message stringlengths 2 20.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 1.95k 109k | cluster float64 17 17 | __index_level_0__ int64 3.91k 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE | instruction | 0 | 6,994 | 17 | 13,988 |
Tags: *special
Correct Solution:
```
__author__ = 'Pavel Mavrin'
a = input().strip()
b = input().strip()
n = len(a) // 2
x = 0
figs = ["[]", "()", "8<"]
for i in range(n):
s1 = figs.index(a[i * 2: (i + 1) * 2])
s2 = figs.index(b[i * 2: (i + 1) * 2])
if s2 == (s1 + 1) % 3:
x += 1
if s1 == (s2 + 1) % 3:
x -= 1
if x > 0:
print("TEAM 1 WINS")
elif x < 0:
print("TEAM 2 WINS")
else:
print("TIE")
``` | output | 1 | 6,994 | 17 | 13,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE | instruction | 0 | 6,995 | 17 | 13,990 |
Tags: *special
Correct Solution:
```
s1 = input()
s2 = input()
num1 = 0
num2 = 0
for i in range(0, len(s1), 2):
c1 = s1[i:i + 2]
c2 = s2[i:i + 2]
if c1 != c2:
if (c1 == "8<" and c2 == "[]"):
num1 += 1
elif (c1 == "()" and c2 == "8<"):
num1 += 1
elif (c1 == "[]" and c2 == "()"):
num1 += 1
else:
num2 += 1
if num1 == num2:
print("TIE")
else:
print("TEAM {} WINS".format(int(num2 > num1) + 1))
``` | output | 1 | 6,995 | 17 | 13,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE | instruction | 0 | 6,996 | 17 | 13,992 |
Tags: *special
Correct Solution:
```
def split_by_n( seq, n ):
while seq:
yield seq[:n]
seq = seq[n:]
def is_first_win(first, second):
if first ==second:
return 0
if first == '8<':
if second == '[]':
return 1
else:
return -1
if first == '[]':
if second == '()':
return 1
else:
return -1
if second == '8<':
return 1
else:
return -1
team1 = list(split_by_n(input(), 2))
team2 = list(split_by_n(input(), 2))
score = 0
for i in range(len(team1)):
score += is_first_win(team1[i], team2[i])
#print(is_first_win(team1[i], team2[i]), team1[i], team2[i])
#print(score)
if score > 0:
print('TEAM 1 WINS')
elif score == 0:
print('TIE')
else:
print('TEAM 2 WINS')
``` | output | 1 | 6,996 | 17 | 13,993 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE | instruction | 0 | 6,997 | 17 | 13,994 |
Tags: *special
Correct Solution:
```
a = input()
a_actions = [(a[i:i+2]) for i in range(0, len(a), 2)]
b = input()
b_actions = [(b[i:i+2]) for i in range(0, len(b), 2)]
# print(a_actions, b_actions)
rock = "()"
paper = "[]"
scissors = "8<"
a_pts = 0
b_pts = 0
for i in range(len(a)//2):
if a_actions[i] == rock:
if b_actions[i] == scissors:
a_pts += 1
elif b_actions[i] == paper:
b_pts += 1
else:
a_pts += 1
b_pts += 1
elif a_actions[i] == paper:
if b_actions[i] == rock:
a_pts += 1
elif b_actions[i] == scissors:
b_pts += 1
else:
a_pts += 1
b_pts += 1
else:
# a_actions[i] == scissors
if b_actions[i] == paper:
a_pts += 1
elif b_actions[i] == rock:
b_pts += 1
else:
a_pts += 1
b_pts += 1
if a_pts > b_pts:
print("TEAM 1 WINS")
elif a_pts < b_pts:
print("TEAM 2 WINS")
else:
print("TIE")
``` | output | 1 | 6,997 | 17 | 13,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE | instruction | 0 | 6,998 | 17 | 13,996 |
Tags: *special
Correct Solution:
```
#!/bin/bash/python
# Date: 2014-04-16
# Author: shijinzhan
# Status:
# Note: '()' < '[]'
# '[]' < '8<'
# '8<' < '()'
team1 = input().replace('(', '6').replace('[', '7')
team2 = input().replace('(', '6').replace('[', '7')
s = 0
for x in range(0, len(team1), 2):
if team1[x] > team2[x]:
s += 1
elif team1[x] < team2[x]:
s -= 1
if team1[x] == '6' and team2[x] == '8':
s += 2
if team1[x] == '8' and team2[x] == '6':
s -= 2
if s > 0:
print("TEAM 1 WINS")
elif s < 0:
print("TEAM 2 WINS")
else:
print("TIE")
``` | output | 1 | 6,998 | 17 | 13,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE | instruction | 0 | 6,999 | 17 | 13,998 |
Tags: *special
Correct Solution:
```
s = input()
t = input()
a = 0
b = 0
for i in range(0, len(s), 2):
if s[i:i+2] == "[]" and t[i:i+2]=="()":
a += 1
if s[i:i+2] == "()" and t[i:i+2]=="8<":
a += 1
if s[i:i+2] == "8<" and t[i:i+2]=="[]":
a += 1
if t[i:i+2] == "[]" and s[i:i+2]=="()":
b += 1
if t[i:i+2] == "()" and s[i:i+2]=="8<":
b += 1
if t[i:i+2] == "8<" and s[i:i+2]=="[]":
b += 1
if a > b:
print("TEAM 1 WINS")
elif a < b:
print("TEAM 2 WINS")
else:
print("TIE")
``` | output | 1 | 6,999 | 17 | 13,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE | instruction | 0 | 7,000 | 17 | 14,000 |
Tags: *special
Correct Solution:
```
cnt=0
t1=input()
t2=input()
for i in range(0,len(t1),2):
p1=t1[i:i+2]
p2=t2[i:i+2]
if(p1=="8<"):
if(p2=="[]"):
cnt+=1
elif(p2=="()"):
cnt-=1
elif(p1=="[]"):
if(p2=="()"):
cnt+=1
elif(p2=="8<"):
cnt-=1
else:
if(p2=="8<"):
cnt+=1
elif(p2=="[]"):
cnt-=1
if(cnt==0):
print("TIE")
elif(cnt>0):
print("TEAM 1 WINS")
else:
print("TEAM 2 WINS")
``` | output | 1 | 7,000 | 17 | 14,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE
Submitted Solution:
```
def a(sMove1, sMove2):
if sMove1 == "8<" and sMove2 == "[]" or sMove1 == "[]" and sMove2 == "()" or sMove1 == "()" and sMove2 == "8<":
return 1
elif sMove2 == "8<" and sMove1 == "[]" or sMove2 == "[]" and sMove1 == "()" or sMove2 == "()" and sMove1 == "8<":
return 2
else:
return 0
t1 = input()
t2 = input()
s = [0, 0, 0]
for i in range(0, len(t1), 2):
s[a(t1[i] + t1[i + 1], t2[i] + t2[i + 1])] += 1
if s[1] > s[2]:
print('TEAM 1 WINS')
elif s[1] < s[2]:
print("TEAM 2 WINS")
else:
print('TIE')
``` | instruction | 0 | 7,001 | 17 | 14,002 |
Yes | output | 1 | 7,001 | 17 | 14,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE
Submitted Solution:
```
s1 = input()
s2 = input()
score1, score2 = 0, 0
for c in range(0, len(s1), 2):
if s1[c] == '8':
if s2[c] == '[':
score1 += 1
if s2[c] == '(':
score2 += 1
if s1[c] == '[':
if s2[c] == '(':
score1 += 1
if s2[c] == '8':
score2 += 1
if s1[c] == '(':
if s2[c] == '8':
score1 += 1
if s2[c] == '[':
score2 += 1
if score1 == score2:
print ('TIE')
else:
print('TEAM {} WINS'.format(2 - int(score1 > score2)))
``` | instruction | 0 | 7,002 | 17 | 14,004 |
Yes | output | 1 | 7,002 | 17 | 14,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE
Submitted Solution:
```
from math import*
def winner(a,b):
if a=="[]":
if b=="()":
return 1
elif b=="[]":
return 0
else:
return -1
elif a=="8<":
if b=="()":
return -1
elif b=="[]":
return 1
else:
return 0
else:
if b=="()":
return 0
elif b=="[]":
return -1
else:
return 1
s1=input()
s2=input()
i=0
ans=0
while i<len(s1):
ans+=winner(s1[i:i+2],s2[i:i+2])
i+=2
if ans==0:
print("TIE")
elif ans<0:
print("TEAM 2 WINS")
else:
print("TEAM 1 WINS")
``` | instruction | 0 | 7,003 | 17 | 14,006 |
Yes | output | 1 | 7,003 | 17 | 14,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE
Submitted Solution:
```
s1 = input()
s2 = input()
def identify(figure):
if figure == '8<':
return 'scissors'
elif figure == '[]':
return 'paper'
else:
return 'stone'
first, second = 0, 0
i = 0
while i < len(s1)-1:
p1 = identify(s1[i:i+2])
p2 = identify(s2[i:i+2])
# print(p1, p2)
if p1 == 'scissors':
if p2 == 'paper':
first += 1
elif p2 == 'stone':
second += 1
elif p1 == 'paper':
if p2 == 'stone':
first += 1
elif p2 == 'scissors':
second += 1
elif p1 == 'stone':
if p2 == 'scissors':
first += 1
elif p2 == 'paper':
second += 1
i += 2
if first == second:
print('TIE')
elif first < second:
print('TEAM 2 WINS')
else:
print('TEAM 1 WINS')
``` | instruction | 0 | 7,004 | 17 | 14,008 |
Yes | output | 1 | 7,004 | 17 | 14,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE
Submitted Solution:
```
s = input()
s1 = input()
print("TIE")
``` | instruction | 0 | 7,005 | 17 | 14,010 |
No | output | 1 | 7,005 | 17 | 14,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE
Submitted Solution:
```
a=input().split('<')
b=input().split('<')
ans=chk=0
for i in range(1,len(a)):
if a[i-1][-1].isdigit() and len(a[i])>1 and (a[i][0]+a[i][1]=='()' or a[i][0]+a[i][1]=='[]'):
ans+=1
for i in range(1,len(b)):
if b[i-1][-1].isdigit() and len(b[i])>1 and (b[i][0]+b[i][1]=='()' or b[i][0]+b[i][1]=='[]'):
chk+=1
print('TEAM 2 WINS') if chk>ans else ('TEAM 1 WINS') if ans<chk else ('TIE')
``` | instruction | 0 | 7,006 | 17 | 14,012 |
No | output | 1 | 7,006 | 17 | 14,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE
Submitted Solution:
```
cnt=0
t1=input()
t2=input()
for i in range(0,len(t1),2):
p1=t1[i:i+2]
p2=t2[i:i+2]
if(p1=="8<"):
if(p2=="[]"):
cnt+=1
elif(p2=="()"):
cnt-=1
elif(p1=="[]"):
if(p2=="()"):
cnt+=1
elif(p1=="8<"):
cnt-=1
else:
if(p2=="8<"):
cnt+=1
elif(p2=="[]"):
cnt-=1
if(cnt==0):
print("TIE")
elif(cnt>0):
print("TEAM 1 WINS")
else:
print("TEAM 2 WINS")
``` | instruction | 0 | 7,007 | 17 | 14,014 |
No | output | 1 | 7,007 | 17 | 14,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two teams meet in The Game World Championship. Some scientists consider this game to be the most intellectually challenging game in the world. You are given two strings describing the teams' actions in the final battle. Figure out who became the champion.
