message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 276 109k | cluster float64 23 23 | __index_level_0__ int64 552 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald got a very curious hexagon for his birthday. The boy found out that all the angles of the hexagon are equal to <image>. Then he measured the length of its sides, and found that each of them is equal to an integer number of centimeters. There the properties of the hexagon ended and Gerald decided to draw on it.
He painted a few lines, parallel to the sides of the hexagon. The lines split the hexagon into regular triangles with sides of 1 centimeter. Now Gerald wonders how many triangles he has got. But there were so many of them that Gerald lost the track of his counting. Help the boy count the triangles.
Input
The first and the single line of the input contains 6 space-separated integers a1, a2, a3, a4, a5 and a6 (1 ≤ ai ≤ 1000) — the lengths of the sides of the hexagons in centimeters in the clockwise order. It is guaranteed that the hexagon with the indicated properties and the exactly such sides exists.
Output
Print a single integer — the number of triangles with the sides of one 1 centimeter, into which the hexagon is split.
Examples
Input
1 1 1 1 1 1
Output
6
Input
1 2 1 2 1 2
Output
13
Note
This is what Gerald's hexagon looks like in the first sample:
<image>
And that's what it looks like in the second sample:
<image>
Submitted Solution:
```
a, b, c = map(int, input().split()[:3])
if a > b:
a, b = b, a
s = x = (a << 1) + 1
for i in range(b - 1):
x += 2
s += x
for i in range(c):
s += x
x -= 2
print(s)
``` | instruction | 0 | 40,028 | 23 | 80,056 |
No | output | 1 | 40,028 | 23 | 80,057 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an image, that can be represented with a 2-d n by m grid of pixels. Each pixel of the image is either on or off, denoted by the characters "0" or "1", respectively. You would like to compress this image. You want to choose an integer k > 1 and split the image into k by k blocks. If n and m are not divisible by k, the image is padded with only zeros on the right and bottom so that they are divisible by k. Each pixel in each individual block must have the same value. The given image may not be compressible in its current state. Find the minimum number of pixels you need to toggle (after padding) in order for the image to be compressible for some k. More specifically, the steps are to first choose k, then the image is padded with zeros, then, we can toggle the pixels so it is compressible for this k. The image must be compressible in that state.
Input
The first line of input will contain two integers n, m (2 ≤ n, m ≤ 2 500), the dimensions of the image.
The next n lines of input will contain a binary string with exactly m characters, representing the image.
Output
Print a single integer, the minimum number of pixels needed to toggle to make the image compressible.
Example
Input
3 5
00100
10110
11001
Output
5
Note
We first choose k = 2.
The image is padded as follows:
001000
101100
110010
000000
We can toggle the image to look as follows:
001100
001100
000000
000000
We can see that this image is compressible for k = 2. | instruction | 0 | 40,112 | 23 | 80,224 |
Tags: brute force
Correct Solution:
```
from sys import stdin as fin
def check(x, y):
if x < 0 or y < 0:
return 0
else:
return table2[x][y]
n, m = [int(x) for x in fin.readline().split()]
table = fin.readlines()
fin.close()
table2 = [[None]*m for _ in range(n)]
table2[0][0] = int(table[0][0])
for __ in range(1, m):
table2[0][__] = int(table[0][__]) + table2[0][ __ -1]
for _ in range(1, n):
table2[_][0] = int(table[_][0]) + table2[_-1][0]
for __ in range(1,m):
table2[_][__] = int(table[_][__]) + table2[_-1][ __] + table2[_][ __ -1] - table2[_-1][ __-1]
list_of_primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163]
list_of_primes += [167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353]
list_of_primes += [359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569]
list_of_primes += [571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773]
list_of_primes += [787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013]
list_of_primes += [1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213]
list_of_primes += [1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429]
list_of_primes += [1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609]
list_of_primes += [1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831]
list_of_primes += [1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011, 2017, 2027, 2029, 2039, 2053]
list_of_primes += [2063, 2069, 2081, 2083, 2087, 2089, 2099, 2111, 2113, 2129, 2131, 2137, 2141, 2143, 2153, 2161, 2179, 2203, 2207, 2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273]
list_of_primes += [2281, 2287, 2293, 2297, 2309, 2311, 2333, 2339, 2341, 2347, 2351, 2357, 2371, 2377, 2381, 2383, 2389, 2393, 2399, 2411, 2417, 2423, 2437, 2441, 2447, 2459, 2467, 2473, 2477]
minimum = float('inf')
for i in [x for x in list_of_primes if x <= max(n, m) + 1]:
current = 0
j = 0
while j < n // i + (n % i != 0):
k = 0
while k < m // i + (m % i != 0):
aofi = table2[min(j*i + i - 1, n-1)][min(k*i + i -1, m-1)] - check(j*i - 1, min(k*i + i - 1, m-1)) - check(min(j*i + i - 1, n-1), k*i -1) + check(min(j*i -1, n - 1 - n%i), min(k*i -1, m - 1 - m%i))
current += min(aofi, i**2 - aofi)
k +=1
j += 1
if current < minimum:
minimum = current
print(minimum)
``` | output | 1 | 40,112 | 23 | 80,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an image, that can be represented with a 2-d n by m grid of pixels. Each pixel of the image is either on or off, denoted by the characters "0" or "1", respectively. You would like to compress this image. You want to choose an integer k > 1 and split the image into k by k blocks. If n and m are not divisible by k, the image is padded with only zeros on the right and bottom so that they are divisible by k. Each pixel in each individual block must have the same value. The given image may not be compressible in its current state. Find the minimum number of pixels you need to toggle (after padding) in order for the image to be compressible for some k. More specifically, the steps are to first choose k, then the image is padded with zeros, then, we can toggle the pixels so it is compressible for this k. The image must be compressible in that state.
Input
The first line of input will contain two integers n, m (2 ≤ n, m ≤ 2 500), the dimensions of the image.
The next n lines of input will contain a binary string with exactly m characters, representing the image.
Output
Print a single integer, the minimum number of pixels needed to toggle to make the image compressible.
Example
Input
3 5
00100
10110
11001
Output
5
Note
We first choose k = 2.
The image is padded as follows:
001000
101100
110010
000000
We can toggle the image to look as follows:
001100
001100
000000
000000
We can see that this image is compressible for k = 2.
Submitted Solution:
```
import math
q=lambda:[int(f)for f in input().split()]
n,m=q()
k=2
a=[]
c=0
for _ in range (n):
a+=[[int(f)for f in input()]]
for x in range(n%k):
a+=[[0]*m]
n+=1
for x in range(m%k):
for y in range(n):
a[y]+=[0]
m+=1
x=k
y=k
while x<n:
while y<m:
z,b=0,0
for i in range(x):
for o in range(y):
if a[i][o]==0:
z+=1
else:b+=1
if b>=z:
c+=z
else:
c+=b
y+=k
x+=2
print(c)
``` | instruction | 0 | 40,115 | 23 | 80,230 |
No | output | 1 | 40,115 | 23 | 80,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an image, that can be represented with a 2-d n by m grid of pixels. Each pixel of the image is either on or off, denoted by the characters "0" or "1", respectively. You would like to compress this image. You want to choose an integer k > 1 and split the image into k by k blocks. If n and m are not divisible by k, the image is padded with only zeros on the right and bottom so that they are divisible by k. Each pixel in each individual block must have the same value. The given image may not be compressible in its current state. Find the minimum number of pixels you need to toggle (after padding) in order for the image to be compressible for some k. More specifically, the steps are to first choose k, then the image is padded with zeros, then, we can toggle the pixels so it is compressible for this k. The image must be compressible in that state.
Input
The first line of input will contain two integers n, m (2 ≤ n, m ≤ 2 500), the dimensions of the image.
The next n lines of input will contain a binary string with exactly m characters, representing the image.
Output
Print a single integer, the minimum number of pixels needed to toggle to make the image compressible.
Example
Input
3 5
00100
10110
11001
Output
5
Note
We first choose k = 2.
The image is padded as follows:
001000
101100
110010
000000
We can toggle the image to look as follows:
001100
001100
000000
000000
We can see that this image is compressible for k = 2.
Submitted Solution:
```
q=lambda:[int(f)for f in input().split()]
n,m=q()
k=2
a=[]
c=0
for _ in range (n):
a+=[[int(f)for f in input()]]
for x in range(n%k):
a+=[[0]*m]
n+=1
for x in range(m%k):
for y in range(n):
a[y]+=[0]
x=k
y=k
while x<n:
while y<m:
z,b=0,0
for i in range(x):
for o in range(y):
if a[i][o]==0:
z+=1
else:b+=1
if b>=z:
c+=z
else:
c+=b
y+=k
x+=2
print(c)
``` | instruction | 0 | 40,117 | 23 | 80,234 |
No | output | 1 | 40,117 | 23 | 80,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an image, that can be represented with a 2-d n by m grid of pixels. Each pixel of the image is either on or off, denoted by the characters "0" or "1", respectively. You would like to compress this image. You want to choose an integer k > 1 and split the image into k by k blocks. If n and m are not divisible by k, the image is padded with only zeros on the right and bottom so that they are divisible by k. Each pixel in each individual block must have the same value. The given image may not be compressible in its current state. Find the minimum number of pixels you need to toggle (after padding) in order for the image to be compressible for some k. More specifically, the steps are to first choose k, then the image is padded with zeros, then, we can toggle the pixels so it is compressible for this k. The image must be compressible in that state.
Input
The first line of input will contain two integers n, m (2 ≤ n, m ≤ 2 500), the dimensions of the image.
The next n lines of input will contain a binary string with exactly m characters, representing the image.
Output
Print a single integer, the minimum number of pixels needed to toggle to make the image compressible.
Example
Input
3 5
00100
10110
11001
Output
5
Note
We first choose k = 2.
The image is padded as follows:
001000
101100
110010
000000
We can toggle the image to look as follows:
001100
001100
000000
000000
We can see that this image is compressible for k = 2.
Submitted Solution:
```
fl=input().split()
n=int(fl[0])
m=int(fl[1])
#on choisi k
k=2
#while n%k==0 and m%k==0:
# k=k+1
#on add des lines
lines=[]
for i in range(n):
add=m%k
lines.append(input())
lines[i]=lines[i]+"0"*add
if n%k!=0:
linessup="0"*m+"0"*(m%k)
lines.append(linessup)
nbpix=int(len(lines[0])*len(lines)/(k*k))
outp=0
for j in range(int((n+n%k)/k)):
for w in range(int((m+m%k)/k)):
counter=0
finalcounter=0
for x in range(k):
for y in range(k):
if lines[k*j+x][k*w+y]!="0":
counter=counter+1
if counter==2:
finalcounter=2
elif counter==0:
finalcounter=0
else:
finalcounter=1
outp=outp+finalcounter
print(outp)
``` | instruction | 0 | 40,118 | 23 | 80,236 |
No | output | 1 | 40,118 | 23 | 80,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of N rows and M columns. The square at the i-th row and j-th column will be denoted as (i,j). Some of the squares contain an object. All the remaining squares are empty. The state of the grid is represented by strings S_1,S_2,\cdots,S_N. The square (i,j) contains an object if S_{i,j}= `#` and is empty if S_{i,j}= `.`.
