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 |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
n = int(input())
list1 = [int(i)for i in input().split()]
dp = []
for x in range(n):
dp.append([10**9,10**9,10**9])
dp[0][0]=1
if list1[0]==0:
dp[0][1]=1
dp[0][2]=1
elif list1[0]==1:
dp[0][1]=0
dp[0][2]=1
elif list1[0]==2:
dp[0][1]=1
dp[0][2]=0
else:
dp[0][1]=0
dp[0][2]=0
for x in range(1,n):
dp[x][0]=1+min([dp[x-1][0],dp[x-1][1],dp[x-1][2]])
if list1[x]==0:
dp[x][1]=1+min([dp[x-1][0],dp[x-1][1],dp[x-1][2]])
dp[x][2]=1+min([dp[x-1][0],dp[x-1][1],dp[x-1][2]])
elif list1[x]==1:
dp[x][1]=min([dp[x-1][0],dp[x-1][2]])
dp[x][2]=1+min([dp[x-1][0],dp[x-1][1],dp[x-1][2]])
elif list1[x]==2:
dp[x][1]=1+min([dp[x-1][0],dp[x-1][1],dp[x-1][2]])
dp[x][2]=min([dp[x-1][0],dp[x-1][1]])
else:
dp[x][1]=min([dp[x-1][0],dp[x-1][2]])
dp[x][2]=min([dp[x-1][0],dp[x-1][1]])
print(min([dp[n-1][0],dp[n-1][1],dp[n-1][2]]))
``` | instruction | 0 | 87,135 | 17 | 174,270 |
Yes | output | 1 | 87,135 | 17 | 174,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
n = int(input())
nums = input()
nums = [int(i) for i in nums.split()]
MAXVAL = 999
dp = [[MAXVAL for i in range(3)] for i in range(n)]
if nums[0] == 0:
dp[0][0] = 1
elif nums[0] == 1:
dp[0][1] = 0
elif nums[0] == 2:
dp[0][2] = 0
else:
dp[0][1] = 0 # if both, CONTEST
dp[0][2] = 0 # if both, GYM
#print(dp)
for i in range(1,n):
if nums[i] == 0:
dp[i][0] = min(dp[i-1]) + 1
elif nums[i] == 1:
# Do contest
dp[i][1] = min(dp[i-1][0], dp[i-1][2]) + 0
# Rest
dp[i][0] = min(dp[i-1][0], dp[i-1][1], dp[i-1][2]) + 1
elif nums[i] == 2:
# Do gym
dp[i][2] = min(dp[i-1][0], dp[i-1][1]) + 0
# Rest
dp[i][0] = min(dp[i-1][0], dp[i-1][1], dp[i-1][2]) + 1
else:
# Rest
dp[i][0] = min(dp[i-1][0], dp[i-1][1], dp[i-1][2]) + 1
# Do gym
dp[i][2] = min(dp[i-1][0], dp[i-1][1]) + 0
# Do contest
dp[i][1] = min(dp[i-1][0], dp[i-1][2]) + 0
#print(dp)
ans = min(dp[n-1])
print(ans)
``` | instruction | 0 | 87,136 | 17 | 174,272 |
Yes | output | 1 | 87,136 | 17 | 174,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
z,zz=input,lambda:list(map(int,z().split()))
fast=lambda:stdin.readline().strip()
zzz=lambda:[int(i) for i in stdin.readline().split()]
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from string import *
from re import *
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
p=ceil(x**.5)+1
for i in range(2,p):
if (x%i==0 and x!=2) or x==0:return 0
return 1
def dfs(u,visit,graph):
visit[u]=True
for i in graph[u]:
if not visit[i]:
dfs(i,visit,graph)
###########################---Test-Case---#################################
"""
"""
###########################---START-CODING---##############################
n=int(z())
arr=zzz()
dp=[[10**9]*3 for _ in range(n+1)]
dp[0][0]=0
for i in range(n):
k=arr[i]
dp[i+1][0]=min(dp[i][2],dp[i][1],dp[i][0])+1
if k==1:
dp[i+1][1]=min(dp[i][0],dp[i][2])
if k==2:
dp[i+1][2]=min(dp[i][0],dp[i][1])
if k==3:
dp[i+1][1]=min(dp[i][0],dp[i][2])
dp[i+1][2]=min(dp[i][0],dp[i][1])
print(min(dp[n]))
``` | instruction | 0 | 87,137 | 17 | 174,274 |
Yes | output | 1 | 87,137 | 17 | 174,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
n = int(input())
c = [int(x) for x in input().split(' ')]
a = [(2, -1)]
b = [(1, -1)]
for i in range(n):
if c[i] == 0: pass
elif c[i] == 1 or c[i] == 2:
if a[-1][0] != c[i]: a.append((c[i], i))
if b[-1][0] != c[i]: b.append((c[i], i))
else:
a.append((2, i) if a[-1][0] == 1 else (1, i))
b.append((2, i) if b[-1][0] == 1 else (1, i))
#print(a, b)
print( n - max(len(a), len(b)) + 1)
``` | instruction | 0 | 87,138 | 17 | 174,276 |
No | output | 1 | 87,138 | 17 | 174,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
n=int(input())
s=input().split()
aflag=0
bflag=0
ans=0
for i in range(n):
if s[i]=='0':
ans+=1;aflag=0;bflag=0
elif s[i]=='1':
if aflag==1:
ans+=1;aflag=0;bflag=0
else:
aflag=1;bflag=0
elif s[i]=='2':
if bflag==1:
ans+=1;aflag=0;bflag=0
else:
bflag=1;aflag=0
else:
if aflag==1:
bflag=1;aflag=0
else:
aflag=1;bflag=0
print(ans)
``` | instruction | 0 | 87,139 | 17 | 174,278 |
No | output | 1 | 87,139 | 17 | 174,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
M = mod = 10**9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
n = val()
l = li()
dp = [[0 for i in range(2)] for j in range(n)]
dp[0][0] = 1 if l[0]>1 else 0
dp[0][1] = 1 if l[0]&1 else 0
ans = 0
if dp[0][0] == 0 and dp[0][1] == 0:ans = 1
currmax = ans
for i in range(1,n):
if not dp[i-1][0] and l[i] > 1:dp[i][0] = 1
if not dp[i-1][1] and l[i]&1:dp[i][1] = 1
if sum(dp[i]) == 0:ans += 1
else:ans = 0
currmax = max(ans,currmax)
print(currmax)
``` | instruction | 0 | 87,140 | 17 | 174,280 |
No | output | 1 | 87,140 | 17 | 174,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:
1. on this day the gym is closed and the contest is not carried out;
2. on this day the gym is closed and the contest is carried out;
3. on this day the gym is open and the contest is not carried out;
4. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input
The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:
* ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
* ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
* ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
* ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
Output
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
* to do sport on any two consecutive days,
* to write the contest on any two consecutive days.
