text stringlengths 1.02k 43.5k | conversation_id int64 853 107k | embedding list | cluster int64 24 24 |
|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Tags: brute force, greedy
Correct Solution:
```
n,k=list(map(int,input().split()))
ans,c=0,0
while k>0 and n>0:
if n % 10 == 0: k -= 1
else: ans += 1
n //= 10
c += 1
print(ans if n>0 else max(0,c-1))
```
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0.1258544921875,
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-0.6... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Tags: brute force, greedy
Correct Solution:
```
n,k=map(int,input().split())
if(n==0):
print(0)
exit()
x=[]
while(n>0):
x.append(n%10)
n=n//10
c1=0
c2=0
for i in range (len(x)):
if(x[i]==0):
c1+=1
if(c1==k):
break
if(x[i]!=0):
c2+=1
#print(c1,c2)
if(c1<k):
print(len(x)-1)
else:
print(c2)
```
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0.120361328125,
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... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Tags: brute force, greedy
Correct Solution:
```
n,k = [x for x in input().split()]
k = int(k)
ans = 0
stack = 0
if k == 0:
print(0)
if n.count('0') < k:
print(len(n)-1)
else:
for i in n[::-1]:
if stack == k:
break
if i=="0":
stack+=1
else:
ans += 1
print(ans)
```
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0.171630859375,
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-0.6376953125,
... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Tags: brute force, greedy
Correct Solution:
```
n,k = input().split()
k = int(k)
n = list(map(int,list(n)))
l = len(n)
j = l-1
while(j>-1 and k>0):
if(n[j]!=0):
del n[j]
else:
k -= 1
j -= 1
if(n[0]==0):
print(l-1)
else:
print(l-len(n))
```
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0.030059814453125,
-0.72900390625,
-0.77197265625,
0.1463623046875,
-0.6796875,
-... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Tags: brute force, greedy
Correct Solution:
```
n, k = input().split()
n, k = n[::-1], int(k)
r = 0
for i in range(len(n)):
if n[i] != '0':
r += 1
if i-r+1 == k:
break
else:
r = len(n)-1
print(r)
```
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0.031280517578125,
-0.7109375,
-0.75732421875,
0.1402587890625,
-0.65625,
-0.71... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Tags: brute force, greedy
Correct Solution:
```
n, k = input().split()
n = list(n)
k = int(k)
zeroes = 0
digits_to_remove = 0
i = len(n) - 1
while True:
if zeroes < k:
if i == 0:
digits_to_remove += len(n) - 1
break
elif n[i] == '0':
zeroes += 1
else:
del n[i]
digits_to_remove += 1
i -= 1
else:
break
print(digits_to_remove)
```
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0.0400390625,
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-0.779296875,
0.147705078125,
-0.66748046875,
-0.6... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Tags: brute force, greedy
Correct Solution:
```
n,m=input().split();n,m=list(n)[::-1],int(m);k=0;i=0;ans=0;p=len(n)
if p<=m:print(p-1)
else:
if n.count('0')<m:print(p-1)
else:
while k!=m and i<p:
if n[i]=='0':k+=1
else:ans+=1
i+=1
print(ans)
```
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0.055328369140625,
-0.71630859375,
-0.78759765625,
0.12225341796875,
-0.65917968... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Tags: brute force, greedy
Correct Solution:
```
n,k=map(str,input().split())
a=list(n)
k=int(k)
if((10**k)>int(n)):
print(len(n)-1)
else:
if((int(n)%(10**k))==0):
print(0)
else:
c=0
z=0
for i in range(len(a)-1,-1,-1):
if(z==k):
break
if(a[i]!='0'):
c=c+1
if(a[i]=='0'):
z=z+1
if(z<k):
print(len(a)-1)
else:
print(c)
```
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0.06695556640625,
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0.11846923828125,
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-... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Submitted Solution:
```
L = input()
L = L.split()
n = L[0]
k = int(L[1])
def Div(N, K):
N = int(N)
if N%10**K == 0:
return True
else:
return False
c = 0
f = len(n)
while f != 1:
f -= 1
if n[f] != '0' and not(Div(n,k)):
b = n[:f] + n[f +1:]
n = b
c += 1
else:
b = n
if int(n) == 0:
print (c)
else:
nd = len(b)
nceros = nd -1
if nceros < k:
c = nceros + c
print (c)
```
Yes
| 100,824 | [
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0.08831787109375,
-0.6728... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Submitted Solution:
```
n, k = list(map(int, input().split()))
n = list(str(n))
a = len(n)
if n.count('0') < k:
print(a-1)
else:
cz = 0
ca = 0
for i in n[::-1]:
if i == '0':
cz += 1
else:
ca += 1
if cz == k:
print(ca)
break
```
Yes
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0.06561279296875,
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... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Submitted Solution:
```
s, k = input().split()
k = int(k)
v, w, l = 0, 0, len(s)
for i in reversed(range(l)):
if s[i] == '0':
v += 1
elif not v >= k:
w += 1
if v >= k:
print(w)
else:
print(l-1)
```
Yes
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0.126220703125,
0.0110931396484375,
-0.002349853515625,
-0.4794921875,
-0.156982421875,
-0.1998291015625,
0.18701171875,
-0.057037353515625,
1.0537109375,
0.609375,
-0.055328369140625,
-0.01873779296875,
-0.7890625,
-0.72021484375,
0.076171875,
-0.6962890625,
-0.69... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Submitted Solution:
```
a = list(input().split())
n = a[0]
k = int(a[1])
c = 0
f = 0
for x in range(-1, -len(n)-1, -1):
if n[x] == "0":
c = c + 1
if c == k:
print( -x-k )
f = 1
break
if f == 0:
print(len(n) - 1)
```
Yes
| 100,827 | [
0.2376708984375,
0.1343994140625,
0.00945281982421875,
-0.0031986236572265625,
-0.46044921875,
-0.1427001953125,
-0.1737060546875,
0.212646484375,
-0.0025615692138671875,
1.0400390625,
0.642578125,
-0.004100799560546875,
-0.01390838623046875,
-0.779296875,
-0.73095703125,
0.08190917968... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Submitted Solution:
```
n,k=list(map(int,input().split()))
ans,c=0,0
while k>0 and n>0:
if n % 10 == 0: k -= 1
else: ans += 1
n //= 10
c += 1
print(ans if n>0 else c-1)
```
No
| 100,828 | [
0.2431640625,
0.1064453125,
0.0292816162109375,
-0.0178985595703125,
-0.46533203125,
-0.1810302734375,
-0.1783447265625,
0.18310546875,
-0.02410888671875,
1.046875,
0.6279296875,
-0.019073486328125,
-0.006961822509765625,
-0.7900390625,
-0.74072265625,
0.09759521484375,
-0.6669921875... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Submitted Solution:
```
n,k=input ().split (); k = int (k); n=n [::-1];
for i in range (0,100) :
if n [:i].count ("0")>=k :
print (i);
break;
else : print (str (len (n)-1))
```
No
| 100,829 | [
0.2425537109375,
0.1151123046875,
0.014373779296875,
-0.0177764892578125,
-0.49169921875,
-0.1697998046875,
-0.1778564453125,
0.189208984375,
-0.038330078125,
1.0341796875,
0.6279296875,
-0.0533447265625,
-0.0202789306640625,
-0.76708984375,
-0.7177734375,
0.097900390625,
-0.69091796... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Submitted Solution:
```
n,k=map(str,input().strip().split())
k=int(k)
num=str(10**k)
if(len(n)<len(num)):
print(len(n)-1)
else:
c,z=0,0
for i in range(len(n)-1,-1,-1):
if(n[i]!='0'):
c+=1
else:
z+=1
if(z==k):
break
print(c)
```
No
| 100,830 | [
0.2152099609375,
0.1151123046875,
0.038482666015625,
-0.0206298828125,
-0.475830078125,
-0.156982421875,
-0.183349609375,
0.16845703125,
-0.04522705078125,
1.01953125,
0.630859375,
-0.058319091796875,
-0.01438140869140625,
-0.763671875,
-0.73046875,
0.0897216796875,
-0.6962890625,
... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k.
In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103 = 1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number n, so that the result is divisible by 10k. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input
The only line of the input contains two integer numbers n and k (0 β€ n β€ 2 000 000 000, 1 β€ k β€ 9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output
Print w β the required minimal number of digits to erase. After removing the appropriate w digits from the number n, the result should have a value that is divisible by 10k. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Examples
Input
30020 3
Output
1
Input
100 9
Output
2
Input
10203049 2
Output
3
Note
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number.
