Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (vovuh)**

```
for i in range(int(input())):
n, m = map(int, input().split())
print('YES' if n % m == 0 else 'NO')
```

Idea: Roms

**Tutorial**

Tutorial is loading...

**Solution (Roms)**

```
for t in range(int(input())):
n = input()
print(*sorted(map(int, input().split()))[::-1])
```

Idea: adedalic

**Tutorial**

Tutorial is loading...

**Solution 1 (adedalic)**

```
fun main() {
val T = readLine()!!.toInt()
testCases@for (tc in 1..T) {
val (_, k) = readLine()!!.split(' ').map { it.toLong() }
val a = readLine()!!.split(' ').map { it.toLong() }.toLongArray()
var maxPower = 1L
while (maxPower < 1e16.toLong())
maxPower *= k
while (maxPower > 0) {
val positions = a.withIndex().filter { it.value >= maxPower }.map { it.index }
if (positions.isNotEmpty()) {
if (positions.size > 1) {
println("NO")
continue@testCases
}
a[positions[0]] -= maxPower
}
maxPower /= k
}
if (a.max()!! > 0L) {
println("NO")
continue@testCases
}
println("YES")
}
}
```

**Solution 2 (adedalic)**

```
fun getMask(a: Long, k: Long): Long? {
var (tmp, res) = listOf(a, 0L)
var cnt = 0
while (tmp > 0) {
if (tmp % k > 1)
return null
res = res or ((tmp % k) shl cnt)
tmp /= k
cnt++
}
return res
}
fun main() {
val T = readLine()!!.toInt()
for (tc in 1..T) {
val (n, k) = readLine()!!.split(' ').map { it.toLong() }
val a = readLine()!!.split(' ').map { getMask(it.toLong(), k) }
val b = a.filterNotNull()
if (b.size < n) {
println("NO")
continue
} else {
val res = b.reduce { acc, l -> if (acc < 0 || (acc and l) > 0) -1 else acc or l }
println(if (res < 0) "NO" else "YES")
}
}
}
```

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
#include <bits/stdc++.h>
using namespace std;
const int N = 200043;
const int MOD = 998244353;
int add(int x, int y)
{
x += y;
while(x >= MOD) x -= MOD;
while(x < 0) x += MOD;
return x;
}
int mul(int x, int y)
{
return (x * 1ll * y) % MOD;
}
int binpow(int x, int y)
{
int z = 1;
while(y)
{
if(y & 1) z = mul(z, x);
x = mul(x, x);
y >>= 1;
}
return z;
}
int inv(int x)
{
return binpow(x, MOD - 2);
}
int divide(int x, int y)
{
return mul(x, inv(y));
}
int fact[N];
void precalc()
{
fact[0] = 1;
for(int i = 1; i < N; i++)
fact[i] = mul(fact[i - 1], i);
}
int C(int n, int k)
{
return divide(fact[n], mul(fact[k], fact[n - k]));
}
int main()
{
precalc();
int n, m;
cin >> n >> m;
int ans = 0;
if(n > 2)
ans = mul(C(m, n - 1), mul(n - 2, binpow(2, n - 3)));
cout << ans << endl;
}
```

Idea: MikeMirzayanov

**Tutorial**

Tutorial is loading...

**Solution (adedalic)**

