A — The Ultimate Square
Authors: Gheal
If $$$n$$$ is odd, it is possible to create a square using all $$$n$$$ blocks.
If $$$n$$$ is even, it is possible to create a square using only the first $$$n-1$$$ blocks, since $$$n-1$$$ is odd. Can we use the last block to create a larger square?
If $$$n$$$ is odd, let $$$k=\frac{n+1}{2}$$$ be the width of the last block. It is possible to create a square of side length $$$k$$$ using every block as follows:
- Line $$$1$$$ contains a $$$1 \times k$$$ block;
- Line $$$2$$$ contains a $$$1 \times 1$$$ block and a $$$1 \times (k-1)$$$ block;
- Line $$$3$$$ contains a $$$1 \times 2$$$ block and a $$$1 \times (k-2)$$$ block;
- $$$\ldots$$$
- Line $$$i$$$ contains a $$$1 \times (i-1)$$$ block and a $$$1 \times (k-i+1)$$$ block;
- $$$\ldots$$$
- Line $$$k$$$ contains a $$$1 \times (k-1)$$$ block and a $$$1 \times 1$$$ block.
Since the area of this square is $$$k^2$$$, and the $$$n+1$$$-th block has a width of $$$k$$$ tiles, the total area of the first $$$n+1$$$ blocks is equal to $$$k^2+k \lt (k+1)^2$$$. Therefore, the answer for $$$n+1$$$ is also $$$k$$$.
In conclusion, the answer for each testcase is $$$\lfloor \frac{n+1}{2} \rfloor$$$.
Time complexity per testcase: $$$O(1)$$$.
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
void testcase(){
ll n;
cin>>n;
cout<<(n+1)/2<<'\n';
}
int main()
{
ll t; cin>>t; while(t--)
testcase();
return 0;
}
B — Balanced Substrings
Author: Gheal
What is the maximum number of distinct characters in a diverse string?
What is the maximum frequency of a character in a diverse string?
What is the maximum possible length a diverse string?
In a diverse string, there are at most $$$10$$$ distinct characters: '0', '1', $$$\ldots$$$, '9'. Therefore, each of these characters can appear at most $$$10$$$ times in a diverse string.
With all this in mind, the maximum possible length of a diverse string is $$$10^2=100$$$. To solve this problem, we only need to check whether each substring of length $$$l \le 100$$$ is diverse.
Time complexity per testcase: $$$O(n \cdot 10^2)$$$
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
void tc(){
ll n;
string s;
cin>>n>>s;
ll ans=0;
for(ll i=0;i<s.size();i++){
int fr[10]{}, distinct=0, max_freq=0;
for(ll j=i;j<s.size() && (++fr[s[j]-'0'])<=10;j++){
max_freq=max(max_freq,fr[s[j]-'0']);
if(fr[s[j]-'0']==1) distinct++;
if(distinct>=max_freq) ans++;
}
}
cout<<ans<<'\n';
}
int main()
{
ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
ll t; cin>>t; while(t--) tc();
return 0;
}
C — Zero Sum Prefixes
What is the answer if $$$a_1=0$$$ and $$$a_i \neq 0$$$ for all $$$2 \le i \le n$$$?
What is the answer if $$$a_i=0$$$ and $$$a_j \neq 0$$$ for all $$$1 \le j \le n$$$, $$$j \neq i$$$?
What is the answer if there are only two indices $$$i$$$ and $$$j$$$ for which $$$a_i=a_j=0$$$?
Let's consider the prefix sum array $$$s=[a_1,a_1+a_2,\ldots,a_1+a_2+\ldots+a_n]$$$.
For every index $$$i$$$ such that $$$a_i=0$$$, if we change the value of $$$a_i$$$ to $$$x$$$, then every element from the suffix $$$[s_i,s_{i+1},\ldots,s_n]$$$ will be increased by $$$x$$$. Therefore, if $$$a_{i_1}=a_{i_2}=\ldots=a_{i_k}=0$$$, we'll partition array $$$s$$$ into multiple subarrays:
- $$$[s_1,s_2,\ldots,s_{i_1-1}]$$$;
- $$$[s_{i_2},s_{i_2+1},\ldots,s_{i_3-1}]$$$;
- $$$\ldots$$$
- $$$[s_{i_k},s_{i_{k+1}},\ldots,s_n]$$$;
Since none of the elements from the first subarray can be changed, it will contribute with the number of occurences of $$$0$$$ in $$$[s_1,s_2,\ldots,s_{i_1-1}]$$$ towards the final answer.
