output_description stringlengths 15 956 | submission_id stringlengths 10 10 | status stringclasses 3 values | problem_id stringlengths 6 6 | input_description stringlengths 9 2.55k | attempt stringlengths 1 13.7k | problem_description stringlengths 7 5.24k | samples stringlengths 2 2.72k |
|---|---|---|---|---|---|---|---|
Print the answer in N+1 lines.
The i-th line (i = 1, \ldots, N+1) should contain the minimum possible value
of S after building railroads for the case K = i-1.
* * * | s242907632 | Wrong Answer | p02604 | Input is given from Standard Input in the following format:
N
X_1 Y_1 P_1
X_2 Y_2 P_2
: : :
X_N Y_N P_N | n = int(input())
xyp = [list(map(int, input().split())) for _ in range(n)]
ans = 0
for i in range(3**n):
for j in range(n):
ans += 1
print(ans)
| Statement
New AtCoder City has an infinite grid of streets, as follows:
* At the center of the city stands a clock tower. Let (0, 0) be the coordinates of this point.
* A straight street, which we will call _East-West Main Street_ , runs east-west and passes the clock tower. It corresponds to the x-axis in the two-dimensional coordinate plane.
* There are also other infinitely many streets parallel to East-West Main Street, with a distance of 1 between them. They correspond to the lines \ldots, y = -2, y = -1, y = 1, y = 2, \ldots in the two-dimensional coordinate plane.
* A straight street, which we will call _North-South Main Street_ , runs north-south and passes the clock tower. It corresponds to the y-axis in the two-dimensional coordinate plane.
* There are also other infinitely many streets parallel to North-South Main Street, with a distance of 1 between them. They correspond to the lines \ldots, x = -2, x = -1, x = 1, x = 2, \ldots in the two-dimensional coordinate plane.
There are N residential areas in New AtCoder City. The i-th area is located at
the intersection with the coordinates (X_i, Y_i) and has a population of P_i.
Each citizen in the city lives in one of these areas.
**The city currently has only two railroads, stretching infinitely, one along
East-West Main Street and the other along North-South Main Street.**
M-kun, the mayor, thinks that they are not enough for the commuters, so he
decides to choose K streets and build a railroad stretching infinitely along
each of those streets.
Let the _walking distance_ of each citizen be the distance from his/her
residential area to the nearest railroad.
M-kun wants to build railroads so that the sum of the walking distances of all
citizens, S, is minimized.
For each K = 0, 1, 2, \dots, N, what is the minimum possible value of S after
building railroads? | [{"input": "3\n 1 2 300\n 3 3 600\n 1 4 800", "output": "2900\n 900\n 0\n 0\n \n\nWhen K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1,\n3, and 1, respectively, to reach a railroad. \nThus, the sum of the walking distances of all citizens, S, is 1 \\times 300 + 3\n\\times 600 + 1 \\times 800 = 2900.\n\nWhen K = 1, if we build a railroad along the street corresponding to the line\ny = 4 in the coordinate plane, the walking distances of the citizens of Area\n1, 2, and 3 become 1, 1, and 0, respectively. \nThen, S = 1 \\times 300 + 1 \\times 600 + 0 \\times 800 = 900. \nWe have many other options for where we build the railroad, but none of them\nmakes S less than 900.\n\nWhen K = 2, if we build a railroad along the street corresponding to the lines\nx = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad\nwith the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.\n\nThe figure below shows the optimal way to build railroads for the cases K = 0,\n1, 2. \nThe street painted blue represents the roads along which we build railroads.\n\n\n\n* * *"}, {"input": "5\n 3 5 400\n 5 3 700\n 5 5 1000\n 5 7 700\n 7 5 400", "output": "13800\n 1600\n 0\n 0\n 0\n 0\n \n\nThe figure below shows the optimal way to build railroads for the cases K = 1,\n2.\n\n\n\n* * *"}, {"input": "6\n 2 5 1000\n 5 2 1100\n 5 5 1700\n -2 -5 900\n -5 -2 600\n -5 -5 2200", "output": "26700\n 13900\n 3200\n 1200\n 0\n 0\n 0\n \n\nThe figure below shows the optimal way to build railroads for the case K = 3.\n\n\n\n* * *"}, {"input": "8\n 2 2 286017\n 3 1 262355\n 2 -2 213815\n 1 -3 224435\n -2 -2 136860\n -3 -1 239338\n -2 2 217647\n -1 3 141903", "output": "2576709\n 1569381\n 868031\n 605676\n 366338\n 141903\n 0\n 0\n 0\n \n\nThe figure below shows the optimal way to build railroads for the case K = 4.\n\n"}] |
Print the answer in N+1 lines.
The i-th line (i = 1, \ldots, N+1) should contain the minimum possible value
of S after building railroads for the case K = i-1.
* * * | s713332813 | Accepted | p02604 | Input is given from Standard Input in the following format:
N
X_1 Y_1 P_1
X_2 Y_2 P_2
: : :
X_N Y_N P_N | n = int(input())
xyp = [list(map(int, input().split())) for i in range(n)]
cost = [10**20] * (1 << n)
cost[0] = 0
for bitt in range(1, 1 << n):
xx = []
yy = []
pp = []
m = 0
for i in range(n):
x, y, p = xyp[i]
if (1 << i) & bitt:
xx.append(x)
yy.append(y)
pp.append(p)
m += 1
for i in range(m):
sx = xx[i]
anss = 0
for x, y, p in zip(xx, yy, pp):
anss += min(abs(y), abs(x), abs(sx - x)) * p
cost[bitt] = min(cost[bitt], anss)
for j in range(m):
sy = yy[j]
anss = 0
for x, y, p in zip(xx, yy, pp):
anss += min(abs(y), abs(x), abs(sy - y)) * p
cost[bitt] = min(cost[bitt], anss)
dp = [
sum(
min(abs(xyp[i][0]), abs(xyp[i][1])) * xyp[i][2]
for i in range(n)
if (1 << i) & j
)
for j in range(1 << n)
]
print(dp[-1])
for j in range(n):
for i in range((1 << n) - 1, 0, -1):
x = i
while x >= 0:
x &= i
dp[i] = min(dp[i], dp[i - x] + cost[x])
x -= 1
print(dp[-1])
| Statement
New AtCoder City has an infinite grid of streets, as follows:
* At the center of the city stands a clock tower. Let (0, 0) be the coordinates of this point.
* A straight street, which we will call _East-West Main Street_ , runs east-west and passes the clock tower. It corresponds to the x-axis in the two-dimensional coordinate plane.
* There are also other infinitely many streets parallel to East-West Main Street, with a distance of 1 between them. They correspond to the lines \ldots, y = -2, y = -1, y = 1, y = 2, \ldots in the two-dimensional coordinate plane.
* A straight street, which we will call _North-South Main Street_ , runs north-south and passes the clock tower. It corresponds to the y-axis in the two-dimensional coordinate plane.
* There are also other infinitely many streets parallel to North-South Main Street, with a distance of 1 between them. They correspond to the lines \ldots, x = -2, x = -1, x = 1, x = 2, \ldots in the two-dimensional coordinate plane.
There are N residential areas in New AtCoder City. The i-th area is located at
the intersection with the coordinates (X_i, Y_i) and has a population of P_i.
Each citizen in the city lives in one of these areas.
**The city currently has only two railroads, stretching infinitely, one along
East-West Main Street and the other along North-South Main Street.**
M-kun, the mayor, thinks that they are not enough for the commuters, so he
decides to choose K streets and build a railroad stretching infinitely along
each of those streets.
Let the _walking distance_ of each citizen be the distance from his/her
residential area to the nearest railroad.
M-kun wants to build railroads so that the sum of the walking distances of all
citizens, S, is minimized.
For each K = 0, 1, 2, \dots, N, what is the minimum possible value of S after
building railroads? | [{"input": "3\n 1 2 300\n 3 3 600\n 1 4 800", "output": "2900\n 900\n 0\n 0\n \n\nWhen K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1,\n3, and 1, respectively, to reach a railroad. \nThus, the sum of the walking distances of all citizens, S, is 1 \\times 300 + 3\n\\times 600 + 1 \\times 800 = 2900.\n\nWhen K = 1, if we build a railroad along the street corresponding to the line\ny = 4 in the coordinate plane, the walking distances of the citizens of Area\n1, 2, and 3 become 1, 1, and 0, respectively. \nThen, S = 1 \\times 300 + 1 \\times 600 + 0 \\times 800 = 900. \nWe have many other options for where we build the railroad, but none of them\nmakes S less than 900.\n\nWhen K = 2, if we build a railroad along the street corresponding to the lines\nx = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad\nwith the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.\n\nThe figure below shows the optimal way to build railroads for the cases K = 0,\n1, 2. \nThe street painted blue represents the roads along which we build railroads.\n\n\n\n* * *"}, {"input": "5\n 3 5 400\n 5 3 700\n 5 5 1000\n 5 7 700\n 7 5 400", "output": "13800\n 1600\n 0\n 0\n 0\n 0\n \n\nThe figure below shows the optimal way to build railroads for the cases K = 1,\n2.\n\n\n\n* * *"}, {"input": "6\n 2 5 1000\n 5 2 1100\n 5 5 1700\n -2 -5 900\n -5 -2 600\n -5 -5 2200", "output": "26700\n 13900\n 3200\n 1200\n 0\n 0\n 0\n \n\nThe figure below shows the optimal way to build railroads for the case K = 3.\n\n\n\n* * *"}, {"input": "8\n 2 2 286017\n 3 1 262355\n 2 -2 213815\n 1 -3 224435\n -2 -2 136860\n -3 -1 239338\n -2 2 217647\n -1 3 141903", "output": "2576709\n 1569381\n 868031\n 605676\n 366338\n 141903\n 0\n 0\n 0\n \n\nThe figure below shows the optimal way to build railroads for the case K = 4.\n\n"}] |
Print the answer in N+1 lines.
The i-th line (i = 1, \ldots, N+1) should contain the minimum possible value
of S after building railroads for the case K = i-1.
* * * | s419473288 | Wrong Answer | p02604 | Input is given from Standard Input in the following format:
N
X_1 Y_1 P_1
X_2 Y_2 P_2
: : :
X_N Y_N P_N | import sys
input = sys.stdin.buffer.readline
# sys.setrecursionlimit(10**9)
# from functools import lru_cache
def RD():
return sys.stdin.read()
def II():
return int(input())
def MI():
return map(int, input().split())
def MF():
return map(float, input().split())
def LI():
return list(map(int, input().split()))
def LF():
return list(map(float, input().split()))
def TI():
return tuple(map(int, input().split()))
# rstrip().decode()
from itertools import combinations
from collections import defaultdict
def main():
n = II()
dx = [defaultdict(lambda: -1) for _ in range(n)]
dy = [defaultdict(lambda: -1) for _ in range(n)]
X = []
Y = []
P = []
for _ in range(n):
x, y, p = MI()
X.append(x)
Y.append(y)
P.append(p)
# print(X,Y,P)
for i in range(n):
aans = 10**15
a = combinations(range(n), i)
for li in list(a):
# print(li)
for j in range(1 << i):
x = [0]
y = [0]
for k in range(i):
if j & 1 << k:
x.append(X[li[k]])
else:
y.append(Y[li[k]])
# print(i,j,k)
# print(x,y)
x = tuple(x)
y = tuple(y)
ans = 0
for ii in range(n):
if dx[ii][x] != -1:
dxx = dx[ii][x]
else:
dxx = 10**6
for xx in x:
dxx = min(dxx, abs(X[ii] - xx))
dx[ii][x] = dxx
if dy[ii][y] != -1:
dyy = dy[ii][y]
else:
dyy = 10**6
for yy in y:
dyy = min(dyy, abs(Y[ii] - yy))
dy[ii][y] = dyy
ans += P[ii] * min(dxx, dyy)
aans = min(aans, ans)
print(aans)
print(0)
if __name__ == "__main__":
main()
| Statement
New AtCoder City has an infinite grid of streets, as follows:
* At the center of the city stands a clock tower. Let (0, 0) be the coordinates of this point.
* A straight street, which we will call _East-West Main Street_ , runs east-west and passes the clock tower. It corresponds to the x-axis in the two-dimensional coordinate plane.
* There are also other infinitely many streets parallel to East-West Main Street, with a distance of 1 between them. They correspond to the lines \ldots, y = -2, y = -1, y = 1, y = 2, \ldots in the two-dimensional coordinate plane.
* A straight street, which we will call _North-South Main Street_ , runs north-south and passes the clock tower. It corresponds to the y-axis in the two-dimensional coordinate plane.
* There are also other infinitely many streets parallel to North-South Main Street, with a distance of 1 between them. They correspond to the lines \ldots, x = -2, x = -1, x = 1, x = 2, \ldots in the two-dimensional coordinate plane.
There are N residential areas in New AtCoder City. The i-th area is located at
the intersection with the coordinates (X_i, Y_i) and has a population of P_i.
Each citizen in the city lives in one of these areas.
**The city currently has only two railroads, stretching infinitely, one along
East-West Main Street and the other along North-South Main Street.**
M-kun, the mayor, thinks that they are not enough for the commuters, so he
decides to choose K streets and build a railroad stretching infinitely along
each of those streets.