Examples
Input
4
1 3 2 0
Output
2
Input
7
1 3 3 2 1 2 3
Output
0
Input
2
2 2
Output
1
Note
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
count = 0
prev = 0
f=0
for i in range(n):
if a[i] == 0:
count+=1
prev = 0
if a[i] == 1 or a[i] == 2:
if prev == a[i]:
count+=1
prev = 0
prev = a[i]
if a[i] == 3:
if prev != 3 and prev != 0:
prev = 3-prev
else:
prev = 3
print(count)
``` | instruction | 0 | 87,141 | 17 | 174,282 |
No | output | 1 | 87,141 | 17 | 174,283 |
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 | 87,222 | 17 | 174,444 |
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 | 87,222 | 17 | 174,445 |
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 | 87,223 | 17 | 174,446 |
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 | 87,223 | 17 | 174,447 |
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:
```
r=lambda:map(int,input().split())
n,k,m=r()
a=list(r())
stck=[]
stck
if k==1:
print(0)
exit(0)
if n==1 and m >=k:
print(m%k)
exit(0)
for i in range(n):#this is team formation possible in a bus itself
if len(stck)==0:
stck.append((a[i],1))
prevV=a[i]
prevL=1
elif a[i]==prevV:
prevL+=1
stck=stck[:len(stck)-prevL+1]
stck+=[(a[i],prevL)]*prevL
else:
stck.append((a[i],1))
prevV=a[i]
prevL=1
if prevL==k:
stck=stck[:len(stck)-k]
last=[stck[len(stck)-1][0],stck[len(stck)-1][1]]
it=0
ans=0
while True:
p=0
r=len(stck)
for i in range(r//2):
if stck[i][0]==stck[r-i-1][0] and stck[i][1]+stck[r-i-1][1]==k:
p+=1
if p==(r//2) and m%2==0 and p!=0:
print(0)
exit(0)
elif p==(r//2) and m%2 !=0 and p!=0:
if r%2!=0:
print(r)
exit(0)
else:
print(0)
exit(0)
elif p!=(r//2) and p>0:
ans+=(m*r-(m-1)*(k*p))
stck=stck[:r-stck[r-1][1]]
elif stck[0][1]<len(stck) and stck[r-1][1]+stck[stck[0][1]][1]==k and stck[r-1][0]==stck[stck[0][1]][0]:
ans+=k*(m-1)
stck=stck[:r-stck[r-1][1]]
elif stck[r-1][1]+last[1]==k and stck[r-1][0]==last[0]:
print(0)
exit(0)
elif p==0:
break
it+=1
if it==0:
print(m*r)
else:
print(ans)
``` | instruction | 0 | 87,224 | 17 | 174,448 |
No | output | 1 | 87,224 | 17 | 174,449 |
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:
```
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 += k
a.pop()
a.pop(0)
else:
s1 += join
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()
``` | instruction | 0 | 87,225 | 17 | 174,450 |
No | output | 1 | 87,225 | 17 | 174,451 |
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:
```
#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
join+=rem
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)
``` | instruction | 0 | 87,226 | 17 | 174,452 |
No | output | 1 | 87,226 | 17 | 174,453 |
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:
```
r=lambda:map(int,input().split())
n,k,m=r()
a=list(r())
stck=[]
stck
if k==1:
print(0)
exit(0)
if n==1 and m >=k:
print(m%k)
exit(0)
for i in range(n):#this is team formation possible in a bus itself
if len(stck)==0:
stck.append((a[i],1))
prevV=a[i]
prevL=1
elif a[i]==prevV:
prevL+=1
stck=stck[:len(stck)-prevL+1]
stck+=[(a[i],prevL)]*prevL
else:
stck.append((a[i],1))
prevV=a[i]
prevL=1
if prevL==k:
stck=stck[:len(stck)-k]
last=[stck[len(stck)-1][0],stck[len(stck)-1][1]]
it=0
ans=0
while True:
p=0
r=len(stck)
for i in range(r//2):
if stck[i][0]==stck[r-i-1][0] and stck[i][1]+stck[r-i-1][1]==k:
p+=1
if p==(r//2) and m%2==0 and p!=0:
print(0)
exit(0)
elif p==(r//2) and m%2 !=0 and p!=0:
if r%2!=0:
print(r)
exit(0)
else:
print(0)
exit(0)
elif p!=(r//2) and p>0:
ans+=(m*r-(m-1)*(k*p))
stck=stck[:r-stck[r-1][1]]
stck2=stck[stck[0][1]:]
elif r!=0 and len(stck2)!=0 and stck[r-1][1]+stck2[0][1]==k and stck[r-1][0]==stck2[0][0]:
ans+=k*(m-1)
stck=stck[:r-stck[r-1][1]]
stck2=stck[stck[0][1]:]
elif stck[r-1][1]+last[1]==k and stck[r-1][0]==last[0]:
print(0)
exit(0)
elif p==0:
break
it+=1
if it==0:
print(m*r)
else:
print(ans)
print(stck)
``` | instruction | 0 | 87,227 | 17 | 174,454 |
No | output | 1 | 87,227 | 17 | 174,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kostya is a progamer specializing in the discipline of Dota 2. Valve Corporation, the developer of this game, has recently released a new patch which turned the balance of the game upside down. Kostya, as the captain of the team, realizes that the greatest responsibility lies on him, so he wants to resort to the analysis of innovations patch from the mathematical point of view to choose the best heroes for his team in every game.
A Dota 2 match involves two teams, each of them must choose some heroes that the players of the team are going to play for, and it is forbidden to choose the same hero several times, even in different teams. In large electronic sports competitions where Kostya's team is going to participate, the matches are held in the Captains Mode. In this mode the captains select the heroes by making one of two possible actions in a certain, predetermined order: pick or ban.
* To pick a hero for the team. After the captain picks, the picked hero goes to his team (later one of a team members will play it) and can no longer be selected by any of the teams.
* To ban a hero. After the ban the hero is not sent to any of the teams, but it still can no longer be selected by any of the teams.
The team captain may miss a pick or a ban. If he misses a pick, a random hero is added to his team from those that were available at that moment, and if he misses a ban, no hero is banned, as if there was no ban.
Kostya has already identified the strength of all the heroes based on the new patch fixes. Of course, Kostya knows the order of picks and bans. The strength of a team is the sum of the strengths of the team's heroes and both teams that participate in the match seek to maximize the difference in strengths in their favor. Help Kostya determine what team, the first one or the second one, has advantage in the match, and how large the advantage is.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of heroes in Dota 2.