Submitted Solution:
```
'''input
100 9
'''
n, k = map(int, input().split())
if str(n).count('0') >= k:
x = str(n)[::-1]
for l in range(1,len(x)):
if x[:l].count('0') == k:
print(l-k)
else:
print(len(str(n))-1)
```
No
| 100,831 | [
0.21875,
0.11126708984375,
0.0266876220703125,
-0.0447998046875,
-0.4462890625,
-0.1746826171875,
-0.17138671875,
0.1712646484375,
0.004108428955078125,
1.05859375,
0.6552734375,
-0.038482666015625,
-0.01232147216796875,
-0.7646484375,
-0.6982421875,
0.0836181640625,
-0.6728515625,
... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Tags: binary search, data structures, greedy
Correct Solution:
```
import sys
import bisect
input = lambda: sys.stdin.readline().rstrip("\r\n")
for _ in range(int(input())):
n=int(input())
l=[]
r=[]
a=[]
for _ in range(n):
L,R=map(int,input().split())
l.append(L)
r.append(R)
a.append([L,R])
l.sort()
r.sort()
ans=n
for i in a:
ans=min(ans,n-bisect.bisect(l,i[1])+bisect.bisect_left(r,i[0]))
print(ans)
```
| 101,416 | [
0.257568359375,
0.2646484375,
0.509765625,
0.46337890625,
-0.58544921875,
-0.7578125,
-0.2467041015625,
-0.03106689453125,
0.404541015625,
0.68603515625,
0.9189453125,
-0.109375,
0.08172607421875,
-0.94091796875,
-0.71923828125,
-0.001392364501953125,
-0.7119140625,
-0.6904296875,
... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Tags: binary search, data structures, greedy
Correct Solution:
```
import sys
import bisect
input=sys.stdin.readline
t=int(input())
for _ in range(t):
n=int(input())
left=[]
right=[]
l=[]
for i in range(n):
p=input().split()
x=int(p[0])
y=int(p[1])
l.append((x,y))
left.append(x)
right.append(y)
left.sort()
right.sort()
mina=10**18
for i in range(n):
y=bisect.bisect_right(left,l[i][1])
y=n-y
z=bisect.bisect_left(right,l[i][0])
mina=min(mina,z+y)
print(mina)
```
| 101,417 | [
0.250244140625,
0.2384033203125,
0.5087890625,
0.429443359375,
-0.58837890625,
-0.76611328125,
-0.248291015625,
-0.0340576171875,
0.401611328125,
0.68798828125,
0.92919921875,
-0.11639404296875,
0.11676025390625,
-0.935546875,
-0.73974609375,
0.003154754638671875,
-0.7421875,
-0.69... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Tags: binary search, data structures, greedy
Correct Solution:
```
from bisect import bisect_left, bisect_right
from math import inf
from sys import stdin
def read_ints():
return map(int, stdin.readline().split())
t_n, = read_ints()
for i_t in range(t_n):
n, = read_ints()
segments = [tuple(read_ints()) for i_segment in range(n)]
ls = sorted(l for l, r in segments)
rs = sorted(r for l, r in segments)
result = +inf
for l, r in segments:
lower_n = bisect_left(rs, l)
higher_n = len(ls) - bisect_right(ls, r)
result = min(result, lower_n + higher_n)
print(result)
```
| 101,418 | [
0.26025390625,
0.258544921875,
0.501953125,
0.42529296875,
-0.5986328125,
-0.7333984375,
-0.2320556640625,
-0.0645751953125,
0.39013671875,
0.7236328125,
0.94384765625,
-0.09429931640625,
0.1435546875,
-0.93359375,
-0.72021484375,
-0.049041748046875,
-0.75830078125,
-0.70166015625,... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Tags: binary search, data structures, greedy
Correct Solution:
```
import sys,math,itertools
from collections import Counter,deque,defaultdict
from bisect import bisect_left,bisect_right
from heapq import heappop,heappush,heapify
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
class BIT:
def __init__(self, n):
self.n = n
self.data = [0]*(n+1)
self.el = [0]*(n+1)
def sum(self, i):
s = 0
while i > 0:
s += self.data[i]
i -= i & -i
return s
def add(self, i, x):
# assert i > 0
self.el[i] += x
while i <= self.n:
self.data[i] += x
i += i & -i
def get(self, i, j=None):
if j is None:
return self.el[i]
return self.sum(j) - self.sum(i)
# n = 6
# a = [1,2,3,4,5,6]
# bit = BIT(n)
# for i,e in enumerate(a):
# bit.add(i+1,e)
# print(bit.get(2,5)) #12 (3+4+5)
for _ in range(inp()):
n = inp()
lr = [inpl() for _ in range(n)]
s = set()
for l,r in lr:
s.add(l); s.add(r)
s = list(s); s.sort()
d = {}
for i,x in enumerate(s):
d[x] = i+1
ln = len(s)
lbit = BIT(ln+10)
rbit = BIT(ln+10)
for i,(l,r) in enumerate(lr):
lr[i][0] = d[l]; lbit.add(d[l]+1,1)
lr[i][1] = d[r]; rbit.add(d[r]+1,1)
res = INF
# print(rbit.get(1,2))
for L,R in lr:
now = rbit.get(0,L) + lbit.get(R+1,ln+10)
res = min(res,now)
print(res)
```
| 101,419 | [
0.2305908203125,
0.2257080078125,
0.447265625,
0.517578125,
-0.55029296875,
-0.73388671875,
-0.226318359375,
-0.1610107421875,
0.469970703125,
0.72509765625,
0.91259765625,
-0.129150390625,
0.104248046875,
-0.95849609375,
-0.74560546875,
-0.023956298828125,
-0.744140625,
-0.7338867... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Tags: binary search, data structures, greedy
Correct Solution:
```
import os,io
from bisect import bisect_left, bisect_right
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
for _ in range (int(input())):
n = int(input())
l = []
r = []
a = []
for i in range (n):
li,ri = [int(i) for i in input().split()]
l.append(li)
r.append(ri)
a.append([li,ri])
l.sort()
r.sort()
cnt = n
for i in range (n):
cnt = min(cnt, n-bisect_right(l,a[i][1])+bisect_left(r,a[i][0]))
print(cnt)
```
| 101,420 | [
0.2353515625,
0.2430419921875,
0.49853515625,
0.44873046875,
-0.59619140625,
-0.77197265625,
-0.221923828125,
-0.042236328125,
0.442626953125,
0.7255859375,
0.95947265625,
-0.0987548828125,
0.11737060546875,
-0.90673828125,
-0.73974609375,
-0.01727294921875,
-0.77490234375,
-0.7001... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Tags: binary search, data structures, greedy
Correct Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from bisect import bisect_left, bisect_right
for _ in range (int(input())):
n = int(input())
l = []
r = []
a = []
for i in range (n):
li,ri = [int(i) for i in input().split()]
l.append(li)
r.append(ri)
a.append([li,ri])
l.sort()
r.sort()
cnt = n
for i in range (n):
cnt = min(cnt, n-bisect_right(l,a[i][1])+bisect_left(r,a[i][0]))
print(cnt)
```
| 101,421 | [
0.31298828125,
0.233642578125,
0.44140625,
0.488037109375,
-0.53662109375,
-0.84228515625,
-0.2452392578125,
0.0712890625,
0.461181640625,
0.69580078125,
0.9384765625,
-0.050506591796875,
0.1163330078125,
-0.93115234375,
-0.6484375,
-0.0144195556640625,
-0.65869140625,
-0.666503906... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Tags: binary search, data structures, greedy
Correct Solution:
```
import bisect
import io, os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
for _ in range(int(input())):
n = int(input())
ls = []
lsl = []
lsr = []
for _ in range(n):
l, r = map(int, input().split())
ls.append([l, r])
lsl.append(l)
lsr.append(r)
lsl.sort()
lsr.sort()
cnt = n
for i in range(n):
cnt = min(cnt, n - bisect.bisect_right(lsl, ls[i][1]) + bisect.bisect_left(lsr, ls[i][0]))
print(cnt)
```
| 101,422 | [
0.253662109375,
0.240478515625,
0.515625,
0.45556640625,
-0.6015625,
-0.7548828125,
-0.1986083984375,
-0.044403076171875,
0.44921875,
0.71142578125,
0.94482421875,
-0.11932373046875,
0.11322021484375,
-0.92236328125,
-0.73876953125,
-0.009185791015625,
-0.75341796875,
-0.6801757812... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Tags: binary search, data structures, greedy
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
#import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=300006, func=lambda a, b: min(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b:a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <=key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.height=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val==0:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
elif val>=1:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
def do(self,temp):
if not temp:
return 0
ter=temp
temp.height=self.do(ter.left)+self.do(ter.right)
if temp.height==0:
temp.height+=1
return temp.height
def query(self, xor):
self.temp = self.root
cur=0
i=31
while(i>-1):
val = xor & (1 << i)
if not self.temp:
return cur
if val>=1:
self.opp = self.temp.right
if self.temp.left:
self.temp = self.temp.left
else:
return cur
else:
self.opp=self.temp.left
if self.temp.right:
self.temp = self.temp.right
else:
return cur
if self.temp.height==pow(2,i):
cur+=1<<(i)
self.temp=self.opp
i-=1
return cur
#-------------------------bin trie-------------------------------------------