```
#include<bits/stdc++.h>
using namespace std;
#define fore(i, l, r) for(int i = int(l); i < int(r); i++)
#define sz(a) int((a).size())
#define x first
#define y second
typedef long long li;
typedef long double ld;
typedef pair<int, int> pt;
template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
return out << "(" << p.x << ", " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
out << "[";
fore(i, 0, sz(v)) {
if(i) out << ", ";
out << v[i];
}
return out << "]";
}
const int INF = int(1e9);
const li INF64 = li(1e18);
const ld EPS = 1e-9;
const int N = 555;
int n, a[N];
inline bool read() {
if(!(cin >> n))
return false;
fore(i, 0, n)
cin >> a[i];
return true;
}
int dp[N][N];
int calcDP(int l, int r) {
assert(l < r);
if(l + 1 == r)
return dp[l][r] = a[l];
if(dp[l][r] != 0)
return dp[l][r];
dp[l][r] = -1;
fore(mid, l + 1, r) {
int lf = calcDP(l, mid);
int rg = calcDP(mid, r);
if(lf > 0 && lf == rg)
return dp[l][r] = lf + 1;
}
return dp[l][r];
}
int dp2[N];
inline void solve() {
fore(i, 0, N)
dp2[i] = INF;
dp2[0] = 0;
fore(i, 0, n) {
fore(j, i + 1, n + 1) {
if(calcDP(i, j) > 0)
dp2[j] = min(dp2[j], dp2[i] + 1);
}
}
cout << dp2[n] << endl;
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
int tt = clock();
#endif
ios_base::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
cout << fixed << setprecision(15);
if(read()) {
solve();
#ifdef _DEBUG
cerr << "TIME = " << clock() - tt << endl;
tt = clock();
#endif
}
return 0;
}
```

Idea: BledDest

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
#include<bits/stdc++.h>
using namespace std;
const int N = 300043;
const int K = 5;
int x, y, z, n;
long long a[N];
typedef vector<vector<int> > state;
map<state, int> d;
int cnt;
int p;
vector<vector<int> > state_log;
int mex(const vector<int>& a)
{
for(int i = 0; i < a.size(); i++)
{
bool f = false;
for(auto x : a)
if(x == i)
f = true;
if(!f)
return i;
}
return a.size();
}
state go(state s)
{
int f1 = mex({s[0][K - x], s[1][K - y], s[2][K - z]});
int f2 = mex({s[0][K - x], s[2][K - z]});
int f3 = mex({s[0][K - x], s[1][K - y]});
state nw = s;
nw[0].push_back(f1);
nw[1].push_back(f2);
nw[2].push_back(f3);
for(int i = 0; i < 3; i++)
nw[i].erase(nw[i].begin());
return nw;
}
void precalc()
{
d.clear();
state cur(3, vector<int>(K, 0));
cnt = 0;
state_log.clear();
while(!d.count(cur))
{
d[cur] = cnt;
state_log.push_back({cur[0].back(), cur[1].back(), cur[2].back()});
cur = go(cur);
cnt++;
}
p = cnt - d[cur];
}
int get_grundy(long long x, int t)
{
if(x < cnt)
return state_log[x][t];
else
{
int pp = cnt - p;
x -= pp;
return state_log[pp + (x % p)][t];
}
}
void read()
{
scanf("%d %d %d %d", &n, &x, &y, &z);
for(int i = 0; i < n; i++)
scanf("%lld", &a[i]);
}
int check(int x, int y)
{
return x == y ? 1 : 0;
}
void solve()
{
precalc();
int ans = 0;
for(int i = 0; i < n; i++)
ans ^= get_grundy(a[i], 0);
int res = 0;
for(int i = 0; i < n; i++)
{
ans ^= get_grundy(a[i], 0);
res += check(ans, get_grundy(max(0ll, a[i] - x), 0));
res += check(ans, get_grundy(max(0ll, a[i] - y), 1));
res += check(ans, get_grundy(max(0ll, a[i] - z), 2));
ans ^= get_grundy(a[i], 0);
}
printf("%d\n", res);
}
int main()
{
int t;
scanf("%d", &t);
for(int i = 0; i < t; i++)
{
read();
solve();
}
}
```

**Tutorial**

Tutorial is loading...