Input
The input contains two strings of equal length (between 2 and 20 characters, inclusive). Each line describes the actions of one team.
Output
Output "TEAM 1 WINS" if the first team won, "TEAM 2 WINS" if the second team won, and "TIE" if there was a tie.
Examples
Input
[]()[]8<
8<[]()8<
Output
TEAM 2 WINS
Input
8<8<()
[]8<[]
Output
TIE
Submitted Solution:
```
def x():
s=input()
return s.count("8<[]()")
t = x()-x()
print("TEAM %d WINS"%(1 if t>0 else 2) if t!=0 else "TIE")
``` | instruction | 0 | 7,008 | 17 | 14,016 |
No | output | 1 | 7,008 | 17 | 14,017 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills. | instruction | 0 | 7,041 | 17 | 14,082 |
Tags: binary search, greedy, implementation
Correct Solution:
```
def main():
from bisect import bisect_left
n, l, x, y = map(int, input().split())
aa, d = list(map(int, input().split())), {}
for z in (x, y, y + x):
for a in aa:
a += z
if a > l:
break
b = aa[bisect_left(aa, a)]
if b <= a:
if b == a:
d[z] = a
break
if len(d) == 2:
break
if d:
if x in d and y in d:
res = []
elif x in d:
res = [y]
elif y in d:
res = [x]
elif y + x in d:
res = [d[y + x] - y]
else:
z, tmp = y - x, []
for a in aa:
a += z
if a > l:
break
b = aa[bisect_left(aa, a)]
if b == a:
tmp.append(a)
for a in tmp:
if a > y:
res = [a - y]
break
elif a + x < l:
res = [a + x]
break
else:
res = [x, y]
print(len(res))
if res:
print(*res)
if __name__ == '__main__':
main()
``` | output | 1 | 7,041 | 17 | 14,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills. | instruction | 0 | 7,042 | 17 | 14,084 |
Tags: binary search, greedy, implementation
Correct Solution:
```
import itertools
import math
def can_measure(a, d):
return any(i + d in a for i in a)
def main():
n, l, x, y = map(int, input().split())
a = set(map(int, input().split()))
can_x = can_measure(a, x)
can_y = can_measure(a, y)
if can_x and can_y:
print(0)
elif can_x:
print(1)
print(y)
elif can_y:
print(1)
print(x)
else:
for i in a:
if i + x + y in a:
print(1)
print(i + x)
break
else:
t = i + x - y in a
if 0 <= i + x <= l and t:
print(1)
print(i + x)
break;
if 0 <= i - y <= l and t:
print(1)
print(i - y)
break;
else:
print(2)
print(x, y)
if __name__ == "__main__":
main()
``` | output | 1 | 7,042 | 17 | 14,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills. | instruction | 0 | 7,043 | 17 | 14,086 |
Tags: binary search, greedy, implementation
Correct Solution:
```
def main():
import sys
tokens = [int(i) for i in sys.stdin.read().split()]
tokens.reverse()
n, l, x, y = [tokens.pop() for i in range(4)]
marks = set(tokens)
flag_x = flag_y = False
index = -1
for i in marks:
if i + x in marks:
flag_x = True
index = y
if i + y in marks:
flag_y = True
index = x
if i + x + y in marks:
index = i + x
if i + y - x in marks and i - x >= 0:
index = i - x
if i + y - x in marks and i + y <= l:
index = i + y
if flag_x and flag_y:
print(0)
elif index != -1:
print(1)
print(index)
else:
print(2)
print(x, y)
main()
``` | output | 1 | 7,043 | 17 | 14,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills. | instruction | 0 | 7,044 | 17 | 14,088 |
Tags: binary search, greedy, implementation
Correct Solution:
```
class CodeforcesTask480BSolution:
def __init__(self):
self.result = ''
self.n_l_x_y = []
self.ruler = []
def read_input(self):
self.n_l_x_y = [int(x) for x in input().split(" ")]
self.ruler = [int(x) for x in input().split(" ")]
def process_task(self):
dists = {}
for a in self.ruler:
dists[a] = True
hasx = False
hasy = False
for a in self.ruler:
try:
if dists[a - self.n_l_x_y[2]]:
hasx = True
except KeyError:
pass
try:
if dists[a + self.n_l_x_y[2]]:
hasx = True
except KeyError:
pass
try:
if dists[a - self.n_l_x_y[3]]:
hasy = True
except KeyError:
pass
try:
if dists[a - self.n_l_x_y[3]]:
hasy = True
except KeyError:
pass
if hasx and hasy:
break
if hasx and hasy:
self.result = "0"
elif hasx:
self.result = "1\n{0}".format(self.n_l_x_y[3])
elif hasy:
self.result = "1\n{0}".format(self.n_l_x_y[2])
else:
res = [0, 0]
sgn = False
dst = self.n_l_x_y[2] + self.n_l_x_y[3]
diff = self.n_l_x_y[3] - self.n_l_x_y[2]
for a in self.ruler:
try:
if dists[a - dst]:
if a - self.n_l_x_y[2] > 0:
sgn = True
res = a - self.n_l_x_y[2]
except KeyError:
pass
try:
if dists[a + dst]:
if a + self.n_l_x_y[2] < self.n_l_x_y[1]:
sgn = True
res = a + self.n_l_x_y[2]
except KeyError:
pass
try:
if dists[a - diff]:
if a + self.n_l_x_y[2] < self.n_l_x_y[1]:
sgn = True
res = a + self.n_l_x_y[2]
except KeyError:
pass
try:
if dists[a + diff]:
if a - self.n_l_x_y[2] > 0:
sgn = True
res = a - self.n_l_x_y[2]
except KeyError:
pass
if sgn:
break
if sgn:
self.result = "1\n{0}".format(res)
else:
self.result = "2\n{0} {1}".format(self.n_l_x_y[2], self.n_l_x_y[3])
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask480BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | output | 1 | 7,044 | 17 | 14,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills. | instruction | 0 | 7,045 | 17 | 14,090 |
Tags: binary search, greedy, implementation
Correct Solution:
```
n, l, x, y = map(int, input().split())
s = set(map(int, input().split()))
def f(d): return any(i + d in s for i in s)
def g():
for i in s:
if i + x + y in s: return i + x
return 0
def h():
for i in s:
if i + y - x in s:
if i - x >= 0: return i - x
if i + y <= l: return i + y
return 0
def e(d):
print(1)
print(d)
if f(x):
if f(y):
print(0)
else:
e(y)
elif f(y):
e(x)
else:
z = g()
if z:
e(z)
else:
z = h()
if z:
e(z)
else:
print(2)
print(x, y)
``` | output | 1 | 7,045 | 17 | 14,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills. | instruction | 0 | 7,046 | 17 | 14,092 |
Tags: binary search, greedy, implementation
Correct Solution:
```
from collections import defaultdict
class LongJumps():
def __init__(self, n, l, x, y, a):
self.n, self.l, self.x, self.y, self.a = n,l,x,y,a
def get_markers(self):
st = defaultdict(set)
req_pts = [self.x,self.y]
exist_check = defaultdict(bool)
value_check = defaultdict(bool)
for v in self.a:
exist_check[v] = True
for v in self.a:
for i in range(len(req_pts)):
if v - req_pts[i] >= 0:
st[v - req_pts[i]].add(i)
if exist_check[v - req_pts[i]]:
value_check[i] = True
if v + req_pts[i] <= l:
st[v+req_pts[i]].add(i)
if exist_check[v + req_pts[i]]:
value_check[i] = True
if value_check[0] and value_check[1]:
print(0)
return
sol_status = 2
status1_marker = None
for v in st:
if len(st[v]) == 2:
sol_status = 1
status1_marker = v
elif len(st[v]) == 1:
if exist_check[v]:
sol_status = 1
status1_marker = req_pts[1-st[v].pop()]
if sol_status == 1:
print(1)
print(status1_marker)
return
else:
print(2)
print(x, y)
n, l, x, y = list(map(int,input().strip(' ').split(' ')))
a = list(map(int,input().strip(' ').split(' ')))
lj = LongJumps(n,l,x,y,a)
lj.get_markers()
``` | output | 1 | 7,046 | 17 | 14,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills. | instruction | 0 | 7,047 | 17 | 14,094 |
Tags: binary search, greedy, implementation
Correct Solution:
```
n, l, x, y = map(int, input().split())
def f(t, q):
i = j = 0
while j < n:
d = t[j] - t[i]
if d < q: j += 1
elif d > q: i += 1
else: return 0
return q
def g(t):
q = x + y
i = j = 0
while j < n:
d = t[j] - t[i]
if d < q: j += 1
elif d > q: i += 1
else: return t[j]
return 0
def h(t):
q = y - x
i = j = 0
while j < n:
d = t[j] - t[i]
if d < q: j += 1
elif d > q: i += 1
else:
a, b = t[i] - x, t[j] + x
if a >= 0: return [a]
if b <= l:return [b]
j += 1
return [x, y]
def e(t):
print(len(t))
print(' '.join(map(str, t)))
t = list(map(int, input().split()))
t.sort()
x = f(t, x)
y = f(t, y)
if x and y:
z = g(t)
if z: e([z - y])
else: e(h(t))
elif x: e([x])
elif y: e([y])
else: e([])
``` | output | 1 | 7,047 | 17 | 14,095 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills. | instruction | 0 | 7,048 | 17 | 14,096 |
Tags: binary search, greedy, implementation
Correct Solution:
```
import itertools
import math
def can_measure(a, d):
return any(i + d in a for i in a)
def main():
n, l, x, y = map(int, input().split())
a = set(map(int, input().split()))
can_x = can_measure(a, x)
can_y = can_measure(a, y)
if can_x and can_y:
print(0)
elif can_x:
print(1)
print(y)
elif can_y:
print(1)
print(x)
else:
for i in a:
if i + x + y in a:
print(1)
print(i + x)
break
else:
t = i + x - y in a
if 0 <= i + x <= l and t:
print(1)
print(i + x)
break;
if 0 <= i - y <= l and t:
print(1)
print(i - y)
break;
else:
print(2)
print(x, y)
if __name__ == "__main__":
main()
# Made By Mostafa_Khaled
``` | output | 1 | 7,048 | 17 | 14,097 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills.
