message stringlengths 2 20.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 1.95k 109k | cluster float64 17 17 | __index_level_0__ int64 3.91k 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Anton likes to play chess, and so does his friend Danik.
Once they have played n games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of games played.
The second line contains a string s, consisting of n uppercase English letters 'A' and 'D' — the outcome of each of the games. The i-th character of the string is equal to 'A' if the Anton won the i-th game and 'D' if Danik won the i-th game.
Output
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Examples
Input
6
ADAAAA
Output
Anton
Input
7
DDDAADA
Output
Danik
Input
6
DADADA
Output
Friendship
Note
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship". | instruction | 0 | 73,798 | 17 | 147,596 |
Tags: implementation, strings
Correct Solution:
```
n = int(input())
line = input()
Anton = 0
Danik = 0
for x in line:
if x == "A":
Anton = Anton + 1
else:
Danik = Danik + 1
if Anton == Danik:
print("Friendship")
elif Anton > Danik:
print("Anton")
else:
print("Danik")
``` | output | 1 | 73,798 | 17 | 147,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Anton likes to play chess, and so does his friend Danik.
Once they have played n games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of games played.
The second line contains a string s, consisting of n uppercase English letters 'A' and 'D' — the outcome of each of the games. The i-th character of the string is equal to 'A' if the Anton won the i-th game and 'D' if Danik won the i-th game.
Output
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Examples
Input
6
ADAAAA
Output
Anton
Input
7
DDDAADA
Output
Danik
Input
6
DADADA
Output
Friendship
Note
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".
Submitted Solution:
```
n = int(input())
res = input()
a = res.count('A')
d = res.count('D')
if a > d:
print('Anton')
elif d > a:
print('Danik')
else:
print('Friendship')
``` | instruction | 0 | 73,800 | 17 | 147,600 |
Yes | output | 1 | 73,800 | 17 | 147,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Anton likes to play chess, and so does his friend Danik.
Once they have played n games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of games played.
The second line contains a string s, consisting of n uppercase English letters 'A' and 'D' — the outcome of each of the games. The i-th character of the string is equal to 'A' if the Anton won the i-th game and 'D' if Danik won the i-th game.
Output
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Examples
Input
6
ADAAAA
Output
Anton
Input
7
DDDAADA
Output
Danik
Input
6
DADADA
Output
Friendship
Note
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".
Submitted Solution:
```
games = int(input())
seq = list(input())
Ds = seq.count("D")
As = seq.count("A")
if As>Ds:
print("Anton")
elif Ds >As:
print("Danik")
else:
print("Friendship")
``` | instruction | 0 | 73,801 | 17 | 147,602 |
Yes | output | 1 | 73,801 | 17 | 147,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Anton likes to play chess, and so does his friend Danik.
Once they have played n games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of games played.
The second line contains a string s, consisting of n uppercase English letters 'A' and 'D' — the outcome of each of the games. The i-th character of the string is equal to 'A' if the Anton won the i-th game and 'D' if Danik won the i-th game.
Output
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Examples
Input
6
ADAAAA
Output
Anton
Input
7
DDDAADA
Output
Danik
Input
6
DADADA
Output
Friendship
Note
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".
Submitted Solution:
```
n = int(input())
result = input()
an = result.count('A')
bor = result.count('D')
if an > bor:
print('Anton')
elif bor > an:
print('Danik')
else:
print('Friendship')
``` | instruction | 0 | 73,802 | 17 | 147,604 |
Yes | output | 1 | 73,802 | 17 | 147,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Anton likes to play chess, and so does his friend Danik.
Once they have played n games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of games played.
The second line contains a string s, consisting of n uppercase English letters 'A' and 'D' — the outcome of each of the games. The i-th character of the string is equal to 'A' if the Anton won the i-th game and 'D' if Danik won the i-th game.
Output
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Examples
Input
6
ADAAAA
Output
Anton
Input
7
DDDAADA
Output
Danik
Input
6
DADADA
Output
Friendship
Note
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".
Submitted Solution:
```
n=int(input())
s=str(input())
if(s.count("D")==s.count("A")):
print('Friendship')
elif(s.count("D")>s.count("A")):
print('Danik')
else:
print('Anton')
``` | instruction | 0 | 73,803 | 17 | 147,606 |
Yes | output | 1 | 73,803 | 17 | 147,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Anton likes to play chess, and so does his friend Danik.
Once they have played n games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of games played.
The second line contains a string s, consisting of n uppercase English letters 'A' and 'D' — the outcome of each of the games. The i-th character of the string is equal to 'A' if the Anton won the i-th game and 'D' if Danik won the i-th game.
Output
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Examples
Input
6
ADAAAA
Output
Anton
Input
7
DDDAADA
Output
Danik
Input
6
DADADA
Output
Friendship
Note
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".
Submitted Solution:
```
n=int(input("enter n"))
A=input()
countA=0
countD=0
for i in A:
if(i=="A"):
countA+=1
elif(i=="D"):
countD+=1
if(countA>countD):
print("Anton")
elif(countD>countA):
print("Danik")
else:
print("Friendship")
``` | instruction | 0 | 73,804 | 17 | 147,608 |
No | output | 1 | 73,804 | 17 | 147,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Anton likes to play chess, and so does his friend Danik.
Once they have played n games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of games played.
The second line contains a string s, consisting of n uppercase English letters 'A' and 'D' — the outcome of each of the games. The i-th character of the string is equal to 'A' if the Anton won the i-th game and 'D' if Danik won the i-th game.
Output
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Examples
Input
6
ADAAAA
Output
Anton
Input
7
DDDAADA
Output
Danik
Input
6
DADADA
Output
Friendship
Note
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".
Submitted Solution:
```
n=int(input("enter n"))
A=input().count("A")
if(n<(A*2)):
print("Danik")
elif(n>(A*2)):
print("Anton")
else:
print("Friendship")
``` | instruction | 0 | 73,805 | 17 | 147,610 |
No | output | 1 | 73,805 | 17 | 147,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Anton likes to play chess, and so does his friend Danik.
Once they have played n games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of games played.
The second line contains a string s, consisting of n uppercase English letters 'A' and 'D' — the outcome of each of the games. The i-th character of the string is equal to 'A' if the Anton won the i-th game and 'D' if Danik won the i-th game.
Output
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Examples
Input
6
ADAAAA
Output
Anton
Input
7
DDDAADA
Output
Danik
Input
6
DADADA
Output
Friendship
Note
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".
Submitted Solution:
```
a=int(input())
b=input()
if(b.count('A')>b.count('D')):
print("Anton")
elif(b.count('D')<b.count('A')):
print("Danik")
else:
print("Friendship")
``` | instruction | 0 | 73,806 | 17 | 147,612 |
No | output | 1 | 73,806 | 17 | 147,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Anton likes to play chess, and so does his friend Danik.
Once they have played n games in a row. For each game it's known who was the winner — Anton or Danik. None of the games ended with a tie.
Now Anton wonders, who won more games, he or Danik? Help him determine this.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of games played.
The second line contains a string s, consisting of n uppercase English letters 'A' and 'D' — the outcome of each of the games. The i-th character of the string is equal to 'A' if the Anton won the i-th game and 'D' if Danik won the i-th game.
Output
If Anton won more games than Danik, print "Anton" (without quotes) in the only line of the output.
If Danik won more games than Anton, print "Danik" (without quotes) in the only line of the output.
If Anton and Danik won the same number of games, print "Friendship" (without quotes).
Examples
Input
6
ADAAAA
Output
Anton
Input
7
DDDAADA
Output
Danik
Input
6
DADADA
Output
Friendship
Note
In the first sample, Anton won 6 games, while Danik — only 1. Hence, the answer is "Anton".
In the second sample, Anton won 3 games and Danik won 4 games, so the answer is "Danik".
In the third sample, both Anton and Danik won 3 games and the answer is "Friendship".
Submitted Solution:
```
n=int(input())
m=input()
t1=['Anton']
t2=['Damik']
t3=['Friendship']
if m.count('A')>m.count('D'):
for arip in t1:
print(arip,end='')
elif m.count('A')<m.count('D'):
for arip in t2:
print(arip,end='')
else:
for arip in t3:
print(arip,end='')
``` | instruction | 0 | 73,807 | 17 | 147,614 |
No | output | 1 | 73,807 | 17 | 147,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A popular reality show is recruiting a new cast for the third season! n candidates numbered from 1 to n have been interviewed. The candidate i has aggressiveness level l_i, and recruiting this candidate will cost the show s_i roubles.
The show host reviewes applications of all candidates from i=1 to i=n by increasing of their indices, and for each of them she decides whether to recruit this candidate or not. If aggressiveness level of the candidate i is strictly higher than that of any already accepted candidates, then the candidate i will definitely be rejected. Otherwise the host may accept or reject this candidate at her own discretion. The host wants to choose the cast so that to maximize the total profit.
The show makes revenue as follows. For each aggressiveness level v a corresponding profitability value c_v is specified, which can be positive as well as negative. All recruited participants enter the stage one by one by increasing of their indices. When the participant i enters the stage, events proceed as follows:
* The show makes c_{l_i} roubles, where l_i is initial aggressiveness level of the participant i.
* If there are two participants with the same aggressiveness level on stage, they immediately start a fight. The outcome of this is:
* the defeated participant is hospitalized and leaves the show.
* aggressiveness level of the victorious participant is increased by one, and the show makes c_t roubles, where t is the new aggressiveness level.
* The fights continue until all participants on stage have distinct aggressiveness levels.
It is allowed to select an empty set of participants (to choose neither of the candidates).
The host wants to recruit the cast so that the total profit is maximized. The profit is calculated as the total revenue from the events on stage, less the total expenses to recruit all accepted participants (that is, their total s_i). Help the host to make the show as profitable as possible.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2000) — the number of candidates and an upper bound for initial aggressiveness levels.
The second line contains n integers l_i (1 ≤ l_i ≤ m) — initial aggressiveness levels of all candidates.
The third line contains n integers s_i (0 ≤ s_i ≤ 5000) — the costs (in roubles) to recruit each of the candidates.
The fourth line contains n + m integers c_i (|c_i| ≤ 5000) — profitability for each aggrressiveness level.
It is guaranteed that aggressiveness level of any participant can never exceed n + m under given conditions.
Output
Print a single integer — the largest profit of the show.