Let the _walking distance_ of each citizen be the distance from his/her
residential area to the nearest railroad.
M-kun wants to build railroads so that the sum of the walking distances of all
citizens, S, is minimized.
For each K = 0, 1, 2, \dots, N, what is the minimum possible value of S after
building railroads? | [{"input": "3\n 1 2 300\n 3 3 600\n 1 4 800", "output": "2900\n 900\n 0\n 0\n \n\nWhen K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1,\n3, and 1, respectively, to reach a railroad. \nThus, the sum of the walking distances of all citizens, S, is 1 \\times 300 + 3\n\\times 600 + 1 \\times 800 = 2900.\n\nWhen K = 1, if we build a railroad along the street corresponding to the line\ny = 4 in the coordinate plane, the walking distances of the citizens of Area\n1, 2, and 3 become 1, 1, and 0, respectively. \nThen, S = 1 \\times 300 + 1 \\times 600 + 0 \\times 800 = 900. \nWe have many other options for where we build the railroad, but none of them\nmakes S less than 900.\n\nWhen K = 2, if we build a railroad along the street corresponding to the lines\nx = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad\nwith the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.\n\nThe figure below shows the optimal way to build railroads for the cases K = 0,\n1, 2. \nThe street painted blue represents the roads along which we build railroads.\n\n\n\n* * *"}, {"input": "5\n 3 5 400\n 5 3 700\n 5 5 1000\n 5 7 700\n 7 5 400", "output": "13800\n 1600\n 0\n 0\n 0\n 0\n \n\nThe figure below shows the optimal way to build railroads for the cases K = 1,\n2.\n\n\n\n* * *"}, {"input": "6\n 2 5 1000\n 5 2 1100\n 5 5 1700\n -2 -5 900\n -5 -2 600\n -5 -5 2200", "output": "26700\n 13900\n 3200\n 1200\n 0\n 0\n 0\n \n\nThe figure below shows the optimal way to build railroads for the case K = 3.\n\n\n\n* * *"}, {"input": "8\n 2 2 286017\n 3 1 262355\n 2 -2 213815\n 1 -3 224435\n -2 -2 136860\n -3 -1 239338\n -2 2 217647\n -1 3 141903", "output": "2576709\n 1569381\n 868031\n 605676\n 366338\n 141903\n 0\n 0\n 0\n \n\nThe figure below shows the optimal way to build railroads for the case K = 4.\n\n"}] |
Print the sum of the shortest distances, modulo 10^9+7.
* * * | s973259681 | Wrong Answer | p03445 | Input is given from Standard Input in the following format:
H W
N
x_1 y_1
x_2 y_2
:
x_N y_N | from collections import deque
mod = 10**9 + 7
h, w = map(int, input().split())
n = int(input())
ans = 0
black = []
row = [0] * h
column = [0] * w
for _ in range(n):
x, y = map(int, input().split())
row[x] += 1
column[y] += 1
black.append([x, y])
cnt = 0
top = 0
bottom = h * w - n
area = []
for i in range(h):
if row[i] == 0:
cnt += 1
if i != h - 1 and row[i + 1] == 0:
top += w
bottom -= w
ans += top * bottom
ans %= mod
else:
area.append([cnt for _ in range(w)])
else:
top += w - row[i]
bottom -= w - row[i]
area.append([1 for _ in range(w)])
for x, y in black:
if x == i:
area[-1][y] = 0
cnt = 0
R = len(area)
cnt = 0
left = 0
right = h * w - n
area2 = []
for j in range(w):
if column[j] == 0:
cnt += 1
if j != w - 1 and column[j + 1] == 0:
left += h
right -= h
ans += left * right
ans %= mod
else:
area2.append([cnt * area[i][j] for i in range(R)])
else:
left += w - column[j]
right -= w - column[j]
area2.append([area[i][j] for i in range(R)])
cnt = 0
C = len(area2)
vec = [[1, 0], [0, 1], [-1, 0], [0, -1]]
def bfs(p, q):
dist = [[10**9 for _ in range(R)] for __ in range(C)]
visited = [[False for _ in range(R)] for __ in range(C)]
dist[p][q] = 0
visited[p][q] = True
q = deque([(p, q)])
while q:
x, y = q.popleft()
for dx, dy in vec:
if 0 <= x + dx < C and 0 <= y + dy < R and area2[x + dx][y + dy] != 0:
if not visited[x + dx][y + dy]:
dist[x + dx][y + dy] = dist[x][y] + 1
visited[x + dx][y + dy] = True
q.append((x + dx, y + dy))
return dist
ans2 = 0
for i in range(C):
for j in range(R):
d = bfs(i, j)
for k in range(C):
for l in range(R):
ans2 += area2[i][j] * area2[k][l] * d[k][l]
ans2 %= mod
ans2 *= pow(2, mod - 2, mod)
print((ans + ans2) % mod)
| Statement
You are given an H \times W grid. The square at the top-left corner will be
represented by (0, 0), and the square at the bottom-right corner will be
represented by (H-1, W-1).
Of those squares, N squares (x_1, y_1), (x_2, y_2), ..., (x_N, y_N) are
painted black, and the other squares are painted white.
Let the shortest distance between white squares A and B be the minimum number
of moves required to reach B from A **visiting only white squares** , where
one can travel to an adjacent square sharing a side (up, down, left or right)
in one move.
Since there are H × W - N white squares in total, there are _{(H×W-N)}C_2 ways
to choose two of the white squares.
For each of these _{(H×W-N)}C_2 ways, find the shortest distance between the
chosen squares, then find the sum of all those distances, modulo 1 000 000
007=10^9+7. | [{"input": "2 3\n 1\n 1 1", "output": "20\n \n\nWe have the following grid (`.`: a white square, `#`: a black square).\n\n \n \n ...\n .#.\n \n\nLet us assign a letter to each white square as follows:\n\n \n \n ABC\n D#E\n \n\nThen, we have:\n\n * dist(A, B) = 1\n * dist(A, C) = 2\n * dist(A, D) = 1\n * dist(A, E) = 3\n * dist(B, C) = 1\n * dist(B, D) = 2\n * dist(B, E) = 2\n * dist(C, D) = 3\n * dist(C, E) = 1\n * dist(D, E) = 4\n\nwhere dist(A, B) is the shortest distance between A and B. The sum of these\ndistances is 20.\n\n* * *"}, {"input": "2 3\n 1\n 1 2", "output": "16\n \n\nLet us assign a letter to each white square as follows:\n\n \n \n ABC\n DE#\n \n\nThen, we have:\n\n * dist(A, B) = 1\n * dist(A, C) = 2\n * dist(A, D) = 1\n * dist(A, E) = 2\n * dist(B, C) = 1\n * dist(B, D) = 2\n * dist(B, E) = 1\n * dist(C, D) = 3\n * dist(C, E) = 2\n * dist(D, E) = 1\n\nThe sum of these distances is 16.\n\n* * *"}, {"input": "3 3\n 1\n 1 1", "output": "64\n \n\n* * *"}, {"input": "4 4\n 4\n 0 1\n 1 1\n 2 1\n 2 2", "output": "268\n \n\n* * *"}, {"input": "1000000 1000000\n 1\n 0 0", "output": "333211937"}] |
Print the sum of the shortest distances, modulo 10^9+7.
* * * | s810414455 | Wrong Answer | p03445 | Input is given from Standard Input in the following format:
H W
N
x_1 y_1
x_2 y_2
:
x_N y_N | print("bandzebo")
| Statement
You are given an H \times W grid. The square at the top-left corner will be
represented by (0, 0), and the square at the bottom-right corner will be
represented by (H-1, W-1).
Of those squares, N squares (x_1, y_1), (x_2, y_2), ..., (x_N, y_N) are
painted black, and the other squares are painted white.
Let the shortest distance between white squares A and B be the minimum number
of moves required to reach B from A **visiting only white squares** , where
one can travel to an adjacent square sharing a side (up, down, left or right)
in one move.
Since there are H × W - N white squares in total, there are _{(H×W-N)}C_2 ways
to choose two of the white squares.
For each of these _{(H×W-N)}C_2 ways, find the shortest distance between the
chosen squares, then find the sum of all those distances, modulo 1 000 000
007=10^9+7. | [{"input": "2 3\n 1\n 1 1", "output": "20\n \n\nWe have the following grid (`.`: a white square, `#`: a black square).\n\n \n \n ...\n .#.\n \n\nLet us assign a letter to each white square as follows:\n\n \n \n ABC\n D#E\n \n\nThen, we have:\n\n * dist(A, B) = 1\n * dist(A, C) = 2\n * dist(A, D) = 1\n * dist(A, E) = 3\n * dist(B, C) = 1\n * dist(B, D) = 2\n * dist(B, E) = 2\n * dist(C, D) = 3\n * dist(C, E) = 1\n * dist(D, E) = 4\n\nwhere dist(A, B) is the shortest distance between A and B. The sum of these\ndistances is 20.\n\n* * *"}, {"input": "2 3\n 1\n 1 2", "output": "16\n \n\nLet us assign a letter to each white square as follows:\n\n \n \n ABC\n DE#\n \n\nThen, we have:\n\n * dist(A, B) = 1\n * dist(A, C) = 2\n * dist(A, D) = 1\n * dist(A, E) = 2\n * dist(B, C) = 1\n * dist(B, D) = 2\n * dist(B, E) = 1\n * dist(C, D) = 3\n * dist(C, E) = 2\n * dist(D, E) = 1\n\nThe sum of these distances is 16.\n\n* * *"}, {"input": "3 3\n 1\n 1 1", "output": "64\n \n\n* * *"}, {"input": "4 4\n 4\n 0 1\n 1 1\n 2 1\n 2 2", "output": "268\n \n\n* * *"}, {"input": "1000000 1000000\n 1\n 0 0", "output": "333211937"}] |
Print the number of the different possible combinations of integers written on
the blackboard after Snuke performs the operations, modulo 1,000,000,007.
* * * | s801583703 | Wrong Answer | p03500 | Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N | #!/usr/bin/env python3
M = 10**9 + 7
def solve(n, k, a):
a_min = min(a)
a_max = max(a)
r = 0
i = 0
p = 1
for i in range(k + 1):
q_min, q_max = a_min // p, a_max // p
if q_min == q_max:
if q_min == 0:
r += 1
r %= M
break
else:
r += min(q_min - q_min // 2, k - i + 1)
r %= M
else:
b = [v % p for v in a]
b.sort()
h_max = k - i
r += min(q_min, k - i) + 1
r %= M
lb = b[0]
for j in range(1, n):
if lb != b[j]:
bj = b[j]
c = 0
h = 0
q = p // 2
f = False
while 0 < q:
if (bj // q) % 2 == 1:
c += q
h += 1
if (lb // q) % 2 != 1:
f = True
break
elif h == h_max:
break
q //= 2
if c <= a_min:
if f:
r += min((a_min - c) // p, k - (i + h)) + 1
r %= M
if (a_max - c) // p == 0:
return r
lb = bj
else:
continue
else:
break
if q_min == 1 and q_max == 2:
break
p *= 2
return r
def main():
n, k = input().split()
n = int(n)
k = int(k)
a = list(map(int, input().split()))
print(solve(n, k, a))
if __name__ == "__main__":
main()
| Statement
There are N non-negative integers written on the blackboard: A_1, ..., A_N.
Snuke can perform the following two operations at most K times in total in any
order:
* Operation A: Replace each integer X on the blackboard with X divided by 2, rounded down to the nearest integer.
* Operation B: Replace each integer X on the blackboard with X minus 1. This operation cannot be performed if one or more 0s are written on the blackboard.
Find the number of the different possible combinations of integers written on
the blackboard after Snuke performs the operations, modulo 1,000,000,007. | [{"input": "2 2\n 5 7", "output": "6\n \n\nThere are six possible combinations of integers on the blackboard: (1, 1), (1,\n2), (2, 3), (3, 5), (4, 6) and (5, 7). For example, (1, 2) can be obtained by\nperforming Operation A and Operation B in this order.\n\n* * *"}, {"input": "3 4\n 10 13 22", "output": "20\n \n\n* * *"}, {"input": "1 100\n 10", "output": "11\n \n\n* * *"}, {"input": "10 123456789012345678\n 228344079825412349 478465001534875275 398048921164061989 329102208281783917 779698519704384319 617456682030809556 561259383338846380 254083246422083141 458181156833851984 502254767369499613", "output": "164286011"}] |
Print the number of the different possible combinations of integers written on
the blackboard after Snuke performs the operations, modulo 1,000,000,007.
* * * | s384858181 | Wrong Answer | p03500 | Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N | # -*- coding: utf-8 -*-
def fun_A(L):
A = list(map(int, [x / 2 for x in L]))
return A
def fun_B(L):
A = [x - 1 for x in L]
return A
def fun_seq(K, func_list):
if len(func_list[-1]) >= K:
return
length = len(func_list[-1])
temp = 2**length
for i in func_list[-temp:]:
func_list.append(i + "A")
func_list.append(i + "B")
fun_seq(K, func_list)
if __name__ == "__main__":
N, K = map(int, input().split())
A = list(map(int, input().split()))
func_list = ["A", "B"]
fun_seq(K, func_list)
# print(len(func_list), func_list)
count = 0
for i in func_list:
for j in i:
if j == "A":
A = fun_A(A)
if j == "B" and 0 not in A:
A = fun_B(A)
else:
continue
count += 1
print(count % 1000000007)
| Statement
There are N non-negative integers written on the blackboard: A_1, ..., A_N.