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
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+
# main code
n,l1,x,y=in_arr()
l=in_arr()
d=Counter(l)
f1=0
f2=0
for i in range(n):
if d[l[i]+x]:
f1=1
if d[l[i]+y]:
f2=1
if f1 and f2:
print 0
elif f1 or f2:
print 1
if f2:
print x
else:
print y
else:
f=0
for i in range(n):
if l[i]+x<=l1 and ( d[l[i]+x+y] or d[l[i]+x-y]):
print 1
#print 'a'
print l[i]+x
exit()
break
if l[i]-x>=0 and (d[l[i]+y-x] or d[l[i]-y-x]):
print 1
print l[i]-x
exit()
print 2
print x,y
``` | instruction | 0 | 7,049 | 17 | 14,098 |
Yes | output | 1 | 7,049 | 17 | 14,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import random
import itertools
import sys
from typing import List
"""
created by shhuan at 2020/1/13 20:48
"""
def solve(N, L, X, Y, A):
vs = set(A)
mx = any([a+X in vs for a in A])
my = any([a+Y in vs for a in A])
if mx and my:
print(0)
elif mx:
print(1)
print(Y)
elif my:
print(1)
print(X)
else:
# try to add 1 mark
for a in vs:
for b, c in [(a + X, Y), (a + Y, X), (a - X, Y), (a - Y, X)]:
if 0 <= b <= L:
if (b + c <= L and b + c in vs) or (b - c >= 0 and b - c in vs):
print(1)
print(b)
return
# add 2 marks
print(2)
print('{} {}'.format(X, Y))
N, L, X, Y = map(int, input().split())
A = [int(x) for x in input().split()]
solve(N, L, X, Y, A)
``` | instruction | 0 | 7,050 | 17 | 14,100 |
Yes | output | 1 | 7,050 | 17 | 14,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills.
Submitted Solution:
```
def main():
n,l,x,y = map(int,input().split())
arr = set(map(int,input().split()))
first = False
second = False
for i in arr:
if i+x in arr:
first = True
if i+y in arr:
second = True
if first and not second:
print(1)
print(y)
return
if second and not first:
print(1)
print(x)
return
if first and second:
print(0)
return
found = False
for i in arr:
if i+x-y in arr and i+x <= l:
found = True
coord = i+x
break
if i+y-x in arr and i+y <= l:
found = True
coord = i+y
break
if i+x+y in arr and i+min(x,y) <= l:
found = True
coord = i+min(x,y)
if i-x-y in arr and i-max(x,y) >= 0:
found = True
coord = i-max(x,y)
if i-x+y in arr and i-x >= 0:
found = True
coord = i-x
break
if i-y+x in arr and i-y >= 0:
found = True
coord = i-y
break
if found:
break
if found:
print(1)
print(coord)
return
print(2)
print(x,y)
main()
``` | instruction | 0 | 7,051 | 17 | 14,102 |
Yes | output | 1 | 7,051 | 17 | 14,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills.
Submitted Solution:
```
def main():
import sys
tokens = [int(i) for i in sys.stdin.read().split()]
tokens.reverse()
n, l, x, y = [tokens.pop() for i in range(4)]
marks = set(tokens)
x_index = y_index = sum_index = sub_index1 = sub_index2 = -1
for i in marks:
if i + x in marks:
x_index = y
if i + y in marks:
y_index = x
if i + x + y in marks:
sum_index = i + x
if i + y - x in marks and i - x >= 0:
sub_index1 = i - x
if i + y - x in marks and i + y <= l:
sub_index2 = i + y
if x_index != -1 and y_index != -1:
print(0)
else:
for i in (x_index, y_index, sum_index, sub_index1, sub_index2):
if i != -1:
print(1)
print(i)
break
else:
print(2)
print(x, y)
main()
``` | instruction | 0 | 7,052 | 17 | 14,104 |
Yes | output | 1 | 7,052 | 17 | 14,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills.
Submitted Solution:
```
n, l, x, y = map(int, input().split())
a = set(map(int, input().split()))
ok1 = ok2 = ok3 = False
for c in a:
if c + x in a:
ok1 = True
if c + y in a:
ok2 = True
if c - x > 0 and c - x + y in a:
ok3 = True
mark = c - x
if c + x < l and c + x - y in a:
ok3 = True
mark = c + x
if c + x + y in a:
ok3 = True
mark = c + x
if c - x - y in a:
ok3 = True
mark = c - x
if ok1 and ok2:
print(0)
elif (not ok1) and (not ok2):
if ok3:
print(1)
print(mark)
else:
print(2)
print(x, y)
else:
print(1)
print(y if ok1 else x)
``` | instruction | 0 | 7,053 | 17 | 14,106 |
Yes | output | 1 | 7,053 | 17 | 14,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills.
Submitted Solution:
```
if __name__ == "__main__":
n, l, x, y = list(map(int, input().split()))
v = list(map(int, input().split()))
s = set(v)
cx = 0
for i in range(n):
if v[i]+x in s:
cx = 1
break
cy = 0
for i in range(n):
if v[i]+y in s:
cy = 1
break
count = 0
ans = []
if cx==0:
count += 1
ans.append(x)
if cy==0:
count += 1
ans.append(y)
if count==2:
for i in range(n):
if (v[i]+x+y in s):
count = 1
ans = [v[i]+x]
break
if count==2:
for i in range(n):
if (v[i]+x-y in s):
if v[i]+x<=l:
count = 1
ans = [v[i]+x]
break
elif v[i]-y<=l:
count = 1
ans = [v[i]-y]
break
print(count)
if count!=0:
print(*ans)
``` | instruction | 0 | 7,054 | 17 | 14,108 |
No | output | 1 | 7,054 | 17 | 14,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills.
Submitted Solution:
```
class CodeforcesTask480BSolution:
def __init__(self):
self.result = ''
self.n_l_x_y = []
self.ruler = []
def read_input(self):
self.n_l_x_y = [int(x) for x in input().split(" ")]
self.ruler = [int(x) for x in input().split(" ")]
def process_task(self):
dists = {}
for a in self.ruler:
dists[a] = True
hasx = False
hasy = False
for a in self.ruler:
try:
if dists[a - self.n_l_x_y[2]]:
hasx = True
except KeyError:
pass
try:
if dists[a + self.n_l_x_y[2]]:
hasx = True
except KeyError:
pass
try:
if dists[a - self.n_l_x_y[3]]:
hasy = True
except KeyError:
pass
try:
if dists[a - self.n_l_x_y[3]]:
hasy = True
except KeyError:
pass
if hasx and hasy:
break
if hasx and hasy:
self.result = "0"
elif hasx:
self.result = "1\n{0}".format(self.n_l_x_y[3])
elif hasy:
self.result = "1\n{1}".format(self.n_l_x_y[2])
else:
res = [0, 0]
sgn = False
dst = self.n_l_x_y[2] + self.n_l_x_y[3]
diff = self.n_l_x_y[3] - self.n_l_x_y[2]
for a in self.ruler:
try:
if dists[a - dst]:
sgn = True
res = a - self.n_l_x_y[2]
except KeyError:
pass
try:
if dists[a + dst]:
sgn = True
res = a + self.n_l_x_y[2]
except KeyError:
pass
try:
if dists[a - diff]:
sgn = True
res = a + self.n_l_x_y[2]
except KeyError:
pass
try:
if dists[a + diff]:
sgn = True
res = a - self.n_l_x_y[2]
except KeyError:
pass
if sgn:
break
if sgn:
self.result = "1\n{0}".format(res)
else:
self.result = "2\n{0} {1}".format(self.n_l_x_y[2], self.n_l_x_y[3])
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask480BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | instruction | 0 | 7,055 | 17 | 14,110 |
No | output | 1 | 7,055 | 17 | 14,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills.
Submitted Solution:
```
def search(a,k,n):
i=0
l=n-1
while(i<=l):
mid=(i+l)//2
if a[mid]==k:
return 1
elif a[mid]<k:
i=mid+1
else:
l=mid-1
return 0
def count(b,x,n):
sum=0
i=0
l=0
while(i<n and l<n):
#print(sum)
if sum<x:
#print(l)
sum=sum+b[l]
l=l+1
if sum>x:
sum=sum-b[i]
i=i+1
if sum==x:
return 1
return 0
def two(a,n):
for i in range(n):
t=a[i]+x
p=a[i]-x
q=a[i]+y
w=a[i]-y
if search(a,t+y,n)!=0 or search(a,abs(t-y),n)!=0:
return t
if search(a,p+y,n)!=0 or search(a,abs(p-y),n)!=0:
return p
if search(a,q+x,n)!=0 or search(a,abs(q-x),n)!=0:
return q
if search(a,w+x,n)!=0 or search(a,abs(w-x),n)!=0:
return w
return 0
n,l,x,y=[int(i) for i in input().split()]
a=[int(i) for i in input().split()]
b=[]
for i in range(n-1):
t=a[i+1]-a[i]
b.append(t)
#print(b)
nl=len(b)
k1=count(b,x,nl)
k2=count(b,y,nl)
#print(k1,k2)
if k1==1 and k2==1:
print(0)
elif k1==0 and k2==0:
z=two(a,n)
if z>l:
z=z-(x+y)
if z<0:
z=0
if z==0:
print(2)
print(x,y)
else:
print(1)
print(z)
elif k1==0:
print(1)
print(x)
else:
print(1)
print(y)
``` | instruction | 0 | 7,056 | 17 | 14,112 |
No | output | 1 | 7,056 | 17 | 14,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler.
Input
The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin.
Output
In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them.
Examples
Input
3 250 185 230
0 185 250
Output
1
230
Input
4 250 185 230
0 20 185 250
Output
0
Input
2 300 185 230
0 300
Output
2
185 230
Note
In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills.
Submitted Solution:
```
if __name__ == "__main__":
n, l, x, y = list(map(int, input().split()))
v = list(map(int, input().split()))
s = set(v)
cx = 0
for i in range(n):
if v[i]+x in s:
cx = 1
break
cy = 0
for i in range(n):
if v[i]+y in s:
cy = 1
break
count = 0
ans = []
if cx==0:
count += 1
ans.append(x)
if cy==0:
count += 1
ans.append(y)
if count==2:
for i in range(n):
if (v[i]+x+y in s) or (v[i]+x-y in s):
ans = [v[i]+x]
break
print(count)
if count!=0:
print(*ans)
``` | instruction | 0 | 7,057 | 17 | 14,114 |
No | output | 1 | 7,057 | 17 | 14,115 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This time the Berland Team Olympiad in Informatics is held in a remote city that can only be reached by one small bus. Bus has n passenger seats, seat i can be occupied only by a participant from the city ai.
Today the bus has completed m trips, each time bringing n participants. The participants were then aligned in one line in the order they arrived, with people from the same bus standing in the order of their seats (i. e. if we write down the cities where the participants came from, we get the sequence a1, a2, ..., an repeated m times).