Examples
Input
5 4
4 3 1 2 1
1 2 1 2 1
1 2 3 4 5 6 7 8 9
Output
6
Input
2 2
1 2
0 0
2 1 -100 -100
Output
2
Input
5 4
4 3 2 1 1
0 2 6 7 4
12 12 12 6 -3 -5 3 10 -4
Output
62
Note
In the first sample case it is optimal to recruit candidates 1, 2, 3, 5. Then the show will pay 1 + 2 + 1 + 1 = 5 roubles for recruitment. The events on stage will proceed as follows:
* a participant with aggressiveness level 4 enters the stage, the show makes 4 roubles;
* a participant with aggressiveness level 3 enters the stage, the show makes 3 roubles;
* a participant with aggressiveness level 1 enters the stage, the show makes 1 rouble;
* a participant with aggressiveness level 1 enters the stage, the show makes 1 roubles, a fight starts. One of the participants leaves, the other one increases his aggressiveness level to 2. The show will make extra 2 roubles for this.
Total revenue of the show will be 4 + 3 + 1 + 1 + 2=11 roubles, and the profit is 11 - 5 = 6 roubles.
In the second sample case it is impossible to recruit both candidates since the second one has higher aggressiveness, thus it is better to recruit the candidate 1. | instruction | 0 | 75,159 | 17 | 150,318 |
Tags: bitmasks, dp
Correct Solution:
```
import sys
input = sys.stdin.readline
n,m=map(int,input().split())
A=list(map(int,input().split()))
C=list(map(int,input().split()))
P=list(map(int,input().split()))
DP=[[-1<<30]*(n+1) for i in range(5001)]
# DP[k][cnt] = Aのmaxがkで, そういう人間がcnt人いるときのprofitの最大値
for i in range(5001):
DP[i][0]=0
for i in range(n-1,-1,-1):
a,c = A[i]-1,C[i]
for j in range(n,-1,-1):
if DP[a][j]==-1<<30:
continue
if DP[a][j] - c + P[a] > DP[a][j+1]:
DP[a][j+1] = DP[a][j] - c + P[a]
x, w=a, j+1
while x+1<n+m:
if DP[x+1][w//2] < DP[x][w] + w//2 * P[x+1]:
DP[x+1][w//2] = DP[x][w] + w//2 * P[x+1]
x,w=x+1,w//2
else:
break
ANS=0
for i in range(5001):
ANS=max(ANS,DP[i][0],DP[i][1])
print(ANS)
``` | output | 1 | 75,159 | 17 | 150,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A popular reality show is recruiting a new cast for the third season! n candidates numbered from 1 to n have been interviewed. The candidate i has aggressiveness level l_i, and recruiting this candidate will cost the show s_i roubles.
The show host reviewes applications of all candidates from i=1 to i=n by increasing of their indices, and for each of them she decides whether to recruit this candidate or not. If aggressiveness level of the candidate i is strictly higher than that of any already accepted candidates, then the candidate i will definitely be rejected. Otherwise the host may accept or reject this candidate at her own discretion. The host wants to choose the cast so that to maximize the total profit.
The show makes revenue as follows. For each aggressiveness level v a corresponding profitability value c_v is specified, which can be positive as well as negative. All recruited participants enter the stage one by one by increasing of their indices. When the participant i enters the stage, events proceed as follows:
* The show makes c_{l_i} roubles, where l_i is initial aggressiveness level of the participant i.
* If there are two participants with the same aggressiveness level on stage, they immediately start a fight. The outcome of this is:
* the defeated participant is hospitalized and leaves the show.
* aggressiveness level of the victorious participant is increased by one, and the show makes c_t roubles, where t is the new aggressiveness level.
* The fights continue until all participants on stage have distinct aggressiveness levels.
It is allowed to select an empty set of participants (to choose neither of the candidates).
The host wants to recruit the cast so that the total profit is maximized. The profit is calculated as the total revenue from the events on stage, less the total expenses to recruit all accepted participants (that is, their total s_i). Help the host to make the show as profitable as possible.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2000) — the number of candidates and an upper bound for initial aggressiveness levels.
The second line contains n integers l_i (1 ≤ l_i ≤ m) — initial aggressiveness levels of all candidates.
The third line contains n integers s_i (0 ≤ s_i ≤ 5000) — the costs (in roubles) to recruit each of the candidates.
The fourth line contains n + m integers c_i (|c_i| ≤ 5000) — profitability for each aggrressiveness level.
It is guaranteed that aggressiveness level of any participant can never exceed n + m under given conditions.
Output
Print a single integer — the largest profit of the show.
Examples
Input
5 4
4 3 1 2 1
1 2 1 2 1
1 2 3 4 5 6 7 8 9
Output
6
Input
2 2
1 2
0 0
2 1 -100 -100
Output
2
Input
5 4
4 3 2 1 1
0 2 6 7 4
12 12 12 6 -3 -5 3 10 -4
Output
62
Note
In the first sample case it is optimal to recruit candidates 1, 2, 3, 5. Then the show will pay 1 + 2 + 1 + 1 = 5 roubles for recruitment. The events on stage will proceed as follows:
* a participant with aggressiveness level 4 enters the stage, the show makes 4 roubles;
* a participant with aggressiveness level 3 enters the stage, the show makes 3 roubles;
* a participant with aggressiveness level 1 enters the stage, the show makes 1 rouble;
* a participant with aggressiveness level 1 enters the stage, the show makes 1 roubles, a fight starts. One of the participants leaves, the other one increases his aggressiveness level to 2. The show will make extra 2 roubles for this.
Total revenue of the show will be 4 + 3 + 1 + 1 + 2=11 roubles, and the profit is 11 - 5 = 6 roubles.
In the second sample case it is impossible to recruit both candidates since the second one has higher aggressiveness, thus it is better to recruit the candidate 1.
Submitted Solution:
```
print("qwq")
``` | instruction | 0 | 75,160 | 17 | 150,320 |
No | output | 1 | 75,160 | 17 | 150,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At many competitions that have a word «cup» in its official name the winner is presented with an actual cup. This time the organizers of one unusual programming competition have decided to please the winner even more and to add a nameplate to the cup with the handle of the winner.
The nameplate is to be rectangular and the text on it will be printed as a table of several rows and columns. Having some measurements done, the organizers have found out that the number a of rows cannot be greater than 5 while the number b of columns cannot exceed 20. Every cell of the table will contain either an asterisk («*») or a letter of user's handle.
Furthermore, the organizers want the rows of the table to be uniform, which means that the number of asterisks used in different rows should differ by at most one (i.e. you can't have two asterisks in the first row and none in the second). The main goal, however, is to obtain the winner's handle precisely when reading the table from top to bottom and from left to right in every row (skipping asterisks).
The organizers want for the nameplate to have as few rows as possible and among all valid tables with the minimum number of rows they want to choose the one that has the minimum number of columns.
The winner is not yet determined so your task is to write a program that, given a certain handle, generates the necessary table.
Input
The only line contains one string s (1 ≤ |s| ≤ 100), comprised of uppercase and lowercase Latin letters, — the handle of the winner.
Output
In the first line output the minimum number a of rows in the table and the minimum number b of columns in an optimal table with rows.
The following a lines should contain b characters each — any valid table.
Examples
Input
tourist
Output
1 7
tourist
Input
MyNameIsLifeIAmForeverByYourSideMyNameIsLife
Output
3 15
MyNameIsLifeIAm
ForeverByYourSi
deMyNameIsL*ife
Submitted Solution:
```
s = input()
a = 1
name_plate = []
while len(s)/a > 20:
a += 1
if len(s) % a == 0:
for i in range(a):
name_plate.append(s[(i * len(s)) // a: ((i + 1) * len(s)) // a])
print(a, len(s) // a)
for i in name_plate:
print(i)
else:
l = ''
asterisk = (len(s) // a + 1) * a - len(s)
j = 0
while j < asterisk:
l += s[(j * (len(s) - asterisk)) // a: ((j + 1) * (len(s) - asterisk)) // a] + '*'
j += 1
l += s[j * (len(s) - asterisk) // a:]
for i in range(a):
name_plate.append(l[(i * len(l)) // a: ((i + 1) * len(l)) // a])
print(a, len(l) // a)
for i in name_plate:
print(i)
``` | instruction | 0 | 76,708 | 17 | 153,416 |
Yes | output | 1 | 76,708 | 17 | 153,417 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At many competitions that have a word «cup» in its official name the winner is presented with an actual cup. This time the organizers of one unusual programming competition have decided to please the winner even more and to add a nameplate to the cup with the handle of the winner.
The nameplate is to be rectangular and the text on it will be printed as a table of several rows and columns. Having some measurements done, the organizers have found out that the number a of rows cannot be greater than 5 while the number b of columns cannot exceed 20. Every cell of the table will contain either an asterisk («*») or a letter of user's handle.
Furthermore, the organizers want the rows of the table to be uniform, which means that the number of asterisks used in different rows should differ by at most one (i.e. you can't have two asterisks in the first row and none in the second). The main goal, however, is to obtain the winner's handle precisely when reading the table from top to bottom and from left to right in every row (skipping asterisks).
The organizers want for the nameplate to have as few rows as possible and among all valid tables with the minimum number of rows they want to choose the one that has the minimum number of columns.
The winner is not yet determined so your task is to write a program that, given a certain handle, generates the necessary table.
Input
The only line contains one string s (1 ≤ |s| ≤ 100), comprised of uppercase and lowercase Latin letters, — the handle of the winner.
Output
In the first line output the minimum number a of rows in the table and the minimum number b of columns in an optimal table with rows.
The following a lines should contain b characters each — any valid table.
Examples
Input
tourist
Output
1 7
tourist
Input
MyNameIsLifeIAmForeverByYourSideMyNameIsLife
Output
3 15
MyNameIsLifeIAm
ForeverByYourSi
deMyNameIsL*ife
Submitted Solution:
```
import math
c = input()
o = 0
s = ''
a = math.ceil(len(c)/20)
b = math.ceil(len(c)/a)
b1 = math.floor(len(c)/a)
print(a, b)
col = a*b-len(c)
for i in range(a):
for j in range(b1):
s = s + c[o+j]
if ((col<(a-i))&(len(c)%a!=0)):
j += 1
s = s + c[o + j]
else:
while (len(s)<b):
s = s+'*'
print(s)
s = ''
o = o + j + 1
``` | instruction | 0 | 76,709 | 17 | 153,418 |
Yes | output | 1 | 76,709 | 17 | 153,419 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At many competitions that have a word «cup» in its official name the winner is presented with an actual cup. This time the organizers of one unusual programming competition have decided to please the winner even more and to add a nameplate to the cup with the handle of the winner.