For each of the other subarrays $$$[s_l,s_{l+1},\ldots,s_r]$$$, let $$$x$$$ be the most frequent element in the subarray, appearing $$$fr[x]$$$ times. Since $$$a_l=0$$$, we can change the value of $$$a_l$$$ to $$$-x$$$. In this case, every $$$x$$$ in this subarray will become equal to $$$0$$$, and our current subarray will contribute with $$$fr[x]$$$ towards the final answer.
Time complexity per testcase: $$$O(NlogN)$$$
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MAXN=2e5+5;
ll a[MAXN];
map<ll,ll> freq;
void tc(){
ll n;
cin>>n;
ll maxfr=0,current_sum=0,ans=0;
bool found_wildcard=0;
freq.clear();
for(ll i=0;i<n;i++){
cin>>a[i];
if(a[i]==0){
if(found_wildcard) ans+=maxfr;
else ans+=freq[0];
found_wildcard=1;
maxfr=0,freq.clear();
}
current_sum+=a[i];
maxfr=max(maxfr,++freq[current_sum]);
}
if(found_wildcard) ans+=maxfr;
else ans+=freq[0];
cout<<ans<<'\n';
}
int main()
{
ios_base::sync_with_stdio(false), cin.tie(0);
ll t; cin>>t; while(t--)
tc();
return 0;
}
D — ConstructOR
Author: Gheal
If at least one of $$$a$$$ and $$$b$$$ is odd, and $$$d$$$ is even, then there are no solutions.
Without loss of generality, we will only consider the case when $$$a$$$ is odd and $$$d$$$ is even. Since the last bit of $$$a$$$ is $$$1$$$, then the last bit of $$$a|x$$$ must also be $$$1$$$.
Therefore, $$$a|x$$$ cannot be a multiple of $$$d$$$, as $$$a|x$$$ is odd, while $$$d$$$ is even.
Note that there are more cases in which no solutions exist, however they are more generalised versions of this case.
If a triplet ($$$a,b,d$$$) has no solutions, then ($$$a\cdot 2,b\cdot 2,d\cdot 2$$$) has no solutions as well.
Combined with the first hint, we can say that a triplet ($$$a,b,d$$$) has no solutions if $$$min(\text{lsb}(a),\text{lsb}(b)) \lt \text{lsb}(d)$$$. Here, $$$\text{lsb}(x)$$$ represents the least significant bit of $$$x$$$.
Since both $$$a|x$$$ and $$$b|x$$$ must be divisible by $$$d$$$, it's better to choose an $$$x$$$ such that $$$a|x=b|x=x$$$.
If $$$d$$$ is odd, since $$$a,b \lt 2^{30}$$$, we can ensure that $$$a|x=b|x=x$$$ by setting the last $$$30$$$ bits of $$$c$$$ to $$$1$$$.
If $$$d$$$ is even, then the last $$$\text{lsb}(d)$$$ bits of $$$x$$$ should be set to $$$0$$$, while the other bits from the last $$$30$$$ bits should be set to $$$1$$$. Here, $$$\text{lsb}(x)$$$ represents the least significant bit of $$$x$$$.
Let $$$k=\text{lsb}(d)$$$, where $$$\text{lsb}(d)$$$ represents the least significant bit of $$$d$$$.
Since $$$a|x$$$ and $$$b|x$$$ are multiples of $$$d$$$, the last $$$k$$$ bits of $$$a$$$ and $$$b$$$ (and also $$$x$$$) must be equal to $$$0$$$.
Otherwise, there are no solutions and we can print $$$-1$$$.
