message stringlengths 2 22.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 145 109k | cluster float64 9 9 | __index_level_0__ int64 290 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The programming competition season has already started and it's time to train for ICPC. Sereja coaches his teams for a number of year and he knows that to get ready for the training session it's not enough to prepare only problems and editorial. As the training sessions lasts for several hours, teams become hungry. Thus, Sereja orders a number of pizzas so they can eat right after the end of the competition.
Teams plan to train for n times during n consecutive days. During the training session Sereja orders exactly one pizza for each team that is present this day. He already knows that there will be ai teams on the i-th day.
There are two types of discounts in Sereja's favourite pizzeria. The first discount works if one buys two pizzas at one day, while the second is a coupon that allows to buy one pizza during two consecutive days (two pizzas in total).
As Sereja orders really a lot of pizza at this place, he is the golden client and can use the unlimited number of discounts and coupons of any type at any days.
Sereja wants to order exactly ai pizzas on the i-th day while using only discounts and coupons. Note, that he will never buy more pizzas than he need for this particular day. Help him determine, whether he can buy the proper amount of pizzas each day if he is allowed to use only coupons and discounts. Note, that it's also prohibited to have any active coupons after the end of the day n.
Input
The first line of input contains a single integer n (1 ≤ n ≤ 200 000) — the number of training sessions.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 10 000) — the number of teams that will be present on each of the days.
Output
If there is a way to order pizzas using only coupons and discounts and do not buy any extra pizzas on any of the days, then print "YES" (without quotes) in the only line of output. Otherwise, print "NO" (without quotes).
Examples
Input
4
1 2 1 2
Output
YES
Input
3
1 0 1
Output
NO
Note
In the first sample, Sereja can use one coupon to buy one pizza on the first and the second days, one coupon to buy pizza on the second and the third days and one discount to buy pizzas on the fourth days. This is the only way to order pizzas for this sample.
In the second sample, Sereja can't use neither the coupon nor the discount without ordering an extra pizza. Note, that it's possible that there will be no teams attending the training sessions on some days.
Submitted Solution:
```
input()
p=0
for x in map(int, input().split()):
if x<p:
p=-1
break
p^=x&1
print('YES' if p==0 else 'NO')
``` | instruction | 0 | 20,792 | 9 | 41,584 |
Yes | output | 1 | 20,792 | 9 | 41,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The programming competition season has already started and it's time to train for ICPC. Sereja coaches his teams for a number of year and he knows that to get ready for the training session it's not enough to prepare only problems and editorial. As the training sessions lasts for several hours, teams become hungry. Thus, Sereja orders a number of pizzas so they can eat right after the end of the competition.
Teams plan to train for n times during n consecutive days. During the training session Sereja orders exactly one pizza for each team that is present this day. He already knows that there will be ai teams on the i-th day.
There are two types of discounts in Sereja's favourite pizzeria. The first discount works if one buys two pizzas at one day, while the second is a coupon that allows to buy one pizza during two consecutive days (two pizzas in total).
As Sereja orders really a lot of pizza at this place, he is the golden client and can use the unlimited number of discounts and coupons of any type at any days.
Sereja wants to order exactly ai pizzas on the i-th day while using only discounts and coupons. Note, that he will never buy more pizzas than he need for this particular day. Help him determine, whether he can buy the proper amount of pizzas each day if he is allowed to use only coupons and discounts. Note, that it's also prohibited to have any active coupons after the end of the day n.
Input
The first line of input contains a single integer n (1 ≤ n ≤ 200 000) — the number of training sessions.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 10 000) — the number of teams that will be present on each of the days.
Output
If there is a way to order pizzas using only coupons and discounts and do not buy any extra pizzas on any of the days, then print "YES" (without quotes) in the only line of output. Otherwise, print "NO" (without quotes).
Examples
Input
4
1 2 1 2
Output
YES
Input
3
1 0 1
Output
NO
Note
In the first sample, Sereja can use one coupon to buy one pizza on the first and the second days, one coupon to buy pizza on the second and the third days and one discount to buy pizzas on the fourth days. This is the only way to order pizzas for this sample.
In the second sample, Sereja can't use neither the coupon nor the discount without ordering an extra pizza. Note, that it's possible that there will be no teams attending the training sessions on some days.
Submitted Solution:
```
n=int(input())
trains=list(map(int,input().split()))
for i in range(n-1):
z=trains[i]
if z>=2:
z=z%2
if z==1:
z=0
trains[i+1]=trains[i+1]-1
trains[i]=z
if trains[n-1]%2==0:
ans='YES'
else:
ans='NO'
if -1 in trains:
ans='NO'
print(ans)
``` | instruction | 0 | 20,793 | 9 | 41,586 |
Yes | output | 1 | 20,793 | 9 | 41,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The programming competition season has already started and it's time to train for ICPC. Sereja coaches his teams for a number of year and he knows that to get ready for the training session it's not enough to prepare only problems and editorial. As the training sessions lasts for several hours, teams become hungry. Thus, Sereja orders a number of pizzas so they can eat right after the end of the competition.
Teams plan to train for n times during n consecutive days. During the training session Sereja orders exactly one pizza for each team that is present this day. He already knows that there will be ai teams on the i-th day.
There are two types of discounts in Sereja's favourite pizzeria. The first discount works if one buys two pizzas at one day, while the second is a coupon that allows to buy one pizza during two consecutive days (two pizzas in total).
As Sereja orders really a lot of pizza at this place, he is the golden client and can use the unlimited number of discounts and coupons of any type at any days.
Sereja wants to order exactly ai pizzas on the i-th day while using only discounts and coupons. Note, that he will never buy more pizzas than he need for this particular day. Help him determine, whether he can buy the proper amount of pizzas each day if he is allowed to use only coupons and discounts. Note, that it's also prohibited to have any active coupons after the end of the day n.
Input
The first line of input contains a single integer n (1 ≤ n ≤ 200 000) — the number of training sessions.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 10 000) — the number of teams that will be present on each of the days.
Output
If there is a way to order pizzas using only coupons and discounts and do not buy any extra pizzas on any of the days, then print "YES" (without quotes) in the only line of output. Otherwise, print "NO" (without quotes).
Examples
Input
4
1 2 1 2
Output
YES
Input
3
1 0 1
Output
NO
Note
In the first sample, Sereja can use one coupon to buy one pizza on the first and the second days, one coupon to buy pizza on the second and the third days and one discount to buy pizzas on the fourth days. This is the only way to order pizzas for this sample.
In the second sample, Sereja can't use neither the coupon nor the discount without ordering an extra pizza. Note, that it's possible that there will be no teams attending the training sessions on some days.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
f=0
s=0
for i in range(n):
if a[i]>2:
f=1
break
if a[i]==0 and s%2!=0:
f=1
break
else:
s+=a[i]
if f==1:
print('NO')
else:
print('YES')
``` | instruction | 0 | 20,794 | 9 | 41,588 |
No | output | 1 | 20,794 | 9 | 41,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The programming competition season has already started and it's time to train for ICPC. Sereja coaches his teams for a number of year and he knows that to get ready for the training session it's not enough to prepare only problems and editorial. As the training sessions lasts for several hours, teams become hungry. Thus, Sereja orders a number of pizzas so they can eat right after the end of the competition.
Teams plan to train for n times during n consecutive days. During the training session Sereja orders exactly one pizza for each team that is present this day. He already knows that there will be ai teams on the i-th day.
There are two types of discounts in Sereja's favourite pizzeria. The first discount works if one buys two pizzas at one day, while the second is a coupon that allows to buy one pizza during two consecutive days (two pizzas in total).
As Sereja orders really a lot of pizza at this place, he is the golden client and can use the unlimited number of discounts and coupons of any type at any days.
Sereja wants to order exactly ai pizzas on the i-th day while using only discounts and coupons. Note, that he will never buy more pizzas than he need for this particular day. Help him determine, whether he can buy the proper amount of pizzas each day if he is allowed to use only coupons and discounts. Note, that it's also prohibited to have any active coupons after the end of the day n.
Input
The first line of input contains a single integer n (1 ≤ n ≤ 200 000) — the number of training sessions.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 10 000) — the number of teams that will be present on each of the days.
Output
If there is a way to order pizzas using only coupons and discounts and do not buy any extra pizzas on any of the days, then print "YES" (without quotes) in the only line of output. Otherwise, print "NO" (without quotes).
Examples
Input
4
1 2 1 2
Output
YES
Input
3
1 0 1
Output
NO
Note
In the first sample, Sereja can use one coupon to buy one pizza on the first and the second days, one coupon to buy pizza on the second and the third days and one discount to buy pizzas on the fourth days. This is the only way to order pizzas for this sample.
In the second sample, Sereja can't use neither the coupon nor the discount without ordering an extra pizza. Note, that it's possible that there will be no teams attending the training sessions on some days.
Submitted Solution:
```
n = int(input())
arr = [0 if int(num) == 0 else 2 if int(num) % 2 == 0 else 1 for num in input().split()]
arr.append(0)
for i in range(len(arr) - 1):
if(arr[i] == 1 and arr[i+1] == 0):
print('NO')
exit()
elif(arr[i] == 1 and arr[i+1] != 2):
arr[i] -= 1
arr[i+1] -= 1
elif(arr[i] == 2):
arr[i] = 0
print("YES")
# 1 2 1 2
# 1 2 -> 0 1
# 2 2 -> 0 2
# 2 1 -> 0 1
# 1 1 -> 0 0
# 1 0 -> bad
# 2 0 ->
# 0 0 -> 0 0
# 0 1 -> 0 1
# 0 2 -> 0 2
``` | instruction | 0 | 20,795 | 9 | 41,590 |
No | output | 1 | 20,795 | 9 | 41,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The programming competition season has already started and it's time to train for ICPC. Sereja coaches his teams for a number of year and he knows that to get ready for the training session it's not enough to prepare only problems and editorial. As the training sessions lasts for several hours, teams become hungry. Thus, Sereja orders a number of pizzas so they can eat right after the end of the competition.
Teams plan to train for n times during n consecutive days. During the training session Sereja orders exactly one pizza for each team that is present this day. He already knows that there will be ai teams on the i-th day.
There are two types of discounts in Sereja's favourite pizzeria. The first discount works if one buys two pizzas at one day, while the second is a coupon that allows to buy one pizza during two consecutive days (two pizzas in total).
As Sereja orders really a lot of pizza at this place, he is the golden client and can use the unlimited number of discounts and coupons of any type at any days.
Sereja wants to order exactly ai pizzas on the i-th day while using only discounts and coupons. Note, that he will never buy more pizzas than he need for this particular day. Help him determine, whether he can buy the proper amount of pizzas each day if he is allowed to use only coupons and discounts. Note, that it's also prohibited to have any active coupons after the end of the day n.
Input
The first line of input contains a single integer n (1 ≤ n ≤ 200 000) — the number of training sessions.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 10 000) — the number of teams that will be present on each of the days.
Output
If there is a way to order pizzas using only coupons and discounts and do not buy any extra pizzas on any of the days, then print "YES" (without quotes) in the only line of output. Otherwise, print "NO" (without quotes).
Examples
Input
4
1 2 1 2
Output
YES
Input
3
1 0 1
Output
NO
Note
In the first sample, Sereja can use one coupon to buy one pizza on the first and the second days, one coupon to buy pizza on the second and the third days and one discount to buy pizzas on the fourth days. This is the only way to order pizzas for this sample.
In the second sample, Sereja can't use neither the coupon nor the discount without ordering an extra pizza. Note, that it's possible that there will be no teams attending the training sessions on some days.
Submitted Solution:
```
n = input()
m = input().split()
hitung = 0
for i in m:
hitung += int(i)
if hitung%2 != 0:
print("NO")
else:
if set(m) == {"0"}:
print("NO")
elif "0" in m:
counter = 0
itung = 0
while counter < len(m):
itung += int(m[counter])
if int(m[counter]) == 0:
if itung%2 == 0:
itung = 0
counter += 1
else:
print("NO")
break
else:
counter += 1
if counter == len(m)-1:
if itung%2 == 0:
print("YES")
else:
print("NO")
else:
print("YES")
``` | instruction | 0 | 20,796 | 9 | 41,592 |
No | output | 1 | 20,796 | 9 | 41,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The programming competition season has already started and it's time to train for ICPC. Sereja coaches his teams for a number of year and he knows that to get ready for the training session it's not enough to prepare only problems and editorial. As the training sessions lasts for several hours, teams become hungry. Thus, Sereja orders a number of pizzas so they can eat right after the end of the competition.
Teams plan to train for n times during n consecutive days. During the training session Sereja orders exactly one pizza for each team that is present this day. He already knows that there will be ai teams on the i-th day.
There are two types of discounts in Sereja's favourite pizzeria. The first discount works if one buys two pizzas at one day, while the second is a coupon that allows to buy one pizza during two consecutive days (two pizzas in total).
