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
spoilerxor, but instead of taking $$$mod 2$$$ I'm taking $$$mod (n - 1)$$$
.