Snuke can perform the following two operations at most K times in total in any
order:
* Operation A: Replace each integer X on the blackboard with X divided by 2, rounded down to the nearest integer.
* Operation B: Replace each integer X on the blackboard with X minus 1. This operation cannot be performed if one or more 0s are written on the blackboard.
Find the number of the different possible combinations of integers written on
the blackboard after Snuke performs the operations, modulo 1,000,000,007. | [{"input": "2 2\n 5 7", "output": "6\n \n\nThere are six possible combinations of integers on the blackboard: (1, 1), (1,\n2), (2, 3), (3, 5), (4, 6) and (5, 7). For example, (1, 2) can be obtained by\nperforming Operation A and Operation B in this order.\n\n* * *"}, {"input": "3 4\n 10 13 22", "output": "20\n \n\n* * *"}, {"input": "1 100\n 10", "output": "11\n \n\n* * *"}, {"input": "10 123456789012345678\n 228344079825412349 478465001534875275 398048921164061989 329102208281783917 779698519704384319 617456682030809556 561259383338846380 254083246422083141 458181156833851984 502254767369499613", "output": "164286011"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s331122487 | Accepted | p03150 | Input is given from Standard Input in the following format:
S | def main():
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**7)
from collections import Counter, deque
from collections import defaultdict
from itertools import combinations, permutations, accumulate, groupby, product
from bisect import bisect_left, bisect_right
from heapq import heapify, heappop, heappush
from math import floor, ceil, pi, factorial
from operator import itemgetter
def I():
return int(input())
def MI():
return map(int, input().split())
def LI():
return list(map(int, input().split()))
def LI2():
return [int(input()) for i in range(n)]
def MXI():
return [[LI()] for i in range(n)]
def SI():
return input().rstrip()
def printns(x):
print("\n".join(x))
def printni(x):
print("\n".join(list(map(str, x))))
inf = 10**17
mod = 10**9 + 7
# main code here!
s = SI()
n = len(s)
for i in range(8):
s1 = s[:i]
s2 = s[n + i - 7 :]
if s1 + s2 == "keyence":
print("YES")
exit()
print("NO")
if __name__ == "__main__":
main()
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s996978817 | Wrong Answer | p03150 | Input is given from Standard Input in the following format:
S | import sys
import heapq
import re
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt, ceil
from functools import lru_cache, reduce
INF = 1 << 60
MOD = 1000000007
sys.setrecursionlimit(10**7)
# UnionFind
class UnionFind:
def __init__(self, n):
self.n = n
self.parents = [-1] * n
def find(self, x):
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def size(self, x):
return -self.parents[self.find(x)]
def same(self, x, y):
return self.find(x) == self.find(y)
def members(self, x):
root = self.find(x)
return [i for i in range(self.n) if self.find(i) == root]
def roots(self):
return [i for i, x in enumerate(self.parents) if x < 0]
def group_count(self):
return len(self.roots())
def all_group_members(self):
return {r: self.members(r) for r in self.roots()}
def __str__(self):
return "\n".join("{}: {}".format(r, self.members(r)) for r in self.roots())
# ダイクストラ
def dijkstra_heap(s, edge, n):
# 始点sから各頂点への最短距離
d = [10**20] * n
used = [True] * n # True:未確定
d[s] = 0
used[s] = False
edgelist = []
for a, b in edge[s]:
heapq.heappush(edgelist, a * (10**6) + b)
while len(edgelist):
minedge = heapq.heappop(edgelist)
# まだ使われてない頂点の中から最小の距離のものを探す
if not used[minedge % (10**6)]:
continue
v = minedge % (10**6)
d[v] = minedge // (10**6)
used[v] = False
for e in edge[v]:
if used[e[1]]:
heapq.heappush(edgelist, (e[0] + d[v]) * (10**6) + e[1])
return d
# 素因数分解
def factorization(n):
arr = []
temp = n
for i in range(2, int(-(-(n**0.5) // 1)) + 1):
if temp % i == 0:
cnt = 0
while temp % i == 0:
cnt += 1
temp //= i
arr.append([i, cnt])
if temp != 1:
arr.append([temp, 1])
if arr == []:
arr.append([n, 1])
return arr
# 2数の最小公倍数
def lcm(x, y):
return (x * y) // gcd(x, y)
# リストの要素の最小公倍数
def lcm_list(numbers):
return reduce(lcm, numbers, 1)
# リストの要素の最大公約数
def gcd_list(numbers):
return reduce(gcd, numbers)
# 素数判定
def is_prime(n):
if n <= 1:
return False
p = 2
while True:
if p**2 > n:
break
if n % p == 0:
return False
p += 1
return True
# limit以下の素数を列挙
def eratosthenes(limit):
A = [i for i in range(2, limit + 1)]
P = []
while True:
prime = min(A)
if prime > sqrt(limit):
break
P.append(prime)
i = 0
while i < len(A):
if A[i] % prime == 0:
A.pop(i)
continue
i += 1
for a in A:
P.append(a)
return P
# 同じものを含む順列
def permutation_with_duplicates(L):
if L == []:
return [[]]
else:
ret = []
# set（集合）型で重複を削除、ソート
S = sorted(set(L))
for i in S:
data = L[:]
data.remove(i)
for j in permutation_with_duplicates(data):
ret.append([i] + j)
return ret
# ここから書き始める
s = input()
x = "keyence"
yes = False
for i in range(8):
left = x[:i]
right = x[i:]
# print("left =", left)
# print("right =", right)
l = s.find(left)
r = s.find(right)
if l != -1 and r != -1 and l + len(left) <= r:
# print(l, r)
yes = True
break
if yes:
print("YES")
else:
print("NO")
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s656786482 | Wrong Answer | p03150 | Input is given from Standard Input in the following format:
S | import re
s = input()
pat = (
"\w*"
+ "k"
+ "\w*"
+ "e"
+ "\w*"
+ "y"
+ "\w*"
+ "e"
+ "\w*"
+ "n"
+ "\w*"
+ "c"
+ "\w*"
+ "e"
+ "\w*"
)
if re.match:
print("YES")
else:
print("NO")
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s419091606 | Runtime Error | p03150 | Input is given from Standard Input in the following format:
S | # coding:utf-8
s = input()
for i in range(n := len(s))
if s[:i] + s[n-7+i:] == 'keyence':
print('YES')
exit()
print('NO') | Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s656669809 | Runtime Error | p03150 | Input is given from Standard Input in the following format:
S | def main():
S = input()
if S[0] != 'k':
return ('NO')
elif S = 'keyence':
return ('YES')
ans = 'keyence'
i = 0
j = 0
while True:
i += 1
j += 1
if S[i] != ans[j]:
break
#残りの文字数
num = 7-j
cnt = 0
while cnt < num:
cnt += 1
#print(ans[-cnt])
if S[-cnt] != ans[-cnt]:
return ('NO')
return ('YES')
print(main())
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s602913367 | Wrong Answer | p03150 | Input is given from Standard Input in the following format:
S | # _*_ coding:utf-8 _*_
# Atcoder_KeyencePrograming_Contest-B
# TODO https://keyence2019.contest.atcoder.jp/tasks/keyence2019_b
import re
def solveProblem(inputStr):
pattern1 = r".*keyence"
pattern2 = r"k.*eyence"
pattern3 = r"ke.*yence"
pattern4 = r"key.*ence"
pattern5 = r"keye.*nce"
pattern6 = r"keyen.*ce"
pattern7 = r"keyenc.*e"
pattern8 = r"keyence.*"
matchOB1 = re.search(pattern1, inputStr)
matchOB2 = re.search(pattern2, inputStr)
matchOB3 = re.search(pattern3, inputStr)
matchOB4 = re.search(pattern4, inputStr)
matchOB5 = re.search(pattern5, inputStr)
matchOB6 = re.search(pattern6, inputStr)
matchOB7 = re.search(pattern7, inputStr)
matchOB8 = re.search(pattern8, inputStr)
if matchOB1 != None:
return "YES"
if matchOB2 != None:
return "YES"
if matchOB3 != None:
return "YES"
if matchOB4 != None:
return "YES"
if matchOB5 != None:
return "YES"
if matchOB6 != None:
return "Yes"
if matchOB7 != None:
return "YES"
if matchOB8 != None:
return "YES"
return "NO"
if __name__ == "__main__":
S = input().strip()
solution = solveProblem(S)
print("{}".format(solution))
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s945813921 | Runtime Error | p03150 | Input is given from Standard Input in the following format:
S | s = input()
search="keyence"
idx=0
flg = False
ans="NO"
if s.find(search)>=0:
print("YES")
elif:
for i in range(7):
x=s.find(search[0:i+1])
print("ps=%s, ns=%s, reduce=%s" %(search[0:i+1], search[i+1:], s[x+i+1:]))
if x == -1:
break
if x == 0:
if s[x+i:].find(search[i+1:])>=0:
ans = "YES"
break
print(ans)
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s857517364 | Runtime Error | p03150 | Input is given from Standard Input in the following format:
S | S=input()
ans='NO'
if 'keyence'in S:
ans='YES'
elif S.find('k')>=0 and S.find('k')<S.rfind('eyence'):
ans='YES'
elif S.find('ke')>=0 and S.find('ke')<S.rfind('yence'):
ans='YES'
elif S.find('key')>=0 and S.find('key')<S.rfind('ence'):
ans='YES
elif S.find('keye')>=0 and S.find('keye')<S.rfind('nce'):
ans='YES'
elif S.find('keyen')>=0 and S.find('keyen')<S.rfind('ce'):
ans='YES'
elif S.find('keyenc')>=0 and S.find('keyenc')<S.rfind('e'):
ans='YES'
print(ans) | Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s642518615 | Runtime Error | p03150 | Input is given from Standard Input in the following format:
S | #include <iostream>
#include <vector>
#include <algorithm>
#include <utility>
#include <string>
#include <queue>
#include <stack>
using namespace std;
typedef long long int ll;
typedef pair<int, int> Pii;
const ll mod = 1000000007;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
string s;
cin >> s;
int n = s.length();
string keyence = "keyence";
int p = 0;
for (int i = 0; i < 7; i++) {
if (s[i] == keyence[p]) p++;
else break;
}
for (int i = n-7+p; i < n; i++) {
if (s[i] == keyence[p]) p++;
else break;
}
if (p == 7) cout << "YES" << endl;
else cout << "NO" << endl;
return 0;
}
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s158540650 | Runtime Error | p03150 | Input is given from Standard Input in the following format:
S | s=str(input())
st=list(s)
l=len(st)
if st[0]="k" and st[1]=="e" and st[2]=="y" and st[3]=="e" st[4]=="n" and st[5]=="c" and st[6]=="e":
print("YES")
elif st[l-7]="k" and st[l-6]=="e" and st[l-5]=="y" and st[l-4]=="e" st[l-3]=="n" and st[l-2]=="c" and st[l-1]=="e":
print("YES")
elif st[0]="k" and st[l-6]=="e" and st[l-5]=="y" and st[l-4]=="e" st[l-3]=="n" and st[l-2]=="c" and st[l-1]=="e":
print("YES")
elif st[0]="k" and st[1]=="e" and st[l-5]=="y" and st[l-4]=="e" st[l-3]=="n" and st[l-2]=="c" and st[l-1]=="e":
print("YES")
elif st[0]="k" and st[1]=="e" and st[2]=="y" and st[l-4]=="e" st[l-3]=="n" and st[l-2]=="c" and st[l-1]=="e":
print("YES")