After that some teams were formed, each consisting of k participants form the same city standing next to each other in the line. Once formed, teams left the line. The teams were formed until there were no k neighboring participants from the same city.
Help the organizers determine how many participants have left in the line after that process ended. We can prove that answer doesn't depend on the order in which teams were selected.
Input
The first line contains three integers n, k and m (1 ≤ n ≤ 105, 2 ≤ k ≤ 109, 1 ≤ m ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105), where ai is the number of city, person from which must take seat i in the bus.
Output
Output the number of remaining participants in the line.
Examples
Input
4 2 5
1 2 3 1
Output
12
Input
1 9 10
1
Output
1
Input
3 2 10
1 2 1
Output
0
Note
In the second example, the line consists of ten participants from the same city. Nine of them will form a team. At the end, only one participant will stay in the line. | instruction | 0 | 7,224 | 17 | 14,448 |
Tags: data structures, implementation
Correct Solution:
```
import sys
from collections import deque
input=sys.stdin.readline
n,k,m=map(int,input().split())
a=list(map(int,input().split()))
r=a[0]
flag=0
for i in range(n):
if r!=a[i]:
flag=1
break
if flag==0:
print((m*n)%k)
sys.exit()
if k>n:
print(m*n)
sys.exit()
curr=a[0]
tmp=1
que=deque([(a[0],1)])
for i in range(1,n):
if a[i]==curr:
tmp+=1
que.append((a[i],tmp))
if tmp==k:
for j in range(k):
que.pop()
if que:
tmp=que[-1][1]
curr=que[-1][0]
else:
curr=-1
else:
tmp=1
curr=a[i]
que.append((a[i],tmp))
quecop=[]
for i in que:
quecop.append(i[0])
leftrem=0
rightrem=0
if not que:
print(0)
sys.exit()
while que[0][0]==que[-1][0]:
r=que[0][0]
count1=0
p=len(que)
count2=p-1
while count1<p and que[count1][0]==r:
count1+=1
if count1==p:
break
while count2>=0 and que[count2][0]==r:
count2-=1
if count1+p-1-count2<k:
break
leftrem+=count1
rightrem+=k-count1
for i in range(count1):
que.popleft()
for i in range(k-count1):
que.pop()
if que:
t=que[0][0]
flag=0
for i in que:
if i[0]!=t:
flag=1
break
if flag:
print(leftrem+rightrem+len(que)*m)
else:
r=[]
for i in range(leftrem):
if r and r[-1][0]==quecop[i]:
r[-1][1]+=1
else:
r.append([quecop[i],1])
if r and r[-1][0]==que[0][0]:
r[-1][0]=(r[-1][0]+(len(que)*m))%k
if r[-1][1]==0:
r.pop()
else:
if (len(que)*m)%k:
r.append([que[0][0],(len(que)*m)%k])
for i in range(len(quecop)-rightrem,len(quecop)):
if r and r[-1][0]==quecop[i]:
r[-1][1]+=1
if r[-1][1]==k:
r.pop()
else:
r.append([quecop[i],1])
finans=0
for i in r:
finans+=i[1]
print(finans)
``` | output | 1 | 7,224 | 17 | 14,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This time the Berland Team Olympiad in Informatics is held in a remote city that can only be reached by one small bus. Bus has n passenger seats, seat i can be occupied only by a participant from the city ai.
Today the bus has completed m trips, each time bringing n participants. The participants were then aligned in one line in the order they arrived, with people from the same bus standing in the order of their seats (i. e. if we write down the cities where the participants came from, we get the sequence a1, a2, ..., an repeated m times).
After that some teams were formed, each consisting of k participants form the same city standing next to each other in the line. Once formed, teams left the line. The teams were formed until there were no k neighboring participants from the same city.
Help the organizers determine how many participants have left in the line after that process ended. We can prove that answer doesn't depend on the order in which teams were selected.
Input
The first line contains three integers n, k and m (1 ≤ n ≤ 105, 2 ≤ k ≤ 109, 1 ≤ m ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105), where ai is the number of city, person from which must take seat i in the bus.
Output
Output the number of remaining participants in the line.
Examples
Input
4 2 5
1 2 3 1
Output
12
Input
1 9 10
1
Output
1
Input
3 2 10
1 2 1
Output
0
Note
In the second example, the line consists of ten participants from the same city. Nine of them will form a team. At the end, only one participant will stay in the line. | instruction | 0 | 7,225 | 17 | 14,450 |
Tags: data structures, implementation
Correct Solution:
```
def process(A, m, k):
n = len(A)
start = n*m
d = []
for i in range(n):
if len(d)==0 or d[-1][0] != A[i]:
d.append([A[i], 1])
else:
d[-1][1]+=1
if d[-1][1]==k:
start-=k*m
d.pop()
if m==1:
return start
p1 = 0
start_d = []
middle_d = []
end_d = []
for x in d:
a, b = x
start_d.append([a, b])
middle_d.append([a, b])
end_d.append([a, b])
if m==2:
middle_d = []
else:
middle_d = [x for x in d]
# print(start_d, middle_d, end_d, start)
#the sequence is start_d + (m-2)* middle_d + end_d
while True:
if p1 >= len(middle_d):
if len(start_d)==0 or p1==len(end_d):
break
v1, c1 = start_d[-1]
v2, c2 = end_d[p1]
if v1==v2 and (c1+c2) >= k:
start-=k
start_d.pop()
if (c1+c2)==k:
p1+=1
else:
end_d[1] = (c1+c2) % k
else:
break
else:
if len(middle_d)-p1==1:
v1, c1 = start_d[-1]
v2, c2 = middle_d[p1]
v3, c3 = end_d[p1]
total_value = c2*(m-2)
if v2==v1:
total_value+=c1
if v2==v3:
total_value+=c3
start-=(total_value-(total_value%k))
if total_value % k==0:
middle_d = []
if v2==v1:
start_d.pop()
if v2==v3:
end_d = end_d
p1+=1
else:
break
else:
v1, c1 = start_d[-1]
v2, c2 = middle_d[p1]
v3, c3 = middle_d[-1]
v4, c4 = end_d[p1]
if v2==v3 and (c2+c3) >= k:
this_value = (c2+c3-(c2+c3) % k)*(m-1)
start-=this_value
middle_d.pop()
start_d.pop()
middle_d[p1][1] = (c2+c3) % k
end_d[p1][1] = (c2+c3) % k
if middle_d[p1][1]==0:
p1+=1
else:
break
# print('start', start_d, start)
# print('middle', middle_d)
# print('end', end_d)
return start
n, k, m = [int(x) for x in input().split()]
A = [int(x) for x in input().split()]
print(process(A, m, k))
``` | output | 1 | 7,225 | 17 | 14,451 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This time the Berland Team Olympiad in Informatics is held in a remote city that can only be reached by one small bus. Bus has n passenger seats, seat i can be occupied only by a participant from the city ai.
Today the bus has completed m trips, each time bringing n participants. The participants were then aligned in one line in the order they arrived, with people from the same bus standing in the order of their seats (i. e. if we write down the cities where the participants came from, we get the sequence a1, a2, ..., an repeated m times).
After that some teams were formed, each consisting of k participants form the same city standing next to each other in the line. Once formed, teams left the line. The teams were formed until there were no k neighboring participants from the same city.
Help the organizers determine how many participants have left in the line after that process ended. We can prove that answer doesn't depend on the order in which teams were selected.
Input
The first line contains three integers n, k and m (1 ≤ n ≤ 105, 2 ≤ k ≤ 109, 1 ≤ m ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105), where ai is the number of city, person from which must take seat i in the bus.
Output
Output the number of remaining participants in the line.
Examples
Input
4 2 5
1 2 3 1
Output
12
Input
1 9 10
1
Output
1
Input
3 2 10
1 2 1
Output
0
Note
In the second example, the line consists of ten participants from the same city. Nine of them will form a team. At the end, only one participant will stay in the line. | instruction | 0 | 7,226 | 17 | 14,452 |
Tags: data structures, implementation
Correct Solution:
```
def main():
_, k, m = [int(x) for x in input().split()]
a = []
last = ("-1", 0)
a.append(last)
for ai in input().split():
if last[0] == ai:
last = (ai, last[1]+1)
a[-1] = last
else:
last = (ai, 1)
a.append(last)
if last[1] == k:
a.pop()
last = a[-1]
a.pop(0)
s1 = 0
while len(a) > 0 and a[0][0] == a[-1][0]:
if len(a) == 1:
s = a[0][1] * m
r1 = s % k
if r1 == 0:
print(s1 % k)
else:
print(r1 + s1)
return
join = a[0][1] + a[-1][1]
if join < k:
break
elif join % k == 0:
s1 += join
a.pop()
a.pop(0)
else:
s1 += (join // k) * k
a[0] = (a[0][0], join % k)
a.pop()
break
s = 0
for ai in a:
s += ai[1]
print(s*m + s1)
if __name__ == "__main__":
main()
``` | output | 1 | 7,226 | 17 | 14,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This time the Berland Team Olympiad in Informatics is held in a remote city that can only be reached by one small bus. Bus has n passenger seats, seat i can be occupied only by a participant from the city ai.
Today the bus has completed m trips, each time bringing n participants. The participants were then aligned in one line in the order they arrived, with people from the same bus standing in the order of their seats (i. e. if we write down the cities where the participants came from, we get the sequence a1, a2, ..., an repeated m times).
After that some teams were formed, each consisting of k participants form the same city standing next to each other in the line. Once formed, teams left the line. The teams were formed until there were no k neighboring participants from the same city.
Help the organizers determine how many participants have left in the line after that process ended. We can prove that answer doesn't depend on the order in which teams were selected.
Input
The first line contains three integers n, k and m (1 ≤ n ≤ 105, 2 ≤ k ≤ 109, 1 ≤ m ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105), where ai is the number of city, person from which must take seat i in the bus.
Output
Output the number of remaining participants in the line.
Examples
Input
4 2 5
1 2 3 1
Output
12
Input
1 9 10
1
Output
1
Input
3 2 10
1 2 1
Output
0
Note
In the second example, the line consists of ten participants from the same city. Nine of them will form a team. At the end, only one participant will stay in the line. | instruction | 0 | 7,227 | 17 | 14,454 |
Tags: data structures, implementation
Correct Solution:
```
#reference sol:-31772413
r=lambda:map(int,input().split())
n,k,m=r()
a=list(r())
stck=[]
for i in range(n):
if len(stck)==0 or stck[-1][0]!=a[i]:
stck.append([a[i],1])
else:
stck[-1][1]+=1
if stck[-1][1]==k:
stck.pop()
rem=0
strt,end=0,len(stck)-1
if m > 1:
while end-strt+1 > 1 and stck[strt][0]==stck[end][0]:
join=stck[strt][1]+stck[end][1]
if join < k:
break
elif join % k==0:
rem+=join
strt+=1
end-=1
else:
stck[strt][1]=join % k
stck[end][1]=0
rem+=(join//k)*k
tr=0
slen=end-strt+1
for el in stck[:slen]:
tr+=el[1]
if slen==0:
print(0)
elif slen==1:
r=(stck[strt][1]*m)%k
if r==0:
print(0)
else:
print(r+rem)
else:
print(tr*m+rem)
``` | output | 1 | 7,227 | 17 | 14,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This time the Berland Team Olympiad in Informatics is held in a remote city that can only be reached by one small bus. Bus has n passenger seats, seat i can be occupied only by a participant from the city ai.
