text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Matrix of given integers
a1,1 a1,2 ... a1, n
a2,1 a2,2 ... a2, n
::
an, 1 an, 2 ... an, n
Then, create a program that outputs the maximum value of the sum of one or more consecutive terms (submatrix) in the vertical and horizontal directions and ends.
Input
The input data is given in the following format.
n
a1,1 a1,2 ... a1, n
a2,1 a2,2 ... a2, n
::
an, 1 an, 2 ... an, n
n is 1 or more and 100 or less, and ai, j is -10000 or more and 10000 or less.
Output
Print the maximum value on one line.
Examples
Input
3
1 -2 3
-4 5 6
7 8 -9
Output
16
Input
4
1 3 -9 2
2 7 -1 5
-8 3 2 -1
5 0 -3 1
Output
15
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0098
"""
import sys
from sys import stdin
input = stdin.readline
def calc_points(n, array):
# 右下から(x, y)までの長方形に含まれる値の和
dp = [[0] * (n + 1) for _ in range(n + 1)]
for y in range(n - 1, -1, -1):
for x in range(n - 1, -1, -1):
dp[y][x] = dp[y+1][x] + dp[y][x+1] - dp[y+1][x+1] + array[y][x]
return dp
def solve(n, array):
ans = []
dp = calc_points(n, array)
for sy in range(n + 1):
for sx in range(n + 1):
for ey in range(sy, n + 1):
for ex in range(sx, n + 1):
s1 = dp[sy][sx] # 右下から始点まで
s2 = dp[sy][ex] # 重なり部分
s3 = dp[ey][sx] # 重なり部分
s4 = dp[ey][ex] # 重なりの重なり部分
s = s1 - s2 -s3 + s4 # (sx, sy) - (ex, ey) 長方形の美味しさポイント
# area = abs(sy - ey) * abs(sx - ex) # 長方形の面積
ans.append(s)
return max(ans)
def main(args):
array = []
# n = 3
# array.append([1, -2, 3])
# array.append([-4, 5, 6])
# array.append([7, 8, -9])
n = int(input())
for _ in range(n):
array.append([int(x) for x in input().split()])
ans = solve(n, array)
print(ans)
if __name__ == '__main__':
main(sys.argv[1:])
```
No
| 4,100 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Matrix of given integers
a1,1 a1,2 ... a1, n
a2,1 a2,2 ... a2, n
::
an, 1 an, 2 ... an, n
Then, create a program that outputs the maximum value of the sum of one or more consecutive terms (submatrix) in the vertical and horizontal directions and ends.
Input
The input data is given in the following format.
n
a1,1 a1,2 ... a1, n
a2,1 a2,2 ... a2, n
::
an, 1 an, 2 ... an, n
n is 1 or more and 100 or less, and ai, j is -10000 or more and 10000 or less.
Output
Print the maximum value on one line.
Examples
Input
3
1 -2 3
-4 5 6
7 8 -9
Output
16
Input
4
1 3 -9 2
2 7 -1 5
-8 3 2 -1
5 0 -3 1
Output
15
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0098
"""
import sys
from sys import stdin
input = stdin.readline
def calc_points(n, array):
# 右下から(x, y)までの長方形に含まれる値の和
dp = [[0] * (n + 1) for _ in range(n + 1)]
for y in range(n - 1, -1, -1):
for x in range(n - 1, -1, -1):
dp[y][x] = dp[y+1][x] + dp[y][x+1] - dp[y+1][x+1] + array[y][x]
return dp
def solve(n, array):
ans = []
dp = calc_points(n, array)
for sy in range(n + 1):
for sx in range(n + 1):
for ey in range(sy+1, n + 1):
for ex in range(sx+1, n + 1):
s1 = dp[sy][sx] # 右下から始点まで
s2 = dp[sy][ex] # 重なり部分
s3 = dp[ey][sx] # 重なり部分
s4 = dp[ey][ex] # 重なりの重なり部分
s = s1 - s2 -s3 + s4 # (sx, sy) - (ex, ey) 長方形の美味しさポイント
# area = abs(sy - ey) * abs(sx - ex) # 長方形の面積
ans.append(s)
return max(ans)
def main(args):
array = []
# n = 3
# array.append([1, -2, 3])
# array.append([-4, 5, 6])
# array.append([7, 8, -9])
n = int(input())
for _ in range(n):
array.append([int(x) for x in input().split()])
ans = solve(n, array)
print(ans)
if __name__ == '__main__':
main(sys.argv[1:])
```
No
| 4,101 |
Provide a correct Python 3 solution for this coding contest problem.
Ninja Atsushi guards the town from the roof of the Ninja Building from early morning to late night every day. This Ninja Building is two adjacent buildings of the same floor, and Atsushi's daily routine is to jump between buildings and head to the rooftop for security.
Because these two buildings are cleaned frequently, there are ladders and slippery areas that can help you climb the building. Moreover, the position of ladders and slippery parts changes every day. Therefore, Atsushi has to think about how to get to the roof every day.
Atsushi jumps over the walls of two buildings of the same floor, aiming for the rooftop of the building. Jumps can be started on the first floor of either building. When jumping to the opposite building, you can jump to the same floor, one floor up, or two floors up.
There are three types of walls, and the movement after jumping to each wall is decided.
* 0. Ordinary wall: Do not move up and down. The next jump will be made from there.
* 1. Ladder: The ladder spans two or more floors and moves to the top of the current ladder. The next jump will be made from there.
* 2. Sliding wall: A normal wall or slides down to the top of the ladder. The next jump will be made from there.
Also, the walls run from the first floor to the top floor just below the roof, and the roof can only be reached from the top floor of the building. Also, the wall on the bottom floor of the building will not be a slippery wall.
Create a program that takes in the number of floors n of the two buildings and the type of wall of the two buildings, and outputs the minimum number of jumps to reach the top floor and reach the rooftop. You may reach the roof of either building. However, if Atsushi cannot reach the roof of either building, please output "NA".
<image>
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 a2 ... an
b1 b2 ... bn
The first line gives the building floor n (3 ≤ n ≤ 100). The second line gives the wall information ai from the first floor to the nth floor of the first building, and the third line gives the wall information bi from the first floor to the nth floor of the second building. ai and bi represent the information of the wall on the i-th floor, 0 is the ordinary wall, 1 is the ladder (straddling the i-th floor and the i + 1-floor), and 2 is the sliding wall.
The number of datasets does not exceed 60.
Output
Outputs the number of jumps on one line for each input dataset.
Example
Input
8
0 0 0 2 2 2 0 0
1 1 1 1 0 0 0 0
4
1 1 2 2
0 0 2 2
0
Output
4
NA
"Correct Solution:
```
def bfs(b):
mem=[[False for i in range(n)]for j in range(2)]
st=[0,0]
for i in range(2):
if b[i][0]!=1:continue
while st[i]<n-1 and b[i][st[i]+1]==1:
st[i]+=1
if st[i]==n-1:return 0
mem[0][st[0]]=True
mem[1][st[1]]=True
que=[[0,st[0],0],[1,st[1],0]]
while len(que)!=0:
d=que.pop(0)
des=(d[0]+1)%2
cst=d[2]+1
for i in range(3):
fl=d[1]+i
if fl>=n:break
state=b[des][fl]
if state!=2 and fl==n-1:
return cst
if state==0:
if mem[des][fl]==False:
que.append([des,fl,cst])
mem[des][fl]=True
elif state==1:
k=1
while True:
if fl+k>=n:return cst
if b[des][fl+k]!=1:
if mem[des][fl+k-1]==False:
que.append([des,fl+k-1,cst])
mem[des][fl+k-1]=True
break
else: k+=1
elif state==2:
k=1
while True:
if b[des][fl-k]!=2:
if mem[des][fl-k]==False:
que.append([des,fl-k,cst])
mem[des][fl-k]=True
break
else:k+=1
return -1
while True:
n=int(input())
if n==0:break
b=[[int(p)for p in input().split(" ")],[int(p)for p in input().split(" ")]]
count=bfs(b)
print(str(count)if count>=0 else "NA")
```
| 4,102 |
Provide a correct Python 3 solution for this coding contest problem.
Ninja Atsushi guards the town from the roof of the Ninja Building from early morning to late night every day. This Ninja Building is two adjacent buildings of the same floor, and Atsushi's daily routine is to jump between buildings and head to the rooftop for security.
Because these two buildings are cleaned frequently, there are ladders and slippery areas that can help you climb the building. Moreover, the position of ladders and slippery parts changes every day. Therefore, Atsushi has to think about how to get to the roof every day.
Atsushi jumps over the walls of two buildings of the same floor, aiming for the rooftop of the building. Jumps can be started on the first floor of either building. When jumping to the opposite building, you can jump to the same floor, one floor up, or two floors up.
There are three types of walls, and the movement after jumping to each wall is decided.
* 0. Ordinary wall: Do not move up and down. The next jump will be made from there.
* 1. Ladder: The ladder spans two or more floors and moves to the top of the current ladder. The next jump will be made from there.
* 2. Sliding wall: A normal wall or slides down to the top of the ladder. The next jump will be made from there.
Also, the walls run from the first floor to the top floor just below the roof, and the roof can only be reached from the top floor of the building. Also, the wall on the bottom floor of the building will not be a slippery wall.
Create a program that takes in the number of floors n of the two buildings and the type of wall of the two buildings, and outputs the minimum number of jumps to reach the top floor and reach the rooftop. You may reach the roof of either building. However, if Atsushi cannot reach the roof of either building, please output "NA".
<image>
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 a2 ... an
b1 b2 ... bn
The first line gives the building floor n (3 ≤ n ≤ 100). The second line gives the wall information ai from the first floor to the nth floor of the first building, and the third line gives the wall information bi from the first floor to the nth floor of the second building. ai and bi represent the information of the wall on the i-th floor, 0 is the ordinary wall, 1 is the ladder (straddling the i-th floor and the i + 1-floor), and 2 is the sliding wall.
The number of datasets does not exceed 60.
Output
Outputs the number of jumps on one line for each input dataset.
Example
Input
8
0 0 0 2 2 2 0 0
1 1 1 1 0 0 0 0
4
1 1 2 2
0 0 2 2
0
Output
4
NA
"Correct Solution:
```
from collections import deque
def bfs(bldg):
dict = {}
que = deque([(bldg[0][0],0,0),(bldg[1][0],1,0)])
while len(que):
h, b, t = que.popleft()
try:
if dict[(h, b)] <= t:
continue
except KeyError:
dict[(h, b)] = t
if h == len(bldg[0]) - 3:
return t
t += 1
if 100 < t:
break
que.append((bldg[b^1][h + 2],b^1,t))
return 'NA'
def building():
bldg = f.readline().split() + ['2', '2']
scaffold = 0
for i in range(len(bldg)):
if '2' == bldg[i]:
bldg[i] = scaffold
else:
scaffold = i
if '0' == bldg[i]:
bldg[i] = i
scaffold = None
for i in reversed(range(len(bldg))):
if '1' == bldg[i]:
if scaffold is None:
scaffold = i
bldg[i] = scaffold
else:
scaffold = None
return bldg
import sys
f = sys.stdin
while True:
n = int(f.readline())
if n == 0:
break
print(bfs([building(),building()]))
```
| 4,103 |
Provide a correct Python 3 solution for this coding contest problem.
Ninja Atsushi guards the town from the roof of the Ninja Building from early morning to late night every day. This Ninja Building is two adjacent buildings of the same floor, and Atsushi's daily routine is to jump between buildings and head to the rooftop for security.
Because these two buildings are cleaned frequently, there are ladders and slippery areas that can help you climb the building. Moreover, the position of ladders and slippery parts changes every day. Therefore, Atsushi has to think about how to get to the roof every day.
Atsushi jumps over the walls of two buildings of the same floor, aiming for the rooftop of the building. Jumps can be started on the first floor of either building. When jumping to the opposite building, you can jump to the same floor, one floor up, or two floors up.
There are three types of walls, and the movement after jumping to each wall is decided.
* 0. Ordinary wall: Do not move up and down. The next jump will be made from there.
* 1. Ladder: The ladder spans two or more floors and moves to the top of the current ladder. The next jump will be made from there.
* 2. Sliding wall: A normal wall or slides down to the top of the ladder. The next jump will be made from there.
Also, the walls run from the first floor to the top floor just below the roof, and the roof can only be reached from the top floor of the building. Also, the wall on the bottom floor of the building will not be a slippery wall.
Create a program that takes in the number of floors n of the two buildings and the type of wall of the two buildings, and outputs the minimum number of jumps to reach the top floor and reach the rooftop. You may reach the roof of either building. However, if Atsushi cannot reach the roof of either building, please output "NA".
<image>
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 a2 ... an
b1 b2 ... bn
The first line gives the building floor n (3 ≤ n ≤ 100). The second line gives the wall information ai from the first floor to the nth floor of the first building, and the third line gives the wall information bi from the first floor to the nth floor of the second building. ai and bi represent the information of the wall on the i-th floor, 0 is the ordinary wall, 1 is the ladder (straddling the i-th floor and the i + 1-floor), and 2 is the sliding wall.
The number of datasets does not exceed 60.
Output
Outputs the number of jumps on one line for each input dataset.
Example
Input
8
0 0 0 2 2 2 0 0
1 1 1 1 0 0 0 0
4
1 1 2 2
0 0 2 2
0
Output
4
NA
"Correct Solution:
```
from collections import deque
def fix(alst, blst, floar, buil, n):
if buil == 0:
lst = alst
else:
lst = blst
if lst[floar] == 0:
return floar
if lst[floar] == 1:
while floar + 1 < n and lst[floar + 1] == 1:
floar += 1
return floar
if lst[floar] == 2:
while lst[floar] == 2:
floar -= 1
return floar
def search(alst, blst, n):
que = deque()
#(cost, floar, buil)
init_floar_a = fix(alst, blst, 0, 0, n)
init_floar_b = fix(alst, blst, 0, 1, n)
if n - 1 in (init_floar_a, init_floar_b):
print(0)
return
que.append((0, init_floar_a, 0))
que.append((0, init_floar_b, 1))
dic = {}
dic[(0, init_floar_a)] = 0
dic[(0, init_floar_b)] = 0
while que:
total, floar, buil = que.popleft()
next_buil = (buil + 1) % 2
for i in range(3):
if floar + i >= n:
break
to = fix(alst, blst, floar + i, next_buil, n)
if to == n - 1:
print(total + 1)
return
if (to, next_buil) not in dic:
dic[(to, next_buil)] = total + 1
que.append((total + 1, to, next_buil))
print("NA")
while True:
n = int(input())
if n == 0:
break
alst = list(map(int, input().split()))
blst = list(map(int, input().split()))
search(alst, blst, n)
```
| 4,104 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ninja Atsushi guards the town from the roof of the Ninja Building from early morning to late night every day. This Ninja Building is two adjacent buildings of the same floor, and Atsushi's daily routine is to jump between buildings and head to the rooftop for security.
Because these two buildings are cleaned frequently, there are ladders and slippery areas that can help you climb the building. Moreover, the position of ladders and slippery parts changes every day. Therefore, Atsushi has to think about how to get to the roof every day.
Atsushi jumps over the walls of two buildings of the same floor, aiming for the rooftop of the building. Jumps can be started on the first floor of either building. When jumping to the opposite building, you can jump to the same floor, one floor up, or two floors up.
There are three types of walls, and the movement after jumping to each wall is decided.
* 0. Ordinary wall: Do not move up and down. The next jump will be made from there.
* 1. Ladder: The ladder spans two or more floors and moves to the top of the current ladder. The next jump will be made from there.
* 2. Sliding wall: A normal wall or slides down to the top of the ladder. The next jump will be made from there.
Also, the walls run from the first floor to the top floor just below the roof, and the roof can only be reached from the top floor of the building. Also, the wall on the bottom floor of the building will not be a slippery wall.
Create a program that takes in the number of floors n of the two buildings and the type of wall of the two buildings, and outputs the minimum number of jumps to reach the top floor and reach the rooftop. You may reach the roof of either building. However, if Atsushi cannot reach the roof of either building, please output "NA".
<image>
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 a2 ... an
b1 b2 ... bn
The first line gives the building floor n (3 ≤ n ≤ 100). The second line gives the wall information ai from the first floor to the nth floor of the first building, and the third line gives the wall information bi from the first floor to the nth floor of the second building. ai and bi represent the information of the wall on the i-th floor, 0 is the ordinary wall, 1 is the ladder (straddling the i-th floor and the i + 1-floor), and 2 is the sliding wall.
The number of datasets does not exceed 60.
Output
Outputs the number of jumps on one line for each input dataset.
Example
Input
8
0 0 0 2 2 2 0 0
1 1 1 1 0 0 0 0
4
1 1 2 2
0 0 2 2
0
Output
4
NA
Submitted Solution:
```
from collections import deque
def fix(alst, blst, floar, buil, n):
if buil == 0:
lst = alst
else:
lst = blst
if lst[floar] == 0:
return floar
if lst[floar] == 1:
while floar + 1 < n and lst[floar + 1] == 1:
floar += 1
return floar
if lst[floar] == 2:
while lst[floar] == 2:
floar -= 1
return floar
def search(alst, blst, n):
que = deque()
#(cost, floar, buil)
init_floar_a = fix(alst, blst, 0, 0, n)
init_floar_b = fix(alst, blst, 0, 1, n)
que.append((0, init_floar_a, 0))
que.append((0, init_floar_b, 1))
dic = {}
dic[(0, init_floar_a)] = 0
dic[(0, init_floar_b)] = 0
while que:
total, floar, buil = que.popleft()
next_buil = (buil + 1) % 2
for i in range(min(3, n - floar)):
to = fix(alst, blst, floar + i, next_buil, n)
if to == n - 1:
print(total + 1)
return
if (to, next_buil) not in dic:
dic[(to, next_buil)] = total + 1
que.append((total + 1, to, next_buil))
print("NA")
while True:
n = int(input())
if n == 0:
break
alst = list(map(int, input().split()))
blst = list(map(int, input().split()))
search(alst, blst, n)
```
No
| 4,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ninja Atsushi guards the town from the roof of the Ninja Building from early morning to late night every day. This Ninja Building is two adjacent buildings of the same floor, and Atsushi's daily routine is to jump between buildings and head to the rooftop for security.
Because these two buildings are cleaned frequently, there are ladders and slippery areas that can help you climb the building. Moreover, the position of ladders and slippery parts changes every day. Therefore, Atsushi has to think about how to get to the roof every day.
Atsushi jumps over the walls of two buildings of the same floor, aiming for the rooftop of the building. Jumps can be started on the first floor of either building. When jumping to the opposite building, you can jump to the same floor, one floor up, or two floors up.
There are three types of walls, and the movement after jumping to each wall is decided.
* 0. Ordinary wall: Do not move up and down. The next jump will be made from there.
* 1. Ladder: The ladder spans two or more floors and moves to the top of the current ladder. The next jump will be made from there.
* 2. Sliding wall: A normal wall or slides down to the top of the ladder. The next jump will be made from there.
Also, the walls run from the first floor to the top floor just below the roof, and the roof can only be reached from the top floor of the building. Also, the wall on the bottom floor of the building will not be a slippery wall.
Create a program that takes in the number of floors n of the two buildings and the type of wall of the two buildings, and outputs the minimum number of jumps to reach the top floor and reach the rooftop. You may reach the roof of either building. However, if Atsushi cannot reach the roof of either building, please output "NA".
<image>
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 a2 ... an
b1 b2 ... bn
The first line gives the building floor n (3 ≤ n ≤ 100). The second line gives the wall information ai from the first floor to the nth floor of the first building, and the third line gives the wall information bi from the first floor to the nth floor of the second building. ai and bi represent the information of the wall on the i-th floor, 0 is the ordinary wall, 1 is the ladder (straddling the i-th floor and the i + 1-floor), and 2 is the sliding wall.
The number of datasets does not exceed 60.
Output
Outputs the number of jumps on one line for each input dataset.
Example
Input
8
0 0 0 2 2 2 0 0
1 1 1 1 0 0 0 0
4
1 1 2 2
0 0 2 2
0
Output
4
NA
Submitted Solution:
```
while True:
n=int(input())
if n==0:break
b=[[int(p)for p in input().split(" ")],[int(p)for p in input().split(" ")]]
mem=[[1145141919 for i in range(n)]for j in range(2)]
mem[1][0]=True
que=[[0,0,0],[1,0,0]]
count=1145141919
while len(que)!=0:
d=que.pop(0)
state=b[d[0]][d[1]]
if state!=2 and d[1]==n-1 and d[2]<count:
count=d[2]
mem[d[0]][d[1]]=d[2]
case=-1
if state==0:case=0
elif state==1:
if b[d[0]][d[1]+1]==1:case=1
else:case=0
elif state==2:case=2
if case==0:
for i in range(3):
if d[1]+i>=n:break
if d[0]==0 and mem[1][d[1]+i]>d[2]+1:que.append([1,d[1]+i,d[2]+1])
elif d[0]==1 and mem[0][d[1]+i]>d[2]+1:que.append([0,d[1]+i,d[2]+1])
elif case==1:
if mem[d[0]][d[1]+1]>d[2]:que.append([d[0],d[1]+1,d[2]])
elif case==2:
if mem[d[0]][d[1]-1]>d[2]:que.append([d[0],d[1]-1,d[2]])
print(str(count)if count!=1145141919 else "NA")
```
No
| 4,106 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ninja Atsushi guards the town from the roof of the Ninja Building from early morning to late night every day. This Ninja Building is two adjacent buildings of the same floor, and Atsushi's daily routine is to jump between buildings and head to the rooftop for security.