Consider placing 1 \times 2 tiles on the grid. Tiles can be placed vertically or horizontally to cover two adjacent empty squares. Tiles must not stick out of the grid, and no two different tiles may intersect. Tiles cannot occupy the square with an object.
Calculate the maximum number of tiles that can be placed and any configulation that acheives the maximum.
Constraints
* 1 \leq N \leq 100
* 1 \leq M \leq 100
* S_i is a string with length M consists of `#` and `.`.
Input
Input is given from Standard Input in the following format:
N M
S_1
S_2
\vdots
S_N
Output
On the first line, print the maximum number of tiles that can be placed.
On the next N lines, print a configulation that achieves the maximum. Precisely, output the strings t_1,t_2,\cdots,t_N constructed by the following way.
* t_i is initialized to S_i.
* For each (i,j), if there is a tile that occupies (i,j) and (i+1,j), change t_{i,j}:=`v`, t_{i+1,j}:=`^`.
* For each (i,j), if there is a tile that occupies (i,j) and (i,j+1), change t_{i,j}:=`>`, t_{i,j+1}:=`<`.
See samples for further information.
You may print any configulation that maximizes the number of tiles.
Examples
Input
3 3
#..
..#
...
Output
3
#><
vv#
^^.
Input
3 3
..
..#
...
Output
3
><
vv#
^^.
Submitted Solution:
```
class maxflow():
def __init__(self, n):
self.pos = []
self.g = [[] for _ in range(n)]
self.n = n
def add_edge(self, v, w, cap):
e1 = [w, 0, cap]
e2 = [v, e1, 0]
e1[1] = e2
self.g[v].append(e1)
self.g[w].append(e2)
def flow(self, s, t, flow_limit = -1):
if flow_limit < 0: flow_limit = 10 ** 100
def bfs():
L = [-1] * self.n
L[s] = 0
Q = [s]
while Q:
v = Q.pop()
for w, _, cap in self.g[v]:
if cap == 0 or L[w] >= 0: continue
L[w] = L[v] + 1
if w == t: return L
Q.append(w)
return L
def dfs(v, up):
if v == s: return up
res = 0
lv = L[v]
for i in range(self.it[v], len(self.g[v])):
w, rev, cap = self.g[v][i]
if lv > L[w] and rev[2] > 0:
d = dfs(w, min(up - res, rev[2]))
if d > 0:
self.g[v][i][2] += d
rev[2] -= d
res += d
if res == up: break
self.it[v] += 1
return res
flow = 0
while flow < flow_limit:
L = bfs()
if L[t] == -1: break
self.it = [0] * self.n
while flow < flow_limit:
f = dfs(t, flow_limit - flow)
if not f: break
flow += f
return flow, self.g
def min_cut(self, s):
visited = [0] * self.n
Q = [s]
while Q:
p = Q.pop()
visited[p] = 1
for e in self.g[p]:
if e[2] and visited[e[0]] == 0:
visited[e[0]] = 1
Q.append(e[0])
return visited
N, M = map(int, input().split())
mf = maxflow(N * M + 2)
A = [[a for a in input()] for _ in range(N)]
X = [[1 if a == "." else 0 for a in aa] for aa in A]
s = N * M
t = s + 1
for i in range(N):
for j in range(M):
if X[i][j]:
if (i ^ j) & 1 == 0:
mf.add_edge(s, i * M + j, 1)
else:
mf.add_edge(i * M + j, t, 1)
for i in range(N - 1):
for j in range(M):
if X[i][j] and X[i+1][j]:
a = i * M + j
b = a + M
if (i ^ j) & 1 == 0:
mf.add_edge(a, b, 1)
else:
mf.add_edge(b, a, 1)
for i in range(N):
for j in range(M - 1):
if X[i][j] and X[i][j+1]:
a = i * M + j
b = a + 1
if (i ^ j) & 1 == 0:
mf.add_edge(a, b, 1)
else:
mf.add_edge(b, a, 1)
fl, g = mf.flow(s, t)
# print("fl =", fl, g)
for v, gv in enumerate(g):
if v >= N * M: continue
for w, _, ca in gv:
if w >= N * M: continue
# if w > v: v, w = w, v
vi, vj = v // M, v % M
wi, wj = w // M, w % M
if (vi ^ vj) % 2 == 0 and ca == 0:
if v > w:
vi, vj, wi, wj = wi, wj, vi, vj
# print("v, w, ca =", v, w, ca)
if vi == wi:
A[vi][vj] = ">"
A[wi][wj] = "<"
else:
A[vi][vj] = "v"
A[wi][wj] = "^"
print(fl)
for x in A:
print("".join(map(str, x)))
``` | instruction | 0 | 40,184 | 23 | 80,368 |
Yes | output | 1 | 40,184 | 23 | 80,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of N rows and M columns. The square at the i-th row and j-th column will be denoted as (i,j). Some of the squares contain an object. All the remaining squares are empty. The state of the grid is represented by strings S_1,S_2,\cdots,S_N. The square (i,j) contains an object if S_{i,j}= `#` and is empty if S_{i,j}= `.`.
Consider placing 1 \times 2 tiles on the grid. Tiles can be placed vertically or horizontally to cover two adjacent empty squares. Tiles must not stick out of the grid, and no two different tiles may intersect. Tiles cannot occupy the square with an object.
Calculate the maximum number of tiles that can be placed and any configulation that acheives the maximum.
Constraints
* 1 \leq N \leq 100
* 1 \leq M \leq 100
* S_i is a string with length M consists of `#` and `.`.
Input
Input is given from Standard Input in the following format:
N M
S_1
S_2
\vdots
S_N
Output
On the first line, print the maximum number of tiles that can be placed.
On the next N lines, print a configulation that achieves the maximum. Precisely, output the strings t_1,t_2,\cdots,t_N constructed by the following way.
* t_i is initialized to S_i.
* For each (i,j), if there is a tile that occupies (i,j) and (i+1,j), change t_{i,j}:=`v`, t_{i+1,j}:=`^`.
* For each (i,j), if there is a tile that occupies (i,j) and (i,j+1), change t_{i,j}:=`>`, t_{i,j+1}:=`<`.
See samples for further information.
You may print any configulation that maximizes the number of tiles.
Examples
Input
3 3
#..
..#
...
Output
3
#><
vv#
^^.
Input
3 3
..
..#
...
Output
3
><
vv#
^^.
Submitted Solution:
```
from collections import deque
class MaxFlow():
def __init__(self, n):
self.n = n
self.graph = [[] for _ in range(n)]
self.pos = []
def add_edge(self, fr, to, cap):
assert 0 <= fr < self.n
assert 0 <= to < self.n
assert 0 <= cap
m = len(self.pos)
self.pos.append((fr, len(self.graph[fr])))
self.graph[fr].append([to, len(self.graph[to]), cap])
self.graph[to].append([fr, len(self.graph[fr]) - 1, 0])
return m
def get_edge(self, idx):
m = len(self.pos)
assert 0 <= idx < m
to, rev, cap = self.graph[self.pos[idx][0]][self.pos[idx][1]]
rev_to, rev_rev, rev_cap = self.graph[to][rev]
return rev_to, to, cap + rev_cap, rev_cap
def edges(self):
m = len(self.pos)
res = []
for i in range(m):
res.append(self.get_edge(i))
return res
def change_edge(self, idx, new_cap, new_flow):
m = len(self.pos)
assert 0 <= idx < m
assert 0 <= new_flow <= new_cap
to, rev, cap = self.graph[self.pos[idx][0]][self.pos[idx][1]]
self.graph[self.pos[idx][0]][self.pos[idx][1]][2] = new_cap - new_flow
self.graph[to][rev][2] = new_flow
def dfs(self, s, v, up):
if v == s: return up
res = 0
lv = self.level[v]
for i in range(self.iter[v], len(self.graph[v])):
to, rev, cap = self.graph[v][i]
if lv <= self.level[to] or self.graph[to][rev][2] == 0: continue
d = self.dfs(s, to, min(up - res, self.graph[to][rev][2]))
if d <= 0: continue
self.graph[v][i][2] += d
self.graph[to][rev][2] -= d
res += d
if res == up: break
return res
def max_flow(self, s, t):
return self.max_flow_with_limit(s, t, 2**63 - 1)
def max_flow_with_limit(self, s, t, limit):
assert 0 <= s < self.n
assert 0 <= t < self.n
flow = 0
while flow < limit:
self.level = [-1] * self.n
self.level[s] = 0
queue = deque()
queue.append(s)
while queue:
v = queue.popleft()
for to, rev, cap in self.graph[v]:
if cap == 0 or self.level[to] >= 0: continue
self.level[to] = self.level[v] + 1
if to == t: break
queue.append(to)
if self.level[t] == -1: break
self.iter = [0] * self.n
while flow < limit:
f = self.dfs(s, t, limit - flow)
if not f: break
flow += f
return flow
def min_cut(self, s):
visited = [0] * self.n
queue = deque()
queue.append(s)
while queue:
p = queue.popleft()
visited[p] = True
for to, rev, cap in self.graph[p]:
if cap and not visited[to]:
visited[to] = True
queue.append(to)
return visited
import sys
input = sys.stdin.readline
N, M = map(int, input().split())
S = [input() for _ in range(N)]
mf = MaxFlow(N * M + 2)
dx = [1, 0, -1, 0]
dy = [0, 1, 0, -1]
res = [['.' for j in range(M)] for i in range(N)]
for i in range(N):
for j in range(M):
if S[i][j] == '#':
res[i][j] = '#'
continue
if (i + j) % 2 == 0:
mf.add_edge(N * M, i * M + j, 1)
for k in range(4):
ni = i + dx[k]
nj = j + dy[k]
if 0 <= ni < N and 0 <= nj < M and S[ni][nj] == '.':
mf.add_edge(i * M + j, ni * M + nj, 1)
else:
mf.add_edge(i * M + j, N * M + 1, 1)
print(mf.max_flow(M * N, M * N + 1))
for fr, to, cap, flow in mf.edges():
if fr == N * M or to == N * M + 1 or flow == 0:
continue
si = fr // M
sj = fr % M
ti = to // M
tj = to % M
if si + 1 == ti:
res[si][sj] = 'v'
res[ti][tj] = '^'
elif sj + 1 == tj:
res[si][sj] = '>'
res[ti][tj] = '<'
elif si == ti + 1:
res[si][sj] = '^'
res[ti][tj] = 'v'
else:
res[si][sj] = '<'
res[ti][tj] = '>'
for s in res:
print(''.join(s))
``` | instruction | 0 | 40,185 | 23 | 80,370 |
Yes | output | 1 | 40,185 | 23 | 80,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of N rows and M columns. The square at the i-th row and j-th column will be denoted as (i,j). Some of the squares contain an object. All the remaining squares are empty. The state of the grid is represented by strings S_1,S_2,\cdots,S_N. The square (i,j) contains an object if S_{i,j}= `#` and is empty if S_{i,j}= `.`.
