message stringlengths 2 39.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 450 109k | cluster float64 2 2 | __index_level_0__ int64 900 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 β€ N β€ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20 | instruction | 0 | 93,483 | 2 | 186,966 |
"Correct Solution:
```
from collections import defaultdict,deque
import sys,heapq,bisect,math,itertools,string,queue,copy,time
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp(): return int(input())
def inpl(): return list(map(int, input().split()))
def inpl_str(): return list(input().split())
N = inp()
if N <= 3:
print(1)
else:
ans = 1
while True:
N = (N+2)//3
ans += 1
if N<= 3:
break
print(ans)
``` | output | 1 | 93,483 | 2 | 186,967 |
Provide a correct Python 3 solution for this coding contest problem.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 β€ N β€ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20 | instruction | 0 | 93,484 | 2 | 186,968 |
"Correct Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import math
print(math.ceil(math.log(int(input()),3)))
``` | output | 1 | 93,484 | 2 | 186,969 |
Provide a correct Python 3 solution for this coding contest problem.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 β€ N β€ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20 | instruction | 0 | 93,485 | 2 | 186,970 |
"Correct Solution:
```
import math
def getN():
return int(input())
def getlist():
return list(map(int, input().split()))
def solve(n):
cnt = 0
while(True):
n = math.ceil(n / 3)
cnt += 1
if n == 1:
return cnt
n = getN()
print(solve(n))
``` | output | 1 | 93,485 | 2 | 186,971 |
Provide a correct Python 3 solution for this coding contest problem.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 β€ N β€ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20 | instruction | 0 | 93,486 | 2 | 186,972 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n = I()
i = 0
while n > 1:
n = math.ceil(n/3)
i += 1
rr.append(i)
break
return '\n'.join(map(str, rr))
print(main())
``` | output | 1 | 93,486 | 2 | 186,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 β€ N β€ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
x = int(input())
count = 1
val = 3
while(val < x):
val *= 3
count += 1
print(count)
``` | instruction | 0 | 93,487 | 2 | 186,974 |
Yes | output | 1 | 93,487 | 2 | 186,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 β€ N β€ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
from math import log
n = int(input())
print(int(log(n - 1, 3)) + 1)
``` | instruction | 0 | 93,488 | 2 | 186,976 |
Yes | output | 1 | 93,488 | 2 | 186,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 β€ N β€ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
N = int(input())
i = 1
while True:
if pow(3, i) >= N:
print(i)
break
i += 1
``` | instruction | 0 | 93,489 | 2 | 186,978 |
Yes | output | 1 | 93,489 | 2 | 186,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 β€ N β€ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque, OrderedDict
from copy import deepcopy
from fractions import gcd
from functools import lru_cache, reduce
from math import ceil, floor
from sys import setrecursionlimit
import heapq
import itertools
import operator
# globals
inf = float('inf')
N = 0
def set_inputs():
global N
N = get_int()
return
def main():
setrecursionlimit(100000)
set_inputs()
# ----------MAIN----------
for i in range(100):
if 3 ** i >= N:
print(i)
return
return
def get_int():
return int(input())
def get_float():
return float(input())
def get_str():
return input().strip()
def get_li():
return [int(i) for i in input().split()]
def get_lf():
return [float(f) for f in input().split()]
def get_lc():
return list(input().strip())
def get_data(n, types, sep=None):
if len(types) == 1:
return [types[0](input()) for _ in range(n)]
return list(zip(*(
[t(x) for t, x in zip(types, input().split(sep=sep))]
for _ in range(n)
)))
if __name__ == '__main__':
main()
``` | instruction | 0 | 93,490 | 2 | 186,980 |
Yes | output | 1 | 93,490 | 2 | 186,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 β€ N β€ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
n = int(input())
c = 0
while n:
n //= 3
c += 1
print(c)
``` | instruction | 0 | 93,491 | 2 | 186,982 |
No | output | 1 | 93,491 | 2 | 186,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 β€ N β€ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
N = int(input())
i = 0
while True:
N //= 3
i += 1
if N == 0:
print(i)
break
``` | instruction | 0 | 93,492 | 2 | 186,984 |
No | output | 1 | 93,492 | 2 | 186,985 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 β€ N β€ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
N = int(input())
n = 1
ans = 1
while True:
n *= 3
if n > N:
print(ans)
break
ans+=1
``` | instruction | 0 | 93,493 | 2 | 186,986 |
No | output | 1 | 93,493 | 2 | 186,987 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Quarkgo Empire Expeditionary Force is an evil organization that plans to invade the Earth. In keeping with the tradition of the invaders, they continued to send monsters at a pace of one every week, targeting the area around Tokyo in Japan. However, each time, five warriors calling themselves the Human Squadron Earth Five appeared, and the monster who was rampaging in the city was easily defeated.
Walzard Thru (Earth name: Genmasa) is a female executive of the Quarkgo Empire Expeditionary Force who is seriously worried about such a situation. She had a headache under a commander who wouldn't learn anything from her weekly defeat, or a genius scientist who repeated some misplaced inventions.
Meanwhile, the next operation was decided to send the blood-sucking monster Dracula to Japan. Dracula is a terrifying monster who turns a blood-sucking human into Dracula. Humans who have been sucked by Dracula also suck the blood of other humans to increase their fellow Dracula. In this way, the strategy is to fill the entire earth with Dracula.
Upon hearing this, Walzard soon realized. This strategy is different from the usual sequel strategy such as taking over a kindergarten bus. It's rare for that bad commander to have devised it, and it has the potential to really conquer the Earth.
The momentum of the game is terrifying. Dracula, who landed on the ground, quickly increased the number of friends. If we went on like this, the invasion of the earth seemed to be just a stone's throw away. However, at that moment, a strong and unpleasant premonition ran through Walzard's mind. No way, if this monster, the original is defeated, his friends will not be wiped out, right?
When I asked the scientist who designed and developed Dracula in a hurry, he was still worried about Walzard. It is designed so that when the original Dracula is destroyed, all humans who have sucked blood will return to their original state. Don't be foolish developer. Why did you add such an extra function!
Walzard jumped to the developer and decided to kick his knee, and immediately started the original recovery work. No matter how much the original and the fake look exactly the same, if nothing is done, it is visible that the rushed Earth Five will see through the original for some reason and be defeated.
According to the developers, all Draculaized humans weigh the same, but the original Dracula is a little heavier. Then you should be able to find the original by using only the balance. You must find and retrieve the original Dracula as soon as possible before the Earth Five appears.
Input
N
The integer N (2 β€ N β€ 2,000,000,000) is written on the first line of the input. This represents the total number of Draculas, both original and fake.
Output
In the worst case, how many times is it enough to use the balance to find one original from the N Draculas using the balance? Output the minimum value. However, comparing the weights of several Draculas placed on the left and right plates of the balance is counted as one time.
Examples
Input
8
Output
2
Input
30
Output
4
Input
2000000000
Output
20
Submitted Solution:
```
#coding:utf-8
import sys
import copy
def main():
line = sys.stdin.readline()
while line.split() != ["0","0"]:
analy(line)
line = sys.stdin.readline()
return
def analy(line):
field = [int(n) for n in line.split()]
num = int(sys.stdin.readline())
data = []
for i in range(num):
a = sys.stdin.readline()
b = [int(n) for n in a.split()]
data.append(b)
print (compute(field, data))
def compute(field, ng):
list = []
for i in range(field[1]):
for j in range(field[0]):
if i == 0 :
if [j+1,i+1] in ng:
list.append(0)
elif j == 0:
#ε·¦γγͺγ
list.append(1)
else:
list.append(list[j-1])
else:
if [j+1,i+1] in ng:
list[j] = 0
elif j != 0:
#no left the same.
list[j] = list[j-1] + list[j]
return list[len(list)-1]
if __name__ == "__main__":
main()
``` | instruction | 0 | 93,494 | 2 | 186,988 |
No | output | 1 | 93,494 | 2 | 186,989 |
Provide a correct Python 3 solution for this coding contest problem.
In 20XX AD, a school competition was held. The tournament has finally left only the final competition. You are one of the athletes in the competition.
The competition you participate in is to compete for the time it takes to destroy all the blue objects placed in the space. Athletes are allowed to bring in competition guns. In the space, there are multiple blue objects, the same number of red objects, and multiple obstacles. There is a one-to-one correspondence between the blue object and the red object, and the blue object must be destroyed by shooting a bullet at the blue object from the coordinates where the red object is placed. The obstacles placed in the space are spherical and the composition is slightly different, but if it is a normal bullet, the bullet will stop there when it touches the obstacle.
The bullet used in the competition is a special bullet called Magic Bullet. This bullet can store magical power, and when the bullet touches an obstacle, it automatically consumes the magical power, and the magic that the bullet penetrates is activated. Due to the difference in the composition of obstacles, the amount of magic required to penetrate and the amount of magic power consumed to activate it are different. Therefore, even after the magic for one obstacle is activated, it is necessary to activate another magic in order to penetrate another obstacle. Also, if the bullet touches multiple obstacles at the same time, magic will be activated at the same time. The amount of magical power contained in the bullet decreases with each magic activation.
While the position and size of obstacles and the amount of magical power required to activate the penetrating magic have already been disclosed, the positions of the red and blue objects have not been disclosed. However, the position of the object could be predicted to some extent from the information of the same competition in the past. You want to save as much magical power as you can, because putting magical power into a bullet is very exhausting. Therefore, assuming the position of the red object and the corresponding blue object, the minimum amount of magical power required to be loaded in the bullet at that time, that is, the magical power remaining in the bullet when reaching the blue object is 0. Let's find the amount of magical power that becomes.
Constraints
* 0 β€ N β€ 50
* 1 β€ Q β€ 50
* -500 β€ xi, yi, zi β€ 500
* 1 β€ ri β€ 1,000
* 1 β€ li β€ 1016
* -500 β€ sxj, syj, szj β€ 500
* -500 β€ dxj, dyj, dzj β€ 500
* Obstacles are never stuck in other obstacles
* The coordinates of the object are not inside or on the surface of the obstacle
* Under each assumption, the coordinates of the red object and the blue object do not match.
Input
All inputs are integers. Each number is separated by a single space.
N Q
x1 y1 z1 r1 l1
::
xN yN zN rN lN
sx1 sy1 sz1 dx1 dy1 dz1
::
sxQ syQ szQ dxQ dyQ dzQ
* N is the number of obstacles, and Q is the number of coordinates of the assumed blue and red objects.
* xi, yi, and zi are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the center of the i-th obstacle, respectively.
* ri is the radius of the i-th obstacle.
* li is the amount of magical power consumed by magic to penetrate the i-th obstacle.