elif st[0]="k" and st[1]=="e" and st[2]=="y" and st[3]=="e" st[l-3]=="n" and st[l-2]=="c" and st[l-1]=="e":
print("YES")
elif st[0]="k" and st[1]=="e" and st[2]=="y" and st[3]=="e" st[4]=="n" and st[l-2]=="c" and st[l-1]=="e":
print("YES")
elif st[0]="k" and st[1]=="e" and st[2]=="y" and st[3]=="e" st[4]=="n" and st[5]=="c" and st[l-1]=="e":
print("YES")
else:
ptint("NO") | Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s580833703 | Runtime Error | p03150 | Input is given from Standard Input in the following format:
S | S = input()
target = "keyence"
length = len(target)
for i in range(length):
first = target[:i]
second = target[i:]
if first == '' and second in S or first != '' first in S and second in S and S.find(first) <= S.find(second):
print("YES")
exit()
print("NO")
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s683139517 | Accepted | p03150 | Input is given from Standard Input in the following format:
S | s = str(input())
st = list(s)
l = len(st)
if (
st[0] == "k"
and st[1] == "e"
and st[2] == "y"
and st[3] == "e"
and st[4] == "n"
and st[5] == "c"
and st[6] == "e"
):
print("YES")
elif (
st[l - 7] == "k"
and st[l - 6] == "e"
and st[l - 5] == "y"
and st[l - 4] == "e"
and st[l - 3] == "n"
and st[l - 2] == "c"
and st[l - 1] == "e"
):
print("YES")
elif (
st[0] == "k"
and st[l - 6] == "e"
and st[l - 5] == "y"
and st[l - 4] == "e"
and st[l - 3] == "n"
and st[l - 2] == "c"
and st[l - 1] == "e"
):
print("YES")
elif (
st[0] == "k"
and st[1] == "e"
and st[l - 5] == "y"
and st[l - 4] == "e"
and st[l - 3] == "n"
and st[l - 2] == "c"
and st[l - 1] == "e"
):
print("YES")
elif (
st[0] == "k"
and st[1] == "e"
and st[2] == "y"
and st[l - 4] == "e"
and st[l - 3] == "n"
and st[l - 2] == "c"
and st[l - 1] == "e"
):
print("YES")
elif (
st[0] == "k"
and st[1] == "e"
and st[2] == "y"
and st[3] == "e"
and st[l - 3] == "n"
and st[l - 2] == "c"
and st[l - 1] == "e"
):
print("YES")
elif (
st[0] == "k"
and st[1] == "e"
and st[2] == "y"
and st[3] == "e"
and st[4] == "n"
and st[l - 2] == "c"
and st[l - 1] == "e"
):
print("YES")
elif (
st[0] == "k"
and st[1] == "e"
and st[2] == "y"
and st[3] == "e"
and st[4] == "n"
and st[5] == "c"
and st[l - 1] == "e"
):
print("YES")
else:
print("NO")
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s315068631 | Runtime Error | p03150 | Input is given from Standard Input in the following format:
S | S = list(input())
if "k" and "e" and "y" and "e" and "n" and "c" and "e" in S:
for i in range(len(S)):
if S[i] == "k":
if S[i + 1] == "e":
if S[i + 2] == "y":
if S[i + 3] == "e":
if S[i + 4] == "n":
if S[i + 5] == "c":
if S[i + 6] == "e":
print("YES")
break
else:
if S[-1] == "e":
print("YES")
break
else:
print("No")
break
else:
if S[-1] == "e" and S[-2] == "c":
print("YES")
break
else:
for j in range(len(S) - i - 5):
if S[i + 5 + j] == "c":
if S[i + 5 + j + 1] == "e":
if i + 5 + j + 2 == len(S):
print("YES")
break
else:
print("NO")
break
else:
print("NO")
break
else:
if S[-1] == "e" and S[-2] == "c" and S[-3] == "n":
print("YES")
break
else:
for j in range(len(S) - i - 4):
if S[i + 4 + j] == "n":
if S[i + 4 + j + 1] == "c":
if S[i + 4 + j + 2] == "e":
if i + 4 + j + 3 == len(S):
print("YES")
break
else:
print("NO")
break
else:
print("NO")
break
else:
print("NO")
break
else:
if (
S[-1] == "e"
and S[-2] == "c"
and S[-3] == "n"
and S[-4] == "e"
):
print("YES")
break
else:
for j in range(len(S) - i - 3):
if S[i + 3 + j] == "e":
if S[i + 3 + j + 1] == "n":
if S[i + 3 + j + 2] == "c":
if S[i + 3 + j + 3] == "e":
if i + 3 + j + 4 == len(S):
print("YES")
break
else:
print("NO")
break
else:
print("NO")
break
else:
print("NO")
break
else:
print("NO")
break
else:
if (
S[-1] == "e"
and S[-2] == "c"
and S[-3] == "n"
and S[-4] == "e"
and S[-5] == "y"
):
print("YES")
break
else:
for j in range(len(S) - i - 2):
if S[i + 2 + j] == "y":
if S[i + 2 + j + 1] == "e":
if S[i + 2 + j + 2] == "n":
if S[i + 2 + j + 3] == "c":
if S[i + 2 + j + 4] == "e":
if i + 2 + j + 5 == len(S):
print("YES")
break
else:
print("NO")
break
else:
print("NO")
break
else:
print("NO")
break
else:
print("NO")
break
else:
print("NO")
break
else:
if (
S[-1] == "e"
and S[-2] == "c"
and S[-3] == "n"
and S[-4] == "e"
and S[-5] == "y"
and S[-6] == "e"
):
print("YES")
break
else:
for j in range(len(S) - i - 1):
if S[i + 1 + j] == "e":
if S[i + 1 + j + 1] == "y":
if S[i + 1 + j + 2] == "e":
if S[i + 1 + j + 3] == "n":
if S[i + 1 + j + 4] == "c":
if S[i + 1 + j + 5] == "e":
if i + 1 + j + 6 == len(S):
print("YES")
break
else:
print("NO")
break
else:
print("NO")
break
else:
print("NO")
break
else:
print("NO")
break
else:
print("NO")
break
else:
print("NO")
break
else:
print("NO")
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s484023755 | Accepted | p03150 | Input is given from Standard Input in the following format:
S | def resolve():
"""
code here
"""
S = input()
is_flag = False
key = "keyence"
for i in range(8):
if i == 0:
if S[-7:] == key:
is_flag = True
elif 1 <= i < 7:
if S[:i] == key[:i] and S[-1 * (7 - i) :] == key[-1 * (7 - i) :]:
is_flag = True
else:
if S[:7] == key:
is_flag = True
print("YES") if is_flag else print("NO")
if __name__ == "__main__":
resolve()
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s574329974 | Wrong Answer | p03150 | Input is given from Standard Input in the following format:
S | S = input()
res = "NO"
for s in range(len(S)):
for t in range(len(S) - s):
if S[:s] + S[t:] == "keyence":
res = "YES"
print(res)
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s627192198 | Wrong Answer | p03150 | Input is given from Standard Input in the following format:
S | S = input()
tgt = "keyence"
res = S == tgt
for i in range(len(S) - 6):
T = S[:i] + S[-7 + i :]
# print(T)
if T == tgt:
res = True
print("YES" if res else "NO")
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s779069457 | Accepted | p03150 | Input is given from Standard Input in the following format:
S | s = list(input())
if s[0] == "k":
li = []
for i in range(6):
li.append(s[len(s) - (6 - i)])
if "".join(li) == "eyence":
print("YES")
quit()
if s[0] + s[1] == "ke":
li = []
for i in range(5):
li.append(s[len(s) - (5 - i)])
if "".join(li) == "yence":
print("YES")
quit()
if s[0] + s[1] + s[2] == "key":
li = []
for i in range(4):
li.append(s[len(s) - (4 - i)])
if "".join(li) == "ence":
print("YES")
quit()
if s[0] + s[1] + s[2] + s[3] == "keye":
li = []
for i in range(3):
li.append(s[len(s) - (3 - i)])
if "".join(li) == "nce":
print("YES")
quit()
if s[0] + s[1] + s[2] + s[3] + s[4] == "keyen":
li = []
for i in range(2):
li.append(s[len(s) - (2 - i)])
if "".join(li) == "ce":
print("YES")
quit()
if s[0] + s[1] + s[2] + s[3] + s[4] + s[5] == "keyenc":
if s[len(s) - 1] == "e":
print("YES")
quit()
lis = []
for i in range(7):
lis.append(s[len(s) - i - 1])
if "".join(lis) == "keyence":
print("YES")
quit()
print("NO")
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s844816205 | Wrong Answer | p03150 | Input is given from Standard Input in the following format:
S | S = input()
S = S.replace("keyence", "A")
if S == "A":
print("YES")
exit()
elif S[0] == "A" and S[len(S) - 1] != "A":
print("YES")
exit()
elif S[len(S) - 1] == "A":
print("YES")
exit()
S1 = S.replace("K", "A")
S1 = S1[::-1]
S1 = S1.replace("ecneye", "B")
S1 = S1[::-1]
if S1[0] == "A" and S1[len(S1) - 1] == "B":
print("YES")
exit()
S2 = S.replace("Ke", "A")
S2 = S2[::-1]
S2 = S2.replace("ecney", "B")
S2 = S2[::-1]
if S2[0] == "A" and S2[len(S2) - 1] == "B":
print("YES")
exit()
S3 = S.replace("Key", "A")
S3 = S3[::-1]
S3 = S3.replace("ecne", "B")
S3 = S3[::-1]
if S3[0] == "A" and S3[len(S3) - 1] == "B":
print("YES")
exit()
S4 = S.replace("Keye", "A")
S4 = S4[::-1]
S4 = S4.replace("ecn", "B")
S4 = S4[::-1]
if S4[0] == "A" and S4[len(S4) - 1] == "B":
print("YES")
exit()
S5 = S.replace("Keyen", "A")
S5 = S5[::-1]
S5 = S5.replace("ec", "B")
S5 = S5[::-1]
if S5[0] == "A" and S5[len(S5) - 1] == "B":
print("YES")
exit()
S6 = S.replace("Keyenc", "A")
S6 = S6.replace("e", "B")
if S6[0] == "A" and S6[len(S6) - 1] == "B":
print("YES")
exit()
print("NO")
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
If S is a KEYENCE string, print `YES`; otherwise, print `NO`.
* * * | s424613060 | Wrong Answer | p03150 | Input is given from Standard Input in the following format:
S | def my_index(l, x, default=False):
if x in l:
return l.index(x)
else:
return -1
a = input().split("!!!")
a = list(a[0])
judge = 0
count = 0
tmp1 = 0
tmp2 = 0
A = []
B = []
A.append(my_index(a, "k"))
tmp1 = my_index(a, "k")
if my_index(a, "k") != -1:
a[my_index(a, "k")] = "AAA"
judge += 1
A.append(my_index(a, "e"))
tmp2 = my_index(a, "e")
if my_index(a, "e") != -1:
a[my_index(a, "e")] = "AAA"
judge += 1
if tmp2 - tmp1 >= 2:
count += 1
tmp1 = tmp2
A.append(my_index(a, "y"))
tmp2 = my_index(a, "y")
if my_index(a, "y") != -1:
a[my_index(a, "y")] = "AAA"
judge += 1
if tmp2 - tmp1 >= 2:
count += 1
tmp1 = tmp2
A.append(my_index(a, "e"))
tmp2 = my_index(a, "e")
if my_index(a, "e") != -1:
a[my_index(a, "e")] = "AAA"
judge += 1
if tmp2 - tmp1 >= 2:
count += 1
tmp1 = tmp2
A.append(my_index(a, "n"))
tmp2 = my_index(a, "n")
if my_index(a, "n") != -1:
a[my_index(a, "n")] = "AAA"
judge += 1
if tmp2 - tmp1 >= 2:
count += 1
tmp1 = tmp2
A.append(my_index(a, "c"))
tmp2 = my_index(a, "c")
if my_index(a, "c") != -1:
a[my_index(a, "c")] = "AAA"
judge += 1
if tmp2 - tmp1 >= 2:
count += 1
tmp1 = tmp2
A.append(my_index(a, "e"))
tmp2 = my_index(a, "e")
if my_index(a, "e") != -1:
a[my_index(a, "e")] = "AAA"
judge += 1
if tmp2 - tmp1 >= 2:
count += 1
tmp1 = tmp2
B = A
A.sort()
if (A == B) and (judge == 7) and (count <= 2):
print("YES")
else:
print("NO")
| Statement
A string is called a KEYENCE string when it can be changed to `keyence` by
removing its contiguous substring (possibly empty) only once.
Given a string S consisting of lowercase English letters, determine if S is a
KEYENCE string. | [{"input": "keyofscience", "output": "YES\n \n\n`keyence` is an abbreviation of `key of science`.\n\n* * *"}, {"input": "mpyszsbznf", "output": "NO\n \n\n* * *"}, {"input": "ashlfyha", "output": "NO\n \n\n* * *"}, {"input": "keyence", "output": "YES"}] |
Print the K-th element.
