message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got an n × m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 ≤ n, m, x, y ≤ 1000; x ≤ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
Submitted Solution:
```
n, m, x, y = [int(t) for t in input().split()]
matrix = []
CostN =[0] * m
costB =[0] * m
for i in range(n):
aux = input()
for i in range(m):
if aux[i] == ".":
costB[i] += 1
else:
CostN[i] += 1
for i in range(m):
matrix.append([-1, -1])
costos = [CostN, costB]
def dp(columna, opcion, cont):
if columna == m-1:
return costos[opcion][columna]
if matrix[columna][opcion] != -1:
return matrix[columna][opcion]
acumulado = 0
while cont < x and columna + cont < m:
acumulado += costos[opcion][columna + cont]
cont += 1
if cont < x:
matrix[columna][opcion] = acumulado
return acumulado
posibles = []
minimo = 10**17
while cont <= y:
if columna + cont < m:
m2 =dp(columna+cont, 1-opcion, 0) + acumulado
if m2 < minimo:
minimo = m2
acumulado += costos[opcion][columna + cont]
cont+= 1
if columna + cont == m -1:
m2=costos[opcion][columna + cont]+acumulado
if m2 < minimo:
minimo = m2
break
else:
break
matrix[columna][opcion] = minimo
return matrix[columna][opcion]
print(min([dp(0,0,0),dp(0,1,0)]))
# 1500673518150
``` | instruction | 0 | 37,117 | 7 | 74,234 |
No | output | 1 | 37,117 | 7 | 74,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've got an n × m pixel picture. Each pixel can be white or black. Your task is to change the colors of as few pixels as possible to obtain a barcode picture.
A picture is a barcode if the following conditions are fulfilled:
* All pixels in each column are of the same color.
* The width of each monochrome vertical line is at least x and at most y pixels. In other words, if we group all neighbouring columns of the pixels with equal color, the size of each group can not be less than x or greater than y.
Input
The first line contains four space-separated integers n, m, x and y (1 ≤ n, m, x, y ≤ 1000; x ≤ y).
Then follow n lines, describing the original image. Each of these lines contains exactly m characters. Character "." represents a white pixel and "#" represents a black pixel. The picture description doesn't have any other characters besides "." and "#".
Output
In the first line print the minimum number of pixels to repaint. It is guaranteed that the answer exists.
Examples
Input
6 5 1 2
##.#.
.###.
###..
#...#
.##.#
###..
Output
11
Input
2 5 1 1
#####
.....
Output
5
Note
In the first test sample the picture after changing some colors can looks as follows:
.##..
.##..
.##..
.##..
.##..
.##..
In the second test sample the picture after changing some colors can looks as follows:
.#.#.
.#.#.
Submitted Solution:
```
def mem(ind,cnt,f):
if(ind == m):
if(cnt >= x): return 0
else: return 1e18
if(dp[ind][cnt][f] != -1): return dp[ind][cnt][f]
ans = 1e18
if(f == 0):
val = n-ct[ind]
else:
val = ct[ind]
if(cnt >= x):
ans = min(ans,mem(ind+1,1,f^1)+val)
if(cnt < y):
ans = min(ans,mem(ind+1,cnt+1,f)+val)
dp[ind][cnt][f] = ans
return dp[ind][cnt][f]
n,m,x,y = map(int,input().split())
s = []
for i in range(n):
s.append(input())
dp = [[[-1,-1] for i in range(y+1)] for j in range(m)]
ct = [0 for i in range(m)]
for i in range(n):
for j in range(m):
if(s[i][j] == '.'):
ct[j] += 1
print(min(mem(0,0,0),mem(0,0,1)))
``` | instruction | 0 | 37,118 | 7 | 74,236 |
No | output | 1 | 37,118 | 7 | 74,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Even polar bears feel cold when lying on the ice. Therefore, a polar bear Alice is going to make a carpet. The carpet can be viewed as a grid with height h and width w. Then the grid is divided into h × w squares. Alice is going to assign one of k different colors to each square. The colors are numbered from 1 to k. She may choose not to use all of the colors.
However, there are some restrictions. For every two adjacent squares (squares that shares an edge) x and y, there is a color constraint in one of the forms:
* color(x) = color(y), or
* color(x) ≠ color(y).
Example of the color constraints:
<image>
Ideally, Alice wants to satisfy all color constraints. But again, life in the Arctic is hard. It is not always possible to satisfy all color constraints. Fortunately, she will still be happy if at least <image> of the color constraints are satisfied.
If she has 4 colors she can color the carpet in the following way:
<image>
And she is happy because <image> of the color constraints are satisfied, and <image>. Your task is to help her color the carpet.
Input
The first line contains three integers h, w, k (2 ≤ h, w ≤ 1000, 1 ≤ k ≤ w·h).
The next 2h - 1 lines describe the color constraints from top to bottom, left to right. They contain w - 1, w, w - 1, w, ..., w - 1 characters respectively. Each color constraint is represented by a character "E" or "N", where "E" means " = " and "N" means " ≠ ".
The color constraints listed in the order they are depicted on the picture.
Output
If there is a coloring that satisfies at least <image> of the color constraints, print "YES" (without quotes) in the first line. In each of the next h lines, print w integers describing the coloring.
Otherwise, print "NO" (without quotes).