Today the bus has completed m trips, each time bringing n participants. The participants were then aligned in one line in the order they arrived, with people from the same bus standing in the order of their seats (i. e. if we write down the cities where the participants came from, we get the sequence a1, a2, ..., an repeated m times).
After that some teams were formed, each consisting of k participants form the same city standing next to each other in the line. Once formed, teams left the line. The teams were formed until there were no k neighboring participants from the same city.
Help the organizers determine how many participants have left in the line after that process ended. We can prove that answer doesn't depend on the order in which teams were selected.
Input
The first line contains three integers n, k and m (1 ≤ n ≤ 105, 2 ≤ k ≤ 109, 1 ≤ m ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105), where ai is the number of city, person from which must take seat i in the bus.
Output
Output the number of remaining participants in the line.
Examples
Input
4 2 5
1 2 3 1
Output
12
Input
1 9 10
1
Output
1
Input
3 2 10
1 2 1
Output
0
Note
In the second example, the line consists of ten participants from the same city. Nine of them will form a team. At the end, only one participant will stay in the line.
Submitted Solution:
```
import queue
def readTuple():
return input().split()
def readInts():
return tuple(map(int, readTuple()))
def solve():
n, k, m = readInts()
nums = []
cnt = 0
for i in readInts():
if nums and i == nums[-1][0]:
nums[-1][1] +=1
if nums[-1][1] == k:
nums.pop()
continue
nums.append([i,1])
i, j = 0, len(nums)-1
comm = 0
while i<j:
if nums[i][0] == nums[j][0] \
and nums[i][1]+nums[j][1] <= k:
if nums[i][1]+nums[j][1] == k:
comm +=1
else:
break
i+=1
j-=1
mid = len(nums)-2*comm
prefix_sum = sum(map(lambda x: x[1], nums[:comm]))
middle_sum = sum(map(lambda x: x[1], nums[comm:comm+mid]))
suffix_sum = sum(map(lambda x: x[1], nums[comm+mid:]))
all_sum = (prefix_sum+suffix_sum+middle_sum)*m
# print(nums)
# print("prefix: ", comm, "SUM: ", prefix_sum)
# print("mid: ", mid, "SUM: ", middle_sum)
# print("suffix: ", comm, "SUM: ", suffix_sum)
# print("ALL: ", all_sum)
# subtract perfect matches
all_sum -= (prefix_sum+suffix_sum)*(m-1)
if mid == 0:
all_sum = 0
elif mid == 1:
if (middle_sum*m)%k == 0:
all_sum = 0
else:
all_sum -= middle_sum*m
all_sum += (middle_sum*m)%k
else:
if nums[comm][0] == nums[-comm-1][0] \
and nums[comm][1]+nums[-comm-1][1] > k:
all_sum -= k*(m-1)
print(all_sum)
if __name__ == '__main__':
solve()
``` | instruction | 0 | 7,228 | 17 | 14,456 |
No | output | 1 | 7,228 | 17 | 14,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This time the Berland Team Olympiad in Informatics is held in a remote city that can only be reached by one small bus. Bus has n passenger seats, seat i can be occupied only by a participant from the city ai.
Today the bus has completed m trips, each time bringing n participants. The participants were then aligned in one line in the order they arrived, with people from the same bus standing in the order of their seats (i. e. if we write down the cities where the participants came from, we get the sequence a1, a2, ..., an repeated m times).
After that some teams were formed, each consisting of k participants form the same city standing next to each other in the line. Once formed, teams left the line. The teams were formed until there were no k neighboring participants from the same city.
Help the organizers determine how many participants have left in the line after that process ended. We can prove that answer doesn't depend on the order in which teams were selected.
Input
The first line contains three integers n, k and m (1 ≤ n ≤ 105, 2 ≤ k ≤ 109, 1 ≤ m ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105), where ai is the number of city, person from which must take seat i in the bus.
Output
Output the number of remaining participants in the line.
Examples
Input
4 2 5
1 2 3 1
Output
12
Input
1 9 10
1
Output
1
Input
3 2 10
1 2 1
Output
0
Note
In the second example, the line consists of ten participants from the same city. Nine of them will form a team. At the end, only one participant will stay in the line.
Submitted Solution:
```
from collections import deque
n,k,m=map(int,input().split())
queue = deque()
A = list(map(int,input().split()))
for j in range(n):
if queue:
per = queue.pop()
if A[j] == per[0]:
per[1] +=1
if per[1] !=k:
queue.append(per)
else:
queue.append(per)
queue.append([A[j],1])
else:
queue.append([A[j],1])
lens = len(queue)
queue = list(queue)
#print(queue)
if lens ==0:
print(0)
elif m == 1:
ans =0
for j in range(lens):
ans+=queue[j][1]
print(ans)
elif lens%2==0:
p = 0
for j in range(lens//2):
if queue[j][0] == queue[-(j+1)][0] and queue[j][1] + queue[-(j+1)][1] == k:
p+=1
#print(p)
ans = 0
l=0
if p != lens//2:
for j in range(lens-1,lens-1-p,-1):
ans-=queue[j][1]
ans-=queue[-(j+1)][1]
ans*=(m-1)
for j in range(lens):
l+=queue[j][1]
ans+=l*m
print(ans)
#print(l*m)
else:
#print('huyznaet')
p = 0
for j in range(lens//2):
if queue[j][0] == queue[-(j+1)][0] and queue[j][1] + queue[-(j+1)][1] == k:
p+=1
#print(p)
ans = 0
l=0
if p != lens//2:
for j in range(lens-1,lens-1-p,-1):
ans-=queue[j][1]
ans-=queue[-(j+1)][1]
ans*=(m-1)
for j in range(lens):
l+=queue[j][1]
ans+=l*m
print(ans)
#print(pizda)
else:
if(m*queue[lens//2][1])%k==0:
print(0)
else:
l=0
for j in range(lens):
l+=queue[j][1]
l-= queue[lens//2][1]
l+=(m*queue[lens//2][1])%k
print(l)
#print(l*m)
``` | instruction | 0 | 7,229 | 17 | 14,458 |
No | output | 1 | 7,229 | 17 | 14,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This time the Berland Team Olympiad in Informatics is held in a remote city that can only be reached by one small bus. Bus has n passenger seats, seat i can be occupied only by a participant from the city ai.
Today the bus has completed m trips, each time bringing n participants. The participants were then aligned in one line in the order they arrived, with people from the same bus standing in the order of their seats (i. e. if we write down the cities where the participants came from, we get the sequence a1, a2, ..., an repeated m times).
After that some teams were formed, each consisting of k participants form the same city standing next to each other in the line. Once formed, teams left the line. The teams were formed until there were no k neighboring participants from the same city.
Help the organizers determine how many participants have left in the line after that process ended. We can prove that answer doesn't depend on the order in which teams were selected.
Input
The first line contains three integers n, k and m (1 ≤ n ≤ 105, 2 ≤ k ≤ 109, 1 ≤ m ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105), where ai is the number of city, person from which must take seat i in the bus.
Output
Output the number of remaining participants in the line.
Examples
Input
4 2 5
1 2 3 1
Output
12
Input
1 9 10
1
Output
1
Input
3 2 10
1 2 1
Output
0
Note
In the second example, the line consists of ten participants from the same city. Nine of them will form a team. At the end, only one participant will stay in the line.
Submitted Solution:
```
n,k,m = input().strip().split()
n = int(n)
k = int(k)
m = int(m)
a = list(map(int, input().strip().split()))
x = []
kol = 1
for i in range(n):
if i ==0:
x.append((a[i],kol))
else:
if a[i] == a[i-1]:
kol +=1
x.append((a[i],kol))
else:
kol = 1
x.append((a[i],kol))
if n>1:
y = []
for i in range(n+1):
y.append(1)
pre = 0
nex = 1
z = {}
tup = {}
tup[0] =1
kol =1
while nex <n:
if nex not in tup:
tup[nex] = nex-1
if a[nex] == a[pre]:
y[nex] = y[pre] + kol
kol = kol + 1
else:
kol = 1
pre = nex
if y[nex] ==k:
kol = 1
z[pre- y[pre] +1] = nex
if pre ==0:
pre = nex
nex = nex + 1
else:
pre = tup[pre - y[pre] + 1]
tup[nex + 1] = nex - y[nex]
nex = nex + 1
i = 0
p =[]
while i <= n-1:
if i in z:
i = z[i] +1
else:
p.append(a[i])
i = i +1
a = p[:]
n = len(a)
i = n-1
s = 2 * n
l=0
if len(a) ==0:
print(0)
else:
if k>=n:
for i in range(n):
if (i>0 and a[i]!=a[i-1]):
l =1
break
if l==1:
print(n*m)
else:
if k <= n*m:
print((n *m) % k)
else:
print(n*m)
else:
l=0
for i in range(n//2):
if a[i] != a[n - i-1]:
l = 1
if l==0:
print(0)
else:
s = 0
i = n-1
while i >= n //2 - 1:
if x[i] == x[n -1 - i + x[i][1] - 1] and x[i][1] * 2 >=k:
if i !=n -1 - i + x[i][1] - 1:
s += ((x[i][1] *2 //k) * k) * (m-1)
else:
s+= x[i][1] * m
i = i - ((x[i][1] *2 //k) * k)//2
else:
i = -1
break
s = n*m-s
print(s)
``` | instruction | 0 | 7,230 | 17 | 14,460 |
No | output | 1 | 7,230 | 17 | 14,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This time the Berland Team Olympiad in Informatics is held in a remote city that can only be reached by one small bus. Bus has n passenger seats, seat i can be occupied only by a participant from the city ai.
Today the bus has completed m trips, each time bringing n participants. The participants were then aligned in one line in the order they arrived, with people from the same bus standing in the order of their seats (i. e. if we write down the cities where the participants came from, we get the sequence a1, a2, ..., an repeated m times).
After that some teams were formed, each consisting of k participants form the same city standing next to each other in the line. Once formed, teams left the line. The teams were formed until there were no k neighboring participants from the same city.
Help the organizers determine how many participants have left in the line after that process ended. We can prove that answer doesn't depend on the order in which teams were selected.