To simplify the construction process, we will try to find some $$$x$$$ such that $$$a|x=b|x=x$$$. Since we already know that the last $$$k$$$ bits of $$$a$$$, $$$b$$$ and $$$x$$$ are $$$0$$$, we will consider that the other $$$30-k$$$ of the $$$30$$$ least significant bits of $$$x$$$ are equal to $$$1$$$:
This gives the following general formula for $$$x$$$:
Now, we'll try to find some $$$p$$$ for which $$$x$$$ is a multiple of $$$d=2^k \cdot d'$$$:
Time complexity per testcase: $$$O(log d)$$$
Note that if $$$a|b$$$ is already a multiple of $$$d$$$, we can consider $$$x=a|b$$$.
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
void tc(){
ll a,b,d,k=0,inverse_of_2,total_inverse=1;
cin>>a>>b>>d;
a|=b;
if(a%d==0){
cout<<a<<'\n';
return;
}
while(a%2==0 && d%2==0)
a/=2,d/=2,k++;
inverse_of_2=(d+1)/2;
if(a%2==1 && d%2==0){
cout<<"-1\n";
return;
}
for(ll i=0;i<30-k;i++)
total_inverse=total_inverse*inverse_of_2%d;
cout<<((total_inverse<<(30-k))-1)*(1ll<<k)<<'\n';
}
int main()
{
ios_base::sync_with_stdio(false), cin.tie(0);
ll t; cin>>t; while(t--)
tc();
}
E — Yet Another Array Counting Problem
Author: Gheal
TBA
TBA
TBA
F — Circular Xor Reversal
How can we perform $$$a_i=a_i \oplus a_j$$$ for any $$$0 \le i,j \lt n$$$, $$$i \neq j$$$?
We can swap the values of $$$a_i$$$ and $$$a_j$$$ by performing the following sequence of xor assignments:
- $$$a_i=a_i \oplus a_j$$$;
- $$$a_j=a_j \oplus a_i$$$;
- $$$a_i=a_i \oplus a_j$$$;
Performing $$$a_i=a_i \oplus a_j$$$ on its own is pretty wasteful, as it requires $$$O(4 \cdot ((j+n-i) \mod n))$$$ operations.
If $$$j=i+3$$$, can we perform $$$a_i=a_i \oplus a_j$$$ and $$$a_{i+1}=a_{i+1} \oplus a_{j-1}$$$ simultaneously?
Building on the idea presented in hint $$$3$$$, can we perform the following $$$\lfloor \frac{n}{2} \rfloor$$$ xor assignments simultaneously if $$$n$$$ is even?
- $$$a_0 = a_0 \oplus a_{n-1}$$$;
- $$$a_1 = a_1 \oplus a_{n-2}$$$;
- $$$a_2 = a_2 \oplus a_{n-3}$$$;
- $$$\ldots$$$
- $$$a_{\frac{n}{2}-1} = a_3 \oplus a_{\frac{n}{2}}$$$
The target number of operations for performing all $$$\lfloor \frac{n}{2} \rfloor$$$ assignments is $$$n \cdot \frac{n}{2}=\frac{n^2}{2}$$$.
Using hints $$$2$$$ and $$$4$$$, we can already solve the problem if $$$n$$$ is even.
If $$$m=dist(i,j)=((j+n-i) \mod n)$$$ and $$$b_k=a_{(i+k) \mod n$$$, let $$$f(i,j)$$$ be a sequence of operations that performs:
- $$$b_0 = b_0 \oplus b_{m-1}$$$;
- $$$b_1 = b_1 \oplus b_{m-2}$$$;
- $$$b_2 = b_2 \oplus b_{m-3}$$$;
- $$$\ldots$$$
- $$$b_{\frac{m}{2}-1} = b_{\frac{n}{2}-1} \oplus b_{\frac{m}{2}}$$$
We can reverse array $$$a$$$ by performing $$$f(0,n-1)$$$, $$$f(\frac{n}{2},\frac{n}{2}-1)$$$ and $f(0,n-1), in this order.
Otherwise, if $$$n$$$ is odd, we can perform $$$f(0,n-1)$$$, $$$f(\frac{n+1}{2},\frac{n-3}{2})$$$ and $f(0,n-1), in this order.
The total number of operations is $$$3\cdot \frac{n^2}{2} \le 3 \cdot \frac{400^2}{2} = 240'000 'lt 250'000$$$.
TBA
TBA
If there is anything wrong or unclear in this editorial, feel free to ping me in the comments.