**Solution (BledDest)**

```
#include<bits/stdc++.h>
using namespace std;
const int N = 1000043;
map<char, int> nxt[N];
bool term[N];
int n, k;
char buf[3];
int dp[N];
int dict[N];
int T[4 * N];
int f[4 * N];
void build(int v, int l, int r)
{
T[v] = int(1e9);
if(l != r - 1)
{
int m = (l + r) / 2;
build(v * 2 + 1, l, m);
build(v * 2 + 2, m, r);
}
}
int getVal(int v)
{
return T[v] + f[v];
}
void push(int v, int l, int r)
{
T[v] += f[v];
if(l != r - 1)
{
f[v * 2 + 1] += f[v];
f[v * 2 + 2] += f[v];
}
f[v] = 0;
}
void upd(int v, int l, int r)
{
if(l != r - 1)
{
T[v] = min(getVal(v * 2 + 1), getVal(v * 2 + 2));
}
}
int get(int v, int l, int r, int L, int R)
{
if(L >= R)
return int(1e9);
if(l == L && r == R)
return getVal(v);
push(v, l, r);
int m = (l + r) / 2;
int ans = min(get(v * 2 + 1, l, m, L, min(m, R)), get(v * 2 + 2, m, r, max(L, m), R));
upd(v, l, r);
return ans;
}
void add(int v, int l, int r, int L, int R, int val)
{
if(L >= R)
return;
if(l == L && r == R)
{
f[v] += val;
return;
}
push(v, l, r);
int m = (l + r) / 2;
add(v * 2 + 1, l, m, L, min(m, R), val);
add(v * 2 + 2, m, r, max(L, m), R, val);
upd(v, l, r);
}
void setVal(int v, int l, int r, int pos, int val)
{
if(l == r - 1)
{
f[v] = 0;
T[v] = val;
return;
}
push(v, l, r);
int m = (l + r) / 2;
if(pos < m)
setVal(v * 2 + 1, l, m, pos, val);
else
setVal(v * 2 + 2, m, r, pos, val);
upd(v, l, r);
}
void dfs(int v, int d, int last)
{
dp[v] = last + 1;
if(term[v])
dp[v] = min(dp[v], get(0, 0, n + 1, 0, d));
setVal(0, 0, n + 1, d, dp[v] + 1);
if(term[v])
add(0, 0, n + 1, 0, d + 1, 1);
for(auto x : nxt[v])
dfs(x.second, d + 1, dp[v]);
setVal(0, 0, n + 1, d, int(1e9));
}
int main()
{
scanf("%d", &n);
for(int i = 0; i < n; i++)
{
int p;
scanf("%d %s", &p, buf);
char c = buf[0];
nxt[p][c] = i + 1;
}
scanf("%d", &k);
for(int i = 0; i < k; i++)
{
scanf("%d", &dict[i]);
term[dict[i]] = true;
}
build(0, 0, n + 1);
dfs(0, 0, -1);
for(int i = 0; i < k; i++)
printf("%d ", dp[dict[i]]);
puts("");
}
```

fast editorial, thanks!

In G we can maintain value "the shortest distance to the smallest line in the lexicographic order from the current prefix" in dfs. To update it just check the distance from the current vertex + 1. And to pass it to the next vertex, we need to add to the value the number of vertices from the S that we went through in other sons. It is O(n). Code.

I solved G by the same way. And This is my code.

Could you plz explain your idea more clearly?

thx :P

Main challenge is speeding up computation of f(u) = min d(v) + rank of u in subtree rooted at v, over all ancestors v of node u. Suppose we have this computed for node u, and wish to update computation for children c of u. The difference is exactly the size of sibling subtrees that come before c. This difference affects subtrees of each ancestor equally, so after adding that to existing f(u), we only need to consider the additional case for a subtree beginning at c so you get something like f(c) = min(f(u)+siblings, d(c)).

Another Example (sorry for all the junk — just look for the dfs method which is 8 LOC)

thx :P

another way to find rank(u) in subtree rooted at v is tin(u)-tin(v)+(isgood(v)) (we increment time only when we encounter a good string)

can anyone explain in problem D why we are multiplying by 2^(n — 3) ?

Each element will appear either before the maximum number, or after it. Therefore each of the N-3 numbers (that aren't the maximum and the number we have to repeat) has two options, so we have to mulitiply by 2, N-3 times.

Since there are n elements we are to choose (n-1) elements and one element will be copy of one of n-1 elements choosen by us. Now we can not make copy of max element as it violates the condition so we are left with n-2 elements so we can choose one of them.Let x be the position of max element so the element which has duplicate must be persent on either side because if it will be persent on single side the sequence can not be strictly increasing or decreasing so we have max element at x and two same elements so now we are left with n-3 elements since there are two choices for every element to either go on right hand side of max element or LHS of that max element . so 2^(n-3) factor is involved. why is it so? Since we can arrange a array in strictly increasing order if we dont have duplicates. Hope it was clarified

we assume one element as left or right side of max .Isn't it possible that element become max for another combination?So why not we use 3^n-3?As there can be three position?

Notice that we have chosen the current set of (n-1) elements already and are then placing the elements in the desired order.

Hence for the current set of elements, only one will be the peak element. For another set, it might be so that an element currently on the left/right becomes the peak, but for that the initial (n-1) elements chosen will be different. Hence the factor of 3 doesn't appear here.

great explanation .Thanks!

We have already selected elements now how can our combo differ?

int mul(int x, int y) { return (x * 1ll * y) % MOD; } For what we use this func, why we cant make x-long long, and y-long long, and multiply them?

