It was three completes from A to C.
** Thoughts ** Just compare
s = input()
t = input()
n = len(t)
if s == t[:n-1]:
print('Yes')
else:
print('No')
** Thoughts ** 2WA () without noticing the deadly typo
a, b, c, k = map(int,input().split())
if a >= k: #all k a
print(k)
elif a + b >= k: #If you take c, it will decrease, so take it from b
print(a)
else:
n = k - (a + b)
ans = a + -1 * n
print(ans)
** Thoughts ** I was impatient with 2WA at B, so I was late to submit it. Withstands a typical bit full search, N is small enough.
n, m, x = map(int,input().split())
ca = [list(map(int,input().split())) for _ in range(n)]
costs = []
for i in range(2 ** n):
flag = True
cost = 0
algos = [0] * m
op = [False] * n
for j in range(n):
if ((i >> j) & 1):
op[n - j - 1] = True
#print(op)
for j in range(n):
if op[j]:
for k, u in enumerate(ca[j][1:]):
algos[k] += u
cost += ca[j][0]
for j in algos:
if j < x:
flag = False
if flag:
costs.append(cost)
if len(costs) == 0:
print(-1)
else:
print(min(costs))
** Thoughts ** I found a loop and sought the remainder, but I couldn't pass it in about 8 cases and WA.
The last three ABCs are WAs in A and B each time, so I have to calm down and solve it. See you again, good night.
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