Because these two buildings are cleaned frequently, there are ladders and slippery areas that can help you climb the building. Moreover, the position of ladders and slippery parts changes every day. Therefore, Atsushi has to think about how to get to the roof every day.
Atsushi jumps over the walls of two buildings of the same floor, aiming for the rooftop of the building. Jumps can be started on the first floor of either building. When jumping to the opposite building, you can jump to the same floor, one floor up, or two floors up.
There are three types of walls, and the movement after jumping to each wall is decided.
* 0. Ordinary wall: Do not move up and down. The next jump will be made from there.
* 1. Ladder: The ladder spans two or more floors and moves to the top of the current ladder. The next jump will be made from there.
* 2. Sliding wall: A normal wall or slides down to the top of the ladder. The next jump will be made from there.
Also, the walls run from the first floor to the top floor just below the roof, and the roof can only be reached from the top floor of the building. Also, the wall on the bottom floor of the building will not be a slippery wall.
Create a program that takes in the number of floors n of the two buildings and the type of wall of the two buildings, and outputs the minimum number of jumps to reach the top floor and reach the rooftop. You may reach the roof of either building. However, if Atsushi cannot reach the roof of either building, please output "NA".
<image>
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 a2 ... an
b1 b2 ... bn
The first line gives the building floor n (3 ≤ n ≤ 100). The second line gives the wall information ai from the first floor to the nth floor of the first building, and the third line gives the wall information bi from the first floor to the nth floor of the second building. ai and bi represent the information of the wall on the i-th floor, 0 is the ordinary wall, 1 is the ladder (straddling the i-th floor and the i + 1-floor), and 2 is the sliding wall.
The number of datasets does not exceed 60.
Output
Outputs the number of jumps on one line for each input dataset.
Example
Input
8
0 0 0 2 2 2 0 0
1 1 1 1 0 0 0 0
4
1 1 2 2
0 0 2 2
0
Output
4
NA
Submitted Solution:
```
def bfs(b):
mem=[[False for i in range(n)]for j in range(2)]
mem[0][0]=True
mem[1][0]=True
que=[[0,0,0],[1,0,0]]
while len(que)!=0:
d=que.pop(0)
state=b[d[0]][d[1]]
if state!=2 and d[1]==n-1:
return d[2]
if state==0:
for i in range(3):
if d[1]+i>=n:break
if d[0]==0 and mem[1][d[1]+i]==False:
que.append([1,d[1]+i,d[2]+1])
mem[1][d[1]+i]=True
elif d[0]==1 and mem[0][d[1]+i]==False:
que.append([0,d[1]+i,d[2]+1])
mem[0][d[1]+i]=True
elif state==1:
k=1
while True:
if d[1]+k>=n:return d[2]
if b[d[0]][d[1]+k]!=1:
for i in range(3):
if d[1]+k+i-1>=n:break
if d[0]==0 and mem[1][d[1]+k+i-1]==False:
que.append([1,d[1]+k+i-1,d[2]+1])
mem[1][d[1]+k+i-1]=True
elif d[0]==1 and mem[0][d[1]+k+i-1]==False:
que.append([0,d[1]+k+i-1,d[2]+1])
mem[0][d[1]+k+i-1]=True
break
else: k+=1
elif state==2:
k=1
while True:
if b[d[0]][d[1]-k]!=2:
for i in range(3):
if d[1]-k+i>=n:break
if d[0]==0 and mem[1][d[1]-k+i]==False:
que.append([1,d[1]-k+i,d[2]+1])
mem[1][d[1]-k+i]=True
elif d[0]==1 and mem[0][d[1]-k+i]==False:
que.append([0,d[1]-k+i,d[2]+1])
mem[0][d[1]-k+i]=True
break
else:k+=1
return 0
while True:
#for t in range(1):
n=int(input())
#n=8
if n==0:break
b=[[int(p)for p in input().split(" ")],[int(p)for p in input().split(" ")]]
#b=[[0,0,0,2,2,2,0,0],[1,1,1,1,0,0,0,0]]
count=bfs(b)
print(str(count)if count!=0 else "NA")
```
No
| 4,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ninja Atsushi guards the town from the roof of the Ninja Building from early morning to late night every day. This Ninja Building is two adjacent buildings of the same floor, and Atsushi's daily routine is to jump between buildings and head to the rooftop for security.
Because these two buildings are cleaned frequently, there are ladders and slippery areas that can help you climb the building. Moreover, the position of ladders and slippery parts changes every day. Therefore, Atsushi has to think about how to get to the roof every day.
Atsushi jumps over the walls of two buildings of the same floor, aiming for the rooftop of the building. Jumps can be started on the first floor of either building. When jumping to the opposite building, you can jump to the same floor, one floor up, or two floors up.
There are three types of walls, and the movement after jumping to each wall is decided.
* 0. Ordinary wall: Do not move up and down. The next jump will be made from there.
* 1. Ladder: The ladder spans two or more floors and moves to the top of the current ladder. The next jump will be made from there.
* 2. Sliding wall: A normal wall or slides down to the top of the ladder. The next jump will be made from there.
Also, the walls run from the first floor to the top floor just below the roof, and the roof can only be reached from the top floor of the building. Also, the wall on the bottom floor of the building will not be a slippery wall.
Create a program that takes in the number of floors n of the two buildings and the type of wall of the two buildings, and outputs the minimum number of jumps to reach the top floor and reach the rooftop. You may reach the roof of either building. However, if Atsushi cannot reach the roof of either building, please output "NA".
<image>
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
n
a1 a2 ... an
b1 b2 ... bn
The first line gives the building floor n (3 ≤ n ≤ 100). The second line gives the wall information ai from the first floor to the nth floor of the first building, and the third line gives the wall information bi from the first floor to the nth floor of the second building. ai and bi represent the information of the wall on the i-th floor, 0 is the ordinary wall, 1 is the ladder (straddling the i-th floor and the i + 1-floor), and 2 is the sliding wall.
The number of datasets does not exceed 60.
Output
Outputs the number of jumps on one line for each input dataset.
Example
Input
8
0 0 0 2 2 2 0 0
1 1 1 1 0 0 0 0
4
1 1 2 2
0 0 2 2
0
Output
4
NA
Submitted Solution:
```
from collections import deque
def fix(alst, blst, floar, buil, n):
if buil == 0:
lst = alst
else:
lst = blst
if lst[floar] == 0:
return floar
if lst[floar] == 1:
while floar + 1 < n and lst[floar + 1] == 1:
floar += 1
return floar
if lst[floar] == 2:
while lst[floar] == 2:
floar -= 1
return floar
def search(alst, blst, n):
que = deque()
#(cost, floar, buil)
que.append((0, 0, 0))
que.append((0, 0, 1))
dic = {}
dic[(0, 0)] = 0
dic[(0, 1)] = 0
while que:
total, floar, buil = que.popleft()
next_buil = (buil + 1) % 2
for i in range(min(3, n - floar)):
to = fix(alst, blst, floar + i, next_buil, n)
if to == n - 1:
print(total + 1)
return
if (to, next_buil) not in dic:
dic[(to, next_buil)] = total + 1
que.append((total + 1, to, next_buil))
print("NA")
while True:
n = int(input())
if n == 0:
break
alst = list(map(int, input().split()))
blst = list(map(int, input().split()))
search(alst, blst, n)
```
No
| 4,108 |
Provide a correct Python 3 solution for this coding contest problem.
You are now examining a unique method to sort a sequence of numbers in increasing order. The method only allows swapping of two numbers that have a common prime factor. For example, a sequence [6, 4, 2, 3, 7] can be sorted using the following steps.
Step 0: 6 4 2 3 7 (given sequence)
Step 1: 2 4 6 3 7 (elements 6 and 2 swapped)
Step 2: 2 6 4 3 7 (elements 4 and 6 swapped)
Step 3: 2 3 4 6 7 (elements 6 and 3 swapped)
Depending on the nature of the sequence, however, this approach may fail to complete the sorting. You have given a name "Coprime sort" to this approach and are now examining if a given sequence is coprime-sortable.
Make a program to determine if a given sequence can be sorted in increasing order by iterating an arbitrary number of swapping operations of two elements that have a common prime number.
Input
The input is given in the following format.
$N$
$a_1$ $a_2$ $...$ $a_N$
The first line provides the number of elements included in the sequence $N$ ($2 \leq N \leq 10^5$). The second line provides an array of integers $a_i$ ($2 \leq a_i \leq 10^5$) that constitute the sequence.
Output
Output "1" if the sequence is coprime-sortable in increasing order, or "0" otherwise.
Examples
Input
5
6 4 2 3 7
Output
1
Input
7
2 9 6 5 6 7 3
Output
0
"Correct Solution:
```
from collections import defaultdict
def main():
def primes(n):
is_prime = [True] * (n + 1)
is_prime[0] = is_prime[1] = False
for i in range(2, int(n ** (1 / 2)) + 1):
if is_prime[i]:
for j in range(i * i, n + 1, i):
is_prime[j] = False
return [i for i in range(n + 1) if is_prime[i]]
prime_list = primes(100000)
solvers = [[] for _ in range(100001)]
for p in prime_list:
for i in range(p, 100001, p):
solvers[i].append(p)
n = int(input())
a_list = list(map(int, input().split()))
parent = [i for i in range(100001)]
def find(x):
if x == parent[x]:
return x
parent[x] = find(parent[x])
return parent[x]
for i, v in enumerate(a_list):
p_rep = find(min(solvers[v]))
for s in solvers[v]:
parent[find(s)] = p_rep
parent[find(v)] = p_rep
dic = defaultdict(set)
for i, v in enumerate(a_list):
dic[find(v)].add(i)
a_list.sort()
for i, v in enumerate(a_list):
if i not in dic[find(v)]:
print(0)
break
else:
print(1)
main()
```
| 4,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In the Indian Puzzle, one is intended to fill out blanks with numbers and operators in a n by n grid in order to make equalities in the grid true (See Figure 1).
<image>
Figure 1
A blank cell should be filled out with a number (from 0 to 9 inclusive) or an operator (+, -, ×, ÷, =), and black cells split equalities. Initially, some cells are already filled with numbers or operators.
The objective of this puzzle is to fill out all the blank cells with numbers (from 0 to 9 inclusive) and operators (+, -, ×, ÷) in a given list. All the numbers and operators in the list must be used to fill out the blank cells.
A equality is organized by more than 2 consecutive left-to-right or top-to-bottom white cells. You can assume that a equality contains exactly one cell with '=' which connects two expressions.
The expressions conform to the order of operations and semantics of conventional four arithmetic operations. First, do all multiplication and division, starting from the left (top). Then, do all addition and subtraction, starting from the left (top). In addition, this puzzle has the following rules:
* Division by zero and division leaving a remainder, are not allowed.
* Leading zeros are not allowed.
* Unary operators like 3×-8=-24 are not allowed.
In a formal description, the equality must conform the syntax in the following BNF:
<Eq> ::= <Ex> = <Ex>
<Ex> ::= <N0> | <Ex> <Op> <N0>
<N0> ::= <N> | 0
<N> ::= <D> | <N> <D> | <N> 0
<Op> ::= + | - | × | ÷
<D> ::= 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
Of course, puzzle creators must make solvable puzzle. Your task is to write a program which verifies whether given puzzle is solvable or not.
Constraints
* Judge data includes at most 100 data sets.
* H, W ≤ 10
* n ≤ 10
Input
Input consists of several datasets. Each dataset consists of:
H W
H × W characters
n
n characters
The integers H and W are the numbers of rows and columns of the grid, respectively. H × W characters denote the grid which contains '.' representing a blank cell, '#' representing a black cell, numbers (from 0 to 9 inclusive), operators('+','-', '*', '/', '=')(× is represented by '*' and ÷ is represented by '/').
The integer n is the number of characters in the list. The last line of a dataset contains n characters indicating numbers and operators for the blank cells.
The end of input is indicated by a line containing two zeros separated by a space.
Output
For each dataset, print "Yes" if the given puzzle is solvable, "No" if not.
Example
Input
5 5
4=..2
+#=#+
.-2=.
=#*#=
.-.=3
6
7 3 1 4 / 8
1 6
8..3=2
2
2 +
0 0
Output
Yes
No
Submitted Solution:
```
import itertools
def myeval(st):
if(not (str(st[0]).isdigit())):
throw()
ok=True
for i in range(len(st)-1):
if((not(str(st[0]).isdigit()))and (not (str(st[i+1]).isdigit()))):
ok=False
if ok:
return eval(st)
else:
throw()
def formula(st):
if(st==""):return True
if(len(st)==1):return True
sts=st.split("=")
if len(sts)==2:
try:
a=myeval(sts[0])
b=myeval(sts[1])
return a==b
except:
return False
else :
return False
def check(field,H,W):
for y in range(H):
row=""
for x in range(W):
row+=str(field[y][x])
sts=row.split("#")
for st in sts:
if (not formula(st)):
return False
else:
a=2
for i in range(W):
col=""
for y in range(H):
col+=str(field[y][x])
sts=col.split("#")
for st in sts:
if (not formula(st)):return False
return True
while True:
H,W=map(int,input().split())
if ( H==0):
break
field=[]
for i in range(H):
st=input()
aa=[]
for x in range(W):
aa.append(st[x])
field.append(aa)
N=int(input())
nums=list(input().split())
unknowns=[]
for y in range(H):
for x in range(W):
if(field[y][x]=='.'):
aa=[x,y]
unknowns.append([x,y])
ok=False
for x in itertools.permutations(nums):
nowfield=field[:]
for i in range(N):
nowfield[unknowns[i][1]][unknowns[i][0]]=x[i]
if check(nowfield,H,W):
ok=True
if ok:
print("Yes")
else:
print("No")
```
No
| 4,110 |
Provide a correct Python 3 solution for this coding contest problem.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
"Correct Solution:
```
#!/usr/bin/env python3
from queue import Queue
def input_split(line):
group, members = line.rstrip('.').split(':')
members = members.split(',')
return group, members
while True:
n = int(input())
if n == 0:
exit()
data = [input() for _ in range(n)]
to = {}
rev = {}
sz = 0
for line in data:
group, members = input_split(line)
if group not in to:
to[group] = sz
rev[sz] = group
sz += 1
for m in members:
if m not in to:
to[m] = sz
rev[sz] = m
sz += 1
g = [[] for _ in range(sz)]
indeg = [0] * sz
for line in data:
group, members = input_split(line)
for m in members:
g[to[m]].append(to[group])
indeg[to[group]] += 1
dp = [set() for i in range(sz)]
q = Queue()
for i in range(sz):
if indeg[i] == 0:
dp[i] |= {i}
q.put(i)
while not q.empty():
cur = q.get()
for nxt in g[cur]:
dp[nxt] |= dp[cur]
indeg[nxt] -= 1
if indeg[nxt] == 0:
q.put(nxt)
print(len(dp[0]))
```
| 4,111 |
Provide a correct Python 3 solution for this coding contest problem.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
"Correct Solution:
```
# coding: utf-8
def addset(args):
global a
for m in args:
if dic[m]==1:
a.add(m)
else:
if m not in used:
used.add(m)
addset(dc[m])
while True:
n=int(input())
if n==0:
break
dic={}
first=[]
dc={}
for i in range(n):
g,m=input().split(':')
m=m.rstrip('.').split(',')
dc[g]=m
if i==0:
first.append(g)
first+=m
dic[g]=2
for mem in m:
if mem not in dic:
dic[mem]=1
a=set()
used=set()
addset(first[1:])
print(len(a))
```
| 4,112 |
Provide a correct Python 3 solution for this coding contest problem.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
"Correct Solution:
```
while True:
N = int(input())
if N == 0:
break
first_group_menbers = set(input()[:-1].split(":")[1].split(","))
d = {}
for i in range(N-1):
group, members = input()[:-1].split(":")
d[group] = set(members.split(","))
for i in range(N-1):
for group, members in d.items():
if group in first_group_menbers:
first_group_menbers.remove(group)
for m in members:
first_group_menbers.add(m)
print(len(first_group_menbers))
```
| 4,113 |
Provide a correct Python 3 solution for this coding contest problem.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
"Correct Solution:
```
while True:
N = int(input())
if N == 0:
break
first_group_menbers = set(input()[:-1].split(":")[1].split(","))
d = {}
for i in range(N-1):
group, members = input()[:-1].split(":")
d[group] = set(members.split(","))
for i in range(N):
for group, members in d.items():
if group in first_group_menbers:
first_group_menbers.remove(group)
for m in members:
first_group_menbers.add(m)
print(len(first_group_menbers))
```
| 4,114 |
Provide a correct Python 3 solution for this coding contest problem.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
"Correct Solution:
```
while True:
N = int(input())
if N == 0:
break
first_group_menbers = set(input()[:-1].split(":")[1].split(","))
d = {}
for i in range(N-1):
group, members = input()[:-1].split(":")
d[group] = set(members.split(","))
for i in range(N-2):
for group, members in d.items():
if group in first_group_menbers:
first_group_menbers.remove(group)
for m in members:
first_group_menbers.add(m)
print(len(first_group_menbers))
```
| 4,115 |
Provide a correct Python 3 solution for this coding contest problem.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
"Correct Solution:
```
import re
N = 1
while True:
N = int(input())
if not N:
break
tmp, groups, firstgroup = [], {}, []
for i in range(N):
tmp = re.split(r"[\,\.\:]", input())
if not i:
firstgroup = tmp[1:]
else:
groups[tmp[0]] = tmp[1:]
while True:
change = False
tmp = []
for i in range(len(firstgroup)):
if firstgroup[i] in groups:
tmp += groups[firstgroup[i]]
change = True
else:
tmp.append(firstgroup[i])
firstgroup = ['']
for i in tmp:
if i not in firstgroup:
firstgroup.append(i)
if not change:
break
print(len(firstgroup) - 1)
```
| 4,116 |
Provide a correct Python 3 solution for this coding contest problem.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n = I()
if n == 0:
break
a = [S().split(':') for _ in range(n)]
ts = set([c[0] for c in a])
ms = {}
for t,m in a:
ms[t] = m[:-1].split(',')
fm = {}
def f(c):
if c in fm:
return fm[c]
if c in ts:
s = set()
for k in ms[c]:
s |= f(k)
fm[c] = s
return s
fm[c] = set([c])
return fm[c]
r = f(a[0][0])
rr.append(len(r))
return '\n'.join(map(str,rr))
print(main())
```
| 4,117 |
Provide a correct Python 3 solution for this coding contest problem.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
"Correct Solution:
```
#!/usr/bin/env python3
import sys
import math
import re
from bisect import bisect_right as br
from bisect import bisect_left as bl
sys.setrecursionlimit(1000000)
from heapq import heappush, heappop,heappushpop
from collections import defaultdict
from itertools import accumulate
from collections import Counter
from collections import deque
from operator import itemgetter
from itertools import permutations
mod = 10**9 + 7
inf = float('inf')
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
while 1:
n = I()
if n == 0:
quit()
f = defaultdict(dict)
t = defaultdict(lambda : False)
ch = defaultdict(lambda : False)
stack = []
for i in range(n):
x = input()[:-1].split(':')
y = x[1].split(',')
x = x[0]
for j in y:
f[x][j] = 1
if i != 0:
continue
stack.append(j)
ans = 0
while stack:
x = stack.pop()
if ch[x]:
continue
ch[x] = True
if len(f[x]) == 0 and not t[x]:
ans += 1
t[x] = True
continue
for i,j in f[x].items():
stack.append(i)
print(ans)
```
| 4,118 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
Submitted Solution:
```
while True:
n = int(input())
if(n==0): break
data = {}
firstname = ''
for i in range(n):
d = input()
name = d.split(':')[0]
if(i==0): firstname = name
data[name] = []
for x in d.split(':')[1].strip('.').split(','):
data[name].append(x)
removed = []
for x in data[firstname]:
if x in data:
removed.append(x)
for y in data[x]:
data[firstname].append(y)
data.pop(x)
edited = set(data[firstname])
for i in removed:
edited.remove(i)
print(len(edited))
```
Yes
| 4,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
Submitted Solution:
```
#!/usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from bisect import bisect_left, bisect_right
import sys, random, itertools, math
sys.setrecursionlimit(10**5)
input = sys.stdin.readline
sqrt = math.sqrt
def LI(): return list(map(int, input().split()))
def LF(): return list(map(float, input().split()))
def LI_(): return list(map(lambda x: int(x)-1, input().split()))
def II(): return int(input())
def IF(): return float(input())
def LS(): return list(map(list, input().split()))
def S(): return list(input().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [input().rstrip()[:-1] for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = 1e10
#solve
def solve():
n = II()
if n == 0:
return False
group = set()
go_group = set()
member = set()
d = defaultdict(int)
g = SR(n)
for gi in g:
group.add(gi.split(":")[0])
d[gi.split(":")[0]] = gi.split(":")[1].split(",")
C = defaultdict(int)
def s(g):
for gi in g:
if gi in group:
if C[gi]:
continue
C[gi] = 1
s(d[gi])
else:
member.add(gi)
s(d[g[0].split(":")[0]])
print(len(member))
return True
#main
if __name__ == '__main__':
while solve():
pass
```
Yes
| 4,120 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
Submitted Solution:
```
def check(dic,index):
ret = []
for c in dic[index]:
if c in dic.keys():
for d in check(dic,c):
if not(d in ret):
ret.append(d)
else:
if not(c in ret):
ret.append(c)
dic[index] = ret
return ret
while True:
n = int(input())
if n == 0:
break
dic = {}
index = ''
for i in range(n):
a,b = [char for char in input()[0:-1].split(':')]
if i == 0:
index = a
b = b.split(',')
dic[a] = b
#print(dic.keys())
print(len(check(dic,index)))
```
Yes
| 4,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
Submitted Solution:
```
def get_num(u, map, vis):
if u in vis:
return 0;
else:
vis[u] = True;
ans = 0;
if u in map:
for v in map[u]:
ans += get_num(v, map, vis);
else:
ans += 1;
return ans;
def main():
while True:
n = int(input());
if n == 0:
break;
edge = {}; name = {}; vis = {};
for i in range(0, n):
str = input();
u = name[i] = str.split(':')[0];
group = str.split(':')[1].split('.')[0].split(',');
for v in group:
if u in edge:
edge[u][v] = True;
else:
edge[u] = {v:True};
print(get_num(name[0], edge, vis));
main();
```
Yes
| 4,122 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
Submitted Solution:
```
from django.template.defaultfilters import first
while True:
N = int(input())
if N == 0:
break
first_group_menbers = set(input()[:-1].split(":")[1].split(","))
d = {}
for i in range(N-1):
group, members = input()[:-1].split(":")
d[group] = set(members.split(","))
for i in range(N):
for group, members in d.items():
if group in first_group_menbers:
first_group_menbers.remove(group)
for m in members:
first_group_menbers.add(m)
print(len(first_group_menbers))
```
No
| 4,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
Submitted Solution:
```
while True:
N = int(input())
if N == 0:
break
first_group_menbers = set(input()[:-1].split(":")[1].split(","))
d = {}
for i in range(N-1):
group, members = input()[:-1].split(":")
d[group] = set(members.split(","))
for i in range(N-3):
for group, members in d.items():
if group in first_group_menbers:
first_group_menbers.remove(group)
for m in members:
first_group_menbers.add(m)
print(len(first_group_menbers))
```
No
| 4,124 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
Submitted Solution:
```
def get_group(str):
return str.split(':')[0];
def get_member(str):
return str.split(':')[1].split('.')[0].split(',');
def get_num(group, map, ans):
for person in map[group]:
if person in map:
get_num(person, map, ans);
else:
ans[person] = True;
def main():
while True:
str = input();
if len(str) == 0:
continue;
n = int(str);
if n == 0:
break;
group = {};
group.clear();
name = {};
name.clear();
for i in range(0, n):
str = input();
name[i] = get_group(str);
group[name[i]] = get_member(str);
ans = {};
ans.clear();
get_num(name[0], group, ans);
print(len(ans));
main();
```
No
| 4,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Peter is a senior manager of Agile Change Management (ACM) Inc., where each employee is a member of one or more task groups. Since ACM is agile, task groups are often reorganized and their members frequently change, so membership management is his constant headache.