As Sereja orders really a lot of pizza at this place, he is the golden client and can use the unlimited number of discounts and coupons of any type at any days.
Sereja wants to order exactly ai pizzas on the i-th day while using only discounts and coupons. Note, that he will never buy more pizzas than he need for this particular day. Help him determine, whether he can buy the proper amount of pizzas each day if he is allowed to use only coupons and discounts. Note, that it's also prohibited to have any active coupons after the end of the day n.
Input
The first line of input contains a single integer n (1 ≤ n ≤ 200 000) — the number of training sessions.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 10 000) — the number of teams that will be present on each of the days.
Output
If there is a way to order pizzas using only coupons and discounts and do not buy any extra pizzas on any of the days, then print "YES" (without quotes) in the only line of output. Otherwise, print "NO" (without quotes).
Examples
Input
4
1 2 1 2
Output
YES
Input
3
1 0 1
Output
NO
Note
In the first sample, Sereja can use one coupon to buy one pizza on the first and the second days, one coupon to buy pizza on the second and the third days and one discount to buy pizzas on the fourth days. This is the only way to order pizzas for this sample.
In the second sample, Sereja can't use neither the coupon nor the discount without ordering an extra pizza. Note, that it's possible that there will be no teams attending the training sessions on some days.
Submitted Solution:
```
n = int(input())
l= list(map(int, input().split()))
if 0 in l:
print("NO")
exit()
flag = 1
for i in range(1,n):
a = l[i-1]
b = l[i]
if a%2==0 and b%2==0:
continue
elif a % 2 == 1 and b % 2 == 1:
b = b-1
elif a % 2 == 0 and b % 2 == 1:
continue
else:
b = b-1
if a==0 or b==0:
flag = 0
break
if flag ==0 or l[-1]%2==1:
print("NO")
else:
print("YES")
``` | instruction | 0 | 20,797 | 9 | 41,594 |
No | output | 1 | 20,797 | 9 | 41,595 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12 | instruction | 0 | 20,962 | 9 | 41,924 |
"Correct Solution:
```
import math
N, K = map(int, input().split())
A = sorted([int(i) for i in input().split()], reverse=True)
F = sorted([int(i) for i in input().split()])
OK = 10**12
NG = -1
while OK-NG > 1:
mid = (OK+NG)//2
cnt = 0
for a, f in zip(A, F):
if a*f > mid:
cnt += math.ceil((a*f-mid)/f)
if cnt <= K:
OK = mid
else:
NG = mid
print(OK)
``` | output | 1 | 20,962 | 9 | 41,925 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12 | instruction | 0 | 20,963 | 9 | 41,926 |
"Correct Solution:
```
n,k = map(int,input().split())
A = list(map(int,input().split()))
F = list(map(int,input().split()))
A.sort()
F.sort(reverse=True)
l,r = 0,10**12+1
while l <r:
mid = (l+r)//2
cnt = 0
for a,f in zip(A,F):
cnt += max(0,a-mid//f)
if cnt <=k:
r = mid
else:
l = mid+1
print(l)
``` | output | 1 | 20,963 | 9 | 41,927 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12 | instruction | 0 | 20,964 | 9 | 41,928 |
"Correct Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
f = list(map(int, input().split()))
a.sort()
f.sort(reverse=True)
left, right = -1, a[-1] * f[0]
while right - left > 1:
mid = (left + right) // 2
ck = 0
for i in range(n): ck += max(0, a[i] - mid // f[i])
if ck <= k: right = mid
else: left = mid
print(right)
``` | output | 1 | 20,964 | 9 | 41,929 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12 | instruction | 0 | 20,965 | 9 | 41,930 |
"Correct Solution:
```
n,k=map(int,input().split())
abilities=sorted(list(map(int,input().split())))
foods=sorted(list(map(int,input().split())),reverse=True)
if sum(abilities)<=k:
print(0)
exit()
def check(score):
ret=0
for a,f in zip(abilities,foods):
ret+=max(0,(a*f-score+f-1)//f)
return ret<=k
ok=10**12+1
ng=0
for _ in range(50):
mid=(ok+ng)//2
if check(mid):
ok=mid
else:
ng=mid
print(ok)
``` | output | 1 | 20,965 | 9 | 41,931 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12 | instruction | 0 | 20,966 | 9 | 41,932 |
"Correct Solution:
```
n,k=map(int,input().split())
a=list(map(int,input().split()))
f=list(map(int,input().split()))
a.sort()
f.sort(reverse=True)
hi=10**12
lo=0
def c(mi):
tmp=0
for i,j in zip(a,f):
if mi//j<i:
tmp+=i-mi//j
if tmp<=k:
return True
return False
while hi>=lo:
mi=(hi+lo)//2
if c(mi):
hi=mi-1
else:
lo=mi+1
print(lo)
``` | output | 1 | 20,966 | 9 | 41,933 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12 | instruction | 0 | 20,967 | 9 | 41,934 |
"Correct Solution:
```
import math
(n,k),a,f = [list(map(int, s.split())) for s in open(0)]
a.sort(reverse=True)
f.sort()
low = 0
high = 0
for x, y in zip(a,f):
high = max(high, x*y)
while high != low:
test = int((high + low) // 2)
tmp = 0
for x, y in zip(a,f):
tmp += max(0, x - math.floor(test / y))
if tmp > k:
low = test + 1
break
else:
high = test
print(high)
``` | output | 1 | 20,967 | 9 | 41,935 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12 | instruction | 0 | 20,968 | 9 | 41,936 |
"Correct Solution:
```
N, K = map(int, input().split())
As = list(map(int, input().split()))
Fs = list(map(int, input().split()))
As.sort()
Fs.sort(reverse=True)
def check(x):
k = 0
for i in range(N):
k += max(0, As[i]-x//Fs[i])
return k <= K
a = 0
b = As[-1]*Fs[0]
while a < b:
i = (a + b)//2
if check(i):
b = i
else:
if a == i:
break
a = i
r = b
print(r)
``` | output | 1 | 20,968 | 9 | 41,937 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12 | instruction | 0 | 20,969 | 9 | 41,938 |
"Correct Solution:
```
import math
N,K=map(int,input().split())
X=[(b,a*b)for a,b in zip(sorted(list(map(int,input().split()))),sorted(list(map(int,input().split())))[::-1])]
m,M,i=0,10**12,1
exec("i,M,m=((i+m)//2,i,m) if sum(math.ceil(max(0,c-i)/b)for b, c in X)<=K else((i+M)//2,M,i);"*50)
print(M)
``` | output | 1 | 20,969 | 9 | 41,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12
Submitted Solution:
```
n,k = map(int,input().split())
A = sorted(list(map(int,input().split())))
F = sorted(list(map(int,input().split())),reverse=True)
low = -1
high = 10**12
while high - low > 1:
score = (low+high)//2
training = 0
for i in range(n):
if A[i]*F[i] > score:
training += A[i] - score//F[i]
if training <= k:
high = score
else:
low = score
print(high)
``` | instruction | 0 | 20,970 | 9 | 41,940 |
Yes | output | 1 | 20,970 | 9 | 41,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12
Submitted Solution:
```
import math
N,K=map(int,input().split())
X=[(b,a*b)for a,b in zip(sorted(list(map(int,input().split()))),sorted(list(map(int,input().split())))[::-1])]
min_i, max_i, index = 0, 10**12, 1
while True:
index, max_i, min_i = ((index-1 + min_i)//2, index-1, min_i) if sum(math.ceil(max(0,c-index)/b)for b, c in X)<=K else((index+1 + max_i)//2, max_i, index+1)
if min_i > max_i:
print(min_i)
break
``` | instruction | 0 | 20,971 | 9 | 41,942 |
Yes | output | 1 | 20,971 | 9 | 41,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12
Submitted Solution:
```
N,K=map(int,input().split())
A=sorted(map(int,input().split()))
F=sorted(map(int,input().split()),reverse=True)
l=-1
r=A[-1]*F[0]
while l+1<r:
x=(l+r)//2
cnt=0
for i in range(N):
cnt+=max(A[i]-x//F[i],0)
if cnt<=K:
r=x
else:
l=x
print(r)
``` | instruction | 0 | 20,972 | 9 | 41,944 |
Yes | output | 1 | 20,972 | 9 | 41,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12
Submitted Solution:
```
n,k=map(int,input().split())
a=sorted(list(map(int,input().split())))
#こっちは変えられない
f=sorted(list(map(int,input().split())),reverse=True)
l,r=0,10**12
def cal(x):
global n,f,a,k
ret=0
for i in range(n):
ret+=max(0,a[i]-(x//f[i]))
#print(ret,x)
return ret
while l+1<r:
x=l+(r-l)//2
if cal(x)<=k:
r=x
else:
l=x
if cal(l)<=k:
print(l)
else:
print(r)
``` | instruction | 0 | 20,973 | 9 | 41,946 |
Yes | output | 1 | 20,973 | 9 | 41,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12
Submitted Solution:
```
n,k = map(int,input().split())
a = list(map(int,input().split()))
f = list(map(int,input().split()))
a.sort()
f.sort(reverse = True)
def is_ok(t):
cost = 0
for aa,ff in zip(a,f):
temp = t // ff
if temp >= aa:
contninue
cost += aa - temp
if cost <= k:
return True
return False
left = 0
right = a[-1]*f[0]
while left < right:
mid = (left + right) // 2
if is_ok(mid):
right = mid
else:
left = mid + 1
print(left)
``` | instruction | 0 | 20,974 | 9 | 41,948 |
No | output | 1 | 20,974 | 9 | 41,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12
Submitted Solution:
```
n,k = map(int,input().split())
a = list(map(int,input().split()))
f = list(map(int,input().split()))
a.sort(reverse=True)
f.sort()
z = zip(a,f)
ok = max(x*y for x,y in z)
ng = -1
while ok-ng>1:
mid = (ok+ng)//2
if sum(max(0,x-mid//y) for x,y in z) <= k:
ok = mid
else:
ng = mid
print(ok)
``` | instruction | 0 | 20,975 | 9 | 41,950 |
No | output | 1 | 20,975 | 9 | 41,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12
Submitted Solution:
```
def resolve():
N,K = map(int,input().split())
A = list(map(int,input().split()))
F = list(map(int,input().split()))
A.sort()
F.sort(reverse=True)
import heapq
q = [((-1)*A[i]*F[i],A[i],F[i]) for i in range(N)]
heapq.heapify(q)
for _ in range(K):
s,a,f = q[0]
if s == 0:
break
heapq.heappushpop(q,((a-1)*f*(-1),a-1,f))
print(q[0][0]*(-1))
resolve()
``` | instruction | 0 | 20,976 | 9 | 41,952 |
No | output | 1 | 20,976 | 9 | 41,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi will take part in an eating contest. Teams of N members will compete in this contest, and Takahashi's team consists of N players numbered 1 through N from youngest to oldest. The consumption coefficient of Member i is A_i.
In the contest, N foods numbered 1 through N will be presented, and the difficulty of Food i is F_i. The details of the contest are as follows:
* A team should assign one member to each food, and should not assign the same member to multiple foods.
* It will take x \times y seconds for a member to finish the food, where x is the consumption coefficient of the member and y is the difficulty of the dish.
* The score of a team is the longest time it takes for an individual member to finish the food.
Before the contest, Takahashi's team decided to do some training. In one set of training, a member can reduce his/her consumption coefficient by 1, as long as it does not go below 0. However, for financial reasons, the N members can do at most K sets of training in total.
What is the minimum possible score of the team, achieved by choosing the amounts of members' training and allocating the dishes optimally?
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* 0 \leq K \leq 10^{18}
* 1 \leq A_i \leq 10^6\ (1 \leq i \leq N)
* 1 \leq F_i \leq 10^6\ (1 \leq i \leq N)
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
F_1 F_2 ... F_N
Output
Print the minimum possible score of the team.
Examples
Input
3 5
4 2 1
2 3 1
Output
2
Input
3 8
4 2 1
2 3 1
Output
0
Input
11 14
3 1 4 1 5 9 2 6 5 3 5
8 9 7 9 3 2 3 8 4 6 2
Output
12
Submitted Solution:
```
N,K=map(int,input().split())
A=list(map(int,input().split()))
F=list(map(int,input().split()))
A=sorted(A,reverse=True)
F=sorted(F)
for i in range(len(A)):
if K>0:
if A[i]-K<0:
K=K-A[i]
A[i]=0
else:
A[i]=A[i]-K
K=K-A[i]
A=sorted(A,reverse=True)
print(max([x * y for (x, y) in zip(A, F)]))
``` | instruction | 0 | 20,977 | 9 | 41,954 |
No | output | 1 | 20,977 | 9 | 41,955 |
Provide a correct Python 3 solution for this coding contest problem.