Consider placing 1 \times 2 tiles on the grid. Tiles can be placed vertically or horizontally to cover two adjacent empty squares. Tiles must not stick out of the grid, and no two different tiles may intersect. Tiles cannot occupy the square with an object.
Calculate the maximum number of tiles that can be placed and any configulation that acheives the maximum.
Constraints
* 1 \leq N \leq 100
* 1 \leq M \leq 100
* S_i is a string with length M consists of `#` and `.`.
Input
Input is given from Standard Input in the following format:
N M
S_1
S_2
\vdots
S_N
Output
On the first line, print the maximum number of tiles that can be placed.
On the next N lines, print a configulation that achieves the maximum. Precisely, output the strings t_1,t_2,\cdots,t_N constructed by the following way.
* t_i is initialized to S_i.
* For each (i,j), if there is a tile that occupies (i,j) and (i+1,j), change t_{i,j}:=`v`, t_{i+1,j}:=`^`.
* For each (i,j), if there is a tile that occupies (i,j) and (i,j+1), change t_{i,j}:=`>`, t_{i,j+1}:=`<`.
See samples for further information.
You may print any configulation that maximizes the number of tiles.
Examples
Input
3 3
#..
..#
...
Output
3
#><
vv#
^^.
Input
3 3
..
..#
...
Output
3
><
vv#
^^.
Submitted Solution:
```
class Weigthed_Digraph:
#入力定義
def __init__(self,vertex=[]):
self.vertex=set(vertex)
self.edge_number=0
self.vertex_number=len(vertex)
self.adjacent_out={v:{} for v in vertex} #出近傍(vが始点)
self.adjacent_in={v:{} for v in vertex} #入近傍(vが終点)
#頂点の追加
def add_vertex(self,*adder):
for v in adder:
if v not in self.vertex:
self.adjacent_in[v]={}
self.adjacent_out[v]={}
self.vertex_number+=1
self.vertex.add(v)
#辺の追加(更新)
def add_edge(self,From,To,weight=1):
for v in [From,To]:
if v not in self.vertex:
self.add_vertex(v)
if To not in self.adjacent_in[From]:
self.edge_number+=1
self.adjacent_out[From][To]=weight
self.adjacent_in[To][From]=weight
#辺を除く
def remove_edge(self,From,To):
for v in [From,To]:
if v not in self.vertex:
self.add_vertex(v)
if To in self.adjacent_out[From]:
del self.adjacent_out[From][To]
del self.adjacent_in[To][From]
self.edge_number-=1
#頂点を除く
def remove_vertex(self,*vertexes):
for v in vertexes:
if v in self.vertex:
self.vertex_number-=1
for u in self.adjacent_out[v]:
del self.adjacent_in[u][v]
self.edge_number-=1
del self.adjacent_out[v]
for u in self.adjacent_in[v]:
del self.adjacent_out[u][v]
self.edge_number-=1
del self.adjacent_in[v]
#Walkの追加
def add_walk(self,*walk):
pass
#Cycleの追加
def add_cycle(self,*cycle):
pass
#頂点の交換
def __vertex_swap(self,p,q):
self.vertex.sort()
#グラフに頂点が存在するか否か
def vertex_exist(self,v):
return v in self.vertex
#グラフに辺が存在するか否か
def edge_exist(self,From,To):
if not(self.vertex_exist(From) and self.vertex_exist(To)):
return False
return To in self.adjacent_out[From]
#近傍
def neighbohood(self,v):
if not self.vertex_exist(v):
return []
return list(self.adjacent[v])
#出次数
def out_degree(self,v):
if not self.vertex_exist(v):
return 0
return len(self.adjacent_out[v])
#入次数
def in_degree(self,v):
if not self.vertex_exist(v):
return 0
return len(self.adjacent_in[v])
#次数
def degree(self,v):
if not self.vertex_exist(v):
return 0
return self.out_degree(v)-self.in_degree(v)
#頂点数
def vertex_count(self):
return len(self.vertex)
#辺数
def edge_count(self):
return self.edge_number
#頂点vを含む連結成分
def connected_component(self,v):
pass
def Flow_by_Dinic(D,source,sink,Mode=False):
from collections import deque
from copy import copy
"""Source sからSink tへの最大流問題をDinic法で解く.\n
Mode:
True---各頂点の流し方も返す
False---解答のみ
(参考元:https://tjkendev.github.io/procon-library/python/max_flow/dinic.html)
"""
def bfs(s,t):
for v in D.vertex:
Level[v]=None
Q=deque([s])
Level[s]=0
while Q:
v=Q.popleft()
for w,cap,_ in H[v]:
if cap and Level[w] is None:
Level[w]=Level[v]+1
Q.append(w)
return Level[t] is not None
def dfs(v,t,f):
if v==t:
return f
for e in H[v]:
w,cap,rev=e
if cap and Level[v]<Level[w]:
d=dfs(w,t,min(f,cap))
if d:
e[1]-=d
rev[1]+=d
return d
return 0
s,t=source,sink
inf=float("inf")
flow=0
Level={v:None for v in D.vertex}
H={v:[] for v in D.vertex}
for u in D.vertex:
for v in D.adjacent_out[u]:
F=[v,D.adjacent_out[u][v],None]
F[2]=B=[u,0,F]
H[u].append(F)
H[v].append(B)
while bfs(s,t):
f=inf
while f:
f=dfs(s,t,inf)
flow+=f
if Mode:
R={}
for v in D.vertex:
R[v]={}
for S in H[v]:
w=S[0]
if w in D.adjacent_out[v]:
R[v][w]=S[-1][1]
return flow,R
else:
return flow
#================================================
N,M=map(int,input().split())
D=Weigthed_Digraph([(y,x) for x in range(1,M+1) for y in range(1,N+1)])
D.add_vertex("s","t")
S=["*"*(M+2) for _ in range(N+2)]
for i in range(1,N+1):
S[i]="*"+input()+"*"
H=[(1,0),(-1,0),(0,1),(0,-1)]
for y in range(1,N+1):
for x in range(1,M+1):
if S[y][x]!=".":
continue
if (x+y)%2:
D.add_edge("s",(y,x),1)
for (a,b) in H:
if S[y+a][x+b]==".":
D.add_edge((y,x),(y+a,x+b),1)
else:
D.add_edge((y,x),"t",1)
flow,F=Flow_by_Dinic(D,"s","t",True)
U=[[S[j][i] for i in range(M+2)] for j in range(N+2)]
for y in range(1,N+1):
for x in range(1,M+1):
for C in F[(y,x)]:
if isinstance(C,str):
continue
elif F[(y,x)][C]==0:
continue
q,p=C
if (y+1==q) and (x==p):
alpha,beta="v","^"
elif (y-1==q) and (x==p):
alpha,beta="^","v"
elif (y==q) and (x+1==p):
alpha,beta=">","<"
elif (y==q) and (x-1==p):
alpha,beta="<",">"
U[y][x]=alpha
U[q][p]=beta
print(flow)
Y=["".join(T[1:-1]) for T in U[1:-1]]
print("\n".join(Y))
``` | instruction | 0 | 40,186 | 23 | 80,372 |
Yes | output | 1 | 40,186 | 23 | 80,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of N rows and M columns. The square at the i-th row and j-th column will be denoted as (i,j). Some of the squares contain an object. All the remaining squares are empty. The state of the grid is represented by strings S_1,S_2,\cdots,S_N. The square (i,j) contains an object if S_{i,j}= `#` and is empty if S_{i,j}= `.`.
Consider placing 1 \times 2 tiles on the grid. Tiles can be placed vertically or horizontally to cover two adjacent empty squares. Tiles must not stick out of the grid, and no two different tiles may intersect. Tiles cannot occupy the square with an object.
Calculate the maximum number of tiles that can be placed and any configulation that acheives the maximum.
Constraints
* 1 \leq N \leq 100
* 1 \leq M \leq 100
* S_i is a string with length M consists of `#` and `.`.
Input
Input is given from Standard Input in the following format:
N M
S_1
S_2
\vdots
S_N
Output
On the first line, print the maximum number of tiles that can be placed.
On the next N lines, print a configulation that achieves the maximum. Precisely, output the strings t_1,t_2,\cdots,t_N constructed by the following way.
* t_i is initialized to S_i.
* For each (i,j), if there is a tile that occupies (i,j) and (i+1,j), change t_{i,j}:=`v`, t_{i+1,j}:=`^`.
* For each (i,j), if there is a tile that occupies (i,j) and (i,j+1), change t_{i,j}:=`>`, t_{i,j+1}:=`<`.
See samples for further information.
You may print any configulation that maximizes the number of tiles.
Examples
Input
3 3
#..
..#
...
Output
3
#><
vv#
^^.
Input
3 3
..
..#
...
Output
3
><
vv#
^^.
Submitted Solution:
```
class maxflow():
def __init__(self, n):
self.pos = []
self.g = [[] for _ in range(n)]
self.n = n
def add_edge(self, v, w, cap):
e1 = [w, 0, cap]
e2 = [v, e1, 0]
e1[1] = e2
self.g[v].append(e1)
self.g[w].append(e2)
def flow(self, s, t, flow_limit = -1):
if flow_limit < 0: flow_limit = 10 ** 100
it = [0] * self.n
def bfs():
L = [-1] * self.n
L[s] = 0
Q = [s]
while Q:
v = Q.pop()
for w, _, cap in self.g[v]:
if cap == 0 or L[w] >= 0: continue
L[w] = L[v] + 1
if w == t: return L
Q.append(w)
return L
def dfs(v, up):
if v == s: return up
res = 0
lv = L[v]
for i in range(it[v], len(self.g[v])):
w, rev, cap = self.g[v][i]
if lv <= L[w] or rev[2] == 0: continue
d = dfs(w, min(up - res, rev[2]))
if d <= 0: continue
self.g[v][i][2] += d
rev[2] -= d
res += d
if res == up: break
return res
flow = 0
while flow < flow_limit:
L = bfs()
if L[t] == -1: break
it = [0] * self.n
while flow < flow_limit:
f = dfs(t, flow_limit - flow)
if not f: break
flow += f
return flow, self.g
def min_cut(self, s):
visited = [0] * self.n
Q = [s]
while Q:
p = Q.pop()
visited[p] = 1
for e in self.g[p]:
if e[2] and visited[e[0]] == 0:
visited[e[0]] = 1
Q.append(e[0])
return visited
N, M = map(int, input().split())
mf = maxflow(N * M + 2)
A = [[a for a in input()] for _ in range(N)]
X = [[1 if a == "." else 0 for a in aa] for aa in A]
s = N * M
t = s + 1
for i in range(N):
for j in range(M):
if X[i][j]:
if (i ^ j) & 1 == 0:
mf.add_edge(s, i * M + j, 1)
else:
mf.add_edge(i * M + j, t, 1)
for i in range(N - 1):
for j in range(M):
if X[i][j] and X[i+1][j]:
a = i * M + j
b = a + M
if (i ^ j) & 1 == 0:
mf.add_edge(a, b, 1)
else:
mf.add_edge(b, a, 1)
for i in range(N):
for j in range(M - 1):
if X[i][j] and X[i][j+1]:
a = i * M + j
b = a + 1
if (i ^ j) & 1 == 0:
mf.add_edge(a, b, 1)
else:
mf.add_edge(b, a, 1)
fl, g = mf.flow(s, t)
# print("fl =", fl, g)
for v, gv in enumerate(g):
if v >= N * M: continue
for w, _, ca in gv:
if w >= N * M: continue
# if w > v: v, w = w, v
vi, vj = v // M, v % M
wi, wj = w // M, w % M
if (vi ^ vj) % 2 == 0 and ca == 0:
if v > w:
vi, vj, wi, wj = wi, wj, vi, vj
# print("v, w, ca =", v, w, ca)
if vi == wi:
A[vi][vj] = ">"
A[wi][wj] = "<"
else:
A[vi][vj] = "v"
A[wi][wj] = "^"
print(fl)
for x in A:
print("".join(map(str, x)))
``` | instruction | 0 | 40,187 | 23 | 80,374 |
Yes | output | 1 | 40,187 | 23 | 80,375 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of N rows and M columns. The square at the i-th row and j-th column will be denoted as (i,j). Some of the squares contain an object. All the remaining squares are empty. The state of the grid is represented by strings S_1,S_2,\cdots,S_N. The square (i,j) contains an object if S_{i,j}= `#` and is empty if S_{i,j}= `.`.