The second line contains n integers s1, s2, ..., sn (1 ≤ si ≤ 106) — the strengths of all the heroes.
The third line contains a single integer m (2 ≤ m ≤ min(n, 20)) — the number of actions the captains of the team must perform.
Next m lines look like "action team", where action is the needed action: a pick (represented as a "p") or a ban (represented as a "b"), and team is the number of the team that needs to perform the action (number 1 or 2).
It is guaranteed that each team makes at least one pick. Besides, each team has the same number of picks and the same number of bans.
Output
Print a single integer — the difference between the strength of the first team and the strength of the second team if the captains of both teams will act optimally well.
Examples
Input
2
2 1
2
p 1
p 2
Output
1
Input
6
6 4 5 4 5 5
4
b 2
p 1
b 1
p 2
Output
0
Input
4
1 2 3 4
4
p 2
b 2
p 1
b 1
Output
-2
Submitted Solution:
```
n = int(input().strip())
arr = [int(x) for x in input().strip().split(' ')]
m = int(input().strip())
actions = []
for i in range(m):
actions.append(input().strip().split(' '))
for i in range(m):
actions[i][1] = int(actions[i][1])
i =0
while i < m-1:
j = 0
while i+j < m-1 and actions[i][1] == actions[i+j+1][1] and actions[i][0] == 'b' and actions[i+j+1][0] == 'p':
j += 1
if j > 0:
actions[i][0] = 'p'
actions[i+j][0] = 'b'
i = i+j+1
else:
i += 1
arr.sort()
arr.reverse()
j = 0
d = {1:0, 2:0}
for i in actions:
if i[0] == 'p':
d[i[1]]+=arr[j]
j += 1
print(d[1]-d[2])
``` | instruction | 0 | 87,915 | 17 | 175,830 |
No | output | 1 | 87,915 | 17 | 175,831 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90 | instruction | 0 | 87,941 | 17 | 175,882 |
Tags: implementation
Correct Solution:
```
f=input()
s=input()
n=int(input())
ans=[]
printed=[]
for i in range(n):
t,ty,m,ca=input().split()
li=[]
if ca=='r':
li.append(ty)
li.append(m)
if li not in printed:
if ty == 'h':
print(f, m, t)
else:
print(s, m, t)
printed.append(li)
else:
li.append(ty)
li.append(m)
if li not in ans:
ans.append(li)
else:
if li not in printed:
if ty == 'h':
print(f, m, t)
else:
print(s, m, t)
printed.append(li)
``` | output | 1 | 87,941 | 17 | 175,883 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90 | instruction | 0 | 87,942 | 17 | 175,884 |
Tags: implementation
Correct Solution:
```
__author__ = 'Administrator'
def inp():
return input()
home = inp()
away = inp()
n = int(inp())
class Node:
time = 0
team = 0
num = 0
color = 0
def __lt__(self, other):
return self.time < other.time
def __init__(self, a, b, c, d):
self.time = a
self.team = b
self.num = c
self.color = d
i = 0
lt = []
while i < n:
temp = inp().split(' ')
o = Node(int(temp[0]), temp[1] == 'h', int(temp[2]), temp[3] == 'r')
lt.append(o)
i += 1
a = [0 for i in range(100)]
b = [0 for i in range(100)]
lt.sort()
i = 0
while i < n:
# print(lt[i].time, lt[i].team, lt[i].num, lt[i].color)
if lt[i].team:
if a[lt[i].num] < 2:
if lt[i].color:
a[lt[i].num] += 2
else:
a[lt[i].num] += 1
if a[lt[i].num] >= 2:
print(home, ' ', lt[i].num, ' ', lt[i].time)
else:
pass
else:
pass
else:
if b[lt[i].num] < 2:
if lt[i].color:
b[lt[i].num] += 2
else:
b[lt[i].num] += 1
if b[lt[i].num] >= 2:
print(away, ' ', lt[i].num, ' ', lt[i].time)
else:
pass
else:
pass
i += 1
# print("---end\n")
``` | output | 1 | 87,942 | 17 | 175,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90 | instruction | 0 | 87,943 | 17 | 175,886 |
Tags: implementation
Correct Solution:
```
import collections
a_name = input()
b_name = input()
n = int(input())
players = [collections.Counter() for i in range(2)]
for i in range(n):
t, team, player, color = input().split()
team = 0 if team == 'h' else 1
player = int(player)
if color == 'y' and players[team][player] == 0:
players[team][player] = 1
elif color == 'r' and 0 <= players[team][player] <= 1 or players[team][player] == 1:
players[team][player] = 2
print(a_name if team == 0 else b_name, player, t)
``` | output | 1 | 87,943 | 17 | 175,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90 | instruction | 0 | 87,944 | 17 | 175,888 |
Tags: implementation
Correct Solution:
```
s = input()
t = input()
n = int(input())
lis=[]
anh=[0]*(101)
ana=[0]*(101)
an=[]
for i in range(n):
n,m,a,b = map(str,input().split())
if m=='h':
lis.append([int(n),s,int(a),b,m])
else:
lis.append([int(n),t,int(a),b,m])
lis.sort()
#print(lis)
for i in lis:
n,m,a,b,c=i
# print(n,m,a,b)
if b=='y':
if c=='h':
if anh[a]!=2:
anh[a]+=1
if anh[a]==2:
an.append([m,a,n])
else:
if ana[a]!=2:
ana[a]+=1
if ana[a]==2:
an.append([m,a,n])
elif b=='r':
if c=='h':
if anh[a]!=2:
anh[a]=2
an.append([m,a,n])
else:
if ana[a]!=2:
ana[a]=2
an.append([m,a,n])
for i in an:
print(*i)
``` | output | 1 | 87,944 | 17 | 175,889 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90 | instruction | 0 | 87,945 | 17 | 175,890 |
Tags: implementation
Correct Solution:
```
from collections import *
s1=input()
s2=input()
n=int(input())
d=defaultdict(int)
for i in range(n):
time,ty,num,col=map(str,input().split())
d[ty+num+col]+=1
if col=='y' and d[ty+num+col]==2 and d[ty+num+'r']==0:
if ty=='a':
print(s2,num,time)
else:
print(s1,num,time)
elif col=='r' and d[ty+num+col]==1 and (d[ty+num+'y']==0 or d[ty+num+'y']==1):
if(ty=='a'):
print(s2,num,time)
else:
print(s1,num,time)
``` | output | 1 | 87,945 | 17 | 175,891 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90 | instruction | 0 | 87,946 | 17 | 175,892 |
Tags: implementation
Correct Solution:
```
team1 = input()
team2 = input()
fouls = int(input())
warn1 = [False] * 100
warn2 = [False] * 100
out1 = [False] * 100
out2 = [False] * 100
warns = []
for i in range(fouls):