Problem A Secret of Chocolate Poles
Wendy, the master of a chocolate shop, is thinking of displaying poles of chocolate disks in the showcase. She can use three kinds of chocolate disks: white thin disks, dark thin disks, and dark thick disks. The thin disks are $1$ cm thick, and the thick disks are $k$ cm thick. Disks will be piled in glass cylinders.
Each pole should satisfy the following conditions for her secret mission, which we cannot tell.
* A pole should consist of at least one disk.
* The total thickness of disks in a pole should be less than or equal to $l$ cm.
* The top disk and the bottom disk of a pole should be dark.
* A disk directly upon a white disk should be dark and vice versa.
As examples, six side views of poles are drawn in Figure A.1. These are the only possible side views she can make when $l = 5$ and $k = 3$.
<image>
Figure A.1. Six chocolate poles corresponding to Sample Input 1
Your task is to count the number of distinct side views she can make for given $l$ and $k$ to help her accomplish her secret mission.
Input
The input consists of a single test case in the following format.
$l$ $k$
Here, the maximum possible total thickness of disks in a pole is $l$ cm, and the thickness of the thick disks is $k$ cm. $l$ and $k$ are integers satisfying $1 \leq l \leq 100$ and $2 \leq k \leq 10$.
Output
Output the number of possible distinct patterns.
Sample Input 1
5 3
Sample Output 1
6
Sample Input 2
9 10
Sample Output 2
5
Sample Input 3
10 10
Sample Output 3
6
Sample Input 4
20 5
Sample Output 4
86
Sample Input 5
100 2
Sample Output 5
3626169232670
Example
Input
5 3
Output
6 | instruction | 0 | 21,106 | 9 | 42,212 |
"Correct Solution:
```
#!/usr/bin/python3
import os
import sys
def main():
L, K = read_ints()
print(solve(L, K))
def solve(L, K):
D = [[0, 0] for _ in range(L + 1)]
D[1][0] = 1
if K <= L:
D[K][0] = 1
for i in range(1, L):
D[i + 1][1] += D[i][0]
D[i + 1][0] += D[i][1]
if i + K <= L:
D[i + K][0] += D[i][1]
return sum([D[i][0] for i in range(1, L + 1)])
###############################################################################
DEBUG = 'DEBUG' in os.environ
def inp():
return sys.stdin.readline().rstrip()
def read_int():
return int(inp())
def read_ints():
return [int(e) for e in inp().split()]
def dprint(*value, sep=' ', end='\n'):
if DEBUG:
print(*value, sep=sep, end=end)
if __name__ == '__main__':
main()
``` | output | 1 | 21,106 | 9 | 42,213 |
Provide a correct Python 3 solution for this coding contest problem.
Problem A Secret of Chocolate Poles
Wendy, the master of a chocolate shop, is thinking of displaying poles of chocolate disks in the showcase. She can use three kinds of chocolate disks: white thin disks, dark thin disks, and dark thick disks. The thin disks are $1$ cm thick, and the thick disks are $k$ cm thick. Disks will be piled in glass cylinders.
Each pole should satisfy the following conditions for her secret mission, which we cannot tell.
* A pole should consist of at least one disk.
* The total thickness of disks in a pole should be less than or equal to $l$ cm.
* The top disk and the bottom disk of a pole should be dark.
* A disk directly upon a white disk should be dark and vice versa.
As examples, six side views of poles are drawn in Figure A.1. These are the only possible side views she can make when $l = 5$ and $k = 3$.
<image>
Figure A.1. Six chocolate poles corresponding to Sample Input 1
Your task is to count the number of distinct side views she can make for given $l$ and $k$ to help her accomplish her secret mission.
Input
The input consists of a single test case in the following format.
$l$ $k$
Here, the maximum possible total thickness of disks in a pole is $l$ cm, and the thickness of the thick disks is $k$ cm. $l$ and $k$ are integers satisfying $1 \leq l \leq 100$ and $2 \leq k \leq 10$.
Output
Output the number of possible distinct patterns.
Sample Input 1
5 3
Sample Output 1
6
Sample Input 2
9 10
Sample Output 2
5
Sample Input 3
10 10
Sample Output 3
6
Sample Input 4
20 5
Sample Output 4
86
Sample Input 5
100 2
Sample Output 5
3626169232670
Example
Input
5 3
Output
6 | instruction | 0 | 21,107 | 9 | 42,214 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
def main():
l, k = map(int, input().split())
dp = [[0] * 2 for i in range(101)]
dp[0][1] = 1
ans = 0
for i in range(1, l+1):
dp[i][0] = dp[i-1][1]
if i-k >= 0:
dp[i][0] += dp[i-k][1]
dp[i][1] = dp[i-1][0]
ans += dp[i][0]
print(ans)
if __name__ == "__main__":
main()
``` | output | 1 | 21,107 | 9 | 42,215 |
Provide a correct Python 3 solution for this coding contest problem.
Problem A Secret of Chocolate Poles
Wendy, the master of a chocolate shop, is thinking of displaying poles of chocolate disks in the showcase. She can use three kinds of chocolate disks: white thin disks, dark thin disks, and dark thick disks. The thin disks are $1$ cm thick, and the thick disks are $k$ cm thick. Disks will be piled in glass cylinders.
Each pole should satisfy the following conditions for her secret mission, which we cannot tell.
* A pole should consist of at least one disk.
* The total thickness of disks in a pole should be less than or equal to $l$ cm.
* The top disk and the bottom disk of a pole should be dark.
* A disk directly upon a white disk should be dark and vice versa.
As examples, six side views of poles are drawn in Figure A.1. These are the only possible side views she can make when $l = 5$ and $k = 3$.
<image>
Figure A.1. Six chocolate poles corresponding to Sample Input 1
Your task is to count the number of distinct side views she can make for given $l$ and $k$ to help her accomplish her secret mission.
Input
The input consists of a single test case in the following format.
$l$ $k$
Here, the maximum possible total thickness of disks in a pole is $l$ cm, and the thickness of the thick disks is $k$ cm. $l$ and $k$ are integers satisfying $1 \leq l \leq 100$ and $2 \leq k \leq 10$.
Output
Output the number of possible distinct patterns.
Sample Input 1
5 3
Sample Output 1
6
Sample Input 2
9 10
Sample Output 2
5
Sample Input 3
10 10
Sample Output 3
6
Sample Input 4
20 5
Sample Output 4
86
Sample Input 5
100 2
Sample Output 5
3626169232670
Example
Input
5 3
Output
6 | instruction | 0 | 21,108 | 9 | 42,216 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
def f(l,k):
r = [0] * (l+1)
r[1] = 1
if l >= k:
r[k] += 1
for i in range(1,l):
if i + 2 <= l:
r[i+2] += r[i]
if i + 1 + k <= l:
r[i+1+k] += r[i]
return sum(r)
while 1:
n,m = LI()
if n == 0:
break
rr.append(f(n,m))
break
return '\n'.join(map(str, rr))
print(main())
``` | output | 1 | 21,108 | 9 | 42,217 |
Provide a correct Python 3 solution for this coding contest problem.
Problem A Secret of Chocolate Poles
Wendy, the master of a chocolate shop, is thinking of displaying poles of chocolate disks in the showcase. She can use three kinds of chocolate disks: white thin disks, dark thin disks, and dark thick disks. The thin disks are $1$ cm thick, and the thick disks are $k$ cm thick. Disks will be piled in glass cylinders.
Each pole should satisfy the following conditions for her secret mission, which we cannot tell.
* A pole should consist of at least one disk.
* The total thickness of disks in a pole should be less than or equal to $l$ cm.
* The top disk and the bottom disk of a pole should be dark.
* A disk directly upon a white disk should be dark and vice versa.
As examples, six side views of poles are drawn in Figure A.1. These are the only possible side views she can make when $l = 5$ and $k = 3$.
<image>
Figure A.1. Six chocolate poles corresponding to Sample Input 1
Your task is to count the number of distinct side views she can make for given $l$ and $k$ to help her accomplish her secret mission.
Input
The input consists of a single test case in the following format.
$l$ $k$
Here, the maximum possible total thickness of disks in a pole is $l$ cm, and the thickness of the thick disks is $k$ cm. $l$ and $k$ are integers satisfying $1 \leq l \leq 100$ and $2 \leq k \leq 10$.
Output
Output the number of possible distinct patterns.
Sample Input 1
5 3
Sample Output 1
6
Sample Input 2
9 10
Sample Output 2
5
Sample Input 3
10 10
Sample Output 3
6
Sample Input 4
20 5
Sample Output 4
86
Sample Input 5
100 2
Sample Output 5
3626169232670
Example
Input
5 3
Output
6 | instruction | 0 | 21,109 | 9 | 42,218 |
"Correct Solution:
```
l,k = map(int,input().split())
dp = [[0,0] for i in range(l+1)]
dp[1][0] = 1
if k <= l:
dp[k][0] = 1
for h in range(1,l):
dp[h+1][0] += dp[h][1]
dp[h+1][1] += dp[h][0]
if h+k <= l:
dp[h+k][1] += dp[h][0]
ans = 0
for b,w in dp:
ans += b
print(ans)
``` | output | 1 | 21,109 | 9 | 42,219 |
Provide a correct Python 3 solution for this coding contest problem.
Problem A Secret of Chocolate Poles
Wendy, the master of a chocolate shop, is thinking of displaying poles of chocolate disks in the showcase. She can use three kinds of chocolate disks: white thin disks, dark thin disks, and dark thick disks. The thin disks are $1$ cm thick, and the thick disks are $k$ cm thick. Disks will be piled in glass cylinders.
Each pole should satisfy the following conditions for her secret mission, which we cannot tell.
* A pole should consist of at least one disk.
* The total thickness of disks in a pole should be less than or equal to $l$ cm.
* The top disk and the bottom disk of a pole should be dark.
* A disk directly upon a white disk should be dark and vice versa.
As examples, six side views of poles are drawn in Figure A.1. These are the only possible side views she can make when $l = 5$ and $k = 3$.
<image>
Figure A.1. Six chocolate poles corresponding to Sample Input 1
Your task is to count the number of distinct side views she can make for given $l$ and $k$ to help her accomplish her secret mission.
Input
The input consists of a single test case in the following format.
$l$ $k$
Here, the maximum possible total thickness of disks in a pole is $l$ cm, and the thickness of the thick disks is $k$ cm. $l$ and $k$ are integers satisfying $1 \leq l \leq 100$ and $2 \leq k \leq 10$.
Output
Output the number of possible distinct patterns.
Sample Input 1
5 3
Sample Output 1
6
Sample Input 2
9 10
Sample Output 2
5
Sample Input 3
10 10
Sample Output 3
6
Sample Input 4
20 5
Sample Output 4
86
Sample Input 5
100 2
Sample Output 5
3626169232670
Example
Input
5 3
Output
6 | instruction | 0 | 21,110 | 9 | 42,220 |
"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,k = LI()
ans = 0
for l in range(1,n+1):
for i in range(1000):
if i*k > l:break
j = l-i*k+i
if j%2:
j //= 2
j += 1
s = 1
for a in range(i):
s *= j-a
s //= a+1
ans += s
print(ans)
return
#B
def B():
return
#C
def C():
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__":
A()
``` | output | 1 | 21,110 | 9 | 42,221 |
Provide a correct Python 3 solution for this coding contest problem.
Problem A Secret of Chocolate Poles
Wendy, the master of a chocolate shop, is thinking of displaying poles of chocolate disks in the showcase. She can use three kinds of chocolate disks: white thin disks, dark thin disks, and dark thick disks. The thin disks are $1$ cm thick, and the thick disks are $k$ cm thick. Disks will be piled in glass cylinders.
Each pole should satisfy the following conditions for her secret mission, which we cannot tell.
* A pole should consist of at least one disk.
* The total thickness of disks in a pole should be less than or equal to $l$ cm.
* The top disk and the bottom disk of a pole should be dark.
* A disk directly upon a white disk should be dark and vice versa.
As examples, six side views of poles are drawn in Figure A.1. These are the only possible side views she can make when $l = 5$ and $k = 3$.
<image>
Figure A.1. Six chocolate poles corresponding to Sample Input 1
Your task is to count the number of distinct side views she can make for given $l$ and $k$ to help her accomplish her secret mission.
Input
The input consists of a single test case in the following format.
$l$ $k$
Here, the maximum possible total thickness of disks in a pole is $l$ cm, and the thickness of the thick disks is $k$ cm. $l$ and $k$ are integers satisfying $1 \leq l \leq 100$ and $2 \leq k \leq 10$.
Output
Output the number of possible distinct patterns.