The nameplate is to be rectangular and the text on it will be printed as a table of several rows and columns. Having some measurements done, the organizers have found out that the number a of rows cannot be greater than 5 while the number b of columns cannot exceed 20. Every cell of the table will contain either an asterisk («*») or a letter of user's handle.
Furthermore, the organizers want the rows of the table to be uniform, which means that the number of asterisks used in different rows should differ by at most one (i.e. you can't have two asterisks in the first row and none in the second). The main goal, however, is to obtain the winner's handle precisely when reading the table from top to bottom and from left to right in every row (skipping asterisks).
The organizers want for the nameplate to have as few rows as possible and among all valid tables with the minimum number of rows they want to choose the one that has the minimum number of columns.
The winner is not yet determined so your task is to write a program that, given a certain handle, generates the necessary table.
Input
The only line contains one string s (1 ≤ |s| ≤ 100), comprised of uppercase and lowercase Latin letters, — the handle of the winner.
Output
In the first line output the minimum number a of rows in the table and the minimum number b of columns in an optimal table with rows.
The following a lines should contain b characters each — any valid table.
Examples
Input
tourist
Output
1 7
tourist
Input
MyNameIsLifeIAmForeverByYourSideMyNameIsLife
Output
3 15
MyNameIsLifeIAm
ForeverByYourSi
deMyNameIsL*ife
Submitted Solution:
```
import math, sys
a=input()
n=len(a)
for i in range(1, 6):
x1=math.ceil(n/i)*i
j=x1//i
if (j>20):
continue
stars=x1-n
l=0
print (i, j)
for k in range (i):
if (k<stars%i):
print ('*'*(stars//i+1), end='')
need=j-(stars//i+1)
while (l<n and need!=0):
print (a[l], end='')
l+=1
need-=1
print()
else:
print ('*'*(stars//i), end='')
need=j-(stars//i)
while (l<n and need!=0):
print (a[l], end='')
l+=1
need-=1
print()
break
``` | instruction | 0 | 76,710 | 17 | 153,420 |
Yes | output | 1 | 76,710 | 17 | 153,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At many competitions that have a word «cup» in its official name the winner is presented with an actual cup. This time the organizers of one unusual programming competition have decided to please the winner even more and to add a nameplate to the cup with the handle of the winner.
The nameplate is to be rectangular and the text on it will be printed as a table of several rows and columns. Having some measurements done, the organizers have found out that the number a of rows cannot be greater than 5 while the number b of columns cannot exceed 20. Every cell of the table will contain either an asterisk («*») or a letter of user's handle.
Furthermore, the organizers want the rows of the table to be uniform, which means that the number of asterisks used in different rows should differ by at most one (i.e. you can't have two asterisks in the first row and none in the second). The main goal, however, is to obtain the winner's handle precisely when reading the table from top to bottom and from left to right in every row (skipping asterisks).
The organizers want for the nameplate to have as few rows as possible and among all valid tables with the minimum number of rows they want to choose the one that has the minimum number of columns.
The winner is not yet determined so your task is to write a program that, given a certain handle, generates the necessary table.
Input
The only line contains one string s (1 ≤ |s| ≤ 100), comprised of uppercase and lowercase Latin letters, — the handle of the winner.
Output
In the first line output the minimum number a of rows in the table and the minimum number b of columns in an optimal table with rows.
The following a lines should contain b characters each — any valid table.
Examples
Input
tourist
Output
1 7
tourist
Input
MyNameIsLifeIAmForeverByYourSideMyNameIsLife
Output
3 15
MyNameIsLifeIAm
ForeverByYourSi
deMyNameIsL*ife
Submitted Solution:
```
import math
s = input()
a = math.ceil(len(s) / 20)
b = math.ceil(len(s) / a)
d = a*b - len(s)
ind = 0
while d > 0:
s = s[:ind] + '*' + s[ind:]
ind += b
if ind >= len(s):
ind = 0
d -= 1
print(a, b)
for i in range(a):
print(s[i*b:(i+1)*b])
``` | instruction | 0 | 76,711 | 17 | 153,422 |
Yes | output | 1 | 76,711 | 17 | 153,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At many competitions that have a word «cup» in its official name the winner is presented with an actual cup. This time the organizers of one unusual programming competition have decided to please the winner even more and to add a nameplate to the cup with the handle of the winner.
The nameplate is to be rectangular and the text on it will be printed as a table of several rows and columns. Having some measurements done, the organizers have found out that the number a of rows cannot be greater than 5 while the number b of columns cannot exceed 20. Every cell of the table will contain either an asterisk («*») or a letter of user's handle.
Furthermore, the organizers want the rows of the table to be uniform, which means that the number of asterisks used in different rows should differ by at most one (i.e. you can't have two asterisks in the first row and none in the second). The main goal, however, is to obtain the winner's handle precisely when reading the table from top to bottom and from left to right in every row (skipping asterisks).
The organizers want for the nameplate to have as few rows as possible and among all valid tables with the minimum number of rows they want to choose the one that has the minimum number of columns.
The winner is not yet determined so your task is to write a program that, given a certain handle, generates the necessary table.
Input
The only line contains one string s (1 ≤ |s| ≤ 100), comprised of uppercase and lowercase Latin letters, — the handle of the winner.
Output
In the first line output the minimum number a of rows in the table and the minimum number b of columns in an optimal table with rows.
The following a lines should contain b characters each — any valid table.
Examples
Input
tourist
Output
1 7
tourist
Input
MyNameIsLifeIAmForeverByYourSideMyNameIsLife
Output
3 15
MyNameIsLifeIAm
ForeverByYourSi
deMyNameIsL*ife
Submitted Solution:
```
from math import *
s = input()
length = ceil(len(s) / 20)
section = ceil(len(s) / length)
current = 0
table = list()
table.append([])
i = 0
while i != len(s):
if i % section == 0 and i != 0:
table.append([])
current += 1
table[current].append(s[i])
i += 1
checked = False
while not checked:
checked = True
for i in range(current):
if len(table[i + 1]) < len(table[i]):
checked = False
table[i + 1].append("*")
break
if checked:
for i in range(current):
if table[i + 1].count("*") > table[i].count("*") + 1:
checked = False
table[i + 1].pop(len(table[i + 1]) - 1)
el = table[i].pop(len(table[i]) - 1)
table[i + 1].insert(0, el)
print(len(table), len(table[0]))
for i in range(len(table)):
print(table[i])
``` | instruction | 0 | 76,712 | 17 | 153,424 |
No | output | 1 | 76,712 | 17 | 153,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
At many competitions that have a word «cup» in its official name the winner is presented with an actual cup. This time the organizers of one unusual programming competition have decided to please the winner even more and to add a nameplate to the cup with the handle of the winner.
The nameplate is to be rectangular and the text on it will be printed as a table of several rows and columns. Having some measurements done, the organizers have found out that the number a of rows cannot be greater than 5 while the number b of columns cannot exceed 20. Every cell of the table will contain either an asterisk («*») or a letter of user's handle.
Furthermore, the organizers want the rows of the table to be uniform, which means that the number of asterisks used in different rows should differ by at most one (i.e. you can't have two asterisks in the first row and none in the second). The main goal, however, is to obtain the winner's handle precisely when reading the table from top to bottom and from left to right in every row (skipping asterisks).
The organizers want for the nameplate to have as few rows as possible and among all valid tables with the minimum number of rows they want to choose the one that has the minimum number of columns.
The winner is not yet determined so your task is to write a program that, given a certain handle, generates the necessary table.
Input
The only line contains one string s (1 ≤ |s| ≤ 100), comprised of uppercase and lowercase Latin letters, — the handle of the winner.
Output
In the first line output the minimum number a of rows in the table and the minimum number b of columns in an optimal table with rows.
The following a lines should contain b characters each — any valid table.