def binarySearchCount1(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
for ik in range(int(input())):
n=int(input())
l=[]
l1=[]
l2=[]
for i in range(n):
a,b=map(int,input().split())
l.append((a,b))
l1.append(b)
l.sort()
l1.sort()
d=defaultdict(list)
inr = defaultdict(int)
for i in range(len(l1)):
d[l1[i]].append(i)
inr[l1[i]]+=1
w=[]
e=[0]*n
s=SegmentTree(e)
for i in range(n):
w.append(l[i][0])
w1=n
for i in range(n):
ind=binarySearchCount1(l1,len(l1),l[i][0])
if ind==n:
ans=0
else:
ind=d[l1[ind]][0]
ans=s.query(ind,n-1)
ans+=binarySearchCount(w,len(w),l[i][1])-i-1
s.__setitem__(d[l[i][1]][inr[l[i][1]]-1],1)
e[d[l[i][1]][inr[l[i][1]]-1]]=1
inr[l[i][1]]-=1
w1=min(w1,n-1-ans)
print(w1)
```
| 101,423 | [
0.318603515625,
0.240234375,
0.441162109375,
0.5029296875,
-0.54248046875,
-0.8349609375,
-0.23291015625,
0.09283447265625,
0.421630859375,
0.6826171875,
0.96923828125,
-0.04052734375,
0.08819580078125,
-0.888671875,
-0.6611328125,
-0.01397705078125,
-0.64306640625,
-0.63818359375,... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Submitted Solution:
```
import sys
import math,bisect,operator
inf,mod = float('inf'),10**9+7
sys.setrecursionlimit(10 ** 5)
from itertools import groupby,accumulate
from heapq import heapify,heappop,heappush
from collections import deque,Counter,defaultdict
I = lambda : int(sys.stdin.readline())
neo = lambda : map(int, sys.stdin.readline().split())
Neo = lambda : list(map(int, sys.stdin.readline().split()))
def overlap(v):
x,y = [],[]
for i,j in v:
x += [i]
y += [j]
x.sort()
y.sort()
Ans = 0
for i in range(n):
p,q = v[i][0],v[i][1]
r = bisect.bisect_right(x,q)
l = bisect.bisect_left(y,p)
Ans = max(Ans,r-l)
return Ans
for _ in range(I()):
n = I()
A = []
for i in range(n):
l,r = neo()
A += [(l,r)]
print(max(n-overlap(A),0))
```
Yes
| 101,424 | [
0.273193359375,
0.3466796875,
0.416259765625,
0.4521484375,
-0.6728515625,
-0.6494140625,
-0.295166015625,
0.045562744140625,
0.402099609375,
0.79052734375,
0.92529296875,
0.024566650390625,
0.107421875,
-0.970703125,
-0.671875,
0.0104217529296875,
-0.73486328125,
-0.7216796875,
... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
for t in range(int(input())):
n=int(input())
lft=[1000000000]
rt=[0]
a=[]
for i in range(n):
l,r=map(int,input().split())
a.append([l,r])
lft.append(l)
rt.append(r)
lft.sort()
rt.sort()
count=0
mini=9999999999999
for i in range(n):
j,k,count=0,n,0
while True:
m=(j+k)//2
if(rt[m]>=a[i][0]):
k=m
else:
j=m
if(k==j+1):
break
count+=j
j,k=0,n
while True:
m=(j+k)//2
if(lft[m]>a[i][1]):
k=m
else:
j=m
if(k==j+1):
break
count+=(n-1-j)
mini=min(mini,count)
print(mini)
```
Yes
| 101,425 | [
0.354248046875,
0.219970703125,
0.291259765625,
0.498046875,
-0.71728515625,
-0.7177734375,
-0.287841796875,
0.112060546875,
0.479248046875,
0.76220703125,
0.859375,
0.046234130859375,
0.11407470703125,
-0.90673828125,
-0.7021484375,
-0.03448486328125,
-0.6953125,
-0.7041015625,
... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Submitted Solution:
```
from sys import stdin, stdout
from bisect import bisect_left, bisect_right
def main():
global dp
global add
for _ in range(int(stdin.readline())):
n = int(stdin.readline())
rangey = []
L = list()
R = list()
for _ in range(n):
l,r = list(map(int, stdin.readline().split()))
rangey.append([l,r])
L.append(l)
R.append(r)
L.sort()
R.sort()
mn = n + 1
for i in range(n):
l,r = rangey[i]
left = bisect_left(R, l)
right = n - bisect_right(L, r)
mn = min(left + right, mn)
stdout.write(str(mn)+"\n")
main()
```
Yes
| 101,426 | [
0.2783203125,
0.2763671875,
0.40380859375,
0.4560546875,
-0.7490234375,
-0.68798828125,
-0.372314453125,
0.0194091796875,
0.374267578125,
0.77685546875,
0.8486328125,
0.018157958984375,
0.0704345703125,
-0.94287109375,
-0.71435546875,
-0.0943603515625,
-0.7578125,
-0.70703125,
-0... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Submitted Solution:
```
"""
pppppppppppppppppppp
ppppp ppppppppppppppppppp
ppppppp ppppppppppppppppppppp
pppppppp pppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppp
pppppppppppppppppppppppppppppppppppppppppppppppp ppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppp
pppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppp
pppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp pppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp ppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppppp
pppppppppppppppppppppppppppppppp
pppppppppppppppppppppp pppppppp
ppppppppppppppppppppp ppppppp
ppppppppppppppppppp ppppp
pppppppppppppppppppp
"""
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
from decimal import Decimal
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(str(var)+"\n")
def outa(*var, end="\n"): sys.stdout.write(' '.join(map(str, var)) + end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
def update(index, value):
while index <= limit:
bit[index] += value
index += index & -index
def query(index):
ret = 0
while index:
ret += bit[index]
index -= index & -index
return ret
for _ in range(int(data())):
n = int(data())
arr = [l() for _ in range(n)]
s = set()
for a, b in arr:
s.add(a)
s.add(b)
s = list(s)
s.sort()
mp = dd(int)
for i in range(len(s)):
mp[s[i]] = i + 1
for i in range(n):
arr[i] = [mp[arr[i][0]], mp[arr[i][1]]]
arr.sort()
limit = n * 2 + 10
bit = [0] * limit
limit -= 1
answer = n
for i, [a, b] in enumerate(arr):
low, high = i, n - 1
index = i
while low <= high:
mid = (low + high) >> 1
if arr[mid][0] <= b:
index = mid
low = mid + 1
else:
high = mid - 1
answer = min(answer, n - (index - i + 1 + query(n * 2 + 2) - query(a - 1)))
update(b, 1)
out(answer)
```
Yes
| 101,427 | [
0.357666015625,
0.251220703125,
0.3876953125,
0.482421875,
-0.689453125,
-0.736328125,
-0.38134765625,
0.130615234375,
0.39453125,
0.72265625,
0.89599609375,
0.00905609130859375,
0.05426025390625,
-0.90185546875,
-0.6884765625,
-0.03668212890625,
-0.72900390625,
-0.646484375,
-0.... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Submitted Solution:
```
import sys
input=sys.stdin.readline
def update(inp,add,ar,n):
while(inp<n):
ar[inp]+=add
inp+=(inp&(-inp))
def fun1(inp,ar,n):
ans=0
while(inp):
ans+=ar[inp]
inp-=(inp&(-inp))
return ans
for _ in range(int(input())):
n=int(input())
ar=[]
se=set({})
for i in range(n):
l,r=map(int,input().split())
ar.append([l,r])
se.add(l)
se.add(r)
se=list(se)
se.sort()
dic={}
le=len(se)
for i in range(le):
dic[se[i]]=i
br=[]
for i in range(n):
br.append([dic[ar[i][0]],dic[ar[i][1]]])
br.sort(key=lambda x:x[0])
le+=1
left=[0]*(le-1)
for i in range(n):
left[br[i][0]]+=1
for i in range(1,le-1):
left[i]+=left[i-1]
right=[0]*le
ans=0
for i in range(n):
xx=left[br[i][1]]-left[br[i][0]]
yy=i-fun1(br[i][0],right,le)
ans=max(xx+yy+1,ans)
update(br[i][1]+1,1,right,le)
print(n-ans)
```
No
| 101,428 | [
0.35546875,
0.256591796875,
0.374267578125,
0.431640625,
-0.6943359375,
-0.7265625,
-0.38330078125,
0.07208251953125,
0.414306640625,
0.73828125,
0.84912109375,
-0.00023627281188964844,
0.07916259765625,
-0.91162109375,
-0.68212890625,
-0.0814208984375,
-0.712890625,
-0.6103515625,... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Submitted Solution:
```
import sys
import bisect,string,math,time,functools,random,fractions
from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations,groupby
rep=range;R=range
def Golf():n,*t=map(int,open(0).read().split())
def I():return int(input())
def S_():return input()
def IS():return input().split()
def LS():return [i for i in input().split()]
def MI():return map(int,input().split())
def LI():return [int(i) for i in input().split()]
def LI_():return [int(i)-1 for i in input().split()]
def NI(n):return [int(input()) for i in range(n)]
def NI_(n):return [int(input())-1 for i in range(n)]
def NLI(n):return [[int(i) for i in input().split()] for i in range(n)]
def NLI_(n):return [[int(i)-1 for i in input().split()] for i in range(n)]
def StoLI():return [ord(i)-97 for i in input()]
def ItoS(n):return chr(n+97)
def LtoS(ls):return ''.join([chr(i+97) for i in ls])
def RA():return map(int,open(0).read().split())
def RLI(n=8,a=1,b=10):return [random.randint(a,b)for i in range(n)]
def RI(a=1,b=10):return random.randint(a,b)
def INP():
N=9
n=random.randint(1,N)
m=random.randint(1,n*n)
A=[random.randint(1,n) for i in range(m)]
B=[random.randint(1,n) for i in range(m)]
G=[[]for i in range(n)];RG=[[]for i in range(n)]