Consider placing 1 \times 2 tiles on the grid. Tiles can be placed vertically or horizontally to cover two adjacent empty squares. Tiles must not stick out of the grid, and no two different tiles may intersect. Tiles cannot occupy the square with an object.
Calculate the maximum number of tiles that can be placed and any configulation that acheives the maximum.
Constraints
* 1 \leq N \leq 100
* 1 \leq M \leq 100
* S_i is a string with length M consists of `#` and `.`.
Input
Input is given from Standard Input in the following format:
N M
S_1
S_2
\vdots
S_N
Output
On the first line, print the maximum number of tiles that can be placed.
On the next N lines, print a configulation that achieves the maximum. Precisely, output the strings t_1,t_2,\cdots,t_N constructed by the following way.
* t_i is initialized to S_i.
* For each (i,j), if there is a tile that occupies (i,j) and (i+1,j), change t_{i,j}:=`v`, t_{i+1,j}:=`^`.
* For each (i,j), if there is a tile that occupies (i,j) and (i,j+1), change t_{i,j}:=`>`, t_{i,j+1}:=`<`.
See samples for further information.
You may print any configulation that maximizes the number of tiles.
Examples
Input
3 3
#..
..#
...
Output
3
#><
vv#
^^.
Input
3 3
..
..#
...
Output
3
><
vv#
^^.
Submitted Solution:
```
import sys
from collections import deque
sys.setrecursionlimit(100000)
input = sys.stdin.readline
class Dinic:
def __init__(self, v):
self.vertice = v
self.inf = int(1e9)
self.graph = [[] for i in range(v)]
self.level = [0]*v
self.iter = [0]*v
def add_edge(self, fr, to, cap, reverse=True):
self.graph[fr].append([to, cap, len(self.graph[to])])
cap2 = cap if reverse else 0
self.graph[to].append([fr, cap2, len(self.graph[fr])-1])
def bfs(self, s):
self.level = [-1]*self.vertice
que = deque([])
self.level[s] = 0
que.append(s)
while que:
v = que.popleft()
for i in range(len(self.graph[v])):
e = self.graph[v][i]
if e[1] > 0 and self.level[e[0]] < 0:
self.level[e[0]] = self.level[v] + 1
que.append(e[0])
def dfs(self, v, t, f):
if v == t:
return f
for i in range(self.iter[v], len(self.graph[v])):
self.iter[v] = i
e = self.graph[v][i]
if e[1] > 0 and self.level[v] < self.level[e[0]]:
d = self.dfs(e[0], t, min(f, e[1]))
if d > 0:
e[1] -= d
self.graph[e[0]][e[2]][1] += d
return d
return 0
def max_flow(self, s, t):
flow = 0
while True:
self.bfs(s)
if self.level[t] < 0:
return flow
self.iter = [0]*self.vertice
f = self.dfs(s, t, self.inf)
while f:
flow += f
f = self.dfs(s, t, self.inf)
def main():
n, m = map(int, input().split())
s = [[1 if x == "#" else 0 for x in input()[:-1]] for i in range(n)]
flow = Dinic(n*m+2)
start = 0
end = n*m+1
direction = [(-1, 0), (1, 0), (0, -1), (0, 1)]
for i in range(n):
for j in range(m):
if s[i][j] == 0:
if (i+j)%2:
flow.add_edge(start, i*m+j+1, 1, reverse=False)
for x, y in direction:
ni, nj = x + i, y + j
if 0 <= ni < n and 0 <= nj < m and s[ni][nj] == 0:
p, q = min(i*m+j+1, ni*m+nj+1), max(i*m+j+1, ni*m+nj+1)
flow.add_edge(p, q, 1, reverse=False)
else:
flow.add_edge(i*m+j+1, end, 1, reverse=False)
res = flow.max_flow(start, end)
print(res)
ans = [["#" if s[i][j] else "." for j in range(n)] for i in range(n)]
flow_list = []
for i, g in enumerate(flow.graph):
for node in g:
fr, to = i, node[0]
if fr == start or fr == end or to == start or to == end:
continue
if fr > to:
continue
cap = node[1]
if cap == 0:
flow_list.append((fr-1, to-1))
for k, v in flow_list:
if m > 1 and v-k == 1:
ans[k//m][k%m] = ">"
ans[v//m][v%m] = "<"
else:
ans[k//m][k%m] = "v"
ans[v//m][v%m] = "^"
for a in ans:
print("".join(a))
if __name__ == "__main__":
main()
``` | instruction | 0 | 40,188 | 23 | 80,376 |
No | output | 1 | 40,188 | 23 | 80,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of N rows and M columns. The square at the i-th row and j-th column will be denoted as (i,j). Some of the squares contain an object. All the remaining squares are empty. The state of the grid is represented by strings S_1,S_2,\cdots,S_N. The square (i,j) contains an object if S_{i,j}= `#` and is empty if S_{i,j}= `.`.
Consider placing 1 \times 2 tiles on the grid. Tiles can be placed vertically or horizontally to cover two adjacent empty squares. Tiles must not stick out of the grid, and no two different tiles may intersect. Tiles cannot occupy the square with an object.
Calculate the maximum number of tiles that can be placed and any configulation that acheives the maximum.
Constraints
* 1 \leq N \leq 100
* 1 \leq M \leq 100
* S_i is a string with length M consists of `#` and `.`.
Input
Input is given from Standard Input in the following format:
N M
S_1
S_2
\vdots
S_N
Output
On the first line, print the maximum number of tiles that can be placed.
On the next N lines, print a configulation that achieves the maximum. Precisely, output the strings t_1,t_2,\cdots,t_N constructed by the following way.
* t_i is initialized to S_i.
* For each (i,j), if there is a tile that occupies (i,j) and (i+1,j), change t_{i,j}:=`v`, t_{i+1,j}:=`^`.
* For each (i,j), if there is a tile that occupies (i,j) and (i,j+1), change t_{i,j}:=`>`, t_{i,j+1}:=`<`.
See samples for further information.
You may print any configulation that maximizes the number of tiles.
Examples
Input
3 3
#..
..#
...
Output
3
#><
vv#
^^.
Input
3 3
..
..#
...
Output
3
><
vv#
^^.
Submitted Solution:
```
code = r"""
# distutils: language=c++
# distutils: include_dirs=[/home/contestant/.local/lib/python3.8/site-packages/numpy/core/include,/opt/atcoder-stl]
# cython: boundscheck=False
# cython: wraparound=False
from libcpp cimport bool
from libcpp.string cimport string
from libcpp.vector cimport vector
cdef extern from "<atcoder/dsu>" namespace "atcoder":
cpdef cppclass dsu:
dsu(int n)
int merge(int a, int b)
bool same(int a, int b)
int leader(int a)
int size(int a)
vector[vector[int]] groups()
cdef class dsu_:
cdef dsu *ptr
def __cinit__(self, int n): self.ptr = new dsu(n)
def __dealloc__(self): del self.ptr
cpdef int merge(self,a,b): return self.ptr.merge(a,b)
cpdef bool same(self,a,b): return self.ptr.same(a,b)
cpdef int leader(self,a): return self.ptr.leader(a)
cpdef int size(self,a): return self.ptr.size(a)
cpdef vector[vector[int]] groups(self): return self.ptr.groups()
cdef extern from "<atcoder/fenwicktree>" namespace "atcoder":
cdef cppclass fenwick_tree[T]:
fenwick_tree()
fenwick_tree(int n)
void add(int p, T x)
T sum(int l, int r)
cdef class fenwick_tree_:
cdef fenwick_tree[int] *ptr
def __cinit__(self, int n): self.ptr = new fenwick_tree[int](n)
def __dealloc__(self): del self.ptr
cpdef void add(self, int p, long x): self.ptr.add(p, x)
cpdef long sum(self, int l, int r): return self.ptr.sum(l, r)
cdef extern from "<atcoder/string>" namespace "atcoder":
vector[int] suffix_array(vector[int] s, int upper)
vector[int] suffix_array(vector[long] s)
vector[int] suffix_array(string s)
vector[int] lcp_array(vector[long] s, vector[int] sa)
vector[int] lcp_array(string s, vector[int] sa)
vector[int] z_algorithm(vector[long] s)
vector[int] z_algorithm(string s)
cpdef vector[int] suffix_array_u(vector[int] s, int upper): return suffix_array(s, upper)
cpdef vector[int] suffix_array_l(vector[long] s): return suffix_array(s)
cpdef vector[int] suffix_array_s(string s): return suffix_array(s)
cpdef vector[int] lcp_array_l(vector[long] s, vector[int] sa): return lcp_array(s, sa)
cpdef vector[int] lcp_array_s(string s, vector[int] sa): return lcp_array(s, sa)
cpdef vector[int] z_algorithm_l(vector[long] s): return z_algorithm(s)
cpdef vector[int] z_algorithm_s(string s): return z_algorithm(s)
cdef extern from "<atcoder/maxflow>" namespace "atcoder":
cdef cppclass mf_graph[Cap]:
mf_graph(int n)
int add_edge(int frm, int to, Cap cap)
Cap flow(int s, int t)
Cap flow(int s, int t, Cap flow_limit)
vector[bool] min_cut(int s)
cppclass edge:
int frm "from", to
Cap cap, flow
edge(edge &e)
edge get_edge(int i)
vector[edge] edges()
void change_edge(int i, Cap new_cap, Cap new_flow)
cdef class mf_graph_:
cdef mf_graph[int] *ptr
def __cinit__(self, int n): self.ptr = new mf_graph[int](n)
def __dealloc__(self): del self.ptr
cpdef int add_edge(self, int frm, int to, int cap): return self.ptr.add_edge(frm, to, cap)
cpdef int flow(self, int s, int t): return self.ptr.flow(s, t)
cpdef int flow_limit(self, int s, int t, int flow_limit): return self.ptr.flow(s, t, flow_limit)
cpdef vector[bool] min_cut(self, int s): return self.ptr.min_cut(s)
cpdef vector[int] get_edge(self, int i):
cdef mf_graph[int].edge *e = new mf_graph[int].edge(self.ptr.get_edge(i))
cdef vector[int] *ret_e = new vector[int]()
ret_e.push_back(e.frm)
ret_e.push_back(e.to)
ret_e.push_back(e.cap)
ret_e.push_back(e.flow)
return ret_e[0]
cpdef vector[vector[int]] edges(self):
cdef vector[mf_graph[int].edge] es = self.ptr.edges()
cdef vector[vector[int]] *ret_es = new vector[vector[int]](es.size())
for i in range(es.size()):
ret_es.at(i).push_back(es.at(i).frm)
ret_es.at(i).push_back(es.at(i).to)
ret_es.at(i).push_back(es.at(i).cap)
ret_es.at(i).push_back(es.at(i).flow)
return ret_es[0]
cpdef void change_edge(self, int i, int new_cap, int new_flow): self.ptr.change_edge(i, new_cap, new_flow)
"""
import os
try:
from atcoder import *
except:
open('atcoder.pyx', 'w').write(code)
os.system('cythonize -i -3 -b atcoder.pyx')
H,W=map(int,input().split())
S=[input() for i in range(H)]
mf=mf_graph_(H*W+2)
s,t=H*W,H*W+1
for i in range(H):
for j in range(W):
if S[i][j]=='#': continue
v=i*W+j
if (i+j)%2:
mf.add_edge(s,v,1)
else:
mf.add_edge(v,t,1);continue
if j<W-1 and S[i][j+1]=='.':
mf.add_edge(v,v+1,1)
if 0<j and S[i][j-1]=='.':
mf.add_edge(v,v-1,1)
if i<H-1 and S[i+1][j]=='.':
mf.add_edge(v,v+W,1)
if 0<i and S[i-1][j]=='.':
mf.add_edge(v,v-W,1)
print(mf.flow(s,t))
r=[[c for c in S[i]] for i in range(H)]
for frm,to,_,flow in mf.edges():
if flow==1 and frm!=s and to!=t:
i,j=frm//W,frm%W
k,l=to//W,to%W
if frm-to==1:
r[i][j],r[k][l]='<','>'
elif to-frm==1:
r[i][j],r[k][l]='>','<'
elif frm-to==W:
r[i][j],r[k][l]='^','v'
else:
r[i][j],r[k][l]='v','^'
print(*[''.join(l) for l in r],sep='\n')
``` | instruction | 0 | 40,189 | 23 | 80,378 |
No | output | 1 | 40,189 | 23 | 80,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of N rows and M columns. The square at the i-th row and j-th column will be denoted as (i,j). Some of the squares contain an object. All the remaining squares are empty. The state of the grid is represented by strings S_1,S_2,\cdots,S_N. The square (i,j) contains an object if S_{i,j}= `#` and is empty if S_{i,j}= `.`.