* * * | s885666443 | Wrong Answer | p02741 | Input is given from Standard Input in the following format:
K | s = input()
print(s[0:4] + " " + s[4:12])
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s837826497 | Runtime Error | p02741 | Input is given from Standard Input in the following format:
K | n = list(map(int, input().split()))
print((n[0] * n[1] + 1) // 2)
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s620431631 | Wrong Answer | p02741 | Input is given from Standard Input in the following format:
K | print(5)
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s230992095 | Wrong Answer | p02741 | Input is given from Standard Input in the following format:
K | N = int(input())
mojis = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j"]
def append_m(moji):
arr = []
for i in mojis[: len(set(moji)) + 1]:
arr.append(moji + i)
return arr
arr = ["a"]
for i in range(N - 1):
tmp = []
for j in arr:
tmp += append_m(j)
arr = tmp
for i in arr:
print(i)
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s706732177 | Runtime Error | p02741 | Input is given from Standard Input in the following format:
K | a = list(input())
b = list(input())
c = list(input())
A, B, C = map(len, (a, b, c))
OFFSET = 4000
L = 8001
ab = [True] * L
ac = [True] * L
bc = [True] * L
for i in range(A):
for j in range(B):
if a[i] != b[j] and a[i] != "?" and b[j] != "?":
ab[i - j + OFFSET] = False # BをAからi-jずらした時を考える
for i in range(A):
for j in range(C):
if a[i] != c[j] and a[i] != "?" and c[j] != "?":
ac[i - j + OFFSET] = False # CをAからi-jずらした時を考える
for i in range(B):
for j in range(C):
if b[i] != c[j] and c[i] != "?" and b[j] != "?":
bc[i - j + OFFSET] = False # CをBからi-jずらした時を考える
ans = float("inf")
for i in range(L): # Bの左端をずらす
for j in range(L): # Cの左端をずらす
k = j - i + OFFSET
if k >= L or k < 0:
continue
if ab[i] and ac[j] and bc[k]:
head = min(i, j, OFFSET)
tail = max(A + OFFSET, B + i, C + j)
ans = min(ans, tail - head)
print(ans)
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s120124488 | Wrong Answer | p02741 | Input is given from Standard Input in the following format:
K | from string import ascii_lowercase
N = int(input())
ans = ["a"]
for i in range(1, N):
ans2 = []
for a in ans:
for b in ascii_lowercase[: len(set(a)) + 1]:
ans2 += [a + b]
ans = ans2
print("\n".join(ans))
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s880274536 | Runtime Error | p02741 | Input is given from Standard Input in the following format:
K | alp = ["a", "b", "c", "d", "e", "f", "g", "h", "i", "j"]
N = int(input())
ans_list = []
for i in range(N):
ans_list.append([])
for i in range(N):
if i == 0:
ans_list[i].append(["a", 1])
else:
for j in range(len(ans_list[i - 1])):
for l in range(ans_list[i - 1][j][1] + 1):
temp = ans_list[i - 1][j][0] + alp[l]
if l == ans_list[i - 1][j][1]:
chk = ans_list[i - 1][j][1] + 1
else:
chk = ans_list[i - 1][j][1]
ans_list[i].append([temp, chk])
for i in range(len(ans_list[N - 1])):
print(ans_list[N - 1][i][0])
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s189703424 | Runtime Error | p02741 | Input is given from Standard Input in the following format:
K | a = input()
a_l = len(a)
b = input()
b_l = len(b)
c = input()
c_l = len(c)
def check(s1, s2):
l1 = len(s1)
l2 = len(s2)
ret = [0] * (l1 + l2 - 1)
for i in range(l1 + l2 - 1):
flag = True
if i - l2 + 1 < 0:
for j in range(i + 1):
if (
s1[j] != "?"
and s2[l2 - i + j - 1] != "?"
and s1[j] != s2[l2 - i + j - 1]
):
flag = False
break
elif i - l2 + 1 >= 0 and i <= l1 - 1:
for j in range(l2):
if (
s1[i - l2 + j + 1] != "?"
and s2[j] != "?"
and s1[i - l2 + j + 1] != s2[j]
):
flag = False
break
else:
for j in range(l1 + l2 - i - 1):
if (
s1[i - l2 + j + 1] != "?"
and s2[j] != "?"
and s1[i - l2 + j + 1] != s2[j]
):
flag = False
break
ret[i] = flag
return ret
ab = check(a, b)
ab_l = len(ab)
ac = check(a, c)
ac_l = len(ac)
bc = check(b, c)
bc_l = len(bc)
ans = a_l + b_l + c_l
for i in range(-3000, ab_l + 3000):
if i < 0 or i >= ab_l or ab[i]:
for j in range(-3000, ac_l + 3000):
if j < 0 or j >= ac_l or ac[j]:
if (
(0 <= j - i + b_l + 1 < bc_l and bc[j - i + b_l + 1])
or j - i + b_l + 1 < 0
or j - i + b_l + 1 >= bc_l
):
l = max(a_l - 1, i, j) - min(0, i - b_l + 1, j - c_l + 1) + 1
ans = min(ans, l)
print(ans)
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s080450355 | Runtime Error | p02741 | Input is given from Standard Input in the following format:
K | def main():
a = input()
b = input()
c = input()
A, B, C = len(a), len(b), len(c)
ab = [True] * 10000
ac = [True] * 10000
bc = [True] * 10000
def check(c1, c2):
if c1 == "?" or c2 == "?" or c1 == c2:
return False
return True
for i in range(A):
for j in range(B):
if check(a[i], b[j]):
ab[i - j + 5000] = False
for i in range(A):
for j in range(C):
if check(a[i], c[j]):
ac[i - j + 5000] = False
for i in range(B):
for j in range(C):
if check(b[i], c[j]):
bc[i - j + 5000] = False
ans = A + B + C
for i in range(-(B + C), A + C):
for j in range(-(B + C), A + B):
if ab[i + 5000] and ac[j + 5000] and bc[j - i + 5000]:
L = min(0, i, j)
R = max(A, i + B, j + C)
ans = min(ans, R - L)
print(ans)
main()
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s689612841 | Runtime Error | p02741 | Input is given from Standard Input in the following format:
K | a = input()
b = input()
c = input()
def common(s1, s2):
l1 = len(s1)
l2 = len(s2)
memo = [[0 for i in range(l1 + 2)] for j in range(l2 + 2)]
for idx1, w1 in enumerate(s1):
for idx2, w2 in enumerate(s2):
if w1 == w2 or w1 == "?" or w2 == "?":
memo[idx2 + 1][idx1 + 1] = memo[idx2][idx1] + 1
return memo
m1 = 0
m2 = 0
m3 = 0
for l in common(a, b):
if max(l) > m1:
m1 = max(l)
for l in common(a, c):
if max(l) > m2:
m2 = max(l)
for l in common(b, c):
if max(l) > m3:
m3 = max(l)
ans = len(a) + len(b) + len(c) - m1 - m2 - m3 + min([m1, m2, m3])
print(ans)
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s182551652 | Accepted | p02741 | Input is given from Standard Input in the following format:
K | print(b" 3 "[int(input()) % 14])
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s006927210 | Runtime Error | p02741 | Input is given from Standard Input in the following format:
K | H, W = map(int, input().split())
if H % 2 != 0:
H = H + 1
if W % 2 != 0:
W = W + 1
print(int(H * W / 2))
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s944812549 | Runtime Error | p02741 | Input is given from Standard Input in the following format:
K | retu = input()
print(int(retu.split(",")[32]))
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s428982842 | Runtime Error | p02741 | Input is given from Standard Input in the following format:
K | x = list(map(int, input().split()))
y = int(input())
print(x)
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s110126828 | Wrong Answer | p02741 | Input is given from Standard Input in the following format:
K | 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s041167222 | Accepted | p02741 | Input is given from Standard Input in the following format:
K | K = "1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51"
k = map(int, K.split(","))
A = list(k)
B = int(input())
print(A[B - 1])
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s530254235 | Runtime Error | p02741 | Input is given from Standard Input in the following format:
K | retu = "1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51"
retu = retu.split(",")
retuint = []
for i in retu:
retuint.append(int(i))
index = input()
print(retuint[int(index - 1)])
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s391268839 | Runtime Error | p02741 | Input is given from Standard Input in the following format:
K | text = "1,1,1,2,1,2,1,5,2,2,1,5,1,2,1,14,1,5,1,5,2,2,1,15,2,2,5,4,1,4,1,51"
textList = text.split(",")
firstInputText = input()
print(textList[int(firstInputText) - 1])
secondInputText = input()
print(textList[int(secondInputText) - 1])
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s647805822 | Runtime Error | p02741 | Input is given from Standard Input in the following format:
K | [H, W] = map(int, input().split())
if (H * W) % 2 == 0:
B = int((H * W) / 2)
print(B)
else:
if (H + W) % 2 == 0:
C = int((H * W + 1) / 2)
print(C)
else:
D = int((H * W - 1) / 2)
print(D)
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the K-th element.
* * * | s709150620 | Runtime Error | p02741 | Input is given from Standard Input in the following format:
K | a, b = map(int, input().split())
c = len([i for i in range(1, a + 1) if i % 2 == 0])
d = len([i for i in range(1, a + 1) if i % 2 != 0])
e = len([i for i in range(1, b + 1) if i % 2 == 0])
f = len([i for i in range(1, b + 1) if i % 2 != 0])
if a % 2 == 0 and b % 2 == 0:
print((a * b) // 2)
elif a % 2 == 0 and b % 2 != 0:
print((a // 2) * b)
elif a % 2 != 0 and b == 0:
print((b // 2) * a)
else:
print(d * f + c * e)
| Statement
Print the K-th element of the following sequence of length 32:
1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51 | [{"input": "6", "output": "2\n \n\nThe 6-th element is 2.\n\n* * *"}, {"input": "27", "output": "5\n \n\nThe 27-th element is 5."}] |
Print the extended image.
* * * | s550160455 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h,w=map(int,input().split())
a=[]
for i in range(h)
a.append(input().split())
for i in range(h)
print(" ".join(a[i]))
print(" ".join(a[i])) | Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s236717882 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | H, W = map(int, input().split())
X = [list(input().split()) for i in range(H)]
for i in range(2 * N):
print(*X[int(i / 2)])
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s547527786 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h,w=map(int,input().split())
a=[]
for i range(h):
a.append=list(map(list,input().split()))
a.append=list(map(list,input().split()))
print(a)
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s129250514 | Accepted | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | N, H = map(int, input().split())
S = list(input() for i in range(N))
for k in range(N):
for u in range(2):
print(S[k])
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s268795463 | Accepted | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | #!/usr/bin/env python3
import sys
# import time
# import math
# import numpy as np
# import scipy.sparse.csgraph as cs # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
# import random # random, uniform, randint, randrange, shuffle, sample
# import string # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
# import re # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
# from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
# from collections import deque # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
# from collections import defaultdict # subclass of dict. defaultdict(facroty)
# from collections import Counter # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
# from datetime import date, datetime # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
# from datetime.datetime import strptime # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
# from datetime import timedelta # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
# from copy import copy, deepcopy # use deepcopy to copy multi-dimentional matrix without reference
# from functools import reduce # reduce(f, iter[, init])
# from functools import lru_cache # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
# from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
# from heapq import nlargest, nsmallest # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
# from itertools import count, cycle, repeat # count(start[,step]), cycle(iter), repeat(elm[,n])
# from itertools import groupby # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
# from itertools import starmap # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
# from itertools import product, permutations # product(iter, repeat=n), permutations(iter[,r])
# from itertools import combinations, combinations_with_replacement
# from itertools import accumulate # accumulate(iter[, f])
# from operator import itemgetter # itemgetter(1), itemgetter('key')
# from fractions import gcd # for Python 3.4 (previous contest @AtCoder)
def main():
mod = 1000000007 # 10^9+7
inf = float("inf") # sys.float_info.max = 1.79...e+308
# inf = 2 ** 64 - 1 # (for fast JIT compile in PyPy) 1.84...e+19
sys.setrecursionlimit(10**6) # 1000 -> 1000000
def input():
return sys.stdin.readline().rstrip()
def ii():
return int(input())
def mi():
return map(int, input().split())
def mi_0():
return map(lambda x: int(x) - 1, input().split())
def lmi():
return list(map(int, input().split()))
def lmi_0():
return list(map(lambda x: int(x) - 1, input().split()))
def li():
return list(input())
h, w = mi()
for _ in range(h):
s = input()
print(s)
print(s)
if __name__ == "__main__":
main()
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s165997740 | Accepted | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | import sys
sys.setrecursionlimit(10**7) # 再帰関数の上限,10**5以上の場合python
import math
from copy import copy, deepcopy
from copy import deepcopy as dcp
from operator import itemgetter
from bisect import bisect_left, bisect, bisect_right # 2分探索
# bisect_left(l,x), bisect(l,x)#aはソート済みである必要あり。aの中からx未満の要素数を返す。rightだと以下
from collections import deque, defaultdict
# deque(l), pop(), append(x), popleft(), appendleft(x)
# q.rotate(n)で → にn回ローテート
from collections import Counter # 文字列を個数カウント辞書に、
# S=Counter(l),S.most_common(x),S.keys(),S.values(),S.items()
from itertools import accumulate, combinations, permutations # 累積和
# list(accumulate(l))
from heapq import heapify, heappop, heappush
# heapify(q),heappush(q,a),heappop(q) #q=heapify(q)としないこと、返り値はNone
# import fractions#古いatcoderコンテストの場合GCDなどはここからimportする
from functools import reduce, lru_cache # pypyでもうごく
# @lru_cache(maxsize = None)#maxsizeは保存するデータ数の最大値、2**nが最も高効率
from decimal import Decimal
def input():
x = sys.stdin.readline()
return x[:-1] if x[-1] == "\n" else x
def printe(*x):
print("## ", *x, file=sys.stderr)
def printl(li):
_ = print(*li, sep="\n") if li else None
def argsort(s, return_sorted=False):
inds = sorted(range(len(s)), key=lambda k: s[k])
if return_sorted:
return inds, [s[i] for i in inds]
return inds
def alp2num(c, cap=False):
return ord(c) - 97 if not cap else ord(c) - 65
def num2alp(i, cap=False):
return chr(i + 97) if not cap else chr(i + 65)
def matmat(A, B):
K, N, M = len(B), len(A), len(B[0])
return [
[sum([(A[i][k] * B[k][j]) for k in range(K)]) for j in range(M)]
for i in range(N)
]
def matvec(M, v):
N, size = len(v), len(M)
return [sum([M[i][j] * v[j] for j in range(N)]) for i in range(size)]
def T(M):
n, m = len(M), len(M[0])
return [[M[j][i] for j in range(n)] for i in range(m)]
def main():
mod = 1000000007
# w.sort(key=itemgetter(1),reversed=True) #二個目の要素で降順並び替え
# N = int(input())
H, W = map(int, input().split())
# A = tuple(map(int, input().split())) #1行ベクトル
# L = tuple(int(input()) for i in range(N)) #改行ベクトル
S = list(tuple(input()) for i in range(H)) # 改行行列
for li in S:
print(*li, sep="")
print(*li, sep="")
if __name__ == "__main__":
main()
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s340580578 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h,w = map(int,input().split())
li = [input() for _ in range(h)]
for i in li:
print(i,i,sep="\n")***. | Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s976906103 | Accepted | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | # from math import factorial,sqrt,ceil,gcd
# from itertools import permutations as permus
# from collections import deque,Counter
# import re
# from functools import lru_cache # 簡単メモ化 @lru_cache(maxsize=1000)
# from decimal import Decimal, getcontext
# # getcontext().prec = 1000
# # eps = Decimal(10) ** (-100)
# import numpy as np
# import networkx as nx
# from scipy.sparse.csgraph import shortest_path, dijkstra, floyd_warshall, bellman_ford, johnson
# from scipy.sparse import csr_matrix
# from scipy.special import comb
# slist = "abcdefghijklmnopqrstuvwxyz"
# 盤面の受け取り
H, W = map(int, input().split())
board = [[] for _ in range(H * 2)]
for i in range(H):
board[2 * i] = list(input())
board[2 * i + 1] = board[2 * i].copy()
# print(*ans) # unpackして出力。間にスペースが入る
for row in board:
print(*row, sep="") # unpackして間にスペース入れずに出力する
# print("{:.10f}".format(ans))
# print("{:0=10d}".format(ans))
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s648317022 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | #
# abc049 b
#
import sys
from io import StringIO
import unittest
class TestClass(unittest.TestCase):
def assertIO(self, input, output):
stdout, stdin = sys.stdout, sys.stdin
sys.stdout, sys.stdin = StringIO(), StringIO(input)
resolve()
sys.stdout.seek(0)
out = sys.stdout.read()[:-1]
sys.stdout, sys.stdin = stdout, stdin
self.assertEqual(out, output)
def test_入力例_1(self):
input = """2 2
*.