Peter updates the membership information whenever any changes occur: for instance, the following line written by him means that Carol and Alice are the members of the Design Group.
design:carol,alice.
The name preceding the colon is the group name and the names following it specify its members.
A smaller task group may be included in a larger one. So, a group name can appear as a member of another group, for instance, as follows.
development:alice,bob,design,eve.
Simply unfolding the design above gives the following membership specification, which is equivalent to the original.
development:alice,bob,carol,alice,eve.
In this case, however, alice occurs twice. After removing one of the duplicates, we have the following more concise specification.
development:alice,bob,carol,eve.
Your mission in this problem is to write a program that, given group specifications, identifies group members.
Note that Peter's specifications can include deeply nested groups. In the following, for instance, the group one contains a single member dave.
one:another.
another:yetanother.
yetanother:dave.
Input
The input is a sequence of datasets, each being in the following format.
n
group1:member1,1,...,member1,m1.
.
.
.
groupi:memberi,1,...,memberi,mi.
.
.
.
groupn:membern,1,...,membern,mn.
The first line contains n, which represents the number of groups and is a positive integer no more than 100. Each of the following n lines contains the membership information of a group: groupi (1 ≤ i ≤ n) is the name of the i-th task group and is followed by a colon (:) and then the list of its mi member s that are delimited by a comma (,) and terminated by a period (.).
Those group names are mutually different. Each mi (1 ≤ i ≤ n) is between 1 and 10, inclusive. A member is another group name if it is one of group1, group2,..., or groupn. Otherwise it is an employee name.
There are no circular (or recursive) definitions of group(s). You may assume that mi member names of a group are mutually different.
Each group or employee name is a non-empty character string of length between 1 and 15, inclusive, and consists of lowercase letters.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, output the number of employees included in the first group of the dataset, that is group1, in a line. No extra characters should occur in the output.
Example
Input
2
development:alice,bob,design,eve.
design:carol,alice.
3
one:another.
another:yetanother.
yetanother:dave.
3
friends:alice,bob,bestfriends,carol,fran,badcompany.
bestfriends:eve,alice.
badcompany:dave,carol.
5
a:b,c,d,e.
b:c,d,e,f.
c:d,e,f,g.
d:e,f,g,h.
e:f,g,h,i.
4
aa:bb.
cc:dd,ee.
ff:gg.
bb:cc.
0
Output
4
1
6
4
2
Submitted Solution:
```
def get_group(str):
return str.split(':')[0];
def get_member(str):
return str.split(':')[1].split('.')[0].split(',');
def get_num(group, map, ans):
for person in map[group]:
if person in map:
get_num(person, map, ans);
else:
ans[person] = True;
def main():
while True:
str = input();
if len(str) == 0:
continue;
n = int(str);
if not str.isdigit():
continue;
if n == 0:
break;
group = {};
group.clear();
name = {};
name.clear();
for i in range(0, n):
str = input();
name[i] = get_group(str);
group[name[i]] = get_member(str);
ans = {};
ans.clear();
get_num(name[0], group, ans);
print(len(ans));
main();
```
No
| 4,126 |
Provide a correct Python 3 solution for this coding contest problem.
A young boy John is playing with eight triangular panels. These panels are all regular triangles of the same size, each painted in a single color; John is forming various octahedra with them.
While he enjoys his playing, his father is wondering how many octahedra can be made of these panels since he is a pseudo-mathematician. Your task is to help his father: write a program that reports the number of possible octahedra for given panels. Here, a pair of octahedra should be considered identical when they have the same combination of the colors allowing rotation.
Input
The input consists of multiple datasets. Each dataset has the following format:
Color1 Color2 ... Color8
Each Colori (1 ≤ i ≤ 8) is a string of up to 20 lowercase alphabets and represents the color of the i-th triangular panel.
The input ends with EOF.
Output
For each dataset, output the number of different octahedra that can be made of given panels.
Example
Input
blue blue blue blue blue blue blue blue
red blue blue blue blue blue blue blue
red red blue blue blue blue blue blue
Output
1
1
3
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
# aa
# 1 2 4 5
# 2 3 1 6
# 3 4 2 7
# 4 1 3 8
# 5 8 6 1
# 6 5 7 2
# 7 6 8 3
# 8 7 5 4
#
def main():
rr = []
k = [
[2, 4, 5],
[3, 1, 6],
[4, 2, 7],
[1, 3, 8],
[8, 6, 1],
[5, 7, 2],
[6, 8, 3],
[7, 5, 4]
]
for i in range(8):
for j in range(3):
k[i][j] -= 1
ptn = []
for i in range(8):
for j in range(3):
t = [-1] * 8
t[0] = i
t[1] = k[i][j]
t[3] = k[i][(j+1)%3]
for l in range(8):
if l == i:
continue
if t[1] in k[l] and t[3] in k[l]:
t[2] = l
break
for l in range(4):
kl = k[t[l]]
for m in range(3):
if kl[m] not in t:
t[l+4] = kl[m]
break
ptn.append(t)
def f(a):
s = set()
r = 0
tc = 0
for t in itertools.permutations(a):
tc += 1
if t in s:
continue
r += 1
for p in ptn:
s.add(tuple(t[p[i]] for i in range(8)))
return r
while True:
a = LS()
if len(a) == 0:
break
rr.append(f(a))
return '\n'.join(map(str,rr))
print(main())
```
| 4,127 |
Provide a correct Python 3 solution for this coding contest problem.
Problem statement
2D, who is good at cooking, is trying to make lunch. Cooking requires all N ingredients a_ {0}, a_ {1},…, a_ {N−1}.
Now, 2D's refrigerator doesn't contain any ingredients, so I have to go to the supermarket to buy it. At the supermarket, you can buy the material a_ {i} for the price x_ {i} yen.
2D is also a wizard and can use M types of magic. The i-th magic can be changed to the material t_ {i} by applying it to the material s_ {i}, and conversely to s_ {i} by applying it to the material t_ {i}. In addition, you can repeatedly use multiple spells on a single material. For example, you can get r from p using the magic of changing from p to q and the magic of changing from q to r.
2D decided to use the power of magic to prepare the materials as cheaply as possible. Find the minimum sum of the prices of the ingredients that 2D needs to buy to complete the dish.
input
The input is given in the following format.
N
a_ {0} x_ {0}
...
a_ {N−1} x_ {N−1}
M
s_ {0} t_ {0}
...
s_ {M−1} t_ {M−1}
Constraint
* All numbers are integers
* All material names consist of at least 1 and no more than 10 lowercase letters.
* If i ≠ j, then a_ {i} ≠ a_ {j}
* 1 \ ≤ x_ {i} \ ≤ 1,000
* 1 \ ≤ N \ ≤ 5,000
* 0 \ ≤ M \ ≤ {\ rm min} (N (N−1) / 2, 1000)
* s_ {i} ≠ t_ {i}
* There is no duplication in the pair of s_ {i}, t_ {i}
* s_ {i}, t_ {i} are included in a_ {0},…, a_ {N−1}
output
Print the answer in one line.
sample
Sample input 1
2
tako 2
yaki 1
1
tako yaki
You can magically turn a cheap yaki into a tako, so buy two yaki.
Sample output 1
2
Sample input 2
Five
a 1
b 2
c 2
d 4
e 3
Five
b a
a c
c d
e b
c b
Sample output 2
Five
As shown below, all materials can be changed from a.
* a: a as it is
* b: a-> c-> b
* c: a-> c
* d: a-> c-> d
* e: a-> b-> e
<image>
Example
Input
2
tako 2
yaki 1
1
tako yaki
Output
2
"Correct Solution:
```
n = int(input())
dic = {}
price = []
for i in range(n):
a, x = input().split()
dic[a] = i
price.append(int(x))
parent = [i for i in range(n)]
def find(x):
if parent[x] == x:return x
parent[x] = find(parent[x])
return parent[x]
m = int(input())
for _ in range(m):
s, t = input().split()
si, ti = dic[s], dic[t]
ps = find(si)
pt = find(ti)
parent[ps] = parent[pt]
groups = {}
for i in range(n):
if find(i) in groups:
min_price, cnt = groups[find(i)]
groups[find(i)] = (min(min_price, price[i]), cnt + 1)
else:
groups[find(i)] = (price[i], 1)
print(sum([v[0] * v[1] for _, v in groups.items()]))
```
| 4,128 |
Provide a correct Python 3 solution for this coding contest problem.
Problem statement
2D, who is good at cooking, is trying to make lunch. Cooking requires all N ingredients a_ {0}, a_ {1},…, a_ {N−1}.
Now, 2D's refrigerator doesn't contain any ingredients, so I have to go to the supermarket to buy it. At the supermarket, you can buy the material a_ {i} for the price x_ {i} yen.
2D is also a wizard and can use M types of magic. The i-th magic can be changed to the material t_ {i} by applying it to the material s_ {i}, and conversely to s_ {i} by applying it to the material t_ {i}. In addition, you can repeatedly use multiple spells on a single material. For example, you can get r from p using the magic of changing from p to q and the magic of changing from q to r.
2D decided to use the power of magic to prepare the materials as cheaply as possible. Find the minimum sum of the prices of the ingredients that 2D needs to buy to complete the dish.
input
The input is given in the following format.
N
a_ {0} x_ {0}
...
a_ {N−1} x_ {N−1}
M
s_ {0} t_ {0}
...
s_ {M−1} t_ {M−1}
Constraint
* All numbers are integers
* All material names consist of at least 1 and no more than 10 lowercase letters.
* If i ≠ j, then a_ {i} ≠ a_ {j}
* 1 \ ≤ x_ {i} \ ≤ 1,000
* 1 \ ≤ N \ ≤ 5,000
* 0 \ ≤ M \ ≤ {\ rm min} (N (N−1) / 2, 1000)
* s_ {i} ≠ t_ {i}
* There is no duplication in the pair of s_ {i}, t_ {i}
* s_ {i}, t_ {i} are included in a_ {0},…, a_ {N−1}
output
Print the answer in one line.
sample
Sample input 1
2
tako 2
yaki 1
1
tako yaki
You can magically turn a cheap yaki into a tako, so buy two yaki.
Sample output 1
2
Sample input 2
Five
a 1
b 2
c 2
d 4
e 3
Five
b a
a c
c d
e b
c b
Sample output 2
Five
As shown below, all materials can be changed from a.
* a: a as it is
* b: a-> c-> b
* c: a-> c
* d: a-> c-> d
* e: a-> b-> e
<image>
Example
Input
2
tako 2
yaki 1
1
tako yaki
Output
2
"Correct Solution:
```
import sys
from collections import defaultdict,deque
def root(x):
if par[x] == x:
return x
par[x] = root(par[x])
return par[x]
def unite(x,y):
x = root(x)
y = root(y)
if rank[x] < rank[y]:
par[x] = y
else:
par[y] = x
if rank[x] == rank[y]:
rank[x] += 1
f = defaultdict(int)
par = defaultdict(int)
rank = defaultdict(int)
n = int(sys.stdin.readline())
for i in range(n):
a,x = sys.stdin.readline().split()
f[a] = int(x)
par[a] = a
rank[a] = 0
m = int(sys.stdin.readline())
for i in range(m):
x,y = sys.stdin.readline().split()
if root(x) != root(y):
unite(x,y)
ans = 0
for i in f.keys():
root(i)
f[par[i]] = min(f[par[i]],f[i])
for i in f.keys():
ans += f[root(i)]
print(ans)
```
| 4,129 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Prepare the Othello board. The upper left is $ (1,1) $ and the lower right is $ (8,8) $. The board to be prepared here is $ (5,4) as follows. ) There is no $ black stone.
........
........
........
... ox ...
.... o ...
........
........
........
Kuroishi: x Shiraishi: o
8x8 board
From this state, Black starts Othello first.
AOR Ika-chan plays $ q $ times according to the following rules.
A rectangular area is given, with the upper left as the cell in the $ a $ and $ b $ columns, and the lower right as the cell in the $ c $ and $ d $ columns.
AOR Ika-chan puts black stones and white stones alternately according to the rules of Othello (reference: Wikipedia Othello) so as to maximize the number of stones contained in this rectangular area.
When no more stones can be placed (when both Shiraishi and Kuroishi have to pass or when all the boards are filled), the game ends.
In each game, output the number of stones when the number of stones contained in the area is maximized.
Since the number of inputs and outputs may increase, it is recommended to use high-speed functions for input and output.
output
In each game, output the number of stones in the area when the number of stones is maximized. Also, output a line break at the end.
Example
Input
3
1 1 8 8
2 4 3 8
8 8 8 8
Output
48
7
1
"Correct Solution:
```
def inpl(): return list(map(int, input().split()))
from collections import defaultdict
G = dict()
for i in range(9):
G[(i, 9)] = 0
G[(9, i)] = 0
G[(9, 9)] = 0
for i in range(8)[::-1]:
a = 8 - i
b = -(-(8 - i)//2)
i= i+1
for j in range(8)[::-1]:
if j % 2 == 1:
j = j+1
G[(i, j)] = G[(i, j+1)] + b
else:
j = j+1
G[(i, j)] = G[(i, j+1)] + a
q = int(input())
for _ in range(q):
a, b, c, d = inpl()
print(G[(a, b)] - G[(c+1, b)] - G[(a, d+1)] + G[(c+1), (d+1)])
```
| 4,130 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Prepare the Othello board. The upper left is $ (1,1) $ and the lower right is $ (8,8) $. The board to be prepared here is $ (5,4) as follows. ) There is no $ black stone.
........
........
........
... ox ...
.... o ...
........
........
........
Kuroishi: x Shiraishi: o
8x8 board
From this state, Black starts Othello first.
AOR Ika-chan plays $ q $ times according to the following rules.
A rectangular area is given, with the upper left as the cell in the $ a $ and $ b $ columns, and the lower right as the cell in the $ c $ and $ d $ columns.
AOR Ika-chan puts black stones and white stones alternately according to the rules of Othello (reference: Wikipedia Othello) so as to maximize the number of stones contained in this rectangular area.
When no more stones can be placed (when both Shiraishi and Kuroishi have to pass or when all the boards are filled), the game ends.
In each game, output the number of stones when the number of stones contained in the area is maximized.
Since the number of inputs and outputs may increase, it is recommended to use high-speed functions for input and output.
output
In each game, output the number of stones in the area when the number of stones is maximized. Also, output a line break at the end.
Example
Input
3
1 1 8 8
2 4 3 8
8 8 8 8
Output
48
7
1
"Correct Solution:
```
#!usr/bin/env python3
from collections import defaultdict
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS():return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = I()
return l
def LIR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LI()
return l
def SR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = S()
return l
def LSR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = SR()
return l
mod = 1000000007
#A
"""
h,w = LI()
a,b = LI()
y = h//a
x = w//b
ans = h*w-a*y*b*x
print(ans)
"""
#B
"""
def m(x,y):
if x == y:
return "T"
if x == "F":
return "T"
return "F"
n = I()
p = input().split()
ans = p[0]
for i in range(1,n):
ans = m(ans,p[i])
print(ans)
"""
#C
s = [[1]*8 for i in range(4)]
for i in range(4):
s.insert(2*i,[1,0,1,0,1,0,1,0])
for j in range(8):
for i in range(1,8):
s[j][i] += s[j][i-1]
for j in range(8):
for i in range(1,8):
s[i][j] += s[i-1][j]
s.insert(0,[0,0,0,0,0,0,0,0,0])
for i in range(1,9):
s[i].insert(0,0)
q = I()
for _ in range(q):
a,b,c,d = LI()
print(s[c][d]-s[c][b-1]-s[a-1][d]+s[a-1][b-1])
#D
"""
n = I()
a = LI()
dp = [float("inf") for i in range(n)]
b = [[] for i in range(n)]
ind = [0 for i in range(100001)]
for i in range(n):
k = bisect.bisect_left(dp,a[i])
dp[k] = a[i]
b[k].append(a[i])
ind[a[i]] = max(i,ind[a[i]])
for i in range(n):
if dp[i] == float("inf"):break
b[i].sort()
b[i] = b[i][::-1]
i -= 1
ans = b[i][0]
now_ind = ind[b[i][0]]
now_v = b[i][0]
while i >= 0:
for j in b[i]:
if ind[j] < now_ind and j < now_v:
ans += j
now_ind = ind[j]
now_v = j
i -= 1
print(ans)
"""
#E
#F
#G
#H
#I
#J
#K
#L
#M
#N
#O
#P
#Q
#R
#S
#T
```
| 4,131 |
Provide a correct Python 3 solution for this coding contest problem.
problem
Prepare the Othello board. The upper left is $ (1,1) $ and the lower right is $ (8,8) $. The board to be prepared here is $ (5,4) as follows. ) There is no $ black stone.
........
........
........
... ox ...
.... o ...
........
........
........
Kuroishi: x Shiraishi: o
8x8 board
From this state, Black starts Othello first.
AOR Ika-chan plays $ q $ times according to the following rules.
A rectangular area is given, with the upper left as the cell in the $ a $ and $ b $ columns, and the lower right as the cell in the $ c $ and $ d $ columns.
AOR Ika-chan puts black stones and white stones alternately according to the rules of Othello (reference: Wikipedia Othello) so as to maximize the number of stones contained in this rectangular area.
When no more stones can be placed (when both Shiraishi and Kuroishi have to pass or when all the boards are filled), the game ends.
In each game, output the number of stones when the number of stones contained in the area is maximized.
Since the number of inputs and outputs may increase, it is recommended to use high-speed functions for input and output.
output
In each game, output the number of stones in the area when the number of stones is maximized. Also, output a line break at the end.
Example
Input
3
1 1 8 8
2 4 3 8
8 8 8 8
Output
48
7
1
"Correct Solution:
```
q = int(input())
li = [input().split() for i in range(q)]
for i in li:
a, b, c, d = map(int, i)
ans = 0
for j in range(a, c+1):
for k in range(b, d+1):
ans += 0 if j%2 == 1 and k%2 == 0 else 1
print(ans)
```
| 4,132 |
Provide a correct Python 3 solution for this coding contest problem.
For given $N$ points in the 2D Euclidean plane, find the distance of the shortest tour that meets the following criteria:
* Visit the points according to the following steps:
1. It starts from the leftmost point (starting point), goes strictly from left to right, and then visits the rightmost point (turn-around point).
2. Then it starts from the turn-around point, goes strictly from right to left, and then back to the starting point.
* Through the processes 1. 2., the tour must visit each point at least once.
Constraints
* $2 \leq N \leq 1000$
* $-1000 \leq x_i, y_i \leq 1000$
* $x_i$ differ from each other
* The given points are already sorted by x-coordinates
Input
The input data is given in the following format:
$N$
$x_1$ $y_1$
$x_2$ $y_2$
...
$x_N$ $y_N$
Output
Print the distance of the shortest tour in a line. The output should not have an error greater than 0.0001.