Examples
Input
tourist
Output
1 7
tourist
Input
MyNameIsLifeIAmForeverByYourSideMyNameIsLife
Output
3 15
MyNameIsLifeIAm
ForeverByYourSi
deMyNameIsL*ife
Submitted Solution:
```
import math
S = input()
a = math.ceil(len(S)/20)
b = math.ceil(len(S)/a)
starscount = a * b - len(S)
print(a,b)
for i in range(a - starscount):
print(S[b * i: b* i +b])
for i in range(a - starscount, a):
print(S[b * i: b* i +b] + '*')
``` | instruction | 0 | 76,715 | 17 | 153,430 |
No | output | 1 | 76,715 | 17 | 153,431 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0. | instruction | 0 | 76,948 | 17 | 153,896 |
Tags: dp, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
S=sorted(map(int,input().split()))
DP=[[0]*n for i in range(n)]
for i in range(n-2,-1,-1):
for j in range(i,n):
DP[i][j]=S[j]-S[i]+min(DP[i+1][j],DP[i][j-1])
print(DP[0][n-1])
``` | output | 1 | 76,948 | 17 | 153,897 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0. | instruction | 0 | 76,949 | 17 | 153,898 |
Tags: dp, greedy
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
a.sort()
dp = [[0 for i in range(n)]for j in range(n)]
for i in range(n-1,-1,-1):
for j in range(i+1,n):
dp[i][j] = a[j]-a[i] + min(dp[i+1][j],dp[i][j-1])
print(dp[0][n-1])
``` | output | 1 | 76,949 | 17 | 153,899 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0. | instruction | 0 | 76,950 | 17 | 153,900 |
Tags: dp, greedy
Correct Solution:
```
import math,sys,bisect,heapq,os
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
from functools import lru_cache
#sys.setrecursionlimit(200000000)
pr = lambda x: x
def input(): return sys.stdin.readline().rstrip('\r\n')
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
aj = lambda: list(map(int, input().split()))
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
def solve():
n, = aj()
a = aj()
a.sort()
dp = [[0 for _ in range(2020)] for _ in range(2020)]
for i in range(n, 0, -1):
for j in range(i+1, n+1):
dp[i][j] = a[j-1]-a[i-1]+min(dp[i+1][j],dp[i][j-1])
print(dp[1][n])
try:
#os.system("online_judge.py")
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
from aj import *
except:
pass
solve()
``` | output | 1 | 76,950 | 17 | 153,901 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0. | instruction | 0 | 76,951 | 17 | 153,902 |
Tags: dp, greedy
Correct Solution:
```
from sys import stdin
input=stdin.readline
def answer():
dp=[0 for i in range(n)]
for i in range(1,n):
for j in range(n-i):
dp[j]=min(dp[j+1],dp[j]) + a[i+j] - a[j]
return dp[0]
n=int(input().strip())
a=sorted(list(map(int,input().strip().split())))
print(answer())
``` | output | 1 | 76,951 | 17 | 153,903 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0. | instruction | 0 | 76,952 | 17 | 153,904 |
Tags: dp, greedy
Correct Solution:
```
n = int(input())
a = input().split()
if n == 1 :
print("0")
else :
for i in range(n) :
a[i] = int(a[i])
b = sorted(a)
oldd = []
for i in range(n - 1) :
oldd.append(b[i + 1] - b[i])
for j in range(2, n) :
newd = []
for i in range(n - j) :
var1 = oldd[i] + (b[i + j] - b[i])
var2 = oldd[i + 1] + (b[i + j] - b[i])
newd.append(min(var1, var2))
oldd = newd
print(oldd[0])
``` | output | 1 | 76,952 | 17 | 153,905 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0. | instruction | 0 | 76,953 | 17 | 153,906 |
Tags: dp, greedy
Correct Solution:
```
import time
def main():
max_int = 10 ** 18
n = i_input()
s = li_input()
s.sort()
if n == 1:
print(0)
return
dp = [max_int] * n
dp[-1] = 0
out = max_int
for i in range(n):
if i:
dp[n - 1] = dp[n - 1] + s[n - 1] - s[i - 1]
for j in range(n - 2, i - 1, -1):
dp[j] = min(
dp[j + 1] + s[j + 1] - s[i],
dp[j] + s[j] - s[i - 1]
)
out = min(out, dp[i])
print(out)
############
def i_input():
return int(input())
def l_input():
return input().split()
def li_input():
return list(map(int, l_input()))
def il_input():
return list(map(int, l_input()))
# endregion
if __name__ == "__main__":
TT = time.time()
main()
# print("\n", time.time() - TT)
``` | output | 1 | 76,953 | 17 | 153,907 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0. | instruction | 0 | 76,954 | 17 | 153,908 |
Tags: dp, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
def do():
n = int(input())
dat = sorted(list(map(int, input().split())))
dp = [0] * ( (n+1) * (n+1) )
for width in range(2, n+1):
for l in range(0, n - width+1):
r = l + width
x = l*n
dp[x + r] = (dat[r-1] - dat[l]) + min(dp[x + r-1], dp[x+n + r])
print(dp[n])
do()
``` | output | 1 | 76,954 | 17 | 153,909 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0. | instruction | 0 | 76,955 | 17 | 153,910 |
Tags: dp, greedy
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split()))
arr.sort()
dp = [[0]*n for i in range(n)]
for i in range(n-2, -1, -1):
for j in range(i+1, n):
dp[i][j] = arr[j] - arr[i] + min(dp[i+1][j], dp[i][j-1])
print(dp[0][n-1])
``` | output | 1 | 76,955 | 17 | 153,911 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0.
Submitted Solution:
```
#import io,os
#input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
arr = list(map(int,input().split()))
arr = sorted(arr)
#dp = [[0 for j in xrange(n)] for i in xrange(n)]
dp = [0 for i in range(n)]
for d in range(1,n):
copydp = dp[:]
for i in range(n-d):
dp[i] = min(copydp[i],copydp[i+1]) + arr[i+d] - arr[i]
# dp[i][i+d] = min(dp[i+1][i+d], dp[i][i+d-1]) + arr[i+d] - arr[i]
# print(dp)
print(dp[0])
``` | instruction | 0 | 76,956 | 17 | 153,912 |
Yes | output | 1 | 76,956 | 17 | 153,913 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0.
Submitted Solution:
```
import sys,functools,collections,bisect,math,heapq
import os
from io import BytesIO, IOBase
from types import GeneratorType
#print = sys.stdout.write
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 = sys.stdin.readline
#
## start here ##
n=int(input())
a=list(map(int,input().split()))
a.sort()
dp=[[0]*n for i in range(n)]
#print(dp)
for i in range(n-1,-1,-1):
for j in range(i+1,n):
dp[i][j] = min(dp[i+1][j],dp[i][j-1])+a[j]-a[i]
print(dp[0][-1])
``` | instruction | 0 | 76,957 | 17 | 153,914 |
Yes | output | 1 | 76,957 | 17 | 153,915 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0.
Submitted Solution:
```
n = int(input())
arr = sorted(list(map(int, input().split())))
dp = [0] * n
for l in range(1, n):
for i in range(n - l):
dp[i] = min(dp[i], dp[i + 1]) + arr[i + l] - arr[i]
print(dp[0])
``` | instruction | 0 | 76,958 | 17 | 153,916 |
Yes | output | 1 | 76,958 | 17 | 153,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0.
Submitted Solution:
```
from functools import lru_cache
import sys
def get_ints(): return list(map(int, sys.stdin.readline().strip().split()))
def solve(N, nums):
nums.sort()
dp = [[0 for i in range(N)] for j in range(N + 1)]
for i in range(N - 1, -1, -1):
for j in range(i + 1, N, 1):
dp[i][j] = nums[j] - nums[i] + min(dp[i + 1][j], dp[i][j - 1])
return dp[0][N - 1]
N = int(input())
nums = get_ints()
print(solve(N, nums))
``` | instruction | 0 | 76,959 | 17 | 153,918 |
Yes | output | 1 | 76,959 | 17 | 153,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0.
Submitted Solution:
```
for _ in range(1):
n=int(input())
#n,m=map(int,input().split())
arr=list(map(int, input().split()))
arr.sort()
ls=[0]
for i in range(1,n):
if arr[i]!=arr[i-1]:
ls.append(i)
aa=10**10
for j in ls:
l=arr[j:]+arr[:j][::-1]
a=0
v1=0
v2=10**10
for i in range(n):
v1 = max(v1, l[i])
v2 = min(v2, l[i])
a += v1 - v2
aa=min(aa,a)
print(aa)
``` | instruction | 0 | 76,960 | 17 | 153,920 |
No | output | 1 | 76,960 | 17 | 153,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0.
Submitted Solution:
```
import sys,functools,collections,bisect,math,heapq
import os
from io import BytesIO, IOBase
from types import GeneratorType
#print = sys.stdout.write
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")
## start here ##
#t = int(input())
for _ in range(1):
n = int(input())
s = list(map(int,input().strip().split()))
s.sort()
ans1 = 0
c = s[0]
for i in s:
ans1 += (i-c)
s.reverse()
ans2 = 0
c = s[0]
for i in s:
ans2 += c-i
ans3 = 0
d = collections.Counter(s)
x = d.most_common()
mi = float('inf')
ma = float('-inf')
for ele,c in x:
if mi > ele:
mi = ele
if ma < ele:
ma = ele
ans3 += ((ma-mi)*c)
print(min(ans1,ans2,ans3))
``` | instruction | 0 | 76,961 | 17 | 153,922 |
No | output | 1 | 76,961 | 17 | 153,923 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0.
Submitted Solution:
```
import sys
#input = sys.stdin.readline
for _ in range(1):
n=int(input())
arr=[int(x) for x in input().split()]
#arr.sort()
d={}
for i in arr:
if i in d:
d[i]+=1
else:
d[i]=1
arr=sorted(list(set(arr)))
n=len(arr)
def helper2():
dp=[[sys.maxsize for i in range(n)] for j in range(n)]
for i in range(n):
dp[i][i]=0
for gap in range(1,n):
for i in range(n):
j=i+gap
if j>=n:
break
dp[i][j]=min(dp[i][j],dp[i][j-1]+(arr[j]-arr[i])*d[arr[j]],dp[i+1][j]+(arr[j]-arr[i])*d[arr[i]])
#print(n,arr)
if n==6 and arr[0]==69:
print(dp)
#print(dp[0][n-1])
return dp[0][n-1]
#print(arr)
print(helper2())
``` | instruction | 0 | 76,962 | 17 | 153,924 |
No | output | 1 | 76,962 | 17 | 153,925 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The student council is preparing for the relay race at the sports festival.
The council consists of n members. They will run one after the other in the race, the speed of member i is s_i. The discrepancy d_i of the i-th stage is the difference between the maximum and the minimum running speed among the first i members who ran. Formally, if a_i denotes the speed of the i-th member who participated in the race, then d_i = max(a_1, a_2, ..., a_i) - min(a_1, a_2, ..., a_i).
You want to minimize the sum of the discrepancies d_1 + d_2 + ... + d_n. To do this, you are allowed to change the order in which the members run. What is the minimum possible sum that can be achieved?
Input
The first line contains a single integer n (1 ≤ n ≤ 2000) — the number of members of the student council.
The second line contains n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 10^9) – the running speeds of the members.
Output
Print a single integer — the minimum possible value of d_1 + d_2 + ... + d_n after choosing the order of the members.
Examples
Input
3
3 1 2
Output
3
Input
1
5
Output
0
Input
6
1 6 3 3 6 3
Output
11
Input
6
104 943872923 6589 889921234 1000000000 69
Output
2833800505
Note
In the first test case, we may choose to make the third member run first, followed by the first member, and finally the second. Thus a_1 = 2, a_2 = 3, and a_3 = 1. We have:
* d_1 = max(2) - min(2) = 2 - 2 = 0.
* d_2 = max(2, 3) - min(2, 3) = 3 - 2 = 1.
* d_3 = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2.
The resulting sum is d_1 + d_2 + d_3 = 0 + 1 + 2 = 3. It can be shown that it is impossible to achieve a smaller value.
In the second test case, the only possible rearrangement gives d_1 = 0, so the minimum possible result is 0.