.*"""
output = """*.
*.
.*
.*"""
self.assertIO(input, output)
def test_入力例_2(self):
input = """1 4
***."""
output = """***.
***."""
self.assertIO(input, output)
def test_入力例_3(self):
input = """9 20
.....***....***.....
....*...*..*...*....
...*.....**.....*...
...*.....*......*...
....*.....*....*....
.....**..*...**.....
.......*..*.*.......
........**.*........
.........**........."""
output = """.....***....***.....
.....***....***.....
....*...*..*...*....
....*...*..*...*....
...*.....**.....*...
...*.....**.....*...
...*.....*......*...
...*.....*......*...
....*.....*....*....
....*.....*....*....
.....**..*...**.....
.....**..*...**.....
.......*..*.*.......
.......*..*.*.......
........**.*........
........**.*........
.........**.........
.........**........."""
self.assertIO(input, output)
def resolve():
H, W = map(int, input().split())
ans = []
for i in range(H):
S = input()
ans.append(S)
ans.append(S)
for i in range(2*H):
print(ans[i])
if __name__ == "__main__":
# unittest.main()
resolve()
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s609962647 | Accepted | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | nums = list(map(int, input().split()))
k = [input() for i in range(nums[0])]
for i in range(nums[0]):
print(k[i])
print(k[i])
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s407816370 | Accepted | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h, w = input().split()
exec('print((input()+chr(10))*2,end="")\n' * int(h))
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s078894513 | Wrong Answer | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | c = input()
if c in ["a", "e", "i", "o", "u"]:
print("vowel")
else:
print("consonant")
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s915639722 | Accepted | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | height, width = [int(i) for i in input().split()]
in_pict = [input() for i in range(height)]
output = []
for i in range(height):
print(in_pict[i] + "\n" + in_pict[i])
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s660781155 | Accepted | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h, w = map(int, input().split())
r = [list(input()) for i in range(h)]
new_h = [0] * w
new_r = [new_h for i in range(2 * h)]
for i in range(0, 2 * h, 2):
new_r[i] = r[(i + 1) // 2]
new_r[i + 1] = r[(i + 1) // 2]
for i in new_r:
print(*i, sep="", end="\n")
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s600457923 | Accepted | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | s, t = map(int, input().split())
A, B = [input().split() for r in range(s)], []
for x in A:
for y in range(2):
B.append(x)
for g in B:
print(g[0])
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s881946566 | Accepted | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | H, W = [int(i) for i in input().split()]
i = 0
while i < H:
hoge = input()
print(hoge)
print(hoge)
i += 1
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s736991983 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h,w = tuple(map(int,input().split()))
s = [input() for_ in range(h)]
for s_i in s:
print(*s_i,sep='')
print(*s_i,sep='')
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s921954307 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | H,W=[int(i) for i in input().split()
ans=[]
for i in range(H):
lis =input()
ans.append(lis)
ans.append(lis)
print(ans) | Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s104489770 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | H,W=[int(i) for i in input().split()
ans=[]
for i in range(H):
lis =input()
ans.append(lis)
ans.append(lis)
print(ans) | Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s067032617 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h,w = map(int, input().split())
listA = [input() for i in range(n)]
for i in range(n):
print(listA[i-1]"\n"listA[i-1])
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s283356273 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | data_num = [int(i) for i in input().split(" ")]
data = [input() for i in range(data_num[0])]
for i in data:
print(i)
print(i)
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s280795735 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h,w = map(int,input().split())
for i in range(h):
c = input()
print(c)
print(c) | Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s966045184 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | w = int(input().split()[1])
c = [input() for _ in [0] * w]
for _c in c:
print(_c)
print(_c)
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s934540050 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} |
H, W = map(int, input().split())
for i in range(H):
s = input()
print(s)
print(s)
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s329395818 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | a,b=input().split();for _ in[0]*int(a):print(input()*2) | Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s082296054 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | a
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s356176497 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} |
H, W = map(int, input().split())
for i in range(H):
s = input()
print(s)
print(s)
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s519281641 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h,w = map(int,input().split())
for i in range(h):
c = input()
print(c)
print(c) | Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s333834596 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h, w = list(map(int, input().split()))
line = [input() for i in range(h)]
for l in line:
print(l)
print(l | Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s567398323 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h,w = tuple(map(int,input().split()))
s = [input() for_ in range(h)]
for s_i in s:
print(s_i)
print(s_i)
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s030195016 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | H, W = map(int,input().strip().split())
for i in range(H):
line = input()
for j in range(2):
print(line) | Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s170626105 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h,w = map(int,input().split())
for _ in range(h):
c = input()
print(c)
print(c)
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s170058606 | Accepted | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h, w = map(int, input().split())
count = 0
prob_list = []
while count < h:
prob_list.append(str(input()))
prob_list.append(prob_list[-1])
count += 1
for row in prob_list:
print("".join(row))
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s587819922 | Accepted | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | p = input().rstrip().split(" ")
l = []
for i in range(0, int(p[0])):
q = input().rstrip()
x = list(q)
l.append(x)
for i in range(0, len(l)):
for j in range(0, 2):
print("".join(l[i]))
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s349502834 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h,w = map(int, input().split())
listA = []
while True:
try:
listA.append(list(map(int,input().split())))
except:
break;
for i in range(n):
print(listA[i-1]"\n"listA[i-1]) | Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s774984504 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | icase = 0
if icase == 0:
w, h = map(int, input().split())
# xy=[[1]*w for i in range(2*h)]
for i in range(h):
cij = input()
print(cij)
print(cij)
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s690443029 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | 9 20
.....***....***.....
....*...*..*...*....
...*.....**.....*...
...*.....*......*...
....*.....*....*....
.....**..*...**.....
.......*..*.*.......
........**.*........
.........**.........
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s946938082 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | from sys import stdin
n, m = int(stdin.readline().rstrip())
data = [stdin.readline().rstrip().split() for _ in range(n)]
for i in range(n):
print(data[i]\n)
print(data[i])
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s934761099 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h,w=map(int,input().split())
c=[]
for _ in range(h):
c.append(list(map(str,input())))
ans=[c[i//2] for i in range(2h)]
for j in ans:
print(''.join(j)) | Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print the extended image.
* * * | s466888195 | Runtime Error | p03853 | The input is given from Standard Input in the following format:
H W
C_{1,1}...C_{1,W}
:
C_{H,1}...C_{H,W} | h,w = map(int, input().split())
listA = [input() for i in range(h)]
for i in range(h):
print(listA[i-1]\nlistA[i-1])
| Statement
There is an image with a height of H pixels and a width of W pixels. Each of
the pixels is represented by either `.` or `*`. The character representing the
pixel at the i-th row from the top and the j-th column from the left, is
denoted by C_{i,j}.
Extend this image vertically so that its height is doubled. That is, print a
image with a height of 2H pixels and a width of W pixels where the pixel at
the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division
is rounded down). | [{"input": "2 2\n *.\n .*", "output": "*.\n *.\n .*\n .*\n \n\n* * *"}, {"input": "1 4\n ***.", "output": "***.\n ***.\n \n\n* * *"}, {"input": "9 20\n .....***....***.....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....*......*...\n ....*.....*....*....\n .....**..*...**.....\n .......*..*.*.......\n ........**.*........\n .........**.........", "output": ".....***....***.....\n .....***....***.....\n ....*...*..*...*....\n ....*...*..*...*....\n ...*.....**.....*...\n ...*.....**.....*...\n ...*.....*......*...\n ...*.....*......*...\n ....*.....*....*....\n ....*.....*....*....\n .....**..*...**.....\n .....**..*...**.....\n .......*..*.*.......\n .......*..*.*.......\n ........**.*........\n ........**.*........\n .........**.........\n .........**........."}] |
Print an integer representing the number of points such that the distance from
the origin is at most D.
* * * | s760192035 | Accepted | p02595 | Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N | (n, d), *z = [map(int, t.split()) for t in open(0)]
print(sum(x * x + y * y <= d * d for x, y in z))
| Statement
We have N points in the two-dimensional plane. The coordinates of the i-th
point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the
origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be
represented as \sqrt{p^2+q^2}. | [{"input": "4 5\n 0 5\n -2 4\n 3 4\n 4 -4", "output": "3\n \n\nThe distance between the origin and each of the given points is as follows:\n\n * \\sqrt{0^2+5^2}=5\n * \\sqrt{(-2)^2+4^2}=4.472\\ldots\n * \\sqrt{3^2+4^2}=5\n * \\sqrt{4^2+(-4)^2}=5.656\\ldots\n\nThus, we have three points such that the distance from the origin is at most\n5.\n\n* * *"}, {"input": "12 3\n 1 1\n 1 1\n 1 1\n 1 1\n 1 2\n 1 3\n 2 1\n 2 2\n 2 3\n 3 1\n 3 2\n 3 3", "output": "7\n \n\nMultiple points may exist at the same coordinates.\n\n* * *"}, {"input": "20 100000\n 14309 -32939\n -56855 100340\n 151364 25430\n 103789 -113141\n 147404 -136977\n -37006 -30929\n 188810 -49557\n 13419 70401\n -88280 165170\n -196399 137941\n -176527 -61904\n 46659 115261\n -153551 114185\n 98784 -6820\n 94111 -86268\n -30401 61477\n -55056 7872\n 5901 -163796\n 138819 -185986\n -69848 -96669", "output": "6"}] |
Print an integer representing the number of points such that the distance from
the origin is at most D.
* * * | s960210061 | Runtime Error | p02595 | Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N | while True:
i = 0
x, y = map(int, input().split())
(x**2 + y**2) ** (1 / 2)
i += 1
print(i)
| Statement
We have N points in the two-dimensional plane. The coordinates of the i-th
point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the
origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be
represented as \sqrt{p^2+q^2}. | [{"input": "4 5\n 0 5\n -2 4\n 3 4\n 4 -4", "output": "3\n \n\nThe distance between the origin and each of the given points is as follows:\n\n * \\sqrt{0^2+5^2}=5\n * \\sqrt{(-2)^2+4^2}=4.472\\ldots\n * \\sqrt{3^2+4^2}=5\n * \\sqrt{4^2+(-4)^2}=5.656\\ldots\n\nThus, we have three points such that the distance from the origin is at most\n5.\n\n* * *"}, {"input": "12 3\n 1 1\n 1 1\n 1 1\n 1 1\n 1 2\n 1 3\n 2 1\n 2 2\n 2 3\n 3 1\n 3 2\n 3 3", "output": "7\n \n\nMultiple points may exist at the same coordinates.\n\n* * *"}, {"input": "20 100000\n 14309 -32939\n -56855 100340\n 151364 25430\n 103789 -113141\n 147404 -136977\n -37006 -30929\n 188810 -49557\n 13419 70401\n -88280 165170\n -196399 137941\n -176527 -61904\n 46659 115261\n -153551 114185\n 98784 -6820\n 94111 -86268\n -30401 61477\n -55056 7872\n 5901 -163796\n 138819 -185986\n -69848 -96669", "output": "6"}] |
Print an integer representing the number of points such that the distance from
the origin is at most D.
