Blog entries - Codeforces

#	User	Rating
1	tourist	3985
2	jiangly	3814
3	jqdai0815	3682
4	Benq	3529
5	orzdevinwang	3526
6	ksun48	3517
7	Radewoosh	3410
8	hos.lyric	3399
9	ecnerwala	3392
9	Um_nik	3392

#	User	Contrib.
1	cry	170
2	maomao90	162
2	Um_nik	162
4	atcoder_official	160
5	djm03178	158
5	-is-this-fft-	158
7	adamant	154
7	Dominater069	154
9	awoo	153
10	luogu_official	152

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

Full text and comments »

ABitVeryScaredOfWomen's blog