Examples
Input
3 4 4
ENE
NNEE
NEE
ENEN
ENN
Output
YES
1 1 2 2
3 4 1 1
3 3 2 4
Submitted Solution:
```
#IN THE NAME OF GOD
h,w,k=map(int,input().split())
lk=[]
for i in range(k):
lk.append(i)
l=[]
output=[]
for i in range(h):
sa=[]
for j in range(w):
sa.append([])
output.append(sa)
output[0][0]=lk[0]
if k>1:
namosavi=lk[1]
else:
pass
for i in range(2*h-1):
l.append(input())
x=0
r=0
mosavi=lk[0]
for i in range(h):
if i>0:
if l[r-1]=='E':
output[i][0]=output[i-1][0]
else:
output[i][0]=lk[1]
b=1
for j in range(len(l[r])):
if l[r][j]=='E':
output[i][b]=(mosavi)
elif l[r][j]=='N':
output[i][b]=(namosavi)
cm=mosavi
mosavi=namosavi
for u in range(len(lk)):
if lk[u]!=cm and lk[u]!=namosavi:
namosavi=lk[u]
if namosavi==cm:
cm=namosavi
if b+1==h+1:
break
else:
b+=1
r+=2
for i in range(h):
print(*output[i])
``` | instruction | 0 | 37,146 | 7 | 74,292 |
No | output | 1 | 37,146 | 7 | 74,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4 | instruction | 0 | 37,213 | 7 | 74,426 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
b = sorted(a)
if b[-1] - b[0] > k:
print("NO")
else:
print("YES")
for v in a:
for i in range(b[0]):
print(end="1 ")
for i in range(v - b[0]):
print((i + 1) % k + 1, end=' ')
print()
``` | output | 1 | 37,213 | 7 | 74,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4 | instruction | 0 | 37,214 | 7 | 74,428 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, k = map(int, input().split())
arr = list(map(int, input().split()))
q = min(arr)
if max(arr) - q > k:
print('NO')
else:
print('YES')
ans = '1 '*(q-1)+'1'
for i in range(1,k+1):
ans += ' '+str(i)
ans = ans.split()
for i in arr:
#print('--->', ans[i*2])
for j in range(i):
print(ans[j], end=' ')
print()
``` | output | 1 | 37,214 | 7 | 74,429 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4 | instruction | 0 | 37,215 | 7 | 74,430 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, k = list(map(int, input().split()))
piles = list(map(int, input().split()))
min_pebbles = min(piles)
pebble_colors = [[1 for j in range(min_pebbles)] for i in range(n)]
# print(f"initial pebble_colors = {pebble_colors}")
for i in range(1, k+1):
# print(f"color = {i}")
for j in range(n):
pebbles_left = piles[j] - len(pebble_colors[j])
if pebbles_left > 0:
pebble_colors[j].append(i)
# print("after using the given color")
# print(pebble_colors)
b = True
# print(f"final pebble_colors = {pebble_colors}")
# print(f"piles = {piles}")
for j in range(n):
if piles[j] != len(pebble_colors[j]):
b = False
print("NO")
break
if b:
print("YES")
for colors in pebble_colors:
for j in colors:
print(j, end=" ")
print()
``` | output | 1 | 37,215 | 7 | 74,431 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4 | instruction | 0 | 37,216 | 7 | 74,432 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
g=input()
n,k=[int(x) for x in g.split()]
r=input()
li=[]
for y in r.split():
li.append(int(y))
m=max(li)
t=min(li)
if m-t>k:
print("NO")
else:
print("YES")
for i in range(n):
lis=[]
u=li[i]
jai=int(u/k)
veer=u%k
while jai>0:
for i in range(k):
lis.append(i+1)
jai-=1
for ihj in range(veer):
lis.append(ihj+1)
lis=sorted(lis)
print(' '.join(map(str, lis)))
``` | output | 1 | 37,216 | 7 | 74,433 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4 | instruction | 0 | 37,217 | 7 | 74,434 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n,k=map(int,input().split())
l=[int(i) for i in input().split()]
if max(l)>min(l)+k:
print('NO')
exit()
j=0
print('YES')
clr=[[0 for j in range(1095)]for i in range(n)]
while j<101:
for c in range(1,k+1):
for i in range(n):
clr[i][j]=c
j+=1
for i in range(n):
for j in range(l[i]):
print(clr[i][j],end=' ')
print()
``` | output | 1 | 37,217 | 7 | 74,435 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4 | instruction | 0 | 37,218 | 7 | 74,436 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
n, k = map(int, input().split())
pilhas = list(map(int, input().split()))
if max(pilhas) - min(pilhas) > k:
print('NO')
else:
print('YES')
for i in range(n):
print(' '.join([str(j % k + 1) for j in range(pilhas[i])]))
``` | output | 1 | 37,218 | 7 | 74,437 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4 | instruction | 0 | 37,219 | 7 | 74,438 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
def solve(n,k,seq) :
ans = []
for outerIndex in range(n) :
maxAllowed = seq[outerIndex] + k
for innerIndex in range(n) :
if maxAllowed < seq[innerIndex] :
print ("NO")
return
for i in seq :
temp = []
start = 1
if i > k :
end = k + 1
remainder = i%k
if remainder > 0 :
temp.extend(list(range(1,remainder+1)))
if i//k > 1 :
theList = list(range(start,end))
for _ in range(1,i//k) :
temp.extend(theList)
else :
end = i + 1
temp.extend(list(range(start,end)))
ans.append(temp)