Consider placing 1 \times 2 tiles on the grid. Tiles can be placed vertically or horizontally to cover two adjacent empty squares. Tiles must not stick out of the grid, and no two different tiles may intersect. Tiles cannot occupy the square with an object.
Calculate the maximum number of tiles that can be placed and any configulation that acheives the maximum.
Constraints
* 1 \leq N \leq 100
* 1 \leq M \leq 100
* S_i is a string with length M consists of `#` and `.`.
Input
Input is given from Standard Input in the following format:
N M
S_1
S_2
\vdots
S_N
Output
On the first line, print the maximum number of tiles that can be placed.
On the next N lines, print a configulation that achieves the maximum. Precisely, output the strings t_1,t_2,\cdots,t_N constructed by the following way.
* t_i is initialized to S_i.
* For each (i,j), if there is a tile that occupies (i,j) and (i+1,j), change t_{i,j}:=`v`, t_{i+1,j}:=`^`.
* For each (i,j), if there is a tile that occupies (i,j) and (i,j+1), change t_{i,j}:=`>`, t_{i,j+1}:=`<`.
See samples for further information.
You may print any configulation that maximizes the number of tiles.
Examples
Input
3 3
#..
..#
...
Output
3
#><
vv#
^^.
Input
3 3
..
..#
...
Output
3
><
vv#
^^.
Submitted Solution:
```
class Weigthed_Digraph:
#入力定義
def __init__(self,vertex=[]):
self.vertex=set(vertex)
self.edge_number=0
self.vertex_number=len(vertex)
self.adjacent_out={v:{} for v in vertex} #出近傍(vが始点)
self.adjacent_in={v:{} for v in vertex} #入近傍(vが終点)
#頂点の追加
def add_vertex(self,*adder):
for v in adder:
if v not in self.vertex:
self.adjacent_in[v]={}
self.adjacent_out[v]={}
self.vertex_number+=1
self.vertex.add(v)
#辺の追加(更新)
def add_edge(self,From,To,weight=1):
for v in [From,To]:
if v not in self.vertex:
self.add_vertex(v)
if To not in self.adjacent_in[From]:
self.edge_number+=1
self.adjacent_out[From][To]=weight
self.adjacent_in[To][From]=weight
#辺を除く
def remove_edge(self,From,To):
for v in [From,To]:
if v not in self.vertex:
self.add_vertex(v)
if To in self.adjacent_out[From]:
del self.adjacent_out[From][To]
del self.adjacent_in[To][From]
self.edge_number-=1
#頂点を除く
def remove_vertex(self,*vertexes):
for v in vertexes:
if v in self.vertex:
self.vertex_number-=1
for u in self.adjacent_out[v]:
del self.adjacent_in[u][v]
self.edge_number-=1
del self.adjacent_out[v]
for u in self.adjacent_in[v]:
del self.adjacent_out[u][v]
self.edge_number-=1
del self.adjacent_in[v]
#Walkの追加
def add_walk(self,*walk):
pass
#Cycleの追加
def add_cycle(self,*cycle):
pass
#頂点の交換
def __vertex_swap(self,p,q):
self.vertex.sort()
#グラフに頂点が存在するか否か
def vertex_exist(self,v):
return v in self.vertex
#グラフに辺が存在するか否か
def edge_exist(self,From,To):
if not(self.vertex_exist(From) and self.vertex_exist(To)):
return False
return To in self.adjacent_out[From]
#近傍
def neighbohood(self,v):
if not self.vertex_exist(v):
return []
return list(self.adjacent[v])
#出次数
def out_degree(self,v):
if not self.vertex_exist(v):
return 0
return len(self.adjacent_out[v])
#入次数
def in_degree(self,v):
if not self.vertex_exist(v):
return 0
return len(self.adjacent_in[v])
#次数
def degree(self,v):
if not self.vertex_exist(v):
return 0
return self.out_degree(v)-self.in_degree(v)
#頂点数
def vertex_count(self):
return len(self.vertex)
#辺数
def edge_count(self):
return self.edge_number
#頂点vを含む連結成分
def connected_component(self,v):
pass
def Flow(D,source,sink,Mode=False):
"""Source sからSink tへの最大流問題を解く.
Mode:
True---各頂点の流し方も返す
False---解答のみ
"""
s,t=source,sink
H={v:[] for v in D.vertex}
for u in D.vertex:
for v in D.adjacent_out[u]:
F=[v,D.adjacent_out[u][v],None]
F[2]=B=[u,0,F]
H[u].append(F)
H[v].append(B)
def dfs(v,t,f):
if v==t:
return f
U[v]=True
for e in H[v]:
w,cap,inv=e
if cap and (not U[w]):
d=dfs(w,t,min(f,cap))
if d:
e[1]-=d
inv[1]+=d
return d
return 0
inf=float("inf")
f=inf
flow=0
while f:
U={v:False for v in D.vertex}
f=dfs(s,t,inf)
flow+=f
if Mode:
R={}
for v in D.vertex:
R[v]={}
for S in H[v]:
w=S[0]
if w in D.adjacent_out[v]:
R[v][w]=S[-1][1]
return flow,R
else:
return flow
#================================================
N,M=map(int,input().split())
D=Weigthed_Digraph([(y,x) for x in range(1,M+1) for y in range(1,N+1)])
D.add_vertex("s","t")
S=["*"*(M+2) for _ in range(N+2)]
for i in range(1,N+1):
S[i]="*"+input()+"*"
H=[(1,0),(-1,0),(0,1),(0,-1)]
for y in range(1,N+1):
for x in range(1,M+1):
if S[y][x]!=".":
continue
if (x+y)%2:
D.add_edge("s",(y,x),1)
for (a,b) in H:
if S[y+a][x+b]==".":
D.add_edge((y,x),(y+a,x+b),1)
else:
D.add_edge((y,x),"t",1)
flow,F=Flow(D,"s","t",True)
U=[[S[j][i] for i in range(M+2)] for j in range(N+2)]
for y in range(1,N+1):
for x in range(1,M+1):
for C in F[(y,x)]:
if isinstance(C,str):
continue
elif F[(y,x)][C]==0:
continue
q,p=C
if (y+1==q) and (x==p):
alpha,beta="v","^"
elif (y-1==q) and (x==p):
alpha,beta="^","v"
elif (y==q) and (x+1==p):
alpha,beta=">","<"
elif (y==q) and (x-1==p):
alpha,beta="<",">"
U[y][x]=alpha
U[q][p]=beta
print(flow)
Y=["".join(T[1:-1]) for T in U[1:-1]]
print("\n".join(Y))
``` | instruction | 0 | 40,190 | 23 | 80,380 |
No | output | 1 | 40,190 | 23 | 80,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid of N rows and M columns. The square at the i-th row and j-th column will be denoted as (i,j). Some of the squares contain an object. All the remaining squares are empty. The state of the grid is represented by strings S_1,S_2,\cdots,S_N. The square (i,j) contains an object if S_{i,j}= `#` and is empty if S_{i,j}= `.`.
Consider placing 1 \times 2 tiles on the grid. Tiles can be placed vertically or horizontally to cover two adjacent empty squares. Tiles must not stick out of the grid, and no two different tiles may intersect. Tiles cannot occupy the square with an object.
Calculate the maximum number of tiles that can be placed and any configulation that acheives the maximum.
Constraints
* 1 \leq N \leq 100
* 1 \leq M \leq 100
* S_i is a string with length M consists of `#` and `.`.
Input
Input is given from Standard Input in the following format:
N M
S_1
S_2
\vdots
S_N
Output
On the first line, print the maximum number of tiles that can be placed.
On the next N lines, print a configulation that achieves the maximum. Precisely, output the strings t_1,t_2,\cdots,t_N constructed by the following way.
* t_i is initialized to S_i.
* For each (i,j), if there is a tile that occupies (i,j) and (i+1,j), change t_{i,j}:=`v`, t_{i+1,j}:=`^`.