for i in range(m):
a,b=A[i]-1,B[i]-1
if a==b:continue
G[a]+=(b,1),;RG[b]+=(a,1),
return n,m,G,RG
def Rtest(T):
case,err=0,0
for i in range(T):
inp=INP()
a1=naive(*inp)
a2=solve(*inp)
if a1==a2:
print(inp)
print('naive',a1)
print('solve',a2)
err+=1
case+=1
print('Tested',case,'case with',err,'errors')
def GI(V,E,ls=None,Directed=False,index=1):
org_inp=[];g=[[] for i in range(V)]
FromStdin=True if ls==None else False
for i in range(E):
if FromStdin:
inp=LI()
org_inp.append(inp)
else:
inp=ls[i]
if len(inp)==2:
a,b=inp;c=1
else:
a,b,c=inp
if index==1:a-=1;b-=1
aa=(a,c);bb=(b,c);g[a].append(bb)
if not Directed:g[b].append(aa)
return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1):
#h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1) # sample usage
mp=[boundary]*(w+2);found={}
for i in R(h):
s=input()
for char in search:
if char in s:
found[char]=((i+1)*(w+2)+s.index(char)+1)
mp_def[char]=mp_def[replacement_of_found]
mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
mp+=[boundary]*(w+2)
return h+2,w+2,mp,found
def TI(n):return GI(n,n-1)
def accum(ls):
rt=[0]
for i in ls:rt+=[rt[-1]+i]
return rt
def bit_combination(n,base=2):
rt=[]
for tb in R(base**n):s=[tb//(base**bt)%base for bt in R(n)];rt+=[s]
return rt
def gcd(x,y):
if y==0:return x
if x%y==0:return y
while x%y!=0:x,y=y,x%y
return y
def YN(x):print(['NO','YES'][x])
def Yn(x):print(['No','Yes'][x])
def show(*inp,end='\n'):
if show_flg:print(*inp,end=end)
mo=10**9+7
#mo=998244353
inf=float('inf')
FourNb=[(-1,0),(1,0),(0,1),(0,-1)];EightNb=[(-1,0),(1,0),(0,1),(0,-1),(1,1),(-1,-1),(1,-1),(-1,1)];compas=dict(zip('WENS',FourNb));cursol=dict(zip('LRUD',FourNb))
l_alp=string.ascii_lowercase
#sys.setrecursionlimit(10**9)
read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()
show_flg=False
show_flg=True
class Comb:
def __init__(self,n):
return
def fact(self,n):
return self.fac[n]
def invf(self,n):
return self.inv[n]
def comb(self,x,y):
if y<0 or y>x:
return 0
return x*(x-1)//2
########################################################################################################################################################################
# Verified by
# https://atcoder.jp/contests/arc033/submissions/me
# https://atcoder.jp/contests/abc174/tasks/abc174_f
#
# speed up TIPS: delete update of el. non-use of getitem, setitem.
#
# Binary Indexed Tree
# Bit.add(i,x) :Add x at i-th value, the following gives the same result
# Bit[i]+=x
# Bit.sum(i) : get sum up to i-th value
# Bit.l_bound(w) get bound of index where x1+x2+...+xi<w
class Bit: # 1-indexed
def __init__(self,n,init=None):
self.size=n
self.m=len(bin(self.size))-2
self.arr=[0]*(2**self.m+1)
self.el=[0]*(2**self.m+1)
if init!=None:
for i in range(len(init)):
self.add(i,init[i])
self.el[i]=init[i]
def __str__(self):
a=[self.sum(i+1)-self.sum(i) for i in range(self.size)]
return str(a)
def add(self,i,x):
if not 0<i<=self.size:return NotImplemented
self.el[i]+=x
while i<=self.size:
self.arr[i]+=x
i+=i&(-i)
return
def sum(self,i):
if not 0<=i<=self.size:return NotImplemented
rt=0
while i>0:
rt+=self.arr[i]
i-=i&(-i)
return rt
def __getitem__(self,key):
return self.el[key]
#return self.sum(key+1)-self.sum(key)
def __setitem__(self,key,value):
self.add(key,value-self.sum(key+1)+self.sum(key))
def l_bound(self,w):
if w<=0:
return 0
x=0
k=2**self.m
while k>0:
if x+k<=self.size and self.arr[x+k]<w:
w-=self.arr[x+k]
x+=k
k>>=1
return x+1
def u_bound(self,w):
if w<=0:
return 0
x=0
k=2**self.m
while k>0:
if x+k<=self.size and self.arr[x+k]<=w:
w-=self.arr[x+k]
x+=k
k>>=1
return x+1
class Bit0(Bit): # 0-indexed
def add(self,j,x):
super().add(j+1,x)
def l_bound(self,w):
return max(super().l_bound(w)-1,0)
def u_bound(self,w):
return max(super().u_bound(w)-1,0)
class Multiset(Bit0):
def __init__(self,max_v):
super().__init__(max_v)
def insert(self,x):
super().add(x,1)
def find(self,x):
return super().l_bound(super().sum(x))
def __str__(self):
return str(self.arr)
def compress(L):
dc={v:i for i,v in enumerate(sorted(set(L)))}
return dc
ans=0
## Segment Tree ##
## Test case: ABC 146 F
## https://atcoder.jp/contests/abc146/tasks/abc146_f
## Initializer Template ##
# Range Sum: sg=SegTree(n)
# Range Minimum: sg=SegTree(n,inf,min,inf)
class SegTree:
def __init__(self,n,init_val=0,function=lambda a,b:a+b,ide=0):
self.size=n
self.ide_ele=ide
self.num=1<<(self.size-1).bit_length()
self.table=[self.ide_ele]*2*self.num
self.index=[0]*2*self.num
self.lazy=[self.ide_ele]*2*self.num
self.func=function
#set_val
if not hasattr(init_val,"__iter__"):
init_val=[init_val]*self.size
for i,val in enumerate(init_val):
self.table[i+self.num-1]=val
self.index[i+self.num-1]=i
#build
for i in range(self.num-2,-1,-1):
self.table[i]=self.func(self.table[2*i+1],self.table[2*i+2])
if self.table[i]==self.table[i*2+1]:
self.index[i]=self.index[i*2+1]
else:
self.index[i]=self.index[i*2+2]
def update(self,k,x):
k+=self.num-1
self.table[k]=x
while k:
k=(k-1)//2
res=self.func(self.table[k*2+1],self.table[k*2+2])
self.table[k]=res
## Remove if index is not needed
if res==self.table[k*2+1]:
self.index[k]=self.index[k*2+1]
else:
self.index[k]=self.index[k*2+2]
## Remove if index is not needed
def evaluate(k,l,r): #ι
ε»Άθ©δΎ‘ε¦η
if lazy[k]!=0:
node[k]+=lazy[k]
if(r-l>1):
lazy[2*k+1]+=lazy[k]//2
lazy[2*k+2]+=lazy[k]//2
lazy[k]=0
def __getitem__(self,key):
if type(key) is slice:
a=None if key.start==None else key.start
b=None if key.stop==None else key.stop
c=None if key.step==None else key.step
return self.table[self.num-1:self.num-1+self.size][slice(a,b,c)]
else:
if 0<=key<self.size:
return self.table[key+self.num-1]
elif -self.size<=key<0:
return self.table[self.size+key+self.num-1]
else:
raise IndexError("list index out of range")
def __setitem__(self,key,value):
if key>=0:
self.update(key,value)
else:
self.update(self.size+key,value)
def query(self,p,q):
if q<=p:
return self.ide_ele
p+=self.num-1
q+=self.num-2
res=self.ide_ele
while q-p>1:
if p&1==0:
res=self.func(res,self.table[p])
if q&1==1:
res=self.func(res,self.table[q])
q-=1
p=p>>1
q=(q-1)>>1
if p==q:
res=self.func(res,self.table[p])
else:
res=self.func(self.func(res,self.table[p]),self.table[q])
return res
def query_id(self,p,q):
if q<=p:
return self.ide_ele
p+=self.num-1
q+=self.num-2
res=self.ide_ele
idx=p
while q-p>1:
if p&1==0:
res=self.func(res,self.table[p])
if res==self.table[p]:
idx=self.index[p]
if q&1==1:
res=self.func(res,self.table[q])
if res==self.table[q]:
idx=self.index[q]
q-=1
p=p>>1
q=(q-1)>>1
if p==q:
res=self.func(res,self.table[p])
if res==self.table[p]:
idx=self.index[p]
else:
res=self.func(self.func(res,self.table[p]),self.table[q])
if res==self.table[p]:
idx=self.index[p]
elif res==self.table[q]:
idx=self.index[q]
return idx
def __str__(self):
# ηι
εγ葨瀺
rt=self.table[self.num-1:self.num-1+self.size]
return str(rt)
for _ in range(I()):
n=I()
p=[]
s=set()
L=[]
R=[]
for i in range(n):
l,r=LI()
p+=(l,r),
R+=r,
L+=l,
L.sort()
R.sort()
for i in range(n):
l,r=p[i]
a=bisect.bisect_left(R,l)
b=n-bisect.bisect_right(L,r)
ans=max(ans,n-1-a-b)
#show((a,b),(l,r),p,L,R)
print(n-1-ans)
```
No
| 101,429 | [
0.29296875,
0.2362060546875,
0.330810546875,
0.52001953125,
-0.63525390625,
-0.64208984375,
-0.323486328125,
-0.01702880859375,
0.494140625,
0.77685546875,
0.83544921875,
-0.053375244140625,
0.044281005859375,
-0.94921875,
-0.69921875,
-0.0330810546875,
-0.7109375,
-0.67529296875,
... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
a=[]
b=[]
for i in range(n):
l,r=map(int,input().split())
a.append([l,r])
b.append([l,r])
a.sort(key=lambda thing: thing[0])
b.sort(key=lambda thing: thing[1])
pointer1=0
pointer2=0
rightborder=0
ans=n-1
for i in range(n):
l=a[i][0]
r=a[i][1]
if r<=rightborder:
continue
rightborder=r
while pointer1+1<n and a[pointer1+1][0]<r:
pointer1+=1
while pointer2<n and b[pointer2][1]<l:
pointer2+=1
ans=min(ans,pointer2+n-pointer1-1)
print(ans)
```
No
| 101,430 | [
0.32177734375,
0.2381591796875,
0.377197265625,
0.428466796875,
-0.666015625,
-0.71728515625,
-0.362060546875,
0.11602783203125,
0.389404296875,
0.740234375,
0.87255859375,
0.021209716796875,
0.05120849609375,
-0.91064453125,
-0.68798828125,
-0.036163330078125,
-0.74169921875,
-0.6... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i β coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.
Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 β€ i β€ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.
Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?
Input
The first line contains a single integer t (1 β€ t β€ 2 β
10^5) β number of test cases. Then t test cases follow.
The first line of each test case contains a single integer n (1 β€ n β€ 2 β
10^5) β the number of segments. This is followed by n lines describing the segments.
Each segment is described by two integers l and r (1 β€ l β€ r β€ 10^9) β coordinates of the beginning and end of the segment, respectively.
It is guaranteed that the sum of n for all test cases does not exceed 2 β
10^5.
Output
For each test case, output a single integer β the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.
Example
Input
4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8
Output
0
1
2
0
Submitted Solution:
```
import bisect
for _ in range(int(input())):
n=int(input())
xx=[0]*n
yy=[0]*n
arr=[(0,0)]*n
for i in range(n):
a,b=map(int,input().split())
xx[i]=a
yy[i]=b
arr[i]=(a,b)
ans=999999999
for i in range(n):
a=bisect.bisect_left(yy,arr[i][0])
b=bisect.bisect_right(xx,arr[i][1])
b=n-b
#print(a,b,arr[i])
ans=min(ans,a+b)
#print(ed)
print(ans)
```
No
| 101,431 | [
0.289794921875,
0.28369140625,
0.377685546875,
0.436767578125,
-0.7099609375,
-0.69970703125,
-0.315185546875,
0.0574951171875,
0.37353515625,
0.77734375,
0.88232421875,
0.0224609375,
0.06610107421875,
-0.9638671875,
-0.72412109375,
-0.0316162109375,
-0.771484375,
-0.6845703125,
... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Tags: dp, sortings
Correct Solution:
```
from sys import stdin
input = stdin.readline
n = int(input())
T = 2001
t, d, p, idx = [], [], [], []
ans = []
arr = []
for i in range(n):
a, b, c = map(int, input().split())
arr.append([a, b, c, i])
arr.sort(key=lambda x: x[1])
for i in arr:
t.append(i[0]); d.append(i[1]); p.append(i[2]); idx.append(i[3])
dp = [[0 for j in range(n)] for i in range(T)]
for time in range(1, T):
for i in range(n):
#dp[time][i] = max(dp[time - 1][i], dp[time][i - 1])
dp[time][i] = dp[time][i - 1]
if d[i] > time >= t[i]:
if i:
dp[time][i] = max(dp[time][i], p[i] + dp[time - t[i]][i - 1])
else:
dp[time][i] = p[i]
b = [0, [0 ,0]]
for i in range(T):
for j in range(n):
if b[0] < dp[i][j]:
b = [dp[i][j], [i, j]]
print(b[0])
b = b[1]
while dp[b[0]][b[1]] and b[0] > -1:
if b[1] and dp[b[0]][b[1]] != dp[b[0]][b[1] - 1]:
ans.append(b[1])
b[0] -= t[b[1]]
elif b[1] == 0:
if dp[b[0]][b[1]]:
ans.append(b[1])
break
b[1] -= 1
print(len(ans))
for i in ans[::-1]:
print(idx[i] + 1, end=' ')
```
| 102,603 | [
0.41015625,
0.1705322265625,
0.1280517578125,
0.30712890625,
-0.373291015625,
-0.369140625,
-0.31201171875,
-0.015594482421875,
0.5302734375,
0.412841796875,
0.86279296875,
-0.09783935546875,
0.348388671875,
-0.5859375,
-0.45068359375,
-0.1407470703125,
-0.5283203125,
-0.3056640625... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Tags: dp, sortings
Correct Solution:
```
n = int(input())
items = []
max_time = 0
for i in range(1,n+1):
t,d,p = map(int,input().split())
max_time = max(max_time, d)
items.append((t,d,p,i))
items.sort(key=lambda x: x[1])
dp = [[(0,[]) for _ in range(n+1)] for _ in range(max_time+1)]
for time in range(1, max_time+1):
for it in range(1, n+1):
if time < items[it-1][0] or time >= items[it-1][1]:
dp[time][it] = max(dp[time][it-1], dp[time-1][it])
else:
pick = dp[time-items[it-1][0]][it-1][0] + items[it-1][2]
if dp[time][it-1][0] > pick :
dp[time][it] = max(dp[time][it-1], dp[time-1][it])
else:
dp[time][it] = (dp[time-items[it-1][0]][it-1][0] + items[it-1][2], list(dp[time-items[it-1][0]][it-1][1]))
dp[time][it][1].append(items[it-1][3])
#print(dp)
res = max(dp[max_time])
print(res[0])
print(len(res[1]))
print(*res[1])
```
| 102,604 | [
0.41015625,
0.1705322265625,
0.1280517578125,
0.30712890625,
-0.373291015625,
-0.369140625,
-0.31201171875,
-0.015594482421875,
0.5302734375,
0.412841796875,
0.86279296875,
-0.09783935546875,
0.348388671875,
-0.5859375,
-0.45068359375,
-0.1407470703125,
-0.5283203125,
-0.3056640625... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Tags: dp, sortings
Correct Solution:
```
from functools import lru_cache
def readints():
return [int(obj) for obj in input().strip().split()]
class Solver:
def main(self):
n = readints()[0]
self.t, self.d, self.p = [], [], []
for i in range(n):
t1, d1, p1 = readints()
self.t.append(t1)
self.d.append(d1)
self.p.append(p1)
self.backtrack = []
sd = max(self.d) + 1
for i in range(n+1):
self.backtrack.append([])
for j in range(sd):
self.backtrack[i].append(0)
triples = zip(self.t, self.d, self.p, range(1, n+1))
triples = sorted(triples, key=lambda x: x[1])
self.t, self.d, self.p, self.indexes = [0], [0], [0], []
for i in range(n):
self.t.append(triples[i][0])
self.d.append(triples[i][1])
self.p.append(triples[i][2])
self.indexes.append(triples[i][3])
self.f = []
for i in range(n+1):
self.f.append([])
for j in range(sd):
self.f[i].append(0)
for i in range(1, n+1):
for d in range(sd):
if d - self.t[i] >= 0 and d < self.d[i] and self.t[i] < self.d[i]:
data = self.f[i - 1][d - self.t[i]]
if data + self.p[i] > self.f[i][d]:
self.f[i][d] = data + self.p[i]
self.backtrack[i][d] = i
data = self.f[i - 1][d]
if data > self.f[i][d]:
self.f[i][d] = data
self.backtrack[i][d] = 0
ans = 0
res = []
best = None
for i in range(sd):
data = self.f[n][i]
if data > ans:
ans = data
best = (n, i)
if best is None:
print('0\n0\n')
return
i = best[0]
s = best[1]
while i > 0:
if self.backtrack[i][s] != 0:
res.append(self.backtrack[i][s])
s -= self.t[self.backtrack[i][s]]
i -= 1
print(ans)
print(len(res))
print(' '.join(str(self.indexes[item - 1]) for item in reversed(res)))
Solver().main()
```
| 102,605 | [
0.41015625,
0.1705322265625,
0.1280517578125,
0.30712890625,
-0.373291015625,
-0.369140625,
-0.31201171875,
-0.015594482421875,
0.5302734375,
0.412841796875,
0.86279296875,
-0.09783935546875,
0.348388671875,
-0.5859375,
-0.45068359375,
-0.1407470703125,
-0.5283203125,
-0.3056640625... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Tags: dp, sortings
Correct Solution:
```
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(2*10**5+10)
write = lambda x: sys.stdout.write(x+"\n")
debug = lambda x: sys.stderr.write(x+"\n")
writef = lambda x: print("{:.12f}".format(x))
n = int(input())
tdp = [list(map(int, input().split())) for _ in range(n)]
index = list(range(n))
index.sort(key=lambda i: tdp[i][1])
# tdp.sort(key=lambda item: (item[1]))
T = max([item[1] for item in tdp])
dp = [0]*(T+1)
ps = []
# for i,(t,d,p) in enumerate(tdp):
for ind in range(n):
i = index[ind]
t,d,p = tdp[i]
ndp = dp[:]
prv = [(k, -1) for k in range(T+1)]
for j in range(T+1):
if j+t<d:
if ndp[j+t]<dp[j]+p:
ndp[j+t] = dp[j]+p
prv[j+t] = (j, i)
dp = ndp
ps.append(prv)
# print(dp)
ans = max(dp)
res = []
for j in range(T+1)[::-1]:
if dp[j]==ans:
for i in range(n)[::-1]:
jj,ii = ps[i][j]
if ii>=0:
res.append(ii+1)
j = jj
break
res = res[::-1]
print(ans)
print(len(res))
write(" ".join(map(str, res)))
```
| 102,606 | [
0.41015625,
0.1705322265625,
0.1280517578125,
0.30712890625,
-0.373291015625,
-0.369140625,
-0.31201171875,
-0.015594482421875,
0.5302734375,
0.412841796875,
0.86279296875,
-0.09783935546875,
0.348388671875,
-0.5859375,
-0.45068359375,
-0.1407470703125,
-0.5283203125,
-0.3056640625... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Tags: dp, sortings
