpectedly difficult problem

#	User	Rating
1	tourist	4009
2	jiangly	3823
3	Benq	3738
4	Radewoosh	3633
5	jqdai0815	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	ksun48	3390
10	gamegame	3386

#	User	Contrib.
1	cry	167
2	maomao90	163
2	Um_nik	163
4	atcoder_official	161
5	adamant	159
6	-is-this-fft-	158
7	awoo	157
8	TheScrasse	154
9	nor	153
9	Dominater069	153

Hi, I was solving this problem a few minutes ago, and the following problem is a subproblem of that one.

Given integer $$$n$$$, $$$ n \ge 3$$$, and array $$$a$$$ consisting of $$$n - 1$$$ integers, all of them are equal to some value $$$b$$$. Later, some integer $$$c$$$, such that $$$c \ne b$$$ is selected and inserted into the array $$$a$$$ at random position. By given $$$n$$$ and $$$a$$$ find $$$c$$$.

This problem itself isn't so difficult, but with a certain restriction I couldn't solve it. The restriction is folowing: you can't use $$$if$$$ and ternary operators in any way.

I came up with the solution, that calls $$$if$$$ only one time.

my solution

#include <bits/stdc++.h>

using namespace std;

int a[100001] = {0};

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, ans = 0;
    cin >> n;
    for(int i = 0; i < n; i++)
        cin >> a[i];
    if(a[0] == a[1])
        for(int i = (n & 1) ^ 1; i < n; i++)
            ans ^= a[i];
    else
        for(int i = 0; i < 3; i++)
            ans ^= a[i];
    cout << ans << "\n";
    return 0;
}

Can you solve this problem with the given restriction?

UPD: In the comments there is one of the solutions, and I've also came up with one.

my final solution

#include <bits/stdc++.h>

using namespace std;

int nxr[32] = {0};

void add(int a, int n) {
    int mask = 1;
    for(int i = 0; i < 31; i++) {
        nxr[i] += (a & mask) >> i;
        nxr[i] %= n;
        mask <<= 1;
    }
    return ;
}

int ans() {
    int mask = 1, ans = 0;
    for(int i = 0; i < 31; i++) {
        ans += nxr[i] * mask;
        mask <<= 1;
    }
    return ans;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, a;
    cin >> n;
    for(int i = 0; i < n; i++) {
        cin >> a;
        add(a, n - 1);
    }
    cout << ans() << "\n";
    return 0;
}

BledDest's solution is shorter, but mine is easier to understand, at least for me, this is just

spoiler

Rev.	By	When	Δ	Comment
en24	tredsused70	2023-01-28 00:38:27	5	(published)
en23	tredsused70	2023-01-28 00:38:18	1028	(saved to drafts)
en22	tredsused70	2023-01-27 23:56:48	0	Tiny change: 'striction?\n' -> 'striction? There is no restriction to time complexity\n'
en21	tredsused70	2023-01-27 22:43:33	0	(published)
en20	tredsused70	2023-01-27 22:41:31	4	Tiny change: 'blem of this one.\n\nG' -> 'blem of that one.\n\nG'
en19	tredsused70	2023-01-27 22:39:12	46
en18	tredsused70	2023-01-27 22:34:44	0
en17	tredsused70	2023-01-27 22:34:44	207
en16	tredsused70	2023-01-27 22:29:54	2
en15	tredsused70	2023-01-27 22:29:23	122
en14	tredsused70	2023-01-27 22:23:31	191
en13	tredsused70	2023-01-27 22:14:54	5	Tiny change: 'h that $c != b$, and b' -> 'h that $c \ne b$, and b'
en12	tredsused70	2023-01-27 22:11:37	28
en11	tredsused70	2023-01-27 22:11:06	20
en10	tredsused70	2023-01-27 22:10:01	256
en9	tredsused70	2023-01-27 22:00:12	51
en8	tredsused70	2023-01-27 21:58:18	13
en7	tredsused70	2023-01-27 21:56:46	159	Tiny change: 'solution">\n~~~~~\n#in' -> 'solution">~~~~~\n#in'
en6	tredsused70	2023-01-27 21:51:24	66
en5	tredsused70	2023-01-27 21:50:44	49	Tiny change: '\n\n<spoil' -> '\n~~~~~\nint n = 1;\ncout << n << "\n";\n~~~~~\n\n\n<spoil'
en4	tredsused70	2023-01-27 21:50:07	5
en3	tredsused70	2023-01-27 21:48:58	25
en2	tredsused70	2023-01-27 21:47:57	256
en1	tredsused70	2023-01-27 21:46:13	122	Initial revision (saved to drafts)

History