Последние 8 часов: Q1–Q18 и алгоритмический минимумThe final 8 hours: Q1–Q18 and the algorithmic minimum

$p=(2y-x)/3=20$ and $q=(2x-y)/3=40$. Both are nonnegative integers, so the answer is Yes.

import sys
x, y = map(int, sys.stdin.read().split())
p_num = 2*y - x
q_num = 2*x - y
ok = p_num >= 0 and q_num >= 0 and p_num % 3 == 0 and q_num % 3 == 0
print("Yes" if ok else "No")

Tests such as $(0,0)$, $(1,1)$, and a point with one coordinate much larger than the other catch almost every formula error.

The most frequent letters by column are b, a, k, e, and r, so the answer is baker.

from collections import Counter
n, m = map(int, input().split())
words = [input().strip() for _ in range(n)]
ans = []
for j in range(m):
    cnt = Counter(word[j] for word in words)
    ans.append(min(cnt, key=lambda ch: (-cnt[ch], ch)))
print(''.join(ans))

The key signal is a sum of coordinate-wise penalties. The global optimum is then assembled from independent local optima.

A subarray sum is the difference of two prefix sums. That difference is divisible by $3$ exactly when the two prefixes have the same remainder, so it is enough to count how often each remainder has already appeared.

The prefix remainders are $0,0,1,2,0,2$. Remainder $0$ occurs three times and contributes $3$ pairs; remainder $2$ occurs twice and contributes one more. The answer is $4$.

from collections import Counter
c=Counter({0:1}); s=ans=0
for x in [3,1,4,1,5]:
    s=(s+x)%3; ans+=c[s]; c[s]+=1

Because every number is positive, the window sum is monotone: moving the right endpoint can only increase it, while moving the left endpoint can only decrease it. Each endpoint therefore moves in one direction, replacing a quadratic search with one linear pass.

The maximum window [1,3,1,1] has sum $6$ and length $4$.

l=s=best=0
for r,x in enumerate(a):
    s+=x
    while s>6: s-=a[l]; l+=1
    best=max(best,r-l+1)

The sliding window works here because every array element is positive.

As $k$ increases, the required time never increases. The predicate “all piles finish within $9$ hours” is therefore false up to one boundary and true afterward, which is exactly the structure needed for binary search on the first feasible value.

At $k=10$, the piles require $10$ hours; at $k=11$, they require $9$. The answer is $11$.

You solved this pattern correctly in today's mock. Preserve the boundary-search template.

The first two columns contain a,e and c,f, the third contains d, and the fifth contains two b characters. The answer is ...../...../...../ac..b/efd.b.

n, m = map(int, input().split())
g = [list(input().strip()) for _ in range(n)]
for j in range(m):
    letters = [g[i][j] for i in range(n) if g[i][j] != '.']
    for i in range(n):
        g[i][j] = '.'
    start = n - len(letters)
    for k, ch in enumerate(letters):
        g[start+k][j] = ch
for row in g:
    print(''.join(row))

The algorithm runs in $O(nm)$ time. Do not sort the letters: falling preserves their vertical order.

The only valid split is 7 11 9 4 1 6 2 10 5 8 3.

The string length determines $n$: for $n\le9$ the length is $n$; otherwise it is $9+2(n-9)=2n-9$.

s = input().strip()
L = len(s)
n = L if L <= 9 else (L + 9) // 2
used = [False] * (n + 1)
path = []

def dfs(i):
    if i == L:
        return len(path) == n
    for size in (1, 2):
        if i + size > L or (size == 2 and s[i] == '0'):
            continue
        value = int(s[i:i+size])
        if 1 <= value <= n and not used[value]:
            used[value] = True
            path.append(value)
            if dfs(i + size):
                return True
            path.pop()
            used[value] = False
    return False

print(*path) if dfs(0) else print(-1)

This search is controlled: each node has at most two branches, and the range and uniqueness checks prune invalid branches early.

The condition “the delivery was on time” restricts the sample space to on-time deliveries only. The denominator is therefore not all $400$ deliveries, but the on-time total already reconstructed in Q5.

North contributes $72$ of the $324$ on-time deliveries. Therefore

Difficulty: 1/4 after Q5, 2/4 on its own. The only trap is leaving all $400$ deliveries in the denominator.

Expectations of a linear combination use the coefficients themselves. For the variance of an independent linear combination, the coefficients are squared—this is the key distinction between the two calculations.

$E\hat\mu=(1+2+2+1)\mu/6=\mu$. By independence,

Difficulty: 2/4. The main trap is that two similar-looking calculations use different rules.

The method of moments equates an observed sample moment to its theoretical counterpart. The given statistic is the sample mean, so compute the first theoretical moment $E_\theta X$ rather than a likelihood.

$E_\theta X=(\theta+1)/(\theta+2)$. Solving $(\theta+1)/(\theta+2)=3/4$ gives $\theta=2$.

Difficulty: 2/4. Do not confuse the method of moments with maximum likelihood.