Examples
Input
3
0 0
1 1
2 0
Output
4.82842712
Input
4
0 1
1 2
2 0
3 1
Output
7.30056308
Input
5
0 0
1 2
2 1
3 2
4 0
Output
10.94427191
"Correct Solution:
```
import sys
# import re
import math
import collections
# import decimal
import bisect
import itertools
import fractions
# import functools
import copy
# import heapq
import decimal
# import statistics
import queue
# import numpy as np
sys.setrecursionlimit(10000001)
INF = 10 ** 16
MOD = 10 ** 9 + 7
# MOD = 998244353
ni = lambda: int(sys.stdin.readline())
ns = lambda: map(int, sys.stdin.readline().split())
na = lambda: list(map(int, sys.stdin.readline().split()))
# ===CODE===
eps = 10 ** (-9)
def main():
n = ni()
e = []
for _ in range(n):
xi, yi = ns()
e.append([xi, yi])
dp = [[INF for _ in range(n)] for _ in range(n)]
dist = lambda p1, p2: math.sqrt((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2)
dp[0][0] = 0
for i in range(n - 1):
for j in range(n):
if dp[i][j] != INF:
dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + dist(e[i], e[i + 1]))
dp[i + 1][i] = min(dp[i + 1][i], dp[i][j] + dist(e[j], e[i + 1]))
for i in range(n):
dp[n - 1][n - 1] = min(dp[n - 1][n - 1], dp[n - 1][i] + dist(e[n - 1], e[i]))
print(dp[n - 1][n - 1])
if __name__ == '__main__':
main()
```
| 4,133 |
Provide a correct Python 3 solution for this coding contest problem.
For given $N$ points in the 2D Euclidean plane, find the distance of the shortest tour that meets the following criteria:
* Visit the points according to the following steps:
1. It starts from the leftmost point (starting point), goes strictly from left to right, and then visits the rightmost point (turn-around point).
2. Then it starts from the turn-around point, goes strictly from right to left, and then back to the starting point.
* Through the processes 1. 2., the tour must visit each point at least once.
Constraints
* $2 \leq N \leq 1000$
* $-1000 \leq x_i, y_i \leq 1000$
* $x_i$ differ from each other
* The given points are already sorted by x-coordinates
Input
The input data is given in the following format:
$N$
$x_1$ $y_1$
$x_2$ $y_2$
...
$x_N$ $y_N$
Output
Print the distance of the shortest tour in a line. The output should not have an error greater than 0.0001.
Examples
Input
3
0 0
1 1
2 0
Output
4.82842712
Input
4
0 1
1 2
2 0
3 1
Output
7.30056308
Input
5
0 0
1 2
2 1
3 2
4 0
Output
10.94427191
"Correct Solution:
```
import sys
from math import sqrt
readline = sys.stdin.readline
write = sys.stdout.write
N = int(readline())
PS = [list(map(int, readline().split())) for i in range(N)]
PS.sort()
def dist(P, Q):
return sqrt((P[0] - Q[0])**2 + (P[1] - Q[1])**2)
INF = 10**18
S = [INF]*N; T = [INF]*N
S[0] = T[0] = dist(PS[0], PS[1])
for k in range(2, N):
s = t = INF
for i in range(k-1):
d = dist(PS[i], PS[k])
s = min(s, T[i] + d)
t = min(t, S[i] + d)
S[k-1] = s; T[k-1] = t
d = dist(PS[k-1], PS[k])
for i in range(k-1):
S[i] += d; T[i] += d
print(min(min(s, t) + dist(PS[i], PS[N-1]) for i, (s, t) in enumerate(zip(S, T))))
```
| 4,134 |
Provide a correct Python 3 solution for this coding contest problem.
For given $N$ points in the 2D Euclidean plane, find the distance of the shortest tour that meets the following criteria:
* Visit the points according to the following steps:
1. It starts from the leftmost point (starting point), goes strictly from left to right, and then visits the rightmost point (turn-around point).
2. Then it starts from the turn-around point, goes strictly from right to left, and then back to the starting point.
* Through the processes 1. 2., the tour must visit each point at least once.
Constraints
* $2 \leq N \leq 1000$
* $-1000 \leq x_i, y_i \leq 1000$
* $x_i$ differ from each other
* The given points are already sorted by x-coordinates
Input
The input data is given in the following format:
$N$
$x_1$ $y_1$
$x_2$ $y_2$
...
$x_N$ $y_N$
Output
Print the distance of the shortest tour in a line. The output should not have an error greater than 0.0001.
Examples
Input
3
0 0
1 1
2 0
Output
4.82842712
Input
4
0 1
1 2
2 0
3 1
Output
7.30056308
Input
5
0 0
1 2
2 1
3 2
4 0
Output
10.94427191
"Correct Solution:
```
import math
from typing import List, Tuple
def dist(p1: Tuple[int, int], p2: Tuple[int, int]) -> float:
return math.sqrt((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2)
if __name__ == "__main__":
N = int(input())
points: List[Tuple[int, int]] = [(0, 0)] * N
for idx in range(N):
x, y = map(lambda x: int(x), input().split())
points[idx] = (x, y)
table1 = [float('inf')] * N
table2 = [float('inf')] * N
table1[0] = table2[0] = dist(points[0], points[1])
for i in range(2, N):
d1 = d2 = float('inf')
for j in range(i - 1):
d = dist(points[i], points[j])
d1 = min(d1, table2[j] + d)
d2 = min(d2, table1[j] + d)
table1[i - 1] = d1
table2[i - 1] = d2
d = dist(points[i - 1], points[i])
for j in range(i - 1):
table1[j] += d
table2[j] += d
ans = min(min(d1, d2) + dist(points[i], points[N - 1])
for i, (d1, d2) in enumerate(zip(table1, table2)))
print(ans)
```
| 4,135 |
Provide a correct Python 3 solution for this coding contest problem.
For given $N$ points in the 2D Euclidean plane, find the distance of the shortest tour that meets the following criteria:
* Visit the points according to the following steps:
1. It starts from the leftmost point (starting point), goes strictly from left to right, and then visits the rightmost point (turn-around point).
2. Then it starts from the turn-around point, goes strictly from right to left, and then back to the starting point.
* Through the processes 1. 2., the tour must visit each point at least once.
Constraints
* $2 \leq N \leq 1000$
* $-1000 \leq x_i, y_i \leq 1000$
* $x_i$ differ from each other
* The given points are already sorted by x-coordinates
Input
The input data is given in the following format:
$N$
$x_1$ $y_1$
$x_2$ $y_2$
...
$x_N$ $y_N$
Output
Print the distance of the shortest tour in a line. The output should not have an error greater than 0.0001.
Examples
Input
3
0 0
1 1
2 0
Output
4.82842712
Input
4
0 1
1 2
2 0
3 1
Output
7.30056308
Input
5
0 0
1 2
2 1
3 2
4 0
Output
10.94427191
"Correct Solution:
```
import math
pts = []
line = input()
N = int(line)
for _ in range(0, N):
line = input()
x, y = list(map(int, line.split()))
pts += [[x, y]]
def dist(i, j):
x1, y1 = pts[i]
x2, y2 = pts[j]
return math.sqrt((x1 - x2)**2 + (y1 - y2)**2)
def btsp():
dp = [ [0] * N for _ in range(0, N)]
for i in range(1, N):
dp[0][i] = dp[0][i - 1] + dist(i - 1, i)
for i in range(1, N):
for j in range(i, N):
if i == j - 1 or i == j:
m = 10 ** 10
for k in range(0, i):
m = min(m, dp[k][i] + dist(j, k))
dp[i][j] = m
else:
dp[i][j] = dp[i][j - 1] + dist(j - 1, j)
return dp[N - 1][N - 1]
print(btsp())
```
| 4,136 |
Provide a correct Python 3 solution for this coding contest problem.
For given $N$ points in the 2D Euclidean plane, find the distance of the shortest tour that meets the following criteria:
* Visit the points according to the following steps:
1. It starts from the leftmost point (starting point), goes strictly from left to right, and then visits the rightmost point (turn-around point).
2. Then it starts from the turn-around point, goes strictly from right to left, and then back to the starting point.
* Through the processes 1. 2., the tour must visit each point at least once.
Constraints
* $2 \leq N \leq 1000$
* $-1000 \leq x_i, y_i \leq 1000$
* $x_i$ differ from each other
* The given points are already sorted by x-coordinates
Input
The input data is given in the following format:
$N$
$x_1$ $y_1$
$x_2$ $y_2$
...
$x_N$ $y_N$
Output
Print the distance of the shortest tour in a line. The output should not have an error greater than 0.0001.
Examples
Input
3
0 0
1 1
2 0
Output
4.82842712
Input
4
0 1
1 2
2 0
3 1
Output
7.30056308
Input
5
0 0
1 2
2 1
3 2
4 0
Output
10.94427191
"Correct Solution:
```
import sys
sys.setrecursionlimit(1000000)
def calc(p1, p2, v):
dists_v = dists[v]
if v == n - 1:
return dists_v[p1] + dists_v[p2]
if visited[p1][p2]:
return dp[p1][p2]
dp[p1][p2] = min(dists_v[p1] + calc(p2, v, v + 1), dists_v[p2] + calc(p1, v, v + 1))
visited[p1][p2] = True
return dp[p1][p2]
n = int(input())
points = [complex(*map(int, input().split())) for _ in range(n)]
dists = [[abs(p1 - p2) for p2 in points] for p1 in points]
visited = [[False] * n for _ in range(n)]
dp = [[0] * n for _ in range(n)]
print(calc(0, 0, 0))
```
| 4,137 |
Provide a correct Python 3 solution for this coding contest problem.
For given $N$ points in the 2D Euclidean plane, find the distance of the shortest tour that meets the following criteria:
* Visit the points according to the following steps:
1. It starts from the leftmost point (starting point), goes strictly from left to right, and then visits the rightmost point (turn-around point).
2. Then it starts from the turn-around point, goes strictly from right to left, and then back to the starting point.
* Through the processes 1. 2., the tour must visit each point at least once.
Constraints
* $2 \leq N \leq 1000$
* $-1000 \leq x_i, y_i \leq 1000$
* $x_i$ differ from each other
* The given points are already sorted by x-coordinates
Input
The input data is given in the following format:
$N$
$x_1$ $y_1$
$x_2$ $y_2$
...
$x_N$ $y_N$
Output
Print the distance of the shortest tour in a line. The output should not have an error greater than 0.0001.
Examples
Input
3
0 0
1 1
2 0
Output
4.82842712
Input
4
0 1
1 2
2 0
3 1
Output
7.30056308
Input
5
0 0
1 2
2 1
3 2
4 0
Output
10.94427191
"Correct Solution:
```
#!python3
import sys
from math import sqrt
sys.setrecursionlimit(1000000)
iim = lambda: map(int, input().rstrip().split())
def resolve():
N = int(input())
P = [list(iim()) for i in range(N)]
def dist(i, j):
x = P[i][0] - P[j][0]
y = P[i][1] - P[j][1]
return sqrt(x*x + y*y)
def tsp(i, j, k):
if i == 0 and j == 0:
return dist(0, 1) + tsp(0, 1, 0)
else:
if dp[i][j]: return dp[i][j]
if k == N: return dist(i, j)
d1 = dist(i, k) + tsp(k, j, k + 1)
d2 = dist(j, k) + tsp(i, k, k + 1)
d = d1 if d1 < d2 else d2
dp[i][j] = d
return d
dp = [[0]*N for i in range(N)]
print(tsp(0,0,0))
if __name__ == "__main__":
resolve()
```
| 4,138 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given $N$ points in the 2D Euclidean plane, find the distance of the shortest tour that meets the following criteria:
* Visit the points according to the following steps:
1. It starts from the leftmost point (starting point), goes strictly from left to right, and then visits the rightmost point (turn-around point).
2. Then it starts from the turn-around point, goes strictly from right to left, and then back to the starting point.
* Through the processes 1. 2., the tour must visit each point at least once.
Constraints
* $2 \leq N \leq 1000$
* $-1000 \leq x_i, y_i \leq 1000$
* $x_i$ differ from each other
* The given points are already sorted by x-coordinates
Input
The input data is given in the following format:
$N$
$x_1$ $y_1$
$x_2$ $y_2$
...
$x_N$ $y_N$
Output
Print the distance of the shortest tour in a line. The output should not have an error greater than 0.0001.
Examples
Input
3
0 0
1 1
2 0
Output
4.82842712
Input
4
0 1
1 2
2 0
3 1
Output
7.30056308
Input
5
0 0
1 2
2 1
3 2
4 0
Output
10.94427191
Submitted Solution:
```
import sys
sys.setrecursionlimit(1000)
def calc(p1, p2, v):
dists_v = dists[v]
if v == n - 1:
return dists_v[p1] + dists_v[p2]
if visited[p1][p2]:
return dp[p1][p2]
dp[p1][p2] = min(dists_v[p1] + calc(p2, v, v + 1), dists_v[p2] + calc(p1, v, v + 1))
visited[p1][p2] = True
return dp[p1][p2]
n = int(input())
points = [complex(*map(int, input().split())) for _ in range(n)]
dists = [[abs(p1 - p2) for p2 in points] for p1 in points]
visited = [[False] * n for _ in range(n)]
dp = [[0] * n for _ in range(n)]
print(calc(0, 0, 0))
```
No
| 4,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given $N$ points in the 2D Euclidean plane, find the distance of the shortest tour that meets the following criteria:
* Visit the points according to the following steps:
1. It starts from the leftmost point (starting point), goes strictly from left to right, and then visits the rightmost point (turn-around point).
2. Then it starts from the turn-around point, goes strictly from right to left, and then back to the starting point.
* Through the processes 1. 2., the tour must visit each point at least once.
Constraints
* $2 \leq N \leq 1000$
* $-1000 \leq x_i, y_i \leq 1000$
* $x_i$ differ from each other
* The given points are already sorted by x-coordinates
Input
The input data is given in the following format:
$N$
$x_1$ $y_1$
$x_2$ $y_2$
...
$x_N$ $y_N$
Output
Print the distance of the shortest tour in a line. The output should not have an error greater than 0.0001.
Examples
Input
3
0 0
1 1
2 0
Output
4.82842712
Input
4
0 1
1 2
2 0
3 1
Output
7.30056308
Input
5
0 0
1 2
2 1
3 2
4 0
Output
10.94427191
Submitted Solution:
```
def calc(p1, p2, v):
dists_v = dists[v]
if v == n - 1:
return dists_v[p1] + dists_v[p2]
if visited[p1][p2]:
return dp[p1][p2]
dp[p1][p2] = min(dists_v[p1] + calc(p2, v, v + 1), dists_v[p2] + calc(p1, v, v + 1))
visited[p1][p2] = True
return dp[p1][p2]
n = int(input())
points = [complex(*map(int, input().split())) for _ in range(n)]
dists = [[abs(p1 - p2) for p2 in points] for p1 in points]
visited = [[False] * n for _ in range(n)]
dp = [[0] * n for _ in range(n)]
print(calc(0, 0, 0))
```
No
| 4,140 |
Provide a correct Python 3 solution for this coding contest problem.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
"Correct Solution:
```
n = num = int(input())
i=2
while i * i <= n:
if n % i ==0:
num = num // i * (i - 1)
while n % i == 0:
n //= i
i += 1
if n != 1 :
num = num // n * (n - 1)
print(num)
```
| 4,141 |
Provide a correct Python 3 solution for this coding contest problem.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
"Correct Solution:
```
def is_prime(x):
# Perform prime number discrimination for odd numbers of 3 or more.
return pow(2, x - 1, x) == 1
def solve():
n = int(input())
ans = n
if n % 2 == 0:
ans -= ans // 2
while n % 2 == 0:
n //= 2
d = 3
while d <= n ** 0.5:
if is_prime(d) and n % d == 0:
ans -= ans // d
while n % d == 0:
n //= d
d += 2
if is_prime(n):
ans -= ans // n
print(ans)
solve()
```
| 4,142 |
Provide a correct Python 3 solution for this coding contest problem.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
"Correct Solution:
```
n = int(input())
def prime_factor(n):
factor = []
for i in range(2, int(n ** 0.5) + 1):
while (n % i == 0):
factor.append(i)
n //= i
if n != 1:
factor.append(n)
return factor
def Euler_phi(n):
divisors = prime_factor(n)
divisors = set(divisors)
res = n
for factor in divisors:
res = res * (factor - 1) // factor
return res
print(Euler_phi(n))
```
| 4,143 |
Provide a correct Python 3 solution for this coding contest problem.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
"Correct Solution:
```
import sys,math
n=m=int(input())
c=0
def factorint(N):
i=2
fct=[]
while i**2<=N:
if N%i!=0:
i+=1
else:
N//=i
fct.append(i)
if N>1:
fct.append(N)
return fct
l1=factorint(n)
l2=list(set(l1))
#print(l1)
#print(l2)
if len(l1)==len(l2):
ans=1
for item in l2:
ans*=(item-1)
else:
l3=[]
x=1
for i in range(len(l2)):
for j in range(i,len(l2)):
if i!=j:
l3.append(l2[i]*l2[j])
x*=l2[i]
l4=[n//item for item in l2]
l5=[n//item for item in l3]
if len(l2)>2:
y=n//x
else:
y=0
ans=(n-(sum(l4)-sum(l5)+y))
# print(x)
# print(l3)
# print(l4)
# print(l5)
print(ans)
```
| 4,144 |
Provide a correct Python 3 solution for this coding contest problem.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
"Correct Solution:
```
def factorize(n):
a = []
while n % 2 == 0:
a.append(2)
n //= 2
f = 3
while f * f <= n:
if n % f == 0:
a.append(f)
n //= f
else:
f += 2
if n != 1:
a.append(n)
return a
def factorize2(n):
plist = factorize(n)
dic = {}
for n in plist:
if not n in dic:
dic[n] = 1
else:
dic[n] += 1
ret = []
for p,n in dic.items():
ret.append([p,n])
return ret
def phi(n):
plist = factorize2(n)
ans = 1
for [p,n] in plist:
ans *= (pow(p,n)-pow(p,n-1))
return ans
Q = int(input())
print(phi(Q))
```
| 4,145 |
Provide a correct Python 3 solution for this coding contest problem.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
"Correct Solution:
```
def soi(n): #?´??????°????§£?????¢?????´?????????
i=2
while True:
if i*i>n:
return n
if n%i==0:
return i
i+=1
a=int(input())
ans=a
count=1
i=2 # ??????????????°
while True:
temp=soi(a)
if a==1: #?´??????°??????????????°
break
while a%temp==0: #?´??????°??§???????¶???????
a/=temp
# print (a)
ans*=(1-1/temp)
#print (ans)
print(int(ans))
def gcd(a,b): #?????§??¬?´???°???????????°??????
r=0
while True:
r=a%b
if r==0:
break
a=b
b=r
return b
```
| 4,146 |
Provide a correct Python 3 solution for this coding contest problem.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
"Correct Solution:
```
import sys,bisect,math
from functools import reduce
sys.setrecursionlimit(15000)
N = int(sys.stdin.readline())
def phi(N):
n = N
s = set()
for i in range(2,math.floor(pow(N,0.5))+1):
while n%i==0:
s.add(i)
n = n//i
if n != 1:
s.add(n)
a1 = reduce(lambda a,b: a*b,[e-1 for e in s])
a2 = reduce(lambda a,b: a*b,[e for e in s])
#print(N,a1,a2)
return a1*N//a2
print(phi(N))
```
| 4,147 |
Provide a correct Python 3 solution for this coding contest problem.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
"Correct Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""
input:
1000000
output:
400000
"""
import sys
def min_prime_factor(n):
k = 2
while True:
if pow(k, 2) > n:
return n
if not n % k:
return k
k += 1
def phi(n):
ans = n
while True:
p = min_prime_factor(n)
if n == 1:
break
while not n % p:
n //= p
ans *= (1 - 1 / p)
return int(ans)
if __name__ == '__main__':
_input = sys.stdin.readlines()
n1 = int(_input[0])
print(phi(n1))
```
| 4,148 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
Submitted Solution:
```
def phi(n):
r = n
i = 2
while i*i <= n:
if n%i==0:
r -= r/i
while n%i==0:
n /= i
i += 1
if n>1: r-= r/n
return r
n = int(input())
print(int(phi(n)))
```
Yes
| 4,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
Submitted Solution:
```
from math import sqrt, floor
from functools import reduce
def divide(p):
global n, a, limit
while not (n % p):
a.add(p)
n = n // p
limit = floor(sqrt(n))
n, p = int(input()), 3
n0 = n
limit = floor(sqrt(n))
a, i = set(), 3
divide(2)
while p <= limit:
divide(p)
p += 2
if n > 1:
a.add(n)
print(reduce(lambda prev, now: prev * (now - 1) // now, a, n0))
```
Yes
| 4,150 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7
def factorize(num: int) -> dict:
""" 素因数分解 """
from math import sqrt
from collections import Counter
d = Counter()
# 終点はルート切り捨て+1
for i in range(2, int(sqrt(num))+1):
# 素因数分解：小さい方から割れるだけ割って素数をカウント
while num % i == 0:
num //= i
d[i] += 1
# 1まで分解したら終了
if num == 1:
break
# 最後に残ったnumは素数(ただし1^1は1^0なので数に入れない)
if num != 1:
d[num] += 1
return d
N = INT()
primes = factorize(N)
ans = N
# Nが持つ各素数について、オイラーのファイ関数を使っていく
for k in primes.keys():
ans *= 1 - 1/k
print(int(ans))
```
Yes
| 4,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
Submitted Solution:
```
def prime_factorization(n) -> list:
if n==1:
return [1]
ret = []
i = 2
while i*i<=n:
while n%i==0:
n //= i
ret.append(i)
i += 1
if n!=1:
ret.append(n)
return ret
n=int(input())
from functools import reduce
from operator import mul
pf=set(prime_factorization(n))
print(int(reduce(mul,[n]+[1-1/p for p in pf])))
```
Yes
| 4,152 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
Submitted Solution:
```
from itertools import compress, count
def prime_factor(n):
p = sieve(int(n ** 0.5))
factor = []
for pi in compress(count(0), p):
while n % pi == 0:
n //= pi
factor.append(pi)
if n != 1:
factor.append(n)
return factor
from sys import stdin
from functools import reduce
from operator import mul
from collections import Counter
readline = stdin.readline
print(reduce(mul, (pow(p, k) - pow(p, k - 1) for p, k in Counter(prime_factor(int(readline()))).items())))
```
No
| 4,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
Submitted Solution:
```
import sys
def gcd(a,b):
if(b == 0):
return a
else:
c = b
b = a%b
a = c
return(gcd(a,b))
x = int(input())
count = 0
for y in range(1,x+1):
if gcd(x,y) == 1:
count+=1
print(count)
```
No
| 4,154 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
Submitted Solution:
```
a=int(input())
li=[0]*10000000
count=1
i=2 # ??????????????°
while i<a:
r=0
ta=a
b=i
if li[i]!=1:
while True:
r=ta%b
if r==0:
break
ta=b
b=r
if b==1:
count+=1
else:
temp=i
l=2
while True:
li[temp*l]=1
l+=1
if temp*l>a:
break
i+=1
print(count)
```
No
| 4,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given integer n, count the totatives of n, that is, the positive integers less than or equal to n that are relatively prime to n.