Sample Input 1
5 3
Sample Output 1
6
Sample Input 2
9 10
Sample Output 2
5
Sample Input 3
10 10
Sample Output 3
6
Sample Input 4
20 5
Sample Output 4
86
Sample Input 5
100 2
Sample Output 5
3626169232670
Example
Input
5 3
Output
6 | instruction | 0 | 21,111 | 9 | 42,222 |
"Correct Solution:
```
l, k = map(int, input().split())
memo = {(l, 0): 1, (l, 1): 0}
def dfs(cur, dark):
if (cur, dark) in memo:
return memo[cur, dark]
res = dfs(cur + 1, dark^1) + (dark^1)
if dark and cur + k <= l:
res += dfs(cur + k, dark^1)
memo[cur, dark] = res
return res
print(dfs(0, 1))
``` | output | 1 | 21,111 | 9 | 42,223 |
Provide a correct Python 3 solution for this coding contest problem.
Problem A Secret of Chocolate Poles
Wendy, the master of a chocolate shop, is thinking of displaying poles of chocolate disks in the showcase. She can use three kinds of chocolate disks: white thin disks, dark thin disks, and dark thick disks. The thin disks are $1$ cm thick, and the thick disks are $k$ cm thick. Disks will be piled in glass cylinders.
Each pole should satisfy the following conditions for her secret mission, which we cannot tell.
* A pole should consist of at least one disk.
* The total thickness of disks in a pole should be less than or equal to $l$ cm.
* The top disk and the bottom disk of a pole should be dark.
* A disk directly upon a white disk should be dark and vice versa.
As examples, six side views of poles are drawn in Figure A.1. These are the only possible side views she can make when $l = 5$ and $k = 3$.
<image>
Figure A.1. Six chocolate poles corresponding to Sample Input 1
Your task is to count the number of distinct side views she can make for given $l$ and $k$ to help her accomplish her secret mission.
Input
The input consists of a single test case in the following format.
$l$ $k$
Here, the maximum possible total thickness of disks in a pole is $l$ cm, and the thickness of the thick disks is $k$ cm. $l$ and $k$ are integers satisfying $1 \leq l \leq 100$ and $2 \leq k \leq 10$.
Output
Output the number of possible distinct patterns.
Sample Input 1
5 3
Sample Output 1
6
Sample Input 2
9 10
Sample Output 2
5
Sample Input 3
10 10
Sample Output 3
6
Sample Input 4
20 5
Sample Output 4
86
Sample Input 5
100 2
Sample Output 5
3626169232670
Example
Input
5 3
Output
6 | instruction | 0 | 21,112 | 9 | 42,224 |
"Correct Solution:
```
l , k = [int(i) for i in input().split()]
INIT = 0
tableB = [INIT] * 110
tableW = [INIT] * 110
tableW[0] = 1
for i in range(1,l+1):
tableB[i] += tableW[i-1]
if i - k >= 0:
tableB[i] += tableW[i-k]
tableW[i+1] = tableB[i]
print(sum(tableB))
#print(tableB)
``` | output | 1 | 21,112 | 9 | 42,225 |
Provide a correct Python 3 solution for this coding contest problem.
Problem A Secret of Chocolate Poles
Wendy, the master of a chocolate shop, is thinking of displaying poles of chocolate disks in the showcase. She can use three kinds of chocolate disks: white thin disks, dark thin disks, and dark thick disks. The thin disks are $1$ cm thick, and the thick disks are $k$ cm thick. Disks will be piled in glass cylinders.
Each pole should satisfy the following conditions for her secret mission, which we cannot tell.
* A pole should consist of at least one disk.
* The total thickness of disks in a pole should be less than or equal to $l$ cm.
* The top disk and the bottom disk of a pole should be dark.
* A disk directly upon a white disk should be dark and vice versa.
As examples, six side views of poles are drawn in Figure A.1. These are the only possible side views she can make when $l = 5$ and $k = 3$.
<image>
Figure A.1. Six chocolate poles corresponding to Sample Input 1
Your task is to count the number of distinct side views she can make for given $l$ and $k$ to help her accomplish her secret mission.
Input
The input consists of a single test case in the following format.
$l$ $k$
Here, the maximum possible total thickness of disks in a pole is $l$ cm, and the thickness of the thick disks is $k$ cm. $l$ and $k$ are integers satisfying $1 \leq l \leq 100$ and $2 \leq k \leq 10$.
Output
Output the number of possible distinct patterns.
Sample Input 1
5 3
Sample Output 1
6
Sample Input 2
9 10
Sample Output 2
5
Sample Input 3
10 10
Sample Output 3
6
Sample Input 4
20 5
Sample Output 4
86
Sample Input 5
100 2
Sample Output 5
3626169232670
Example
Input
5 3
Output
6 | instruction | 0 | 21,113 | 9 | 42,226 |
"Correct Solution:
```
import math
l, k = map(int, input().split())
c = 0
for i in range(1, int((l + 1) / 2) + 1):
for j in range(i + 1):
if 2 * i - 1 + j * (k - 1) <= l:
c += math.factorial(i) / math.factorial(j) / math.factorial(i - j)
else:
break
print(int(c))
``` | output | 1 | 21,113 | 9 | 42,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem A Secret of Chocolate Poles
Wendy, the master of a chocolate shop, is thinking of displaying poles of chocolate disks in the showcase. She can use three kinds of chocolate disks: white thin disks, dark thin disks, and dark thick disks. The thin disks are $1$ cm thick, and the thick disks are $k$ cm thick. Disks will be piled in glass cylinders.
Each pole should satisfy the following conditions for her secret mission, which we cannot tell.
* A pole should consist of at least one disk.
* The total thickness of disks in a pole should be less than or equal to $l$ cm.
* The top disk and the bottom disk of a pole should be dark.
* A disk directly upon a white disk should be dark and vice versa.
As examples, six side views of poles are drawn in Figure A.1. These are the only possible side views she can make when $l = 5$ and $k = 3$.
<image>
Figure A.1. Six chocolate poles corresponding to Sample Input 1
Your task is to count the number of distinct side views she can make for given $l$ and $k$ to help her accomplish her secret mission.
Input
The input consists of a single test case in the following format.
$l$ $k$
Here, the maximum possible total thickness of disks in a pole is $l$ cm, and the thickness of the thick disks is $k$ cm. $l$ and $k$ are integers satisfying $1 \leq l \leq 100$ and $2 \leq k \leq 10$.
Output
Output the number of possible distinct patterns.
Sample Input 1
5 3
Sample Output 1
6
Sample Input 2
9 10
Sample Output 2
5
Sample Input 3
10 10
Sample Output 3
6
Sample Input 4
20 5
Sample Output 4
86
Sample Input 5
100 2
Sample Output 5
3626169232670
Example
Input
5 3
Output
6
Submitted Solution:
```
l, k = map(int, input().split())
dpW = [0] * 101
dpB = [0] * 101
dpW[0] = 1
ans = 0
for i in range(1, l + 1):
if i >= k:
dpB[i] += dpW[i - k]
ans += dpW[i - k]
dpB[i] += dpW[i - 1]
dpW[i] += dpB[i - 1]
ans += dpW[i - 1]
print(ans)
``` | instruction | 0 | 21,114 | 9 | 42,228 |
Yes | output | 1 | 21,114 | 9 | 42,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem A Secret of Chocolate Poles
Wendy, the master of a chocolate shop, is thinking of displaying poles of chocolate disks in the showcase. She can use three kinds of chocolate disks: white thin disks, dark thin disks, and dark thick disks. The thin disks are $1$ cm thick, and the thick disks are $k$ cm thick. Disks will be piled in glass cylinders.
Each pole should satisfy the following conditions for her secret mission, which we cannot tell.
* A pole should consist of at least one disk.
* The total thickness of disks in a pole should be less than or equal to $l$ cm.
* The top disk and the bottom disk of a pole should be dark.
* A disk directly upon a white disk should be dark and vice versa.
As examples, six side views of poles are drawn in Figure A.1. These are the only possible side views she can make when $l = 5$ and $k = 3$.
<image>
Figure A.1. Six chocolate poles corresponding to Sample Input 1
Your task is to count the number of distinct side views she can make for given $l$ and $k$ to help her accomplish her secret mission.
Input
The input consists of a single test case in the following format.
$l$ $k$
Here, the maximum possible total thickness of disks in a pole is $l$ cm, and the thickness of the thick disks is $k$ cm. $l$ and $k$ are integers satisfying $1 \leq l \leq 100$ and $2 \leq k \leq 10$.
Output
Output the number of possible distinct patterns.
Sample Input 1
5 3
Sample Output 1
6
Sample Input 2
9 10
Sample Output 2
5
Sample Input 3
10 10
Sample Output 3
6
Sample Input 4
20 5
Sample Output 4
86
Sample Input 5
100 2
Sample Output 5
3626169232670
Example
Input
5 3
Output
6
Submitted Solution:
```
from functools import lru_cache # おまじない
l,k=map(int,input().split())
@lru_cache() # python 素晴らしい
def rec(i,isBlack):
if i < 0: return 0
if i == 0: return 1 if not isBlack else 0 # black at bottom
if isBlack: # top is black
return rec(i-1, False) + rec(i-k, False)
else: # top is white
return rec(i-1, True)
ans = sum ( rec(i,True) for i in range(1, l+1))
print(ans)
``` | instruction | 0 | 21,115 | 9 | 42,230 |
Yes | output | 1 | 21,115 | 9 | 42,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem A Secret of Chocolate Poles
Wendy, the master of a chocolate shop, is thinking of displaying poles of chocolate disks in the showcase. She can use three kinds of chocolate disks: white thin disks, dark thin disks, and dark thick disks. The thin disks are $1$ cm thick, and the thick disks are $k$ cm thick. Disks will be piled in glass cylinders.
Each pole should satisfy the following conditions for her secret mission, which we cannot tell.
* A pole should consist of at least one disk.
* The total thickness of disks in a pole should be less than or equal to $l$ cm.
* The top disk and the bottom disk of a pole should be dark.
* A disk directly upon a white disk should be dark and vice versa.
As examples, six side views of poles are drawn in Figure A.1. These are the only possible side views she can make when $l = 5$ and $k = 3$.
<image>
Figure A.1. Six chocolate poles corresponding to Sample Input 1
Your task is to count the number of distinct side views she can make for given $l$ and $k$ to help her accomplish her secret mission.
Input
The input consists of a single test case in the following format.
$l$ $k$
Here, the maximum possible total thickness of disks in a pole is $l$ cm, and the thickness of the thick disks is $k$ cm. $l$ and $k$ are integers satisfying $1 \leq l \leq 100$ and $2 \leq k \leq 10$.
Output
Output the number of possible distinct patterns.