Input
The first line contains three integers n, k and m (1 ≤ n ≤ 105, 2 ≤ k ≤ 109, 1 ≤ m ≤ 109).
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105), where ai is the number of city, person from which must take seat i in the bus.
Output
Output the number of remaining participants in the line.
Examples
Input
4 2 5
1 2 3 1
Output
12
Input
1 9 10
1
Output
1
Input
3 2 10
1 2 1
Output
0
Note
In the second example, the line consists of ten participants from the same city. Nine of them will form a team. At the end, only one participant will stay in the line.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import random
import itertools
import sys
"""
created by shhuan at 2017/10/28 16:57
"""
N, K, M = map(int, input().split())
A = [int(x) for x in input().split()]
def group(a, k):
if not a:
return []
if len(a) < k:
return a
ans = []
p = a[0]
c = 1
for v in a[1:]:
if v == p:
c += 1
if c == k:
c = 0
else:
ans += [p] * c
p = v
c = 1
ans += [p] * c
return ans
A = group(A, K)
rem = []
found = True
while A and found:
found = False
if len(set(A)) == 1:
A = [A[0]] * ((len(A)*M) % K)
M %= K
break
else:
if K > 2*len(A):
break
i = 2
a = group(A*i, K)
if a and len(a) != len(A)*i:
rem = A * (M%i) + rem
A = a
M //= i
found = True
break
A = group(A*M+rem, K)
if A:
print(len(A))
else:
print(0)
``` | instruction | 0 | 7,231 | 17 | 14,462 |
No | output | 1 | 7,231 | 17 | 14,463 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n football teams in the world.
The Main Football Organization (MFO) wants to host at most m games. MFO wants the i-th game to be played between the teams a_i and b_i in one of the k stadiums.
Let s_{ij} be the numbers of games the i-th team played in the j-th stadium. MFO does not want a team to have much more games in one stadium than in the others. Therefore, for each team i, the absolute difference between the maximum and minimum among s_{i1}, s_{i2}, …, s_{ik} should not exceed 2.
Each team has w_i — the amount of money MFO will earn for each game of the i-th team. If the i-th team plays l games, MFO will earn w_i ⋅ l.
MFO needs to find what games in what stadiums they need to host in order to earn as much money as possible, not violating the rule they set.
However, this problem is too complicated for MFO. Therefore, they are asking you to help them.
Input
The first line contains three integers n, m, k (3 ≤ n ≤ 100, 0 ≤ m ≤ 1 000, 1 ≤ k ≤ 1 000) — the number of teams, the number of games, and the number of stadiums.
The second line contains n integers w_1, w_2, …, w_n (1 ≤ w_i ≤ 1 000) — the amount of money MFO will earn for each game of the i-th game.
Each of the following m lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the teams that can play the i-th game. It is guaranteed that each pair of teams can play at most one game.
Output
For each game in the same order, print t_i (1 ≤ t_i ≤ k) — the number of the stadium, in which a_i and b_i will play the game. If the i-th game should not be played, t_i should be equal to 0.
If there are multiple answers, print any.
Example
Input
7 11 3
4 7 8 10 10 9 3
6 2
6 1
7 6
4 3
4 6
3 1
5 3
7 5
7 3
4 2
1 4
Output
3
2
1
1
3
1
2
1
2
3
2
Note
One of possible solutions to the example is shown below:
<image> | instruction | 0 | 7,666 | 17 | 15,332 |
Tags: graphs
Correct Solution:
```
import random
import math
def set_color(game, color):
color_count[game[0]][game[2]] -= 1
color_count[game[1]][game[2]] -= 1
game[2] = color
color_count[game[0]][game[2]] += 1
color_count[game[1]][game[2]] += 1
def fix(node):
minimum = math.inf
maximum = 0
for i in range(k):
minimum = min(minimum, color_count[node][i])
maximum = max(maximum, color_count[node][i])
if maximum - minimum <= 2:
return False
rand = 0
for game in games:
if (game[0] == node or game[1] == node) and color_count[node][game[2]] == maximum:
rand = r(1,k)
set_color(game, rand % k)
return True
return False
n, m, k = map(int,input().split())
games = [[0 for _ in range(4)] for _ in range(m)]
color_count = [[0 for _ in range(k)] for _ in range(n)]
answers = [0 for _ in range(m)]
_ = list(map(int,input().split()))
color = 0
r = lambda x,y : random.randint(x,y)
for i in range(m):
a, b = map(int,input().split())
color = r(1,k) % k
games[i] = [a-1,b-1,color,i]
color_count[games[i][0]][color] += 1
color_count[games[i][1]][color] += 1
bad = True
while bad:
random.shuffle(games)
bad = False
for i in range(n):
while(fix(i)):
bad = True
for game in games:
answers[game[3]] = game[2] + 1
for i in range(m):
print(answers[i])
``` | output | 1 | 7,666 | 17 | 15,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n football teams in the world.
The Main Football Organization (MFO) wants to host at most m games. MFO wants the i-th game to be played between the teams a_i and b_i in one of the k stadiums.
Let s_{ij} be the numbers of games the i-th team played in the j-th stadium. MFO does not want a team to have much more games in one stadium than in the others. Therefore, for each team i, the absolute difference between the maximum and minimum among s_{i1}, s_{i2}, …, s_{ik} should not exceed 2.
Each team has w_i — the amount of money MFO will earn for each game of the i-th team. If the i-th team plays l games, MFO will earn w_i ⋅ l.
MFO needs to find what games in what stadiums they need to host in order to earn as much money as possible, not violating the rule they set.
However, this problem is too complicated for MFO. Therefore, they are asking you to help them.
Input
The first line contains three integers n, m, k (3 ≤ n ≤ 100, 0 ≤ m ≤ 1 000, 1 ≤ k ≤ 1 000) — the number of teams, the number of games, and the number of stadiums.
The second line contains n integers w_1, w_2, …, w_n (1 ≤ w_i ≤ 1 000) — the amount of money MFO will earn for each game of the i-th game.
Each of the following m lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the teams that can play the i-th game. It is guaranteed that each pair of teams can play at most one game.
Output
For each game in the same order, print t_i (1 ≤ t_i ≤ k) — the number of the stadium, in which a_i and b_i will play the game. If the i-th game should not be played, t_i should be equal to 0.
If there are multiple answers, print any.
Example
Input
7 11 3
4 7 8 10 10 9 3
6 2
6 1
7 6
4 3
4 6
3 1
5 3
7 5
7 3
4 2
1 4
Output
3
2
1
1
3
1
2
1
2
3
2
Note
One of possible solutions to the example is shown below:
<image>
Submitted Solution:
```
print("submit")
``` | instruction | 0 | 7,667 | 17 | 15,334 |
No | output | 1 | 7,667 | 17 | 15,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n football teams in the world.
The Main Football Organization (MFO) wants to host at most m games. MFO wants the i-th game to be played between the teams a_i and b_i in one of the k stadiums.
Let s_{ij} be the numbers of games the i-th team played in the j-th stadium. MFO does not want a team to have much more games in one stadium than in the others. Therefore, for each team i, the absolute difference between the maximum and minimum among s_{i1}, s_{i2}, …, s_{ik} should not exceed 2.
Each team has w_i — the amount of money MFO will earn for each game of the i-th team. If the i-th team plays l games, MFO will earn w_i ⋅ l.
MFO needs to find what games in what stadiums they need to host in order to earn as much money as possible, not violating the rule they set.
However, this problem is too complicated for MFO. Therefore, they are asking you to help them.
Input
The first line contains three integers n, m, k (3 ≤ n ≤ 100, 0 ≤ m ≤ 1 000, 1 ≤ k ≤ 1 000) — the number of teams, the number of games, and the number of stadiums.
The second line contains n integers w_1, w_2, …, w_n (1 ≤ w_i ≤ 1 000) — the amount of money MFO will earn for each game of the i-th game.
Each of the following m lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the teams that can play the i-th game. It is guaranteed that each pair of teams can play at most one game.
Output
For each game in the same order, print t_i (1 ≤ t_i ≤ k) — the number of the stadium, in which a_i and b_i will play the game. If the i-th game should not be played, t_i should be equal to 0.
If there are multiple answers, print any.
Example
Input
7 11 3
4 7 8 10 10 9 3
6 2
6 1
7 6
4 3
4 6
3 1
5 3
7 5
7 3
4 2
1 4
Output
3
2
1
1
3
1
2
1
2
3
2
Note
One of possible solutions to the example is shown below:
<image>
Submitted Solution:
```
def football(tgs, total_money_game, game):
played = []
stadion_played = []
same_stadion = []
stadion_r = tgs[2]
s_stadion = 1
stadion = []
for i in range(0, tgs[1]):
if game[i] not in played:
played.append(game[i])
for j in range(tgs[2],0,-1):
if j not in stadion_played:
stadion_played.append(j)
stadion.append(j)
break
else:
if len(stadion_played) == tgs[2]:
if same_stadion.count(s_stadion) > 1:
same_stadion = []
s_stadion += 1
if stadion_r == 0:
stadion_r = 3
stadion_played.remove(stadion_r)
stadion_played.remove(s_stadion)
stadion_played.append(s_stadion)
stadion.append(s_stadion)
same_stadion.append(s_stadion)
if stadion_r >= 1:
stadion_r -= 1
else:
stadion_r = tgs[2]
break
return stadion
tgs = [7, 11, 3]
total_money_game = [4, 7, 8, 10, 10, 9, 3]
game = [[6, 2],[6, 1],[7, 6],[4, 3],[4, 6],[3, 1],[5, 3],[7, 5],[7, 3],[4, 2],[1, 4]]
stadions = football(tgs, total_money_game, game)
``` | instruction | 0 | 7,668 | 17 | 15,336 |
No | output | 1 | 7,668 | 17 | 15,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n football teams in the world.
The Main Football Organization (MFO) wants to host at most m games. MFO wants the i-th game to be played between the teams a_i and b_i in one of the k stadiums.
Let s_{ij} be the numbers of games the i-th team played in the j-th stadium. MFO does not want a team to have much more games in one stadium than in the others. Therefore, for each team i, the absolute difference between the maximum and minimum among s_{i1}, s_{i2}, …, s_{ik} should not exceed 2.
Each team has w_i — the amount of money MFO will earn for each game of the i-th team. If the i-th team plays l games, MFO will earn w_i ⋅ l.
MFO needs to find what games in what stadiums they need to host in order to earn as much money as possible, not violating the rule they set.
However, this problem is too complicated for MFO. Therefore, they are asking you to help them.
Input
The first line contains three integers n, m, k (3 ≤ n ≤ 100, 0 ≤ m ≤ 1 000, 1 ≤ k ≤ 1 000) — the number of teams, the number of games, and the number of stadiums.
The second line contains n integers w_1, w_2, …, w_n (1 ≤ w_i ≤ 1 000) — the amount of money MFO will earn for each game of the i-th game.
