AtCoder Beginners Contest Past Question Challenge 5

AtCoder Beginners Contest Past Question Challenge 5

ABC134D - Preparing Boxes

If you decide in order while decreasing i from i = N, there is nothing particularly difficult.

N = int(input())
a = list(map(int, input().split()))

t = [0] * N
for i in range(N - 1, -1, -1):
    t[i] = (sum(t[2 * (i + 1) - 1::i + 1]) % 2) ^ a[i]

print(sum(t))
print(*[i + 1 for i in range(N) if t[i] == 1])

ABC080D - Recording

At first glance, it looks like the imos method, but since there is S -0.5, if you use the imos method without thinking about anything, the number of consecutive recordings on the same channel will be two only during 0.5.

The easiest solution is to quit the imos method. If = 1 instead of + = 1, it's okay to double, and in fact it's okay in terms of processing time.

N, C = map(int, input().split())

tt = [[0] * (10 ** 5 + 1) for _ in range(C)]
for _ in range(N):
    s, t, c = map(int, input().split())
    for i in range(s - 1, t):
        tt[c - 1][i] = 1

ct = [0] * (10 ** 5 + 1)
for i in range(C):
    for j in range(10 ** 5 + 1):
        ct[j] += tt[i][j]

print(max(ct))

It's not difficult to do with the imos method, but it's surprisingly difficult to write elegantly. Is it easiest to sort and change the movement if it is continuous?

from operator import itemgetter

N, C = map(int, input().split())
stc = [list(map(int, input().split())) for _ in range(N)]
stc.sort(key=itemgetter(2, 0))

cs = [0] * (10 ** 5 + 1)
pt, pc = -1, -1
for s, t, c in stc:
    if pt == s and pc == c:
        cs[s] += 1
    else:
        cs[s - 1] += 1
    cs[t] -= 1
    pt, pc = t, c

for i in range(1, 10 ** 5 + 1):
    cs[i] += cs[i - 1]

print(max(cs))

By the way, in theory, it can be moss, but if I was passing through the channel that it would not hit so much, I was sniped firmly. Questioner, as expected …….

ABC121D - XOR World

f (A, B) is the same as f (0, A-1) xor f (0, B), so g (N) = f (0, N), where g is Just write it. You can count the number of 1 for each digit of the binary number and process it as an even number or an odd number.

def h(A, n):
    if A == -1:
        return 0
    return A // (2 * n) * n + max(A % (2 * n) - (n - 1), 0)


def g(A):
    result = 0
    for i in range(48):
        t = 1 << i
        if h(A, t) % 2 == 1:
            result += t
    return result


def f(A, B):
    return g(A - 1) ^ g(B)


A, B = map(int, input().split())

print(f(A, B))

In fact, for any even n,n xor (n + 1)is 1, so there is no need to count for each digit.

def g(A):
    t = ((A + 1) // 2) % 2  #N for any even n^ (n + 1) == 1
    if A % 2 == 0:
        return A ^ t
    return t


def f(A, B):
    return g(A - 1) ^ g(B)


A, B = map(int, input().split())

print(f(A, B))

ABC117D - XXOR

For each bit of A, add up the number of standing bits and the number of non-standing bits, and if there are more standing numbers, that bit is 0, and if there are more non-standing numbers, that bit is 1 X. All you have to do is. However, since X has a condition of K or less, if it exceeds K, give up the bit.

N, K = map(int, input().split())
A = list(map(int, input().split()))

bcs = [0] * 41
for i in range(N):
    a = A[i]
    for j in range(41):
        if a & (1 << j) != 0:
            bcs[j] += 1

X = 0
for i in range(40, -1, -1):
    if bcs[i] >= N - bcs[i]:
        continue
    t = 1 << i
    if X + t <= K:
        X += t

result = 0
for i in range(N):
    result += X ^ A[i]
print(result)

ABC094D - Binomial Coefficients

a i </ sub> is the maximum number. A j </ sub> is the next largest number, but a i </ sub> </ sub> C < sub> a j </ sub> </ sub> == a i </ sub> </ sub> C a i </ sub> -a j </ sub> </ sub> and look for the order.

n = int(input())
a = list(map(int, input().split()))

a.sort()
x = a[-1]
b = [(min(e, x - e), e) for e in a[:-1]]
b.sort()
print(x, b[-1][1])

ABC098D - Xor Sum 2

Addition is a monotonous increase, but XOR is not a monotonous increase because 1 xor 1 is 0. By the way, since 0 xor N == 0 + N, "XOR <= addition". In other words, the result of XOR is less than the result of addition even once. Then, no matter how much you increase the number of r from there, it will not be the same forever. Also, if a certain l <r has the same value, the combination of l + 1 and r will also have the same value. You can see that the exclusive method is good.