Correct Solution:
```
P = [0] * 2001
S = [[] for i in range(2001)]
q = [list(map(int, input().split())) + [str(i + 1)] for i in range(int(input()))]
q.sort(key=lambda q: q[1])
for t, d, p, i in q:
for x in range(t, d)[::-1]:
if P[x] < P[x - t] + p:
P[x] = P[x - t] + p
S[x] = S[x - t] + [i]
k = P.index(max(P))
print('\n'.join([str(P[k]), str(len(S[k])), ' '.join(S[k])]))
```
| 102,607 | [
0.41015625,
0.1705322265625,
0.1280517578125,
0.30712890625,
-0.373291015625,
-0.369140625,
-0.31201171875,
-0.015594482421875,
0.5302734375,
0.412841796875,
0.86279296875,
-0.09783935546875,
0.348388671875,
-0.5859375,
-0.45068359375,
-0.1407470703125,
-0.5283203125,
-0.3056640625... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Tags: dp, sortings
Correct Solution:
```
from sys import stdin, stdout
T = 2001
INF = int(1e9)
n = int(stdin.readline())
items = []
for i in range(n):
c, d, t = map(int, stdin.readline().split())
items.append((d, c, t, i+1))
items.sort()
dp = [[None for i in range(T)] for j in range(n)]
def solve(pos, time):
if pos >= n or time >= T:
return 0
if dp[pos][time] is not None:
return dp[pos][time]
d, c, t, i = items[pos]
ans = solve(pos+1, time)
if time + c < d:
ans = max(ans, solve(pos+1, time + c) + t)
dp[pos][time] = ans
return ans
ans = []
def recover(pos, time):
global ans
if pos >= n or time >= T:
return
d, c, t, i = items[pos]
if solve(pos+1, time) == solve(pos, time):
recover(pos+1, time)
else:
ans.append(i)
recover(pos+1, time + c)
stdout.write(str(solve(0, 0)) + '\n')
recover(0, 0)
stdout.write(str(len(ans)) + '\n')
for i in ans:
stdout.write(str(i) + ' ')
stdout.write('\n')
```
| 102,608 | [
0.41015625,
0.1705322265625,
0.1280517578125,
0.30712890625,
-0.373291015625,
-0.369140625,
-0.31201171875,
-0.015594482421875,
0.5302734375,
0.412841796875,
0.86279296875,
-0.09783935546875,
0.348388671875,
-0.5859375,
-0.45068359375,
-0.1407470703125,
-0.5283203125,
-0.3056640625... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Tags: dp, sortings
Correct Solution:
```
n = int(input())
a = []
for i in range(n):
t,d,p = map(int,input().split())
a.append([t,d,p,i+1])
a.sort(key = lambda x: x[1])
d = {0: [0,[]]}
for i in a:
e = {}
for j in d:
if d[j][0] + i[0] < i[1]:
if j + i[2] in d:
if d[j][0]+i[0] < d[j+i[2]][0]:
e[j+i[2]] = [d[j][0]+i[0],d[j][1]+[i[3]]]
else:
e[j+i[2]] = [d[j][0]+i[0],d[j][1]+[i[3]]]
d.update(e)
t = max(d)
print(t)
k = d[t][1]
print(len(k))
k = list(map(str,k))
print(' '.join(k))
```
| 102,609 | [
0.41015625,
0.1705322265625,
0.1280517578125,
0.30712890625,
-0.373291015625,
-0.369140625,
-0.31201171875,
-0.015594482421875,
0.5302734375,
0.412841796875,
0.86279296875,
-0.09783935546875,
0.348388671875,
-0.5859375,
-0.45068359375,
-0.1407470703125,
-0.5283203125,
-0.3056640625... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Tags: dp, sortings
Correct Solution:
```
import os
import sys
import re
from collections import OrderedDict
if 'PYCHARM' in os.environ:
sys.stdin = open('in', 'r')
n = int(input())
things = []
for i in range(n):
t, d, p = map(int, input().split())
things.append((d, t, p, i + 1))
things.sort()
D = 2001
f = [[0] * D]
w = [[False] * D]
for i in range(n):
thing = things[i]
w.append([False] * D)
f.append(list(f[i]))
for j in range(D):
ni = j + thing[1]
nv = f[i][j] + thing[2]
if ni < thing[0]:
if f[i + 1][ni] < nv:
f[i + 1][ni] = nv
w[i + 1][ni] = True
ind = 0
for i in range(D):
if f[n][i] > f[n][ind]:
ind = i
print(f[n][ind])
ans = []
for i in range(n, 0, -1):
if w[i][ind]:
ind -= things[i - 1][1]
ans.append(things[i - 1][3])
print(len(ans))
print(*reversed(ans))
```
| 102,610 | [
0.41015625,
0.1705322265625,
0.1280517578125,
0.30712890625,
-0.373291015625,
-0.369140625,
-0.31201171875,
-0.015594482421875,
0.5302734375,
0.412841796875,
0.86279296875,
-0.09783935546875,
0.348388671875,
-0.5859375,
-0.45068359375,
-0.1407470703125,
-0.5283203125,
-0.3056640625... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Submitted Solution:
```
import copy
n = int(input())
arr = []
m = 0
for er in range(n):
temp = list(map(int,input().split(" ")))
temp.append(er+1)
if(temp[0]<temp[1]):
arr.append(temp)
if(temp[1]>m):
m = temp[1]
else:
n-=1
arr.sort(key = lambda x:x[1])
#print(arr)
temp=[]
for i in range(n):
temp.append([0])
com = []
com.append(temp)
total = [0,0]
for i in range(1,m+1):
temp = []
for j in range(n+1):
if(j==0):
temp.append([0])
else:
p=arr[j-1]
if(p[0]>i or p[1]<=i):
temp.append(temp[j-1])
else:
te = i - p[0]
ty = copy.deepcopy(com[te][j-1])
ty[0]+=p[2]
ty.append(j)
if(total[0]<ty[0]):
total = ty
if(temp[j-1][0]<ty[0]):
temp.append(ty)
else:
temp.append(temp[j-1])
#print(temp , " temp")
com.append(temp)
#print(com)
print(total[0])
if(total[0]>0):
print(len(total)-1)
for i in range(1,len(total)):
print(arr[total[i]-1][3] , end = " ")
else:
print("0")
```
Yes
| 102,611 | [
0.49072265625,
0.25390625,
0.1114501953125,
0.29248046875,
-0.43017578125,
-0.2357177734375,
-0.382568359375,
0.032196044921875,
0.54248046875,
0.402587890625,
0.8125,
-0.09027099609375,
0.2802734375,
-0.58203125,
-0.426513671875,
-0.178955078125,
-0.46044921875,
-0.33349609375,
... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
mod = 10 ** 9 + 7
mod1 = 998244353
#setrecursionlimit(300000)
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
sys.setrecursionlimit(300000)
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=0, func=lambda a, b: max(a, b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
n=int(input())
ma=0
l=[]
time=[]
for i in range(n):
s=list(map(int,input().split()))
ma=max(ma,s[1])
l.append(s+[i+1])
time.append(s[1])
l=sort_list(l,time)
dp=[[[] for i in range(ma+1)]for j in range(n)]
dp1=[[0 for i in range(ma+1)]for j in range(n)]
for i in range(ma+1):
if l[0][0]<i<=l[0][1]:
dp1[0][i]=l[0][2]
dp[0][i]=[l[0][-1]]
for i in range(1,n):
for j in range(ma+1):
if j-l[i][0]>0 and j<=l[i][1]:
dp1[i][j]=max(dp1[i-1][j],dp1[i-1][j-l[i][0]]+l[i][2])
if dp1[i][j]==dp1[i-1][j]:
dp[i][j]=dp[i-1][j]+[]
else:
dp[i][j]=dp[i-1][j-l[i][0]]+[l[i][-1]]
else:
dp1[i][j]=dp1[i-1][j]
dp[i][j]=dp[i-1][j]+[]
ans=max(dp1[-1])
print(ans)
ind=dp1[-1].index(ans)
print(len(dp[-1][ind]))
print(*dp[-1][ind])
```
Yes
| 102,612 | [
0.49072265625,
0.25390625,
0.1114501953125,
0.29248046875,
-0.43017578125,
-0.2357177734375,
-0.382568359375,
0.032196044921875,
0.54248046875,
0.402587890625,
0.8125,
-0.09027099609375,
0.2802734375,
-0.58203125,
-0.426513671875,
-0.178955078125,
-0.46044921875,
-0.33349609375,
... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Submitted Solution:
```
from sys import stdin
n = int(stdin.readline())
items = []
for i in range(n):
t,d,p = [int(x) for x in stdin.readline().split()]
items.append((d,t,p,i))
items.sort()
mem = [{} for x in range(n)]
mem2 = [{} for x in range(n)]
def best(time,x):
time += items[x][1]
if time in mem[x]:
return mem[x][time]
if time >= items[x][0]:
mem[x][time] = 0
mem2[x][time] = -1
return 0
top = 0
top2 = -1
for i in range(x+1,n):
temp = best(time,i)
if temp > top:
top = temp
top2 = i
mem[x][time] = top+items[x][2]
mem2[x][time] = top2
return mem[x][time]
top = -1
l = []
for x in range(n):
b = best(0,x)
if b > top:
top = b
#print('new',x,top)
l = []
c = x
time = 0
while c != -1:
time += items[c][1]
l.append(items[c][3])
if time in mem2[c]:
c = mem2[c][time]
else:
c = -1
print(top)
if top != 0:
print(len(l))