* For each (i,j), if there is a tile that occupies (i,j) and (i,j+1), change t_{i,j}:=`>`, t_{i,j+1}:=`<`.
See samples for further information.
You may print any configulation that maximizes the number of tiles.
Examples
Input
3 3
#..
..#
...
Output
3
#><
vv#
^^.
Input
3 3
..
..#
...
Output
3
><
vv#
^^.
Submitted Solution:
```
# Dinic's algorithm
from collections import deque
class Dinic:
def __init__(self, N):
self.N = N
self.G = [[] for i in range(N)]
def add_edge(self, fr, to, cap):
forward = [to, cap, None]
forward[2] = backward = [fr, 0, forward]
self.G[fr].append(forward)
self.G[to].append(backward)
def add_multi_edge(self, v1, v2, cap1, cap2):
edge1 = [v2, cap1, None]
edge1[2] = edge2 = [v1, cap2, edge1]
self.G[v1].append(edge1)
self.G[v2].append(edge2)
def bfs(self, s, t):
self.level = level = [None]*self.N
deq = deque([s])
level[s] = 0
G = self.G
while deq:
v = deq.popleft()
lv = level[v] + 1
for w, cap, _ in G[v]:
if cap and level[w] is None:
level[w] = lv
deq.append(w)
return level[t] is not None
def dfs(self, v, t, f):
if v == t:
return f
level = self.level
for e in self.it[v]:
w, cap, rev = e
if cap and level[v] < level[w]:
d = self.dfs(w, t, min(f, cap))
if d:
e[1] -= d
rev[1] += d
return d
return 0
def flow(self, s, t):
flow = 0
INF = 10**9 + 7
G = self.G
while self.bfs(s, t):
*self.it, = map(iter, self.G)
f = INF
while f:
f = self.dfs(s, t, INF)
flow += f
return flow
N,M = map(int,input().split())
S = [input() for i in range(N)]
G = Dinic(N*M+2)
for i in range(N):
for j in range(M):
node = i*M + j + 1
if (i+j)%2 and S[i][j]==".":
G.add_edge(0,node,1)
if i!=0 and S[i-1][j]==".":
G.add_edge(node,node-M,1)
if i!=N-1 and S[i+1][j]==".":
G.add_edge(node,node+M,1)
if j!=0 and S[i][j-1]==".":
G.add_edge(node,node-1,1)
if j!=M-1 and S[i][j+1]==".":
G.add_edge(node,node+1,1)
else:
G.add_edge(node,N*M+1,1)
ans = G.flow(0,N*M+1)
print(ans)
res = [[S[i][j] for j in range(M)] for i in range(N)]
for i in range(N):
for j in range(M):
node = i*M + j + 1
if (i+j)%2 and S[i][j]==".":
for w,cap,rev in G.G[node]:
if cap==0:
ni,nj = (w-1)//M,(w-1)%M
if ni==i-1:
res[ni][nj] = "v"
res[i][j] = "^"
elif ni==i+1:
res[i][j] = "v"
res[ni][nj] = "^"
elif nj==j-1:
res[ni][nj] = ">"
res[i][j] = "<"
else:
res[i][j] = ">"
res[ni][nj] = "<"
for i in range(N):
print("".join(res[i]))
``` | instruction | 0 | 40,191 | 23 | 80,382 |
No | output | 1 | 40,191 | 23 | 80,383 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}.
Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
Constraints
* C_{i,j}(1 \leq i \leq 2, 1 \leq j \leq 3) is a lowercase English letter.
Input
Input is given from Standard Input in the following format:
C_{11}C_{12}C_{13}
C_{21}C_{22}C_{23}
Output
Print `YES` if this grid remains the same when rotated 180 degrees; print `NO` otherwise.
Examples
Input
pot
top
Output
YES
Input
tab
bet
Output
NO
Input
eye
eel
Output
NO
Submitted Solution:
```
from collections import deque
n = int(input())
mins = [1000 for i in range(n+1)]
used = [0 for i in range(n+1)]
deq = deque()
deq.append((1,1))
while deq:
x,cnt = deq.popleft()
mins[x] = min(mins[x],cnt)
used[x] = 1
if used[10*x%n] == 0:
deq.appendleft((10*x%n,cnt))
if used[(x+1)%n] == 0:
deq.append(((x+1)%n,cnt+1))
print(mins[0])
``` | instruction | 0 | 40,291 | 23 | 80,582 |
No | output | 1 | 40,291 | 23 | 80,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60. | instruction | 0 | 40,479 | 23 | 80,958 |
Tags: constructive algorithms
Correct Solution:
```
import sys
import math
from collections import defaultdict,deque
n=int(sys.stdin.readline())
arr=[[0 for _ in range(n)] for x in range(n)]
count=0
num=0
for i in range(n//2):
for j in range(n//2):
arr[i][j]=num
num+=1
arr[i][j]=arr[i][j]*4+count
count=1
num=0
for i in range(n//2,n):
for j in range(n//2):
arr[i][j]=num
num+=1
arr[i][j]=arr[i][j]*4+count
count=2
num=0
for i in range(n//2):
for j in range(n//2,n):
arr[i][j]=num
arr[i][j]=arr[i][j]*4+count
num+=1
count=3
num=0
for i in range(n//2,n):
for j in range(n//2,n):
arr[i][j]=num
arr[i][j]=arr[i][j]*4+count
num+=1
for i in range(n):
print(*arr[i])
``` | output | 1 | 40,479 | 23 | 80,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60. | instruction | 0 | 40,480 | 23 | 80,960 |
Tags: constructive algorithms
Correct Solution:
```
import functools
import math
import sys
from collections import defaultdict
# input = sys.stdin.readline
rt = lambda: map(int, input().split())
ri = lambda: int(input())
rl = lambda: list(map(int, input().split()))
def main():
n = ri()
pattern = [[0, 4, 8, 12], [13, 9, 5, 1], [2, 6, 10, 14], [15, 11, 7, 3]]
a = [[0 for _ in range(n)] for _ in range(n)]
blocks = n*n//16
posX = 0
posY = 0
for x in range(blocks):
for i in range(4):
for j in range(4):
a[posX+i][posY+j] = pattern[i][j]+x*16
posX += 4
if posX >= n:
posX = 0
posY += 4
for row in a:
print(' '.join(map(str, row)))
if __name__ == '__main__':
main()
``` | output | 1 | 40,480 | 23 | 80,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60. | instruction | 0 | 40,481 | 23 | 80,962 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
x = 0
for q in range(0, n, 4):
a = [[], [], [], []]
for q1 in range(0, n, 4):
for q2 in range(4):
for q3 in range(4):
a[q2].append(x+q2*4+q3)
x += 16
for q1 in a:
print(*q1)
``` | output | 1 | 40,481 | 23 | 80,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60. | instruction | 0 | 40,482 | 23 | 80,964 |
Tags: constructive algorithms
Correct Solution:
```
# Contest: Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2) (https://codeforces.com/contest/1208)
# Problem: C: Magic Grid (https://codeforces.com/contest/1208/problem/C)
def rint():
return int(input())
def rints():
return list(map(int, input().split()))
n = rint()
m = [[0] * n for _ in range(n)]
for y in range(0, n, 2):
for x in range(0, n, 2):
b = y * n + 2 * x
m[y][x] = b
m[y][x + 1] = b + 1
m[y + 1][x] = b + 3
m[y + 1][x + 1] = b + 2
for row in m:
print(' '.join(map(str, row)))
``` | output | 1 | 40,482 | 23 | 80,965 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60. | instruction | 0 | 40,483 | 23 | 80,966 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
t = [[0]*n for _ in range(n)]
n2 = n//2
c = 0
for i in range(n2):
for j in range(n2):
t[i][j] = c
t[i+n2][j] = c+1
t[i][j+n2] = c+2
t[i+n2][j+n2] = c+3
c += 4
for r in t:
print(' '.join((str(v) for v in r)))
``` | output | 1 | 40,483 | 23 | 80,967 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60. | instruction | 0 | 40,484 | 23 | 80,968 |
Tags: constructive algorithms
Correct Solution:
```
from sys import stdin,stdout
n=int(stdin.readline().strip())
x=0
y=0
acum=0
mt=[[0 for i in range(n)] for j in range(n)]
for i in range((n//4)*(n//4)):
for j in range(4):
for k in range(4):
mt[x+j][y+k]=acum
acum+=1
y+=4
if(y==n):
y=0
x+=4
for i in range(n):
for j in range(n):
stdout.write("%d " % mt[i][j])
stdout.write("\n")
``` | output | 1 | 40,484 | 23 | 80,969 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60. | instruction | 0 | 40,485 | 23 | 80,970 |
Tags: constructive algorithms
Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().strip()
arr = []
n = int(input())
for i in range(n//4):
cur = n*4*i
for j in range(4):
arr.append([])
for k in range(n):
arr[-1].append(cur+j+4*k)
for i in arr:
for j in i:
print(j, end=' ')
print()
``` | output | 1 | 40,485 | 23 | 80,971 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60. | instruction | 0 | 40,486 | 23 | 80,972 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
ans = []
for i in range(n):
ans.append([0 for i in range(n)])
c = 0
for k in range(n//4):
for i in range(n):
for j in range(4):
ans[i][k*4 + j] = c
c+=1
for i in ans:
print(*i)
``` | output | 1 | 40,486 | 23 | 80,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60.
Submitted Solution:
```
num = int(input())
ans = [[0] * num for _ in range(num)]
for i in range(num):
base = i // 2 * num * 2 + i % 2
for j in range(num):
ans[i][j] = base + j * 2
for line in ans:
print(" ".join([str(item) for item in line]))
``` | instruction | 0 | 40,487 | 23 | 80,974 |
Yes | output | 1 | 40,487 | 23 | 80,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60.
Submitted Solution:
```
n=int(input())//2
a=[[0]*(2*n) for i in range(2*n)]
num=0
for i in range(n):
for j in range(n):
a[i][j]=a[i+n][j]=a[i][j+n]=a[i+n][j+n]=num*4
a[i][j]+=1
a[i+n][j]+=2
a[i][j+n]+=3
num+=1
for i in range(n*2):
print(*a[i])
``` | instruction | 0 | 40,488 | 23 | 80,976 |
Yes | output | 1 | 40,488 | 23 | 80,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60.
Submitted Solution:
```
n = int(input())
gap = 4*n
a = [[-1] * n for _ in range(n)]
for j in range(n):
base = gap * (j // 4) + j % 4
for i in range(n):
a[i][j] = str(base + i * 4)
for i in range(n):
print(' '.join(a[i]))
``` | instruction | 0 | 40,489 | 23 | 80,978 |
Yes | output | 1 | 40,489 | 23 | 80,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60.