Of course, we can, there is no difference: long long 1.08 s and int 1.08s for calculation of $$$2^{28}!$$$. Only one think that you should do is declare

`mod`

as constant. If no const — long long 3.33s. It is because division by a constant can be calculated without integer division.Hey Bro can you elaborate this further

It is because division by a constant can be calculated without integer divisionBook: Hacker's Delight, 2nd Edition, Chapter 10: Integer Division By Constants. Main idea in few words: change $$$\dfrac{x}{y}$$$ by multiplication $$$x \cdot y^{-1}$$$, where $$$y$$$ is constant.

Can anyone please explain E problem it would be great help. Thanks in advance :)

I'll try to go along the lines of the solution:

Let $$$dp[i][j]$$$ be the value of the single remaining element, when subarray $$$[i:j]$$$ is reduced using the given operation. If there is no way to reduce this subarray into a single element, $$$dp[i][j]$$$ will be $$$-1$$$.

How do you compute $$$dp[i][j]$$$? Consider index $$$k$$$, such that $$$i \le k \lt j$$$. We divide subarray $$$[i:j]$$$ into two halves: $$$[i:k]$$$ and $$$[k+1:j]$$$.

Base case: when subarray is of size 1 $$$(i=j)$$$, answer will obviously be the number in that subarray. Now for any $$$k$$$, if $$$dp[i][k] = dp[k+1][j]$$$, i.e. we can divide subarray into two halves such that both halves are reduced to the same number, we can combine the two numbers, hence $$$dp[i][j] = dp[i][k] + 1$$$.

This is one half of the solution. We now compute the minimum number of partitions that can be reduced to size 1.

Let $$$dp2[i]$$$ : minimum number of partitions required for array $$$[1:i]$$$. We compute this in this way: Take index $$$k$$$, $$$1 \le k \le i$$$. If $$$[k:i]$$$ can be reduced to a single element, i.e. $$$dp[k][i] \ne -1$$$, then we can find optimal partitioning of subarray $$$[1:k-1]$$$ and just add one to the answer. Formally, $$$dp2[i] = min(dp2[i], dp2[k] + 1)$$$, for all $$$k$$$ such that $$$dp[k][i] \ne -1$$$.

The required answer is $$$dp[n]$$$. Here's my solution for reference: 72913085. Hope this helped.

Can you explain why will dp[i][j] reduce to a single unique value?

I do feel it's intuitive but still why is it impossible for having two different ways of combining elements of a subarray and reaching two different final merged values?

consider sequence $$$ 2^{a_1}, 2^{a_2}, ..., 2^{a_n} $$$ combining two adjacent equal values $$$ a_i = a_{i+1} $$$ is equivalent to merging two values in to one value $$$ 2^{a'} = 2^{a_i} + 2^{a_{i+1}} = 2 ^ {a_i + 1} $$$. So if [l, r] can be reduced to a single value, it must be unique.

This proof is from tmwilliamlin168 explanation

How come converting into a sum helped to prove uniqueness?

I think he means that, the "merging" operation is akin to adding 2 similar powers of 2, sum of a few powers of 2,if reduced to a single power of 2, then that term should be unique

We can prove that a subarray $$$[i:j]$$$ can be divided in atmost one way, i.e. there can be only one such $$$k$$$, for which $$$dp[i][k] = dp[k+1][j]$$$.

For this, we define a

fully reducible subarrayas a subarray which can be reduced to a single element. We claim two things: removal of elements from a fully reducible subarray will only reduce the value of the remaining element, providedthe remaining subarray is a fully reducible subarray. Similarly, addition of elements will only increase the value of the remaining element (you can try this out).Hence, if $$$dp[i][k] = dp[k+1][j]$$$, there is no possible way to transfer elements from subarray $$$[i:k]$$$ to $$$[k+1:j]$$$, or vice versa, such that $$$dp[i][k'] = dp[k'+1][j]$$$, for another $$$k'$$$.

We can also prove by contradiction,

Lets say there are two values of K, i.e K1 and K2, such that dp(i, K1) = dp(K1 + 1,j) = X and dp(i, K2) = dp(K2 + 1, j) = Y.