Submitted Solution:
```
n=int(input())
l1=list(map(int,input().split()))
l2=[]
l3=[]
l1.sort()
k=l1[0]
l1.pop(0)
l1.append(k)
for i in range(len(l1)):
l2.append(l1[i])
a=min(l2)
b=max(l2)
c=b-a
l3.append(c)
print(sum(l3))
``` | instruction | 0 | 76,963 | 17 | 153,926 |
No | output | 1 | 76,963 | 17 | 153,927 |
Provide tags and a correct Python 3 solution for this coding contest problem.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4 | instruction | 0 | 76,964 | 17 | 153,928 |
Tags: data structures, dfs and similar, dp, implementation, trees
Correct Solution:
```
import sys
input = sys.stdin.readline
k = int(input())
s = input().strip()
a = ["*"] + list(s[::-1])
dp = [0]*(1<<k)
for i in range((1<<k)-1,0,-1):
if 2*i < 1<<k:
if a[i] == "1":
dp[i] = dp[2*i]
elif a[i] == "0":
dp[i] = dp[2*i+1]
else:
dp[i] = dp[2*i] + dp[2*i+1]
else:
if a[i] == "?":
dp[i] = 2
else:
dp[i] = 1
def solve():
idx,char = input().split()
idx = int(idx)
idx = (1<<k) - idx
a[idx] = char
cur = idx
while cur:
if 2*cur >= (1<<k):
if a[cur] == "?":
dp[cur] = 2
else:
dp[cur] = 1
cur //= 2
continue
if a[cur] == "1":
dp[cur] = dp[2*cur]
elif a[cur] == "0":
dp[cur] = dp[2*cur + 1]
else:
dp[cur] = dp[2*cur] + dp[2*cur + 1]
cur //= 2
cur = 1
ans = 1
while 2*cur < (1<<k):
if a[cur] == "1":
cur = 2*cur
elif a[cur] == "0":
cur = 2*cur + 1
else:
break
print(dp[cur])
for nq in range(int(input())):
solve()
``` | output | 1 | 76,964 | 17 | 153,929 |
Provide tags and a correct Python 3 solution for this coding contest problem.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4 | instruction | 0 | 76,965 | 17 | 153,930 |
Tags: data structures, dfs and similar, dp, implementation, trees
Correct Solution:
```
"""
ID: happyn61
LANG: PYTHON3
PROB: loan
"""
from itertools import product
import itertools
#from collections import defaultdict
import sys
import math
import heapq
from collections import deque
MOD=1000000000007
#fin = open ('loan.in', 'r')
#fout = open ('loan.out', 'w')
#print(dic["4734"])
def find(parent,i):
if parent[i] != i:
parent[i]=find(parent,parent[i])
return parent[i]
# A utility function to do union of two subsets
def union(parent,rank,xx,yy):
x=find(parent,xx)
y=find(parent,yy)
if rank[x]>rank[y]:
parent[y]=x
elif rank[y]>rank[x]:
parent[x]=y
else:
parent[y]=x
rank[x]+=1
ans=0
#NK=sys.stdin.readline().strip().split()
K=int(sys.stdin.readline().strip())
#N=int(NK[0])
#K=int(NK[1])
#M=int(NK[2])
ol=list(sys.stdin.readline().strip())
#d={0:0,1:0}
#ol.reverse()
k=2**K
l=[1 for i in range(k*2)]
for i in range(k-1):
j=k-1-i
if ol[i]=="?":
l[j]=l[j*2]+l[j*2+1]
elif ol[i]=="0":
l[j]=l[j*2+1]
else:
l[j]=l[j*2]
Q=int(sys.stdin.readline().strip())
for i in range(Q):
p,c=sys.stdin.readline().strip().split()
p=int(p)
j=k-p
ol[p-1]=c
while j>0:
if ol[p-1]=="?":
l[j]=l[j*2]+l[j*2+1]
elif ol[p-1]=="0":
l[j]=l[j*2+1]
else:
l[j]=l[j*2]
j=j//2
p=k-j
print(l[1])
``` | output | 1 | 76,965 | 17 | 153,931 |
Provide tags and a correct Python 3 solution for this coding contest problem.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4 | instruction | 0 | 76,966 | 17 | 153,932 |
Tags: data structures, dfs and similar, dp, implementation, trees
Correct Solution:
```
import os, sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
k=int(input())
a=input()
b=[-1]+[i for i in reversed(a)]
ans=[-1]*(len(b))+[1]*(2**k)
for j in range(len(b)-1,0,-1):
if b[j]=='?':
ans[j]=ans[2*j]+ans[2*j+1]
elif b[j]=='0':
ans[j]=ans[2*j+1]
else:
ans[j]=ans[2*j]
def update(i,v):
j=len(b)-i
b[j]=v
while j:
if b[j] == '?':
ans[j] = ans[2 * j] + ans[2 * j + 1]
elif b[j] == '0':
ans[j] = ans[2 * j + 1]
else:
ans[j] = ans[2 * j]
j//=2
for _ in range(int(input())):
i,v=input().split()
i=int(i)
update(i,v)
print(ans[1])
``` | output | 1 | 76,966 | 17 | 153,933 |
Provide tags and a correct Python 3 solution for this coding contest problem.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4 | instruction | 0 | 76,967 | 17 | 153,934 |
Tags: data structures, dfs and similar, dp, implementation, trees
Correct Solution:
```
import sys
# sys.setrecursionlimit(10**5)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LI1(): return list(map(int1, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
inf = 10**16
# md = 998244353
md = 10**9+7
k = II()
n = 1 << k
s = SI()
vv = [2]*n
tt = [2]*n
for i, c in enumerate(s, 1):
i = n-i
if c == "?":
tt[i] = 2
if i >= n//2: vv[i] = 2
else: vv[i] = vv[i*2]+vv[i*2+1]
else:
tt[i] = int(c)
if i >= n//2: vv[i] = 1
else: vv[i] = vv[i*2+1-int(c)]
# print(tt)
# print(vv)
for _ in range(II()):
p, c = SI().split()
p = n-int(p)
if c == "?": tt[p] = 2
else: tt[p] = int(c)
while p:
if p >= n//2:
if tt[p] == 2:
vv[p] = 2
else:
vv[p] = 1
else:
if tt[p] == 2:
vv[p] = vv[p*2]+vv[p*2+1]
else:
vv[p] = vv[p*2+1-tt[p]]
p //= 2
print(vv[1])
``` | output | 1 | 76,967 | 17 | 153,935 |
Provide tags and a correct Python 3 solution for this coding contest problem.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4 | instruction | 0 | 76,968 | 17 | 153,936 |
Tags: data structures, dfs and similar, dp, implementation, trees
Correct Solution:
```
import sys
input = sys.stdin.readline
k = int(input())
s = list(input().rstrip())
n = len(s) + 1
x = [0]
p2 = 1
d = dict()
for _ in range(k):
for i in range(2 * p2, p2, -1):
x.append(s[-(i - 1)])
d[(-(i - 1)) % n] = len(x) - 1
p2 *= 2
cnt = [1] * (2 * n)
for i in range(n - 1, 0, -1):
if x[i] == "?":
cnt[i] = cnt[2 * i] + cnt[2 * i + 1]
else:
cnt[i] = cnt[2 * i + int(x[i])]
q = int(input())
for _ in range(q):
p, c = input().rstrip().split()
u = d[int(p)]
x[u] = c
if c == "?":
cnt[u] = cnt[2 * u] + cnt[2 * u + 1]
else:
cnt[u] = cnt[2 * u + int(c)]
while u ^ 1:
u //= 2
if x[u] == "?":
cnt[u] = cnt[2 * u] + cnt[2 * u + 1]
else:
cnt[u] = cnt[2 * u + int(x[u])]
ans = cnt[1]
print(ans)
``` | output | 1 | 76,968 | 17 | 153,937 |
Provide tags and a correct Python 3 solution for this coding contest problem.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4 | instruction | 0 | 76,969 | 17 | 153,938 |
Tags: data structures, dfs and similar, dp, implementation, trees
Correct Solution:
```
def divisors(M):
d=[]
i=1
while M>=i**2:
if M%i==0:
d.append(i)
if i**2!=M:
d.append(M//i)
i=i+1
return d
def popcount(x):
x = x - ((x >> 1) & 0x55555555)
x = (x & 0x33333333) + ((x >> 2) & 0x33333333)
x = (x + (x >> 4)) & 0x0f0f0f0f
x = x + (x >> 8)
x = x + (x >> 16)
return x & 0x0000007f
def eratosthenes(n):
res=[0 for i in range(n+1)]
prime=set([])
for i in range(2,n+1):
if not res[i]:
prime.add(i)
for j in range(1,n//i+1):
res[i*j]=1
return prime
def factorization(n):
res=[]
for p in prime:
if n%p==0:
while n%p==0:
n//=p
res.append(p)
if n!=1:
res.append(n)
return res
def euler_phi(n):
res = n
for x in range(2,n+1):
if x ** 2 > n:
break
if n%x==0:
res = res//x * (x-1)
while n%x==0:
n //= x
if n!=1:
res = res//n * (n-1)
return res
def ind(b,n):
res=0
while n%b==0:
res+=1
n//=b
return res
def isPrimeMR(n):
d = n - 1
d = d // (d & -d)
L = [2, 3, 5, 7, 11, 13, 17]
for a in L:
t = d
y = pow(a, t, n)
if y == 1: continue
while y != n - 1:
y = (y * y) % n
if y == 1 or t == n - 1: return 0
t <<= 1