print ("YES")
for arr in ans :
for ele in arr :
print (ele,end = " ")
print (" ")
n,k = list(map(int,input().split()))
seq = list(map(int,input().split()))
solve(n,k,seq)
``` | output | 1 | 37,219 | 7 | 74,439 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4 | instruction | 0 | 37,220 | 7 | 74,440 |
Tags: constructive algorithms, greedy, implementation
Correct Solution:
```
# ========= /\ /| |====/|
# | / \ | | / |
# | /____\ | | / |
# | / \ | | / |
# ========= / \ ===== |/====|
# code
if __name__ == "__main__":
n,k = map(int,input().split())
a = [int(i) for i in input().split()]
l = min(a)
h = max(a)
if k < h - l:
print('NO')
else:
print('YES')
for i in a:
p = [1]*l
y = i - l
j = 1
while y > 0:
y -= 1
p.append(j)
j += 1
print(' '.join([str(m) for m in p]))
``` | output | 1 | 37,220 | 7 | 74,441 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
n,k = list(map(int, input().split()))
a = list(map(int, input().split()))
c=1
maxi = max(a)
out = [[] for i in range(n)]
while maxi>0:
mini = min(a)
for i in range(n):
if a[i]<=(mini+1):
out[i].extend([c]*(a[i]))
a[i]=0
else:
out[i].extend([c]*(mini+1))
a[i]-=(mini+1)
c+=1
maxi-=(mini+1)
if c<=(k+1):
print("YES")
for row in out:
print(*row)
else:
print("NO")
``` | instruction | 0 | 37,221 | 7 | 74,442 |
Yes | output | 1 | 37,221 | 7 | 74,443 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4
Submitted Solution:
```
def main():
n,k= map(int,input().split())
arr = list(map(int,input().split()))
if max(arr)-min(arr)>k:
print("NO")
else:
print("YES")
l=[]
for i in range(k):
l.append(i+1)
p=[*l]*100
l+=p
for x in arr:
print(*l[:x])
if __name__ == '__main__':
main()
``` | instruction | 0 | 37,222 | 7 | 74,444 |
Yes | output | 1 | 37,222 | 7 | 74,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4
Submitted Solution:
```
# -*- coding: utf-8 -*-
n, k = map(int, input().split())
piles = list(map(int, input().split()))
if (max(piles) - min(piles) <= k):
print("YES")
for pile in piles:
answer, c = [], 1
for i in range(pile):
answer.append(c)
c = 1 if (c == k) else c + 1
print(' '.join(map(str, sorted(answer))))
else:
print("NO")
``` | instruction | 0 | 37,223 | 7 | 74,446 |
Yes | output | 1 | 37,223 | 7 | 74,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4
Submitted Solution:
```
def ans(v1,v2):
x=int(v1/v2)
y=int(v1%v2)
s=""
for i in range(0,x):
for j in range(1,v2+1):
s+=str(j)+" "
for i in range(0,y):
s+=str(i+1)+" "
return s
n,k=map(int,input().split(' '))
a=list(map(int,input().split(' ')))
b=a[:]
b.sort()
if (b[-1]-b[0])>k:
print("NO")
else:
print("YES")
for j in a:
print(ans(j,k))
``` | instruction | 0 | 37,224 | 7 | 74,448 |
Yes | output | 1 | 37,224 | 7 | 74,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4
Submitted Solution:
```
n,k=map(int,input().split(" "))
arr=list(map(int,input().split(" ")))
mi=min(arr)
ma=max(arr)
if ma-mi>k:
print("NO")
else:
print("YES")
for x in arr:
for y in range(0,x):
print(y%k+1,end="")
print()
``` | instruction | 0 | 37,225 | 7 | 74,450 |
No | output | 1 | 37,225 | 7 | 74,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4
Submitted Solution:
```
import math
n,m = map(int,input().split())
x = [int(x) for x in input().split()]
k = n%m
p = math.ceil(n/m)
if(k!=0 and p>2):
print("NO")
else:
print("YES")
for i in x:
f=0
t=0
while(t<p and f<i):
g=1
while(g<=m and t<=p and f<i):
print(g,end=" ")
g+=1
f+=1
t+=1
print()
``` | instruction | 0 | 37,226 | 7 | 74,452 |
No | output | 1 | 37,226 | 7 | 74,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4
Submitted Solution:
```
import sys
inf = float("inf")
# sys.setrecursionlimit(10000000)
# abc='abcdefghijklmnopqrstuvwxyz'
# abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod, MOD = 1000000007, 998244353
# vow=['a','e','i','o','u']
# dx,dy=[-1,1,0,0],[0,0,1,-1]
# import random
# from collections import deque, Counter, OrderedDict,defaultdict
# from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
# from math import ceil,floor,log,sqrt,factorial
# from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
n,k = get_ints()
Arr = get_array()
mini = min(Arr)
maxi = max(Arr)
if maxi>=mini*k +2:
print('NO')
exit()
print('YES')
for i in range(n):
times = Arr[i]
for i in range(times):
if i!=times-1:
print(i%k +1 , end = ' ')
else:
print(i%k +1)
``` | instruction | 0 | 37,227 | 7 | 74,454 |
No | output | 1 | 37,227 | 7 | 74,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.
In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.
Input
The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.
The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.
Output
If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .
Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.