* * * | s513702632 | Runtime Error | p02595 | Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N | z = [map(int, t.split()) for t in open(0)]
print(sum(x * x + y * y <= z[0][1] ** 2 for x, y in z))
| Statement
We have N points in the two-dimensional plane. The coordinates of the i-th
point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the
origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be
represented as \sqrt{p^2+q^2}. | [{"input": "4 5\n 0 5\n -2 4\n 3 4\n 4 -4", "output": "3\n \n\nThe distance between the origin and each of the given points is as follows:\n\n * \\sqrt{0^2+5^2}=5\n * \\sqrt{(-2)^2+4^2}=4.472\\ldots\n * \\sqrt{3^2+4^2}=5\n * \\sqrt{4^2+(-4)^2}=5.656\\ldots\n\nThus, we have three points such that the distance from the origin is at most\n5.\n\n* * *"}, {"input": "12 3\n 1 1\n 1 1\n 1 1\n 1 1\n 1 2\n 1 3\n 2 1\n 2 2\n 2 3\n 3 1\n 3 2\n 3 3", "output": "7\n \n\nMultiple points may exist at the same coordinates.\n\n* * *"}, {"input": "20 100000\n 14309 -32939\n -56855 100340\n 151364 25430\n 103789 -113141\n 147404 -136977\n -37006 -30929\n 188810 -49557\n 13419 70401\n -88280 165170\n -196399 137941\n -176527 -61904\n 46659 115261\n -153551 114185\n 98784 -6820\n 94111 -86268\n -30401 61477\n -55056 7872\n 5901 -163796\n 138819 -185986\n -69848 -96669", "output": "6"}] |
Print an integer representing the number of points such that the distance from
the origin is at most D.
* * * | s500036617 | Accepted | p02595 | Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N | import os
import sys
from io import BytesIO, IOBase
# from collections import defaultdict as dd
# from collections import deque as dq
# import itertools as it
# from math import sqrt, log, log2
# from fractions import Fraction
def main():
t = 1
for _ in range(t):
# n = int(input())
n, d = map(int, input().split())
c = 0
for i in range(n):
x, y = map(int, input().split())
if (x**2 + y**2) <= d**2:
c += 1
print(c)
# nums = list(map(int, input().split()))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
| Statement
We have N points in the two-dimensional plane. The coordinates of the i-th
point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the
origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be
represented as \sqrt{p^2+q^2}. | [{"input": "4 5\n 0 5\n -2 4\n 3 4\n 4 -4", "output": "3\n \n\nThe distance between the origin and each of the given points is as follows:\n\n * \\sqrt{0^2+5^2}=5\n * \\sqrt{(-2)^2+4^2}=4.472\\ldots\n * \\sqrt{3^2+4^2}=5\n * \\sqrt{4^2+(-4)^2}=5.656\\ldots\n\nThus, we have three points such that the distance from the origin is at most\n5.\n\n* * *"}, {"input": "12 3\n 1 1\n 1 1\n 1 1\n 1 1\n 1 2\n 1 3\n 2 1\n 2 2\n 2 3\n 3 1\n 3 2\n 3 3", "output": "7\n \n\nMultiple points may exist at the same coordinates.\n\n* * *"}, {"input": "20 100000\n 14309 -32939\n -56855 100340\n 151364 25430\n 103789 -113141\n 147404 -136977\n -37006 -30929\n 188810 -49557\n 13419 70401\n -88280 165170\n -196399 137941\n -176527 -61904\n 46659 115261\n -153551 114185\n 98784 -6820\n 94111 -86268\n -30401 61477\n -55056 7872\n 5901 -163796\n 138819 -185986\n -69848 -96669", "output": "6"}] |
Print an integer representing the number of points such that the distance from
the origin is at most D.
* * * | s395327589 | Accepted | p02595 | Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N | import sys
from math import hypot
def input():
return sys.stdin.readline().strip()
def list2d(a, b, c):
return [[c] * b for i in range(a)]
def list3d(a, b, c, d):
return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e):
return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1):
return int(-(-x // y))
def INT():
return int(input())
def MAP():
return map(int, input().split())
def LIST(N=None):
return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes():
print("Yes")
def No():
print("No")
def YES():
print("YES")
def NO():
print("NO")
sys.setrecursionlimit(10**9)
INF = 10**19
MOD = 10**9 + 7
EPS = 10**-10
N, D = MAP()
ans = 0
for i in range(N):
x, y = MAP()
d = hypot(x, y)
if d < D + EPS:
ans += 1
print(ans)
| Statement
We have N points in the two-dimensional plane. The coordinates of the i-th
point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the
origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be
represented as \sqrt{p^2+q^2}. | [{"input": "4 5\n 0 5\n -2 4\n 3 4\n 4 -4", "output": "3\n \n\nThe distance between the origin and each of the given points is as follows:\n\n * \\sqrt{0^2+5^2}=5\n * \\sqrt{(-2)^2+4^2}=4.472\\ldots\n * \\sqrt{3^2+4^2}=5\n * \\sqrt{4^2+(-4)^2}=5.656\\ldots\n\nThus, we have three points such that the distance from the origin is at most\n5.\n\n* * *"}, {"input": "12 3\n 1 1\n 1 1\n 1 1\n 1 1\n 1 2\n 1 3\n 2 1\n 2 2\n 2 3\n 3 1\n 3 2\n 3 3", "output": "7\n \n\nMultiple points may exist at the same coordinates.\n\n* * *"}, {"input": "20 100000\n 14309 -32939\n -56855 100340\n 151364 25430\n 103789 -113141\n 147404 -136977\n -37006 -30929\n 188810 -49557\n 13419 70401\n -88280 165170\n -196399 137941\n -176527 -61904\n 46659 115261\n -153551 114185\n 98784 -6820\n 94111 -86268\n -30401 61477\n -55056 7872\n 5901 -163796\n 138819 -185986\n -69848 -96669", "output": "6"}] |
Print an integer representing the number of points such that the distance from
the origin is at most D.
* * * | s251147248 | Accepted | p02595 | Input is given from Standard Input in the following format:
N D
X_1 Y_1
\vdots
X_N Y_N | import sys
def vfunc(f):
return lambda *lst, **kwargs: list(map(lambda *x: f(*x, **kwargs), *lst))
def rvfunc(f):
return lambda *lst, **kwargs: list(
map(
lambda *x: (
rvfunc(f)(*x, **kwargs)
if all(vfunc(lambda y: isinstance(y, list))(x))
else f(*x, **kwargs)
),
*lst
)
)
def reader():
return vfunc(lambda l: l.split())(sys.stdin.readlines())
ls = rvfunc(int)(reader())
# print(ls)
n, d = ls[0]
dst = vfunc(lambda ab: (ab[0] ** 2 + ab[1] ** 2) ** 0.5 <= d)(ls[1:])
print(sum(dst))
| Statement
We have N points in the two-dimensional plane. The coordinates of the i-th
point are (X_i,Y_i).
Among them, we are looking for the points such that the distance from the
origin is at most D. How many such points are there?
We remind you that the distance between the origin and the point (p, q) can be
represented as \sqrt{p^2+q^2}. | [{"input": "4 5\n 0 5\n -2 4\n 3 4\n 4 -4", "output": "3\n \n\nThe distance between the origin and each of the given points is as follows:\n\n * \\sqrt{0^2+5^2}=5\n * \\sqrt{(-2)^2+4^2}=4.472\\ldots\n * \\sqrt{3^2+4^2}=5\n * \\sqrt{4^2+(-4)^2}=5.656\\ldots\n\nThus, we have three points such that the distance from the origin is at most\n5.\n\n* * *"}, {"input": "12 3\n 1 1\n 1 1\n 1 1\n 1 1\n 1 2\n 1 3\n 2 1\n 2 2\n 2 3\n 3 1\n 3 2\n 3 3", "output": "7\n \n\nMultiple points may exist at the same coordinates.\n\n* * *"}, {"input": "20 100000\n 14309 -32939\n -56855 100340\n 151364 25430\n 103789 -113141\n 147404 -136977\n -37006 -30929\n 188810 -49557\n 13419 70401\n -88280 165170\n -196399 137941\n -176527 -61904\n 46659 115261\n -153551 114185\n 98784 -6820\n 94111 -86268\n -30401 61477\n -55056 7872\n 5901 -163796\n 138819 -185986\n -69848 -96669", "output": "6"}] |
If there is no permutation that can generate a tree isomorphic to Takahashi's
favorite tree, print `-1`. If it exists, print the lexicographically smallest
such permutation, with spaces in between.
* * * | s365832542 | Wrong Answer | p03384 | Input is given from Standard Input in the following format:
n
v_1 w_1
v_2 w_2
:
v_{n-1} w_{n-1} | print(-1)
| Statement
Takahashi has an ability to generate a tree using a permutation
(p_1,p_2,...,p_n) of (1,2,...,n), in the following process:
First, prepare Vertex 1, Vertex 2, ..., Vertex N. For each i=1,2,...,n,
perform the following operation:
* If p_i = 1, do nothing.
* If p_i \neq 1, let j' be the largest j such that p_j < p_i. Span an edge between Vertex i and Vertex j'.
Takahashi is trying to make his favorite tree with this ability. His favorite
tree has n vertices from Vertex 1 through Vertex n, and its i-th edge connects
Vertex v_i and w_i. Determine if he can make a tree isomorphic to his favorite
tree by using a proper permutation. If he can do so, find the
lexicographically smallest such permutation. | [{"input": "6\n 1 2\n 1 3\n 1 4\n 1 5\n 5 6", "output": "1 2 4 5 3 6\n \n\nIf the permutation (1, 2, 4, 5, 3, 6) is used to generate a tree, it looks as\nfollows:\n\n\n\nThis is isomorphic to the given graph.\n\n* * *"}, {"input": "6\n 1 2\n 2 3\n 3 4\n 1 5\n 5 6", "output": "1 2 3 4 5 6\n \n\n* * *"}, {"input": "15\n 1 2\n 1 3\n 2 4\n 2 5\n 3 6\n 3 7\n 4 8\n 4 9\n 5 10\n 5 11\n 6 12\n 6 13\n 7 14\n 7 15", "output": "-1"}] |
If there is no permutation that can generate a tree isomorphic to Takahashi's
favorite tree, print `-1`. If it exists, print the lexicographically smallest
such permutation, with spaces in between.
* * * | s499964797 | Wrong Answer | p03384 | Input is given from Standard Input in the following format:
n
v_1 w_1
v_2 w_2
:
v_{n-1} w_{n-1} | #!/usr/bin/env python3
def solve(n, adj_list, d):
s = []
path_adj_list = [[] for _ in range(n)]
for v in range(n):
if 1 < d[v]:
p = path_adj_list[v]
for w in adj_list[v]:
if 1 < d[w]:
p.append(w)
if 2 < len(p):
print(-1)
return
if len(p) == 1:
s.append(v)
if len(s) == 0:
ans = [1] + [v for v in range(3, n + 1)] + [2]
print(" ".join(list(map(str, ans))))
return
visited = [False] * n
v, w = s
while v != w and d[v] == d[w]:
visited[v] = True
visited[w] = True
f = False
for nv in path_adj_list[v]:
if not visited[nv]:
f = True
v = nv
break
if not f:
break
f = False
for nw in path_adj_list[w]:
if not visited[nw]:
f = True
w = nw
break
if not f:
break
if d[v] > d[w]:
v = s[1]
else:
v = s[0]
visited = [False] * n
visited[v] = True
ans = [1] + [w for w in range(3, d[v] + 1)] + [2]
c = d[v]
v = path_adj_list[v][0]
while True:
visited[v] = True
ans += [w for w in range(c + 2, c + d[v])] + [c + 1]
c += d[v] - 1
f = False
for nv in path_adj_list[v]:
if not visited[nv]:
f = True
v = nv
break
if not f:
break
ans += [n]
print(" ".join(list(map(str, ans))))
return
def main():
n = input()
n = int(n)
adj_list = [[] for _ in range(n)]
d = [0] * n
for _ in range(n - 1):
v, w = input().split()
v = int(v) - 1
w = int(w) - 1
adj_list[v].append(w)
adj_list[w].append(v)
d[v] += 1
d[w] += 1
solve(n, adj_list, d)
if __name__ == "__main__":
main()
| Statement
Takahashi has an ability to generate a tree using a permutation
(p_1,p_2,...,p_n) of (1,2,...,n), in the following process:
First, prepare Vertex 1, Vertex 2, ..., Vertex N. For each i=1,2,...,n,
perform the following operation:
* If p_i = 1, do nothing.
* If p_i \neq 1, let j' be the largest j such that p_j < p_i. Span an edge between Vertex i and Vertex j'.