* sxj, syj, and szj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the red object in the jth assumption, respectively.
* dxj, dyj, and dzj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the blue object in the jth assumption, respectively.
Output
Assuming the position of each pair of red objects and the corresponding blue objects, the amount of magical power to be loaded in the bullet is output on one line, assuming that there are only obstacles, red objects, and one pair of blue objects in space. Let's do it. The bullet is supposed to fly in a straight line from the position of the red object to the position of the blue object, and since the size of the bullet is very small, it is treated as a point.
Examples
Input
5 1
0 10 0 5 2
0 20 0 5 12
0 30 0 5 22
0 40 0 5 32
0 50 0 5 42
0 0 0 0 60 0
Output
110
Input
1 1
10 5 0 5 9
0 0 0 9 12 0
Output
9
Input
5 5
-38 -71 -293 75 1
-158 -38 -405 66 1
-236 -303 157 266 1
316 26 411 190 1
207 -312 -27 196 1
-50 292 -375 -401 389 -389
460 278 409 -329 -303 411
215 -220 -200 309 -474 300
261 -494 -87 -300 123 -463
386 378 486 -443 -64 299
Output
0
2
1
3
0 | instruction | 0 | 93,495 | 2 | 186,990 |
"Correct Solution:
```
import math
def sub(a, b):
return [x-y for x,y in zip(a,b)]
def dot(a, b):
return sum(x*y for x,y in zip(a,b))
def cross(a, b):
return [a[1]*b[2]-a[2]*b[1], a[2]*b[0]-a[0]*b[2], a[0]*b[1]-a[1]*b[0]]
def abs2(a):
return sum([x*x for x in a])
def deg(a, b):
return (math.acos(dot(a,b)/math.sqrt(abs2(a)*abs2(b))))
N,Q = map(int,input().split())
ls = [list(map(int,input().split())) for i in range(N)]
for i in range(Q):
sx,sy,sz,dx,dy,dz = map(int,input().split())
cost = 0
s = [sx,sy,sz]
d = [dx,dy,dz]
sd = sub(d,s)
ds = sub(s,d)
for v in ls:
if abs(deg(sub(v[:3],s),sd)) < math.pi/2 and abs(deg(sub(v[:3],d), ds)) < math.pi/2:
if abs2(cross(sub(v[:3],s),sd)) <= abs2(sd)*v[3]**2:
cost += v[4]
else:
if min(abs2(sub(v[:3],s)), abs2(sub(v[:3],d))) <= v[3]**2:
cost += v[4]
print(cost)
``` | output | 1 | 93,495 | 2 | 186,991 |
Provide a correct Python 3 solution for this coding contest problem.
In 20XX AD, a school competition was held. The tournament has finally left only the final competition. You are one of the athletes in the competition.
The competition you participate in is to compete for the time it takes to destroy all the blue objects placed in the space. Athletes are allowed to bring in competition guns. In the space, there are multiple blue objects, the same number of red objects, and multiple obstacles. There is a one-to-one correspondence between the blue object and the red object, and the blue object must be destroyed by shooting a bullet at the blue object from the coordinates where the red object is placed. The obstacles placed in the space are spherical and the composition is slightly different, but if it is a normal bullet, the bullet will stop there when it touches the obstacle.
The bullet used in the competition is a special bullet called Magic Bullet. This bullet can store magical power, and when the bullet touches an obstacle, it automatically consumes the magical power, and the magic that the bullet penetrates is activated. Due to the difference in the composition of obstacles, the amount of magic required to penetrate and the amount of magic power consumed to activate it are different. Therefore, even after the magic for one obstacle is activated, it is necessary to activate another magic in order to penetrate another obstacle. Also, if the bullet touches multiple obstacles at the same time, magic will be activated at the same time. The amount of magical power contained in the bullet decreases with each magic activation.
While the position and size of obstacles and the amount of magical power required to activate the penetrating magic have already been disclosed, the positions of the red and blue objects have not been disclosed. However, the position of the object could be predicted to some extent from the information of the same competition in the past. You want to save as much magical power as you can, because putting magical power into a bullet is very exhausting. Therefore, assuming the position of the red object and the corresponding blue object, the minimum amount of magical power required to be loaded in the bullet at that time, that is, the magical power remaining in the bullet when reaching the blue object is 0. Let's find the amount of magical power that becomes.
Constraints
* 0 β€ N β€ 50
* 1 β€ Q β€ 50
* -500 β€ xi, yi, zi β€ 500
* 1 β€ ri β€ 1,000
* 1 β€ li β€ 1016
* -500 β€ sxj, syj, szj β€ 500
* -500 β€ dxj, dyj, dzj β€ 500
* Obstacles are never stuck in other obstacles
* The coordinates of the object are not inside or on the surface of the obstacle
* Under each assumption, the coordinates of the red object and the blue object do not match.
Input
All inputs are integers. Each number is separated by a single space.
N Q
x1 y1 z1 r1 l1
::
xN yN zN rN lN
sx1 sy1 sz1 dx1 dy1 dz1
::
sxQ syQ szQ dxQ dyQ dzQ
* N is the number of obstacles, and Q is the number of coordinates of the assumed blue and red objects.
* xi, yi, and zi are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the center of the i-th obstacle, respectively.
* ri is the radius of the i-th obstacle.
* li is the amount of magical power consumed by magic to penetrate the i-th obstacle.
* sxj, syj, and szj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the red object in the jth assumption, respectively.
* dxj, dyj, and dzj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the blue object in the jth assumption, respectively.
Output
Assuming the position of each pair of red objects and the corresponding blue objects, the amount of magical power to be loaded in the bullet is output on one line, assuming that there are only obstacles, red objects, and one pair of blue objects in space. Let's do it. The bullet is supposed to fly in a straight line from the position of the red object to the position of the blue object, and since the size of the bullet is very small, it is treated as a point.
Examples
Input
5 1
0 10 0 5 2
0 20 0 5 12
0 30 0 5 22
0 40 0 5 32
0 50 0 5 42
0 0 0 0 60 0
Output
110
Input
1 1
10 5 0 5 9
0 0 0 9 12 0
Output
9
Input
5 5
-38 -71 -293 75 1
-158 -38 -405 66 1
-236 -303 157 266 1
316 26 411 190 1
207 -312 -27 196 1
-50 292 -375 -401 389 -389
460 278 409 -329 -303 411
215 -220 -200 309 -474 300
261 -494 -87 -300 123 -463
386 378 486 -443 -64 299
Output
0
2
1
3
0 | instruction | 0 | 93,496 | 2 | 186,992 |
"Correct Solution:
```
#!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS():return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = I()
return l
def LIR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LI()
return l
def SR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = S()
return l
def LSR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LS()
return l
sys.setrecursionlimit(1000000)
mod = 1000000007
#A
def A():
n = I()
s = 0
ans = "YES"
for i in range(n):
k = input()
if k == "Un":
s -= 1
else:
s += 1
if s < 0:
ans = "NO"
if s:
print("NO")
else:
print(ans)
return
#B
def B():
return
#C
def C():
n,q = LI()
p = LIR(n)
for _ in range(q):
x0,y0,z0,s,t,u = LI()
a,b,c = (s-x0,t-y0,u-z0)
ans = 0
su = a**2+b**2+c**2
for xr,yr,zr,r,l in p:
x = (su-a**2)/su*x0-a/su*(b*y0+c*z0-a*xr-b*yr-c*zr)
y = (su-b**2)/su*y0-b/su*(a*x0+c*z0-a*xr-b*yr-c*zr)
z = (su-c**2)/su*z0-c/su*(b*y0+a*x0-a*xr-b*yr-c*zr)
if min(x0,s)-1e-4 <= x <= max(x0,s)+1e-4 and min(y0,t)-1e-4 <= y <= max(y0,t)+1e-4 and min(z0,u)-1e-4 <= z <= max(z0,u)+1e-4:
x -= xr
y -= yr
z -= zr
r2 = x**2+y**2+z**2
if r2 <= r**2+1e-4:
ans += l
print(ans)
return
#D
def D():
return
#E
def E():
return
#F
def F():
return
#G
def G():
return
#H
def H():
return
#I
def I_():
return
#J
def J():
return
#Solve
if __name__ == "__main__":
C()
``` | output | 1 | 93,496 | 2 | 186,993 |
Provide a correct Python 3 solution for this coding contest problem.
In 20XX AD, a school competition was held. The tournament has finally left only the final competition. You are one of the athletes in the competition.
The competition you participate in is to compete for the time it takes to destroy all the blue objects placed in the space. Athletes are allowed to bring in competition guns. In the space, there are multiple blue objects, the same number of red objects, and multiple obstacles. There is a one-to-one correspondence between the blue object and the red object, and the blue object must be destroyed by shooting a bullet at the blue object from the coordinates where the red object is placed. The obstacles placed in the space are spherical and the composition is slightly different, but if it is a normal bullet, the bullet will stop there when it touches the obstacle.
The bullet used in the competition is a special bullet called Magic Bullet. This bullet can store magical power, and when the bullet touches an obstacle, it automatically consumes the magical power, and the magic that the bullet penetrates is activated. Due to the difference in the composition of obstacles, the amount of magic required to penetrate and the amount of magic power consumed to activate it are different. Therefore, even after the magic for one obstacle is activated, it is necessary to activate another magic in order to penetrate another obstacle. Also, if the bullet touches multiple obstacles at the same time, magic will be activated at the same time. The amount of magical power contained in the bullet decreases with each magic activation.
While the position and size of obstacles and the amount of magical power required to activate the penetrating magic have already been disclosed, the positions of the red and blue objects have not been disclosed. However, the position of the object could be predicted to some extent from the information of the same competition in the past. You want to save as much magical power as you can, because putting magical power into a bullet is very exhausting. Therefore, assuming the position of the red object and the corresponding blue object, the minimum amount of magical power required to be loaded in the bullet at that time, that is, the magical power remaining in the bullet when reaching the blue object is 0. Let's find the amount of magical power that becomes.
Constraints
* 0 β€ N β€ 50
* 1 β€ Q β€ 50
* -500 β€ xi, yi, zi β€ 500
* 1 β€ ri β€ 1,000
* 1 β€ li β€ 1016
* -500 β€ sxj, syj, szj β€ 500
* -500 β€ dxj, dyj, dzj β€ 500
* Obstacles are never stuck in other obstacles
* The coordinates of the object are not inside or on the surface of the obstacle
* Under each assumption, the coordinates of the red object and the blue object do not match.
Input
All inputs are integers. Each number is separated by a single space.