a, b, c, d = map(str, input().split(' '))
a, c = int(a), int(c)
if b == 'a' and d == 'r' and out2[c] == False:
warns.append([team2, c, a])
out2[c] = True
elif b == 'h' and d == 'r' and out1[c] == False:
warns.append([team1, c, a])
out1[c] = True
elif b == 'a' and d == 'y' and warn2[c] and out2[c] == False:
warns.append([team2, c, a])
out2[c] = True
elif b == 'a' and d == 'y' and not warn2[c]:
warn2[c] = True
elif b == 'h' and d == 'y' and warn1[c] and out1[c] == False:
warns.append([team1, c, a])
out1[c] = True
elif b == 'h' and d == 'y' and not warn1[c]:
warn1[c] = True
warns.sort(key = lambda x:x[2])
for a in warns:
print(' '.join([str(x) for x in a]))
``` | output | 1 | 87,946 | 17 | 175,893 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90 | instruction | 0 | 87,947 | 17 | 175,894 |
Tags: implementation
Correct Solution:
```
if __name__ == '__main__':
h_team_name = input()
a_team_name = input()
h=dict()
a1 = dict()
ans = []
for _ in range (int(input())):
a,b,c,d=input().split()
if d == 'r':
if b == 'h':
h.setdefault(c,0)
h[c]+=2
if h[c] == 2 or h[c]==3:
ans.append([h_team_name,c,a])
else:
a1.setdefault(c,0)
a1[c]+=2
if a1[c]==2 or a1[c]==3:
ans.append([a_team_name,c,a])
else:
if b == 'h':
h.setdefault(c,0)
h[c]+=1
if h[c] == 2:
ans.append([h_team_name,c,a])
else:
a1.setdefault(c,0)
a1[c]+=1
if a1[c]==2:
ans.append([a_team_name,c,a])
for i in ans:
print(*i)
``` | output | 1 | 87,947 | 17 | 175,895 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90 | instruction | 0 | 87,948 | 17 | 175,896 |
Tags: implementation
Correct Solution:
```
__author__ = 'PrimuS'
s1 = input().strip("\n")
s2 = input().strip("\n")
n = int(input())
d1 = {}
d2 = {}
for i in range(n):
ss = input().split()
t = int(ss[0])
x = int(ss[2])
ss[3].strip("\n")
if ss[1] == 'h':
if x in d1 and d1[x] == 1:
print(s1, x, t)
d1[x] = 2
elif x not in d1:
if ss[3] == 'r':
print(s1, x,t)
d1[x] = 2
else:
d1[x] = 1
else:
if x in d2 and d2[x] == 1:
print(s2, x, t)
d2[x] = 2
elif x not in d2:
if ss[3] == 'r':
print(s2, x, t)
d2[x] = 2
else:
d2[x] = 1
``` | output | 1 | 87,948 | 17 | 175,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90
Submitted Solution:
```
s1 = input()
s2 = input()
n = int(input())
h = []
a = []
for i in range(n):
k = input().split()
if k[1] == 'h':
if (h.count(k[2]) == 0) and k[3]=='y':
h.append(k[2])
elif h.count(k[2]) == 1:
h.append(k[2])
print(s1,k[2],k[0])
if (h.count(k[2]) == 0) and k[3]=='r':
h.append(k[2])
h.append(k[2])
print(s1,k[2],k[0])
else:
if (a.count(k[2]) == 0) and k[3]=='y':
a.append(k[2])
elif a.count(k[2]) == 1:
a.append(k[2])
print(s2,k[2],k[0])
if (a.count(k[2]) == 0) and k[3]=='r':
a.append(k[2])
a.append(k[2])
print(s2,k[2],k[0])
``` | instruction | 0 | 87,949 | 17 | 175,898 |
Yes | output | 1 | 87,949 | 17 | 175,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90
Submitted Solution:
```
from collections import defaultdict
n1, n2 = input(), input()
n = int(input())
#mp = defaultdict(list)
mp = [[0 for i in range(100)], [0 for i in range(100)]]
for i in range(n):
line = input().split()
t = int(line[0])
m = int(line[2])
if line[1] == 'h':
mp[0][m] += 1
if line[3] == 'r':
if mp[0][m] < 3:
print(n1, m, t)
mp[0][m] = 3
continue
if mp[0][m] == 2:
print(n1, m, t)
mp[0][m] = 3
else:
mp[1][m] += 1
if line[3] == 'r':
if mp[1][m] < 3:
print(n2, m, t)
mp[1][m] = 3
continue
if mp[1][m] == 2:
print(n2, m, t)
mp[1][m] = 3
``` | instruction | 0 | 87,950 | 17 | 175,900 |
Yes | output | 1 | 87,950 | 17 | 175,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90
Submitted Solution:
```
from collections import defaultdict
def main(t1,t2,n,t,ha,m,yr):
reds = {}
yellows = {} # defaultdict(int)
teams = [t1,t2]
for i in range(n):
player = (ha[i], m[i])
if player not in reds:
if player not in yellows:
yellows[player] = 0
if yr[i] == 'y':
yellows[player] += 1
if yellows[player] >= 2 or yr[i] == 'r':
reds[player] = 1
print(teams[ha[i]=='a'],m[i],t[i])
def main_input():
t1 = input()
t2 = input()
n = int(input())
t = list(range(n))
ha = list(range(n))
m = list(range(n))
yr = list(range(n))
for i in range(n):
t[i], ha[i], m[i], yr[i] = input().split()
t = [int(x) for x in t]
m = [int(x) for x in m]
#print(t1,t2,n)
#print(t,ha,m,yr)
main(t1,t2,n,t,ha,m,yr)
if __name__ == "__main__":
main_input()
``` | instruction | 0 | 87,951 | 17 | 175,902 |
Yes | output | 1 | 87,951 | 17 | 175,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90
Submitted Solution:
```
from collections import *
home = input()
away = input()
n = int(input())
hd = defaultdict(int)
ad = defaultdict(int)
for i in range(n):
a = list(map(str, input().split()))
if a[1] == "h":
if hd[a[2]] != -1:
if a[-1] == "r":
hd[a[2]] = -1
print(home, a[2], a[0])
else:
hd[a[2]] += 1
if hd[a[2]] == 2:
hd[a[2]] = -1
print(home, a[2], a[0])
else:
if ad[a[2]] != -1:
if a[-1] == "r":
ad[a[2]] = -1
print(away, a[2], a[0])
else:
ad[a[2]] += 1
if ad[a[2]] == 2:
ad[a[2]] = -1
print(away, a[2], a[0])
``` | instruction | 0 | 87,952 | 17 | 175,904 |
Yes | output | 1 | 87,952 | 17 | 175,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90
Submitted Solution:
```
homename= input()
awayname= input()
fouls= int(input())
a= []
for foul in range(fouls):
print(foul)
print("hello")
data= input().split()
print(data)
if data[3]== "r":
if data[1]== "h":
print(" ".join([homename, data[2], data[0]]))
elif data[1]== "a":
print(" ".join([awayname, data[2], data[0]]))
elif data[3]== "y":
#print("Hell")
for i in range(len(a)):
#print(a[i][1:], data[1:])
if a[i][1:]== data[1:]:
if data[1]== "h":
print(" ".join([homename, data[2], data[0]]))
elif data[1]== "a":
print(" ".join([awayname, data[2], data[0]]))
