Break through in 1 minute. Just write.

```
N = int(input())
if N % 2 == 0:
print('White')
else:
print('Black')
```

Break through in 4 minutes. It took a long time to change the formula to subtract the total from A to A-1 because the formula to calculate the total from A to B did not come out.

```
N = int(input())
result = 0
for _ in range(N):
A, B = map(int, input().split())
result += B * (B + 1) // 2
result -= (A - 1) * A // 2
print(result)
```

When I calmed down, the ceremony came out properly (laughs).

```
N = int(input())
result = 0
for _ in range(N):
A, B = map(int, input().split())
result += (A + B) * (B - A + 1) // 2
print(result)
```

Break through in 19 minutes. Start by google the formula of the straight line passing through the two points. N ≤ 10 ^{ 2 </ sup> So * O * (* N * 3 </ sup>) is OK It took me a while to notice. In the input / output example, division by zero came out and I remember the x = a type straight line (laughs). Although the formula was slightly wrong, I was lucky enough to be picked up in the input / output example. Can be fixed, no mistakes AC.}

```
N = int(input())
xy = [tuple(map(int, input().split())) for _ in range(N)]
for i in range(N - 1):
xi, yi = xy[i]
for j in range(i + 1, N):
xj, yj = xy[j]
for k in range(j + 1, N):
x, y = xy[k]
if xj == xi:
if x == xi:
print('Yes')
exit()
elif yj == yi:
if y == yi:
print('Yes')
exit()
elif abs(y - (yj - yi) / (xj - xi) * (x - xi) - yi) < 0.000001:
print('Yes')
exit()
print('No')
```

Addendum: If you multiply both sides by the division term, you can calculate accurately with an integer, and there was no division by zero ...

```
N = int(input())
xy = [tuple(map(int, input().split())) for _ in range(N)]
for i in range(2, N):
xi, yi = xy[i]
for j in range(1, i):
xj, yj = xy[j]
for k in range(j):
x, y = xy[k]
if (y - yi) * (xj - xi) == (yj - yi) * (x - xi):
print('Yes')
exit()
print('No')
```

It breaks through in 13 and a half minutes. If you notice that digits over 1000 are automatically multiples of 8, it's a problem that ends immediately. So, I was able to AC with the code below, but there seems to be a bug. 3 digits At the above time, I didn't use '008' or '016' as a tree, and I thought that it might be a check for a pattern in which the characters left over in 4 digits are 0. ,safe.

```
from collections import Counter
S = input()
if len(S) == 1:
if S == '8':
print('Yes')
else:
print('No')
elif len(S) == 2:
if ''.join(sorted(S)) in ['16', '24', '23', '04', '48', '56', '46', '27', '08', '88', '69']:
print('Yes')
else:
print('No')
else:
c = Counter(S)
for x in range(104, 1000, 8):
d = Counter(str(x))
for k in d:
if k not in c:
break
if d[k] > c[k]:
break
else:
print('Yes')
exit()
print('No')
```

ABC181E - Traveling Salesman among Aerial Cities

Break through in 36 and a half minutes. WA2. There is no choice but to try each pair with the teacher one by one. After trying, the total is calculated every time and it becomes * O * (* N * ^{ 2 </ sup>) and TLE. So, if you take the cumulative sum from the front and back, you can make it * O * (* N *). The best transformation form of the teacher is even, I thought that * O * (1) would not be required, but it is troublesome. So I skipped with * O * (log N </ i>) and got a total of * O * ( N </ i> log N </ i>). Implementation calories Was expensive but not difficult.}

```
from itertools import accumulate
from bisect import bisect_left
from collections import Counter
INF = float('inf')
N, M = map(int, input().split())
H = list(map(int, input().split()))
W = list(map(int, input().split()))
W.sort()
c = Counter(H)
h = sorted(k for k in c if c[k] % 2 == 1)
n = len(h)
l = list(accumulate(h[i + 1] - h[i] for i in range(0, n - 1, 2)))
r = list(accumulate(h[i] - h[i - 1] for i in range(n - 1, 0, -2)))
result = INF
for i in range(n):
t = 0
a = i // 2 - 1
if a != -1:
t += l[a]
a = (n - 1 - i) // 2 - 1
if a != -1:
t += r[a]
j = bisect_left(W, h[i])
u = INF
if j != M:
u = min(u, W[j] - h[i])
if j != 0:
u = min(u, h[i] - W[j - 1])
t += u
if i % 2 == 1:
t += h[i + 1] - h[i - 1]
result = min(result, t)
print(result)
```

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