N Q
x1 y1 z1 r1 l1
::
xN yN zN rN lN
sx1 sy1 sz1 dx1 dy1 dz1
::
sxQ syQ szQ dxQ dyQ dzQ
* N is the number of obstacles, and Q is the number of coordinates of the assumed blue and red objects.
* xi, yi, and zi are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the center of the i-th obstacle, respectively.
* ri is the radius of the i-th obstacle.
* li is the amount of magical power consumed by magic to penetrate the i-th obstacle.
* sxj, syj, and szj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the red object in the jth assumption, respectively.
* dxj, dyj, and dzj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the blue object in the jth assumption, respectively.
Output
Assuming the position of each pair of red objects and the corresponding blue objects, the amount of magical power to be loaded in the bullet is output on one line, assuming that there are only obstacles, red objects, and one pair of blue objects in space. Let's do it. The bullet is supposed to fly in a straight line from the position of the red object to the position of the blue object, and since the size of the bullet is very small, it is treated as a point.
Examples
Input
5 1
0 10 0 5 2
0 20 0 5 12
0 30 0 5 22
0 40 0 5 32
0 50 0 5 42
0 0 0 0 60 0
Output
110
Input
1 1
10 5 0 5 9
0 0 0 9 12 0
Output
9
Input
5 5
-38 -71 -293 75 1
-158 -38 -405 66 1
-236 -303 157 266 1
316 26 411 190 1
207 -312 -27 196 1
-50 292 -375 -401 389 -389
460 278 409 -329 -303 411
215 -220 -200 309 -474 300
261 -494 -87 -300 123 -463
386 378 486 -443 -64 299
Output
0
2
1
3
0 | instruction | 0 | 93,497 | 2 | 186,994 |
"Correct Solution:
```
N, Q = map(int, input().split())
x = [None] * N
y = [None] * N
z = [None] * N
r = [None] * N
l = [None] * N
for i in range(N):
x[i], y[i], z[i], r[i], l[i] = map(int, input().split())
for _ in range(Q):
ans = 0
sx, sy, sz, dx, dy, dz = map(int, input().split())
vx = dx - sx
vy = dy - sy
vz = dz - sz
for i in range(N):
t = 0
t += vx * (x[i] - sx)
t += vy * (y[i] - sy)
t += vz * (z[i] - sz)
t /= vx**2 + vy**2 + vz**2
len2 = 0
len2 += (sx + vx * t - x[i])**2
len2 += (sy + vy * t - y[i])**2
len2 += (sz + vz * t - z[i])**2
if 0 < t < 1 and len2 <= r[i]**2 + 1e-9:
ans += l[i]
print(ans)
``` | output | 1 | 93,497 | 2 | 186,995 |
Provide a correct Python 3 solution for this coding contest problem.
In 20XX AD, a school competition was held. The tournament has finally left only the final competition. You are one of the athletes in the competition.
The competition you participate in is to compete for the time it takes to destroy all the blue objects placed in the space. Athletes are allowed to bring in competition guns. In the space, there are multiple blue objects, the same number of red objects, and multiple obstacles. There is a one-to-one correspondence between the blue object and the red object, and the blue object must be destroyed by shooting a bullet at the blue object from the coordinates where the red object is placed. The obstacles placed in the space are spherical and the composition is slightly different, but if it is a normal bullet, the bullet will stop there when it touches the obstacle.
The bullet used in the competition is a special bullet called Magic Bullet. This bullet can store magical power, and when the bullet touches an obstacle, it automatically consumes the magical power, and the magic that the bullet penetrates is activated. Due to the difference in the composition of obstacles, the amount of magic required to penetrate and the amount of magic power consumed to activate it are different. Therefore, even after the magic for one obstacle is activated, it is necessary to activate another magic in order to penetrate another obstacle. Also, if the bullet touches multiple obstacles at the same time, magic will be activated at the same time. The amount of magical power contained in the bullet decreases with each magic activation.
While the position and size of obstacles and the amount of magical power required to activate the penetrating magic have already been disclosed, the positions of the red and blue objects have not been disclosed. However, the position of the object could be predicted to some extent from the information of the same competition in the past. You want to save as much magical power as you can, because putting magical power into a bullet is very exhausting. Therefore, assuming the position of the red object and the corresponding blue object, the minimum amount of magical power required to be loaded in the bullet at that time, that is, the magical power remaining in the bullet when reaching the blue object is 0. Let's find the amount of magical power that becomes.
Constraints
* 0 β€ N β€ 50
* 1 β€ Q β€ 50
* -500 β€ xi, yi, zi β€ 500
* 1 β€ ri β€ 1,000
* 1 β€ li β€ 1016
* -500 β€ sxj, syj, szj β€ 500
* -500 β€ dxj, dyj, dzj β€ 500
* Obstacles are never stuck in other obstacles
* The coordinates of the object are not inside or on the surface of the obstacle
* Under each assumption, the coordinates of the red object and the blue object do not match.
Input
All inputs are integers. Each number is separated by a single space.
N Q
x1 y1 z1 r1 l1
::
xN yN zN rN lN
sx1 sy1 sz1 dx1 dy1 dz1
::
sxQ syQ szQ dxQ dyQ dzQ
* N is the number of obstacles, and Q is the number of coordinates of the assumed blue and red objects.
* xi, yi, and zi are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the center of the i-th obstacle, respectively.
* ri is the radius of the i-th obstacle.
* li is the amount of magical power consumed by magic to penetrate the i-th obstacle.
* sxj, syj, and szj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the red object in the jth assumption, respectively.
* dxj, dyj, and dzj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the blue object in the jth assumption, respectively.
Output
Assuming the position of each pair of red objects and the corresponding blue objects, the amount of magical power to be loaded in the bullet is output on one line, assuming that there are only obstacles, red objects, and one pair of blue objects in space. Let's do it. The bullet is supposed to fly in a straight line from the position of the red object to the position of the blue object, and since the size of the bullet is very small, it is treated as a point.
Examples
Input
5 1
0 10 0 5 2
0 20 0 5 12
0 30 0 5 22
0 40 0 5 32
0 50 0 5 42
0 0 0 0 60 0
Output
110
Input
1 1
10 5 0 5 9
0 0 0 9 12 0
Output
9
Input
5 5
-38 -71 -293 75 1
-158 -38 -405 66 1
-236 -303 157 266 1
316 26 411 190 1
207 -312 -27 196 1
-50 292 -375 -401 389 -389
460 278 409 -329 -303 411
215 -220 -200 309 -474 300
261 -494 -87 -300 123 -463
386 378 486 -443 -64 299
Output
0
2
1
3
0 | instruction | 0 | 93,498 | 2 | 186,996 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-1,1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
n,q = LI()
a = [LI() for _ in range(n)]
b = [LI() for _ in range(q)]
rr = []
def k(a,b):
return sum([(a[i]-b[i]) ** 2 for i in range(3)]) ** 0.5
def f(a,b,c,r):
ab = k(a,b)
ac = k(a,c)
bc = k(b,c)
if ac <= r or bc <= r:
return True
at = (ac ** 2 - r ** 2)
bt = (bc ** 2 - r ** 2)
t = max(at,0) ** 0.5 + max(bt,0) ** 0.5
return ab >= t - eps
for x1,y1,z1,x2,y2,z2 in b:
tr = 0
ta = (x1,y1,z1)
tb = (x2,y2,z2)
for x,y,z,r,l in a:
if f(ta,tb,(x,y,z),r):
tr += l
rr.append(tr)
return '\n'.join(map(str,rr))
print(main())
``` | output | 1 | 93,498 | 2 | 186,997 |
Provide a correct Python 3 solution for this coding contest problem.
In 20XX AD, a school competition was held. The tournament has finally left only the final competition. You are one of the athletes in the competition.
The competition you participate in is to compete for the time it takes to destroy all the blue objects placed in the space. Athletes are allowed to bring in competition guns. In the space, there are multiple blue objects, the same number of red objects, and multiple obstacles. There is a one-to-one correspondence between the blue object and the red object, and the blue object must be destroyed by shooting a bullet at the blue object from the coordinates where the red object is placed. The obstacles placed in the space are spherical and the composition is slightly different, but if it is a normal bullet, the bullet will stop there when it touches the obstacle.
The bullet used in the competition is a special bullet called Magic Bullet. This bullet can store magical power, and when the bullet touches an obstacle, it automatically consumes the magical power, and the magic that the bullet penetrates is activated. Due to the difference in the composition of obstacles, the amount of magic required to penetrate and the amount of magic power consumed to activate it are different. Therefore, even after the magic for one obstacle is activated, it is necessary to activate another magic in order to penetrate another obstacle. Also, if the bullet touches multiple obstacles at the same time, magic will be activated at the same time. The amount of magical power contained in the bullet decreases with each magic activation.
While the position and size of obstacles and the amount of magical power required to activate the penetrating magic have already been disclosed, the positions of the red and blue objects have not been disclosed. However, the position of the object could be predicted to some extent from the information of the same competition in the past. You want to save as much magical power as you can, because putting magical power into a bullet is very exhausting. Therefore, assuming the position of the red object and the corresponding blue object, the minimum amount of magical power required to be loaded in the bullet at that time, that is, the magical power remaining in the bullet when reaching the blue object is 0. Let's find the amount of magical power that becomes.
Constraints
* 0 β€ N β€ 50
* 1 β€ Q β€ 50
* -500 β€ xi, yi, zi β€ 500
* 1 β€ ri β€ 1,000
* 1 β€ li β€ 1016
* -500 β€ sxj, syj, szj β€ 500
* -500 β€ dxj, dyj, dzj β€ 500
* Obstacles are never stuck in other obstacles
* The coordinates of the object are not inside or on the surface of the obstacle
* Under each assumption, the coordinates of the red object and the blue object do not match.
Input
All inputs are integers. Each number is separated by a single space.
N Q
x1 y1 z1 r1 l1
::
xN yN zN rN lN
sx1 sy1 sz1 dx1 dy1 dz1
::
sxQ syQ szQ dxQ dyQ dzQ
* N is the number of obstacles, and Q is the number of coordinates of the assumed blue and red objects.
* xi, yi, and zi are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the center of the i-th obstacle, respectively.
* ri is the radius of the i-th obstacle.
* li is the amount of magical power consumed by magic to penetrate the i-th obstacle.
* sxj, syj, and szj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the red object in the jth assumption, respectively.
* dxj, dyj, and dzj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the blue object in the jth assumption, respectively.
Output
Assuming the position of each pair of red objects and the corresponding blue objects, the amount of magical power to be loaded in the bullet is output on one line, assuming that there are only obstacles, red objects, and one pair of blue objects in space. Let's do it. The bullet is supposed to fly in a straight line from the position of the red object to the position of the blue object, and since the size of the bullet is very small, it is treated as a point.