a.remove(a[i])
break
else:
a.append(data)
#print(a)
``` | instruction | 0 | 87,953 | 17 | 175,906 |
No | output | 1 | 87,953 | 17 | 175,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90
Submitted Solution:
```
hom_tm = input()
awy_tm = input()
fouls = int(input())
home = {}
away = {}
for i in range(fouls):
time,land,player,card = input().split()
if land=='h':
if card=='y':
if player not in home:
home[player] = 1
else:
home[player] += 1
if home[player]==2:
home[player] = 0
print(hom_tm,player,time)
else:
if player in home:
home[player] = 0
print(hom_tm,player,time)
else:
if card=='y':
if player not in away:
away[player] = 1
else:
away[player] += 1
if away[player]==2:
away[player] = 0
print(awy_tm,player,time)
else:
if player in away:
away[player] = 0
print(awy_tm,player,time)
``` | instruction | 0 | 87,954 | 17 | 175,908 |
No | output | 1 | 87,954 | 17 | 175,909 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90
Submitted Solution:
```
team1 = input()
team2 = input()
fouls = int(input())
warn1 = [False] * 100
warn2 = [False] * 100
warns = []
for i in range(fouls):
a, b, c, d = map(str, input().split(' '))
a, c = int(a), int(c)
if b == 'a' and d == 'r' and warn2[c] == False:
warns.append([team2, c, a])
warn2[c] = True
elif b == 'h' and d == 'r' and warn1[c] == False:
warns.append([team1, c, a])
warn1[c] = True
elif b == 'a' and d == 'y' and warn2[c]:
warns.append([team2, c, a])
elif b == 'a' and d == 'y' and not warn2[c]:
warn2[c] = True
elif b == 'h' and d == 'y' and warn1[c]:
warns.append([team1, c, a])
elif b == 'h' and d == 'y' and not warn1[c]:
warn1[c] = True
warns.sort(key = lambda x:x[2])
for a in warns:
print(' '.join([str(x) for x in a]))
``` | instruction | 0 | 87,955 | 17 | 175,910 |
No | output | 1 | 87,955 | 17 | 175,911 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only the first moment of time when he would receive a red card from Vasya.
Input
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.
Next follows number n (1 ≤ n ≤ 90) — the number of fouls.
Each of the following n lines contains information about a foul in the following form:
* first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
* then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
* then goes the player's number m (1 ≤ m ≤ 99);
* then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
Output
For each event when a player received his first red card in a chronological order print a string containing the following information:
* The name of the team to which the player belongs;
* the player's number in his team;
* the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
Examples
Input
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
Output
MC 25 70
MC 42 82
CSKA 13 90
Submitted Solution:
```
import sys
hom = sys.stdin.readline()[:-1]
away = sys.stdin.readline()[:-1]
numcases = int(sys.stdin.readline())
ht=[0]*100
at=[0]*100
def solve(a):
p=int(a.split()[2])
t=int(a.split()[0])
ah = a.split()[1]
if ah=='h':
n = hom
if a.split()[3]=='y':
ht[p]+=1
else:
ht[p]+=2
if ht[p]==2:
x =[(t,n,p)]
return x
else:
n = away
if a.split()[3]=='y':
at[p]+=1
else:
at[p]+=2
if at[p]==2:
x =[(t,n,p)]
return x
r=[]
for casenum in range(1,numcases+1):
x =solve(sys.stdin.readline())
if x!= None:
r= r+ x
for i in list(sorted(r)):
print (i[1], i[2],i[0])
``` | instruction | 0 | 87,956 | 17 | 175,912 |
No | output | 1 | 87,956 | 17 | 175,913 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position. | instruction | 0 | 88,372 | 17 | 176,744 |
Tags: greedy
Correct Solution:
```
I=lambda:map(int,input().split())
n,d=I()
r,p=list(I()),list(I())
q,w,e,t=r[d-1]+p[0],d-2,1,0
while w>=0 and e<n:
if r[w]+p[e]<=q:w-=1;t+=1
e+=1
print(d-t)
``` | output | 1 | 88,372 | 17 | 176,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position. | instruction | 0 | 88,373 | 17 | 176,746 |
Tags: greedy
Correct Solution:
```
n,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
x=k
j=n-1
love=a[k-1]+b[0]
for i in range(k-1):
if love>=(a[i]+b[j]):
x-=1
j-=1
print(x)
``` | output | 1 | 88,373 | 17 | 176,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position. | instruction | 0 | 88,374 | 17 | 176,748 |
Tags: greedy
Correct Solution:
```
n,d=map(int,input().split())
s=[int(x) for x in input().split()]
p=[int(X) for X in input().split()]
z=[]
tt=s[d-1]+p[0]
z.append(s[d-1]+p[0])
p[0]=-1
s[d-1]=-1
i=0
j=1
v=n-1
s.sort(reverse=True)
p.sort()
while (i<n-1 and j<=v ):
if (s[i]+p[j])<=tt:
z.append(s[i]+p[j])
i+=1
j+=1
else:
z.append(s[i]+p[v])
i+=1
v-=1
z.sort(reverse=True)
print(z.index(tt)+1)
``` | output | 1 | 88,374 | 17 | 176,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position. | instruction | 0 | 88,375 | 17 | 176,750 |
Tags: greedy
Correct Solution:
```
a = input().split()
a = [int(i) for i in a]
b = input().split()
b = [int(i) for i in b]
c = input().split()
c = [int(i) for i in c]
cha = []
for i in range(0, a[1] - 1):
cha.append(b[i] -b[a[1] - 1])
cao = []
for i in range(1, a[1]):
cao.append(c[0] - c[-i])
dui = 0
cha.sort()
cao.sort()
i = 0
j = 0
while(j < a[1] - 1):
if cha[i] <= cao[j]:
dui += 1
i += 1
j += 1
else:
j += 1
print (a[1] - dui)
``` | output | 1 | 88,375 | 17 | 176,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position. | instruction | 0 | 88,376 | 17 | 176,752 |
Tags: greedy
Correct Solution:
```
n,d = map(int,input().split())
s = list(map(int,input().split()))
p = list(map(int,input().split()))
maxPoints = s[d-1]+p[0]
pos = n-1
points = 1
count = 0
while (pos>=0 and points<n):
if (pos == d-1):
pos -= 1
else:
if (s[pos]+p[points]<=maxPoints):
count += 1
points += 1
pos -= 1
else:
points += 1
print(n-count)