Sample Input 1
5 3
Sample Output 1
6
Sample Input 2
9 10
Sample Output 2
5
Sample Input 3
10 10
Sample Output 3
6
Sample Input 4
20 5
Sample Output 4
86
Sample Input 5
100 2
Sample Output 5
3626169232670
Example
Input
5 3
Output
6
Submitted Solution:
```
import math
def combinations_count(n, r):
return math.factorial(n) // (math.factorial(n - r) * math.factorial(r))
def main():
l = list(map(int, input().split()))
combinate_a = []
combinate_b = []
ans = 0
for a in range(l[0]):
for b in range(l[0]):
if (l[1] + 1) * a + 2 * b <= l[0] + 1:
combinate_a.append(a)
combinate_b.append(b)
if (l[1] + 1) * a + 2 * b > l[0] + 1:
break
for i in range(1, len(combinate_a)):
if combinate_a[i] == 0:
ans += 1
else:
ans += combinations_count(combinate_a[i]+combinate_b[i], combinate_a[i])
print(ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 21,116 | 9 | 42,232 |
No | output | 1 | 21,116 | 9 | 42,233 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252 | instruction | 0 | 21,788 | 9 | 43,576 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub
sys.setrecursionlimit(100000)
input = sys.stdin.readline
INF = 2**62-1
EPS = 1.0e-8
def read_int():
return int(input())
def read_int_n():
return list(map(int, input().split()))
def read_float():
return float(input())
def read_float_n():
return list(map(float, input().split()))
def read_str():
return input().strip()
def read_str_n():
return list(map(str, input().split()))
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.time()
ret = f(*args, **kwargs)
e = time.time()
error_print(e - s, 'sec')
return ret
return wrap
class Circle:
def __init__(self, x, y, r):
self.x = x
self.y = y
self.r = r
def center(self):
return Point(self.x, self.y)
def intersection(self, c):
d = self.center().distance(c.center())
if d > self.r + c.r:
return None
# One circle within other
if d < abs(self.r - c.r):
return None
# coincident circles
if d == 0 and self.r == c.r:
return None
else:
a = (self.r**2 - c.r**2 + d**2) / (2*d)
h = math.sqrt(self.r**2-a**2 + sys.float_info.epsilon)
x2 = self.x+a*(c.x-self.x)/d
y2 = self.y+a*(c.y-self.y)/d
x3 = x2+h*(c.y-self.y)/d
y3 = y2-h*(c.x-self.x)/d
x4 = x2-h*(c.y-self.y)/d
y4 = y2+h*(c.x-self.x)/d
return Point(x3, y3), Point(x4, y4)
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def distance(self, p):
return math.sqrt((self.x - p.x)**2 + (self.y - p.y)**2)
class Bisect:
def __init__(self, func):
self.__func = func
def bisect_left(self, x, lo, hi):
while hi - lo > EPS:
mid = (lo+hi)/2
if self.__func(mid) < x:
lo = mid
else:
hi = mid
return lo
@mt
def slv(N, K, XYC):
def f(t):
cs = []
for x, y, c in XYC:
cs.append(Circle(x, y, t / c))
cand = [c.center() for c in cs]
for i, j in combinations(range(N), r=2):
p = cs[i].intersection(cs[j])
if p:
cand.append(p[0])
cand.append(p[1])
ans = 0
for p in cand:
tmp = 0
for c in cs:
d = c.center().distance(p)
if d - EPS < c.r:
tmp += 1
ans = max(ans, tmp)
return ans
b = Bisect(f)
return b.bisect_left(K, 0, 10000000)
def main():
N, K = read_int_n()
XYC = [read_int_n() for _ in range(N)]
print(slv(N, K, XYC))
if __name__ == '__main__':
main()
``` | output | 1 | 21,788 | 9 | 43,577 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252 | instruction | 0 | 21,789 | 9 | 43,578 |
"Correct Solution:
```
import sys
readline = sys.stdin.buffer.readline
EPS = 1e-7
class Circle:
def __init__(self, cx, cy, r):
self.center = (cx, cy)
self.radius = r
def cross(self, C2):
x1, y1 = self.center
x2, y2 = C2.center
R1 = self.radius
R2 = C2.radius
if (R2-R1)**2 + EPS <= (x2-x1)**2 + (y2-y1)**2 <= (R2+R1)**2 - EPS:
dx, dy = x2-x1, y2-y1
a = (R2**2 - R1**2 + x1**2 - x2**2 + y1**2 - y2**2)/2
if abs(dy) > EPS:
#交点はdx*X + dy*Y + a = 0 上
D = dx**2 + dy**2
bx = a*dx+dx*dy*y1-x1*dy**2
cx = dy**2*(x1**2-R1**2) + (a+dy*y1)**2
cx1 = (-bx + (bx**2-D*cx)**0.5)/D
cx2 = (-bx - (bx**2-D*cx)**0.5)/D
cy1 = -(dx*cx1+a)/dy
cy2 = -(dx*cx2+a)/dy
else:
k = (R1**2-R2**2+dx**2)/2/dx
cx1 = cx2 = x1+k
cy1 = y1 + (R1**2-k**2)**0.5
cy2 = y1 - (R1**2-k**2)**0.5
return [(cx1, cy1), (cx2, cy2)]
else:
return []
def isinside(self, x, y):
x1, y1 = self.center
return (x1-x)**2 + (y1-y)**2 - EPS < self.radius**2
N, K = map(int, readline().split())
Points = [tuple(map(int, readline().split())) for _ in range(N)]
ok = 1e6
ng = 0
candidate = [(x, y) for x, y, _ in Points]
for _ in range(40):
med = (ok+ng)/2
Cs = [Circle(x, y, med/c) for x, y, c in Points]
candidate2 = candidate[:]
for i in range(N):
for j in range(i):
candidate2.extend(Cs[i].cross(Cs[j]))
for px, py in candidate2:
res = 0
for C in Cs:
if C.isinside(px, py):
res += 1
if res >= K:
ok = med
break
else:
ng = med
print(ok)
``` | output | 1 | 21,789 | 9 | 43,579 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252 | instruction | 0 | 21,790 | 9 | 43,580 |
"Correct Solution:
```
import sys
# input = sys.stdin.readline
input = sys.stdin.buffer.readline
import math
from itertools import combinations
EPS = 10**-7
class P2():
def __init__(self, x, y):
self.x = x
self.y = y
self.norm2 = (self.x**2 + self.y**2)**0.5
def __add__(self, other):
return P2(self.x + other.x, self.y + other.y)
def __sub__(self, other):
return P2(self.x - other.x, self.y - other.y)
def __mul__(self, other: float):
return P2(self.x * other, self.y * other)
def __truediv__(self, other: float):
return P2(self.x / other, self.y / other)
def dot(self, other):
return self.x * other.x + self.y * other.y
def det(self, other):
return self.x * other.y - self.y * other.x
class Circle():
def __init__(self, p, r):
self.p = p
self.r = r
n, k = map(int, input().split())
p = []
c = []
for _ in range(n):
x, y, cc = map(float, input().split())
p.append(P2(x, y))
c.append(cc)
dist = [[(p[i] - p[j]).norm2 for j in range(n)] for i in range(n)]
def check(r):
circles = [Circle(p[i], r / c[i]) for i in range(n)]
# 半径rの円２つの交点の列挙
cand = [pi for pi in p]
for c_i, c_j in combinations(circles, r=2):
d_ij = c_i.p - c_j.p
if d_ij.norm2 < c_i.r + c_j.r - EPS:
# 点iと点jを中心とする円が交点を持つ時
x = (c_j.r**2 - c_i.r**2 + d_ij.norm2**2) / (2 * d_ij.norm2)
h = math.sqrt(max(c_j.r**2 - x**2, 0.0))
v = P2(-d_ij.y, d_ij.x) * (h / d_ij.norm2)
dx = c_j.p + d_ij * (x / d_ij.norm2)
cand.append(dx + v)
cand.append(dx - v)
for cand_p in cand:
cnt = 0
for c_i in circles:
if (c_i.p - cand_p).norm2 < c_i.r + EPS:
cnt += 1
if cnt >= k:
return True
return False
lb = 0.0 # False
ub = 3000 * 100 # True
while ub - lb > EPS:
mid = (ub + lb) / 2
if check(mid):
ub = mid
else:
lb = mid
print(ub)
``` | output | 1 | 21,790 | 9 | 43,581 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252 | instruction | 0 | 21,791 | 9 | 43,582 |
"Correct Solution:
```
#!/usr/bin/env python3
import sys
import cmath
import math
import copy
INF = float("inf")
EPS = 1e-9
def ctc(p1, r1, p2, r2):
# 二つの円の共通接線の数
# common tangent count?
# common tangent of circle?
# 2円の関係性は、
# 同一(=INF), 離れている(=4)、外接(=3)、交わる(=2)、内接(=1)、内包(=0)がある
# r1 <= r2として一般性を失わない
if r2 < r1:
p1, p2 = p2, p1
r1, r2 = r2, r1
d = abs(p1-p2)
if p1 == p2 and d < EPS: # 同一の円
return INF
elif (d - (r2+r1)) > EPS: # 離れている
return 4
elif (d - (r2+r1)) > -EPS: # 外接している
return 3
elif (d-(r2-r1)) > EPS: # 交わっている
return 2
elif (d-(r2-r1)) > -EPS: # 内接している
return 1
else: # 内包している
return 0
def inner(p1, c1, p2, c2):
coeff1 = c1/(c1+c2)
coeff2 = c2/(c1+c2)
return p1*coeff1+p2*coeff2
def intersection(p1, r1, p2, r2):
status = ctc(p1, r1, p2, r2)
d = abs(p1-p2)
if not (1 <= status <= 3):
return []
elif status == 3: # 外接
return [inner(p1, r1, p2, r2)]
elif status == 1: # 内接
return [p1 + r1*(p1-p2)/d]
else: # 交わっている場合
if r1 > r2:
p1, p2 = p2, p1
r1, r2 = r2, r1
d = abs(p1-p2)
cos = (r1**2+d**2-r2**2)/(2*r1*d)
angle = math.acos(cos)
phi = cmath.phase(p2-p1)
return [
p1 + cmath.rect(r1, phi+angle),
p1 + cmath.rect(r1, phi-angle)
]
def circle_in(p, o, r1, allow_on_edge=True):
if allow_on_edge:
return abs(o-p)-r1 < EPS
else:
return abs(o-p)-r1 < -EPS
def solve(N: int, K: int, x: "List[int]", y: "List[int]", c: "List[int]"):
# 時刻Tの時点でK枚以上の肉が焼ける熱源の配置はあるか
comp = [complex(xx, yy) for xx, yy in zip(x, y)]
left = 0
right = 2000*max(c)
while right - left > EPS:
mid = (left + right)/2
yakiniku = 0
b = copy.copy(comp)
for i in range(N):
for j in range(i+1, N):
b.extend(intersection(comp[i], mid/c[i], comp[j], mid/c[j]))
for p in b:
count = [circle_in(p, comp[k], mid/c[k])
for k in range(N)]
yakiniku = max(yakiniku, sum(count))
if yakiniku >= K:
right = mid
else:
left = mid
print(mid)
return
def main():
def iterate_tokens():
for line in sys.stdin:
for word in line.split():
yield word
tokens = iterate_tokens()
N = int(next(tokens)) # type: int
K = int(next(tokens)) # type: int
x = [int()] * (N) # type: "List[int]"
y = [int()] * (N) # type: "List[int]"
c = [int()] * (N) # type: "List[int]"
for i in range(N):
x[i] = int(next(tokens))
y[i] = int(next(tokens))
c[i] = int(next(tokens))
solve(N, K, x, y, c)
if __name__ == '__main__':
main()
``` | output | 1 | 21,791 | 9 | 43,583 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252 | instruction | 0 | 21,792 | 9 | 43,584 |
"Correct Solution:
```
N,K = map(int,input().split())
XYC = [tuple(map(int,input().split())) for i in range(N)]
from math import hypot,acos,atan2,cos,sin
def circles_cross_points(x1,y1,r1,x2,y2,r2):
d = hypot(x1-x2, y1-y2)
if r1 + r2 < d or abs(r1 - r2) >= d: return []
x = (r1**2 + d**2 - r2**2) / (2*r1*d)
if not -1 <= x <= 1: return []
a = acos(x)
t = atan2(y2-y1, x2-x1)
return [
(x1+cos(t+a)*r1, y1+sin(t+a)*r1),
(x1+cos(t-a)*r1, y1+sin(t-a)*r1)
]
eps = 10e-10
def is_ok(t):
cands = []
for x,y,_ in XYC:
cands.append((x,y))
for i,(x1,y1,c1) in enumerate(XYC[:-1]):
for x2,y2,c2 in XYC[i+1:]:
ps = circles_cross_points(x1,y1,t/c1,x2,y2,t/c2)
cands += ps
for cx,cy in cands:
k = 0
for x,y,c in XYC:
d = hypot(cx-x,cy-y)
if c*d <= t + eps:
k += 1
if k >= K:
return True
return False
ok = 10**10
ng = 0
for _ in range(100):
m = (ok+ng)/2
if is_ok(m):
ok = m
else:
ng = m
if ok-ng < 10e-7: break
print(ok)