Input
n
An integer n (1 ≤ n ≤ 1000000000).
Output
The number of totatives in a line.
Examples
Input
6
Output
2
Input
1000000
Output
400000
Submitted Solution:
```
p=1;a=[]
n=int(input());s=n**.5
while n%2==0 and n>3:a+=[2];n//=2
d=3
while s>d and n>d:
if n%d==0:a+=[d];n//=d
else:d+=2
a+=[n]
for f in set(a):
p*=f**a.count(f)*(1-1/f)
print(int(p)
```
No
| 4,156 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Tags: greedy, math
Correct Solution:
```
n, m = map(int, input().split())
sdst = 0
for i in range(n):
sdst += abs(i - n//2)
bdst = 0
for i in range(n):
bdst += abs(i)
ans = 0
for i in range(m):
x,d = map(int, input().split())
if d < 0:
s = x*n + sdst*d
else:
s = x*n + bdst*d
ans += s
ans /=n
print("%.8f" % ans)
```
| 4,157 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Tags: greedy, math
Correct Solution:
```
from sys import stdin
from collections import deque
mod = 10**9 + 7
# def rl():
# return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# la = []
# s = list(input())
# n = len(s)
#
# i = 0
# ans = []
# while i<n:
# # print(i)
# if s[i] == '1':
# cnt1 = 0
# cnt2 = 0
# while s[i] == '1' or s[i] == '0':
# if s[i] == '1':
# cnt2+=1
# if s[i] == '0':
# cnt1+=1
#
# i+=1
# if i == n:
# break
#
# for x in range(cnt1):
# ans.append('0')
# for y in range(cnt2):
# ans.append('1')
#
# if i == n:
# break
# elif s[i] == '2':
# cnt1 = 0
# cnt2 = 0
# cnt3 = 0
# st = -1
# ba = -1
# while s[i] == '2' or s[i] == '1' or s[i] == '0':
# if s[i] == '2':
# cnt2+=1
# elif s[i] == '1':
# cnt1+=1
# ba = i
# else:
# if ba == -1:
# st = i
# cnt3+=1
# else:
# break
#
#
# i+=1
# if i == n:
# break
#
# for x in range(cnt1):
# ans.append('1')
# for y in range(cnt2):
# ans.append('2')
# for y in range(cnt3):
# ans.append('0')
#
# if i == n:
# break
#
#
#
# else:
# ans.append('0')
# i+=1
#
#
#
#
# print(''.join(ans))
#
n,m = map(int,input().split())
ans = 0
for i in range(m):
x,d = map(int,input().split())
ans+=n*x
if d>=0:
z = n-1
ans+=d*((z*(z+1))//2)
else:
z = n-1
z1 = z//2
z2 = z-z//2
ans+=d*((z1*(z1+1))//2)
ans+=d*((z2*(z2+1))//2)
# print(ans)
print(ans/n)
```
| 4,158 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Tags: greedy, math
Correct Solution:
```
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
S = 0
for _ in range(m):
x, d = map(int, input().split())
S += n*x
if d>=0:
S += n*(n-1)//2*d
else:
m = n//2
S += 2*(m+1)*m//2*d
if n%2==0:
S -= m*d
print(S/n)
```
| 4,159 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Tags: greedy, math
Correct Solution:
```
def auxFun(x):
return x * (x + 1) // 2
n, m = (int(x) for x in input().split())
answer = 0
aux = (n + 1) // 2
mx = auxFun(n - 1)
mn = auxFun(aux - 1) + auxFun(n - aux)
for i in range(m):
x, d = (int(x) for x in input().split())
if d < 0: aux = mn
else: aux = mx
answer += n * x + d * aux
print("%.12f" % (answer / n))
```
| 4,160 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Tags: greedy, math
Correct Solution:
```
n, m = map(int, input().split())
res = 0
for i in range(m):
x, d = map(int, input().split())
res += x
if d < 0:
if n % 2 == 0:
k = n // 2 - 1
res += d * (k + 1) * (k + 1) / n
else:
k = n // 2
res += d * k * (k + 1) / n
else:
res += d * (n - 1) / 2
if ((n==99999 and m==99998) or(n==99995 and m==99994)):
print(0.0000100)
elif n==19999 and m==99999:
print(0.0000000)
elif n==99997 and m==100000:
print(-333.329999899997006)
else:
print(res)
```
| 4,161 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Tags: greedy, math
Correct Solution:
```
n, m = map(int, input().split())
sum = 0
for i in range(m):
x, d = map(int, input().split())
sum += n*x
if d > 0:
sum += d * ((n * (n - 1)) // 2)
elif d < 0:
h = (n + 1) // 2
p = h * (h - 1) // 2 + (n + 1 - h) * (n - h) // 2
sum += p * d
print("{0:.7f}".format(sum/n))
```
| 4,162 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Tags: greedy, math
Correct Solution:
```
n,m = map(int,input().split())
s=0
a = n*(n-1)//2
nHalf = n//2
b = nHalf**2
if n%2:
b+=nHalf
for i in range(m):
x,y = map(int,input().split())
s+=n*x
s+=y*(a if y>0 else b)
print(s/n)
```
| 4,163 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Tags: greedy, math
Correct Solution:
```
def sum_to_n(n):
return n * (n + 1) // 2
(n, m) = map(int, input().split())
mid = 0
if n % 2 == 1:
mid = 2 * sum_to_n(n // 2)
else:
mid = 2 * sum_to_n(n // 2) - n // 2
edge = sum_to_n(n - 1)
sum = 0
for i in range(m):
(x, d) = map(int, input().split())
sum += x * n
if d > 0:
sum += edge * d
else:
sum += mid * d
print(sum / n)
```
| 4,164 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Submitted Solution:
```
n, m = [int(i) for i in input().split()]
max_sum = (n * (n - 1)) // 2
min_sum = ((n - 1) // 2 + 1) * ((n - 1) // 2) // 2 + (n - 1 - (n - 1) // 2 + 1) * (n - 1 - (n - 1) // 2) // 2
ans = 0
#print(n, max_sum, min_sum)
for i in range(m):
data = [int(j) for j in input().split()]
if data[1] > 0:
wk1 = max_sum
else:
wk1 = min_sum
ans += data[0] * n + data[1] * wk1
print(ans / n)
```
Yes
| 4,165 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Submitted Solution:
```
#Bhargey Mehta (Junior)
#DA-IICT, Gandhinagar
import sys, math, queue
#sys.stdin = open('input.txt', 'r')
MOD = 998244353
sys.setrecursionlimit(1000000)
n, m = map(int, input().split())
x, y, z = 0, 0, 0
for i in range(m):
xi, di = map(int, input().split())
x += xi
if di > 0: y += di
else: z += di
mid = n//2
a = [x for i in range(n)]
for i in range(n):
a[i] += i*y + abs(mid-i)*z
print(sum(a)/n)
```
Yes
| 4,166 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Submitted Solution:
```
n,m=map(int,input().split())
d=(n*(n-1))//2
d1=(n-1)//2
d1=d1*(d1+1)+(1-n%2)*n//2
#print(d,d1)
s=0
for i in range(m):
x,y=map(int,input().split())
s+=n*x
if y>0:
s+=d*y
else:
s+=d1*y
print(s/n)
```
Yes
| 4,167 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Submitted Solution:
```
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
info = [list(map(int, input().split())) for i in range(m)]
ans = 0
tmp1 = (n - 1) // 2
tmp2 = -((-(n - 1)) // 2)
for x, d in info:
ans += n * x
if d > 0:
ans += d * (n - 1) * (n) // 2
if d < 0:
ans += d * (tmp1) * (tmp1 + 1) // 2
ans += d * (tmp2) * (tmp2 + 1) // 2
print(ans / n)
```
Yes
| 4,168 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Submitted Solution:
```
n, m = map(int, input().split())
k, x, y = 0, (n - 1) / 2, sum(abs(i - (n + 1) // 2) for i in range(1, n + 1)) / n
for i in range(m):
a, b = map(int, input().split())
k += a + (b * (x if b > 0 else y))
print(k)
```
No
| 4,169 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Submitted Solution:
```
[n, m] = input().split()
n, m = int(n), int(m)
avg = 0
for i in range(m):
[x, d] = input().split()
x, d = int(x), int(d)
avg += n * x
if n > 1:
if d > 0:
avg += d * n * (n - 1) / 2
else:
avg += d * ((n - 1) / 2)**2
if n % 2 == 0:
avg += d * n / 2
print(float(avg) / n)
```
No
| 4,170 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Submitted Solution:
```
n, m = map(int, input().split())
res = 0
mx = (n-1)*n//2
mn = 0
if n&1:
mn = (n//2)*(n/2+1)
else:
mn = n*n//4
for i in range(m):
x, d = map(int, input().split())
res += x*n
if d > 0:
res += mx*d
else:
res += mn*d
print('%.10f'%(res/n))
```
No
| 4,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice got an array of length n as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen m changes of the following form. For some integer numbers x and d, he chooses an arbitrary position i (1 ≤ i ≤ n) and for every j ∈ [1, n] adds x + d ⋅ dist(i, j) to the value of the j-th cell. dist(i, j) is the distance between positions i and j (i.e. dist(i, j) = |i - j|, where |x| is an absolute value of x).
For example, if Alice currently has an array [2, 1, 2, 2] and Bob chooses position 3 for x = -1 and d = 2 then the array will become [2 - 1 + 2 ⋅ 2,~1 - 1 + 2 ⋅ 1,~2 - 1 + 2 ⋅ 0,~2 - 1 + 2 ⋅ 1] = [5, 2, 1, 3]. Note that Bob can't choose position i outside of the array (that is, smaller than 1 or greater than n).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of elements of the array and the number of changes.
Each of the next m lines contains two integers x_i and d_i (-10^3 ≤ x_i, d_i ≤ 10^3) — the parameters for the i-th change.
Output
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed 10^{-6}.
Examples
Input
2 3
-1 3
0 0
-1 -4
Output
-2.500000000000000
Input
3 2
0 2
5 0
Output
7.000000000000000
Submitted Solution:
```
from decimal import *
n, m = [int(_) for _ in input().split()]
n2 = n/2
getcontext().prec = 6000
avg = Decimal(0.0)
negative = Decimal(0.0)
positive = Decimal(0.0)
for c in range(m):
x, d = [int(_) for _ in input().split()]
avg += Decimal(x) * Decimal(n)
if d < 0:
negative += d
else:
positive += d
# Fix the negative part
avg += Decimal(n2*(n2+1)) * negative
if n % 2 == 0:
avg -= Decimal(n2) * negative
# Fix the positive part
avg += Decimal(n *(n-1) / 2) * positive
print ("%.20lf" % (avg / Decimal(n)))
```
No
| 4,172 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers a and b. There are two possible operations:
1. multiply one of the numbers by some prime p;
2. divide one of the numbers on its prime factor p.
What is the minimum number of operations required to obtain two integers having the same number of divisors? You are given several such pairs, you need to find the answer for each of them.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of pairs of integers for which you are to find the answer.
Each of the next t lines contain two integers a_i and b_i (1 ≤ a_i, b_i ≤ 10^6).
Output
Output t lines — the i-th of them should contain the answer for the pair a_i, b_i.
Example
Input
8
9 10
100 17
220 70
17 19
4 18
32 20
100 32
224 385
Output
1
3
1
0
1
0
1
1
Note
These are the numbers with equal number of divisors, which are optimal to obtain in the sample test case:
* (27, 10), 4 divisors
* (100, 1156), 9 divisors
* (220, 140), 12 divisors
* (17, 19), 2 divisors
* (12, 18), 6 divisors
* (50, 32), 6 divisors
* (224, 1925), 12 divisors
Note that there can be several optimal pairs of numbers.
Submitted Solution:
```
def find(num1, num2):
first = []
second = []
for i in range(2, num1):
if(num1%i==0):
first.append(i)
for i in range(2, num2):
if(num2%i==0):
second.append(i)
return abs(len(first) - len(second))
n = int(input())
ans = []
for i in range(n):
x, y = map(int, input().split())
ans.append(find(x, y))
for i in ans:
print(i)
```
No
| 4,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers a and b. There are two possible operations:
1. multiply one of the numbers by some prime p;
2. divide one of the numbers on its prime factor p.
What is the minimum number of operations required to obtain two integers having the same number of divisors? You are given several such pairs, you need to find the answer for each of them.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of pairs of integers for which you are to find the answer.
Each of the next t lines contain two integers a_i and b_i (1 ≤ a_i, b_i ≤ 10^6).
Output
Output t lines — the i-th of them should contain the answer for the pair a_i, b_i.
Example
Input
8
9 10
100 17
220 70
17 19
4 18
32 20
100 32
224 385
Output
1
3
1
0
1
0
1
1
Note
These are the numbers with equal number of divisors, which are optimal to obtain in the sample test case:
* (27, 10), 4 divisors
* (100, 1156), 9 divisors
* (220, 140), 12 divisors
* (17, 19), 2 divisors
* (12, 18), 6 divisors
* (50, 32), 6 divisors
* (224, 1925), 12 divisors
Note that there can be several optimal pairs of numbers.
Submitted Solution:
```
def primfacs(n):
i = 2
primfac = []
while i * i <= n:
while n % i == 0:
primfac.append(i)
n = n / i
i = i + 1
if n > 1:
primfac.append(n)
return primfac
def prime(n):
t = []
m = primfacs(n)
for a in set(m):
t.append((a, m.count(a)))
return t
def ans(a, b):
ps = [x[1] for x in prime(a)]
qs = [x[1] for x in prime(b)]
while len(ps) > len(qs):
qs.append(0)
while len(qs) > len(ps):
ps.append(0)
return(sum([abs(ps[i]-qs[i]) for i in range(len(qs))]))
h = []
for i in range(int(input())):
a, b = map(int, input().split())
h.append(ans(a,b))
for j in h:
print(j)
```
No
| 4,174 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers a and b. There are two possible operations:
1. multiply one of the numbers by some prime p;
2. divide one of the numbers on its prime factor p.
What is the minimum number of operations required to obtain two integers having the same number of divisors? You are given several such pairs, you need to find the answer for each of them.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of pairs of integers for which you are to find the answer.
Each of the next t lines contain two integers a_i and b_i (1 ≤ a_i, b_i ≤ 10^6).
Output
Output t lines — the i-th of them should contain the answer for the pair a_i, b_i.
Example
Input
8
9 10
100 17
220 70
17 19
4 18
32 20
100 32
224 385
Output
1
3
1
0
1
0
1
1
Note
These are the numbers with equal number of divisors, which are optimal to obtain in the sample test case:
* (27, 10), 4 divisors
* (100, 1156), 9 divisors
* (220, 140), 12 divisors
* (17, 19), 2 divisors
* (12, 18), 6 divisors
* (50, 32), 6 divisors
* (224, 1925), 12 divisors
Note that there can be several optimal pairs of numbers.
Submitted Solution:
```
def pr(n):
factors = {}
d = 2
factors[n] = 1
while d * d <= n:
if n % d == 0:
factors[d] = 1
n//=d
else:
d += 1
if n > 1:
factors[n] = 1
else:
break
return(factors.keys())
k = int(input())
for i in range(k):
r = input()
a,b = r.split(" ")
z = int(a)
x = int(b)
print(abs(len(pr(z))-len(pr(x))))
```
No
| 4,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two positive integers a and b. There are two possible operations:
1. multiply one of the numbers by some prime p;
2. divide one of the numbers on its prime factor p.
What is the minimum number of operations required to obtain two integers having the same number of divisors? You are given several such pairs, you need to find the answer for each of them.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of pairs of integers for which you are to find the answer.
Each of the next t lines contain two integers a_i and b_i (1 ≤ a_i, b_i ≤ 10^6).
Output
Output t lines — the i-th of them should contain the answer for the pair a_i, b_i.
Example
Input
8
9 10
100 17
220 70
17 19
4 18
32 20
100 32
224 385
Output
1
3
1
0
1
0
1
1
Note
These are the numbers with equal number of divisors, which are optimal to obtain in the sample test case:
* (27, 10), 4 divisors
* (100, 1156), 9 divisors
* (220, 140), 12 divisors
* (17, 19), 2 divisors
* (12, 18), 6 divisors
* (50, 32), 6 divisors
* (224, 1925), 12 divisors
Note that there can be several optimal pairs of numbers.
Submitted Solution:
```
import math
def numb(a):
res=0
for i in range(a):
if a%(i+1)==0:
res+=1
print (i+2)
return res
def check(a,b):
return abs(numb(a)-numb(b))
n=int(input())
for i in range(n):
a,b=map(int, input().split())
print(check(a,b))
```
No
| 4,176 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pasha is a young technician, nevertheless, he has already got a huge goal: to assemble a PC. The first task he has to become familiar with is to assemble an electric scheme.
The scheme Pasha assembled yesterday consists of several wires. Each wire is a segment that connects two points on a plane with integer coordinates within the segment [1, 10^9].
There are wires of two colors in the scheme:
* red wires: these wires are horizontal segments, i.e. if such a wire connects two points (x_1, y_1) and (x_2, y_2), then y_1 = y_2;
* blue wires: these wires are vertical segments, i.e. if such a wire connects two points (x_1, y_1) and (x_2, y_2), then x_1 = x_2.
Note that if a wire connects a point to itself, it may be blue, and it can be red. Also, in Pasha's scheme no two wires of the same color intersect, i.e. there are no two wires of same color that have common points.
The imperfection of Pasha's scheme was that the wires were not isolated, so in the points where two wires of different colors intersect, Pasha saw sparks. Pasha wrote down all the points where he saw sparks and obtained a set of n distinct points. After that he disassembled the scheme.
Next morning Pasha looked at the set of n points where he had seen sparks and wondered how many wires had he used. Unfortunately, he does not remember that, so he wonders now what is the smallest number of wires he might have used in the scheme. Help him to determine this number and place the wires in such a way that in the resulting scheme the sparks occur in the same places.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of points where Pasha saw sparks.
Each of the next n lines contain two integers x and y (1 ≤ x, y ≤ 10^9) — the coordinates of a point with sparks. It is guaranteed that all points are distinct.
Output
Print the description of the scheme in the following format.
In the first line print h — the number of horizontal red wires (0 ≤ h). In each of the next h lines print 4 integers x_1, y_1, x_2, y_2 — the coordinates of two points (x_1, y_1) and (x_2, y_2) that are connected with this red wire. The segmenst must be horizontal, i.e. y_1 = y_2 must hold. Also, the constraint 1 ≤ x_1, y_1, x_2, y_2 ≤ 10^9 must hold.
After that print v — the number of vertical blue wires (0 ≤ v). In each of the next v lines print 4 integers x_1, y_1, x_2, y_2 — the coordinates of two points (x_1, y_1) and (x_2, y_2) that are connected with this blue wire. The segmenst must be vertical, i.e. x_1 = x_2 shomustuld hold. Also, the constraint 1 ≤ x_1, y_1, x_2, y_2 ≤ 10^9 must hold.
No two segments of the same color should have common points. The set of points where sparks appear should be the same as given in the input.
The number of segments (h + v) should be minimum possible. It's easy to see that the answer always exists. If there are multiple possible answers, print any.
Examples
Input
4
2 2
2 4
4 2
4 4
Output
2
5 2 1 2
1 4 5 4
2
2 1 2 5
4 5 4 1
Input
4
2 1
3 2
2 3
1 2
Output
4
2 1 2 1
3 2 3 2
1 2 1 2
2 3 2 3
3
1 2 1 2
3 2 3 2
2 3 2 1
Note
In the first example Pasha could have assembled the following scheme:
<image>
In this scheme there are 2 wires of each color: red ones connecting (5, 2) with (1, 2) and (1, 4) with (5, 4), blue ones connecting (2, 1) with (2, 5) and (4, 5) with (4, 1). Note that the sparks appear in the points that are described in the input, they are shown in yellow on the picture. For example, Pasha will see the spark in the point (2, 4) because the second red wire and the first blue wire intersect there. It is possible to show that we can't have less than 4 wires to produce a scheme with same sparks positions.
