Daily AtCoder # 14 in Python

Introduction

#14 Today is B.

** Thoughts ** ABC081-B ABC081-B divides $ A_i $ until it is divisible by 2 when it is an even number. If it is an odd number, it will be quit immediately.

n = int(input())
a = list(map(int,input().split()))
counter = []
for i in range(n):
    count = 0
    if a[i] % 2 == 0:
        while a[i] % 2 == 0:
            a[i] //= 2
            count += 1
        counter.append(count)
    else:
        print(0)
        quit()

print(min(counter))

ABC087-B ABC087-B has a triple loop with a stop of thinking.

a = int(input())
b = int(input())
c = int(input())
x = int(input())

ans = 0
for i in range(a+1):
    for j in range(b+1):
        for k in range(c+1):
            price = 500 * i + 100 * j + 50 * k
            if price == x:
                ans += 1

print(ans)

ABC083-B ABC083-B has N of about $ 10 ^ 4 $, so I'm checking all N's. $ 1 \ leq i \ leq n $ i is set to str, totaled for each digit, and divided by if.

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

ans = 0
for i in range(n+1):
    i = str(i)
    k = 0
    for j in range(len(i)):
        k += int(i[j])
    if k <= b and k >= a:
        ans += int(i)

print(ans)

ABC088-B ABC088-B works best with each other, so take the largest card left. So, I sort and add up one by one. You don't have to divide it by player, but you can make a difference for each player.

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

a.sort(reverse=True)
alice = 0
bob = 0
for i in range(n):
    if i % 2 == 0:
        alice += a[i]
    else:
        bob += a[i]

print(alice-bob)

ABC085-B ABC085-B cannot put rice cakes of the same size, so they are sorted with no duplication.

n = int(input())
d = {int(input()) for _ in range(n)}

d = list(d)
d.sort()
print(len(d))

Summary

About B can be solved. From tomorrow's C is the production! see you