return 1
def findFactorRho(n):
from math import gcd
m = 1 << n.bit_length() // 8
for c in range(1, 99):
f = lambda x: (x * x + c) % n
y, r, q, g = 2, 1, 1, 1
while g == 1:
x = y
for i in range(r):
y = f(y)
k = 0
while k < r and g == 1:
ys = y
for i in range(min(m, r - k)):
y = f(y)
q = q * abs(x - y) % n
g = gcd(q, n)
k += m
r <<= 1
if g == n:
g = 1
while g == 1:
ys = f(ys)
g = gcd(abs(x - ys), n)
if g < n:
if isPrimeMR(g): return g
elif isPrimeMR(n // g): return n // g
return findFactorRho(g)
def primeFactor(n):
i = 2
ret = {}
rhoFlg = 0
while i*i <= n:
k = 0
while n % i == 0:
n //= i
k += 1
if k: ret[i] = k
i += 1 + i % 2
if i == 101 and n >= 2 ** 20:
while n > 1:
if isPrimeMR(n):
ret[n], n = 1, 1
else:
rhoFlg = 1
j = findFactorRho(n)
k = 0
while n % j == 0:
n //= j
k += 1
ret[j] = k
if n > 1: ret[n] = 1
if rhoFlg: ret = {x: ret[x] for x in sorted(ret)}
return ret
def divisors(n):
res = [1]
prime = primeFactor(n)
for p in prime:
newres = []
for d in res:
for j in range(prime[p]+1):
newres.append(d*p**j)
res = newres
res.sort()
return res
def xorfactorial(num):
if num==0:
return 0
elif num==1:
return 1
elif num==2:
return 3
elif num==3:
return 0
else:
x=baseorder(num)
return (2**x)*((num-2**x+1)%2)+function(num-2**x)
def xorconv(n,X,Y):
if n==0:
res=[(X[0]*Y[0])%mod]
return res
x=[X[i]+X[i+2**(n-1)] for i in range(2**(n-1))]
y=[Y[i]+Y[i+2**(n-1)] for i in range(2**(n-1))]
z=[X[i]-X[i+2**(n-1)] for i in range(2**(n-1))]
w=[Y[i]-Y[i+2**(n-1)] for i in range(2**(n-1))]
res1=xorconv(n-1,x,y)
res2=xorconv(n-1,z,w)
former=[(res1[i]+res2[i])*inv for i in range(2**(n-1))]
latter=[(res1[i]-res2[i])*inv for i in range(2**(n-1))]
former=list(map(lambda x:x%mod,former))
latter=list(map(lambda x:x%mod,latter))
return former+latter
def merge_sort(A,B):
pos_A,pos_B = 0,0
n,m = len(A),len(B)
res = []
while pos_A < n and pos_B < m:
a,b = A[pos_A],B[pos_B]
if a < b:
res.append(a)
pos_A += 1
else:
res.append(b)
pos_B += 1
res += A[pos_A:]
res += B[pos_B:]
return res
class UnionFindVerSize():
def __init__(self, N):
self._parent = [n for n in range(0, N)]
self._size = [1] * N
self.group = N
def find_root(self, x):
if self._parent[x] == x: return x
self._parent[x] = self.find_root(self._parent[x])
stack = [x]
while self._parent[stack[-1]]!=stack[-1]:
stack.append(self._parent[stack[-1]])
for v in stack:
self._parent[v] = stack[-1]
return self._parent[x]
def unite(self, x, y):
gx = self.find_root(x)
gy = self.find_root(y)
if gx == gy: return
self.group -= 1
if self._size[gx] < self._size[gy]:
self._parent[gx] = gy
self._size[gy] += self._size[gx]
else:
self._parent[gy] = gx
self._size[gx] += self._size[gy]
def get_size(self, x):
return self._size[self.find_root(x)]
def is_same_group(self, x, y):
return self.find_root(x) == self.find_root(y)
class WeightedUnionFind():
def __init__(self,N):
self.parent = [i for i in range(N)]
self.size = [1 for i in range(N)]
self.val = [0 for i in range(N)]
self.flag = True
self.edge = [[] for i in range(N)]
def dfs(self,v,pv):
stack = [(v,pv)]
new_parent = self.parent[pv]
while stack:
v,pv = stack.pop()
self.parent[v] = new_parent
for nv,w in self.edge[v]:
if nv!=pv:
self.val[nv] = self.val[v] + w
stack.append((nv,v))
def unite(self,x,y,w):
if not self.flag:
return
if self.parent[x]==self.parent[y]:
self.flag = (self.val[x] - self.val[y] == w)
return
if self.size[self.parent[x]]>self.size[self.parent[y]]:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[x] += self.size[y]
self.val[y] = self.val[x] - w
self.dfs(y,x)
else:
self.edge[x].append((y,-w))
self.edge[y].append((x,w))
self.size[y] += self.size[x]
self.val[x] = self.val[y] + w
self.dfs(x,y)
class Dijkstra():
class Edge():
def __init__(self, _to, _cost):
self.to = _to
self.cost = _cost
def __init__(self, V):
self.G = [[] for i in range(V)]
self._E = 0
self._V = V
@property
def E(self):
return self._E
@property
def V(self):
return self._V
def add_edge(self, _from, _to, _cost):
self.G[_from].append(self.Edge(_to, _cost))
self._E += 1
def shortest_path(self, s):
import heapq
que = []
d = [10**15] * self.V
d[s] = 0
heapq.heappush(que, (0, s))
while len(que) != 0:
cost, v = heapq.heappop(que)
if d[v] < cost: continue
for i in range(len(self.G[v])):
e = self.G[v][i]
if d[e.to] > d[v] + e.cost:
d[e.to] = d[v] + e.cost
heapq.heappush(que, (d[e.to], e.to))
return d
#Z[i]:length of the longest list starting from S[i] which is also a prefix of S
#O(|S|)
def Z_algorithm(s):
N = len(s)
Z_alg = [0]*N
Z_alg[0] = N
i = 1
j = 0
while i < N:
while i+j < N and s[j] == s[i+j]:
j += 1
Z_alg[i] = j
if j == 0:
i += 1
continue
k = 1
while i+k < N and k + Z_alg[k]<j:
Z_alg[i+k] = Z_alg[k]
k += 1
i += k
j -= k
return Z_alg
class BIT():
def __init__(self,n,mod=0):
self.BIT = [0]*(n+1)
self.num = n
self.mod = mod
def query(self,idx):
res_sum = 0
mod = self.mod
while idx > 0:
res_sum += self.BIT[idx]
if mod:
res_sum %= mod
idx -= idx&(-idx)
return res_sum
#Ai += x O(logN)
def update(self,idx,x):
mod = self.mod
while idx <= self.num:
self.BIT[idx] += x
if mod:
self.BIT[idx] %= mod
idx += idx&(-idx)
return
class dancinglink():
def __init__(self,n,debug=False):
self.n = n
self.debug = debug
self._left = [i-1 for i in range(n)]
self._right = [i+1 for i in range(n)]
self.exist = [True for i in range(n)]
def pop(self,k):
if self.debug:
assert self.exist[k]
L = self._left[k]
R = self._right[k]
if L!=-1:
if R!=self.n:
self._right[L],self._left[R] = R,L
else:
self._right[L] = self.n
elif R!=self.n:
self._left[R] = -1
self.exist[k] = False
def left(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._left[res]
if res==-1:
break
k -= 1
return res
def right(self,idx,k=1):
if self.debug:
assert self.exist[idx]
res = idx
while k:
res = self._right[res]
if res==self.n:
break
k -= 1
return res
class SparseTable():
def __init__(self,A,merge_func,ide_ele):
N=len(A)
n=N.bit_length()
self.table=[[ide_ele for i in range(n)] for i in range(N)]
self.merge_func=merge_func
for i in range(N):
self.table[i][0]=A[i]
for j in range(1,n):
for i in range(0,N-2**j+1):
f=self.table[i][j-1]
s=self.table[i+2**(j-1)][j-1]
self.table[i][j]=self.merge_func(f,s)
def query(self,s,t):
b=t-s+1
m=b.bit_length()-1
return self.merge_func(self.table[s][m],self.table[t-2**m+1][m])
class BinaryTrie:
class node:
def __init__(self,val):
self.left = None
self.right = None
self.max = val
def __init__(self):
self.root = self.node(-10**15)
def append(self,key,val):
pos = self.root
for i in range(29,-1,-1):
pos.max = max(pos.max,val)
if key>>i & 1:
if pos.right is None:
pos.right = self.node(val)
pos = pos.right
else:
pos = pos.right
else:
if pos.left is None:
pos.left = self.node(val)
pos = pos.left
else:
pos = pos.left
pos.max = max(pos.max,val)
def search(self,M,xor):
res = -10**15
pos = self.root
for i in range(29,-1,-1):
if pos is None:
break
if M>>i & 1:
if xor>>i & 1:
if pos.right:
res = max(res,pos.right.max)
pos = pos.left
else:
if pos.left:
res = max(res,pos.left.max)
pos = pos.right
else:
if xor>>i & 1:
pos = pos.right
else:
pos = pos.left
if pos:
res = max(res,pos.max)
return res
def solveequation(edge,ans,n,m):
#edge=[[to,dire,id]...]