Lets assume K1 < K2,

If that is the case then according to the dp definition, the subarray (i, K1) reduced to the number X, and the subarray (K1 + 1, j) also reduced to X whereas the subarrays (i, K2) and (K2 + 1, j) reduced to Y,

Since K1 < K2, the subarray (i, K1) is a prefix of (i, K2) thus X < Y. And (K2 + 1, j) is a suffix of (K1 + 1, j) which suggests that X > Y.

This leads to a contradiction.

Amazing, thanks a lot :)

Thanks a lot!

Thanks a lot :)

This explanation was better than the one in editorial. Thanks a lot!

Thanks a lot to all in this thread This explanation is much better than that of editorial

in problem F, why are five lines enough to determine all the other values?

Let's consider that the number of remaining soldiers is i.Because x,y,z<=5.You only need the answer to[i-5,i-1] to update i.When you know the answer to i,you delete the answer to (i-5),and add the new answer to the end of the vector.In other words,the vector is scrolling. Because the period is at most 36,you can use brute force to find it. I hope this helps. ：）

Can anyone explain how we get string "ieh" in first example in two seconds? After getting string "i" "ieh" is lexicographically second after "i", so we need at least 3 seconds, isn't it?

How to solve problem

Cusingbitmasks. Kindly help me.You can make each number of the array v into a number based on k.If the number only consists of 0 and 1, it must be able to become 0.Then, mark the position of each 1.If the number consists of other numbers or the position of some 1 is already marked, the array won't be able to be filled with 0 and the answer is "NO". After you finish that operation for each number in the array v, the answer will be "YES".This is my solusion.

I am a Chinese and I am sorry that my English is poor. if there is something wrong in my words, please tell me and I will repair it as fast as posiible.

Can you please explain What is the use of

stepvariable in your code?How we will check if this i power is used or not?Is it posssible to solve G on Java?

See my solution, I had to make my recursion iterative in order to bypass stack overflow

awoo Why the same solutions receive completely different times?:

72925974 4321 ms = 72925981 2885 ms

72927822 1668 ms = 72927856 920 ms

Can anyone explain B. Bogo Sort Problem??

In this problem it was quite an observation that if we sort the array in descending(or non increasing order to be precise) then A[i]-i can never be equal to A[j]-j for i<j. Because if i<j that implies A[i]>A[j] since we sorted the array in non increasing order.Now we are subtracting larger value by smaller index example array of 5 4 will become 5-1 != 4-2. hope that helps. Here's the link to my solution

If we sort the array in non-increasing order, then we have

$$$ \begin{equation} i \lt j \end{equation}\tag{1} $$$

$$$ \begin{equation} a_i \ge a_j \end{equation}\tag{2} $$$

Rewrite $$$(1)$$$ to

$$$ \begin{equation} -i \gt -j \end{equation}\tag{3} $$$

and add $$$a_i$$$ to both side

$$$ \begin{equation} a_i - i \gt a_i -j \end{equation}\tag{4} $$$

Add $$$-j$$$ to both side to $$$(2)$$$

$$$ \begin{equation} a_i - j \ge a_j - j \end{equation}\tag{5} $$$

From $$$(4)$$$ and $$$(5)$$$, we have

$$$ \begin{equation} a_i - i \gt a_i - j \ge a_j - j \end{equation}\tag{6} $$$

which means

$$$ \begin{equation} a_i - i \gt a_j - j \end{equation}\tag{7} $$$

$$$ \begin{equation} i - a_i \ne j - a_j \end{equation}\tag{8} $$$

$$$ \begin{equation} j - a_j \ne i - a_i \end{equation}\tag{9} $$$

Therefore, after sorting the array in non-increasing order, for each pair of indexes $$$ i < j $$$ , the condition $$$ j - a_j \ne i - a_i$$$ holds.

Can anyone explain me C please.

convert given numbers into k-base system. now, do addition of k-base numbers without k-base system. now, you can observe that, if there is any bit is >= 2 then it's not possible to make given numbers using power of k. because, you can use k^some power only one time.

great way!

can u please explain in more details?

if a number is in k-base system, then if number n = 1111 = k^0 + k^1 + k^2 + k^3 , and if n = 2111 = k^0 + k^1 + k^2 + 2*k^3, means we need to add k^3 two times. so we can't make this number.

lets assume we have to check for no. X ,if X is summation of power of k or not then X=k^n+k^m+k^l....,where n>m>l..,so if we take k^n common and divide X by k^n then resulting no. should be divisible by k, X=k^n(1+k^(m-n)+k^(l-n)) X/(k^n)-1=k^(m-n)+k^(l-n)..