Examples
Input
5 1
0 10 0 5 2
0 20 0 5 12
0 30 0 5 22
0 40 0 5 32
0 50 0 5 42
0 0 0 0 60 0
Output
110
Input
1 1
10 5 0 5 9
0 0 0 9 12 0
Output
9
Input
5 5
-38 -71 -293 75 1
-158 -38 -405 66 1
-236 -303 157 266 1
316 26 411 190 1
207 -312 -27 196 1
-50 292 -375 -401 389 -389
460 278 409 -329 -303 411
215 -220 -200 309 -474 300
261 -494 -87 -300 123 -463
386 378 486 -443 -64 299
Output
0
2
1
3
0 | instruction | 0 | 93,499 | 2 | 186,998 |
"Correct Solution:
```
eps = 1.0 / 10**10
def LI(): return [int(x) for x in input().split()]
def main():
n,q = LI()
na = [LI() for _ in range(n)]
qa = [LI() for _ in range(q)]
rr = []
def k(a,b):
return sum([(a[i]-b[i]) ** 2 for i in range(3)]) ** 0.5
def f(a,b,c,r):
ab = k(a,b)
ac = k(a,c)
bc = k(b,c)
if ac <= r or bc <= r:
return True
at = (ac ** 2 - r ** 2) ** 0.5
bt = (bc ** 2 - r ** 2) ** 0.5
return ab >= at + bt - eps
for x1,y1,z1,x2,y2,z2 in qa:
tr = 0
for x,y,z,r,l in na:
if f((x1,y1,z1),(x2,y2,z2),(x,y,z),r):
tr += l
rr.append(tr)
return '\n'.join(map(str,rr))
print(main())
``` | output | 1 | 93,499 | 2 | 186,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In 20XX AD, a school competition was held. The tournament has finally left only the final competition. You are one of the athletes in the competition.
The competition you participate in is to compete for the time it takes to destroy all the blue objects placed in the space. Athletes are allowed to bring in competition guns. In the space, there are multiple blue objects, the same number of red objects, and multiple obstacles. There is a one-to-one correspondence between the blue object and the red object, and the blue object must be destroyed by shooting a bullet at the blue object from the coordinates where the red object is placed. The obstacles placed in the space are spherical and the composition is slightly different, but if it is a normal bullet, the bullet will stop there when it touches the obstacle.
The bullet used in the competition is a special bullet called Magic Bullet. This bullet can store magical power, and when the bullet touches an obstacle, it automatically consumes the magical power, and the magic that the bullet penetrates is activated. Due to the difference in the composition of obstacles, the amount of magic required to penetrate and the amount of magic power consumed to activate it are different. Therefore, even after the magic for one obstacle is activated, it is necessary to activate another magic in order to penetrate another obstacle. Also, if the bullet touches multiple obstacles at the same time, magic will be activated at the same time. The amount of magical power contained in the bullet decreases with each magic activation.
While the position and size of obstacles and the amount of magical power required to activate the penetrating magic have already been disclosed, the positions of the red and blue objects have not been disclosed. However, the position of the object could be predicted to some extent from the information of the same competition in the past. You want to save as much magical power as you can, because putting magical power into a bullet is very exhausting. Therefore, assuming the position of the red object and the corresponding blue object, the minimum amount of magical power required to be loaded in the bullet at that time, that is, the magical power remaining in the bullet when reaching the blue object is 0. Let's find the amount of magical power that becomes.
Constraints
* 0 β€ N β€ 50
* 1 β€ Q β€ 50
* -500 β€ xi, yi, zi β€ 500
* 1 β€ ri β€ 1,000
* 1 β€ li β€ 1016
* -500 β€ sxj, syj, szj β€ 500
* -500 β€ dxj, dyj, dzj β€ 500
* Obstacles are never stuck in other obstacles
* The coordinates of the object are not inside or on the surface of the obstacle
* Under each assumption, the coordinates of the red object and the blue object do not match.
Input
All inputs are integers. Each number is separated by a single space.
N Q
x1 y1 z1 r1 l1
::
xN yN zN rN lN
sx1 sy1 sz1 dx1 dy1 dz1
::
sxQ syQ szQ dxQ dyQ dzQ
* N is the number of obstacles, and Q is the number of coordinates of the assumed blue and red objects.
* xi, yi, and zi are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the center of the i-th obstacle, respectively.
* ri is the radius of the i-th obstacle.
* li is the amount of magical power consumed by magic to penetrate the i-th obstacle.
* sxj, syj, and szj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the red object in the jth assumption, respectively.
* dxj, dyj, and dzj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the blue object in the jth assumption, respectively.
Output
Assuming the position of each pair of red objects and the corresponding blue objects, the amount of magical power to be loaded in the bullet is output on one line, assuming that there are only obstacles, red objects, and one pair of blue objects in space. Let's do it. The bullet is supposed to fly in a straight line from the position of the red object to the position of the blue object, and since the size of the bullet is very small, it is treated as a point.
Examples
Input
5 1
0 10 0 5 2
0 20 0 5 12
0 30 0 5 22
0 40 0 5 32
0 50 0 5 42
0 0 0 0 60 0
Output
110
Input
1 1
10 5 0 5 9
0 0 0 9 12 0
Output
9
Input
5 5
-38 -71 -293 75 1
-158 -38 -405 66 1
-236 -303 157 266 1
316 26 411 190 1
207 -312 -27 196 1
-50 292 -375 -401 389 -389
460 278 409 -329 -303 411
215 -220 -200 309 -474 300
261 -494 -87 -300 123 -463
386 378 486 -443 -64 299
Output
0
2
1
3
0
Submitted Solution:
```
from math import sin, acos
N, Q = map(int, input().split())
obs = [tuple(map(int, input().split())) for _ in range(N)]
def dist(v):
return sum(x ** 2 for x in v)
def cos(v1, v2):
return sum(x * y for x, y in zip(v1, v2)) / pow(dist(v1), 0.5) / pow(dist(v2), 0.5)
for _ in range(Q):
sx, sy, sz, dx, dy, dz = map(int, input().split())
v_oa = (dx - sx, dy - sy, dz - sz)
ans = 0
for x, y, z, r, l in obs:
v_or = (x - sx, y - sy, z - sz)
# print(pow(dist(v_or), 0.5), sin(acos(cos(v_or, v_oa))), abs(pow(dist(v_or), 0.5) * sin(acos(cos(v_or, v_oa)))), r + 1e-9)
if cos(v_or, v_oa) >= 0 and abs(pow(dist(v_or), 0.5) * sin(acos(cos(v_or, v_oa)))) <= r + 1e-9:
ans += l
print(ans)
``` | instruction | 0 | 93,500 | 2 | 187,000 |
No | output | 1 | 93,500 | 2 | 187,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In 20XX AD, a school competition was held. The tournament has finally left only the final competition. You are one of the athletes in the competition.
The competition you participate in is to compete for the time it takes to destroy all the blue objects placed in the space. Athletes are allowed to bring in competition guns. In the space, there are multiple blue objects, the same number of red objects, and multiple obstacles. There is a one-to-one correspondence between the blue object and the red object, and the blue object must be destroyed by shooting a bullet at the blue object from the coordinates where the red object is placed. The obstacles placed in the space are spherical and the composition is slightly different, but if it is a normal bullet, the bullet will stop there when it touches the obstacle.
The bullet used in the competition is a special bullet called Magic Bullet. This bullet can store magical power, and when the bullet touches an obstacle, it automatically consumes the magical power, and the magic that the bullet penetrates is activated. Due to the difference in the composition of obstacles, the amount of magic required to penetrate and the amount of magic power consumed to activate it are different. Therefore, even after the magic for one obstacle is activated, it is necessary to activate another magic in order to penetrate another obstacle. Also, if the bullet touches multiple obstacles at the same time, magic will be activated at the same time. The amount of magical power contained in the bullet decreases with each magic activation.
While the position and size of obstacles and the amount of magical power required to activate the penetrating magic have already been disclosed, the positions of the red and blue objects have not been disclosed. However, the position of the object could be predicted to some extent from the information of the same competition in the past. You want to save as much magical power as you can, because putting magical power into a bullet is very exhausting. Therefore, assuming the position of the red object and the corresponding blue object, the minimum amount of magical power required to be loaded in the bullet at that time, that is, the magical power remaining in the bullet when reaching the blue object is 0. Let's find the amount of magical power that becomes.
Constraints
* 0 β€ N β€ 50
* 1 β€ Q β€ 50
* -500 β€ xi, yi, zi β€ 500
* 1 β€ ri β€ 1,000
* 1 β€ li β€ 1016
* -500 β€ sxj, syj, szj β€ 500
* -500 β€ dxj, dyj, dzj β€ 500
* Obstacles are never stuck in other obstacles
* The coordinates of the object are not inside or on the surface of the obstacle
* Under each assumption, the coordinates of the red object and the blue object do not match.
Input
All inputs are integers. Each number is separated by a single space.
N Q
x1 y1 z1 r1 l1
::
xN yN zN rN lN
sx1 sy1 sz1 dx1 dy1 dz1
::
sxQ syQ szQ dxQ dyQ dzQ
* N is the number of obstacles, and Q is the number of coordinates of the assumed blue and red objects.
* xi, yi, and zi are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the center of the i-th obstacle, respectively.
* ri is the radius of the i-th obstacle.
* li is the amount of magical power consumed by magic to penetrate the i-th obstacle.
* sxj, syj, and szj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the red object in the jth assumption, respectively.
* dxj, dyj, and dzj are the x-coordinate, y-coordinate, and z-coordinate that represent the position of the blue object in the jth assumption, respectively.
Output
Assuming the position of each pair of red objects and the corresponding blue objects, the amount of magical power to be loaded in the bullet is output on one line, assuming that there are only obstacles, red objects, and one pair of blue objects in space. Let's do it. The bullet is supposed to fly in a straight line from the position of the red object to the position of the blue object, and since the size of the bullet is very small, it is treated as a point.