x=[0]*m
used=[False]*n
for v in range(n):
if used[v]:
continue
y = dfs(v)
if y!=0:
return False
return x
def dfs(v):
used[v]=True
r=ans[v]
for to,dire,id in edge[v]:
if used[to]:
continue
y=dfs(to)
if dire==-1:
x[id]=y
else:
x[id]=-y
r+=y
return r
class SegmentTree:
def __init__(self, init_val, segfunc, ide_ele):
n = len(init_val)
self.segfunc = segfunc
self.ide_ele = ide_ele
self.num = 1 << (n - 1).bit_length()
self.tree = [ide_ele] * 2 * self.num
self.size = n
for i in range(n):
self.tree[self.num + i] = init_val[i]
for i in range(self.num - 1, 0, -1):
self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])
def update(self, k, x):
k += self.num
self.tree[k] = x
while k > 1:
self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
k >>= 1
def query(self, l, r):
if r==self.size:
r = self.num
res = self.ide_ele
l += self.num
r += self.num
while l < r:
if l & 1:
res = self.segfunc(res, self.tree[l])
l += 1
if r & 1:
res = self.segfunc(res, self.tree[r - 1])
l >>= 1
r >>= 1
return res
def bisect_l(self,l,r,x):
l += self.num
r += self.num
Lmin = -1
Rmin = -1
while l<r:
if l & 1:
if self.tree[l] <= x and Lmin==-1:
Lmin = l
l += 1
if r & 1:
if self.tree[r-1] <=x:
Rmin = r-1
l >>= 1
r >>= 1
if Lmin != -1:
pos = Lmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
elif Rmin != -1:
pos = Rmin
while pos<self.num:
if self.tree[2 * pos] <=x:
pos = 2 * pos
else:
pos = 2 * pos +1
return pos-self.num
else:
return -1
import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import gcd,log
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
k = int(input())
s = [c for c in input()]
segment_val = [1 for i in range(2**(k+1))]
game_idx_to_segment_idx = [-1 for i in range(2**k)]
segment_idx_to_game_idx = [-1 for i in range(2**k)]
next_idx = 0
for i in range(k-1,-1,-1):
for n in range(2**i,2**(i+1)):
game_idx_to_segment_idx[next_idx] = n
segment_idx_to_game_idx[n] = next_idx
next_idx += 1
def init():
for i in range(2**k-1,0,-1):
game_idx = segment_idx_to_game_idx[i]
t = s[game_idx]
if t=="0":
segment_val[i] = segment_val[2*i]
elif t=="1":
segment_val[i] = segment_val[2*i+1]
else:
segment_val[i] = segment_val[2*i] + segment_val[2*i+1]
def update(k,c):
i = game_idx_to_segment_idx[k-1]
s[k-1] = c
while i:
game_idx = segment_idx_to_game_idx[i]
t = s[game_idx]
if t=="0":
segment_val[i] = segment_val[2*i]
elif t=="1":
segment_val[i] = segment_val[2*i+1]
else:
segment_val[i] = segment_val[2*i] + segment_val[2*i+1]
i >>= 1
init()
for _ in range(int(input())):
k,c = input().split()
k = int(k)
update(k,c)
print(segment_val[1])
``` | output | 1 | 76,969 | 17 | 153,939 |
Provide tags and a correct Python 3 solution for this coding contest problem.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4 | instruction | 0 | 76,970 | 17 | 153,940 |
Tags: data structures, dfs and similar, dp, implementation, trees
Correct Solution:
```
import sys
input=sys.stdin.readline
class segtree:
global s
def __init__(self, init_val, segfunc, k):
n = len(init_val)
self.segfunc = segfunc
self.ide_ele = 0
self.k2 = 2**k
self.k3=2**(k-1)
self.num = self.k2
self.tree = [0] * self.num
def update(self, k, x):
self.tree[k] = x
if k<self.k3:
t=s[self.k2-k-1]
if t=="?":
self.tree[k] = self.tree[k*2] + self.tree[(k*2) ^ 1]
elif t=="1":
self.tree[k] = self.tree[min(k*2,(k*2) ^ 1)]
else:
self.tree[k] = self.tree[max(k*2,(k*2) ^ 1)]
while k > 1:
t=s[self.k2-(k >> 1)-1]
if t=="?":
self.tree[k >> 1] = self.tree[k] + self.tree[k ^ 1]
elif t=="1":
self.tree[k >> 1] = self.tree[min(k,k^1)]
else:
self.tree[k >> 1] = self.tree[max(k,k^1)]
k >>= 1
def get(self, k):
return self.tree[k]
def deb(self):
return self.tree
def segf(x,y):
return x+y
k=int(input())
s=list(input())
q=int(input())
num2=2**(k-1)
n=2**k
seg=segtree([0]*(2**k),segf,k)
for i in range(num2):
if s[i]=="?":
seg.update(n-i-1,2)
else:
seg.update(n-i-1,1)
for i in range(q):
p,c=input().split()
p=int(p)
s[p-1]=c
if p<=num2:
if c=="?":
seg.update(n-p,2)
else:
seg.update(n-p,1)
seg.update(n-p,seg.get(n-p))
print(seg.get(1))
``` | output | 1 | 76,970 | 17 | 153,941 |
Provide tags and a correct Python 3 solution for this coding contest problem.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4 | instruction | 0 | 76,971 | 17 | 153,942 |
Tags: data structures, dfs and similar, dp, implementation, trees
Correct Solution:
```
import sys
input=sys.stdin.readline
k=int(input())
s=list(input().rstrip())
q=int(input())
query=[input().rstrip().split() for i in range(q)]
a=[" "]+s[::-1]
n=len(a)
cnt=[0]*n
for i in range(n//2,n):
if a[i]=="?":
cnt[i]=2
else:
cnt[i]=1
for i in range(1,n//2)[::-1]:
if a[i]=="?":
cnt[i]=cnt[i*2]+cnt[i*2+1]
elif a[i]=="1":
cnt[i]=cnt[i*2]
else:
cnt[i]=cnt[i*2+1]
for j in range(q):
p,c=query[j]
p=int(p)
i=n-p
a[i]=c
if i>=n//2:
if a[i]=="?":
cnt[i]=2
else:
cnt[i]=1
i//=2
while i>=1:
if a[i]=="?":
cnt[i]=cnt[i*2]+cnt[i*2+1]
elif a[i]=="1":
cnt[i]=cnt[i*2]
else:
cnt[i]=cnt[i*2+1]
i//=2
print(cnt[1])
``` | output | 1 | 76,971 | 17 | 153,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4
Submitted Solution:
```
import math, sys
from itertools import permutations
from collections import defaultdict, Counter, deque
from heapq import heapify, heappush, heappop
MOD = int(1e9) + 7
INF = float('inf')
class SegTree:
def __init__(self, k):
self.n = 2 ** k
self.size = self.n * 2 - 1
self.arr = [0 for i in range(self.size)]
def get(self):
return self.arr[0]
def update(self, j, val, decision, mapped_index1, mapped_index2):
j = mapped_index2[j]
while j >= 0:
index = mapped_index1[j]
if decision[index] == '1':
self.arr[j] = self.arr[j * 2 + 2]
elif decision[index] == '0':
self.arr[j] = self.arr[j * 2 + 1]
else:
self.arr[j] = self.arr[j * 2 + 1] + self.arr[j * 2 + 2]
j = (j - 1) // 2
def __set(self, l, r, i, j, val, decision, mapped_index):
if r - l == 1:
self.arr[j] = 1
return
m = (l + r) // 2
if i < m:
self.__set(l, m, i, j * 2 + 1, val, decision, mapped_index)
else:
self.__set(m, r, i, j * 2 + 2, val, decision, mapped_index)
index = mapped_index[j]
if decision[index] == '1':
self.arr[j] = self.arr[j * 2 + 2]
elif decision[index] == '0':
self.arr[j] = self.arr[j * 2 + 1]
else:
self.arr[j] = self.arr[j * 2 + 1] + self.arr[j * 2 + 2]
def set(self, i, val, decision, mapped_index):
self.__set(0, self.n, i, 0, val, decision, mapped_index)
def solve():
k = int(input())
s = list(input())
mapped_index1 = [0 for i in range(len(s))]
mapped_index2 = {}
idx = 0
i, p = 0, 1
while i < len(s):
tmp = []
stop = i + p
add = p - 1
while i < len(s) and i < stop:
mapped_index1[i] = len(s) - (i + add) - 1
add -= 2
i += 1
p *= 2
st = SegTree(k)
for i in range(2 ** k):
st.set(i, 1, s, mapped_index1)
for i in range(len(mapped_index1)):
mapped_index2[mapped_index1[i]] = i
q = int(input())
for _ in range(q):
j, val = input().split()
j = int(j) - 1
s[j] = val
st.update(j, val, s, mapped_index1, mapped_index2)
print(st.get())
def input():
return sys.stdin.readline().rstrip('\n').strip()
def print(*args, sep=' ', end='\n'):
first = True
for arg in args:
if not first:
sys.stdout.write(sep)
sys.stdout.write(str(arg))
first = False
sys.stdout.write(end)
ts = 1
# ts = int(input())
for t in range(1, ts + 1):
solve()
``` | instruction | 0 | 76,972 | 17 | 153,944 |
Yes | output | 1 | 76,972 | 17 | 153,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4
Submitted Solution:
```
import sys
from math import gcd
input = sys.stdin.readline
def solve():
k = int(input())
s1 = list(input().strip())
a = [1]*(len(s1)*2+1)
j = 0
s = [None]*len(s1)
p = [None]*len(s1)
for i in range(k):
w = (1<<(k-i-1))-1
#print(w,1<<(k-1-i))
for z in range(1<<(k-1-i)):
p[j] = w
s[w] = s1[j]
j += 1
w += 1
#print(s)
for i in range(len(s1)-1,-1,-1):
if s[i] == '1':
a[i] = a[i*2+2]
elif s[i] == '0':
a[i] = a[i*2+1]
else:
a[i] = a[i*2+1]+a[i*2+2]
#print(s)
#print(p)
for i in range(int(input())):
x, c = input().split()
x = p[int(x)-1]
s[x] = c
while True:
if s[x] == '1':
a[x] = a[x*2+2]
elif s[x] == '0':
a[x] = a[x*2+1]
else:
a[x] = a[x*2+1]+a[x*2+2]
if x == 0:
break
x = (x-1)//2
print(a[0])
solve()
``` | instruction | 0 | 76,973 | 17 | 153,946 |
Yes | output | 1 | 76,973 | 17 | 153,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4
Submitted Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
s=list(input())[::-1]
s=s[1:]
#print(s)
temp=[1 for i in range(2000000)]
for ind in range((1<<n)-2,-1,-1):
#print(ind)
if s[ind]=='0':
temp[ind]=temp[2*ind+2]
elif s[ind]=='1':
temp[ind]=temp[2*ind+1]
else:
temp[ind]=temp[2*ind+1]+temp[2*ind+2]
#helper(0)
#print(temp)
for _ in range(int(input())):
a,b=[(x) for x in input().split()]
a=int(a)-1
a=(1<<n)-2-a
s[a]=b
ind=a
while ind>=0:
#if ind==0:
if s[ind]=='0':
temp[ind]=temp[2*ind+2]
elif s[ind]=='1':
temp[ind]=temp[2*ind+1]
else:
temp[ind]=temp[2*ind+1]+temp[2*ind+2]
if ind==0:
break
ind=((ind-1)//2)
#helper2(a)
print(temp[0])
``` | instruction | 0 | 76,974 | 17 | 153,948 |
Yes | output | 1 | 76,974 | 17 | 153,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
import math
import collections
import functools
import itertools
# from fractions import *
import heapq
import bisect
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
ALPHA='abcde'
M = 10**9 + 7
EPS = 1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def INT():return int(input())
def STR():return input()
def INTs():return tuple(map(int,input().split()))
def ARRINT():return [int(i) for i in input().split()]
def ARRSTR():return [i for i in input().split()]
#-------------------------code---------------------------#
k = INT()
n = 2**k-1
s = list(STR())
s = s[::-1]
SegmentTree = [0]*n
for i in range(n-1, -1, -1):
if 2*i+1 >= n:
if s[i] == '?':
SegmentTree[i] = 2
else:
SegmentTree[i] = 1
else:
if s[i] == '1' or s[i] == '?':
SegmentTree[i] += SegmentTree[2*i+1]