code for above explanation

Solution for Problem C is not quit clear can anyone help me out with that??? adedalic

see this comment : https://codeforces.com/blog/entry/74640?#comment-587595

Could you tell me what does this

return dp[l][r] = a[l]?I think it first assigns the value of a[I] to dp[l][r] and then returns the value of dp[l][r];:^)

I have an $$$O(n \log n)$$$ solution to problem E: https://codeforces.com/blog/entry/74656

Can someone please explain how time complexity of problem E is n^3 ? Thanks.

Figured it out after rethinking the problem. Let's think of the main function for loop and calcDP separately. So calcDP is an N^3 function while our for loop runs in N^2. They are independent entities. So when you call calcDP from inside your N^2 for loop, their complexities don't affect each other.

There is another solution to problem D with complexity $$$O(n \log P)$$$.

Welcome to see in https://codeforces.com/blog/entry/74685. For the Chinese version https://andylizf.github.io/2020/03/10/CF1312D-Count-the-Arrays.

Can anyone prove it is equal to the normal solution?

can anyone explain c better plz?

The question states that you have an array ( which is initially all zeros ), and you apply some number of operations, say $$$M$$$.

In $$$i$$$th operation ( $$$0 \le i \lt M$$$ ), you can either pass and go to step $$$i+1$$$, or add $$$k^i$$$ to one of the elements of the array.

Notice that, each power of $$$k$$$ is used

atmostonce, if at all. Thus, we simply convert each number to base $$$k$$$, and see if any power of $$$k$$$ is repeated.Another possible solution of 1312E — Array Shrinking. Complexity is O(V * N) where V is limit of

`a`

. The solution is here.To solve F, I used the heuristic that the period is equal to $$$smallest+largest$$$ out of $$$x,y,z$$$ It only has 2 exceptions (considering $$$y$$$ and $$$z$$$ equivalent): $$$1,3,4$$$ (period=7) and $$$4,1,2$$$ (period=3).

I have no clue on why this works, wondering if it just somehow worked because of small constraints on the values. Any insights?

Link to my code: 73467272

what is the time complexity of D,is it O(m log p) or (n log p)??

I think the time complexity is only O(log p).

For problem G,we don't need any data struct .Simple dfs is enough.

The code is very short:

CodeThough there are many details,it's still much simpler than solutions using data structures.Also,it's very fast.

In problemA, if we consider case n = 8 and m = 4 then how can a convex polygon with 4 sides can have a same center.

Thanks, in advance

Number the vertex from 0 to n-1 (7) take 0th and 2nd and 4th and 6th vertex

How do you prove that we can have the vertices of smaller polygon in common with larger polygon?

I don't have a math-heavy perfect proof but you can think in this way.

Lets call polygon of m sides be polygonIn and polygon of n sides be polygonOut. select one side of polygonIn and connect it to the center of polygon the angle will be (2*pi/m) also note that this angle is equal to x*(2*pi/n) (x is some integer) because the end points of a side of polygonIn must be end points of some x continous sides of polygonOut.

Example in case of hexagon and triangle x = 2 because end points of one side of triangle is also end points of 2 continous sides of hexagon.

so we get x*(2*pi/n) = 2*pi/m

x*m = n

n%m = 0

How can we check the period is at most 36 by brute force ?

Is there a theoretical proof for the bound on period ?

In problem E, would the single representative of a subarray always be unique (to cache beforehand), would appreciate if someone can give a proof.

I think This is really good .

Got it, thank you

Can any one help me with my solution (81649940) to Problem E.

my approach :

dp[i][j][0] — min length of which can be obtained from subarray [i,j]

dp[i][j][1]- after reduction of this([i,j]) subarray what is the left most value eg if array is 6 6 3 3 5 then dp[i][j][2] = 7;(from reduced array- 7 4 5)

dp[i][j][2]- after reduction of this subarray what is the right most value eg. if array is 6 6 3 3 5 then dp[i][j][2] = 5;(from reduced array- 7 4 5)

can someone explain me problem C. It would be great help. Thanks

resolved

can anyone explain Problem A in detail?

no

This tutorial is not linked in the problems, for example 1312A - Two Regular Polygons Maybe someone can fix this.

fixed