The likelihood is the density of the observed sample viewed as a function of the parameter. Taking logs preserves the maximizer and turns the product into an easy-to-differentiate sum; the candidate must still satisfy $\theta>-1$.

$\ell'(\theta)=4/(\theta+1)-2$. Setting this to zero gives $\theta+1=2$, so $\hat\theta=1.0000$. The second derivative is negative, and the estimate satisfies $\theta>-1$.

Template for Q13: $\hat\theta=-n/\sum\ln X_i-1$.

The median is not the parameter itself but a function of it. Derive that function from the CDF first, and only then substitute the already obtained estimate $\hat\theta$.

$F_\theta(x)=x^{\theta+1}$. The median satisfies $m^{\theta+1}=1/2$, hence $m=2^{-1/(\theta+1)}$. At $\hat\theta=1$, this gives $1/\sqrt2=0.7071$.

BFS explores the graph in layers: first distance $1$, then distance $2$, and so on. The first recorded distance to a vertex is therefore minimal. Mark a vertex when it enters the queue, not when it leaves, to avoid adding it several times.

A shortest path is $2\to4\to5\to6$, with length $3$.

from collections import deque
q=deque([2]); dist={2:0}
while q:
    v=q.popleft()
    for u in g[v]:
        if u not in dist:
            dist[u]=dist[v]+1; q.append(u)

BFS finds shortest paths in an unweighted graph. A weighted graph may require a different algorithm.

The problem has two distinct stages. The first white ball updates the urn probabilities by Bayes' rule. Returning the ball restores the composition, but “that same urn” preserves the updated source choice for the second draw.

After the first white ball, the posterior probabilities of urns I and II are $40/43$ and $3/43$. Since the ball was replaced,

Difficulty: 3/4. This combines two steps: a Bayesian update and a prediction. The main trap is reverting to the original urn probabilities.

Convert every percentage to a count first. The unknown rates then disappear, and the problem becomes a table with rows for hubs and columns for on-time and late deliveries. The ratio $2:3:5$ and the total $400$ determine the row totals.

The hubs have $80$, $120$, and $200$ couriers. North has $72$ on-time deliveries. North and Central together have $174$, so Central has $102$ on-time and $18$ late deliveries. There are $324$ on-time deliveries overall, so South has $150$ on-time and $50$ late deliveries. The ratio is $18:50=9:25$.

Difficulty: 3/4. There is little formal machinery, but the four-step count reconstruction is sensitive to a single arithmetic error.

Integration by parts is not being used to evaluate every integral separately. Its purpose is to lower the exponent of $x^n$ by one, turning the remaining integral into $B_{n-1}$ and producing a recurrence.

$B_n=[x^ne^x]_0^1-n\int_0^1x^{n-1}e^x dx=e-nB_{n-1}$. Since $B_0=e-1$, we obtain $B_1=1$, $B_2=e-2$, and $B_3=6-2e$. Their sum is $5-e=2.2817$.

Difficulty: 2/4 computationally, 3/4 in written form. You already used this technique correctly on the first attempt today.

For a continuous i.i.d. sample, multiply the densities. Taking logs turns the product into $10\ln(\theta+1)+\theta\sum\ln X_i$; the maximum must still lie in the allowed domain $\theta>-1$.

$\ell'(\theta)=10/(\theta+1)-4=0$, so $\hat\theta=1.5000$. The second derivative is negative.

Difficulty: 3/4 for you right now. The algebra is short, but today's s2002 showed that this chain is not yet automatic.

A plug-in estimate means deriving the target characteristic as a function of the unknown parameter first, then substituting $\hat\theta$. For the median, integrate the density to obtain the CDF and solve $F_\theta(m)=1/2$.

$m(\theta)=2^{-1/(\theta+1)}$. At $\hat\theta=1.5$, the estimate is $\hat m=2^{-0.4}=0.7579$.

Difficulty: 3/4. This composes two methods, exactly the kind of chain that was least stable today.

Q15 already provides the denominator $1+v^Tu=3$. Sherman–Morrison preserves the rank-one structure, and $\operatorname{tr}(uv^T)=v^Tu$, so there is no need to write every entry of the inverse.

$A^{-1}=I-uv^T/3$. Therefore $\operatorname{tr}(A^{-1})=3-2/3=7/3=2.3333$.

Difficulty: 2/4 after Q15, 3/4 on its own. You solved an analogous problem correctly today; no ladder is needed.

The repeated factor $x^2$ requires both $A/x$ and $B/x^2$; omitting the second term makes the decomposition impossible. After integration, combine the logarithms before taking the improper limit: separately they diverge, but their difference has a finite limit.

$1/[x^2(x+1)]=-1/x+1/x^2+1/(x+1)$. Evaluating the antiderivative $-\ln x-1/x+\ln(x+1)$ at the improper limit gives $1-\ln2=0.3069$.

Difficulty: 2/4 conceptually, 3/4 for accuracy. You solved the basic version immediately today.