Examples
Input
4 4
1 2 3 4
Output
YES
1
1 4
1 2 4
1 2 3 4
Input
5 2
3 2 4 1 3
Output
NO
Input
5 4
3 2 4 3 5
Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4
Submitted Solution:
```
R = lambda:list(map(int,input().split()))
n,k = R()
a = R()
r = [[] for _ in range(n)]
x = min(a)
m = 1
for i in range(n):
for j in range(x):
r[i].append(m)
a[i] -= 1
flg = True
xfl = not (1 in a)
while max(a) > 0:
if xfl:
m += 1
xfl = False
if m > k:
flg = False
break
for i in range(n):
if a[i] > 0:
r[i].append(m)
a[i] -= 1
if a[i] == 0:
xfl = True
print('YES' if flg else 'NO')
if flg:
for i in range(n):
print( " ".join([str(u) for u in r[i]]) )
``` | instruction | 0 | 37,228 | 7 | 74,456 |
No | output | 1 | 37,228 | 7 | 74,457 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs. | instruction | 0 | 37,259 | 7 | 74,518 |
Tags: implementation
Correct Solution:
```
import sys
import string
from collections import Counter, defaultdict
from math import fsum, sqrt, gcd, ceil, factorial
from operator import add
inf = float("inf")
# input = sys.stdin.readline
flush = lambda: sys.stdout.flush
comb = lambda x, y: (factorial(x) // factorial(y)) // factorial(x - y)
# inputs
# ip = lambda : input().rstrip()
ip = lambda: input()
ii = lambda: int(input())
r = lambda: map(int, input().split())
rr = lambda: list(r())
a, b, c = r()
x, y, z = r()
extra = 0
if a > x:
extra += (a-x)//2
else:
extra -= x-a
a = x
if b > y:
extra += (b-y)//2
else:
extra -= y-b
y = b
if c > z:
extra += (c-z)//2
else:
extra -= z-c
c = z
print("Yes" if a >= x and b >= y and c >= z and extra >=0 else "No")
``` | output | 1 | 37,259 | 7 | 74,519 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs. | instruction | 0 | 37,260 | 7 | 74,520 |
Tags: implementation
Correct Solution:
```
a,b,c=map(int,input().split())
x,y,z=map(int,input().split())
a=a-x;b=b-y;c=c-z;
frag=0
if(a>0):
frag+=a//2;
else:
frag+=a;
if(b>0):
frag+=b//2;
else:
frag+=b;
if(c>0):
frag+=c//2;
else:
frag+=c;
print("Yes" if frag>=0 else"No")
``` | output | 1 | 37,260 | 7 | 74,521 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs. | instruction | 0 | 37,261 | 7 | 74,522 |
Tags: implementation
Correct Solution:
```
import sys
import math
input=sys.stdin.readline
a,b,c=(int(i) for i in input().split())
x,y,z=(int(i) for i in input().split())
req=0
hv=0
if(a<x):
req+=(x-a)
else:
hv+=(a-x)//2
if(b<y):
req+=(y-b)
else:
hv+=(b-y)//2
if(c<z):
req+=(z-c)
else:
hv+=(c-z)//2
if(hv>=req):
print("Yes")
else:
print("No")
``` | output | 1 | 37,261 | 7 | 74,523 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs. | instruction | 0 | 37,262 | 7 | 74,524 |
Tags: implementation
Correct Solution:
```
#!/usr/bin/env python3
try:
while True:
a, b, c = map(int, input().split())
x, y, z = map(int, input().split())
a -= x
b -= y
c -= z
if sum(t >> 1 for t in (a, b, c) if t > 0) >= sum(-t for t in (a, b, c) if t < 0):
print("Yes")
else:
print("No")
except EOFError:
pass
``` | output | 1 | 37,262 | 7 | 74,525 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs. | instruction | 0 | 37,263 | 7 | 74,526 |
Tags: implementation
Correct Solution:
```
#!/usr/bin/env python3
# 606A_spheres.py - Codeforces.com/problemset/problem/606/A by Sergey 2015
import unittest
import sys
###############################################################################
# Spheres Class (Main Program)
###############################################################################
class Spheres:
""" Spheres representation """
def __init__(self, test_inputs=None):
""" Default constructor """
it = iter(test_inputs.split("\n")) if test_inputs else None
def uinput():
return next(it) if it else sys.stdin.readline().rstrip()
# Reading a single line of multiple elements
self.numa = list(map(int, uinput().split()))
self.numb = list(map(int, uinput().split()))
self.delta = [a - b for (a, b) in zip(self.numa, self.numb)]
pos = [d // 2 for d in self.delta if d > 0]
neg = [d for d in self.delta if d < 0]
self.result = 1
if sum(pos) + sum(neg) < 0:
self.result = 0
def calculate(self):
""" Main calcualtion function of the class """
return "Yes" if self.result else "No"
###############################################################################
# Unit Tests
###############################################################################
class unitTests(unittest.TestCase):
def test_single_test(self):
""" Spheres class testing """
# Constructor test
test = "4 4 0\n2 1 2"
d = Spheres(test)
self.assertEqual(d.numa, [4, 4, 0])
self.assertEqual(d.numb, [2, 1, 2])
self.assertEqual(d.delta, [2, 3, -2])
# Sample test
self.assertEqual(Spheres(test).calculate(), "Yes")
# Sample test
test = "5 6 1\n2 7 2"
self.assertEqual(Spheres(test).calculate(), "No")
# Sample test
test = "3 3 3\n2 2 2"
self.assertEqual(Spheres(test).calculate(), "Yes")
# Time limit test
# self.time_limit_test(5000)
def time_limit_test(self, nmax):
""" Timelimit testing """
import random
import timeit
# Random inputs
test = str(nmax) + " " + str(nmax) + "\n"
numnums = [str(i) + " " + str(i+1) for i in range(nmax)]
test += "\n".join(numnums) + "\n"
nums = [random.randint(1, 10000) for i in range(nmax)]
test += " ".join(map(str, nums)) + "\n"
# Run the test
start = timeit.default_timer()
d = Spheres(test)
calc = timeit.default_timer()
d.calculate()
stop = timeit.default_timer()
print("\nTimelimit Test: " +
"{0:.3f}s (init {1:.3f}s calc {2:.3f}s)".