N = int(input())
A = list(map(int, input().split()))

result = 0
r = 0
xs = A[0]
cs = A[0]
for l in range(N):
    while r < N - 1 and xs ^ A[r + 1] == cs + A[r + 1]:
        r += 1
        xs ^= A[r]
        cs += A[r]
    result += r - l + 1
    xs ^= A[l]
    cs -= A[l]
print(result)

Recommended Posts

AtCoder Beginners Contest Past Question Challenge 6
AtCoder Beginners Contest Past Question Challenge 9
AtCoder Beginners Contest Past Question Challenge 7
AtCoder Beginners Contest Past Question Challenge 10
AtCoder Beginners Contest Past Question Challenge 5
AtCoder Grand Contest Past Question Challenge 2
AtCoder Grand Contest Past Question Challenge 1
AtCoder Past Question Review (12 / 6,7)
AtCoder Past Question Review (12/5)
Challenge AtCoder
# 2 Python beginners challenge AtCoder! ABC085C --Otoshidama
AtCoder Beginner Contest 102 Review of past questions
AtCoder Beginner Contest 085 Review of past questions
AtCoder Beginner Contest 062 Review of past questions
AtCoder Beginner Contest 113 Review of past questions
AtCoder Beginner Contest 074 Review of past questions
AtCoder Beginner Contest 051 Review of past questions
AtCoder Beginner Contest 119 Review of past questions
AtCoder Beginner Contest 151 Review of past questions
AtCoder Beginner Contest 075 Review of past questions
AtCoder Beginner Contest 054 Review of past questions
AtCoder Beginner Contest 110 Review of past questions
AtCoder Beginner Contest 117 Review of past questions
AtCoder Beginner Contest 070 Review of past questions
AtCoder Beginner Contest 105 Review of past questions
AtCoder Beginner Contest 112 Review of past questions
AtCoder Beginner Contest 076 Review of past questions
AtCoder Beginner Contest 089 Review of past questions
AtCoder Beginner Contest 069 Review of past questions
AtCoder Beginner Contest 079 Review of past questions
AtCoder Beginner Contest 056 Review of past questions
AtCoder Beginner Contest 087 Review of past questions
AtCoder Beginner Contest 093 Review of past questions
AtCoder Beginner Contest 046 Review of past questions
AtCoder Beginner Contest 123 Review of past questions
AtCoder Beginner Contest 049 Review of past questions
AtCoder Beginner Contest 078 Review of past questions
AtCoder Beginner Contest 047 Review of past questions
AtCoder Beginner Contest 104 Review of past questions
AtCoder Beginner Contest 057 Review of past questions
AtCoder Beginner Contest 121 Review of past questions
AtCoder Beginner Contest 090 Review of past questions
AtCoder Beginner Contest 103 Review of past questions
AtCoder Beginner Contest 061 Review of past questions
AtCoder Beginner Contest 083 Review of past questions
AtCoder Beginner Contest 124 Review of past questions
AtCoder Beginner Contest 116 Review of past questions
AtCoder Beginner Contest 097 Review of past questions
AtCoder Beginner Contest 092 Review of past questions
AtCoder Beginner Contest 099 Review of past questions
AtCoder Beginner Contest 065 Review of past questions
AtCoder Beginner Contest 053 Review of past questions
AtCoder Beginner Contest 094 Review of past questions
AtCoder Beginner Contest 063 Review of past questions
AtCoder Beginner Contest 107 Review of past questions
AtCoder Beginner Contest: D Question Answers Python
AtCoder Beginner Contest 071 Review of past questions
AtCoder Beginner Contest 064 Review of past questions
AtCoder Beginner Contest 082 Review of past questions
AtCoder Beginner Contest 084 Review of past questions
AtCoder Beginner Contest 058 Review of past questions