yukicoder contest 264 entry record

A 1217 Sesuseso

Just go around the loop, find the unmatched indexes, and print XtoY.

S = input()

for i in range(26):
    a = chr(i + ord('a'))
    if S[i] == a:
        continue
    print('%sto%s' % (a, S[i]))
    break

B 1218 Something Like a Theorem

Even if I searched all over without thinking about anything, it didn't become TLE, and that was it.

n, z = map(int, input().split())

for x in range(1, z + 1):
    for y in range(x, z + 1):
        if x ** n + y ** n == z ** n:
            print('Yes')
            exit()
print('No')

C 1219 Mancala Combo

As an operation, A _{i = i on the far left is operated and repeated, but of course if you implement it naively to simulate it, it will be TLE (Note that it is implemented with heapq and experienced Orz). If you think about it carefully, add up the number of times you erased A _{i + 1 to A _{N to the right of A _{i . Therefore, when the stacking is c, c + A _{i should be a multiple of i. When scanning from the right while stacking c, * O * (* N *) is displayed. Can be solved.}}}}}

N, *A = map(int, open(0).read().split())

c = 0
for i in range(N - 1, -1, -1):
    if (A[i] + c) % (i + 1) != 0:
        print('No')
        break
    c += (A[i] + c) // (i + 1)
else:
    print('Yes')