format(stop-start, calc-start, stop-calc))
if __name__ == "__main__":
# Avoiding recursion limitaions
sys.setrecursionlimit(100000)
if sys.argv[-1] == "-ut":
unittest.main(argv=[" "])
# Print the result string
sys.stdout.write(Spheres().calculate())
``` | output | 1 | 37,263 | 7 | 74,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs. | instruction | 0 | 37,264 | 7 | 74,528 |
Tags: implementation
Correct Solution:
```
a, b, c = map(int, input().split())
x, y, z = map(int, input().split())
pos = 0
neg = 0
if x > a:
pos += x-a
else:
neg += (a-x)//2
if y > b:
pos += y-b
else:
neg += (b-y)//2
if z > c:
pos += z-c
else:
neg += (c-z)//2
if pos <= neg:
print("Yes")
else:
print('No')
``` | output | 1 | 37,264 | 7 | 74,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs. | instruction | 0 | 37,265 | 7 | 74,530 |
Tags: implementation
Correct Solution:
```
u = list(map(int,input().split()))
v = list(map(int,input().split()))
r = sorted([a-b for a, b in zip(u, v)])
print('Yes' if sum([(x>>1) for x in r if x > 0]) >= sum([(-x) for x in r if x < 0]) else 'No')
``` | output | 1 | 37,265 | 7 | 74,531 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs. | instruction | 0 | 37,266 | 7 | 74,532 |
Tags: implementation
Correct Solution:
```
'''
二维数组转置,排序,lambda表达式函数
'''
mylist = [list(map(int, input().split())),list(map(int, input().split()))]
mylist = list(map(list,zip(*mylist)))
for my in mylist:
if my[0]<my[1]:
my[1]-=my[0]
my[0]=0
else:
my[0]-=my[1]
my[1]=0
# for i in range(len(mylist)):
# if mylist[i][0]<mylist[i][1]:
# mylist[i][1]-=mylist[i][0]
# mylist[i][0]=0
# else:
# mylist[i][0]-=mylist[i][1]
# mylist[i][1]=0
mylist.sort(key=lambda my:my[0], reverse=True)
i=0 #初始剩余资源
j=0 #目标资源量
while j<3 and i<3:
while mylist[j][1]>0 and i<3:
if mylist[i][0]<=mylist[j][1]*2:
mylist[j][1]-=mylist[i][0]//2
i+=1
else:
mylist[i][0]-=mylist[j][1]*2
mylist[j][1]=0
j+=1
if mylist[2][1]+mylist[1][1]+mylist[0][1]>0:
print("No")
else:
print("Yes")
``` | output | 1 | 37,266 | 7 | 74,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.
Submitted Solution:
```
a = list(map(int,input().split()))
b = list(map(int,input().split()))
sm1 = sm2 = 0
for i in range(0 , 3):
if a[i] > b[i] :
sm1 += (a[i] - b[i]) // 2
else:
sm2 += (b[i] - a[i])
if (sm1 >= sm2):
print('Yes')
else:
print('No')
``` | instruction | 0 | 37,267 | 7 | 74,534 |
Yes | output | 1 | 37,267 | 7 | 74,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.
Submitted Solution:
```
a, b, c = [int(i) for i in input().split()]
x, y, z = [int(j) for j in input().split()]
a, b, c = a - x, b - y, c - z
result = 0
if a > 0:
result += a // 2
else:
result += a
if b > 0:
result += b // 2
else:
result += b
if c > 0:
result += c // 2
else:
result += c
if result >= 0:
print("Yes")
else:
print("No")
``` | instruction | 0 | 37,268 | 7 | 74,536 |
Yes | output | 1 | 37,268 | 7 | 74,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.
Submitted Solution:
```
a = list(map(int,input().split()))
b = list(map(int,input().split()))
s = 0
for i in range(3):
if a[i] >= b[i]:
s += (a[i]-b[i])//2
else:
s -= (b[i]-a[i])
if s >= 0:
print('Yes')
else:
print('No')
``` | instruction | 0 | 37,269 | 7 | 74,538 |
Yes | output | 1 | 37,269 | 7 | 74,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.
Submitted Solution:
```
l=input().split()
s=input().split()
r=[0]*3
for i in range(3):
l[i]=int(l[i])
s[i]=int(s[i])
r[i]=l[i]-s[i]
k=0
m,n=0,0
for i in range(3):
if(r[i]>=0):
m+=r[i]//2
else:
k+=1
r[i]=-r[i]
n+=r[i]
if(k==3):
print("No")
else:
if(m>=n):
print("Yes")
else:
print("No")
``` | instruction | 0 | 37,270 | 7 | 74,540 |
Yes | output | 1 | 37,270 | 7 | 74,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.