``` | output | 1 | 21,792 | 9 | 43,585 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252 | instruction | 0 | 21,793 | 9 | 43,586 |
"Correct Solution:
```
import math
def intersectionCC0(x1, y1, r1, r2):
r = x1**2 + y1**2
t = (r + r1**2 - r2**2) / (2*r)
dd = r1**2/r - t**2
if - (10**-8) < dd < 0: dd = 0
if dd < 0: return []
x, y = t * x1, t * y1
if dd == 0: return [(x, y)]
sq = math.sqrt(dd)
dx, dy = y1 * sq, -x1 * sq
return [(x+dx, y+dy), (x-dx, y-dy)]
def intersectionCC(x1, y1, x2, y2, r1, r2):
return [(x1+x, y1+y) for x, y in intersectionCC0(x2-x1, y2-y1, r1, r2)]
def chk(t):
L = [a for a in L0]
for i, (x1, y1, c1) in enumerate(X):
for x2, y2, c2 in X[:i]:
L += intersectionCC(x1, y1, x2, y2, t/c1, t/c2)
for x0, y0 in L:
cnt = 0
for x, y, c in X:
if math.sqrt((x-x0) ** 2 + (y-y0) ** 2) * c <= t + 10**-8:
cnt += 1
if cnt >= K: return 1
return 0
N, K = map(int, input().split())
X = []
L0 = []
for _ in range(N):
x, y, c = map(int, input().split())
X.append((x, y, c))
L0.append((x, y))
l, r = 0, 150000
while (r-l) * (1<<20) > max(1, l):
m = (l+r) / 2
if chk(m):
r = m
else:
l = m
print((l+r)/2)
``` | output | 1 | 21,793 | 9 | 43,587 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252 | instruction | 0 | 21,794 | 9 | 43,588 |
"Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
n, k = map(int, input().split())
XY = []
C = []
kouho = []
for i in range(n):
x, y, c = map(int, input().split())
XY.append((x, y))
C.append(c)
kouho.append(complex(x, y))
ct = 100
eps = 10 ** (-7)
class circle:
def __init__(self, center, r):
self.center = center
self.r = r
def circle_cross(c1, c2):
r1 = c1.r
o1 = c1.center
r2 = c2.r
o2 = c2.center
d = abs(o1 - o2)
if d > r1 + r2:
return 0, 0, 0
elif r1 + r2 < d + eps and r1 + r2 > d - eps:
# d==r1+r2
return 1, (r2 * o1 + r1 * o2) / (r1 + r2), 0
elif d < r1 + r2 and d > abs(r1 - r2):
# abs(r1-r2)<d<r1+r2
o21 = o2 - o1
oh_nagasa = (d ** 2 + r1 ** 2 - r2 ** 2) / (2 * d)
if oh_nagasa < eps:
return 1, (r2 * o1 + r1 * o2) / (r1 + r2), 0
oh = oh_nagasa * (o21 / d)
housen = (o21 / d) * complex(0, 1)
suisen = max(0, (r1 ** 2 - oh_nagasa ** 2) ** (1 / 2))
ans1 = o1 + oh + housen * suisen
ans2 = o1 + oh - housen * suisen
return 2, ans1, ans2
elif abs(r1 - r2) > d - eps and abs(r1 - r2) < d + eps:
# d=abs(r1-r2)
ans = (-r2 * o1 + r1 * o2) / (r1 - r2)
return 1, ans, 0
else:
# d<abs(r1-r2)
return 0, 0, 0
if n == 1:
print(0)
exit()
if k == 1:
print(0)
exit()
ok = 10 ** 10
ng = 0
while ct > 0:
mid = (ok + ng) / 2
kouho_temp = kouho[::]
for i in range(n - 1):
a, b = XY[i]
p = C[i]
A = circle(complex(a, b), mid / p)
for j in range(i + 1, n):
c, d = XY[j]
q = C[j]
B = circle(complex(c, d), mid / q)
t1, t2, t3 = circle_cross(A, B)
if t1 == 0:
continue
if t1 == 1:
kouho_temp.append(t2)
if t1 == 2:
kouho_temp.append(t2)
kouho_temp.append(t3)
res = 0
for z in kouho_temp:
cnt = 0
for i in range(n):
x, y = XY[i]
p = C[i]
z1 = complex(x, y)
if abs(z - z1) < mid / p + eps:
cnt += 1
res = max(cnt, res)
if res >= k:
ok = mid
else:
ng = mid
ct -= 1
print(mid)
``` | output | 1 | 21,794 | 9 | 43,589 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252 | instruction | 0 | 21,795 | 9 | 43,590 |
"Correct Solution:
```
from math import atan2, cos, sin
EPS = 1e-7
def hypot(a):
return sum(x ** 2 for x in a) ** 0.5
def scale(s, v):
assert isinstance(s, int) or isinstance(s, float)
return [s * v[0], s * v[1], s * v[2]]
def dot(a, b):
return a[0] * b[0] + a[1] * b[1] + a[2] * b[2]
def add(a, b):
return [a[0] + b[0], a[1] + b[1], a[2] + b[2]]
def sub(a, b):
return [a[0] - b[0], a[1] - b[1], a[2] - b[2]]
def matmul(mat, v):
return [dot(row, v) for row in mat]
def rotZ(theta):
return [[cos(theta), -sin(theta), 0], [sin(theta), cos(theta), 0], [0, 0, 1]]
def _circleCircleIntersection(R, d, r):
# For circles centered at (0, 0) and (d, 0) with radius R and r respectively
# Equations are:
# x^2 + y^2 = R^2
# (x - d)^2 + y^2 = r^2
# Subtracting and expanding:
# x^2 - (x^2 - 2 * d * x + d^2) = R^2 - r^2
# 2 * d * x - d^2 = R^2 - r^2
# x = (R^2 - r^2 + d^2) / (2*d)
if d == 0:
return [(0, 0)]
x = (R ** 2 - r ** 2 + d ** 2) / (2 * d)
ySq = R ** 2 - x ** 2
if ySq == 0:
return [(x, 0)]
elif ySq > 0:
y = ySq ** 0.5
# assert abs((x ** 2 + y ** 2) - R ** 2) < EPS
# assert abs(((x - d) ** 2 + y ** 2) - r ** 2) < EPS
return [(x, y), (x, -y)]
else:
raise AssertionError()
def circleCircleIntersection(x1, y1, r1, x2, y2, r2):
# Either 1 point, 2 points, or
# 0 points because one circle contains the other
# 0 points because the circles don't intersect
v1 = [x1, y1, 0]
v2 = [x2, y2, 0]
d = hypot(sub(v2, v1))
if r1 + r2 < d:
# Circles don't intersect
return []
if min(r1, r2) + d < max(r1, r2):
# One circle contains the other
return []
# Translate
offset = list(v1)
v1 = sub(v1, offset)
v2 = sub(v2, offset)
assert v1[0] == 0 and v1[1] == 0
# Rotate
angle = -atan2(v2[1], v2[0])
v2 = matmul(rotZ(angle), v2)
assert abs(v2[1]) < EPS
ret = []
for pt in _circleCircleIntersection(r1, v2[0], r2):
pt = [pt[0], pt[1], 0]
# Un-rotate
pt = matmul(rotZ(-angle), pt)
# Un-translate
pt = add(pt, offset)
ret.append((pt[0], pt[1]))
return ret
def circleCircleIntersection(x1, y1, r1, x2, y2, r2):
# Alternative implementation
v1 = [x1, y1, 0]
v2 = [x2, y2, 0]
d = hypot(sub(v2, v1))
if r1 + r2 < d:
# Circles don't intersect
return []
if min(r1, r2) + d < max(r1, r2):
# One circle contains the other
return []
if d == 0:
return [(x1, y1)]
# Unit vector in the direction between the centers
u = scale(1 / d, sub(v2, v1))
# Distance from x1, y1 to the midpoint of the chord
a = (r1 ** 2 - r2 ** 2 + d ** 2) / (2 * d)
# Get the midpoint on the chord
mid = add(v1, scale(a, u))
# Unit vector in direction of the chord
uPerpendicular = [u[1], -u[0], 0]
# 2 * h is the length of the chord
h = (r1 ** 2 - a ** 2) ** 0.5
# Get the two solutions
ret1 = add(mid, scale(h, uPerpendicular))
ret2 = add(mid, scale(-h, uPerpendicular))
return [[ret1[0], ret1[1]], [ret2[0], ret2[1]]]
N, K = list(map(int, input().split()))
XYC = []
for i in range(N):
XYC.append(list(map(int, input().split())))
def numMeat(t):
# At time t, each x, y can be heated by anything within a circle of radius radius t/c
# Gather all intersecting points, some of which must be part of the region with most overlaps
intersect = []
for i in range(len(XYC)):
x1, y1, c1 = XYC[i]
r1 = t / c1
for j in range(i + 1, len(XYC)):
x2, y2, c2 = XYC[j]
r2 = t / c2
pts = circleCircleIntersection(x1, y1, r1, x2, y2, r2)
intersect.extend(pts)
intersect.append((x1, y1))
if False:
import matplotlib.pyplot as plt
plt.figure()
ax = plt.gca()
ax.cla()
bound = t // 5
ax.set_xlim((-bound, bound))
ax.set_ylim((-bound, bound))
for i in range(len(XYC)):
x1, y1, c1 = XYC[i]
r1 = t / c1
ax.add_artist(plt.Circle((x1, y1), r1, color="blue", fill=False))
for x, y in intersect:
ax.plot([x], [y], ".", color="red")
plt.show()
best = 0
for pt in intersect:
count = 0
for x, y, c in XYC:
if c * ((x - pt[0]) ** 2 + (y - pt[1]) ** 2) ** 0.5 <= t + EPS:
count += 1
best = max(best, count)
return best
lo = 0
# Get upperbound (by arbitrarily placing heat source at origin)
hi = max(c * (x ** 2 + y ** 2) ** 0.5 for x, y, c in XYC)
while hi - lo > EPS:
mid = (hi + lo) / 2
if numMeat(mid) >= K:
hi = mid
else:
lo = mid
print(hi)
``` | output | 1 | 21,795 | 9 | 43,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252
Submitted Solution:
```
import cmath
import itertools
import math
import os
import random
import sys
INF = float("inf")
PI = cmath.pi
TAU = cmath.pi * 2
EPS = 1e-10
if os.getenv("LOCAL"):
sys.stdin = open("_in.txt", "r")
sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# MOD = 998244353
class Point:
"""
2次元空間上の点
"""
# 反時計回り側にある
CCW_COUNTER_CLOCKWISE = 1
# 時計回り側にある
CCW_CLOCKWISE = -1
# 線分の後ろにある
CCW_ONLINE_BACK = 2
# 線分の前にある
CCW_ONLINE_FRONT = -2
# 線分上にある
CCW_ON_SEGMENT = 0
def __init__(self, x: float, y: float):
self.c = complex(x, y)
@property
def x(self):
return self.c.real
@property
def y(self):
return self.c.imag
@staticmethod
def from_complex(c: complex):
return Point(c.real, c.imag)
@staticmethod
def from_polar(r: float, phi: float):
c = cmath.rect(r, phi)
return Point(c.real, c.imag)
def __add__(self, p):
"""
:param Point p:
"""
c = self.c + p.c
return Point(c.real, c.imag)
def __iadd__(self, p):
"""
:param Point p:
"""
self.c += p.c
return self
def __sub__(self, p):
"""
:param Point p:
"""
c = self.c - p.c
return Point(c.real, c.imag)
def __isub__(self, p):
"""
:param Point p:
"""
self.c -= p.c
return self
def __mul__(self, f: float):
c = self.c * f
return Point(c.real, c.imag)
def __imul__(self, f: float):
self.c *= f
return self
def __truediv__(self, f: float):
c = self.c / f
return Point(c.real, c.imag)
def __itruediv__(self, f: float):
self.c /= f
return self
def __repr__(self):
return "({}, {})".format(round(self.x, 10), round(self.y, 10))
def __neg__(self):
c = -self.c
return Point(c.real, c.imag)
def __eq__(self, p):
return abs(self.c - p.c) < EPS
def __abs__(self):
return abs(self.c)
@staticmethod
def ccw(a, b, c):
"""
線分 ab に対する c の位置
線分上にあるか判定するだけなら on_segment とかのが速い
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_1_C&lang=ja
:param Point a:
:param Point b:
:param Point c:
"""
b = b - a
c = c - a
det = b.det(c)
if det > EPS:
return Point.CCW_COUNTER_CLOCKWISE
if det < -EPS:
return Point.CCW_CLOCKWISE
if b.dot(c) < -EPS:
return Point.CCW_ONLINE_BACK
if b.dot(b - c) < -EPS:
return Point.CCW_ONLINE_FRONT
return Point.CCW_ON_SEGMENT
def dot(self, p):
"""
内積
:param Point p:
:rtype: float
"""
return self.x * p.x + self.y * p.y
def det(self, p):
"""
外積
:param Point p:
:rtype: float
"""
return self.x * p.y - self.y * p.x
def dist(self, p):
"""
距離
:param Point p:
:rtype: float
"""
return abs(self.c - p.c)
def norm(self):
"""
原点からの距離
:rtype: float
"""
return abs(self.c)
def phase(self):
"""
原点からの角度
:rtype: float
"""
return cmath.phase(self.c)
def angle(self, p, q):
"""
p に向いてる状態から q まで反時計回りに回転するときの角度
-pi <= ret <= pi
:param Point p:
:param Point q:
:rtype: float
"""
return (cmath.phase(q.c - self.c) - cmath.phase(p.c - self.c) + PI) % TAU - PI