Examples
Input
5 1
0 10 0 5 2
0 20 0 5 12
0 30 0 5 22
0 40 0 5 32
0 50 0 5 42
0 0 0 0 60 0
Output
110
Input
1 1
10 5 0 5 9
0 0 0 9 12 0
Output
9
Input
5 5
-38 -71 -293 75 1
-158 -38 -405 66 1
-236 -303 157 266 1
316 26 411 190 1
207 -312 -27 196 1
-50 292 -375 -401 389 -389
460 278 409 -329 -303 411
215 -220 -200 309 -474 300
261 -494 -87 -300 123 -463
386 378 486 -443 -64 299
Output
0
2
1
3
0
Submitted Solution:
```
from math import sin, acos
N, Q = map(int, input().split())
obs = [tuple(map(int, input().split())) for _ in range(N)]
def dist(v):
return sum(x ** 2 for x in v)
def cos(v1, v2):
return sum(x * y for x, y in zip(v1, v2)) / pow(dist(v1), 0.5) / pow(dist(v2), 0.5)
for _ in range(Q):
sx, sy, sz, dx, dy, dz = map(int, input().split())
v_oa = (dx - sx, dy - sy, dz - sz)
ans = 0
for x, y, z, r, l in obs:
v_or = (x - sx, y - sy, z - sz)
# print(pow(dist(v_or), 0.5), sin(acos(cos(v_or, v_oa))), abs(pow(dist(v_or), 0.5) * sin(acos(cos(v_or, v_oa)))), r + 1e-9)
if abs(pow(dist(v_or), 0.5) * sin(acos(cos(v_or, v_oa)))) <= r + 1e-9:
ans += l
print(ans)
``` | instruction | 0 | 93,501 | 2 | 187,002 |
No | output | 1 | 93,501 | 2 | 187,003 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6. | instruction | 0 | 93,564 | 2 | 187,128 |
Tags: binary search, brute force, divide and conquer, math
Correct Solution:
```
from bisect import bisect_left, bisect_right
import sys
def input():
return sys.stdin.readline().rstrip()
def slv():
n, k, A, B = map(int, input().split())
avengers = list(map(int, input().split()))
avengers.sort()
def get_count(l, r):
"""
return the number of hero in range[l,r).
"""
if l >= r:
return 0
return bisect_right(avengers, r - 1) - bisect_right(avengers, l - 1)
def query(L, R):
assert L <= R
if L == R:
cnt = get_count(L, R + 1)
if cnt:
return B * cnt
return A
else:
med = (R + L + 1)//2
#[L,R] -> [L,med] [med,R]
cnt = get_count(L, R + 1)
if cnt:
res = B * cnt * (R - L + 1)
else:
res = A
return res
return min(query(L, med - 1) + query(med, R), res)
ans = query(1, (1 << n))
print(ans)
return
def main():
t = 1
for i in range(t):
slv()
return
if __name__ == "__main__":
main()
``` | output | 1 | 93,564 | 2 | 187,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6. | instruction | 0 | 93,565 | 2 | 187,130 |
Tags: binary search, brute force, divide and conquer, math
Correct Solution:
```
import sys
from bisect import bisect_left as bl, bisect_right as br
def min_cost(l, r):
cnt = br(pos, r) - bl(pos, l)
if cnt == 0:
return a
curr = b * cnt
if l == r:
return curr
m = (l + r) // 2
return min(curr * (r - l + 1), min_cost(l, m) + min_cost(m + 1, r))
n, k, a, b = map(int, input().split())
pos = [int(x) for x in input().split()]
pos.sort()
print(min_cost(1, 2 ** n))
``` | output | 1 | 93,565 | 2 | 187,131 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6. | instruction | 0 | 93,566 | 2 | 187,132 |
Tags: binary search, brute force, divide and conquer, math
Correct Solution:
```
from bisect import bisect_left as lb, bisect_right as ub
n, k, A, B = map(int, input().split(' '))
a = sorted(map(int, input().split(' ')))
def heihei(l, r):
cnt = ub(a, r) - lb(a, l)
if cnt == 0:
return A
if l == r:
return A if cnt == 0 else B * cnt
mid = (l + r) >> 1
lr, rr = heihei(l, mid), heihei(mid + 1, r)
return min(B * cnt * (r - l + 1), lr + rr)
print(heihei(1, 2 ** n))
``` | output | 1 | 93,566 | 2 | 187,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6. | instruction | 0 | 93,567 | 2 | 187,134 |
Tags: binary search, brute force, divide and conquer, math
Correct Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 2/11/19
"""
import collections
import time
import os
import sys
import bisect
import heapq
N, K, A, B = map(int, input().split())
positions = [int(x) for x in input().split()]
positions.sort()
def count(l, r):
a = bisect.bisect_left(positions, l)
b = bisect.bisect_right(positions, r)
return b-a
def solve(l, r):
a, b = bisect.bisect_left(positions, l), bisect.bisect_right(positions, r)
x = b - a
ans = 0
if x == 0:
ans = A
else:
ans = B * x * (r-l+1)
if l == r or x == 0:
return ans
m = (l+r) // 2
ans = min(ans, solve(l, m) + solve(m+1, r))
return ans
print(solve(1, 1 << N))
``` | output | 1 | 93,567 | 2 | 187,135 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6. | instruction | 0 | 93,568 | 2 | 187,136 |
Tags: binary search, brute force, divide and conquer, math
Correct Solution:
```
import bisect
n, k, A, B = map(int, input().split())
X = sorted([int(a)-1 for a in input().split()])
def calc(l, r):
def cnt(a, b):
return bisect.bisect_left(X, b) - bisect.bisect_left(X, a)
c = cnt(l, r)
if c == 0:
ret = A
elif r-l == 1:
ret = B * c
else:
m = (l+r) // 2
a = cnt(l, m)
b = c - a
if a == 0:
ret = min(A + calc(m, r), B * c * (r - l))
elif b == 0:
ret = min(A + calc(l, m), B * c * (r - l))
else:
ret = calc(l, m) + calc(m, r)
return ret
print(calc(0, 2**n))
``` | output | 1 | 93,568 | 2 | 187,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6. | instruction | 0 | 93,569 | 2 | 187,138 |
Tags: binary search, brute force, divide and conquer, math
Correct Solution:
```
n, k, A, B = map(int, input().split())
X = sorted([int(a)-1 for a in input().split()])
def calc(l, r, S):
if len(S) == 0:
ret = A
elif r-l == 1:
ret = B * len(S)
else:
m = (l+r) // 2
SL = []
SR = []
for s in S:
if s < m:
SL.append(s)
else:
SR.append(s)
ret = min(B * len(S) * (r - l), calc(l, m, SL) + calc(m, r, SR))
return ret
print(calc(0, 2**n, X))
``` | output | 1 | 93,569 | 2 | 187,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6. | instruction | 0 | 93,570 | 2 | 187,140 |
Tags: binary search, brute force, divide and conquer, math
Correct Solution:
```
def I(): return(list(map(int,input().split())))
def rec(l,r):
numavens=getnums(l,r)
# print(numavens,l,r)
if numavens==0:return a
if l==r:return b*numavens
mid=(r-l)//2+l
return min(((r-l)+1)*numavens*b,rec(l,mid)+rec(mid+1,r))
def getnums(l,r):
if l==0:return get(r)
return get(r)-get(l-1)
def get(t):
l=0
r=k-1
# print(x)
while(l<=r):
mid=(r-l)//2+l
if x[mid]<=t:l=mid+1
else: r=mid-1
# print(l,t)
return l
n,k,a,b=I()
x=I()
x.sort()
l=1
r=2**n
print(rec(l,r))
``` | output | 1 | 93,570 | 2 | 187,141 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6. | instruction | 0 | 93,571 | 2 | 187,142 |
Tags: binary search, brute force, divide and conquer, math
Correct Solution:
```
import bisect
n, k, A, B = map(int, input().split())
X = sorted([int(a)-1 for a in input().split()])
def calc(l, r, S):
cl = bisect.bisect_left(S, l)
cr = bisect.bisect_left(S, r)
c = cr - cl
if c == 0:
ret = A
elif r-l == 1:
ret = B * c
else:
m = (l+r) // 2
cm = bisect.bisect_left(S, m)
a = cm - cl
b = cr - cm
SL = []
SR = []
for s in S:
if s < m:
SL.append(s)
else:
SR.append(s)
if a == 0:
ret = min(A + calc(m, r, SR), B * c * (r - l))
elif b == 0:
ret = min(A + calc(l, m, SL), B * c * (r - l))
else:
ret = calc(l, m, SL) + calc(m, r, SR)
return ret
print(calc(0, 2**n, X))
``` | output | 1 | 93,571 | 2 | 187,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6.
Submitted Solution:
```
n,k,A,B = map(int,input().split())
f = list(sorted(list(map(int, input().split()))))
def min_cost(l, r, full):
if len(full) == 0:
return A
elif l == r:
return B * len(full)
else:
mid = (l + r) // 2
left = []
right = []
for i in full:
if i <= mid:
left.append(i)
else:
right.append(i)
return min(len(full) * (r - l + 1) * B, min_cost(l, mid, left) + min_cost(mid + 1, r, right))
print(min_cost(1, 2 ** n, f))
``` | instruction | 0 | 93,572 | 2 | 187,144 |
Yes | output | 1 | 93,572 | 2 | 187,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6.
Submitted Solution:
```
n,k,A,B = map(int,input().split())
avg = [int(i) for i in input().split()]
avg.sort()
def get(l,h,a):
if len(a) == 0: return A
elif l==h: return B * len(a)
else:
mid = (l+h)//2
left = []
right = []
for i in a:
if i <= mid: left.append(i)
else: right.append(i)
pow_a = len(a)*(h-l+1)*B
pow_b = get(l,mid,left) + get(mid+1,h,right)
return min(pow_a,pow_b)
print(get(1,2**n,avg))
``` | instruction | 0 | 93,573 | 2 | 187,146 |
Yes | output | 1 | 93,573 | 2 | 187,147 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6.
Submitted Solution:
```
n,k,a,b=map(int,input().split())
from bisect import bisect_right as br
from bisect import bisect_left as bl
p=list(map(int,input().split()))
p.sort()
def dp(i,j,nb):
if i==j:
if nb==0:return(a)
return(nb*b)
if nb==0:c=a;return(c)
else:c=b*nb*(j-i+1)
r=i+(j-i)//2
l=r+1
nbl=br(p,r)-bl(p,i)
nbr=nb-nbl
return(min(c,dp(i,r,nbl)+dp(l,j,nbr)))
print(dp(1,2**n,k))
``` | instruction | 0 | 93,574 | 2 | 187,148 |
Yes | output | 1 | 93,574 | 2 | 187,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6.
Submitted Solution:
```
import bisect
# Turn list(str) into list(int)
def getIntList(lst):
assert type(lst) is list
rep = []
for i in lst:rep.append(int(i))
return rep
n,k,A,B = getIntList(input().split())
pos = getIntList(input().split())
def calc(l,r,heros):
assert type(heros) is list
if len(heros) == 0: return A
if r-l == 0: return B*len(heros)
mid = bisect.bisect_right(heros,(l+r)//2)
tans = min(
calc(l,(l+r)//2,heros[:mid])+calc((l+r)//2+1,r,heros[mid:]),
B*len(heros)*(r-l+1)
)
return tans
pos.sort()
ans = calc(1,2**n,pos)
print(ans)
``` | instruction | 0 | 93,575 | 2 | 187,150 |
Yes | output | 1 | 93,575 | 2 | 187,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6.