Submitted Solution:
```
import sys
h = sys.stdin.readline().split(' ')
n = sys.stdin.readline().split(' ')
h = [int(i) for i in h]
n = [int(i) for i in n]
d = [h[i] - n[i] for i in range(0,3)]
left = 0
need = 0
for i in range(0,3):
if d[i] > 0:
left += d[i]
for i in range(0,3):
if d[i] < 0:
need -= d[i]
if left // 2 >= need:
print('Yes')
else:
print('No')
``` | instruction | 0 | 37,271 | 7 | 74,542 |
No | output | 1 | 37,271 | 7 | 74,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.
Submitted Solution:
```
''' ===============================
-- @uthor : Kaleab Asfaw
-- Handle : kaleabasfaw2010
-- Bio : High-School Student
==============================='''
# Fast IO
import sys
import os
from io import BytesIO, IOBase
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")
# Others
# from math import floor, ceil, gcd
# from decimal import Decimal as d
mod = 10**9+7
def lcm(x, y): return (x * y) / (gcd(x, y))
def fact(x, mod=mod):
ans = 1
for i in range(1, x+1): ans = (ans * i) % mod
return ans
def arr2D(n, m, default=0): return [[default for j in range(m)] for i in range(n)]
def arr3D(n, m, r, default=0): return [[[default for k in range(r)] for j in range(m)] for i in range(n)]
def sortDictV(x): return {k: v for k, v in sorted(x.items(), key = lambda item : item[1])}
class DSU:
def __init__(self, length): self.length = length; self.parent = [-1] * self.length # O(log(n))
def getParent(self, node, start): # O(log(n))
if node >= self.length: return False
if self.parent[node] < 0:
if start != node: self.parent[start] = node
return node
return self.getParent(self.parent[node], start)
def union(self, node1, node2): # O(log(n))
parent1 = self.getParent(node1, node1); parent2 = self.getParent(node2, node2)
if parent1 == parent2: return False
elif self.parent[parent1] <= self.parent[parent2]: self.parent[parent1] += self.parent[parent2]; self.parent[parent2] = parent1
else: self.parent[parent2] += self.parent[parent1]; self.parent[parent1] = parent2
return True
def getCount(self, node): return -self.parent[self.getParent(node, node)] # O(log(n))
from math import ceil
a, b, c = list(map(int, input().split()))
d, e, f = list(map(int, input().split()))
diff = abs(a - d) + abs(b - e) + abs(c - f)
diff = ceil(diff / 4)
if a + b + c - diff >= d + e + f:
print("Yes")
else:
print("No")
``` | instruction | 0 | 37,272 | 7 | 74,544 |
No | output | 1 | 37,272 | 7 | 74,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.
Submitted Solution:
```
haz=map(int,input().split())
want=map(int,input().split())
stash=0
for x,y in zip(haz,want):
if x>y:
stash+=(x-y)//2
need=0
for x,y in zip(haz,want):
if x<y:
need+=y-x
if stash>need:
print("Yes")
else:
print("No")
``` | instruction | 0 | 37,273 | 7 | 74,546 |
No | output | 1 | 37,273 | 7 | 74,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Carl is a beginner magician. He has a blue, b violet and c orange magic spheres. In one move he can transform two spheres of the same color into one sphere of any other color. To make a spell that has never been seen before, he needs at least x blue, y violet and z orange spheres. Can he get them (possible, in multiple actions)?
Input
The first line of the input contains three integers a, b and c (0 ≤ a, b, c ≤ 1 000 000) — the number of blue, violet and orange spheres that are in the magician's disposal.
The second line of the input contains three integers, x, y and z (0 ≤ x, y, z ≤ 1 000 000) — the number of blue, violet and orange spheres that he needs to get.
Output
If the wizard is able to obtain the required numbers of spheres, print "Yes". Otherwise, print "No".
Examples
Input
4 4 0
2 1 2
Output
Yes
Input
5 6 1
2 7 2
Output
No
Input
3 3 3
2 2 2
Output
Yes
Note
In the first sample the wizard has 4 blue and 4 violet spheres. In his first action he can turn two blue spheres into one violet one. After that he will have 2 blue and 5 violet spheres. Then he turns 4 violet spheres into 2 orange spheres and he ends up with 2 blue, 1 violet and 2 orange spheres, which is exactly what he needs.
Submitted Solution:
```
a, b, c = map(int, input().split())
x, y, z = map(int, input().split())
a = a - x
b = b - y
c = c - z
a = a % 2 * -(a < 0) + a # (a % 2) * -(a < 0) + a
b = b % 2 * -(b < 0) + b
c = c % 2 * -(c < 0) + c
if a + b + c > 0:
print("Yes")
else:
print("No")
``` | instruction | 0 | 37,274 | 7 | 74,548 |
No | output | 1 | 37,274 | 7 | 74,549 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | instruction | 0 | 37,459 | 7 | 74,918 |
Tags: implementation
Correct Solution:
```
n = int(input())
a = input()
flag = 0
for i in range(len(a)):
if(i ==0 or i == n - 1) and a[i] == '?' :
flag = 1
for i in range(1, len(a) - 1):
if a[i] == '?' and (a[i-1] == a[i+1] or a[i-1] == '?' or a[i+1] == '?'):
flag = 1
for i in range(1, len(a)):
if a[i] == a[i-1] and a[i] != '?':
flag = 0
break
if flag:
print('Yes')
else:
print('No')
``` | output | 1 | 37,459 | 7 | 74,919 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | instruction | 0 | 37,460 | 7 | 74,920 |
Tags: implementation
Correct Solution:
```
n = int(input())
a = input()
a = list(a)
t=0
g =a.count("?")