if s[i] == '0' or s[i] == '?':
SegmentTree[i] += SegmentTree[2*i+2]
for _ in range(INT()):
p, c = ARRSTR()
p = n - int(p)
s[p] = c
while True:
SegmentTree[p] = 0
if 2*p+1 >= 2**k-1:
if s[p] == '?':
SegmentTree[p] = 2
else:
SegmentTree[p] = 1
else:
if s[p] == '1' or s[p] == '?':
SegmentTree[p] += SegmentTree[2*p+1]
if s[p] == '0' or s[p] == '?':
SegmentTree[p] += SegmentTree[2*p+2]
if p == 0:
break
p = (p-1)//2
print(SegmentTree[0])
``` | instruction | 0 | 76,975 | 17 | 153,950 |
Yes | output | 1 | 76,975 | 17 | 153,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4
Submitted Solution:
```
import sys
import math
input = sys.stdin.readline
imp = 'IMPOSSIBLE'
k = int(input())
s = input()[:-1]
n = len(s)
g = {}
le_old = 1
ind = 0
for i in range(n):
le = k - math.floor(math.log2(2 ** k - 1 - i))
if le == le_old:
ind += 1
else:
ind = 1
le_old = le
if le == 1:
if s[i] == '?':
g[i + 1] = 2
else:
g[i + 1] = 1
else:
ind2 = i - ind - 2 ** (k - le + 1) + (ind - 1) * 2 + 1
if s[i] == '?':
g[i + 1] = g[ind2 + 1] + g[ind2 + 2]
elif s[i] == '0':
g[i + 1] = g[ind2 + 1]
else:
g[i + 1] = g[ind2 + 2]
q = int(input())
s = [s[i] for i in range(n)]
for i in range(q):
if i == q - 1:
p, c = input().split(' ')
else:
p, c = input()[:-1].split(' ')
p = int(p)
odohrane = 0
s[p - 1] = c
for le in range(1, k + 1):
odohrane = odohrane + 2 ** (k - le)
if odohrane < p:
continue
else:
ind = p - odohrane + 2 ** (k - le)
if le == 1:
if c == '?':
g[p] = 2
else:
g[p] = 1
else:
ind2 = p - ind - 2 ** (k - le + 1) + (ind - 1) * 2
g1 = g[ind2 + 1]
g2 = g[ind2 + 2]
#print(ind2, g1, g2)
if s[p - 1] == '?':
g[p] = g1 + g2
elif s[p - 1] == '0':
g[p] = g1
else:
g[p] = g2
#print(p, g[p])
p = odohrane + math.ceil(ind / 2)
print(g[2 ** k - 1])
#for i in g:
# print(g[i])
#print()
# print('CASE #' + str(test + 1) + ': ' + str(res))
``` | instruction | 0 | 76,976 | 17 | 153,952 |
No | output | 1 | 76,976 | 17 | 153,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4
Submitted Solution:
```
import sys
input = sys.stdin.readline
k = int(input())
s = input().strip()
a = ["*"] + list(s[::-1])
dp = [0]*(1<<k)
for i in range((1<<k)-1,0,-1):
if 2*i < 1<<k:
if a[i] == "1":
dp[i] = dp[2*i]
elif a[i] == "0":
dp[i] = dp[2*i+1]
else:
dp[i] = dp[2*i] + dp[2*i+1]
else:
if a[i] == "?":
dp[i] = 2
else:
dp[i] = 1
def solve():
idx,char = input().split()
idx = int(idx)
idx = (1<<k) - idx
a[idx] = char
cur = idx
while cur:
if 2*cur >= (1<<k):
if a[cur] == "?":
dp[cur] = 2
else:
dp[cur] = 1
cur //= 2
continue
if a[cur] == "1":
dp[cur] = dp[2*cur]
elif a[cur] == "0":
dp[cur] = dp[2*cur + 1]
else:
dp[cur] = dp[2*cur] + dp[2*cur + 1]
cur //= 2
cur = 1
ans = 1
while 2*cur < (1<<k):
if a[cur] == "1":
cur = 2*cur
elif a[cur] == "0":
cur = 2*cur + 1
else:
ans = dp[cur]
break
print(ans)
for nq in range(int(input())):
solve()
``` | instruction | 0 | 76,977 | 17 | 153,954 |
No | output | 1 | 76,977 | 17 | 153,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4
Submitted Solution:
```
#from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
#from bisect import bisect_right as br #c++ upperbound br(array,element)
#from __future__ import print_function, division #while using python2
# from itertools import accumulate
# from collections import defaultdict, Counter
def modinv(n,p):
return pow(n,p-2,p)
import math
def main():
#sys.stdin = open('input.txt', 'r')
#sys.stdout = open('output.txt', 'w')
k = int(input())
s = input()
arr = [-1]
for c in s:
if(c == '?'):arr.append(2)
else: arr.append(int(c))
p = [1 for i in range(len(arr))]
p[0] = -1
ranges = []
lst = 1
for j in range(k-1, -1,-1):
ranges.append([lst, lst+2**j-1])
lst += 2**j
def update(ind, val):
range_ind = 0
while not ranges[range_ind][0] <= ind <= ranges[range_ind][1]:
range_ind +=1
lower_ind, upper_ind = [-1, -1]
if range_ind > 0:
l, r = ranges[range_ind]
x = ind - l
lower_ind = ranges[range_ind-1][0] + 2*x
upper_ind = lower_ind + 1
if val == 0:
if lower_ind != -1:
p[ind] = p[lower_ind]
elif val == 1:
if upper_ind != -1:
p[ind] = p[upper_ind]
else:
if lower_ind != -1:
p[ind] = p[lower_ind] + p[upper_ind]
root = 2 ** k - 1
l, r = [-1, -1]
while True and range_ind < len(ranges):
if ind % 2 == 1:
l = ind
else:
l = ind-1
r = l + 1
if r >= root:
break
diff = (l - ranges[range_ind][0])//2
# print("diff", diff)
ind = r+1 + diff
range_ind += 1
if p[ind] == 1:
p[ind] = p[r]
elif p[ind] == 0:
p[ind] = p[l]
else:
p[ind] = p[l] + p[r]
for i in range(1, len(arr)):
if arr[i] == 2:
update(i, 2)
# print(p)
q = int(input())
for j in range(q):
ind, c = [x for x in input().split()]
if c == '?':
c = 2
else:
c = int(c)
update(int(ind), c)
print(p[-1])
#------------------ Python 2 and 3 footer by Pajenegod and c1729-----------------------------------------
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
import os, sys
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s:self.buffer.write(s.encode('ascii'))
self.read = lambda:self.buffer.read().decode('ascii')
self.readline = lambda:self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
if __name__ == '__main__':
main()
``` | instruction | 0 | 76,978 | 17 | 153,956 |
No | output | 1 | 76,978 | 17 | 153,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.
For example, this picture describes the chronological order of games with k = 3:
<image>
Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:
* if s_i is 0, then the team with lower index wins the i-th game;
* if s_i is 1, then the team with greater index wins the i-th game;
* if s_i is ?, then the result of the i-th game is unknown (any team could win this game).
Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.
You are given the initial state of the string s. You have to process q queries of the following form:
* p c — replace s_p with character c, and print f(s) as the result of the query.
Input
The first line contains one integer k (1 ≤ k ≤ 18).
The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.
The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.
Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.
Output
For each query, print one integer — f(s).
Example
Input
3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1
Output
1
2
3
3
5
4
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**5)
def II():
return int(sys.stdin.readline())
def LI():
return list(map(int, sys.stdin.readline().split()))
def MI():
return map(int, sys.stdin.readline().split())
def SI():
#return sys.stdin.readline().strip()
return input()
k = II()
s = list(SI())
s.reverse()
q = II()
d = [0 for i in range(len(s)//2)]
d+=[1 for i in range(len(s)//2,len(s))]
for i in range(len(s)-1, -1,-1):
ind = i
while ind!=0 and (s[(ind-1)//2] == '?' or ind%2==(int(s[(ind-1)//2]))%2):
if s[(ind-1)//2] == '?':
d[(ind-1)//2] += d[ind]
else:
d[(ind-1)//2] = d[ind]
ind-=1
ind//=2
def get(i):
global s
global d
if i >= len(s):
return 1
return d[i]
for _ in range(q):
p,c = SI().split()
p = int(p)
s[len(s)-p] = c
ind = len(s)-p
if c == '?':
d[ind] = get(ind*2+1)+get(ind*2+2)
elif c == '1':
d[ind] = get(ind*2+1)
else:
d[ind] = get(ind*2+2)
while ind!=0:
nextInd = (ind-1)//2
d[nextInd] = get(nextInd*2+1)+get(nextInd*2+2)
if s[nextInd] == '0':
d[nextInd]-=get(nextInd*2+1)
elif s[nextInd] == '1':
d[nextInd]-=get(nextInd*2+2)
ind = nextInd
print(d[0])
``` | instruction | 0 | 76,979 | 17 | 153,958 |
No | output | 1 | 76,979 | 17 | 153,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The football season has just ended in Berland. According to the rules of Berland football, each match is played between two teams. The result of each match is either a draw, or a victory of one of the playing teams. If a team wins the match, it gets w points, and the opposing team gets 0 points. If the game results in a draw, both teams get d points.
The manager of the Berland capital team wants to summarize the results of the season, but, unfortunately, all information about the results of each match is lost. The manager only knows that the team has played n games and got p points for them.
You have to determine three integers x, y and z — the number of wins, draws and loses of the team. If there are multiple answers, print any of them. If there is no suitable triple (x, y, z), report about it.
Input
The first line contains four integers n, p, w and d (1 ≤ n ≤ 10^{12}, 0 ≤ p ≤ 10^{17}, 1 ≤ d < w ≤ 10^{5}) — the number of games, the number of points the team got, the number of points awarded for winning a match, and the number of points awarded for a draw, respectively. Note that w > d, so the number of points awarded for winning is strictly greater than the number of points awarded for draw.
Output
If there is no answer, print -1.
Otherwise print three non-negative integers x, y and z — the number of wins, draws and losses of the team. If there are multiple possible triples (x, y, z), print any of them. The numbers should meet the following conditions:
* x ⋅ w + y ⋅ d = p,
* x + y + z = n.
Examples
Input
30 60 3 1
Output
17 9 4
Input
10 51 5 4
Output
-1
Input
20 0 15 5
Output
0 0 20
Note
One of the possible answers in the first example — 17 wins, 9 draws and 4 losses. Then the team got 17 ⋅ 3 + 9 ⋅ 1 = 60 points in 17 + 9 + 4 = 30 games.
In the second example the maximum possible score is 10 ⋅ 5 = 50. Since p = 51, there is no answer.
In the third example the team got 0 points, so all 20 games were lost. | instruction | 0 | 77,625 | 17 | 155,250 |
Tags: brute force, math, number theory
Correct Solution:
```
#code
n,p,w,d = map(int, input().split())
f=0
for i in range(w+1):
if ((p-(i*d))%w==0):
j = (p-(i*d))//w
q = n-i-j
if (j>=0 and q>=0):
#print(j,i,q)
f=1
break
if f==1:
print(j,i,q)
else:
print(-1)
``` | output | 1 | 77,625 | 17 | 155,251 |
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