AtCoder Beginner Contest 182 题解

​​比赛地址​​

A - twiblr

直接输出2A+100-B即可，时间复杂度O(1)。

a, b = map(int, input().split())
print(2 * a + 100 - b)

Problem B - Almost GCD

暴力穷举即可。

IA = lambda: map(int, input().split())
 
N = 1005
cnt = [0 for i in range(N)]
n = int(input())
a = list(IA())
maxx = -1
res = -1
for it in a:
    for i in range(2, it + 1):
        if it % i == 0:
            cnt[i] += 1
            if maxx < cnt[i]:
                maxx = cnt[i]
                res = i
print(res)

Problem C - To 3

题意：

题解

利用一个数能被3整除当且仅当其各位之和sum能被3整除。

• 如果sum本身能被3整除，则不需要删除。
• 否则统计原数的每一位数%3后的个数，比较%3 =1与%3 =2 的个数，有两种方法可以使其sum变为 %3 =0：

1. %3=1 与%3=2，相互抵消，还剩下的差值即为答案。
2. %3=1 与%3=2，先内部消化，%3 =1的 三个一消除，%3 =2的三个一消除，最后再相互抵消，差值即为答案。

IA = lambda: map(int, input().strip().split())
 
s = input()
n = len(s)
num = [0 for i in range(3)]
summ = 0
for it in s:
    x = int(it)
    num[x % 3] += 1
    summ = (summ + x) % 3
 
if summ % 3 == 0:
    print(0)
else:
    # print(num)
    cha = abs(num[1] - num[2])
    num[1] %= 3
    num[2] %= 3
    cha = min(cha, abs(num[1] - num[2]))
    if cha == n:
        print(-1)
    else:
        print(cha)

Problem D - Wandering

模拟即可。
记录最远位置maxx，当前位置res，前缀和sum，以及前缀和的最大值r。

在每一轮中，首先更新前缀和，然后更新前缀和的最大值，本轮能达到的最大值显然是res+r，用其更新maxx，再用res+sum更新res。

IA = lambda: map(int, input().strip().split())
n = int(input())
a = list(IA())
summ = 0
res = 0
maxx = res
r = -1e9
for i in range(n):
    r = max(summ, r)
    maxx = max(maxx, res + r)
    summ += a[i]
    res += summ
    maxx = max(maxx, res)

print(maxx)

Problem E - Akari

题意

题解
将所有灯和墙都放到矩形中，然后逐行从左到右扫描一遍，再从右到左扫描一遍；逐列从上到下扫描一遍，再从下到上扫描一遍。最后统计亮着的格子即可。

时间复杂度O(HW)。

IA = lambda: map(int, input().strip().split())
h, w, n, m = map(int, input().split())
field = [[0] * w for _ in range(h)]
for _ in range(n):
    a, b = map(int, input().split())
    field[a - 1][b - 1] = 1
for _ in range(m):
    a, b = map(int, input().split())
    field[a - 1][b - 1] = -1

# for l in field:
#     print(l)

# print()
for i in range(h):
    for j in range(1, w):
        if (field[i][j - 1] == 1 or field[i][j - 1] == 2) and field[i][j] == 0:
            field[i][j] = 2
    for j in range(w - 2, -1, -1):
        if (field[i][j + 1] == 1 or field[i][j + 1] == 2) and field[i][j] == 0:
            field[i][j] = 2
for j in range(w):
    for i in range(1, h):
        if (field[i - 1][j] == 1
                or field[i - 1][j] == 3) and field[i][j] != -1:
            field[i][j] = 3
    for i in range(h - 2, -1, -1):
        if (field[i + 1][j] == 1
                or field[i + 1][j] == 3) and field[i][j] != -1:
            field[i][j] = 3
ans = 0
for l in field:
    ans += l.count(1)
    ans += l.count(2)
    ans += l.count(3)
    # print(l)
print(ans)

python超时，但C++稳过

IA = lambda: map(int, input().strip().split())
H, W, n, m = IA()
a = [[0] * (W + 1) for i in range(H + 1)]
for i in range(n):
    x, y = IA()
    a[x][y] = 1

for i in range(m):
    x, y = IA()
    a[x][y] = -1

for i in range(1, H + 1):
    light = False
    for j in range(1, W + 1):
        if a[i][j] == 1:
            light = True
        elif a[i][j] == -1:
            light = False
        elif light:
            a[i][j] = 2

    light = False
    for j in range(W, 0, -1):
        if a[i][j] == 1:
            light = True
        elif a[i][j] == -1:
            light = False
        elif light:
            a[i][j] = 2

for j in range(1, W + 1):
    light = False
    for i in range(1, H + 1):
        if a[i][j] == 1:
            light = True
        elif a[i][j] == -1:
            light = False
        elif light:
            a[i][j] = 2
    light = False
    for i in range(H, 0, -1):
        if a[i][j] == 1:
            light = True
        elif a[i][j] == -1:
            light = False
        elif light:
            a[i][j] = 2

res = 0
for i in range(1, H + 1):
    for j in range(1, W + 1):
        # print(a[i][j], end=" ")
        res += a[i][j] > 0
    # print()
print(res)

F – Valid payments

题意