if g==0:
print("No")
exit()
for i in range(n-1):
j = i+1
if (a[i]==a[j] and a[i]!="?") :
print("No")
exit()
for i in range(1,n-1):
if a[i-1]!=a[i+1] and a[i]=="?" and (a[i-1]!="?" and a[i+1]!="?" ):
t+=1
if t>=g and t!=0:
print("no")
exit()
print("Yes")
``` | output | 1 | 37,460 | 7 | 74,921 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | instruction | 0 | 37,461 | 7 | 74,922 |
Tags: implementation
Correct Solution:
```
n = int(input())
s = input()
if(s.find('CC')!=-1 or s.find('YY')!=-1 or s.find('MM')!=-1):
print('No')
elif(s.find("??")!=-1):
print("Yes")
else:
c = 0
p = []
f = 0
for i in range(len(s)):
if(s[i]=="?"):
p.append(i)
for y in p:
if(y==0 or y==n-1):
print("Yes")
f = 1
break
else:
if(s[y-1]==s[y+1]):
c+=2
break
if(c>=2 and f!=1):
print("Yes")
elif(c<2 and f!=1):
print("No")
``` | output | 1 | 37,461 | 7 | 74,923 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | instruction | 0 | 37,462 | 7 | 74,924 |
Tags: implementation
Correct Solution:
```
def findq(s, i):
start = s.find('?', i)
if start != -1:
stop = start
while stop < len(s) and s[stop] == '?':
stop += 1
return start, stop - start
else:
return None, None
def main():
n = int(input())
s = input()
# does it obey the rule now ?
no = False
for x in 'CMY':
if x * 2 in s:
no = True
if no:
print('No')
return
else:
# is there more than two options?
two_ways = False
i, l = findq(s, 0)
while i is not None:
if l == 1 and 0 <= i-1 and i+l < n and s[i-1] != s[i+l]:
pass
else:
two_ways = True
break
i, l = findq(s, i+l)
if two_ways:
print('Yes')
else:
print('No')
if __name__ == '__main__':
main()
``` | output | 1 | 37,462 | 7 | 74,925 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | instruction | 0 | 37,463 | 7 | 74,926 |
Tags: implementation
Correct Solution:
```
import sys
from functools import wraps
def memoize(function):
memo = {}
@wraps(function)
def wrapper(*args):
if args in memo:
return memo[args]
else:
res = function(*args)
memo[args] = res
return res
return wrapper
@memoize
def count(canvas, pos, last):
if pos == len(canvas):
return 1
res = 0
for color in "CYM":
if not (color == last or (canvas[pos] != '?' and color != canvas[pos])):
res += count(canvas, pos + 1, color)
return res
def main(args):
n = int(input())
canvas = input()
print("Yes" if count(canvas, 0, 'X') >= 2 else "No")
if __name__ == '__main__':
sys.exit(main(sys.argv))
``` | output | 1 | 37,463 | 7 | 74,927 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | instruction | 0 | 37,464 | 7 | 74,928 |
Tags: implementation
Correct Solution:
```
def run():
input()
s = input()
for i in range(len(s)):
if s[i] != '?':
if i > 0 and s[i-1] == s[i]:
return False
if i < len(s) - 1 and s[i+1] == s[i]:
return False
continue
for i in range(len(s)):
if s[i] != '?': continue
if i == 0:
return True
if i == len(s) -1:
return True
if s[i-1] == s[i] or s[i+1] == s[i]:
return True
if s[i-1] == s[i+1]:
return True
return False
print('Yes' if run() else 'No')
``` | output | 1 | 37,464 | 7 | 74,929 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | instruction | 0 | 37,465 | 7 | 74,930 |
Tags: implementation
Correct Solution:
```
n = int(input())
s = list(input().strip())
for i in range(1, n):
if s[i] != '?' and s[i-1] != '?' and s[i] == s[i-1]:
print('No')
exit(0)
ok = False if s[0] != '?' and s[n-1] != '?' else True
p = 1
while p+1 < n:
while p+1 < n and s[p] != '?':
p += 1
q = p
while q+1 < n and s[q] == '?':
q += 1
l = q - p
if l > 1:
ok = True
else:
if q+1 < n and s[p-1] == s[q]:
ok = True
p = q
if ok:
print('Yes')
else:
print('No')
``` | output | 1 | 37,465 | 7 | 74,931 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | instruction | 0 | 37,466 | 7 | 74,932 |
Tags: implementation
Correct Solution:
```
n = int(input())
colors = [i for i in input()]
maners = []
if(colors.count("?") == 0):
print("No")
elif(n == 1):
print("Yes")
else:
i = 0
chave = True
while i<n-1:
if(colors[i] == colors[i+1] and colors[i] != "?"):
chave = False
break
i+=1
if(chave):
string = "".join(colors)
inter = 0
inter += string.count("C?M")
inter += string.count("C?Y")
inter += string.count("Y?C")
inter += string.count("Y?M")
inter += string.count("M?C")
inter += string.count("M?Y")
if inter == string.count("?"):
print("No")
else:
print("Yes")
else:
print("No")
``` | output | 1 | 37,466 | 7 | 74,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example.