Each of the following m lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the teams that can play the i-th game. It is guaranteed that each pair of teams can play at most one game.
Output
For each game in the same order, print t_i (1 ≤ t_i ≤ k) — the number of the stadium, in which a_i and b_i will play the game. If the i-th game should not be played, t_i should be equal to 0.
If there are multiple answers, print any.
Example
Input
7 11 3
4 7 8 10 10 9 3
6 2
6 1
7 6
4 3
4 6
3 1
5 3
7 5
7 3
4 2
1 4
Output
3
2
1
1
3
1
2
1
2
3
2
Note
One of possible solutions to the example is shown below:
<image>
Submitted Solution:
```
import operator
import collections
import sys
#User enters file line of input as n m k which gets split into the different variables
n, m, k = input().split(' ') #input("Enter number of teams, matches and stadiums: ").split(' ')
#User enters second line of input as value1 value2 value3......valuen which gets split as an array
money = input().split(' ') # input("Enter amount of money earned per team in the order of the teams: ").split(' ')
#Defining an empty array to get the list of matches played between the teams.
matches_list = []
if int(m) == 0:
#print("Sorry. No matches this year :(")
sys.exit(0)
for i in range(int(m)):
#User enters team input as team1 team2 which gets split into a tuple which is further appended into the "matches_list"
#team1, team2 = input(f"Enter teams that play match {str(i+1)}: ").split(' ')
team1, team2 = input().split(' ') #input("Enter teams that play match: ").split(' ')
#team1, team2 = input(str(i+1))
matches_list.append((int(team1), int(team2)))
#It is guaranteed that each pair of teams can play at most one game hence not checking if input is unique
#Generating stadium numbers as an array
stadiums = [i for i in range(1, int(k)+1)]
#Sorting the matches played in the order of most valuable to least valuable matches based on the amount earned from the induvidual teams.
#Doing this will ensure that the matches with the most earning potential will not be left out in case that particular team cannot play all the given matches because of the "not exceeds 2" rule.
matches_on_revenue = {}
for match in matches_list:
revenue = int(money[match[0]-1]) + int(money[match[1]-1])
matches_on_revenue[match] = revenue
matches_on_revenue = sorted(matches_on_revenue.items(), key=operator.itemgetter(1), reverse=True)
#Making a note of matches played per stadium so far, number of times a team has played at a stadium and stadium assigned for each match.
#Initial values would be 0 or empty and will be updated as we loop through each match while assigning stadiums
matches_per_stadium = {i : 0 for i in stadiums}
teams_at_stadium = {i : [] for i in stadiums}
stadium_for_match = {i[0] : 0 for i in matches_on_revenue}
#Create a function which assigns stadiums for each match. The dictionary "stadium_for_match" will be passed to this fucntion
def assign_stadium(stadium_for_match):
#looping through each match in the dictionary in the order of most valuable to least valuable match. Using Python 3.7 so expecting the order of the dictionary to be preserved.
for key, value in stadium_for_match.items():
#Assiging variable n to 0 as stadium has not yet been assigned
n = 0
#If a stadium is not yet assigned for a particular match, the fucntion will proceed with the logic for that match else it will skip to next match
if not value:
try:
#Try to fetch a stadium where no matches have been played so far. The "matches per stadium" dictionary comes in handy for this
stadium_no = list(matches_per_stadium.keys())[list(matches_per_stadium.values()).index(0)]
#if stadium available, setting n to 1 as stadium has been found
n=1
except:
#if no stadium has 0 matches played, sort the stadiums in the order of least to highest matches played and loop through each stadium
stadiums_by_matches = sorted(matches_per_stadium.items(), key=operator.itemgetter(1))
for stadium_by_match in stadiums_by_matches:
#Get the stadium number
stadium_no = stadium_by_match[0]
#For that stadium, find the team that has played the lowest number of matches and get the value of how many matches played by that team.
# Ignore the matches played by the teams that are present in the current match because even if we add them, it will be a plus one on both ends and so the difference will still be the same
least_common = [i for i in collections.Counter(teams_at_stadium[stadium_no]).most_common() if i[0] != key[0] and i[0] != key[1]][-1][1]
#Check the following conditions
# 1a) If in current match, team1 has played a match in this stadium.
# 1b) If yes, whether the difference between number of matches played by team1 and the number of matches played by the team that has played the lowest in this stadium is less than 2.
# 2a) If in current match, team2 has played a match in this stadium.
# 2b) If yes, whether the difference between number of matches played by team2 and the number of matches played by the team that has played the lowest in this stadium is less than 2.
if ((teams_at_stadium[stadium_no].count(key[0]) == 0) or (abs(teams_at_stadium[stadium_no].count(key[0]) - least_common) < 2)) and ((teams_at_stadium[stadium_no].count(key[1]) == 0) or (abs(teams_at_stadium[stadium_no].count(key[1]) - least_common) < 2) ):
#If the above conditions are satisfied, pick this stadium and break the loop. Else move on to the next stadium.
n=1
break
# If a stadium is found, assign it to the match.
# Increment the value of this stadium in the "matches_per_stadium" dictionary by 1.
# Add the team numbers in the match to the teams_at_stadium list of this stadium again.
if n:
stadium_for_match[key] = stadium_no
matches_per_stadium[stadium_no] = matches_per_stadium[stadium_no] + 1
teams_at_stadium[stadium_no].append(key[0])
teams_at_stadium[stadium_no].append(key[1])
#If no stadium is found for this match the value will remain 0.
#Move to the next match.
#Return the final dictionary and also if there were any stadiums found for any matches
return stadium_for_match, n
#Initially assuming there were stadiums found for matches
n = 1
while n:
#Call the assign stadium function till there were no changes made to any matches.
#The reason for this instead of a single function call is:
#Lets say a match was assigned value 0 because it could not fit into any stadium due to the "exceeds 2 rule"
#After iterating thorugh the entire dictionary, it may be so that this difference had come down from 2. In those cases, there would be a stadium avalible for a match which was previously not available.
stadium_for_match, n = assign_stadium(stadium_for_match)
#Once the stadiums are finalized, print the output stadium numbers in the same order as the matches provided in the input.
for i in matches_list:
print(stadium_for_match[i])
``` | instruction | 0 | 7,669 | 17 | 15,338 |
No | output | 1 | 7,669 | 17 | 15,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n football teams in the world.
The Main Football Organization (MFO) wants to host at most m games. MFO wants the i-th game to be played between the teams a_i and b_i in one of the k stadiums.
Let s_{ij} be the numbers of games the i-th team played in the j-th stadium. MFO does not want a team to have much more games in one stadium than in the others. Therefore, for each team i, the absolute difference between the maximum and minimum among s_{i1}, s_{i2}, …, s_{ik} should not exceed 2.
Each team has w_i — the amount of money MFO will earn for each game of the i-th team. If the i-th team plays l games, MFO will earn w_i ⋅ l.
MFO needs to find what games in what stadiums they need to host in order to earn as much money as possible, not violating the rule they set.
However, this problem is too complicated for MFO. Therefore, they are asking you to help them.
Input
The first line contains three integers n, m, k (3 ≤ n ≤ 100, 0 ≤ m ≤ 1 000, 1 ≤ k ≤ 1 000) — the number of teams, the number of games, and the number of stadiums.
The second line contains n integers w_1, w_2, …, w_n (1 ≤ w_i ≤ 1 000) — the amount of money MFO will earn for each game of the i-th game.
Each of the following m lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the teams that can play the i-th game. It is guaranteed that each pair of teams can play at most one game.
Output
For each game in the same order, print t_i (1 ≤ t_i ≤ k) — the number of the stadium, in which a_i and b_i will play the game. If the i-th game should not be played, t_i should be equal to 0.
If there are multiple answers, print any.
Example
Input
7 11 3
4 7 8 10 10 9 3
6 2
6 1
7 6
4 3
4 6
3 1
5 3
7 5
7 3
4 2
1 4
Output
3
2
1
1
3
1
2
1
2
3
2
Note
One of possible solutions to the example is shown below:
<image>
Submitted Solution:
```
s = input()
if '0' * 7 in s or '1' * 7 in s:
print('YES')
else:
print('NO')
``` | instruction | 0 | 7,670 | 17 | 15,340 |
No | output | 1 | 7,670 | 17 | 15,341 |
Provide a correct Python 3 solution for this coding contest problem.
B: Dansunau www --Dance Now!-
story
Last lab life! Daigakuin! !! Dosanko Snow has won 9th place in the event "Master Idol World", which can be said to be the outpost of the biggest competition "Lab Life" where master idols compete. The sharp dance is ridiculed as "9th place dance", and the whole body's deciding pose is also ridiculed as "9th place stance", which causes deep emotional wounds. In the upcoming Lab Life Preliminary Qualifying, we can never take 9th place ... Dosanko Snow, who renewed his determination, refined his special skill "self-control" and headed for the battlefield. ..
problem
A tournament "Lab Life" will be held in which N groups of units compete. In this tournament, the three divisions of Smile Pure Cool will be played separately, and the overall ranking will be decided in descending order of the total points scored in each game. The ranking of the smile category is determined in descending order of the smile value of each unit. Similarly, in the pure / cool category, the ranking is determined in descending order of pure / cool value. The score setting according to the ranking is common to all divisions, and the unit ranked i in a division gets r_i points in that division.
Here, if there are a plurality of units having the same value as the unit with the rank i, they are regarded as the same rate i rank, and the points r_i are equally obtained. More specifically, when k units are in the same ratio i position, k units get equal points r_i, and no unit gets points from r_ {i + 1} to r_ {i + k-1}. Also, the next largest unit (s) gets the score r_ {i + k}. As a specific example, consider a case where there are five units with smile values of 1, 3, 2, 3, and 2, and 10, 8, 6, 4, and 2 points are obtained in descending order of rank. At this time, the 2nd and 4th units will get 10 points, the 3rd and 5th units will get 6 points, and the 1st unit will get 2 points.
Unit Dosanko Snow, who participates in the Lab Life Preliminary Qualifying, thinks that "Lab Life is not a play", so he entered the tournament first and became the first unit. However, when we obtained information on the smile value, pure value, and cool value (hereinafter referred to as 3 values) of all N groups participating in the tournament, we found that their overall ranking was (equal rate) 9th. Dosanko Snow can raise any one of the three values by the special skill "Self Control", but if you raise it too much, you will get tired and it will affect the main battle, so you want to make the rise value as small as possible. Dosanko Snow wants to get out of 9th place anyway, so when you raise any one of the 3 values by self-control so that it will be 8th or higher at the same rate, find the minimum value that needs to be raised.
Input format
The input is given in the following format.