Submitted Solution:
```
n, k, A, B = list(map(int, input().split()))
n = 2 ** n
a = sorted(list(map(int, input().split())))
def recurs(start, end, sum_all, array):
if sum_all == 0:
return A
else:
energy = B*(end - start)*sum_all
if end - start == 1:
return energy
c, d = [], []
for x in array:
if x < ((start + end)//2):
c += [x]
else:
d += [x]
first = sum(c)
second = sum_all - first
tmp = recurs(start, (start + end)//2, first, c)
tmp += recurs((start + end)//2, end, second, d)
return (energy < tmp) * energy + (1 - (energy < tmp))*tmp
print (recurs(0, n, k, a))
# import numpy as np
# print('ΠΠ²Π΅Π΄ΠΈΡΠ΅ ΠΊΠΎΠ»ΠΈΡΠ΅ΡΡΠ²ΠΎ ΡΡΠ°Π²Π½Π΅Π½ΠΈΠΉ:')
# n = int(input())
# print('ΠΠ²Π΅Π΄ΠΈΡΠ΅ Π»ΠΈΠ½Π΅ΠΉΠ½ΡΠ΅ ΠΊΠΎΡΡΠΈΡΠΈΠ΅Π½ΡΡ:')
# a = np.array([list(float(t) for t in input().split()) for i in range(n)])
# print(a.shape)
``` | instruction | 0 | 93,576 | 2 | 187,152 |
No | output | 1 | 93,576 | 2 | 187,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6.
Submitted Solution:
```
def get(tl, tr, a):
global A, B
if len(a) == 0:
return A
if tl == tr:
return B
tm = (tl + tr) // 2
c, d = [], []
for x in a:
if x <= tm:
c += [x]
else:
d += [x]
return min(B * len(a) * (tr - tl + 1), get(tl, tm, c) + get(tm + 1, tr, d))
n, k, A, B = map(int, input().split())
a = list(sorted(list(map(int, input().split()))))
print(get(1, 1 << n, a))
``` | instruction | 0 | 93,577 | 2 | 187,154 |
No | output | 1 | 93,577 | 2 | 187,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6.
Submitted Solution:
```
nkAB=list(map(int,input().split()))
K=list(map(int,input().split()))
n=nkAB[0]
k=nkAB[1]
A=nkAB[2]
B=nkAB[3]
posn_avn=[]
for i in range(pow(2,n)):
posn_avn.append(0)
for i in K:
posn_avn[i-1]=1
def result(l,a,b):
m=0
if(l.count(0)==len(l))and(len(l)>=1):
m+=a
if(len(l)==1)and(l[0]==1):
m+=b
else:
if(len(l)>1)and(l.count(0)!=len(l)):
l1=l[0:int(len(l)/2)]
l2=l[int(len(l)/2):len(l)]
m+=result(l1,a,b)
m+=result(l2,a,b)
return m
print(result(posn_avn,A,B))
``` | instruction | 0 | 93,578 | 2 | 187,156 |
No | output | 1 | 93,578 | 2 | 187,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.
Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:
* if the current length is at least 2, divide the base into 2 equal halves and destroy them separately, or
* burn the current base. If it contains no avenger in it, it takes A amount of power, otherwise it takes his B β
n_a β
l amount of power, where n_a is the number of avengers and l is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input
The first line contains four integers n, k, A and B (1 β€ n β€ 30, 1 β€ k β€ 10^5, 1 β€ A,B β€ 10^4), where 2^n is the length of the base, k is the number of avengers and A and B are the constants explained in the question.
The second line contains k integers a_{1}, a_{2}, a_{3}, β¦, a_{k} (1 β€ a_{i} β€ 2^n), where a_{i} represents the position of avenger in the base.
Output
Output one integer β the minimum power needed to destroy the avengers base.
Examples
Input
2 2 1 2
1 3
Output
6
Input
3 2 1 2
1 7
Output
8
Note
Consider the first example.
One option for Thanos is to burn the whole base 1-4 with power 2 β
2 β
4 = 16.
Otherwise he can divide the base into two parts 1-2 and 3-4.
For base 1-2, he can either burn it with power 2 β
1 β
2 = 4 or divide it into 2 parts 1-1 and 2-2.
For base 1-1, he can burn it with power 2 β
1 β
1 = 2. For 2-2, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 1-2 is 2 + 1 = 3, which is less than 4.
Similarly, he needs 3 power to destroy 3-4. The total minimum power needed is 6.
Submitted Solution:
```
def explosion(a, x, y, q, l, r):
if r - l == 1:
q1 = q
while q1 < len(a) and a[q1] < r:
q1 += 1
if q1 == q:
return x
else:
return (q1-q)*y
q1 = q
while q1 < len(a) and a[q1] < (l+r)//2:
q1 += 1
q2 = q1
while q2 < len(a) and a[q2] < r:
q2 += 1
if q1-q == 0 and q2-q1 == 0:
return x
if q2-q1 != 0 and q1-q != 0:
return explosion(a, x, y, q, l, (l+r)//2)+explosion(a, x, y, q1, (l+r)//2, r)
if q1 != q:
d = explosion(a, x, y, q, l, (l+r)//2)
if d > x:
return x+explosion(a, x, y, q, l, (l+r)//2)
else:
return (r-l)*(q1-q)*y
else:
d = explosion(a, x, y, q1, (l+r)//2, r)
if d > x:
return x+explosion(a, x, y, q1, (l+r)//2, r)
else:
return (r-l)*(q2-q1)*y
n, k, x, y = map(int, input().split())
a = sorted(list(map(int, input().split())))
print(explosion(a, x, y, 0, 0, 2**n))
``` | instruction | 0 | 93,579 | 2 | 187,158 |
No | output | 1 | 93,579 | 2 | 187,159 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted. | instruction | 0 | 93,747 | 2 | 187,494 |
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
from itertools import permutations as p
from collections import Counter,defaultdict as dd
from sys import stdin,stdout
inp=stdin.readline
for t in range(int(inp())):
s=inp().rstrip('\n')
l=set(s)
l1=[i for i in p(l)]
d=Counter(s)
c=0
ans=l.copy()
for i in l1:
x=0
d1=dd(int)
sh=dd(int)
for j in i:
d1[j]=x
x+=d[j]
x=0
a=0
for j in s:
x+=d1[j]+sh[j]-a
a+=1
d1[j]+=1
for k in i:
if k==j:
break
sh[k]+=1
if x>c:
c=x
ans=list(i)
ansf=[]
for i in ans:
ansf.append(i*d[i])
stdout.write(''.join(ansf)+'\n')
``` | output | 1 | 93,747 | 2 | 187,495 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted. | instruction | 0 | 93,748 | 2 | 187,496 |
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
import time
#start_time = time.time()
#def TIME_(): print(time.time()-start_time)
import os, sys
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string, heapq as h
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.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")
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def getBin(): return list(map(int,list(input())))
def isInt(s): return '0' <= s[0] <= '9'
def ceil_(a,b): return (a+b-1)//b
MOD = 10**9 + 7
"""
"""
import itertools
def mergeSort(arr, n):
# A temp_arr is created to store
# sorted array in merge function
temp_arr = [0]*n
return _mergeSort(arr, temp_arr, 0, n-1)
def _mergeSort(arr, temp_arr, left, right):
# A variable inv_count is used to store
# inversion counts in each recursive call
inv_count = 0
# We will make a recursive call if and only if
# we have more than one elements
if left < right:
# mid is calculated to divide the array into two subarrays
# Floor division is must in case of python
mid = (left + right)//2
# It will calculate inversion counts in the left subarray
inv_count += _mergeSort(arr, temp_arr, left, mid)
# It will calculate inversion counts in right subarray
inv_count += _mergeSort(arr, temp_arr, mid + 1, right)
# It will merge two subarrays in a sorted subarray
inv_count += merge(arr, temp_arr, left, mid, right)
return inv_count
# This function will merge two subarrays in a single sorted subarray
def merge(arr, temp_arr, left, mid, right):
i = left # Starting index of left subarray
j = mid + 1 # Starting index of right subarray
k = left # Starting index of to be sorted subarray
inv_count = 0
# Conditions are checked to make sure that i and j don't exceed their
# subarray limits.
while i <= mid and j <= right:
# There will be no inversion if arr[i] <= arr[j]
if arr[i] <= arr[j]:
temp_arr[k] = arr[i]
k += 1
i += 1
else:
# Inversion will occur.
temp_arr[k] = arr[j]
inv_count += (mid-i + 1)
k += 1
j += 1
# Copy the remaining elements of left subarray into temporary array
while i <= mid:
temp_arr[k] = arr[i]
k += 1
i += 1
# Copy the remaining elements of right subarray into temporary array
while j <= right:
temp_arr[k] = arr[j]
k += 1
j += 1
# Copy the sorted subarray into Original array
for loop_var in range(left, right + 1):
arr[loop_var] = temp_arr[loop_var]
return inv_count
ops = list(range(4))
letters = ['A','N','T','O']
def solve():
S = listStr()
N = len(S)
apps = [[] for _ in range(4)]
for i,s in enumerate(S):
if s == 'A': apps[0].append(i)
elif s == 'N': apps[1].append(i)
elif s == 'T': apps[2].append(i)
else: apps[3].append(i)
best = 0
ans = S
for perm in itertools.permutations(ops):
tmp = []
for i in range(4):
tmp += apps[perm[i]]
curr = mergeSort(tmp,N)
if curr > best:
best = curr
ans = []
for i in range(4):
ans += [letters[perm[i]]]*len(apps[perm[i]])
return ''.join(ans)
for _ in range(getInt()):
print(solve())
#solve()
#TIME_()
``` | output | 1 | 93,748 | 2 | 187,497 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted. | instruction | 0 | 93,749 | 2 | 187,498 |
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
import itertools
TC = int(input())
m = {"A": 0, "N": 1, "O": 2, "T": 3}
st = "ANOT"
for _ in range(TC):
s = input()
arr = []
for ch in s:
arr.append(m[ch])
cnt = [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
cnt1 = [0, 0, 0, 0]
for i in arr:
for j in range(4):
cnt[j][i] += cnt1[j]
cnt1[i]+=1
val=-1
ans=[]
for perm in itertools.permutations([0,1,2,3]):
curr=0
for i in range(4):
for j in range(i+1,4):
curr+=cnt[perm[j]][perm[i]]
if curr>val:
val=curr
ans=perm
for i in range(4):
for j in range(cnt1[ans[i]]):
print(st[ans[i]],end="")
print()
``` | output | 1 | 93,749 | 2 | 187,499 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted. | instruction | 0 | 93,750 | 2 | 187,500 |
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
from itertools import permutations
def f(pos, k):
m = len(pos)
a1, a2 = [0], [0]
for i in range(m):
a1.append(a1[-1]+pos[i]-i)
for i in range(1, m+1):
a2.append(a2[-1]+k-i-pos[m-i])
a2 = a2[::-1]
ind = max(range(m+1), key=lambda i: a1[i]+a2[i])
return ind, a1[ind] + a2[ind]
def solve(s):
n = len(s)
ma = -1
ans = []
for p in permutations(["A","N","O","T"]):
cur = 0
ans1, ans2 = [], []
s1 = s[::]
for c in p:
s2 = []
pos = []
for i in range(len(s1)):
if s1[i] == c:
pos.append(i)
else:
s2.append(s1[i])
x, y = f(pos, len(s1))
cur += y
ans1.extend([c]*x)
ans2.extend([c]*(len(pos)-x))
s1 = s2
ans1.extend(ans2[::-1])
if cur > ma:
ma = cur
ans = ans1
return "".join(ans)
import sys
input = lambda: sys.stdin.readline().rstrip()
t = int(input())
for i in range(t):
s = list(input())
print(solve(s))
``` | output | 1 | 93,750 | 2 | 187,501 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted. | instruction | 0 | 93,751 | 2 | 187,502 |
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
from itertools import permutations
class SegmentTree(): # adapted from https://www.geeksforgeeks.org/segment-tree-efficient-implementation/
def __init__(self,arr,func,initialRes=0):
self.f=func
self.N=len(arr)
self.tree=[0 for _ in range(2*self.N)]
self.initialRes=initialRes
for i in range(self.N):
self.tree[self.N+i]=arr[i]
for i in range(self.N-1,0,-1):
self.tree[i]=self.f(self.tree[i<<1],self.tree[i<<1|1])
def updateTreeNode(self,idx,value): #update value at arr[idx]
self.tree[idx+self.N]=value
idx+=self.N
i=idx
while i>1:
self.tree[i>>1]=self.f(self.tree[i],self.tree[i^1])
i>>=1
def query(self,l,r): #get sum (or whatever function) on interval [l,r] inclusive
r+=1
res=self.initialRes
l+=self.N
r+=self.N
while l<r:
if l&1:
res=self.f(res,self.tree[l])
l+=1
if r&1:
r-=1
res=self.f(res,self.tree[r])
l>>=1
r>>=1
return res
def getMaxSegTree(arr):
return SegmentTree(arr,lambda a,b:max(a,b),initialRes=-float('inf'))
def getMinSegTree(arr):
return SegmentTree(arr,lambda a,b:min(a,b),initialRes=float('inf'))
def getSumSegTree(arr):
return SegmentTree(arr,lambda a,b:a+b,initialRes=0)
def getMinMoves(arr,s):
n=len(s)
charIndexes=dict() # store indexes of each char in s in descending order
for i in range(n-1,-1,-1):
if s[i] not in charIndexes.keys():
charIndexes[s[i]]=[]
charIndexes[s[i]].append(i)
elmCnts=getSumSegTree([1]*n) # to track how many indexes ahead
nMoves=0
for c in arr:
i=charIndexes[c].pop()
nMoves+=(elmCnts.query(0,i)-1)
elmCnts.updateTreeNode(i,0)
return nMoves
def main():
n=int(input())
allans=[]
for _ in range(n):
s=input()
# My guess is that the solution involves grouping all the same characters together
chars=[]
charCnts=dict()
for c in s:
if c not in charCnts.keys():
chars.append(c)
charCnts[c]=0
charCnts[c]+=1
currMax=-1
ans=[]
for p in permutations(chars):
arr=[]
for c in p:
for __ in range(charCnts[c]):
arr.append(c)
nMoves=getMinMoves(arr,s)
if nMoves>currMax:
ans=arr
currMax=nMoves
allans.append(''.join(ans))
multiLineArrayPrint(allans)
return
import sys
# input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
# def readFloatArr():
# return [float(x) for x in input().split()]
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
dv=defaultValFactory;da=dimensionArr
if len(da)==1:return [dv() for _ in range(da[0])]
else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
def queryInteractive(i,j):
print('? {} {}'.format(i,j))
sys.stdout.flush()
return int(input())
def answerInteractive(ans):
print('! {}'.format(' '.join([str(x) for x in ans])))
sys.stdout.flush()
inf=float('inf')
MOD=10**9+7
# MOD=998244353
for _abc in range(1):
main()
``` | output | 1 | 93,751 | 2 | 187,503 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted. | instruction | 0 | 93,752 | 2 | 187,504 |
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
import sys
from itertools import permutations
def I(): return int(sys.stdin.readline().rstrip())
def MI(): return map(int,sys.stdin.readline().rstrip().split())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def LI2(): return list(map(int,sys.stdin.readline().rstrip()))
def S(): return sys.stdin.readline().rstrip()
def LS(): return list(sys.stdin.readline().rstrip().split())
def LS2(): return list(sys.stdin.readline().rstrip())
t = I()
C = ['A','N','T','O']
for _ in range(t):
X = S()
count1 = [0]*4
count2 = [[0]*4 for _ in range(4)]
for x in X:
if x == 'A':
count1[0] += 1
for i in range(4):
if i == 0:
continue
count2[i][0] += count1[i]
elif x == 'N':
count1[1] += 1
for i in range(4):
if i == 1:
continue
count2[i][1] += count1[i]
elif x == 'T':
count1[2] += 1
for i in range(4):
if i == 2:
continue
count2[i][2] += count1[i]
else:
count1[3] += 1
for i in range(4):
if i == 3:
continue
count2[i][3] += count1[i]
M = -1
ans = (0,0,0,0)
for A in list(permutations([0,1,2,3],4)):
cnt = 0
for i in range(4):
a = A[i]
for j in range(i+1,4):
b = A[j]
cnt += count2[b][a]
if M < cnt:
M = cnt
ans = A
ANS = [C[ans[0]]]*count1[ans[0]]+[C[ans[1]]]*count1[ans[1]]+[C[ans[2]]]*count1[ans[2]]+[C[ans[3]]]*count1[ans[3]]
ANS = ''.join(ANS)
print(ANS)
``` | output | 1 | 93,752 | 2 | 187,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted. | instruction | 0 | 93,753 | 2 | 187,506 |
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq,bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
from itertools import permutations
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
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")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary-----------------------------------
def merge(arr, temp, left, mid, right):
inv_count = 0
i = left
j = mid
k = left
while ((i <= mid - 1) and (j <= right)):
if (arr[i] <= arr[j]):
temp[k] = arr[i]
k += 1
i += 1
else:
temp[k] = arr[j]
k += 1
j += 1
inv_count = inv_count + (mid - i)
while (i <= mid - 1):
temp[k] = arr[i]
k += 1
i += 1
while (j <= right):
temp[k] = arr[j]
k += 1
j += 1
for i in range(left, right + 1, 1):
arr[i] = temp[i]
return inv_count
def _mergeSort(arr, temp, left, right):
inv_count = 0
if (right > left):
mid = int((right + left) / 2)
inv_count = _mergeSort(arr, temp, left, mid)
inv_count += _mergeSort(arr, temp, mid + 1, right)
inv_count += merge(arr, temp, left, mid + 1, right)
return inv_count
def countSwaps(arr, n):
temp = [0 for i in range(n)]
return _mergeSort(arr, temp, 0, n - 1)
for ik in range(int(input())):
s=list(input())
n=len(s)
ind=defaultdict(int)
c=defaultdict(int)
for i in range(n-1,-1,-1):
c[s[i]]+=1
ind = defaultdict(list)
for i in range(n):
ind[s[i]].append(i)
def find(d,tr):
tr=list(tr)
fi=""
for i in tr:
fi+=i*c[i]
ans=[]
for i in tr:
for j in ind[i]:
ans.append(j)
return (countSwaps(ans,len(ans)),fi)
perm = [''.join(p) for p in permutations("ANTO")]
ans=(0,"")
for i in perm:
ans=max(ans,find(s,i))
print(ans[1])
``` | output | 1 | 93,753 | 2 | 187,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted. | instruction | 0 | 93,754 | 2 | 187,508 |
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import Counter, deque, defaultdict
from itertools import permutations
class Fenwick:
def __init__(self, n):
self.tree = [0] * (n + 1)
self.n = n
def update(self, x, val):
while x <= self.n:
self.tree[x] += val
x += x & -x
def query(self, x):
ans = 0
while x:
ans += self.tree[x]
x -= x & -x
return ans
for _ in range(int(input())):
s = input()[:-1]
n = len(s)
cand = set(s)
m = len(cand)
cnt = Counter(s)
ans = -1
for idx in permutations(cand):
cur = ""
for x in idx:
cur += x * cnt[x]
dic = defaultdict(deque)
for i, c in enumerate(s, 1):
dic[c].append(i)
A = []
for c in cur:
A.append(dic[c].popleft())
bit = Fenwick(n)
tot = 0
for a in A:
tot += bit.query(n) - bit.query(a)
bit.update(a, 1)
if tot > ans:
ans = tot
P = cur
print(P)
``` | output | 1 | 93,754 | 2 | 187,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
from itertools import permutations
t = int(input())
perms = list(permutations(range(4)))
def inversion(d, L):
if len(d) == 1:
return (g for g in d), 1, 0