print(' '.join([str(x+1) for x in l]))
else:
print(0)
print()
```
Yes
| 102,613 | [
0.49072265625,
0.25390625,
0.1114501953125,
0.29248046875,
-0.43017578125,
-0.2357177734375,
-0.382568359375,
0.032196044921875,
0.54248046875,
0.402587890625,
0.8125,
-0.09027099609375,
0.2802734375,
-0.58203125,
-0.426513671875,
-0.178955078125,
-0.46044921875,
-0.33349609375,
... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Submitted Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
import io
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
from collections import Counter
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
BUFSIZE = 8192
'''
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
'''
#-------------------game starts now-----------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
prime = [True for i in range(10001)]
prime[0]=prime[1]=False
#pp=[0]*10000
def SieveOfEratosthenes(n=10000):
p = 2
c=0
while (p <= n):
if (prime[p] == True):
c+=1
for i in range(p, n+1, p):
#pp[i]=1
prime[i] = False
p += 1
#-----------------------------------DSU--------------------------------------------------
class DSU:
def __init__(self, R, C):
#R * C is the source, and isn't a grid square
self.par = range(R*C + 1)
self.rnk = [0] * (R*C + 1)
self.sz = [1] * (R*C + 1)
def find(self, x):
if self.par[x] != x:
self.par[x] = self.find(self.par[x])
return self.par[x]
def union(self, x, y):
xr, yr = self.find(x), self.find(y)
if xr == yr: return
if self.rnk[xr] < self.rnk[yr]:
xr, yr = yr, xr
if self.rnk[xr] == self.rnk[yr]:
self.rnk[xr] += 1
self.par[yr] = xr
self.sz[xr] += self.sz[yr]
def size(self, x):
return self.sz[self.find(x)]
def top(self):
# Size of component at ephemeral "source" node at index R*C,
# minus 1 to not count the source itself in the size
return self.size(len(self.sz) - 1) - 1
#---------------------------------Lazy Segment Tree--------------------------------------
# https://github.com/atcoder/ac-library/blob/master/atcoder/lazysegtree.hpp
class LazySegTree:
def __init__(self, _op, _e, _mapping, _composition, _id, v):
def set(p, x):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
_d[p] = x
for i in range(1, _log + 1):
_update(p >> i)
def get(p):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
return _d[p]
def prod(l, r):
assert 0 <= l <= r <= _n
if l == r:
return _e
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push(r >> i)
sml = _e
smr = _e
while l < r:
if l & 1:
sml = _op(sml, _d[l])
l += 1
if r & 1:
r -= 1
smr = _op(_d[r], smr)
l >>= 1
r >>= 1
return _op(sml, smr)
def apply(l, r, f):
assert 0 <= l <= r <= _n
if l == r:
return
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push((r - 1) >> i)
l2 = l
r2 = r
while l < r:
if l & 1:
_all_apply(l, f)
l += 1
if r & 1:
r -= 1
_all_apply(r, f)
l >>= 1
r >>= 1
l = l2
r = r2
for i in range(1, _log + 1):
if ((l >> i) << i) != l:
_update(l >> i)
if ((r >> i) << i) != r:
_update((r - 1) >> i)
def _update(k):
_d[k] = _op(_d[2 * k], _d[2 * k + 1])
def _all_apply(k, f):
_d[k] = _mapping(f, _d[k])
if k < _size:
_lz[k] = _composition(f, _lz[k])
def _push(k):
_all_apply(2 * k, _lz[k])
_all_apply(2 * k + 1, _lz[k])
_lz[k] = _id
_n = len(v)
_log = _n.bit_length()
_size = 1 << _log
_d = [_e] * (2 * _size)
_lz = [_id] * _size
for i in range(_n):
_d[_size + i] = v[i]
for i in range(_size - 1, 0, -1):
_update(i)
self.set = set
self.get = get
self.prod = prod
self.apply = apply
MIL = 1 << 20
def makeNode(total, count):
# Pack a pair into a float
return (total * MIL) + count
def getTotal(node):
return math.floor(node / MIL)
def getCount(node):
return node - getTotal(node) * MIL
nodeIdentity = makeNode(0.0, 0.0)
def nodeOp(node1, node2):
return node1 + node2
# Equivalent to the following:
return makeNode(
getTotal(node1) + getTotal(node2), getCount(node1) + getCount(node2)
)
identityMapping = -1
def mapping(tag, node):
if tag == identityMapping:
return node
# If assigned, new total is the number assigned times count
count = getCount(node)
return makeNode(tag * count, count)
def composition(mapping1, mapping2):
# If assigned multiple times, take first non-identity assignment
return mapping1 if mapping1 != identityMapping else mapping2
#---------------------------------Pollard rho--------------------------------------------
def memodict(f):
"""memoization decorator for a function taking a single argument"""
class memodict(dict):
def __missing__(self, key):
ret = self[key] = f(key)
return ret
return memodict().__getitem__
def pollard_rho(n):
"""returns a random factor of n"""
if n & 1 == 0:
return 2
if n % 3 == 0:
return 3
s = ((n - 1) & (1 - n)).bit_length() - 1
d = n >> s
for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
p = pow(a, d, n)
if p == 1 or p == n - 1 or a % n == 0:
continue
for _ in range(s):
prev = p
p = (p * p) % n
if p == 1:
return math.gcd(prev - 1, n)
if p == n - 1:
break
else:
for i in range(2, n):
x, y = i, (i * i + 1) % n
f = math.gcd(abs(x - y), n)
while f == 1:
x, y = (x * x + 1) % n, (y * y + 1) % n
y = (y * y + 1) % n
f = math.gcd(abs(x - y), n)
if f != n:
return f
return n
@memodict
def prime_factors(n):
"""returns a Counter of the prime factorization of n"""
if n <= 1:
return Counter()
f = pollard_rho(n)
return Counter([n]) if f == n else prime_factors(f) + prime_factors(n // f)
def distinct_factors(n):
"""returns a list of all distinct factors of n"""
factors = [1]
for p, exp in prime_factors(n).items():
factors += [p**i * factor for factor in factors for i in range(1, exp + 1)]
return factors
def all_factors(n):
"""returns a sorted list of all distinct factors of n"""
small, large = [], []
for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
if not n % i:
small.append(i)
large.append(n // i)
if small[-1] == large[-1]:
large.pop()
large.reverse()
small.extend(large)
return small
#---------------------------------Binary Search------------------------------------------
def binarySearch(arr, n,i, key):
left = 0
right = n-1
mid = 0
res=n
while (left <= right):
mid = (right + left)//2
if (arr[mid][i] > key):
res=mid
right = mid-1
else:
left = mid + 1
return res
def binarySearch1(arr, n,i, key):
left = 0
right = n-1
mid = 0
res=-1
while (left <= right):
mid = (right + left)//2
if (arr[mid][i] > key):
right = mid-1
else:
res=mid
left = mid + 1
return res
#---------------------------------running code------------------------------------------
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
import io
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
from collections import Counter
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
prime = [True for i in range(10001)]
prime[0]=prime[1]=False
#pp=[0]*10000
def SieveOfEratosthenes(n=10000):
p = 2
c=0
while (p <= n):
if (prime[p] == True):
c+=1
for i in range(p, n+1, p):
#pp[i]=1
prime[i] = False
p += 1
#-----------------------------------DSU--------------------------------------------------
class DSU:
def __init__(self, R, C):
#R * C is the source, and isn't a grid square
self.par = range(R*C + 1)
self.rnk = [0] * (R*C + 1)
self.sz = [1] * (R*C + 1)
def find(self, x):
if self.par[x] != x:
self.par[x] = self.find(self.par[x])
return self.par[x]
def union(self, x, y):
xr, yr = self.find(x), self.find(y)
if xr == yr: return
if self.rnk[xr] < self.rnk[yr]:
xr, yr = yr, xr
if self.rnk[xr] == self.rnk[yr]:
self.rnk[xr] += 1
self.par[yr] = xr
self.sz[xr] += self.sz[yr]
def size(self, x):