Submitted Solution:
```
n=int(input())
x=[]
y=[]
X={}
Y={}
for i in range(n):
a,b=map(int,input().split())
x.append(a)
y.append(b)
s1=list(set(x))
s2=list(set(x))
s1.sort()
s2.sort()
for i in s1:
X[i]=[]
for i in s2:
Y[i]=[]
for i in range(n):
X[x[i]].append(y[i])
Y[y[i]].append(x[i])
for i in s1:
X[i].sort()
for i in s2:
Y[i].sort()
print(len(s2))
for i in s2:
k=len(Y[i])
print(Y[i][0],i,Y[i][k-1],i)
print(len(s1))
for i in s1:
h=len(X[i])
print(i,X[i][0],i,X[i][h-1])
```
No
| 4,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Pasha is a young technician, nevertheless, he has already got a huge goal: to assemble a PC. The first task he has to become familiar with is to assemble an electric scheme.
The scheme Pasha assembled yesterday consists of several wires. Each wire is a segment that connects two points on a plane with integer coordinates within the segment [1, 10^9].
There are wires of two colors in the scheme:
* red wires: these wires are horizontal segments, i.e. if such a wire connects two points (x_1, y_1) and (x_2, y_2), then y_1 = y_2;
* blue wires: these wires are vertical segments, i.e. if such a wire connects two points (x_1, y_1) and (x_2, y_2), then x_1 = x_2.
Note that if a wire connects a point to itself, it may be blue, and it can be red. Also, in Pasha's scheme no two wires of the same color intersect, i.e. there are no two wires of same color that have common points.
The imperfection of Pasha's scheme was that the wires were not isolated, so in the points where two wires of different colors intersect, Pasha saw sparks. Pasha wrote down all the points where he saw sparks and obtained a set of n distinct points. After that he disassembled the scheme.
Next morning Pasha looked at the set of n points where he had seen sparks and wondered how many wires had he used. Unfortunately, he does not remember that, so he wonders now what is the smallest number of wires he might have used in the scheme. Help him to determine this number and place the wires in such a way that in the resulting scheme the sparks occur in the same places.
Input
The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of points where Pasha saw sparks.
Each of the next n lines contain two integers x and y (1 ≤ x, y ≤ 10^9) — the coordinates of a point with sparks. It is guaranteed that all points are distinct.
Output
Print the description of the scheme in the following format.
In the first line print h — the number of horizontal red wires (0 ≤ h). In each of the next h lines print 4 integers x_1, y_1, x_2, y_2 — the coordinates of two points (x_1, y_1) and (x_2, y_2) that are connected with this red wire. The segmenst must be horizontal, i.e. y_1 = y_2 must hold. Also, the constraint 1 ≤ x_1, y_1, x_2, y_2 ≤ 10^9 must hold.
After that print v — the number of vertical blue wires (0 ≤ v). In each of the next v lines print 4 integers x_1, y_1, x_2, y_2 — the coordinates of two points (x_1, y_1) and (x_2, y_2) that are connected with this blue wire. The segmenst must be vertical, i.e. x_1 = x_2 shomustuld hold. Also, the constraint 1 ≤ x_1, y_1, x_2, y_2 ≤ 10^9 must hold.
No two segments of the same color should have common points. The set of points where sparks appear should be the same as given in the input.
The number of segments (h + v) should be minimum possible. It's easy to see that the answer always exists. If there are multiple possible answers, print any.
Examples
Input
4
2 2
2 4
4 2
4 4
Output
2
5 2 1 2
1 4 5 4
2
2 1 2 5
4 5 4 1
Input
4
2 1
3 2
2 3
1 2
Output
4
2 1 2 1
3 2 3 2
1 2 1 2
2 3 2 3
3
1 2 1 2
3 2 3 2
2 3 2 1
Note
In the first example Pasha could have assembled the following scheme:
<image>
In this scheme there are 2 wires of each color: red ones connecting (5, 2) with (1, 2) and (1, 4) with (5, 4), blue ones connecting (2, 1) with (2, 5) and (4, 5) with (4, 1). Note that the sparks appear in the points that are described in the input, they are shown in yellow on the picture. For example, Pasha will see the spark in the point (2, 4) because the second red wire and the first blue wire intersect there. It is possible to show that we can't have less than 4 wires to produce a scheme with same sparks positions.
Submitted Solution:
```
def intersect(x, y, x_line, y_line):
return (x_line[0] <= y <= x_line[1]) and (y_line[0] <= x <= y_line[1])
n = int(input())
scs = []
for i in range(n):
l = input().split()
scs.append((int(l[0]), int(l[1])))
red_wires = {} # horizontal
blue_wires = {} # vertical
for sc in scs:
if not sc[0] in blue_wires.keys():
blue_wires[sc[0]] = []
blue_wires[sc[0]].append(sc[1])
if not sc[1] in red_wires.keys():
red_wires[sc[1]] = []
red_wires[sc[1]].append(sc[0])
for y in red_wires.keys():
red_wires[y].sort()
min_x = min(red_wires[y])
max_x = max(red_wires[y])
red_wires[y].clear()
red_wires[y].append((min_x, max_x))
for x in blue_wires.keys():
blue_wires[x].sort()
min_y = min(blue_wires[x])
max_y = max(blue_wires[x])
blue_wires[x].clear()
blue_wires[x].append((min_y, max_y))
for y in red_wires.keys():
for x in red_wires.keys():
# if any lines intersect
for red_line in red_wires[y]:
for blue_line in blue_wires[x]:
if intersect(x, y, blue_line, red_line) and not (x, y) in scs:
red_wires[y].append((x + 1, red_line[1]))
red_wires[y].append((red_line[0], x - 1))
red_wires[y].remove(red_line)
rsize = 0
bsize = 0
for y in red_wires.keys():
rsize += len(red_wires[y])
for x in red_wires.keys():
bsize += len(blue_wires[x])
print(rsize)
for y in red_wires.keys():
for line in red_wires[y]:
print(str(line[0]) + str(y) + str(line[1]) + str(y))
print(bsize)
for x in blue_wires.keys():
for line in blue_wires[x]:
print(str(x) + str(line[0]) + str(x) + str(line[1]))
```
No
| 4,178 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova has taken his summer practice this year and now he should write a report on how it went.
Vova has already drawn all the tables and wrote down all the formulas. Moreover, he has already decided that the report will consist of exactly n pages and the i-th page will include x_i tables and y_i formulas. The pages are numbered from 1 to n.
Vova fills the pages one after another, he can't go filling page i + 1 before finishing page i and he can't skip pages.
However, if he draws strictly more than k tables in a row or writes strictly more than k formulas in a row then he will get bored. Vova wants to rearrange tables and formulas in each page in such a way that he doesn't get bored in the process. Vova can't move some table or some formula to another page.
Note that the count doesn't reset on the start of the new page. For example, if the page ends with 3 tables and the next page starts with 5 tables, then it's counted as 8 tables in a row.
Help Vova to determine if he can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.
Input
The first line contains two integers n and k (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ 10^6).
The second line contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^6) — the number of tables on the i-th page.
The third line contains n integers y_1, y_2, ..., y_n (1 ≤ y_i ≤ 10^6) — the number of formulas on the i-th page.
Output
Print "YES" if Vova can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.
Otherwise print "NO".
Examples
Input
2 2
5 5
2 2
Output
YES
Input
2 2
5 6
2 2
Output
NO
Input
4 1
4 1 10 1
3 2 10 1
Output
YES
Note
In the first example the only option to rearrange everything is the following (let table be 'T' and formula be 'F'):
* page 1: "TTFTTFT"
* page 2: "TFTTFTT"
That way all blocks of tables have length 2.
In the second example there is no way to fit everything in such a way that there are no more than 2 tables in a row and 2 formulas in a row.
Tags: dp, greedy
Correct Solution:
```
def get(px, py, x, y):
ans = 10**18
global k
if px <= k:
cnt = (px + x + k - 1) // k - 1
if y == cnt:
ans = min(ans, px + x - cnt * k)
elif y > cnt and y <= x * k:
ans = min(ans, 1)
if py <= k:
cnt = (x + k - 1) // k - 1
if y == cnt:
ans = min(ans, x - cnt * k)
elif y > cnt and y <= (x - 1) * k + (k - py):
ans = min(ans, 1)
return ans
n, k = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
dp = [[10**18] * (n + 1), [10**18] * (n + 1)]
dp[0][0], dp[1][0] = 0, 0
for i in range(n):
dp[0][i + 1] = get(dp[0][i], dp[1][i], x[i], y[i])
dp[1][i + 1] = get(dp[1][i], dp[0][i], y[i], x[i])
print('YES' if min(dp[0][-1], dp[1][-1]) <= k else 'NO')
```
| 4,179 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova has taken his summer practice this year and now he should write a report on how it went.
Vova has already drawn all the tables and wrote down all the formulas. Moreover, he has already decided that the report will consist of exactly n pages and the i-th page will include x_i tables and y_i formulas. The pages are numbered from 1 to n.
Vova fills the pages one after another, he can't go filling page i + 1 before finishing page i and he can't skip pages.
However, if he draws strictly more than k tables in a row or writes strictly more than k formulas in a row then he will get bored. Vova wants to rearrange tables and formulas in each page in such a way that he doesn't get bored in the process. Vova can't move some table or some formula to another page.
Note that the count doesn't reset on the start of the new page. For example, if the page ends with 3 tables and the next page starts with 5 tables, then it's counted as 8 tables in a row.
Help Vova to determine if he can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.
Input
The first line contains two integers n and k (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ 10^6).
The second line contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^6) — the number of tables on the i-th page.
The third line contains n integers y_1, y_2, ..., y_n (1 ≤ y_i ≤ 10^6) — the number of formulas on the i-th page.
Output
Print "YES" if Vova can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.
Otherwise print "NO".
Examples
Input
2 2
5 5
2 2
Output
YES
Input
2 2
5 6
2 2
Output
NO
Input
4 1
4 1 10 1
3 2 10 1
Output
YES
Note
In the first example the only option to rearrange everything is the following (let table be 'T' and formula be 'F'):
* page 1: "TTFTTFT"
* page 2: "TFTTFTT"
That way all blocks of tables have length 2.
In the second example there is no way to fit everything in such a way that there are no more than 2 tables in a row and 2 formulas in a row.
Tags: dp, greedy
Correct Solution:
```
def max(a, b):
if a > b:
return a
return b
n, k = map(int, input().split())
o = [int(t) for t in (input()+' '+input()).split()]
f, s = 0, 0
for i in range(n):
f = max(0, o[i] + f - k * o[i+n])
s = max(0, o[i+n] + s - k * o[i])
if f > k or s > k:
print('NO')
exit(0)
print('YES')
```
| 4,180 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova has taken his summer practice this year and now he should write a report on how it went.
Vova has already drawn all the tables and wrote down all the formulas. Moreover, he has already decided that the report will consist of exactly n pages and the i-th page will include x_i tables and y_i formulas. The pages are numbered from 1 to n.
Vova fills the pages one after another, he can't go filling page i + 1 before finishing page i and he can't skip pages.
However, if he draws strictly more than k tables in a row or writes strictly more than k formulas in a row then he will get bored. Vova wants to rearrange tables and formulas in each page in such a way that he doesn't get bored in the process. Vova can't move some table or some formula to another page.
Note that the count doesn't reset on the start of the new page. For example, if the page ends with 3 tables and the next page starts with 5 tables, then it's counted as 8 tables in a row.
Help Vova to determine if he can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.
Input
The first line contains two integers n and k (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ 10^6).
The second line contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^6) — the number of tables on the i-th page.
The third line contains n integers y_1, y_2, ..., y_n (1 ≤ y_i ≤ 10^6) — the number of formulas on the i-th page.
Output
Print "YES" if Vova can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.
Otherwise print "NO".
Examples
Input
2 2
5 5
2 2
Output
YES
Input
2 2
5 6
2 2
Output
NO
Input
4 1
4 1 10 1
3 2 10 1
Output
YES
Note
In the first example the only option to rearrange everything is the following (let table be 'T' and formula be 'F'):
* page 1: "TTFTTFT"
* page 2: "TFTTFTT"
That way all blocks of tables have length 2.
In the second example there is no way to fit everything in such a way that there are no more than 2 tables in a row and 2 formulas in a row.
Tags: dp, greedy
Correct Solution:
```
n, k = map(int, input().split())
o = [int(t) for t in (input()+' '+input()).split()]
f, s = 0, 0
for i in range(n):
f = max(0, o[i] + f - k * o[i+n])
s = max(0, o[i+n] + s - k * o[i])
if f > k or s > k:
print('NO')
exit(0)
print('YES')
```
| 4,181 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova has taken his summer practice this year and now he should write a report on how it went.
Vova has already drawn all the tables and wrote down all the formulas. Moreover, he has already decided that the report will consist of exactly n pages and the i-th page will include x_i tables and y_i formulas. The pages are numbered from 1 to n.
Vova fills the pages one after another, he can't go filling page i + 1 before finishing page i and he can't skip pages.
However, if he draws strictly more than k tables in a row or writes strictly more than k formulas in a row then he will get bored. Vova wants to rearrange tables and formulas in each page in such a way that he doesn't get bored in the process. Vova can't move some table or some formula to another page.
Note that the count doesn't reset on the start of the new page. For example, if the page ends with 3 tables and the next page starts with 5 tables, then it's counted as 8 tables in a row.
Help Vova to determine if he can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.
Input
The first line contains two integers n and k (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ 10^6).
The second line contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^6) — the number of tables on the i-th page.
The third line contains n integers y_1, y_2, ..., y_n (1 ≤ y_i ≤ 10^6) — the number of formulas on the i-th page.
Output
Print "YES" if Vova can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.
Otherwise print "NO".
Examples
Input
2 2
5 5
2 2
Output
YES
Input
2 2
5 6
2 2
Output
NO
Input
4 1
4 1 10 1
3 2 10 1
Output
YES
Note
In the first example the only option to rearrange everything is the following (let table be 'T' and formula be 'F'):
* page 1: "TTFTTFT"
* page 2: "TFTTFTT"
That way all blocks of tables have length 2.
In the second example there is no way to fit everything in such a way that there are no more than 2 tables in a row and 2 formulas in a row.
Tags: dp, greedy
Correct Solution:
```
def max(a, b):
if a > b:
return a
else:
return b
n, k = map(int, input().split())
x = [int(t) for t in input().split()]
y = [int(t) for t in input().split()]
f, s = 0, 0
for i in range(n):
f = max(0, x[i] + f - k * y[i])
s = max(0, y[i] + s - k * x[i])
if f > k or s > k:
print('NO')
exit(0)
print('YES')
```
| 4,182 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova has taken his summer practice this year and now he should write a report on how it went.
Vova has already drawn all the tables and wrote down all the formulas. Moreover, he has already decided that the report will consist of exactly n pages and the i-th page will include x_i tables and y_i formulas. The pages are numbered from 1 to n.
Vova fills the pages one after another, he can't go filling page i + 1 before finishing page i and he can't skip pages.
However, if he draws strictly more than k tables in a row or writes strictly more than k formulas in a row then he will get bored. Vova wants to rearrange tables and formulas in each page in such a way that he doesn't get bored in the process. Vova can't move some table or some formula to another page.
Note that the count doesn't reset on the start of the new page. For example, if the page ends with 3 tables and the next page starts with 5 tables, then it's counted as 8 tables in a row.
Help Vova to determine if he can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.
Input
The first line contains two integers n and k (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ 10^6).
The second line contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^6) — the number of tables on the i-th page.
The third line contains n integers y_1, y_2, ..., y_n (1 ≤ y_i ≤ 10^6) — the number of formulas on the i-th page.
Output
Print "YES" if Vova can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.
Otherwise print "NO".
Examples
Input
2 2
5 5
2 2
Output
YES
Input
2 2
5 6
2 2
Output
NO
Input
4 1
4 1 10 1
3 2 10 1
Output
YES
Note
In the first example the only option to rearrange everything is the following (let table be 'T' and formula be 'F'):
* page 1: "TTFTTFT"
* page 2: "TFTTFTT"
That way all blocks of tables have length 2.
In the second example there is no way to fit everything in such a way that there are no more than 2 tables in a row and 2 formulas in a row.
Submitted Solution:
```
n, k = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
podrnastr = 0
for tabl, formul in zip(x, y):
if max(tabl, formul) - min(tabl, formul) == 1:
continue
podrnastr = max(tabl, formul) // min(tabl, formul) + 1
if podrnastr > k:
print('NO')
raise SystemExit
print('YES')
```
No
| 4,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova has taken his summer practice this year and now he should write a report on how it went.
Vova has already drawn all the tables and wrote down all the formulas. Moreover, he has already decided that the report will consist of exactly n pages and the i-th page will include x_i tables and y_i formulas. The pages are numbered from 1 to n.
Vova fills the pages one after another, he can't go filling page i + 1 before finishing page i and he can't skip pages.
However, if he draws strictly more than k tables in a row or writes strictly more than k formulas in a row then he will get bored. Vova wants to rearrange tables and formulas in each page in such a way that he doesn't get bored in the process. Vova can't move some table or some formula to another page.
Note that the count doesn't reset on the start of the new page. For example, if the page ends with 3 tables and the next page starts with 5 tables, then it's counted as 8 tables in a row.
Help Vova to determine if he can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.
Input
The first line contains two integers n and k (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ k ≤ 10^6).
The second line contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ 10^6) — the number of tables on the i-th page.
The third line contains n integers y_1, y_2, ..., y_n (1 ≤ y_i ≤ 10^6) — the number of formulas on the i-th page.
Output
Print "YES" if Vova can rearrange tables and formulas on each page in such a way that there is no more than k tables in a row and no more than k formulas in a row.
Otherwise print "NO".
Examples
Input
2 2
5 5
2 2
Output
YES
Input
2 2
5 6
2 2
Output
NO
Input
4 1
4 1 10 1
3 2 10 1
Output
YES
Note
In the first example the only option to rearrange everything is the following (let table be 'T' and formula be 'F'):
* page 1: "TTFTTFT"
* page 2: "TFTTFTT"
That way all blocks of tables have length 2.
In the second example there is no way to fit everything in such a way that there are no more than 2 tables in a row and 2 formulas in a row.
Submitted Solution:
```
n, k = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
podrnastr = 0
for tabl, formul in zip(x, y):
if max(tabl, formul) - min(tabl, formul) == 1:
continue
podrnastr = max(tabl, formul) // min(tabl, formul)
if podrnastr > k:
print('NO')
raise SystemExit
print('YES')
```
No
| 4,184 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alex decided to try his luck in TV shows. He once went to the quiz named "What's That Word?!". After perfectly answering the questions "How is a pseudonym commonly referred to in the Internet?" ("Um... a nick?"), "After which famous inventor we name the unit of the magnetic field strength?" ("Um... Nikola Tesla?") and "Which rock band performs "How You Remind Me"?" ("Um... Nickelback?"), he decided to apply to a little bit more difficult TV show: "What's in This Multiset?!".
The rules of this TV show are as follows: there are n multisets numbered from 1 to n. Each of them is initially empty. Then, q events happen; each of them is in one of the four possible types:
* 1 x v — set the x-th multiset to a singleton \\{v\}.
* 2 x y z — set the x-th multiset to a union of the y-th and the z-th multiset. For example: \{1, 3\}∪\{1, 4, 4\}=\{1, 1, 3, 4, 4\}.
* 3 x y z — set the x-th multiset to a product of the y-th and the z-th multiset. The product A × B of two multisets A, B is defined as \{ \gcd(a, b) ∣ a ∈ A, b ∈ B \}, where \gcd(p, q) is the greatest common divisor of p and q. For example: \{2, 2, 3\} × \{1, 4, 6\}=\{1, 2, 2, 1, 2, 2, 1, 1, 3\}.
* 4 x v — the participant is asked how many times number v occurs in the x-th multiset. As the quiz turned out to be too hard in the past, participants should now give the answers modulo 2 only.
Note, that x, y and z described above are not necessarily different. In events of types 2 and 3, the sum or the product is computed first, and then the assignment is performed.
Alex is confused by the complicated rules of the show. Can you help him answer the requests of the 4-th type?
Input
The first line contains two integers n and q (1 ≤ n ≤ 10^5, 1 ≤ q ≤ 10^6) — the number of multisets and the number of events.
Each of the following q lines describes next event in the format given in statement. It's guaranteed that 1 ≤ x,y,z ≤ n and 1 ≤ v ≤ 7000 always holds.
It's guaranteed that there will be at least one event of the 4-th type.
Output
Print a string which consists of digits 0 and 1 only, and has length equal to the number of events of the 4-th type. The i-th digit of the string should be equal to the answer for the i-th query of the 4-th type.
Example
Input
4 13
1 1 1
1 2 4
1 3 6
4 4 4
1 4 4
2 2 1 2
2 3 3 4
4 4 4
3 2 2 3
4 2 1
4 2 2
4 2 3
4 2 4
Output
010101
Note
Here is how the multisets look in the example test after each of the events; i is the number of queries processed so far:
<image>
Submitted Solution:
```
from math import gcd
from collections import Counter
from itertools import repeat, product
n, q = map(int, input().split())
c = [Counter() for _ in repeat(None, n+1)]
ans=[]
for _ in repeat(None, q):
com, args = input().split(maxsplit=1)
args = list(map(int, args.split()))
if com=='1':
x, v = args
c[x]=Counter([v])
elif com=='2':
x, y, z = args
c[x]=c[y]+c[z]
elif com=='3':
x, y, z = args
c[x]=Counter(gcd(q,w) for q,w in product(c[y], c[z]))
elif com=='4':
x, v = args
ans.append(c[x][v]%2)
print(''.join(map(str, ans)))
```
No
| 4,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alex decided to try his luck in TV shows. He once went to the quiz named "What's That Word?!". After perfectly answering the questions "How is a pseudonym commonly referred to in the Internet?" ("Um... a nick?"), "After which famous inventor we name the unit of the magnetic field strength?" ("Um... Nikola Tesla?") and "Which rock band performs "How You Remind Me"?" ("Um... Nickelback?"), he decided to apply to a little bit more difficult TV show: "What's in This Multiset?!".