``` | output | 1 | 88,376 | 17 | 176,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position. | instruction | 0 | 88,377 | 17 | 176,754 |
Tags: greedy
Correct Solution:
```
n,d=map(int,input().split())
s=[*map(int,input().split())]
p=[*map(int,input().split())]
p=p[::-1]
item=s[d-1]+p[-1]
p.pop()
res=1
from bisect import bisect as bis
for i,x in enumerate(s):
if i!=d-1:
diff=item-x
if diff<0:res+=1;p.pop()
else:
a=bis(p,diff)
if a==0:p.pop();res+=1
else:p.pop(a-1)
print(res)
``` | output | 1 | 88,377 | 17 | 176,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position. | instruction | 0 | 88,378 | 17 | 176,756 |
Tags: greedy
Correct Solution:
```
from bisect import bisect_right
n,d = map(int,input().split())
s = list(map(int,input().split()))
p = list(map(int,input().split()))
t_wyn = []
wyn = s[d-1]+p[0]
t_wyn.append(wyn)
del s[d-1]
s_sort = sorted(s)
p_sort = sorted(p)
p_sort.pop()
for x in range(0,len(s),+1):
szuk = wyn-s_sort[x]-1
if szuk < 0:
t_wyn.append(s_sort[x]+p_sort[-1])
else:
pos = bisect_right(p_sort,szuk)
if pos > 0:
t_wyn.append(s_sort[x] + p_sort[pos-1])
del p_sort[pos-1]
continue
else:
t_wyn.append(s_sort[x] + p_sort[-1])
p_sort.pop()
t_wyn = sorted(t_wyn)
pos = bisect_right(t_wyn,wyn)
print(n-(pos-1))
``` | output | 1 | 88,378 | 17 | 176,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position. | instruction | 0 | 88,379 | 17 | 176,758 |
Tags: greedy
Correct Solution:
```
from bisect import bisect_right
num_racers, selected_racer = map(int, input().split())
selected_racer -= 1
racers_points = list(map(int, input().split()))
selected_racer_points = racers_points[selected_racer]
awards = list(map(int, input().split()))
awards.sort(reverse=True)
selected_racer_points += awards[0]
remaining_awards = awards[1:]
del racers_points[selected_racer]
remaining_racers_points = sorted(racers_points, reverse=True)
best_award_pos = 0
worst_award_pos = len(remaining_awards) - 1
for pos, racer_points in enumerate(remaining_racers_points):
if racer_points + remaining_awards[worst_award_pos] <= selected_racer_points:
remaining_racers_points[pos] += remaining_awards[worst_award_pos]
worst_award_pos -= 1
else:
remaining_racers_points[pos] += remaining_awards[best_award_pos]
best_award_pos += 1
remaining_racers_points.append(selected_racer_points)
remaining_racers_points.sort()
print(num_racers + 1 - bisect_right(remaining_racers_points, selected_racer_points))
``` | output | 1 | 88,379 | 17 | 176,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position.
Submitted Solution:
```
from os import path
import sys,time
# mod = int(1e9 + 7)
# import re
from math import ceil, floor,gcd,log,log2 ,factorial
from collections import *
# from bisect import *
maxx = float('inf')
#----------------------------INPUT FUNCTIONS------------------------------------------#
I = lambda :int(sys.stdin.buffer.readline())
tup= lambda : map(int , sys.stdin.buffer.readline().split())
lint = lambda :[int(x) for x in sys.stdin.buffer.readline().split()]
S = lambda: sys.stdin.readline().replace('\n', '').strip()
def grid(r, c): return [lint() for i in range(r)]
stpr = lambda x : sys.stdout.write(f'{x}' + '\n')
star = lambda x: print(' '.join(map(str, x)))
localsys = 0
start_time = time.time()
if (path.exists('input.txt')):
sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
#left shift --- num*(2**k) --(k - shift)
n , p = tup()
a = lint()
b= lint()
x = a[p-1] + b[0]
for i in range(p-1):
if a[i] + b[-1] <= x:
b.pop()
p-=1
print(p)
if localsys:
print("\n\nTime Elased :",time.time() - start_time,"seconds")
``` | instruction | 0 | 88,380 | 17 | 176,760 |
Yes | output | 1 | 88,380 | 17 | 176,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position.
Submitted Solution:
```
from collections import Counter
import string
import math
import sys
from fractions import Fraction
def array_int():
return [int(i) for i in sys.stdin.readline().split()]
def vary(arrber_of_variables):
if arrber_of_variables==1:
return int(sys.stdin.readline())
if arrber_of_variables>=2:
return map(int,sys.stdin.readline().split())
def makedict(var):
return dict(Counter(var))
# i am noob wanted to be better and trying hard for that
def printDivisors(n):
divisors=[]
# Note that this loop runs till square root
i = 1
while i <= math.sqrt(n):
if (n % i == 0) :
# If divisors are equal, print only one
if (n//i == i) :
divisors.append(i)
else :
# Otherwise print both
divisors.extend((i,n//i))
i = i + 1
return divisors
def countTotalBits(num):
# convert number into it's binary and
# remove first two characters 0b.
binary = bin(num)[2:]
return(len(binary))
def isPrime(n) :
# Corner cases
if (n <= 1) :
return False
if (n <= 3) :
return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
# print(math.gcd(3,2))
""" def dfs(node,val):
global tree,visited
visited[node]=1
ans[node]=val
val^=1
for i in tree[node]:
if visited[i]==-1:
dfs(i,val) """
n,d=vary(2)
intial=array_int()
winnig=array_int()
intial[d-1]=intial[d-1]+winnig[0]
j=n-1
count=0
for i in range(d-1):
if intial[i]+winnig[j]<=intial[d-1]:
count+=1
j-=1
print(d-count)
``` | instruction | 0 | 88,381 | 17 | 176,762 |
Yes | output | 1 | 88,381 | 17 | 176,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position.
Submitted Solution:
```
n,d=[int(x) for x in input().split()]
sk=[int(x) for x in input().split()]
pk=[int(x) for x in input().split()]
maxm=sk[d-1]+max(pk)
dip=0
pkk=1
for k in range(0,d):
if(sk[k]+pk[-pkk]<=maxm):
dip+=1
pkk+=1
print(d-dip+1)
``` | instruction | 0 | 88,382 | 17 | 176,764 |
Yes | output | 1 | 88,382 | 17 | 176,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position.