Submitted Solution:
```
'''input
12
'''
import sys
def ri(flag=0):
if flag==0:
return [int(i) for i in sys.stdin.readline().split()]
else:
return int(sys.stdin.readline())
n=ri(1)
mat=[[0 for i in range(n)] for j in range(n)]
bits=[0 for i in range(21)]
# for i in range(n*n):
# for j in range(21):
# if(i&(1<<j)):
# bits[j]+=1
cnt=0
for i in range(0,n,4):
for j in range(0,n,4):
for p in range(4):
for q in range(4):
mat[i+p][j+q]=cnt
cnt+=1
for i in mat:
print(*i)
size = n*n
# for i in range(20):
# if i%2==0:
# vis = dd(int)
# for j in range(n):
# for k in range(n):
# if bits[i]>=4 and j+1<n and k+1<n and vis[j,k]==0:
# vis[j,k]=1
# vis[j+1,k]=1
# vis[j,k+1]=1
# vis[j+1,j+1]=1
# mat[j][k]+= (1<<i)
# mat[j+1][k]+= (1<<i)
# mat[j][k+1]+= (1<<i)
# mat[j+1][k+1]+= (1<<i)
# bits[i]-=4
``` | instruction | 0 | 40,490 | 23 | 80,980 |
Yes | output | 1 | 40,490 | 23 | 80,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60.
Submitted Solution:
```
n = int(input())
for i in range(n):
print(" ".join(map(str, [k for k in range(n*i, n*i+n)])))
``` | instruction | 0 | 40,491 | 23 | 80,982 |
No | output | 1 | 40,491 | 23 | 80,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60.
Submitted Solution:
```
#########################################################################################################\
#########################################################################################################
###################################The_Apurv_Rathore#####################################################
#########################################################################################################
#########################################################################################################
import sys,os,io
from sys import stdin
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop,heapify
from bisect import bisect_left , bisect_right
import math
# sys.setrecursionlimit(100000)
alphabets = list('abcdefghijklmnopqrstuvwxyz')
def isPrime(x):
for i in range(2,x):
if i*i>x:
break
if (x%i==0):
return False
return True
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
def power(x, y, p) :
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
def si():
return input()
def divideCeil(n,x):
if (n%x==0):
return n//x
return n//x+1
def ii():
return int(input())
def li():
return list(map(int,input().split()))
#__________________________TEMPLATE__________________OVER_______________________________________________________
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
else:
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def solve():
n = ii()
l = [[0 for i in range(4)] for j in range(4)]
c = 0
f = 0
for i in range(4):
if f==0:
f = 1
for j in range(4):
l[i][j]=c
c+=1
else:
f = 0
for j in range(4-1,-1,-1):
l[i][j]=c
c+=1
ans = [[0 for i in range(n)] for j in range(n)]
i = 0
j = 0
prev = 0
while(i<n):
# print(i,j,n)
# for ie in ans:
# print(*ie)
for iii in range(i,i+4):
for jjj in range(j,j+4):
# print(iii,jjj)
ans[iii][jjj]=l[iii-i][jjj-j]+prev
prev = ans[i+3][j+3]
j+=4
if j==n:
j = 0
i+=4
for i in ans:
print(*i)
t = 1
# t = int(input())
for _ in range(t):
solve()
``` | instruction | 0 | 40,492 | 23 | 80,984 |
No | output | 1 | 40,492 | 23 | 80,985 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60.
Submitted Solution:
```
n = int(input())
j = 0
for i in range(n):
row = []
s = ''
for j in range(j, j + n):
s += str(j) + ' '
j += 1
print(s)
``` | instruction | 0 | 40,493 | 23 | 80,986 |
No | output | 1 | 40,493 | 23 | 80,987 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions.
* All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once.
* [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column.
You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.
Input
The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.
Output
Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.
If there are multiple answers, print any. We can show that an answer always exists.
Examples
Input
4
Output
8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14
Input
8
Output
19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61
Note
In the first example, XOR of each row and each column is 13.
In the second example, XOR of each row and each column is 60.
Submitted Solution:
```
class MagicGrid:
def __init__(self):
self.N = 0
def computeMagicGrid(self):
self.N = int(input())
for x in range(0,self.N*self.N,4):
print(x,x+1,x+2,x+3)
mg = MagicGrid()
mg.computeMagicGrid()
``` | instruction | 0 | 40,494 | 23 | 80,988 |
No | output | 1 | 40,494 | 23 | 80,989 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image>
Submitted Solution:
```
# 20200118 19:26 ~ 19:47 ~ 20:20
n = int(input())
arr = list(map(int, input().split()))
ans_1 = 0
ans_2 = 0
for i in range(n):
if i%2 == 1:
ans_1+=int(arr[i]/2) + arr[i]%2
ans_2+=int(arr[i]/2)
else:
ans_1+=int(arr[i]/2)
ans_2+=int(arr[i]/2) + arr[i]%2
print(min(ans_1, ans_2))
``` | instruction | 0 | 40,519 | 23 | 81,038 |
Yes | output | 1 | 40,519 | 23 | 81,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image>
Submitted Solution:
```
import sys
import math
input = sys.stdin.readline
n=int(input())
arr=list(map(int,input().split()))
b=0
w=0
for i in range(n):
if i%2==0:
b+=math.ceil(arr[i]/2)
w+=arr[i]//2
else:
w+=math.ceil(arr[i]/2)
b+=arr[i]//2
print(min(b,w))
``` | instruction | 0 | 40,520 | 23 | 81,040 |
Yes | output | 1 | 40,520 | 23 | 81,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image>
Submitted Solution:
```
n = int(input())
L = list(map(int,input().split()))
col = "white"
w = 0
b = 0
for i in range(n) :
if col == "white" :
w += ( L[i] + 1 )//2
b += L[i]//2
col = "black"
else :
b += (L[i] + 1) // 2
w += L[i] // 2
col = "white"
print(min(w,b))
``` | instruction | 0 | 40,521 | 23 | 81,042 |
Yes | output | 1 | 40,521 | 23 | 81,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image>
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sun Dec 22 00:32:52 2019
@author: sj
"""
n=int(input())
a=input().split(" ")
for i in range(n):
a[i]=int(a[i])
w=a[0]//2
b=a[0]-w
c=1
def no(x):
if x==1:
return 0
return 1
for i in range(1,len(a)):
if a[i-1]%2==0:
if a[i]%2==0:
w+=a[i]//2
b+=a[i]//2
c=no(c)
else:
if c==1:
w+=a[i]//2
b+=a[i]-a[i]//2
else:
b+=a[i]//2
w+=a[i]-a[i]//2
else:
if a[i]%2==0:
w+=a[i]//2
b+=a[i]//2
else:
if c==1:
b+=a[i]//2
w+=a[i]-a[i]//2
else:
w+=a[i]//2
b+=a[i]-a[i]//2
c=no(c)
print(min(b,w))
``` | instruction | 0 | 40,522 | 23 | 81,044 |
Yes | output | 1 | 40,522 | 23 | 81,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image>
Submitted Solution:
```
n = input()
L = map(int, input().split())
total = 0
gaps = 0
prev_was_odd = False
for x in L:
total += x
if x % 2 == 0:
prev_was_odd = False
else:
if prev_was_odd:
gaps -= 1
prev_was_odd = False
else:
gaps += 1
prev_was_odd = True
print((total - gaps) // 2)
``` | instruction | 0 | 40,523 | 23 | 81,046 |
No | output | 1 | 40,523 | 23 | 81,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image>
Submitted Solution:
```
n = int(input())
A = list(map(int, input().split()))
ans = 0
flag = False
for x in A:
ans += x // 2
if x & 1:
if flag:
ans += 1
flag = flag ^ True
else:
flag = False
print(ans)
``` | instruction | 0 | 40,524 | 23 | 81,048 |
No | output | 1 | 40,524 | 23 | 81,049 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image>
Submitted Solution:
```
from sys import stdin, gettrace
if not gettrace():
def input():
return next(stdin)[:-1]
def main():
n = int(input())
aa = [int(a) for a in input().split()]
count = 0
rem = 0
for a in aa:
count += a//2
if rem == 1 and a%2 == 1:
count +=1
rem = 0
else:
rem = count%2
print(count)
if __name__ == "__main__":
main()
``` | instruction | 0 | 40,525 | 23 | 81,050 |
No | output | 1 | 40,525 | 23 | 81,051 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image>
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
ans=0
for i in range(n):
ans+=a[i]//2
if i!=0:
if a[i]%2==1 and a[i-1]%2==1:
ans+=1
a[i]-=1
print(ans)
``` | instruction | 0 | 40,526 | 23 | 81,052 |
No | output | 1 | 40,526 | 23 | 81,053 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:
* y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
* y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
* x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Output
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
Examples
Input
4
1 2
1 2
1 0
0 1
0 2
Output
NO
Input
2
1 3
1 0
-1 3
Output
YES
Input
2
1 3
1 0
0 2
Output
YES
Input
2
1 3
1 0
0 3
Output
NO
Note
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
<image> | instruction | 0 | 40,796 | 23 | 81,592 |
Tags: geometry, sortings
Correct Solution:
```
n=int(input())
X,Y=input().split()
X,Y=float(X)+1e-10,float(Y)-1e-10
L=[list(map(float,input().split())) for _ in range(n)]
print('NO' if [i for x,i in sorted((k*X+b,i) for i,(k,b) in enumerate(L))] == [i for x,i in sorted((k*Y+b,i) for i,(k,b) in enumerate(L))] else 'YES')
``` | output | 1 | 40,796 | 23 | 81,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:
* y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
* y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
* x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Output
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
Examples
Input
4
1 2
1 2
1 0
0 1
0 2
Output
NO
Input
2
1 3
1 0
-1 3
Output
YES
Input
2
1 3
1 0
0 2
Output
YES
Input
2
1 3
1 0
0 3
Output
NO
Note
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
<image> | instruction | 0 | 40,797 | 23 | 81,594 |
Tags: geometry, sortings
Correct Solution:
```
#!/usr/bin/env python
# coding: utf-8
# In[33]:
ertekek=[]
n=int(input())
x1, x2 = [int(i) for i in input().split()]
for i in range(n):
k, b = [int(i) for i in input().split()]
bal=k*x1+b
jobb=k*x2+b
ertekek.append([bal,jobb])
ertekek.sort()
szamlalo=0
for i in range(len(ertekek)-1):
if ertekek[i][0]<ertekek[i+1][0] and ertekek[i][1]>ertekek[i+1][1]:
szamlalo+=1
if szamlalo>0:
print("YES")
else:
print("NO")
# In[ ]:
``` | output | 1 | 40,797 | 23 | 81,595 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:
* y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
* y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
* x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Output
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
Examples
Input
4
1 2
1 2
1 0
0 1
0 2
Output
NO
Input
2
1 3
1 0
-1 3
Output
YES
Input
2
1 3
1 0
0 2
Output
YES
Input
2
1 3
1 0
0 3
Output
NO
Note
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
<image> | instruction | 0 | 40,798 | 23 | 81,596 |
Tags: geometry, sortings
Correct Solution:
```
n = int(input())
x1, x2 = map(int, input().split())
ps = [map(int, input().split()) for _ in range(n)]
ys = [0] * n
for i, (k, b) in enumerate(ps):
ys[i] = k * x1 + b, k * x2 + b
ys.sort()
max_y = -float('inf')
ok = False
for y1, y2 in ys:
if y2 < max_y:
ok = True
break
max_y = max(max_y, y2)
print('YES' if ok else 'NO')