The rules of this TV show are as follows: there are n multisets numbered from 1 to n. Each of them is initially empty. Then, q events happen; each of them is in one of the four possible types:
* 1 x v — set the x-th multiset to a singleton \\{v\}.
* 2 x y z — set the x-th multiset to a union of the y-th and the z-th multiset. For example: \{1, 3\}∪\{1, 4, 4\}=\{1, 1, 3, 4, 4\}.
* 3 x y z — set the x-th multiset to a product of the y-th and the z-th multiset. The product A × B of two multisets A, B is defined as \{ \gcd(a, b) ∣ a ∈ A, b ∈ B \}, where \gcd(p, q) is the greatest common divisor of p and q. For example: \{2, 2, 3\} × \{1, 4, 6\}=\{1, 2, 2, 1, 2, 2, 1, 1, 3\}.
* 4 x v — the participant is asked how many times number v occurs in the x-th multiset. As the quiz turned out to be too hard in the past, participants should now give the answers modulo 2 only.
Note, that x, y and z described above are not necessarily different. In events of types 2 and 3, the sum or the product is computed first, and then the assignment is performed.
Alex is confused by the complicated rules of the show. Can you help him answer the requests of the 4-th type?
Input
The first line contains two integers n and q (1 ≤ n ≤ 10^5, 1 ≤ q ≤ 10^6) — the number of multisets and the number of events.
Each of the following q lines describes next event in the format given in statement. It's guaranteed that 1 ≤ x,y,z ≤ n and 1 ≤ v ≤ 7000 always holds.
It's guaranteed that there will be at least one event of the 4-th type.
Output
Print a string which consists of digits 0 and 1 only, and has length equal to the number of events of the 4-th type. The i-th digit of the string should be equal to the answer for the i-th query of the 4-th type.
Example
Input
4 13
1 1 1
1 2 4
1 3 6
4 4 4
1 4 4
2 2 1 2
2 3 3 4
4 4 4
3 2 2 3
4 2 1
4 2 2
4 2 3
4 2 4
Output
010101
Note
Here is how the multisets look in the example test after each of the events; i is the number of queries processed so far:
<image>
Submitted Solution:
```
print("wl")
```
No
| 4,186 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alex decided to try his luck in TV shows. He once went to the quiz named "What's That Word?!". After perfectly answering the questions "How is a pseudonym commonly referred to in the Internet?" ("Um... a nick?"), "After which famous inventor we name the unit of the magnetic field strength?" ("Um... Nikola Tesla?") and "Which rock band performs "How You Remind Me"?" ("Um... Nickelback?"), he decided to apply to a little bit more difficult TV show: "What's in This Multiset?!".
The rules of this TV show are as follows: there are n multisets numbered from 1 to n. Each of them is initially empty. Then, q events happen; each of them is in one of the four possible types:
* 1 x v — set the x-th multiset to a singleton \\{v\}.
* 2 x y z — set the x-th multiset to a union of the y-th and the z-th multiset. For example: \{1, 3\}∪\{1, 4, 4\}=\{1, 1, 3, 4, 4\}.
* 3 x y z — set the x-th multiset to a product of the y-th and the z-th multiset. The product A × B of two multisets A, B is defined as \{ \gcd(a, b) ∣ a ∈ A, b ∈ B \}, where \gcd(p, q) is the greatest common divisor of p and q. For example: \{2, 2, 3\} × \{1, 4, 6\}=\{1, 2, 2, 1, 2, 2, 1, 1, 3\}.
* 4 x v — the participant is asked how many times number v occurs in the x-th multiset. As the quiz turned out to be too hard in the past, participants should now give the answers modulo 2 only.
Note, that x, y and z described above are not necessarily different. In events of types 2 and 3, the sum or the product is computed first, and then the assignment is performed.
Alex is confused by the complicated rules of the show. Can you help him answer the requests of the 4-th type?
Input
The first line contains two integers n and q (1 ≤ n ≤ 10^5, 1 ≤ q ≤ 10^6) — the number of multisets and the number of events.
Each of the following q lines describes next event in the format given in statement. It's guaranteed that 1 ≤ x,y,z ≤ n and 1 ≤ v ≤ 7000 always holds.
It's guaranteed that there will be at least one event of the 4-th type.
Output
Print a string which consists of digits 0 and 1 only, and has length equal to the number of events of the 4-th type. The i-th digit of the string should be equal to the answer for the i-th query of the 4-th type.
Example
Input
4 13
1 1 1
1 2 4
1 3 6
4 4 4
1 4 4
2 2 1 2
2 3 3 4
4 4 4
3 2 2 3
4 2 1
4 2 2
4 2 3
4 2 4
Output
010101
Note
Here is how the multisets look in the example test after each of the events; i is the number of queries processed so far:
<image>
Submitted Solution:
```
n, q = input().strip().split(' ')
n, q = [int(n), int(q)]
multisets = [[] for i in range(n+1)]
ans = []
def gcd(a, b):
if b > a:
return gcd(b, a)
if a % b == 0:
return b
return gcd(b, a % b)
for i in range(q):
in_row = input().strip().split(' ')
in_row = list(map(int, in_row))
if in_row[0] == 1:
# event 1
multisets[in_row[1]] = [in_row[2]]
print(multisets)
elif in_row[0] == 2:
# event 2
multisets[in_row[1]] = multisets[in_row[2]] + multisets[in_row[3]]
print(multisets)
elif in_row[0] == 3:
# event 3
result = []
for i1 in multisets[in_row[2]]:
for i2 in multisets[in_row[3]]:
result.append(gcd(i1,i2))
multisets[in_row[1]] = result
elif in_row[0] == 4:
#event 4
# print("Multisets:", multisets[in_row[1]])
ans.append((multisets[in_row[1]].count(in_row[2]))%2)
for i in ans:
print(i, end="")
```
No
| 4,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alex decided to try his luck in TV shows. He once went to the quiz named "What's That Word?!". After perfectly answering the questions "How is a pseudonym commonly referred to in the Internet?" ("Um... a nick?"), "After which famous inventor we name the unit of the magnetic field strength?" ("Um... Nikola Tesla?") and "Which rock band performs "How You Remind Me"?" ("Um... Nickelback?"), he decided to apply to a little bit more difficult TV show: "What's in This Multiset?!".
The rules of this TV show are as follows: there are n multisets numbered from 1 to n. Each of them is initially empty. Then, q events happen; each of them is in one of the four possible types:
* 1 x v — set the x-th multiset to a singleton \\{v\}.
* 2 x y z — set the x-th multiset to a union of the y-th and the z-th multiset. For example: \{1, 3\}∪\{1, 4, 4\}=\{1, 1, 3, 4, 4\}.
* 3 x y z — set the x-th multiset to a product of the y-th and the z-th multiset. The product A × B of two multisets A, B is defined as \{ \gcd(a, b) ∣ a ∈ A, b ∈ B \}, where \gcd(p, q) is the greatest common divisor of p and q. For example: \{2, 2, 3\} × \{1, 4, 6\}=\{1, 2, 2, 1, 2, 2, 1, 1, 3\}.
* 4 x v — the participant is asked how many times number v occurs in the x-th multiset. As the quiz turned out to be too hard in the past, participants should now give the answers modulo 2 only.
Note, that x, y and z described above are not necessarily different. In events of types 2 and 3, the sum or the product is computed first, and then the assignment is performed.
Alex is confused by the complicated rules of the show. Can you help him answer the requests of the 4-th type?
Input
The first line contains two integers n and q (1 ≤ n ≤ 10^5, 1 ≤ q ≤ 10^6) — the number of multisets and the number of events.
Each of the following q lines describes next event in the format given in statement. It's guaranteed that 1 ≤ x,y,z ≤ n and 1 ≤ v ≤ 7000 always holds.
It's guaranteed that there will be at least one event of the 4-th type.
Output
Print a string which consists of digits 0 and 1 only, and has length equal to the number of events of the 4-th type. The i-th digit of the string should be equal to the answer for the i-th query of the 4-th type.
Example
Input
4 13
1 1 1
1 2 4
1 3 6
4 4 4
1 4 4
2 2 1 2
2 3 3 4
4 4 4
3 2 2 3
4 2 1
4 2 2
4 2 3
4 2 4
Output
010101
Note
Here is how the multisets look in the example test after each of the events; i is the number of queries processed so far:
<image>
Submitted Solution:
```
ch=input()
l=ch.split()
n=int(l[0])
q=int(l[1])
ch=''
def pgcd(a,b):
while b!=0:
a,b=b,a%b
return a
cha=[]
res=[[],[],[],[]]
for i in range (q):
ch=input()
l=ch.split()
if l[0]==1:
res[0]=[l[2]]
elif l[0]==2:
res[l[1]]=res[l[2]]+res[l[3]]
elif l[0]==3:
f=[]
for j in range(len(res[l[2]])):
for k in range(len(res[l[3]])):
f.append(pgcd(res[l[2]][j],res[l[3]][k]))
res[l[2]]=f
elif l[0]==4:
d=res[l[1]]
cha.append(d.count(l[2])%2)
pr=''
for m in range(len(cha)):
pr+=str(cha[m])
print(pr)
```
No
| 4,188 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an undirected tree of n vertices.
Some vertices are colored blue, some are colored red and some are uncolored. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
You choose an edge and remove it from the tree. Tree falls apart into two connected components. Let's call an edge nice if neither of the resulting components contain vertices of both red and blue colors.
How many nice edges are there in the given tree?
Input
The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of vertices in the tree.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2) — the colors of the vertices. a_i = 1 means that vertex i is colored red, a_i = 2 means that vertex i is colored blue and a_i = 0 means that vertex i is uncolored.
The i-th of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the edges of the tree. It is guaranteed that the given edges form a tree. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
Output
Print a single integer — the number of nice edges in the given tree.
Examples
Input
5
2 0 0 1 2
1 2
2 3
2 4
2 5
Output
1
Input
5
1 0 0 0 2
1 2
2 3
3 4
4 5
Output
4
Input
3
1 1 2
2 3
1 3
Output
0
Note
Here is the tree from the first example:
<image>
The only nice edge is edge (2, 4). Removing it makes the tree fall apart into components \{4\} and \{1, 2, 3, 5\}. The first component only includes a red vertex and the second component includes blue vertices and uncolored vertices.
Here is the tree from the second example:
<image>
Every edge is nice in it.
Here is the tree from the third example:
<image>
Edge (1, 3) splits the into components \{1\} and \{3, 2\}, the latter one includes both red and blue vertex, thus the edge isn't nice. Edge (2, 3) splits the into components \{1, 3\} and \{2\}, the former one includes both red and blue vertex, thus the edge also isn't nice. So the answer is 0.
Tags: dfs and similar, trees
Correct Solution:
```
import sys
input=sys.stdin.readline
n = int(input())
a = [int(t) for t in input().split(' ')]
mx = [[] for _ in range(n)]
for i in range(n-1):
v1, v2 = map(int,input().split())
mx[v1-1].append(v2-1)
mx[v2-1].append(v1-1)
count = [[0, 0] for _ in range(n)]
total = [a.count(1), a.count(2)]
answer = 0
OBSERVE = 0
CHECK = 1
stack = [(OBSERVE, 0, -1)]
while len(stack):
#print(stack,count)
state, vertex, parent = stack.pop()
if state == OBSERVE:
stack.append((CHECK, vertex, parent))
for child in mx[vertex]:
#print(nv,v,from_)
if child != parent:
stack.append((OBSERVE, child, vertex))
else:
for child in mx[vertex]:
if child != parent:
#print(child,parent,count)
if count[child][0] == total[0] and count[child][1] == 0 or count[child][1] == total[1] and count[child][0] == 0:
answer += 1
count[vertex][0] += count[child][0]
count[vertex][1] += count[child][1]
if a[vertex] != 0:
#print(count)
count[vertex][a[vertex]-1] += 1
#print(count)
print(answer)
```
| 4,189 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an undirected tree of n vertices.
Some vertices are colored blue, some are colored red and some are uncolored. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
You choose an edge and remove it from the tree. Tree falls apart into two connected components. Let's call an edge nice if neither of the resulting components contain vertices of both red and blue colors.
How many nice edges are there in the given tree?
Input
The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of vertices in the tree.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2) — the colors of the vertices. a_i = 1 means that vertex i is colored red, a_i = 2 means that vertex i is colored blue and a_i = 0 means that vertex i is uncolored.
The i-th of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the edges of the tree. It is guaranteed that the given edges form a tree. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
Output
Print a single integer — the number of nice edges in the given tree.
Examples
Input
5
2 0 0 1 2
1 2
2 3
2 4
2 5
Output
1
Input
5
1 0 0 0 2
1 2
2 3
3 4
4 5
Output
4
Input
3
1 1 2
2 3
1 3
Output
0
Note
Here is the tree from the first example:
<image>
The only nice edge is edge (2, 4). Removing it makes the tree fall apart into components \{4\} and \{1, 2, 3, 5\}. The first component only includes a red vertex and the second component includes blue vertices and uncolored vertices.
Here is the tree from the second example:
<image>
Every edge is nice in it.
Here is the tree from the third example:
<image>
Edge (1, 3) splits the into components \{1\} and \{3, 2\}, the latter one includes both red and blue vertex, thus the edge isn't nice. Edge (2, 3) splits the into components \{1, 3\} and \{2\}, the former one includes both red and blue vertex, thus the edge also isn't nice. So the answer is 0.
Tags: dfs and similar, trees
Correct Solution:
```
import sys
n = int(input())
a = [int(t) for t in input().split(' ')]
mx = [[] for _ in range(n)]
lines = sys.stdin.readlines()
for i in range(n-1):
v1, v2 = (int(t) - 1 for t in lines[i].split(' '))
mx[v1].append(v2)
mx[v2].append(v1)
count = [[0, 0] for _ in range(n)]
total = [a.count(1), a.count(2)]
answer = 0
OBSERVE = 0
CHECK = 1
stack = [(OBSERVE, 0, -1)]
while len(stack):
state, v, from_ = stack.pop()
if state == OBSERVE:
stack.append((CHECK, v, from_))
for nv in mx[v]:
if nv != from_:
stack.append((OBSERVE, nv, v))
else:
for nv in mx[v]:
if nv != from_:
if count[nv][0] == total[0] and count[nv][1] == 0 or count[nv][1] == total[1] and count[nv][0] == 0:
answer += 1
count[v][0] += count[nv][0]
count[v][1] += count[nv][1]
if a[v] != 0:
count[v][a[v]-1] += 1
print(answer)
```
| 4,190 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an undirected tree of n vertices.
Some vertices are colored blue, some are colored red and some are uncolored. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
You choose an edge and remove it from the tree. Tree falls apart into two connected components. Let's call an edge nice if neither of the resulting components contain vertices of both red and blue colors.
How many nice edges are there in the given tree?
Input
The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of vertices in the tree.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2) — the colors of the vertices. a_i = 1 means that vertex i is colored red, a_i = 2 means that vertex i is colored blue and a_i = 0 means that vertex i is uncolored.
The i-th of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the edges of the tree. It is guaranteed that the given edges form a tree. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
Output
Print a single integer — the number of nice edges in the given tree.
Examples
Input
5
2 0 0 1 2
1 2
2 3
2 4
2 5
Output
1
Input
5
1 0 0 0 2
1 2
2 3
3 4
4 5
Output
4
Input
3
1 1 2
2 3
1 3
Output
0
Note
Here is the tree from the first example:
<image>
The only nice edge is edge (2, 4). Removing it makes the tree fall apart into components \{4\} and \{1, 2, 3, 5\}. The first component only includes a red vertex and the second component includes blue vertices and uncolored vertices.
Here is the tree from the second example:
<image>
Every edge is nice in it.
Here is the tree from the third example:
<image>
Edge (1, 3) splits the into components \{1\} and \{3, 2\}, the latter one includes both red and blue vertex, thus the edge isn't nice. Edge (2, 3) splits the into components \{1, 3\} and \{2\}, the former one includes both red and blue vertex, thus the edge also isn't nice. So the answer is 0.
Tags: dfs and similar, trees
Correct Solution:
```
#!/usr/bin/env python3
"""
This file is part of https://github.com/cheran-senthil/PyRival
Copyright 2019 Cheran Senthilkumar <hello@cheran.io>
"""
import os
import sys
from atexit import register
from io import BytesIO
input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
sys.__stdout__ = BytesIO()
register(lambda: os.write(1, sys.__stdout__.getvalue()))
res = 0
def main():
n = int(input())
a = input().split()
red_cnt = a.count(b'1')
blue_cnt = a.count(b'2')
tree = [[] for _ in range(n)]
for _ in range(n - 1):
v, u = map(int, input().split())
tree[v - 1].append(u - 1)
tree[u - 1].append(v - 1)
dp, visited = [[0, 0] for _ in range(n)], [False] * n
def dfs(node):
global res
finished = [False] * n
stack = [node]
while stack:
node = stack[-1]
node_cnt = dp[node]
if not visited[node]:
visited[node] = True
else:
stack.pop()
node_cnt[0] += a[node] == b'1'
node_cnt[1] += a[node] == b'2'
finished[node] = True
for child in tree[node]:
if not visited[child]:
stack.append(child)
elif finished[child]:
child_cnt = dp[child]
node_cnt[0] += child_cnt[0]
node_cnt[1] += child_cnt[1]
if ((child_cnt[0] == red_cnt) and (child_cnt[1] == 0)) or ((child_cnt[0] == 0) and
(child_cnt[1] == blue_cnt)):
res += 1
dfs(0)
print(res)
if __name__ == '__main__':
main()
```
| 4,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an undirected tree of n vertices.
Some vertices are colored blue, some are colored red and some are uncolored. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
You choose an edge and remove it from the tree. Tree falls apart into two connected components. Let's call an edge nice if neither of the resulting components contain vertices of both red and blue colors.
How many nice edges are there in the given tree?
Input
The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of vertices in the tree.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2) — the colors of the vertices. a_i = 1 means that vertex i is colored red, a_i = 2 means that vertex i is colored blue and a_i = 0 means that vertex i is uncolored.
The i-th of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the edges of the tree. It is guaranteed that the given edges form a tree. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
Output
Print a single integer — the number of nice edges in the given tree.
Examples
Input
5
2 0 0 1 2
1 2
2 3
2 4
2 5
Output
1
Input
5
1 0 0 0 2
1 2
2 3
3 4
4 5
Output
4
Input
3
1 1 2
2 3
1 3
Output
0
Note
Here is the tree from the first example:
<image>
The only nice edge is edge (2, 4). Removing it makes the tree fall apart into components \{4\} and \{1, 2, 3, 5\}. The first component only includes a red vertex and the second component includes blue vertices and uncolored vertices.
Here is the tree from the second example:
<image>
Every edge is nice in it.
Here is the tree from the third example:
<image>
Edge (1, 3) splits the into components \{1\} and \{3, 2\}, the latter one includes both red and blue vertex, thus the edge isn't nice. Edge (2, 3) splits the into components \{1, 3\} and \{2\}, the former one includes both red and blue vertex, thus the edge also isn't nice. So the answer is 0.
Tags: dfs and similar, trees
Correct Solution:
```
from sys import stdin,stdout
from collections import *
from math import gcd,floor,ceil
st=lambda:list(stdin.readline().strip())
li=lambda:list(map(int,stdin.readline().split()))
mp=lambda:map(int,stdin.readline().split())
inp=lambda:int(stdin.readline())
pr=lambda n: stdout.write(str(n)+"\n")
mod=1000000007
INF=float('inf')
ans=0
n=inp()
l=[0]+li()
edge=[li() for i in range(n-1)]
d={i:[] for i in range(n+1)}
for i in edge:
a,b=i
d[a].append(b)
d[b].append(a)
red,blue=0,0
for i in l:
red+= (i==1)
blue+= (i==2)
stack=[1]
v=[0 for i in range(n+1)]
RED=[1 if i==1 else 0 for i in l ]
BLUE=[1 if i==2 else 0 for i in l ]
par=[-1 for i in range(n+1)]
v[1]=1
while stack:
a=stack[-1]
f=0
for i in d[a]:
if not v[i]:
v[i]=1
stack.append(i)
f=1
par[i]=a
if f==0:
z=stack.pop()
if stack:
RED[par[z]]+=RED[z]
BLUE[par[z]]+=BLUE[z]
for i in range(1,n+1):
if par[i]!=-1:
if RED[i]==red and BLUE[i]==0 or RED[i]==0 and BLUE[i]==blue:
ans+=1
pr(ans)
```
| 4,192 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an undirected tree of n vertices.
Some vertices are colored blue, some are colored red and some are uncolored. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
You choose an edge and remove it from the tree. Tree falls apart into two connected components. Let's call an edge nice if neither of the resulting components contain vertices of both red and blue colors.
How many nice edges are there in the given tree?
Input
The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of vertices in the tree.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2) — the colors of the vertices. a_i = 1 means that vertex i is colored red, a_i = 2 means that vertex i is colored blue and a_i = 0 means that vertex i is uncolored.
The i-th of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the edges of the tree. It is guaranteed that the given edges form a tree. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
Output
Print a single integer — the number of nice edges in the given tree.
Examples
Input
5
2 0 0 1 2
1 2
2 3
2 4
2 5
Output
1
Input
5
1 0 0 0 2
1 2
2 3
3 4
4 5
Output
4
Input
3
1 1 2
2 3
1 3
Output
0
Note
Here is the tree from the first example:
<image>
The only nice edge is edge (2, 4). Removing it makes the tree fall apart into components \{4\} and \{1, 2, 3, 5\}. The first component only includes a red vertex and the second component includes blue vertices and uncolored vertices.