Submitted Solution:
```
import sys
input = sys.stdin.readline
'''
'''
def solve(n, d, cur, race):
if d == 0:
return 1
#result = [0] * (d+1)
pts = cur[d] + race[0]
#result[d] = pts
comp_index = d - 1
point_index = 1
while comp_index != -1:
comp_pts = cur[comp_index]
while point_index < n and comp_pts + race[point_index] > pts:
point_index += 1
if point_index == n:
return comp_index+2
else:
point_index += 1
#if point_index < n:
#print(comp_index+1, race[point_index])
comp_index -= 1
return 1
n, d = map(int, input().split())
d -= 1
cur = list(map(int,input().split()))
race = list(map(int, input().split()))
print(solve(n, d, cur, race))
``` | instruction | 0 | 88,383 | 17 | 176,766 |
Yes | output | 1 | 88,383 | 17 | 176,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position.
Submitted Solution:
```
n,d = map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
ps=[]
cs = a[d-1] +b[0]
b=b[1:]
f=0
a.pop(d-1)
src=[]
j=n-2
i=0
for k in range(n-1):
if a[k]>=cs :
src.append(a[k]+b[i])
i+=1
else:
src.append(a[k]+b[j])
j-=1
src.sort(reverse=True)
for j in range(n-1):
if cs>=src[j]:
print(j+1)
f=1
break
if f==0:
ans1=n
``` | instruction | 0 | 88,384 | 17 | 176,768 |
No | output | 1 | 88,384 | 17 | 176,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position.
Submitted Solution:
```
l1=input().split(' ')
l2=input().split(' ')
l3=input().split(' ')
n=int(l1[0])
p=int(l1[1])
pt1=int(l2[p-1])
mpt=pt1+int(l3[0])
c=0
for i in l2:
if int(i)>mpt:
c+=1
l=c
l4=[]
for i in range(c,p):
k=1
for j in range(1,n):
if j not in l4:
t=int(l2[i])+int(l3[j])
if t<mpt :
k=0
l4.append(j)
break
if t==mpt:
v=j
k=2
if k==2:
l4.append(v)
elif k==1:
l+=1
print (l+1)
``` | instruction | 0 | 88,385 | 17 | 176,770 |
No | output | 1 | 88,385 | 17 | 176,771 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position.
Submitted Solution:
```
import sys
def get_ints(): return map(int, sys.stdin.readline().strip().split())
n,d=get_ints()
s=list(map(int,input().split()))
p=list(map(int,input().split()))
m=s.pop(d-1)
a=p.pop(0)
p.reverse()
for i in range(n-1):
s[i]+=p[i]
s.append(m+a)
s.sort()
#print(s,m+a)
print(s.index(m+a)+1)
``` | instruction | 0 | 88,386 | 17 | 176,772 |
No | output | 1 | 88,386 | 17 | 176,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Formula 1 officials decided to introduce new competition. Cars are replaced by space ships and number of points awarded can differ per race.
Given the current ranking in the competition and points distribution for the next race, your task is to calculate the best possible ranking for a given astronaut after the next race. It's guaranteed that given astronaut will have unique number of points before the race.
Input
The first line contains two integer numbers N (1 ≤ N ≤ 200000) representing number of F1 astronauts, and current position of astronaut D (1 ≤ D ≤ N) you want to calculate best ranking for (no other competitor will have the same number of points before the race).
The second line contains N integer numbers S_k (0 ≤ S_k ≤ 10^8, k=1...N), separated by a single space, representing current ranking of astronauts. Points are sorted in non-increasing order.
The third line contains N integer numbers P_k (0 ≤ P_k ≤ 10^8, k=1...N), separated by a single space, representing point awards for the next race. Points are sorted in non-increasing order, so winner of the race gets the maximum number of points.
Output
Output contains one integer number — the best possible ranking for astronaut after the race. If multiple astronauts have the same score after the race, they all share the best ranking.
Example
Input
4 3
50 30 20 10
15 10 7 3
Output
2
Note
If the third ranked astronaut wins the race, he will have 35 points. He cannot take the leading position, but he can overtake the second position if the second ranked astronaut finishes the race at the last position.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n, d = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
score = a[d-1] + b[0]
current_pos = 1
j = n - 1
k = 1
i = 0
while i != d:
if a[i] > score:
current_pos += 1
k += 1
i += 1
else:
if a[i] + b[j] > score:
current_pos += 1
break
elif a[i] + b[j] <= score:
j -= 1
i += 1
print(current_pos)
``` | instruction | 0 | 88,387 | 17 | 176,774 |
No | output | 1 | 88,387 | 17 | 176,775 |
Provide tags and a correct Python 3 solution for this coding contest problem.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) — the first boy enters two pairs;
* (1, 2) and (2, 2) — the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 ≤ a, b, k ≤ 2 ⋅ 10^5) — the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, … a_k. (1 ≤ a_i ≤ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, … b_k. (1 ≤ b_i ≤ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 ⋅ 10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer — the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1). | instruction | 0 | 88,623 | 17 | 177,246 |
Tags: combinatorics, graphs, math
Correct Solution:
```
#pyrival orz
import os
import sys
import math
from io import BytesIO, IOBase
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
############ ---- Dijkstra with path ---- ############
def dijkstra(start, distance, path, n):
# requires n == number of vertices in graph,
# adj == adjacency list with weight of graph
visited = [False for _ in range(n)] # To keep track of vertices that are visited
distance[start] = 0 # distance of start node from itself is 0
for i in range(n):
v = -1 # Initialize v == vertex from which its neighboring vertices' distance will be calculated
for j in range(n):
# If it has not been visited and has the lowest distance from start
if not visited[v] and (v == -1 or distance[j] < distance[v]):
v = j
if distance[v] == math.inf:
break
visited[v] = True # Mark as visited
for edge in adj[v]:
destination = edge[0] # Neighbor of the vertex
weight = edge[1] # Its corresponding weight
if distance[v] + weight < distance[destination]: # If its distance is less than the stored distance
distance[destination] = distance[v] + weight # Update the distance
path[destination] = v # Update the path
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a%b)
def lcm(a, b):
return (a*b)//gcd(a, b)
def ncr(n, r):
return math.factorial(n)//(math.factorial(n-r)*math.factorial(r))
def npr(n, r):
return math.factorial(n)//math.factorial(n-r)
def seive(n):
primes = [True]*(n+1)
ans = []
for i in range(2, n):
if not primes[i]:
continue
j = 2*i
while j <= n:
primes[j] = False
j += i
for p in range(2, n+1):
if primes[p]:
ans += [p]
return ans
def factors(n):
factors = []
x = 1
while x*x <= n:
if n % x == 0:
if n // x == x:
factors.append(x)
else:
factors.append(x)