Takahashi is trying to make his favorite tree with this ability. His favorite
tree has n vertices from Vertex 1 through Vertex n, and its i-th edge connects
Vertex v_i and w_i. Determine if he can make a tree isomorphic to his favorite
tree by using a proper permutation. If he can do so, find the
lexicographically smallest such permutation. | [{"input": "6\n 1 2\n 1 3\n 1 4\n 1 5\n 5 6", "output": "1 2 4 5 3 6\n \n\nIf the permutation (1, 2, 4, 5, 3, 6) is used to generate a tree, it looks as\nfollows:\n\n\n\nThis is isomorphic to the given graph.\n\n* * *"}, {"input": "6\n 1 2\n 2 3\n 3 4\n 1 5\n 5 6", "output": "1 2 3 4 5 6\n \n\n* * *"}, {"input": "15\n 1 2\n 1 3\n 2 4\n 2 5\n 3 6\n 3 7\n 4 8\n 4 9\n 5 10\n 5 11\n 6 12\n 6 13\n 7 14\n 7 15", "output": "-1"}] |
If there is no permutation that can generate a tree isomorphic to Takahashi's
favorite tree, print `-1`. If it exists, print the lexicographically smallest
such permutation, with spaces in between.
* * * | s963971297 | Wrong Answer | p03384 | Input is given from Standard Input in the following format:
n
v_1 w_1
v_2 w_2
:
v_{n-1} w_{n-1} | import sys
readline = sys.stdin.readline
def parorder(Edge, p):
N = len(Edge)
par = [0] * N
par[p] = -1
stack = [p]
order = []
visited = set([p])
ast = stack.append
apo = order.append
while stack:
vn = stack.pop()
apo(vn)
for vf in Edge[vn]:
if vf in visited:
continue
visited.add(vf)
par[vf] = vn
ast(vf)
return par, order
def getcld(p):
res = [[] for _ in range(len(p))]
for i, v in enumerate(p[1:], 1):
res[v].append(i)
return res
def dfs(St):
dist = [0] * N
stack = St[:]
used = [False] * N
for s in St:
used[s] = True
while stack:
vn = stack.pop()
for vf in Edge[vn]:
if not used[vf]:
used[vf] = True
dist[vf] = 1 + dist[vn]
stack.append(vf)
return dist
N = int(readline())
Edge = [[] for _ in range(N)]
for _ in range(N - 1):
a, b = map(int, readline().split())
a -= 1
b -= 1
Edge[a].append(b)
Edge[b].append(a)
dist0 = dfs([0])
fs = dist0.index(max(dist0))
distfs = dfs([fs])
en = distfs.index(max(distfs))
disten = dfs([en])
Dia = distfs[en]
if Dia <= 2:
print(*list(range(1, N + 1)))
else:
path = []
for i in range(N):
if distfs[i] + disten[i] == Dia:
path.append(i)
if max(dfs(path)) > 1:
print(-1)
else:
path.sort(key=lambda x: distfs[x])
cnt = 1
hold = 0
perm1 = [None] * N
onpath = set(path)
idx = 0
for i in range(Dia + 1):
vn = path[i]
hold = 0
for vf in Edge[vn]:
if vf in onpath:
continue
hold += 1
perm1[idx] = cnt + hold
idx += 1
perm1[idx] = cnt
idx += 1
cnt = cnt + hold + 1
cnt = 1
hold = 0
perm2 = [None] * N
onpath = set(path)
idx = 0
for i in range(Dia + 1):
vn = path[Dia - i]
hold = 0
for vf in Edge[vn]:
if vf in onpath:
continue
hold += 1
perm2[idx] = cnt + hold
idx += 1
perm2[idx] = cnt
idx += 1
cnt = cnt + hold + 1
print(*min(perm1, perm2))
| Statement
Takahashi has an ability to generate a tree using a permutation
(p_1,p_2,...,p_n) of (1,2,...,n), in the following process:
First, prepare Vertex 1, Vertex 2, ..., Vertex N. For each i=1,2,...,n,
perform the following operation:
* If p_i = 1, do nothing.
* If p_i \neq 1, let j' be the largest j such that p_j < p_i. Span an edge between Vertex i and Vertex j'.
Takahashi is trying to make his favorite tree with this ability. His favorite
tree has n vertices from Vertex 1 through Vertex n, and its i-th edge connects
Vertex v_i and w_i. Determine if he can make a tree isomorphic to his favorite
tree by using a proper permutation. If he can do so, find the
lexicographically smallest such permutation. | [{"input": "6\n 1 2\n 1 3\n 1 4\n 1 5\n 5 6", "output": "1 2 4 5 3 6\n \n\nIf the permutation (1, 2, 4, 5, 3, 6) is used to generate a tree, it looks as\nfollows:\n\n\n\nThis is isomorphic to the given graph.\n\n* * *"}, {"input": "6\n 1 2\n 2 3\n 3 4\n 1 5\n 5 6", "output": "1 2 3 4 5 6\n \n\n* * *"}, {"input": "15\n 1 2\n 1 3\n 2 4\n 2 5\n 3 6\n 3 7\n 4 8\n 4 9\n 5 10\n 5 11\n 6 12\n 6 13\n 7 14\n 7 15", "output": "-1"}] |
If there is no permutation that can generate a tree isomorphic to Takahashi's
favorite tree, print `-1`. If it exists, print the lexicographically smallest
such permutation, with spaces in between.
* * * | s524488437 | Wrong Answer | p03384 | Input is given from Standard Input in the following format:
n
v_1 w_1
v_2 w_2
:
v_{n-1} w_{n-1} | import sys
input = sys.stdin.readline
sys.setrecursionlimit(2 * 10**5)
n = int(input())
edge = [[] for i in range(n)]
for i in range(n - 1):
v, w = map(int, input().split())
edge[v - 1].append(w - 1)
edge[w - 1].append(v - 1)
if n == 2:
exit(print(1, 2))
leafcnt = [0] * n
for v in range(n):
for nv in edge[v]:
if len(edge[nv]) == 1:
leafcnt[v] += 1
used = [False] * n
line = []
def line_check(v):
used[v] = True
line.append(leafcnt[v])
flag = False
for nv in edge[v]:
if not used[nv] and len(edge[nv]) != 1:
if not flag:
line_check(nv)
flag = True
else:
return False
return True
for v in range(n):
if not used[v] and len(edge[v]) == 1 + leafcnt[v] and len(edge[v]) != 1:
if not line:
check = line_check(v)
if not check:
exit(print(-1))
else:
exit(print(-1))
line_rev = line[::-1]
res = min(line, line_rev)
res = [0] + res + [0]
res[1] -= 1
res[-2] -= 1
ans = []
L = 1
for val in res:
R = L + val
for i in range(L + 1, R + 1):
ans.append(i)
ans.append(L)
L = R + 1
print(*ans)
| Statement
Takahashi has an ability to generate a tree using a permutation
(p_1,p_2,...,p_n) of (1,2,...,n), in the following process:
First, prepare Vertex 1, Vertex 2, ..., Vertex N. For each i=1,2,...,n,
perform the following operation:
* If p_i = 1, do nothing.
* If p_i \neq 1, let j' be the largest j such that p_j < p_i. Span an edge between Vertex i and Vertex j'.
Takahashi is trying to make his favorite tree with this ability. His favorite
tree has n vertices from Vertex 1 through Vertex n, and its i-th edge connects
Vertex v_i and w_i. Determine if he can make a tree isomorphic to his favorite
tree by using a proper permutation. If he can do so, find the
lexicographically smallest such permutation. | [{"input": "6\n 1 2\n 1 3\n 1 4\n 1 5\n 5 6", "output": "1 2 4 5 3 6\n \n\nIf the permutation (1, 2, 4, 5, 3, 6) is used to generate a tree, it looks as\nfollows:\n\n\n\nThis is isomorphic to the given graph.\n\n* * *"}, {"input": "6\n 1 2\n 2 3\n 3 4\n 1 5\n 5 6", "output": "1 2 3 4 5 6\n \n\n* * *"}, {"input": "15\n 1 2\n 1 3\n 2 4\n 2 5\n 3 6\n 3 7\n 4 8\n 4 9\n 5 10\n 5 11\n 6 12\n 6 13\n 7 14\n 7 15", "output": "-1"}] |
Print Tak's total accommodation fee.
* * * | s321257494 | Accepted | p04011 | The input is given from Standard Input in the following format:
N
K
X
Y | # 044_a
# A - 高橋君とホテルイージー
N = int(input()) # 連泊数
K = int(input()) # 料金変動泊数
X = int(input()) # 通常価格
Y = int(input()) # 変動価格
if (
((1 <= N & N <= 10000) & (1 <= K & K <= 10000))
& (1 <= X & X <= 10000)
& (1 <= Y & Y <= 10000)
):
if X > Y:
price = 0
if N < K:
for i in range(1, N + 1):
price += X
if N >= K:
for i in range(1, K + 1):
price += X
for j in range(K + 1, N + 1):
price += Y
print(price)
| Statement
There is a hotel with the following accommodation fee:
* X yen (the currency of Japan) per night, for the first K nights
* Y yen per night, for the (K+1)-th and subsequent nights
Tak is staying at this hotel for N consecutive nights. Find his total
accommodation fee. | [{"input": "5\n 3\n 10000\n 9000", "output": "48000\n \n\nThe accommodation fee is as follows:\n\n * 10000 yen for the 1-st night\n * 10000 yen for the 2-nd night\n * 10000 yen for the 3-rd night\n * 9000 yen for the 4-th night\n * 9000 yen for the 5-th night\n\nThus, the total is 48000 yen.\n\n* * *"}, {"input": "2\n 3\n 10000\n 9000", "output": "20000"}] |
Print Tak's total accommodation fee.
* * * | s483526877 | Wrong Answer | p04011 | The input is given from Standard Input in the following format:
N
K
X
Y | N = [int(input()) for i in range(4)]
if N[0] >= N[1]:
for i in range(N[0]):
print(str(i + 1) + "泊目は" + str(N[2]) + "円です")
else:
for i in range(N[0]):
print(str(i + 1) + "泊目は" + str(N[2]) + "円です")
for i in range(N[1] - N[0]):
print(str(N[0] + i + 1) + "泊目は" + str(N[3]) + "円です")
| Statement
There is a hotel with the following accommodation fee:
* X yen (the currency of Japan) per night, for the first K nights
* Y yen per night, for the (K+1)-th and subsequent nights
Tak is staying at this hotel for N consecutive nights. Find his total
accommodation fee. | [{"input": "5\n 3\n 10000\n 9000", "output": "48000\n \n\nThe accommodation fee is as follows:\n\n * 10000 yen for the 1-st night\n * 10000 yen for the 2-nd night\n * 10000 yen for the 3-rd night\n * 9000 yen for the 4-th night\n * 9000 yen for the 5-th night\n\nThus, the total is 48000 yen.\n\n* * *"}, {"input": "2\n 3\n 10000\n 9000", "output": "20000"}] |
Print Tak's total accommodation fee.
* * * | s313851401 | Runtime Error | p04011 | The input is given from Standard Input in the following format:
N
K
X
Y | a = int(input())
b = int(input())
c = int(input())
d = int(input())
e = 0
for i in range(1,a+1):
if i >= b+1:
e += d
else:
e += c
print(e) | Statement
There is a hotel with the following accommodation fee:
* X yen (the currency of Japan) per night, for the first K nights
* Y yen per night, for the (K+1)-th and subsequent nights
Tak is staying at this hotel for N consecutive nights. Find his total
accommodation fee. | [{"input": "5\n 3\n 10000\n 9000", "output": "48000\n \n\nThe accommodation fee is as follows:\n\n * 10000 yen for the 1-st night\n * 10000 yen for the 2-nd night\n * 10000 yen for the 3-rd night\n * 9000 yen for the 4-th night\n * 9000 yen for the 5-th night\n\nThus, the total is 48000 yen.\n\n* * *"}, {"input": "2\n 3\n 10000\n 9000", "output": "20000"}] |
Print Tak's total accommodation fee.
* * * | s049448303 | Runtime Error | p04011 | The input is given from Standard Input in the following format:
N
K
X
Y | total_stay = int(input())
normal_stay = int(input())
normal_cost = int(input())
discounted_cost = int(input())
if total_stay <= normal_stay:
total_cost = total_stay * normal_cost
else:
total_cost = (
normal_stay * normal_cost + (total_stay - normal_stay) * discounted_cost
)
print(P)
| Statement
There is a hotel with the following accommodation fee:
* X yen (the currency of Japan) per night, for the first K nights
* Y yen per night, for the (K+1)-th and subsequent nights
Tak is staying at this hotel for N consecutive nights. Find his total
accommodation fee. | [{"input": "5\n 3\n 10000\n 9000", "output": "48000\n \n\nThe accommodation fee is as follows:\n\n * 10000 yen for the 1-st night\n * 10000 yen for the 2-nd night\n * 10000 yen for the 3-rd night\n * 9000 yen for the 4-th night\n * 9000 yen for the 5-th night\n\nThus, the total is 48000 yen.\n\n* * *"}, {"input": "2\n 3\n 10000\n 9000", "output": "20000"}] |
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