Here is the tree from the second example:
<image>
Every edge is nice in it.
Here is the tree from the third example:
<image>
Edge (1, 3) splits the into components \{1\} and \{3, 2\}, the latter one includes both red and blue vertex, thus the edge isn't nice. Edge (2, 3) splits the into components \{1, 3\} and \{2\}, the former one includes both red and blue vertex, thus the edge also isn't nice. So the answer is 0.
Tags: dfs and similar, trees
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import defaultdict, deque, Counter, OrderedDict
import threading
def main():
n = int(input())
a = [0]+[*map(int,input().split())]
adj = [[] for _ in range(n+1)]
for i in range(n-1):
x,y = map(int,input().split())
adj[x].append(y)
adj[y].append(x)
num = [[0,0] for _ in range(n+1)]
q=deque([[1,-1]])
Eq = deque()
while q:
x, par = q.popleft()
Eq.appendleft([x,par])
for nx in adj[x]:
if nx == par:continue
q.append([nx,x])
while Eq:
x, par = Eq.popleft()
if a[x] == 1: num[x][0] += 1
if a[x] == 2: num[x][1] += 1
for nx in adj[x]:
if nx == par: continue
num[x][0] += num[nx][0]
num[x][1] += num[nx][1]
ans = 0
for i in range(2,n+1):
if num[i][0] == num[1][0] and num[i][1] == 0:ans+=1
if num[i][0] == 0 and num[i][1] == num[1][1]: ans+=1
print(ans)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
"""threading.stack_size(40960000)
thread = threading.Thread(target=main)
thread.start()"""
main()
```
| 4,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an undirected tree of n vertices.
Some vertices are colored blue, some are colored red and some are uncolored. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
You choose an edge and remove it from the tree. Tree falls apart into two connected components. Let's call an edge nice if neither of the resulting components contain vertices of both red and blue colors.
How many nice edges are there in the given tree?
Input
The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of vertices in the tree.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2) — the colors of the vertices. a_i = 1 means that vertex i is colored red, a_i = 2 means that vertex i is colored blue and a_i = 0 means that vertex i is uncolored.
The i-th of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the edges of the tree. It is guaranteed that the given edges form a tree. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
Output
Print a single integer — the number of nice edges in the given tree.
Examples
Input
5
2 0 0 1 2
1 2
2 3
2 4
2 5
Output
1
Input
5
1 0 0 0 2
1 2
2 3
3 4
4 5
Output
4
Input
3
1 1 2
2 3
1 3
Output
0
Note
Here is the tree from the first example:
<image>
The only nice edge is edge (2, 4). Removing it makes the tree fall apart into components \{4\} and \{1, 2, 3, 5\}. The first component only includes a red vertex and the second component includes blue vertices and uncolored vertices.
Here is the tree from the second example:
<image>
Every edge is nice in it.
Here is the tree from the third example:
<image>
Edge (1, 3) splits the into components \{1\} and \{3, 2\}, the latter one includes both red and blue vertex, thus the edge isn't nice. Edge (2, 3) splits the into components \{1, 3\} and \{2\}, the former one includes both red and blue vertex, thus the edge also isn't nice. So the answer is 0.
Tags: dfs and similar, trees
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
def main():
n = int(input())
a = list(map(int,input().split()))
red = a.count(1)
blue = a.count(2)
path = [[] for _ in range(n+1)]
for _ in range(n-1):
u1,v1 = map(int,input().split())
path[u1].append(v1)
path[v1].append(u1)
st = [1]
parent = [0]*(n+1)
parent[1] = 1
dp = [[0,0] for _ in range(n+1)]
# red ; blue
ans = 0
while len(st):
while len(path[st[-1]]) and parent[path[st[-1]][-1]]:
path[st[-1]].pop()
if not len(path[st[-1]]):
x = st.pop()
if a[x-1] == 1:
dp[x][0] += 1
elif a[x-1] == 2:
dp[x][1] += 1
if (dp[x][0] == red and not dp[x][1]) or (dp[x][1] == blue and not dp[x][0]):
ans += 1
if len(st):
dp[st[-1]][0] += dp[x][0]
dp[st[-1]][1] += dp[x][1]
continue
i = path[st[-1]].pop()
parent[i] = 1
st.append(i)
print(ans)
#Fast IO Region
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")
if __name__ == '__main__':
main()
```
| 4,194 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an undirected tree of n vertices.
Some vertices are colored blue, some are colored red and some are uncolored. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
You choose an edge and remove it from the tree. Tree falls apart into two connected components. Let's call an edge nice if neither of the resulting components contain vertices of both red and blue colors.
How many nice edges are there in the given tree?
Input
The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of vertices in the tree.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2) — the colors of the vertices. a_i = 1 means that vertex i is colored red, a_i = 2 means that vertex i is colored blue and a_i = 0 means that vertex i is uncolored.
The i-th of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the edges of the tree. It is guaranteed that the given edges form a tree. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
Output
Print a single integer — the number of nice edges in the given tree.
Examples
Input
5
2 0 0 1 2
1 2
2 3
2 4
2 5
Output
1
Input
5
1 0 0 0 2
1 2
2 3
3 4
4 5
Output
4
Input
3
1 1 2
2 3
1 3
Output
0
Note
Here is the tree from the first example:
<image>
The only nice edge is edge (2, 4). Removing it makes the tree fall apart into components \{4\} and \{1, 2, 3, 5\}. The first component only includes a red vertex and the second component includes blue vertices and uncolored vertices.
Here is the tree from the second example:
<image>
Every edge is nice in it.
Here is the tree from the third example:
<image>
Edge (1, 3) splits the into components \{1\} and \{3, 2\}, the latter one includes both red and blue vertex, thus the edge isn't nice. Edge (2, 3) splits the into components \{1, 3\} and \{2\}, the former one includes both red and blue vertex, thus the edge also isn't nice. So the answer is 0.
Tags: dfs and similar, trees
Correct Solution:
```
from collections import deque
import sys
input = sys.stdin.readline
class Graph(object):
"""docstring for Graph"""
def __init__(self,n,d): # Number of nodes and d is True if directed
self.n = n
self.graph = [[] for i in range(n)]
self.parent = [-1 for i in range(n)]
self.directed = d
def addEdge(self,x,y):
self.graph[x].append(y)
if not self.directed:
self.graph[y].append(x)
def bfs(self, root): # NORMAL BFS
self.parent = [-1 for i in range(self.n)]
queue = [root]
queue = deque(queue)
vis = [0]*self.n
while len(queue)!=0:
element = queue.popleft()
vis[element] = 1
for i in self.graph[element]:
if vis[i]==0:
queue.append(i)
self.parent[i] = element
def dfs(self, root, ans): # Iterative DFS
stack=[root]
vis=[0]*self.n
stack2=[]
while len(stack)!=0: # INITIAL TRAVERSAL
element = stack.pop()
if vis[element]:
continue
vis[element] = 1
stack2.append(element)
for i in self.graph[element]:
if vis[i]==0:
self.parent[i] = element
stack.append(i)
while len(stack2)!=0: # BACKTRACING. Modify the loop according to the question
element = stack2.pop()
m = [0,0]
for i in self.graph[element]:
if i!=self.parent[element]:
m[0] += ans[i][0]
m[1] += ans[i][1]
if arr[element] == 1:
m[0] += 1
elif arr[element] == 2:
m[1] += 1
ans[element] = m
return ans
def shortestpath(self, source, dest): # Calculate Shortest Path between two nodes
self.bfs(source)
path = [dest]
while self.parent[path[-1]]!=-1:
path.append(parent[path[-1]])
return path[::-1]
def ifcycle(self):
self.bfs(0)
queue = [0]
vis = [0]*n
queue = deque(queue)
while len(queue)!=0:
element = queue.popleft()
vis[element] = 1
for i in graph[element]:
if vis[i]==1 and i!=parent[element]:
return True
if vis[i]==0:
queue.append(i)
vis[i] = 1
return False
def reroot(self, root, ans):
stack = [root]
count = 0
vis = [0]*self.n
while len(stack)!=0:
e = stack[-1]
# print (e,ans)
if vis[e]:
stack.pop()
if self.parent[e]!=-1:
ans[self.parent[e]][0] += ans[e][0]
ans[self.parent[e]][1] += ans[e][1]
if self.parent[self.parent[e]]!=-1:
ans[self.parent[e]][0] -= ans[self.parent[self.parent[e]]][0]
ans[self.parent[e]][1] -= ans[self.parent[self.parent[e]]][1]
continue
vis[e]=1
for i in self.graph[e]:
if not vis[i]:
stack.append(i)
if self.parent[e]==-1:
continue
ans[self.parent[e]][0] -= ans[e][0]
ans[self.parent[e]][1] -= ans[e][1]
if self.parent[self.parent[e]]!=-1:
ans[self.parent[e]][0] += ans[self.parent[self.parent[e]]][0]
ans[self.parent[e]][1] += ans[self.parent[self.parent[e]]][1]
if 0 in ans[e] and 0 in ans[self.parent[e]]:
count+=1
return count
n = int(input())
g = Graph(n,False)
arr = list(map(int,input().split()))
for i in range(n-1):
x,y = map(int,input().split())
g.addEdge(x-1,y-1)
ans = [[0,0] for i in range(n)]
a = g.dfs(0,ans)
print (g.reroot(0,a))
```
| 4,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an undirected tree of n vertices.
Some vertices are colored blue, some are colored red and some are uncolored. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
You choose an edge and remove it from the tree. Tree falls apart into two connected components. Let's call an edge nice if neither of the resulting components contain vertices of both red and blue colors.
How many nice edges are there in the given tree?
Input
The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of vertices in the tree.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2) — the colors of the vertices. a_i = 1 means that vertex i is colored red, a_i = 2 means that vertex i is colored blue and a_i = 0 means that vertex i is uncolored.
The i-th of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the edges of the tree. It is guaranteed that the given edges form a tree. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
Output
Print a single integer — the number of nice edges in the given tree.
Examples
Input
5
2 0 0 1 2
1 2
2 3
2 4
2 5
Output
1
Input
5
1 0 0 0 2
1 2
2 3
3 4
4 5
Output
4
Input
3
1 1 2
2 3
1 3
Output
0
Note
Here is the tree from the first example:
<image>
The only nice edge is edge (2, 4). Removing it makes the tree fall apart into components \{4\} and \{1, 2, 3, 5\}. The first component only includes a red vertex and the second component includes blue vertices and uncolored vertices.
Here is the tree from the second example:
<image>
Every edge is nice in it.
Here is the tree from the third example:
<image>
Edge (1, 3) splits the into components \{1\} and \{3, 2\}, the latter one includes both red and blue vertex, thus the edge isn't nice. Edge (2, 3) splits the into components \{1, 3\} and \{2\}, the former one includes both red and blue vertex, thus the edge also isn't nice. So the answer is 0.
Tags: dfs and similar, trees
Correct Solution:
```
import os, sys
from io import IOBase, BytesIO
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
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')
# Cout implemented in Python
import sys
class ostream:
def __lshift__(self,a):
sys.stdout.write(str(a))
return self
cout = ostream()
endl = '\n'
from collections import deque
n=int(input())
a=list(map(int,input().split()))
tr=[[] for i in range(n)]
for i in range(n-1):
u,v=map(int,input().split())
tr[u-1].append(v-1)
tr[v-1].append(u-1)
r={}
b={}
s=[0]
s=deque()
s.append(0)
p={0:-1}
m=[]
while s:
x=s.popleft()
if a[x]==1:r[x]=1;b[x]=0
elif a[x]==2:r[x]=0;b[x]=1
else:r[x]=0;b[x]=0
for j in tr[x]:
if j==p[x]:continue
s.append(j)
p[j]=x
m.append(x)
for i in range(n-1,0,-1):
r[p[m[i]]]+=r[m[i]]
b[p[m[i]]]+=b[m[i]]
red=a.count(1)
blue=a.count(2)
ans=0
for i in range(1,n):
if (r[i]==red and b[i]==0) or (r[i]==0 and b[i]==blue):ans+=1
print(ans)
```
| 4,196 |
Provide tags and a correct Python 2 solution for this coding contest problem.
You are given an undirected tree of n vertices.
Some vertices are colored blue, some are colored red and some are uncolored. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
You choose an edge and remove it from the tree. Tree falls apart into two connected components. Let's call an edge nice if neither of the resulting components contain vertices of both red and blue colors.
How many nice edges are there in the given tree?
Input
The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of vertices in the tree.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2) — the colors of the vertices. a_i = 1 means that vertex i is colored red, a_i = 2 means that vertex i is colored blue and a_i = 0 means that vertex i is uncolored.
The i-th of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the edges of the tree. It is guaranteed that the given edges form a tree. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
Output
Print a single integer — the number of nice edges in the given tree.
Examples
Input
5
2 0 0 1 2
1 2
2 3
2 4
2 5
Output
1
Input
5
1 0 0 0 2
1 2
2 3
3 4
4 5
Output
4
Input
3
1 1 2
2 3
1 3
Output
0
Note
Here is the tree from the first example:
<image>
The only nice edge is edge (2, 4). Removing it makes the tree fall apart into components \{4\} and \{1, 2, 3, 5\}. The first component only includes a red vertex and the second component includes blue vertices and uncolored vertices.
Here is the tree from the second example:
<image>
Every edge is nice in it.
Here is the tree from the third example:
<image>
Edge (1, 3) splits the into components \{1\} and \{3, 2\}, the latter one includes both red and blue vertex, thus the edge isn't nice. Edge (2, 3) splits the into components \{1, 3\} and \{2\}, the former one includes both red and blue vertex, thus the edge also isn't nice. So the answer is 0.
Tags: dfs and similar, trees
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
raw_input = stdin.readline
pr = stdout.write
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
for i in arr:
stdout.write(str(i)+' ')
stdout.write('\n')
range = xrange # not for python 3.0+
n=int(raw_input())
l=in_arr()
d=[[] for i in range(n+1)]
b=l.count(2)
r=l.count(1)
for i in range(n-1):
u,v=in_arr()
d[u].append(v)
d[v].append(u)
q=[1]
vis=[0]*(n+1)
pos=0
vis[1]=1
while pos<n:
x=q[pos]
pos+=1
for i in d[x]:
if not vis[i]:
vis[i]=1
q.append(i)
dp=[[0,0] for i in range(n+1)]
while q:
x=q.pop()
vis[x]=0
if l[x-1]==1:
dp[x][0]=1
elif l[x-1]==2:
dp[x][1]=1
for i in d[x]:
if not vis[i]:
dp[x][0]+=dp[i][0]
dp[x][1]+=dp[i][1]
q=[1]
vis[1]=1
ans=0
while q:
x=q.pop(0)
for i in d[x]:
if not vis[i] and (dp[i][0] or dp[i][1]):
vis[i]=1
q.append(i)
if ((dp[i][1]==b) and (dp[i][0]==0)) or (dp[i][1]==0 and dp[i][0]==r):
#print x,i,b-dp[i][1],r-dp[i][0]
ans+=1
pr_num(ans)
```
| 4,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an undirected tree of n vertices.
Some vertices are colored blue, some are colored red and some are uncolored. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
You choose an edge and remove it from the tree. Tree falls apart into two connected components. Let's call an edge nice if neither of the resulting components contain vertices of both red and blue colors.
How many nice edges are there in the given tree?
Input
The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of vertices in the tree.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2) — the colors of the vertices. a_i = 1 means that vertex i is colored red, a_i = 2 means that vertex i is colored blue and a_i = 0 means that vertex i is uncolored.
The i-th of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the edges of the tree. It is guaranteed that the given edges form a tree. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
Output
Print a single integer — the number of nice edges in the given tree.
Examples
Input
5
2 0 0 1 2
1 2
2 3
2 4
2 5
Output
1
Input
5
1 0 0 0 2
1 2
2 3
3 4
4 5
Output
4
Input
3
1 1 2
2 3
1 3
Output
0
Note
Here is the tree from the first example:
<image>
The only nice edge is edge (2, 4). Removing it makes the tree fall apart into components \{4\} and \{1, 2, 3, 5\}. The first component only includes a red vertex and the second component includes blue vertices and uncolored vertices.
Here is the tree from the second example:
<image>
Every edge is nice in it.
Here is the tree from the third example:
<image>
Edge (1, 3) splits the into components \{1\} and \{3, 2\}, the latter one includes both red and blue vertex, thus the edge isn't nice. Edge (2, 3) splits the into components \{1, 3\} and \{2\}, the former one includes both red and blue vertex, thus the edge also isn't nice. So the answer is 0.
Submitted Solution:
```
from collections import deque
import sys
input = sys.stdin.readline
def bfs(s):
q = deque()
q.append(s)
visit = [0] * (n + 1)
visit[s] = 1
parent = [-1] * (n + 1)
p = []
while q:
i = q.popleft()
p.append(i)
for j in G[i]:
if not visit[j]:
q.append(j)
visit[j] = 1
parent[j] = i
return parent, p
n = int(input())
a = list(map(int, input().split()))
cnt = [0] * 3
for i in a:
cnt[i] += 1
G = [[] for _ in range(n + 1)]
for _ in range(n - 1):
u, v = map(int, input().split())
G[u].append(v)
G[v].append(u)
parent, p = bfs(1)
dpr, dpb = [0] * (n + 1), [0] * (n + 1)
ans = 0
for _ in range(n - 1):
i = p.pop()
j = parent[i]
if a[i - 1] == 1:
dpr[i] += 1
elif a[i - 1] == 2:
dpb[i] += 1
if (not dpr[i] and dpb[i] == cnt[2]) or (not dpb[i] and dpr[i] == cnt[1]):
ans += 1
dpr[j] += dpr[i]
dpb[j] += dpb[i]
print(ans)
```
Yes
| 4,198 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an undirected tree of n vertices.
Some vertices are colored blue, some are colored red and some are uncolored. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
You choose an edge and remove it from the tree. Tree falls apart into two connected components. Let's call an edge nice if neither of the resulting components contain vertices of both red and blue colors.
How many nice edges are there in the given tree?
Input
The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the number of vertices in the tree.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2) — the colors of the vertices. a_i = 1 means that vertex i is colored red, a_i = 2 means that vertex i is colored blue and a_i = 0 means that vertex i is uncolored.
The i-th of the next n - 1 lines contains two integers v_i and u_i (1 ≤ v_i, u_i ≤ n, v_i ≠ u_i) — the edges of the tree. It is guaranteed that the given edges form a tree. It is guaranteed that the tree contains at least one red vertex and at least one blue vertex.
Output
Print a single integer — the number of nice edges in the given tree.
Examples
Input
5
2 0 0 1 2
1 2
2 3
2 4
2 5
Output
1
Input
5
1 0 0 0 2
1 2
2 3
3 4
4 5
Output
4
Input
3
1 1 2
2 3
1 3
Output
0
Note
Here is the tree from the first example:
<image>
The only nice edge is edge (2, 4). Removing it makes the tree fall apart into components \{4\} and \{1, 2, 3, 5\}. The first component only includes a red vertex and the second component includes blue vertices and uncolored vertices.
Here is the tree from the second example:
<image>
Every edge is nice in it.
Here is the tree from the third example:
<image>
Edge (1, 3) splits the into components \{1\} and \{3, 2\}, the latter one includes both red and blue vertex, thus the edge isn't nice. Edge (2, 3) splits the into components \{1, 3\} and \{2\}, the former one includes both red and blue vertex, thus the edge also isn't nice. So the answer is 0.
Submitted Solution:
```
import sys
from collections import defaultdict
n = int(input())
#n,k = [int(__) for __ in raw_input().split()]
arr = [int(__) for __ in input().split()]
sactive = set()
sactive.add(0)
nums = [0] * n
d1 = defaultdict(set)
d2 = defaultdict(set)
d = defaultdict(set)
lines = sys.stdin.readlines()
for i in range(n-1):
sactive.add(i+1)
s,f = [int(__) for __ in lines[i].strip().split()]
s -= 1
f -= 1
d[s].add(f)
d[f].add(s)
nums[f] += 1
nums[s] += 1
leaves = set()
for i in range(n):
if nums[i] == 1:
leaves.add(i)
while len(leaves):
x = leaves.pop()
if arr[x] == 0:
sactive.remove(x)
nums[x] -= 1
targ = d[x].pop()
nums[targ] -= 1
d[targ].remove(x)
if nums[targ] == 1:
leaves.add(targ)
sactive1 = sactive.copy()
for targ in d:
d1[targ] = d[targ].copy()
nums1 = nums[:]
nums2 = nums[:]
for i in range(n):
if nums1[i] == 1:
leaves.add(i)
while len(leaves):
x = leaves.pop()
if arr[x] != 1:
sactive1.remove(x)
nums1[x] -= 1
targ = d1[x].pop()
nums1[targ] -= 1
d1[targ].remove(x)
if nums1[targ] == 1:
leaves.add(targ)
sactive2 = sactive.copy()
for targ in d:
d2[targ] = d[targ].copy()
for i in range(n):
if nums2[i] == 1:
leaves.add(i)
while len(leaves):
x = leaves.pop()
if arr[x] != 2:
sactive2.remove(x)
nums2[x] -= 1
targ = d2[x].pop()
nums2[targ] -= 1
d2[targ].remove(x)
if nums2[targ] == 1:
leaves.add(targ)
if len(sactive1 & sactive2) > 0:
print(0)
else:
print(len(sactive) - len(sactive1) - len(sactive2) + 1)
#print(nums)
#print('both',sactive)
#print('1',sactive1)
#print('2',sactive2)
#print(d)
```
Yes
| 4,199 |
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