a, A, b = inversion(d[:L//2], L//2)
c, C, e = inversion(d[L//2:], (L+1)//2)
k = 0
s = []
a_ = next(a)
c_ = next(c)
while a_ != -1 != c_:
if a_ < c_:
s.append(a_)
a_ = next(a, -1)
A -= 1
else:
s.append(c_)
c_ = next(c, -1)
k += A
while a_ != -1:
s.append(a_)
a_ = next(a, -1)
while c_ != -1:
s.append(c_)
c_ = next(c, -1)
return (g for g in s), L, b+e+k
for _ in range(t):
s = input().strip()
pos = [[], [], [], []]
for i, c in enumerate(s):
pos["ANTO".index(c)].append(i)
siz = [len(a) for a in pos]
m = 0
best = (0, 1, 2, 3)
for k in perms:
d = []
for j in range(4):
d += pos[k[j]]
z = inversion(d, len(d))[2]
if m < z:
best = k
m = z
res = ""
for x in best:
res += "ANTO"[x] * siz[x]
print(res)
``` | instruction | 0 | 93,755 | 2 | 187,510 |
Yes | output | 1 | 93,755 | 2 | 187,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
import os
import 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")
#######################################
from itertools import permutations as p
from collections import Counter,defaultdict as dd
for t in range(int(input())):
s=input()
l=set(s)
l1=[i for i in p(l)]
d=Counter(s)
c=0
ans=l.copy()
for i in l1:
x=0
d1=dd(int)
sh=dd(int)
for j in i:
d1[j]=x
x+=d[j]
x=0
a=0
for j in s:
x+=d1[j]+sh[j]-a
a+=1
d1[j]+=1
for k in i:
if k==j:
break
sh[k]+=1
if x>c:
c=x
ans=list(i)
ansf=[]
for i in ans:
ansf.append(i*d[i])
print(''.join(ansf))
``` | instruction | 0 | 93,756 | 2 | 187,512 |
Yes | output | 1 | 93,756 | 2 | 187,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
#!/usr/bin/env python3
import sys
import getpass # not available on codechef
import math, random
import functools, itertools, collections, heapq, bisect
from collections import Counter, defaultdict, deque
input = sys.stdin.readline # to read input quickly
# available on Google, AtCoder Python3, not available on Codeforces
# import numpy as np
# import scipy
M9 = 10**9 + 7 # 998244353
yes, no = "YES", "NO"
# d4 = [(1,0),(0,1),(-1,0),(0,-1)]
# d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
# d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)] # hexagonal layout
MAXINT = sys.maxsize
# if testing locally, print to terminal with a different color
OFFLINE_TEST = getpass.getuser() == "hkmac"
# OFFLINE_TEST = False # codechef does not allow getpass
def log(*args):
if OFFLINE_TEST:
print('\033[36m', *args, '\033[0m', file=sys.stderr)
def solve(*args):
# screen input
if OFFLINE_TEST:
log("----- solving ------")
log(*args)
log("----- ------- ------")
return solve_(*args)
def read_matrix(rows):
return [list(map(int,input().split())) for _ in range(rows)]
def read_strings(rows):
return [input().strip() for _ in range(rows)]
def minus_one(arr):
return [x-1 for x in arr]
def minus_one_matrix(mrr):
return [[x-1 for x in row] for row in mrr]
# ---------------------------- template ends here ----------------------------
class FenwickTree:
# also known as Binary Indexed Tree
# binarysearch.com/problems/Virtual-Array
# https://leetcode.com/problems/create-sorted-array-through-instructions
# may need to be implemented again to reduce constant factor
def __init__(self, bits=31):
self.c = defaultdict(int)
self.LARGE = 2**bits
def update(self, x, increment):
x += 1 # to avoid infinite loop at x > 0
while x <= self.LARGE:
# increase by the greatest power of two that divides x
self.c[x] += increment
x += x & -x
def query(self, x):
x += 1 # to avoid infinite loop at x > 0
res = 0
while x > 0:
# decrease by the greatest power of two that divides x
res += self.c[x]
x -= x & -x
return res
def count_inversions(perm):
res = 0
t = FenwickTree(bits = len(bin(max(perm))))
for i,x in enumerate(perm):
cnt = t.query(x)
res += i-cnt
t.update(x, 1)
return res
def solve_(srr):
if len(srr) == 1:
return srr
c1 = defaultdict(list)
for i,x in enumerate(srr):
c1[x].append(i)
c = Counter(srr)
maxval = -1
maxres = None
for comb in itertools.permutations(c.keys()):
locs = []
for k in comb:
locs.extend(c1[k])
val = count_inversions(locs)
if val > maxval:
maxval = val
maxres = comb
log(maxval)
trr = ""
for k in maxres:
trr += k*len(c1[k])
# log(maxval)
return trr
# for case_num in [0]: # no loop over test case
# for case_num in range(100): # if the number of test cases is specified
for case_num in range(int(input())):
# read line as an integer
# k = int(input())
# read line as a string
srr = input().strip()
# read one line and parse each word as a string
# lst = input().split()
# read one line and parse each word as an integer
# a,b,c = list(map(int,input().split()))
# lst = list(map(int,input().split()))
# lst = minus_one(lst)
# read multiple rows
# arr = read_strings(k) # and return as a list of str
# mrr = read_matrix(k) # and return as a list of list of int
# mrr = minus_one_matrix(mrr)
res = solve(srr) # include input here
# print length if applicable
# print(len(res))
# parse result
# res = " ".join(str(x) for x in res)
# res = "\n".join(str(x) for x in res)
# res = "\n".join(" ".join(str(x) for x in row) for row in res)
# print result
# print("Case #{}: {}".format(case_num+1, res)) # Google and Facebook - case number required
print(res)
``` | instruction | 0 | 93,757 | 2 | 187,514 |
Yes | output | 1 | 93,757 | 2 | 187,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
from itertools import permutations
class BIT():
def __init__(self, init):
if type(init) == int:
self.n = init + 1
self.X = [0] * self.n
else:
self.n = len(init) + 1
self.X = [0] + init
for i in range(1, self.n):
if i + (i & -i) < self.n:
self.X[i + (i & -i)] += self.X[i]
def add(self, i, x=1):
i += 1
while i < self.n:
self.X[i] += x
i += i & (-i)
def getsum(self, i):
ret = 0
while i != 0:
ret += self.X[i]
i -= i & (-i)
return ret
def getrange(self, l, r):
return self.getsum(r) - self.getsum(l)
def inversion_number(L):
bit = BIT(len(L))
ret = 0
for i, a in enumerate(L):
ret += i - bit.getsum(a + 1)
bit.add(a)
return ret
def inversion_number(L): # οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
ll = sorted(list(set(L)))
dd = {a: i for i, a in enumerate(ll)}
nL = [dd[l] for l in L]
bit = BIT(len(nL))
ret = 0
for i, a in enumerate(nL):
ret += i - bit.getsum(a + 1)
bit.add(a)
return ret
def calc(A):
n = len(A)
C = [0] * 4
for a in A:
C[a] += 1
ma = 0
pma = [0, 1, 2, 3]
for p in permutations([0, 1, 2, 3]):
D = [0] * 4
c = 0
for i, a in enumerate(p):
D[a] = c
c += C[a]
B = [0] * n
for i, a in enumerate(A):
B[i] = D[a]
D[a] += 1
# print("p, B =", p, B)
ivn = inversion_number(B)
if ivn > ma:
ma = ivn
pma = p
# print("ma, p =", ma, p)
re = []
for a in pma:
re.append("ANTO"[a] * C[a])
return "".join(re)
T = int(input())
for _ in range(T):
A = [0 if a == "A" else 1 if a == "N" else 2 if a == "T" else 3 for a in input()]
print(calc(A))
``` | instruction | 0 | 93,758 | 2 | 187,516 |
Yes | output | 1 | 93,758 | 2 | 187,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
t=int(input());
for i in range(t):
w=input();
nw=w[1:]+w[:0]
print(nw)
``` | instruction | 0 | 93,759 | 2 | 187,518 |
No | output | 1 | 93,759 | 2 | 187,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
import sys
input = sys.stdin.readline
from itertools import permutations
# all operations are log(n)
class bit:
def __init__(self, n):
self.n = n+1
self.mx_p = 0
while 1 << self.mx_p < self.n:
self.mx_p += 1
self.a = [0]*(self.n)
self.tot = 0
def add(self, idx, val):
self.tot += val
idx += 1
while idx < self.n:
self.a[idx] += val
idx += idx & -idx
def sum_prefix(self, idx):
tot = 0
idx += 1
while idx > 0:
tot += self.a[idx]
idx -= idx & -idx
return tot
# inversions of a permutation 0..len(a)-1
def inversions(a):
tot = 0
loc = [0]*len(a)
for i in range(len(a)):
loc[a[i]] = i
bt = bit(len(a))
for i in range(len(a)-1,-1,-1):
tot += bt.sum_prefix(loc[i])
bt.add(loc[i],1)
return tot
# creates array c where c[i] = position of a[i] in b
def create_c(a,b):
c = [0]*len(a)
idx = {}
positions = {}
for i in range(len(b)):
if b[i] in positions:
positions[b[i]].append(i)
else:
positions[b[i]] = [i]
for i in range(len(a)):
if a[i] in idx:
idx[a[i]] += 1
else:
idx[a[i]] = 0
c[i] = positions[a[i]][idx[a[i]]]
return c
for _ in range(int(input())):
a = input().strip()
if len(a) <= 4:
print(a)
continue
counts = dict(zip("ANTO",[0]*4))
for i in a:
counts[i] += 1
best_perm = None
best_perm_cost = -1
for order in permutations("ANTO"):
final_perm = ''.join([char*counts[char] for char in order])
cost = inversions(create_c(a,final_perm))
if cost > best_perm_cost:
best_perm = final_perm
best_perm_cost = cost
print(best_perm)
``` | instruction | 0 | 93,760 | 2 | 187,520 |
No | output | 1 | 93,760 | 2 | 187,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 β€ t β€ 100000) β the number of testcases.
The first and only line of each testcase contains 1 string a (1 β€ |a| β€ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA β NNOAT β NNAOT β NANOT β NANTO β ANNTO β ANTNO β ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
#from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write(' '.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
#from decimal import Decimal
#from fractions import Fraction
#sys.setrecursionlimit(100000)
INF = float('inf')
mod = int(1e9)+7
for t in range(int(data())):
a = data()
m = [[0]*26 for i in range(26)]
d = dd(int)
for i in a:
s = ord(i)-65
for j in range(26):
if s!=j:
m[s][j] += m[j][j]
m[s][s] += 1
d[i] += 1
l = sorted(d.keys())
total = 0
for i in range(len(l)-1):
s1 = ord(l[i]) - 65
for j in range(i+1,len(l)):
s2 = ord(l[j]) - 65
total += m[s1][s2]
flag = True
while flag:
flag = False
for i in range(1,len(l)):
ind = i
total1 = total
s1 = ord(l[i]) - 65
for j in range(i-1,-1,-1):
s2 = ord(l[j]) - 65
total1 += m[s1][s2] - m[s2][s1]
if total1>total:
total = total1
flag = True
while ind!=j:
l[ind],l[ind-1] = l[ind-1], l[ind]
ind -= 1
ans = ''
for i in l:
ans += i*d[i]
out(ans)
``` | instruction | 0 | 93,761 | 2 | 187,522 |
No | output | 1 | 93,761 | 2 | 187,523 |
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