def area(self, p, q):
"""
p, q となす三角形の面積
:param Point p:
:param Point q:
:rtype: float
"""
return abs((p - self).det(q - self) / 2)
def projection_point(self, p, q, allow_outer=False):
"""
線分 pq を通る直線上に垂線をおろしたときの足の座標
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_1_A&lang=ja
:param Point p:
:param Point q:
:param allow_outer: 答えが線分の間になくても OK
:rtype: Point|None
"""
diff_q = q - p
# 答えの p からの距離
r = (self - p).dot(diff_q) / abs(diff_q)
# 線分の角度
phase = diff_q.phase()
ret = Point.from_polar(r, phase) + p
if allow_outer or (p - ret).dot(q - ret) < EPS:
return ret
return None
def reflection_point(self, p, q):
"""
直線 pq を挟んで反対にある点
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_1_B&lang=ja
:param Point p:
:param Point q:
:rtype: Point
"""
# 距離
r = abs(self - p)
# pq と p-self の角度
angle = p.angle(q, self)
# 直線を挟んで角度を反対にする
angle = (q - p).phase() - angle
return Point.from_polar(r, angle) + p
def on_segment(self, p, q, allow_side=True):
"""
点が線分 pq の上に乗っているか
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_1_C&lang=ja
:param Point p:
:param Point q:
:param allow_side: 端っこでギリギリ触れているのを許容するか
:rtype: bool
"""
if not allow_side and (self == p or self == q):
return False
# 外積がゼロ: 面積がゼロ == 一直線
# 内積がマイナス: p - self - q の順に並んでる
return abs((p - self).det(q - self)) < EPS and (p - self).dot(q - self) < EPS
@staticmethod
def circumcenter_of(p1, p2, p3):
"""
外心
:param Point p1:
:param Point p2:
:param Point p3:
:rtype: Point|None
"""
if abs((p2 - p1).det(p3 - p1)) < EPS:
# 外積がゼロ == 一直線
return None
# https://ja.wikipedia.org/wiki/外接円
a = (p2.x - p3.x) ** 2 + (p2.y - p3.y) ** 2
b = (p3.x - p1.x) ** 2 + (p3.y - p1.y) ** 2
c = (p1.x - p2.x) ** 2 + (p1.y - p2.y) ** 2
num = p1 * a * (b + c - a) + p2 * b * (c + a - b) + p3 * c * (a + b - c)
den = a * (b + c - a) + b * (c + a - b) + c * (a + b - c)
return num / den
@staticmethod
def incenter_of(p1, p2, p3):
"""
内心
:param Point p1:
:param Point p2:
:param Point p3:
"""
# https://ja.wikipedia.org/wiki/三角形の内接円と傍接円
d1 = p2.dist(p3)
d2 = p3.dist(p1)
d3 = p1.dist(p2)
return (p1 * d1 + p2 * d2 + p3 * d3) / (d1 + d2 + d3)
class Circle:
def __init__(self, o, r):
"""
:param Point o:
:param float r:
"""
self.o = o
self.r = r
def __eq__(self, other):
return self.o == other.o and abs(self.r - other.r) < EPS
def ctc(self, c):
"""
共通接線 common tangent の数
4: 離れてる
3: 外接
2: 交わってる
1: 内接
0: 内包
inf: 同一
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_7_A&lang=ja
:param Circle c:
:rtype: int
"""
if self.o == c.o:
return INF if abs(self.r - c.r) < EPS else 0
# 円同士の距離
d = self.o.dist(c.o) - self.r - c.r
if d > EPS:
return 4
elif d > -EPS:
return 3
# elif d > -min(self.r, c.r) * 2:
elif d + min(self.r, c.r) * 2 > EPS:
return 2
elif d + min(self.r, c.r) * 2 > -EPS:
return 1
return 0
def has_point_on_edge(self, p):
"""
指定した点が円周上にあるか
:param Point p:
:rtype: bool
"""
return abs(self.o.dist(p) - self.r) < EPS
def contains(self, p, allow_on_edge=True):
"""
指定した点を含むか
:param Point p:
:param bool allow_on_edge: 辺上の点を許容するか
"""
if allow_on_edge:
# return self.o.dist(p) <= self.r
return self.o.dist(p) - self.r < EPS
else:
# return self.o.dist(p) < self.r
return self.o.dist(p) - self.r < -EPS
def area(self):
"""
面積
"""
return self.r ** 2 * PI
def circular_segment_area(self, angle):
"""
弓形⌓の面積
:param float angle: 角度ラジアン
"""
# 扇形の面積
sector_area = self.area() * angle / TAU
# 三角形部分を引く
return sector_area - self.r ** 2 * math.sin(angle) / 2
def intersection_points(self, other, allow_outer=False):
"""
:param Segment|Circle other:
:param bool allow_outer:
"""
if isinstance(other, Segment):
return self.intersection_points_with_segment(other, allow_outer=allow_outer)
if isinstance(other, Circle):
return self.intersection_points_with_circle(other)
raise NotImplementedError()
def intersection_points_with_segment(self, s, allow_outer=False):
"""
線分と交差する点のリスト
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_7_D&lang=ja
:param Segment s:
:param bool allow_outer: 線分の間にない点を含む
:rtype: list of Point
"""
# 垂線の足
projection_point = self.o.projection_point(s.p1, s.p2, allow_outer=True)
# 線分との距離
dist = self.o.dist(projection_point)
# if dist > self.r:
if dist - self.r > EPS:
return []
if dist - self.r > -EPS:
if allow_outer or s.has_point(projection_point):
return [projection_point]
else:
return []
# 足から左右に diff だけ動かした座標が答え
diff = Point.from_polar(math.sqrt(self.r ** 2 - dist ** 2), s.phase())
ret1 = projection_point + diff
ret2 = projection_point - diff
ret = []
if allow_outer or s.has_point(ret1):
ret.append(ret1)
if allow_outer or s.has_point(ret2):
ret.append(ret2)
return ret
def intersection_points_with_circle(self, other):
"""
円と交差する点のリスト
Verify: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_7_E&langja
:param circle other:
:rtype: list of Point
"""
ctc = self.ctc(other)
if not 1 <= ctc <= 3:
return []
if ctc == 3:
# 外接
return [Point.from_polar(self.r, (other.o - self.o).phase()) + self.o]
if ctc == 1:
# 内接
if self.r > other.r:
return [Point.from_polar(self.r, (other.o - self.o).phase()) + self.o]
else:
return [Point.from_polar(self.r, (self.o - other.o).phase()) + self.o]
# 2つ交点がある
assert ctc == 2
a = other.r
b = self.r
c = self.o.dist(other.o)
# 余弦定理で cos(a) を求めます
cos_a = (b ** 2 + c ** 2 - a ** 2) / (2 * b * c)
angle = math.acos(cos_a)
phi = (other.o - self.o).phase()
return [
self.o + Point.from_polar(self.r, phi + angle),
self.o + Point.from_polar(self.r, phi - angle),
]
def tangent_points_with_point(self, p):
"""
p を通る接線との接点
:param Point p:
:rtype: list of Point
"""
dist = self.o.dist(p)
# if dist < self.r:
if dist - self.r < -EPS:
# p が円の内部にある
return []
if dist - self.r < EPS:
# p が円周上にある
return [Point(p.x, p.y)]
a = math.sqrt(dist ** 2 - self.r ** 2)
b = self.r
c = dist
# 余弦定理で cos(a) を求めます
cos_a = (b ** 2 + c ** 2 - a ** 2) / (2 * b * c)
angle = math.acos(cos_a)
phi = (p - self.o).phase()
return [
self.o + Point.from_polar(self.r, phi + angle),
self.o + Point.from_polar(self.r, phi - angle),
]
def tangent_points_with_circle(self, other):
"""
other との共通接線との接点
:param Circle other:
:rtype: list of Point
"""
ctc = self.ctc(other)
if ctc > 4:
raise ValueError('2つの円が同一です')
if ctc == 0:
return []
if ctc == 1:
return self.intersection_points_with_circle(other)
assert ctc in (2, 3, 4)
ret = []
# 共通外接線を求める
# if self.r == other.r:
if abs(self.r - other.r) < EPS:
# 半径が同じ == 2つの共通外接線が並行
phi = (other.o - self.o).phase()
ret.append(self.o + Point.from_polar(self.r, phi + PI / 2))
ret.append(self.o + Point.from_polar(self.r, phi - PI / 2))
else:
# 2つの共通外接線の交点から接線を引く
intersection = self.o + (other.o - self.o) / (self.r - other.r) * self.r
ret += self.tangent_points_with_point(intersection)
# 共通内接線を求める
# 2つの共通内接線の交点から接線を引く
intersection = self.o + (other.o - self.o) / (self.r + other.r) * self.r
ret += self.tangent_points_with_point(intersection)
return ret
@staticmethod
def circumscribed_of(p1, p2, p3):
"""
p1・p2・p3 のなす三角形の外接円
Verify:
:param Point p1:
:param Point p2:
:param Point p3:
"""
if p1.on_segment(p2, p3):
return Circle((p2 + p3) / 2, p2.dist(p3) / 2)
if p2.on_segment(p1, p3):
return Circle((p1 + p3) / 2, p1.dist(p3) / 2)
if p3.on_segment(p1, p2):
return Circle((p1 + p2) / 2, p1.dist(p2) / 2)
o = Point.circumcenter_of(p1, p2, p3)
return Circle(o, o.dist(p1))
@staticmethod
def min_enclosing(points):
"""
points をすべて含む最小の円
計算量の期待値は O(N)
https://www.jaist.ac.jp/~uehara/course/2014/i481f/pdf/ppt-7.pdf
Verify: https://www.spoj.com/problems/QCJ4/
:param list of Point points:
:rtype: Circle
"""
points = points[:]
random.shuffle(points)
# 前から順番に決めて大きくしていく
ret = Circle(points[0], 0)
for i, p in enumerate(points):
if ret.contains(p):
continue
ret = Circle(p, 0)
for j, q in enumerate(points[:i]):
if ret.contains(q):
continue
# 2点を直径とする円
ret = Circle((p + q) / 2, abs(p - q) / 2)
for k, r in enumerate(points[:j]):
if ret.contains(r):
continue
# 3点に接する円
ret = Circle.circumscribed_of(p, q, r)
return ret
def argsort(li, key=None, reverse=False):
return [i for _, i in sorted(
[(a, i) for i, a in enumerate(li)], key=(lambda t: key(t[0])) if key else None, reverse=reverse
)]
N, K = list(map(int, sys.stdin.buffer.readline().split()))
XYC = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(N)]
def test(t: float):
# http://tottoripaper.hatenablog.jp/entry/2015/03/10/231142
# K こ以上の肉を焼けるか
circles = []
for x, y, c in XYC:
circles.append(Circle(Point(x, y), t /c))
# 交点
points = []
for c1, c2 in itertools.combinations(circles, r=2):
points += c1.intersection_points_with_circle(c2)
for c in circles:
points.append(c.o)
for p in points:
cnt = 0
for c in circles:
cnt += c.contains(p)
if cnt >= K:
return True
return False
#
# # T: iとjが重なる時間
# T = []
# for i, j in itertools.combinations(range(N), r=2):
# xi, yi, ci = XYC[i]
# xj, yj, cj = XYC[j]
# d = (xi - xj) ** 2 + (yi - yj) ** 2
# t = d / (ci + cj)
# T.append((t, i, j))
# T.sort()
# print(T)
# ans = INF
# for t, i, j in T:
# xi, yi, ci = XYC[i]
# xj, yj, cj = XYC[j]
# p = Point(xi, yi)
# q = Point(xj, yj)
# test_p = Point.from_polar(ci * t, (q - p).phase())
# if test(test_p, t):
# ans = t
# break
# print(ans)
ok = 1 << 100
ng = 0
while abs(ok - ng) > EPS:
mid = (ok + ng) / 2
if test(mid):
ok = mid
else:
ng = mid
print(ok)
``` | instruction | 0 | 21,796 | 9 | 43,592 |
Yes | output | 1 | 21,796 | 9 | 43,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252
Submitted Solution:
```
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
import math
eps = 1e-12
def calc_intersection_points(circle0, circle1):
x0, y0, r0 = circle0
x1, y1, r1 = circle1
d = math.hypot(x0 - x1, y0 - y1)
if d > r0 + r1 or d < abs(r0 - r1):
return []
ret = []
if r0 + r1 == d:
a = r0
h = 0.0
else:
a = (r0**2 - r1**2 + d**2) / (2.0 * d)
h = math.sqrt(r0**2 - a**2)
x2 = x0 + a * (x1 - x0) / d
y2 = y0 + a * (y1 - y0) / d
x3 = x2 + h * (y1 - y0) / d
y3 = y2 - h * (x1 - x0) / d
ret.append((x3, y3))
if h == 0:
return ret
x4 = x2 - h * (y1 - y0) / d
y4 = y2 + h * (x1 - x0) / d
ret.append((x4, y4))
return ret
n, k = map(int, input().split())