Submitted Solution:
```
import sys
def input(): return sys.stdin.readline().rstrip("\r\n")
def List(): return list(map(int, input().split()))
def Num(): return int(input())
n = Num()
s = input()
is_cons = False
for i in range(1, n):
if s[i] != "?" and s[i] == s[i - 1]:
is_cons = True
if is_cons:
print("NO")
else:
array, ch = [], ["Y", "C", "M"]
if n == 1:
if s in ch:
print("NO")
else:
print("YES")
elif n == 2:
if s.count("?"):
print("YES")
else:
print("NO")
else:
if s[0] == "?":
array.append(2)
elif s[n - 1] == "?":
array.append(2)
else:
for i in range(1, n - 1):
if s[i] == "?":
if s[i - 1] in ch and s[i + 1] in ch:
if s[i - 1] != s[i + 1]:
array.append(1)
else:
array.append(2)
else:
array.append(2)
ans = 1
for i in array:
ans *= i
print("YES" if ans >= 2 else "NO")
``` | instruction | 0 | 37,467 | 7 | 74,934 |
Yes | output | 1 | 37,467 | 7 | 74,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example.
Submitted Solution:
```
#troi
n=int(input())
line=input()
trig=True
while trig:
for i in range(len(line)-1):
if line[i]==line[i+1]!='?':
print('No')
trig=False
break
if trig==False: break
if line[0]=='?' or line[-1]=='?':
print('Yes')
trig=False
break
for i in range(1,len(line)-1):
if line[i]==line[i+1]=='?' or (line[i-1]==line[i+1]!='?' and line[i]=='?'):
print('Yes')
trig=False
break
if trig==False: break
print('No')
break
``` | instruction | 0 | 37,468 | 7 | 74,936 |
Yes | output | 1 | 37,468 | 7 | 74,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example.
Submitted Solution:
```
n = int(input())
s = input()
if ((s.find('??') >= 0 or s[0] == '?' or s[-1] == '?'
or any(s.find(c + '?' + c) >= 0 for c in ['C', 'M', 'Y']))
and all(s.find(x) < 0 for x in ['CC', 'MM', 'YY'])):
print("Yes")
else:
print("No")
``` | instruction | 0 | 37,469 | 7 | 74,938 |
Yes | output | 1 | 37,469 | 7 | 74,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example.
Submitted Solution:
```
#from collections import Counter,defaultdict
#get= lambda : map(int,input().split())
import re
input()
s=input()
if re.findall('C{2,}|M{2,}|Y{2,}',s):
print ("No")
exit()
if s.startswith('?') or s.endswith('?'):
print("Yes")
exit()
s=' '+s+' '
l=0
for i,j in enumerate(s):
if '?'==j:
l+=1
else:
if l==1:
if s[i-2]==s[i]:
print("Yes")
break
if l>=2:
print("Yes")
break
l=0
else:
print("No")
#http://codeforces.com/problemset/problem/957/A
``` | instruction | 0 | 37,470 | 7 | 74,940 |
Yes | output | 1 | 37,470 | 7 | 74,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example.
Submitted Solution:
```
n = int(input())
lst = list(input())
s=0
z=0
#print(len(lst))
lst1 = ["C", "Y", "M"]
n=0
for i in range(len(lst)-1):
if lst[i]=="?":
n+=1
for i in range(len(lst)-1):
if lst[i]=='?':
s+=1
if i==0:
s+=1
elif i==len(lst)-1:
s+=1
elif i>0 and i<len(lst)-1:
if lst[i-1]==lst[i+1]:
s+=1
else:
z+=2
s+=1
if lst[i]==lst[i+1]:
z=1
#print(z,s)
if n==len(lst)-1:
print("Yes")
elif z==1:
print("No")
elif s!=0:
if z%s==0 and z!=0:
print("No")
elif s==0:
print("Yes")
else:
print("Yes")
``` | instruction | 0 | 37,471 | 7 | 74,942 |
No | output | 1 | 37,471 | 7 | 74,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example.
Submitted Solution:
```
n = int(input())
s = input()
flag=0
for i in range(n):
if i==0 and s[0] == '?':
print('Yes')
flag=1
break
elif i==n-1 and s[n-1] == '?':
print('Yes')
flag=1
break
else:
if s[i-1]=='?' or s[i+1]=='?':
print('Yes')
flag=1
break
elif s[i-1] == s[i+1]:
print('Yes')
flag=1
break
for i in range(n-1):
if s[i] == s[i+1] and s[i] != '?':
flag=0
if flag==0:
print('No')
``` | instruction | 0 | 37,472 | 7 | 74,944 |
No | output | 1 | 37,472 | 7 | 74,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into n consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input
The first line contains a single positive integer n (1 ≤ n ≤ 100) — the length of the canvas.
The second line contains a string s of n characters, the i-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Examples
Input
5
CY??Y
Output
Yes
Input
5
C?C?Y
Output
Yes
Input
5
?CYC?
Output
Yes
Input
5
C??MM
Output
No
Input
3
MMY
Output
No
Note
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example.
Submitted Solution:
```
n = int (input ())
a = list (input ())
b = 0; t = 1
for I in range(n-1):
if a[I]==a[I+1] and a [I]!="?": t = 0; break
if a.count("?") > 1 and t: print ("Yes")
else : print ("No")
``` | instruction | 0 | 37,473 | 7 | 74,946 |
No | output | 1 | 37,473 | 7 | 74,947 |
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