factors.append(n//x)
x += 1
return factors
# Functions: list of factors, seive of primes, gcd of two numbers,
# lcm of two numbers, npr, ncr
def main():
try:
for _ in range(inp()):
la, lb, k = invr()
a = inlt()
b = inlt()
da = {}
db = {}
for i in range(k):
if a[i] not in da:
da[a[i]] = 0
if b[i] not in db:
db[b[i]] = 0
da[a[i]] += 1
db[b[i]] += 1
ans = 0
for i in range(k):
ans += k - da[a[i]] - db[b[i]] + 1 - i
da[a[i]] -= 1
db[b[i]] -= 1
print(ans)
except Exception as e:
print(e)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | output | 1 | 88,623 | 17 | 177,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) — the first boy enters two pairs;
* (1, 2) and (2, 2) — the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 ≤ a, b, k ≤ 2 ⋅ 10^5) — the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, … a_k. (1 ≤ a_i ≤ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, … b_k. (1 ≤ b_i ≤ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 ⋅ 10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer — the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1). | instruction | 0 | 88,624 | 17 | 177,248 |
Tags: combinatorics, graphs, math
Correct Solution:
```
import sys
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from bisect import insort
from collections import Counter
from collections import deque
from heapq import heappush,heappop,heapify
from itertools import permutations,combinations
from itertools import accumulate as ac
mod = int(1e9)+7
#mod = 998244353
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin.readline().rstrip()
out = lambda x : stdout.write(str(x)+"\n")
t = ip()
for _ in range(t):
a,b,k = inp()
x = list(inp())
y = list(inp())
dica = Counter()
dicb = Counter()
ch = dd(int)
ans = (k*(k-1))//2
for i in range(k):
if i == 0:
dica[x[i]] += 1
dicb[y[i]] += 1
ch[(x[i],y[i])] += 1
else:
xx = x[i]
yy = y[i]
cal = dica[xx]
cal += dicb[yy]
cal -= ch[(xx,yy)]
ans -= cal
dica[xx] += 1
dicb[yy] += 1
ch[(xx,yy)] += 1
print(ans)
``` | output | 1 | 88,624 | 17 | 177,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) — the first boy enters two pairs;
* (1, 2) and (2, 2) — the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 ≤ a, b, k ≤ 2 ⋅ 10^5) — the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, … a_k. (1 ≤ a_i ≤ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, … b_k. (1 ≤ b_i ≤ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 ⋅ 10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer — the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1). | instruction | 0 | 88,625 | 17 | 177,250 |
Tags: combinatorics, graphs, math
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------------------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=-10**6, func=lambda a, b: max(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=10**6, func=lambda a, b: min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
prime = [True for i in range(50001)]
pp=[]
def SieveOfEratosthenes(n=50000):
# Create a boolean array "prime[0..n]" and initialize
# all entries it as true. A value in prime[i] will
# finally be false if i is Not a prime, else true.
p = 2
while (p * p <= n):
# If prime[p] is not changed, then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for i in range(50001):
if prime[i]:
pp.append(i)
#---------------------------------running code------------------------------------------
for _ in range (int(input())):
x,y,k=map(int,input().split())
d1=defaultdict(int)
d2=defaultdict(int)
s=[]
a=list(map(int,input().split()))
b=list(map(int,input().split()))
for i in range (k):
s.append((a[i],b[i]))
d1[a[i]]+=1
d2[b[i]]+=1
res=0
for i in s:
res+=k-d1[i[0]]-d2[i[1]]+1
res//=2
print(res)
``` | output | 1 | 88,625 | 17 | 177,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
At the school where Vasya is studying, preparations are underway for the graduation ceremony. One of the planned performances is a ball, which will be attended by pairs of boys and girls.
Each class must present two couples to the ball. In Vasya's class, a boys and b girls wish to participate. But not all boys and not all girls are ready to dance in pairs.
Formally, you know k possible one-boy-one-girl pairs. You need to choose two of these pairs so that no person is in more than one pair.
For example, if a=3, b=4, k=4 and the couples (1, 2), (1, 3), (2, 2), (3, 4) are ready to dance together (in each pair, the boy's number comes first, then the girl's number), then the following combinations of two pairs are possible (not all possible options are listed below):
* (1, 3) and (2, 2);
* (3, 4) and (1, 3);
But the following combinations are not possible:
* (1, 3) and (1, 2) — the first boy enters two pairs;
* (1, 2) and (2, 2) — the second girl enters two pairs;
Find the number of ways to select two pairs that match the condition above. Two ways are considered different if they consist of different pairs.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains three integers a, b and k (1 ≤ a, b, k ≤ 2 ⋅ 10^5) — the number of boys and girls in the class and the number of couples ready to dance together.
The second line of each test case contains k integers a_1, a_2, … a_k. (1 ≤ a_i ≤ a), where a_i is the number of the boy in the pair with the number i.
The third line of each test case contains k integers b_1, b_2, … b_k. (1 ≤ b_i ≤ b), where b_i is the number of the girl in the pair with the number i.
It is guaranteed that the sums of a, b, and k over all test cases do not exceed 2 ⋅ 10^5.
It is guaranteed that each pair is specified at most once in one test case.
Output
For each test case, on a separate line print one integer — the number of ways to choose two pairs that match the condition above.
Example
Input
3
3 4 4
1 1 2 3
2 3 2 4
1 1 1
1
1
2 2 4
1 1 2 2
1 2 1 2
Output
4
0
2
Note
In the first test case, the following combinations of pairs fit:
* (1, 2) and (3, 4);
* (1, 3) and (2, 2);
* (1, 3) and (3, 4);
* (2, 2) and (3, 4).
There is only one pair in the second test case.
In the third test case, the following combinations of pairs fit:
* (1, 1) and (2, 2);
* (1, 2) and (2, 1). | instruction | 0 | 88,626 | 17 | 177,252 |
Tags: combinatorics, graphs, math
Correct Solution:
```
for tests in range(int(input())):
a,b,k=map(int,input().split())
la=list(map(int,input().split()))
lb=list(map(int,input().split()))
d={}
g={}
for i in range(k):
if d.get(la[i])==None:
d[la[i]]=1
else:
d[la[i]]+=1
if g.get(lb[i])==None:
g[lb[i]]=1
else:
g[lb[i]]+=1
sum=0
for i in range(k):
x=la[i]
y=lb[i]
sum+=k-i-d[x]-g[y]+1
d[x]-=1
g[y]-=1
print(sum)
``` | output | 1 | 88,626 | 17 | 177,253 |
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