niku = []
for _ in range(n):
x, y, c = map(int, input().split())
niku.append((x, y, c))
l = 0.0; r = 10**8
for _ in range(64):
mid = (l + r) / 2.0
circles = []
for x, y, c in niku:
circles.append((x, y, mid / c))
points = []
for i in range(n):
for j in range(i+1, n):
points += calc_intersection_points(circles[i], circles[j])
for x, y, _ in niku:
points.append((x, y))
cooked = 0
for x, y in points:
ret = 0
for cx, cy, cr in circles:
if math.hypot(x - cx, y - cy) <= cr + eps:
ret += 1
if ret > cooked:
cooked = ret
if cooked >= k:
r = mid
else:
l = mid
print(l)
``` | instruction | 0 | 21,797 | 9 | 43,594 |
Yes | output | 1 | 21,797 | 9 | 43,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252
Submitted Solution:
```
import sys
input = sys.stdin.readline
from math import sqrt
N,K=map(int,input().split())
M=[tuple(map(int,input().split())) for i in range(N)]
def cross_coor(x0,y0,c0,x1,y1,c1): # 二円の交点
if (c1-c0)**2<=(x1-x0)**2+(y1-y0)**2<=(c0+c1)**2:
x2,y2=x1-x0,y1-y0
a=(x2**2+y2**2+c0**2-c1**2)/2
xa=(a*x2+y2*sqrt((x2**2+y2**2)*(c0**2)-a**2))/(x2**2+y2**2)+x0
ya=(a*y2-x2*sqrt((x2**2+y2**2)*(c0**2)-a**2))/(x2**2+y2**2)+y0
xb=(a*x2-y2*sqrt((x2**2+y2**2)*(c0**2)-a**2))/(x2**2+y2**2)+x0
yb=(a*y2+x2*sqrt((x2**2+y2**2)*(c0**2)-a**2))/(x2**2+y2**2)+y0
return [(xa,ya),(xb,yb)]
else:
return False
OK=1<<31
NG=0
while OK-NG>10**(-8):
mid=(OK+NG)/2
CANDI=[]
for i in range(N):
x0,y0,c0=M[i]
CANDI.append((x0,y0))
for j in range(i+1,N):
x1,y1,c1=M[j]
CR=cross_coor(x0,y0,mid/c0,x1,y1,mid/c1)
if CR:
CANDI.extend(CR)
for z,w in CANDI:
count=0
for x,y,c in M:
if (c**2)*((x-z)**2+(y-w)**2)<=(mid+10**(-8))**2:
count+=1
if count>=K:
OK=mid
break
else:
NG=mid
print((OK+NG)/2)
``` | instruction | 0 | 21,798 | 9 | 43,596 |
Yes | output | 1 | 21,798 | 9 | 43,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252
Submitted Solution:
```
#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from itertools import permutations
import sys
import math
import bisect
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
res = list(sys.stdin.readline())
if res[-1] == "\n":
return res[:-1]
return res
def IR(n):
return [I() for i in range(n)]
def LIR(n):
return [LI() for i in range(n)]
def SR(n):
return [S() for i in range(n)]
def LSR(n):
return [LS() for i in range(n)]
sys.setrecursionlimit(1000000)
mod = 1000000007
def solve():
# 2円の中心間の距離の2乗
def D2(x1,y1,x2,y2):
return (x1-x2)**2+(y1-y2)**2
# 2円が交点を持つか
def check(R1,R2):
x1,y1,r1 = R1
x2,y2,r2 = R2
d2 = D2(x1,y1,x2,y2)
if (r1-r2)**2 < d2 <= (r1+r2)**2:
return 1
else:
return 0
# 2円の交点を出力
def point(R1,R2):
if not check(R1,R2):
return []
else:
x1,y1,r1 = R1
x2,y2,r2 = R2
x2 -= x1
y2 -= y1
r = x2**2+y2**2
a = (r+r1**2-r2**2)/2
d = (r*r1**2-a**2)**0.5
X,Y = a*x2/r, a*y2/r
k = d/r
X1,Y1 = X+y2*k+x1, Y-x2*k+y1
X2,Y2 = X-y2*k+x1, Y+x2*k+y1
return [(X1,Y1),(X2,Y2)]
# 円に点が入っているか
def inR(p,R):
x2,y2 = p
x1,y1,r1 = R
d2 = D2(x1,y1,x2,y2)
if d2 <= r1**2+1e-3:
return 1
else:
return 0
n,k = LI()
p = LIR(n)
l = 0
r = 1000000
while r-l > 1e-6:
t = (l+r) / 2
P = []
for i in range(n):
xi,yi,ci = p[i]
ri = t/ci
P.append((xi,yi))
for j in range(i):
xj,yj,cj = p[j]
rj = t/cj
ps = point((xi,yi,ri),(xj,yj,rj))
for pi in ps:
P.append(pi)
for pi in P:
s = 0
for x,y,c in p:
r1 = t/c
if inR(pi,(x,y,r1)):
s += 1
if s >= k:
break
if s >= k:
r = t
break
else:
l = t
print(l)
return
#Solve
if __name__ == "__main__":
solve()
``` | instruction | 0 | 21,799 | 9 | 43,598 |
Yes | output | 1 | 21,799 | 9 | 43,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252
Submitted Solution:
```
import numpy as np
from scipy.optimize import basinhopping, minimize
import sys
input = sys.stdin.buffer.readline
def main() -> None:
N, K = map(int, input().split())
A = np.array(list(map(int, sys.stdin.read().split())))
X, Y, C = A[::3], A[1::3], A[2::3]
def f(x, X=X, Y=Y, C=C):
return np.sort(C * np.sqrt(np.power(X-x[0], 2) + np.power(Y-x[1], 2)))[K-1]
# res = minimize(f, [0, 0], method='Nelder-Mead', tol=1e-7)
res = basinhopping(f, [0, 0], niter=100)
print("{:.10f}".format(f(res.x)))
if __name__ == '__main__':
main()
``` | instruction | 0 | 21,800 | 9 | 43,600 |
No | output | 1 | 21,800 | 9 | 43,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252
Submitted Solution:
```
N,K = map(int,input().split())
niku = [tuple(map(int,input().split())) for _ in range(N)]
ans = 10**18
for i in range(N-1):
x1,y1,c1 = niku[i]
for j in range(i+1,N):
x2,y2,c2 = niku[j]
X = c1*x1/(c1+c2)+c2*x2/(c1+c2)
Y = c1*y1/(c1+c2)+c2*y2/(c1+c2)
time = []
for k in range(N):
xk,yk,ck = niku[k]
time.append(ck*((X-xk)**2+(Y-yk)**2)**0.5)
time.sort()
ans = min(ans,time[K-1])
for i in range(N):
X,Y,_ = niku[i]
time = []
for k in range(N):
xk,yk,ck = niku[k]
time.append(ck*((X-xk)**2+(Y-yk)**2)**0.5)
time.sort()
ans = min(ans,time[K-1])
for i in range(N-1):
x1,y1,c1 = niku[i]
for j in range(i+1,N):
x2,y2,c2 = niku[j]
if c1 == c2:
continue
X = c1*x1/(c1-c2)-c2*x2/(c1-c2)
Y = c1*y1/(c1-c2)-c2*y2/(c1-c2)
time = []
for k in range(N):
xk,yk,ck = niku[k]
time.append(ck*((X-xk)**2+(Y-yk)**2)**0.5)
time.sort()
ans = min(ans,time[K-1])
for i in range(N-2):
x1,y1,c1 = niku[i]
for j in range(i+1,N-1):
x2,y2,c2 = niku[j]
for k in range(j+1,N):
x3,y3,c3 = niku[k]
X = c1*x1/(c1+c2+c3)+c2*x2/(c1+c2+c3)+c3*x3/(c1+c2+c3)
Y = c1*y1/(c1+c2+c3)+c2*y2/(c1+c2+c3)+c3*y3/(c1+c2+c3)
time = []
for l in range(N):
xk,yk,ck = niku[l]
time.append(ck*((X-xk)**2+(Y-yk)**2)**0.5)
time.sort()
ans = min(ans,time[K-1])
print(ans)
``` | instruction | 0 | 21,801 | 9 | 43,602 |
No | output | 1 | 21,801 | 9 | 43,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**9)
N, K = map(int,input().split())
niku = list(list(map(int,input().split()))for _ in range(N))
ans = float('inf')
def func(l,r):
if r-l <= 10**(-8) or r-l < r*(10**(-8)):
return l
mid = (l+r)/2
hoge = [[0]*N for _ in range(N)]
count = N
for i in range(N):
hoge[i][i] = 1
for j in range(i+1,N):
if ((niku[i][1]-niku[j][1])**2 + (niku[i][0]-niku[j][0])**2) <= (mid/niku[i][2] + mid/niku[j][2])**2:
hoge[i][j] = 1
hoge[j][i] = 1
while 1:
dex = 0
num = 0
zero = 0
for i in range(N):
fuga = N-sum(hoge[i])
if fuga > num:
num = fuga
dex = i
zero += fuga
if zero == 0:
break
for i in range(N):
if hoge[i][dex] == 0:
hoge[i][dex] = 1
hoge[dex][i] = 1
zero -= 2
count -= 1
if count >= K:
return func(l,mid)
else:
return func(mid,r)
print(func(0,10**9))
``` | instruction | 0 | 21,802 | 9 | 43,604 |
No | output | 1 | 21,802 | 9 | 43,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi wants to grill N pieces of meat on a grilling net, which can be seen as a two-dimensional plane. The coordinates of the i-th piece of meat are \left(x_i, y_i\right), and its hardness is c_i.
Takahashi can use one heat source to grill the meat. If he puts the heat source at coordinates \left(X, Y\right), where X and Y are real numbers, the i-th piece of meat will be ready to eat in c_i \times \sqrt{\left(X - x_i\right)^2 + \left(Y-y_i\right)^2} seconds.
Takahashi wants to eat K pieces of meat. Find the time required to have K or more pieces of meat ready if he put the heat source to minimize this time.
Constraints
* All values in input are integers.
* 1 \leq N \leq 60
* 1 \leq K \leq N
* -1000 \leq x_i , y_i \leq 1000
* \left(x_i, y_i\right) \neq \left(x_j, y_j\right) \left(i \neq j \right)
* 1 \leq c_i \leq 100
Input
Input is given from Standard Input in the following format:
N K
x_1 y_1 c_1
\vdots
x_N y_N c_N
Output
Print the answer.
It will be considered correct if its absolute or relative error from our answer is at most 10^{-6}.
Examples
Input
4 3
-1 0 3
0 0 3
1 0 2
1 1 40
Output
2.4
Input
10 5
-879 981 26
890 -406 81
512 859 97
362 -955 25
128 553 17
-885 763 2
449 310 57
-656 -204 11
-270 76 40
184 170 16
Output
7411.2252
Submitted Solution:
```
def i1():
return int(input())
def i2():
return [int(i) for i in input().split()]
import math
[n,k]=i2()
d=[]
for i in range(n):
d.append(i2())
def check(t):
for x1,y1,c1 in d:
for x2,y2,c2 in d:
if x1==x2 and y1==y2:
continue
aa=2*(x1-x2)
bb=2*(y1-y2)
cc=t**2*(1/c2**2-1/c1**2)+x1**2-x2**2+y1**2-y2**2
if aa==0:
continue
ss=bb**2/aa**2+1
tt=bb/aa*(x1-cc/aa)-y1
uu=(x1-cc/aa)**2+y1**2-(t/c1)**2
if tt**2-ss*uu<0 or ss==0:
continue
Y1=-tt-math.sqrt(tt**2-ss*uu)
Y1/=ss
X1=(cc-bb*Y1)/aa
count=0
for x,y,c in d:
if c*math.sqrt((X1-x)**2+(Y1-y)**2)<=t:
count+=1
if count>=k:
return True
Y2=-tt+math.sqrt(tt**2-ss*uu)
Y2/=ss
X2=(cc-bb*Y2)/aa
count=0
for x,y,c in d:
if c*math.sqrt((X2-x)**2+(Y2-y)**2)<=t:
count+=1
if count>=k:
return True
for X,Y,C in d:
count=0
for x,y,c in d:
if c*math.sqrt((X-x)**2+(Y-y)**2)<=t:
count+=1
if count>=k:
return True
return False
low=0
high=100*3000
eps=10**(-7)
while high-low>eps:
mid=(low+high)/2
if check(mid):
high=mid
else:
low=mid
print(high)
``` | instruction | 0 | 21,803 | 9 | 43,606 |
No | output | 1 | 21,803 | 9 | 43,607 |
Provide a correct Python 3 solution for this coding contest problem.
E869120's and square1001's 16-th birthday is coming soon.
Takahashi from AtCoder Kingdom gave them a round cake cut into 16 equal fan-shaped pieces.
E869120 and square1001 were just about to eat A and B of those pieces, respectively,
when they found a note attached to the cake saying that "the same person should not take two adjacent pieces of cake".
Can both of them obey the instruction in the note and take desired numbers of pieces of cake?
Constraints
* A and B are integers between 1 and 16 (inclusive).
* A+B is at most 16.
Input
Input is given from Standard Input in the following format:
A B
Output
If both E869120 and square1001 can obey the instruction in the note and take desired numbers of pieces of cake, print `Yay!`; otherwise, print `:(`.
Examples
Input
5 4
Output
Yay!
Input
8 8
Output
Yay!
Input
11 4
Output
:( | instruction | 0 | 21,852 | 9 | 43,704 |
"Correct Solution:
```
a,b=map(int,input().split())
print('Yay!' if (a<9 and b<9) else ':(')
``` | output | 1 | 21,852 | 9 | 43,705 |
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