``` | output | 1 | 40,798 | 23 | 81,597 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:
* y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
* y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
* x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Output
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
Examples
Input
4
1 2
1 2
1 0
0 1
0 2
Output
NO
Input
2
1 3
1 0
-1 3
Output
YES
Input
2
1 3
1 0
0 2
Output
YES
Input
2
1 3
1 0
0 3
Output
NO
Note
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
<image> | instruction | 0 | 40,799 | 23 | 81,598 |
Tags: geometry, sortings
Correct Solution:
```
# Author : nitish420 --------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# mod=10**9+7
# sys.setrecursionlimit(10**6)
# mxm=sys.maxsize
# from functools import lru_cache
def main():
n=int(input())
x,y=map(int,input().split())
arr=[]
for _ in range(n):
k,b=map(int,input().split())
arr.append([k*x+b,k*y+b])
arr.sort()
ans='NO'
for i in range(n-1):
a,b=arr[i],arr[i+1]
if a[0]<b[0] and a[1]>b[1]:
ans="YES"
break
print(ans)
# 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()
``` | output | 1 | 40,799 | 23 | 81,599 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:
* y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
* y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
* x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Output
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
Examples
Input
4
1 2
1 2
1 0
0 1
0 2
Output
NO
Input
2
1 3
1 0
-1 3
Output
YES
Input
2
1 3
1 0
0 2
Output
YES
Input
2
1 3
1 0
0 3
Output
NO
Note
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
<image> | instruction | 0 | 40,800 | 23 | 81,600 |
Tags: geometry, sortings
Correct Solution:
```
n = int(input())
x1, x2 = map(int, input().split())
points = [list(map(int, input().split())) for _ in range(n)]
one = list()
two = list()
# y = 1 * x + 2
# at x = 1, y = 3
# at x = 2, y = 4
for i in range(n):
k, b = points[i]
one.append((k * (x1 + 1e-10) + b, i))
two.append((k * (x2 - 1e-10) + b, i))
one.sort()
two.sort()
overlap = False
for i in range(n):
if one[i][1] != two[i][1]:
overlap = True
break
if overlap:
print("YES")
else:
print("NO")
``` | output | 1 | 40,800 | 23 | 81,601 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:
* y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
* y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
* x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Output
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
Examples
Input
4
1 2
1 2
1 0
0 1
0 2
Output
NO
Input
2
1 3
1 0
-1 3
Output
YES
Input
2
1 3
1 0
0 2
Output
YES
Input
2
1 3
1 0
0 3
Output
NO
Note
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
<image> | instruction | 0 | 40,801 | 23 | 81,602 |
Tags: geometry, sortings
Correct Solution:
```
n = int(input())
x1, x2 = map(float, input().split())
x1 += 1e-10
x2 -= 1e-10
lines = [tuple(map(float, input().split())) for _ in range(n)]
l1 = sorted((k * x1 + b, i) for i, (k, b) in enumerate(lines))
l2 = sorted((k * x2 + b, i) for i, (k, b) in enumerate(lines))
if [i for x, i in l1] == [i for x, i in l2]:
print('NO')
else:
print('YES')
``` | output | 1 | 40,801 | 23 | 81,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:
* y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
* y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
* x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Output
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
Examples
Input
4
1 2
1 2
1 0
0 1
0 2
Output
NO
Input
2
1 3
1 0
-1 3
Output
YES
Input
2
1 3
1 0
0 2
Output
YES
Input
2
1 3
1 0
0 3
Output
NO
Note
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
<image> | instruction | 0 | 40,802 | 23 | 81,604 |
Tags: geometry, sortings
Correct Solution:
```
def equation(k, x, b):
return k * x + b
num = int(input())
ans = []
x1, x2 = map(int, input().split())
for i in range(0,num):
k, b = map(int, input().split())
ans.append((equation(k, x1, b), equation(k, x2, b)))
ans.sort()
for i in range(1, num):
if (ans[i][0] > ans[i - 1][0] and ans[i][1] < ans[i - 1][1]) or (ans[i - 1][0] < ans[i][0] and ans[i - 1][1] > ans[i][1]):
print("YES")
exit()
print("NO")
``` | output | 1 | 40,802 | 23 | 81,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:
* y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
* y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
* x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Output
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
Examples
Input
4
1 2
1 2
1 0
0 1
0 2
Output
NO
Input
2
1 3
1 0
-1 3
Output
YES
Input
2
1 3
1 0
0 2
Output
YES
Input
2
1 3
1 0
0 3
Output
NO
Note
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
<image> | instruction | 0 | 40,803 | 23 | 81,606 |
Tags: geometry, sortings
Correct Solution:
```
import sys
#sys.stdin = open("input.txt")
#sys.stdout = open("output.txt", "w")
n = int(input())
k = []
b = []
x1, x2 = (int(i) for i in input().split())
for i in range(n):
k1, b1 = (int(j) for j in input().split())
k.append(k1)
b.append(b1)
zn = [(k[i]*x1 + b[i], k[i]*x2 + b[i]) for i in range(n)]
#print(zn)
zn.sort()
found = False;
for i in range(n - 1):
if zn[i][1] > zn[i+1][1]:
found = True
break
if found:
print("YES")
else:
print("NO")
``` | output | 1 | 40,803 | 23 | 81,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:
* y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
* y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
* x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Output
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
Examples
Input
4
1 2
1 2
1 0
0 1
0 2
Output
NO
Input
2
1 3
1 0
-1 3
Output
YES
Input
2
1 3
1 0
0 2
Output
YES
Input
2
1 3
1 0
0 3
Output
NO
Note
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
<image>
Submitted Solution:
```
n = int(input())
b = [0 for i in range(n)]
k = [0 for i in range(n)]
x1, x2 = (int(x) for x in input().split())
for i in range(n):
k[i], b[i] = (int(x) for x in input().split())
y1 = [ x1 * k[i] + b[i] for i in range(n)]
y2 = [ x2 * k[i] + b[i] for i in range(n)]
y3 = [ (y1[i], y2[i], i) for i in range(n)]
y3.sort()
r1 = [0 for i in range(n)]
for i in range(n):
r1[y3[i][-1]] = i
y4 = [ (y2[i], r1[i], i) for i in range(n)]
y4.sort()
r2 = [0 for i in range(n)]
for i in range(n):
r2[y4[i][-1]] = i
ans = 0
for i in range(n):
if r1[i] < r2[i]:
ans = 1
break
#print(y1)
#print(y2)
#print(y3)
#print(y4)
#print(r1)
#print(r2)
if ans:
print("Yes")
else:
print("No")
``` | instruction | 0 | 40,804 | 23 | 81,608 |
Yes | output | 1 | 40,804 | 23 | 81,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:
* y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
* y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
* x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Output
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
Examples
Input
4
1 2
1 2
1 0
0 1
0 2
Output
NO
Input
2
1 3
1 0
-1 3
Output
YES
Input
2
1 3
1 0
0 2
Output
YES
Input
2
1 3
1 0
0 3
Output
NO
Note
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
<image>
Submitted Solution:
```
def main():
n = int(input())
x1, x2 = map(int, input().split())
res = []
for _ in range(n):
k, b = map(int, input().split())
res.append((k * x1 + b, k * x2 + b))
res.sort()
_, a = res[0]
for _, b in res:
if a > b:
print("YES")
return
a = b
print("NO")
if __name__ == '__main__':
main()
``` | instruction | 0 | 40,805 | 23 | 81,610 |
Yes | output | 1 | 40,805 | 23 | 81,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:
* y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
* y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
* x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Output
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
Examples
Input
4
1 2
1 2
1 0
0 1
0 2
Output
NO
Input
2
1 3
1 0
-1 3
Output
YES
Input
2
1 3
1 0
0 2
Output
YES
Input
2
1 3
1 0
0 3
Output
NO
Note
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
<image>
Submitted Solution:
```
n = int(input())
x1, x2 = map(int, input().split())
x1 += 10**-8
x2 -= 10**-8
lines = []
for i in range(n):
lines.append(list(map(int, input().split())))
ord_left = []
ord_right = []
for i in range(n):
ord_left.append(lines[i][0] * x1 + lines[i][1])
ord_right.append(lines[i][0] * x2 + lines[i][1])
enum_l = list(range(n))
enum_r = list(range(n))
enum_l.sort(key=lambda ord: ord_left[ord])
enum_r.sort(key=lambda ord: ord_right[ord])
if enum_l == enum_r:
print("NO")
else:
print("YES")
# for i in range(len(ord_right)):
# line = ord_left[i][0]
``` | instruction | 0 | 40,806 | 23 | 81,612 |
Yes | output | 1 | 40,806 | 23 | 81,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The teacher gave Anton a large geometry homework, but he didn't do it (as usual) as he participated in a regular round on Codeforces. In the task he was given a set of n lines defined by the equations y = ki·x + bi. It was necessary to determine whether there is at least one point of intersection of two of these lines, that lays strictly inside the strip between x1 < x2. In other words, is it true that there are 1 ≤ i < j ≤ n and x', y', such that:
* y' = ki * x' + bi, that is, point (x', y') belongs to the line number i;
* y' = kj * x' + bj, that is, point (x', y') belongs to the line number j;
* x1 < x' < x2, that is, point (x', y') lies inside the strip bounded by x1 < x2.
You can't leave Anton in trouble, can you? Write a program that solves the given task.
Input
The first line of the input contains an integer n (2 ≤ n ≤ 100 000) — the number of lines in the task given to Anton. The second line contains integers x1 and x2 ( - 1 000 000 ≤ x1 < x2 ≤ 1 000 000) defining the strip inside which you need to find a point of intersection of at least two lines.
The following n lines contain integers ki, bi ( - 1 000 000 ≤ ki, bi ≤ 1 000 000) — the descriptions of the lines. It is guaranteed that all lines are pairwise distinct, that is, for any two i ≠ j it is true that either ki ≠ kj, or bi ≠ bj.
Output
Print "Yes" (without quotes), if there is at least one intersection of two distinct lines, located strictly inside the strip. Otherwise print "No" (without quotes).
Examples
Input
4
1 2
1 2
1 0
0 1
0 2
Output
NO
Input
2
1 3
1 0
-1 3
Output
YES
Input
2
1 3
1 0
0 2
Output
YES
Input
2
1 3
1 0
0 3
Output
NO
Note
In the first sample there are intersections located on the border of the strip, but there are no intersections located strictly inside it.
<image>
Submitted Solution:
```
n = int(input())
x, y = map(int,input().split())
u, v = [], []
x += 0.0000001
y -= 0.0000001
for i in range(n):
k, m = map(int, input().split())
u += [(k*x+m, i)]
v += [(k*y+m, i)]
u, v = sorted(u), sorted(v)
for i in range(n):
if u[i][1] != v[i][1]:
print('YES')
break
else:
print('NO')
``` | instruction | 0 | 40,807 | 23 | 81,614 |
Yes | output | 1 | 40,807 | 23 | 81,615 |
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