N
r_1 ... r_N
s_1 p_1 c_1
...
s_N p_N c_N
The first line is given the integer N, which represents the number of units, in one line. The second line that follows is given N integers separated by blanks. The i (1 \ leq i \ leq N) th integer represents the score r_i that can be obtained when the ranking is i in each division. Of the following Nth line, the jth line is given three integers s_j, p_j, c_j. These represent the smile value s_j, pure value p_j, and cool value c_j of the jth unit, respectively. Dosanko Snow is the first unit.
Constraint
* 9 \ leq N \ leq 100
* 100 \ geq r_1> ...> r_N \ geq 1
* 1 \ leq s_j, p_j, c_j \ leq 100 (1 \ leq j \ leq N)
* Before self-control, Dosanko Snow was 9th (sometimes 9th, but never more than 8th)
Output format
By increasing any of the three values by x by self-control, output the minimum x that makes Dosanko Snow 8th or higher at the same rate on one line. However, if self-control does not raise the ranking no matter how much you raise it, output "Saiko".
Input example 1
9
9 8 7 6 5 4 3 2 1
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
Output example 1
2
For example, if you raise the smile value by 2 by self-control, the rankings in each division of Dosanko Snow will be 7th, 9th, and 9th, respectively, and you will get 3 + 1 + 1 = 5 points. On the other hand, the ranking of the second unit in each division is 9th, 8th, and 8th, respectively, and 1 + 2 + 2 = 5 points are obtained. Since the other units get 6 points or more, these two units will be ranked 8th at the same rate, satisfying the conditions.
Input example 2
9
9 8 7 6 5 4 3 2 1
1 1 1
2 6 9
6 9 2
9 2 6
3 5 8
5 8 3
8 3 5
4 7 4
7 4 7
Output example 2
Saiko
No matter how much you raise the value, Dosanko Snow can only get up to 11 points, but other units can always get 14 points or more, so Dosanko Snow is immovable in 9th place.
Example
Input
9
9 8 7 6 5 4 3 2 1
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
Output
2 | instruction | 0 | 8,357 | 17 | 16,714 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10**6)
input = sys.stdin.readline
n = int(input())
R = list(map(int,input().split()))
SPC = []
S = []
P = []
C = []
ans = 10**5
for i in range(n):
s,p,c = map(int,input().split())
S.append(s)
P.append(p)
C.append(c)
SPC.append([s,p,c])
you = list(SPC[0])
for delta in range(1,102-you[0]):
S[0] = you[0] + delta
SPC[0][0] = S[0]
Ssort = sorted(S,reverse = True)
Psort = sorted(P,reverse = True)
Csort = sorted(C,reverse = True)
pointS = {}
pointP = {}
pointC = {}
for r,s,p,c in zip(R,Ssort,Psort,Csort):
if s not in pointS:pointS[s] = r
if p not in pointP:pointP[p] = r
if c not in pointC:pointC[c] = r
point = [pointS[s]+pointP[p]+pointC[c] for s,p,c in SPC]
if sorted(point,reverse = True).index(point[0]) != 8:
ans = min(ans,delta)
break
S[0] = you[0]
SPC[0] = you.copy()
for delta in range(1,102-you[1]):
P[0] = you[1] + delta
SPC[0][1] = P[0]
Ssort = sorted(S,reverse = True)
Psort = sorted(P,reverse = True)
Csort = sorted(C,reverse = True)
pointS = {}
pointP = {}
pointC = {}
for r,s,p,c in zip(R,Ssort,Psort,Csort):
if s not in pointS:pointS[s] = r
if p not in pointP:pointP[p] = r
if c not in pointC:pointC[c] = r
point = [pointS[s]+pointP[p]+pointC[c] for s,p,c in SPC]
if sorted(point,reverse = True).index(point[0]) != 8:
ans = min(ans,delta)
break
P[0] = you[1]
SPC[0] = you.copy()
for delta in range(1,102-you[2]):
C[0] = you[2] + delta
SPC[0][2] = C[0]
Ssort = sorted(S,reverse = True)
Psort = sorted(P,reverse = True)
Csort = sorted(C,reverse = True)
pointS = {}
pointP = {}
pointC = {}
for r,s,p,c in zip(R,Ssort,Psort,Csort):
if s not in pointS:pointS[s] = r
if p not in pointP:pointP[p] = r
if c not in pointC:pointC[c] = r
point = [pointS[s]+pointP[p]+pointC[c] for s,p,c in SPC]
if sorted(point,reverse = True).index(point[0]) != 8:
ans = min(ans,delta)
break
print(ans if ans != 10**5 else 'Saiko')
``` | output | 1 | 8,357 | 17 | 16,715 |
Provide a correct Python 3 solution for this coding contest problem.
In Group C of the 3rd year, we decided to use the "class flag" used at the sports festival on November 10, 2007 at future class reunions. So, in order to decide which students to keep the "class flag", I decided to play the following game using a large amount of candy that the teacher gave me the other day.
* Each student will take one candy in the order of student number.
* If candy remains after one round, continue to take candy in order from the person with the first student number.
* The person who picks up the last candy will be the student who keeps the "class flag".
There are 39 students in the 3rd grade C class. Their student numbers are 3C01 to 3C39. For example, if you have 50 candies and everyone in the class finishes taking the first candy, you will have 11 candies left. If you take it again in student number order, the last one will be taken by the 3C11 student. That is, the 3C11 student is the student who keeps the "class flag".
Create a program that takes the number of candies as input and outputs the student number of the student who stores the "class flag".
Input
Multiple test cases are given. Each test case is given in the following format. For each test case, an integer a (1 ≤ a ≤ 10000) representing the number of candies is given on one line. Process until the end of input (EOF).
The number of datasets does not exceed 20.
Output
For each test case, output the student number (half-width alphanumeric characters) of the student who stores the "class flag" on one line.
Example
Input
50
5576
5577
5578
Output
3C11
3C38
3C39
3C01 | instruction | 0 | 8,971 | 17 | 17,942 |
"Correct Solution:
```
while 1:
try:
n = int(input())
except:
break
if n % 39 == 0:
ans = 39
else:
ans = n % 39
print("3C" + str(ans).zfill(2))
``` | output | 1 | 8,971 | 17 | 17,943 |
Provide a correct Python 3 solution for this coding contest problem.
In Group C of the 3rd year, we decided to use the "class flag" used at the sports festival on November 10, 2007 at future class reunions. So, in order to decide which students to keep the "class flag", I decided to play the following game using a large amount of candy that the teacher gave me the other day.
* Each student will take one candy in the order of student number.
* If candy remains after one round, continue to take candy in order from the person with the first student number.
* The person who picks up the last candy will be the student who keeps the "class flag".
There are 39 students in the 3rd grade C class. Their student numbers are 3C01 to 3C39. For example, if you have 50 candies and everyone in the class finishes taking the first candy, you will have 11 candies left. If you take it again in student number order, the last one will be taken by the 3C11 student. That is, the 3C11 student is the student who keeps the "class flag".
Create a program that takes the number of candies as input and outputs the student number of the student who stores the "class flag".
Input
Multiple test cases are given. Each test case is given in the following format. For each test case, an integer a (1 ≤ a ≤ 10000) representing the number of candies is given on one line. Process until the end of input (EOF).
The number of datasets does not exceed 20.
Output
For each test case, output the student number (half-width alphanumeric characters) of the student who stores the "class flag" on one line.
Example
Input
50
5576
5577
5578
Output
3C11
3C38
3C39
3C01 | instruction | 0 | 8,972 | 17 | 17,944 |
"Correct Solution:
```
try:
while True:
ame = int(input())
num1 = ame % 39
if num1 == 0:
num1 = 39
if num1 < 10:
print("3C0" + str(num1))
else:
print("3C" + str(num1))
except EOFError as e:
num = 0
``` | output | 1 | 8,972 | 17 | 17,945 |
Provide a correct Python 3 solution for this coding contest problem.
In Group C of the 3rd year, we decided to use the "class flag" used at the sports festival on November 10, 2007 at future class reunions. So, in order to decide which students to keep the "class flag", I decided to play the following game using a large amount of candy that the teacher gave me the other day.
* Each student will take one candy in the order of student number.
* If candy remains after one round, continue to take candy in order from the person with the first student number.
* The person who picks up the last candy will be the student who keeps the "class flag".
There are 39 students in the 3rd grade C class. Their student numbers are 3C01 to 3C39. For example, if you have 50 candies and everyone in the class finishes taking the first candy, you will have 11 candies left. If you take it again in student number order, the last one will be taken by the 3C11 student. That is, the 3C11 student is the student who keeps the "class flag".
Create a program that takes the number of candies as input and outputs the student number of the student who stores the "class flag".
Input
Multiple test cases are given. Each test case is given in the following format. For each test case, an integer a (1 ≤ a ≤ 10000) representing the number of candies is given on one line. Process until the end of input (EOF).
The number of datasets does not exceed 20.
Output
For each test case, output the student number (half-width alphanumeric characters) of the student who stores the "class flag" on one line.
Example
Input
50
5576
5577
5578
Output
3C11
3C38
3C39
3C01 | instruction | 0 | 8,973 | 17 | 17,946 |
"Correct Solution:
```
while True:
try:
a = int(input())
except:break
if a%39 == 0:
print("3C39")
else:
b = a // 39
c = a - 39 * b
if c < 10:
print("3C0",c,sep='')
else:
print("3C",c,sep='')
``` | output | 1 | 8,973 | 17 | 17,947 |
Provide a correct Python 3 solution for this coding contest problem.
In Group C of the 3rd year, we decided to use the "class flag" used at the sports festival on November 10, 2007 at future class reunions. So, in order to decide which students to keep the "class flag", I decided to play the following game using a large amount of candy that the teacher gave me the other day.
* Each student will take one candy in the order of student number.
* If candy remains after one round, continue to take candy in order from the person with the first student number.
* The person who picks up the last candy will be the student who keeps the "class flag".
There are 39 students in the 3rd grade C class. Their student numbers are 3C01 to 3C39. For example, if you have 50 candies and everyone in the class finishes taking the first candy, you will have 11 candies left. If you take it again in student number order, the last one will be taken by the 3C11 student. That is, the 3C11 student is the student who keeps the "class flag".
Create a program that takes the number of candies as input and outputs the student number of the student who stores the "class flag".
Input
Multiple test cases are given. Each test case is given in the following format. For each test case, an integer a (1 ≤ a ≤ 10000) representing the number of candies is given on one line. Process until the end of input (EOF).
The number of datasets does not exceed 20.
Output
For each test case, output the student number (half-width alphanumeric characters) of the student who stores the "class flag" on one line.
Example
Input
50
5576
5577
5578
Output
3C11
3C38
3C39
3C01 | instruction | 0 | 8,974 | 17 | 17,948 |
"Correct Solution:
```
while True:
try:
a = int(input())
except:
break
a = str(a%39)
if a == "0":
a = "39"
elif len(a) == 1:
a = "0"+a
print("3C"+a)
``` | output | 1 | 8,974 | 17 | 17,949 |
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