return self.sz[self.find(x)]
def top(self):
# Size of component at ephemeral "source" node at index R*C,
# minus 1 to not count the source itself in the size
return self.size(len(self.sz) - 1) - 1
#---------------------------------Lazy Segment Tree--------------------------------------
# https://github.com/atcoder/ac-library/blob/master/atcoder/lazysegtree.hpp
class LazySegTree:
def __init__(self, _op, _e, _mapping, _composition, _id, v):
def set(p, x):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
_d[p] = x
for i in range(1, _log + 1):
_update(p >> i)
def get(p):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
return _d[p]
def prod(l, r):
assert 0 <= l <= r <= _n
if l == r:
return _e
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push(r >> i)
sml = _e
smr = _e
while l < r:
if l & 1:
sml = _op(sml, _d[l])
l += 1
if r & 1:
r -= 1
smr = _op(_d[r], smr)
l >>= 1
r >>= 1
return _op(sml, smr)
def apply(l, r, f):
assert 0 <= l <= r <= _n
if l == r:
return
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push((r - 1) >> i)
l2 = l
r2 = r
while l < r:
if l & 1:
_all_apply(l, f)
l += 1
if r & 1:
r -= 1
_all_apply(r, f)
l >>= 1
r >>= 1
l = l2
r = r2
for i in range(1, _log + 1):
if ((l >> i) << i) != l:
_update(l >> i)
if ((r >> i) << i) != r:
_update((r - 1) >> i)
def _update(k):
_d[k] = _op(_d[2 * k], _d[2 * k + 1])
def _all_apply(k, f):
_d[k] = _mapping(f, _d[k])
if k < _size:
_lz[k] = _composition(f, _lz[k])
def _push(k):
_all_apply(2 * k, _lz[k])
_all_apply(2 * k + 1, _lz[k])
_lz[k] = _id
_n = len(v)
_log = _n.bit_length()
_size = 1 << _log
_d = [_e] * (2 * _size)
_lz = [_id] * _size
for i in range(_n):
_d[_size + i] = v[i]
for i in range(_size - 1, 0, -1):
_update(i)
self.set = set
self.get = get
self.prod = prod
self.apply = apply
MIL = 1 << 20
def makeNode(total, count):
# Pack a pair into a float
return (total * MIL) + count
def getTotal(node):
return math.floor(node / MIL)
def getCount(node):
return node - getTotal(node) * MIL
nodeIdentity = makeNode(0.0, 0.0)
def nodeOp(node1, node2):
return node1 + node2
# Equivalent to the following:
return makeNode(
getTotal(node1) + getTotal(node2), getCount(node1) + getCount(node2)
)
identityMapping = -1
def mapping(tag, node):
if tag == identityMapping:
return node
# If assigned, new total is the number assigned times count
count = getCount(node)
return makeNode(tag * count, count)
def composition(mapping1, mapping2):
# If assigned multiple times, take first non-identity assignment
return mapping1 if mapping1 != identityMapping else mapping2
#---------------------------------Pollard rho--------------------------------------------
def memodict(f):
"""memoization decorator for a function taking a single argument"""
class memodict(dict):
def __missing__(self, key):
ret = self[key] = f(key)
return ret
return memodict().__getitem__
def pollard_rho(n):
"""returns a random factor of n"""
if n & 1 == 0:
return 2
if n % 3 == 0:
return 3
s = ((n - 1) & (1 - n)).bit_length() - 1
d = n >> s
for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
p = pow(a, d, n)
if p == 1 or p == n - 1 or a % n == 0:
continue
for _ in range(s):
prev = p
p = (p * p) % n
if p == 1:
return math.gcd(prev - 1, n)
if p == n - 1:
break
else:
for i in range(2, n):
x, y = i, (i * i + 1) % n
f = math.gcd(abs(x - y), n)
while f == 1:
x, y = (x * x + 1) % n, (y * y + 1) % n
y = (y * y + 1) % n
f = math.gcd(abs(x - y), n)
if f != n:
return f
return n
@memodict
def prime_factors(n):
"""returns a Counter of the prime factorization of n"""
if n <= 1:
return Counter()
f = pollard_rho(n)
return Counter([n]) if f == n else prime_factors(f) + prime_factors(n // f)
def distinct_factors(n):
"""returns a list of all distinct factors of n"""
factors = [1]
for p, exp in prime_factors(n).items():
factors += [p**i * factor for factor in factors for i in range(1, exp + 1)]
return factors
def all_factors(n):
"""returns a sorted list of all distinct factors of n"""
small, large = [], []
for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
if not n % i:
small.append(i)
large.append(n // i)
if small[-1] == large[-1]:
large.pop()
large.reverse()
small.extend(large)
return small
#---------------------------------Binary Search------------------------------------------
def binarySearch(arr, n,i, key):
left = 0
right = n-1
mid = 0
res=n
while (left <= right):
mid = (right + left)//2
if (arr[mid][i] > key):
res=mid
right = mid-1
else:
left = mid + 1
return res
def binarySearch1(arr, n,i, key):
left = 0
right = n-1
mid = 0
res=-1
while (left <= right):
mid = (right + left)//2
if (arr[mid][i] > key):
right = mid-1
else:
res=mid
left = mid + 1
return res
#---------------------------------running code------------------------------------------
t=1
#t=int(input())
for _ in range (t):
n=int(input())
#n,x=map(int,input().split())
#a=list(map(int,input().split()))
#b=list(map(int,input().split()))
#s=input()
#n=len(s)
a=[]
m=0
for i in range (n):
tm,d,p=map(int,input().split())
a.append([d,tm,p,i+1])
m=max(m,d)
a.sort()
dp=[0]*(m+1)
result=[set()]*(m+1)
for i in range (n):
for j in range (a[i][0]-1,a[i][1]-1,-1):
curr=dp[j-a[i][1]]+a[i][2]
if curr>dp[j]:
dp[j]=curr
result[j]=result[j-a[i][1]].copy()
result[j].add(a[i][3])
#print(dp)
#print(result)
mx=max(dp)
ans=None
for i in range (m+1):
if dp[i]==mx:
ans=result[i]
print(mx)
print(len(ans))
print(*ans)
```
Yes
| 102,614 | [
0.49072265625,
0.25390625,
0.1114501953125,
0.29248046875,
-0.43017578125,
-0.2357177734375,
-0.382568359375,
0.032196044921875,
0.54248046875,
0.402587890625,
0.8125,
-0.09027099609375,
0.2802734375,
-0.58203125,
-0.426513671875,
-0.178955078125,
-0.46044921875,
-0.33349609375,
... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Submitted Solution:
```
import os
import sys
import re
from collections import OrderedDict
if 'PYCHARM' in os.environ:
sys.stdin = open('in', 'r')
n = int(input())
things = []
for i in range(n):
t, d, p = map(int, input().split())
things.append((d, t, p, i + 1))
things.sort()
D = 2001
f = [0] * D
p = [-1] * D
pi = [-1] * D
for thing in things:
for i in reversed(range(D)):
ni = i + thing[1]
nv = f[i] + thing[2]
if ni <= thing[0]:
if f[ni] < nv:
f[ni] = nv
p[ni] = thing[3]
pi[ni] = i
ind = 0
for i in range(0, D):
if f[i] > f[ind]:
ind = i
print(f[ind])
ans = []
while ind:
ans.append(p[ind])
ind = pi[ind]
print(len(ans))
print(*reversed(ans))
```
No
| 102,615 | [
0.49072265625,
0.25390625,
0.1114501953125,
0.29248046875,
-0.43017578125,
-0.2357177734375,
-0.382568359375,
0.032196044921875,
0.54248046875,
0.402587890625,
0.8125,
-0.09027099609375,
0.2802734375,
-0.58203125,
-0.426513671875,
-0.178955078125,
-0.46044921875,
-0.33349609375,
... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import random
"""
created by shhuan at 2017/10/4 21:10
"""
N = int(input())
M = []
for i in range(N):
M.append([int(x) for x in input().split()])
N = len(M)
T = [0] + [x[0] for x in M]
D = [0] + [x[1] for x in M]
P = [0] + [x[2] for x in M]
dmax = max(D)
# dp[t][i]ε³ζΆι΄tε
θ½ε€ζ―ζηεiδΈͺη©εηζ倧η©εδ»·εΌ
dp = [[0 for _ in range(N+1)] for _ in range(dmax)]
track = [[0 for _ in range(N+1)] for _ in range(dmax)]
for t in range(dmax):
for i in range(1, N+1):
ti = t-T[i]
if T[i] <= t < D[i] and ti >= 0:
dp[t][i] = max(dp[t][i-1], dp[ti][i-1] + P[i])
if dp[t][i-1] > dp[ti][i-1]+P[i]:
track[t][i] = (t, i-1, -1)
else:
track[t][i] = (ti, i-1, i)
else:
dp[t][i] = dp[t][i-1]
track[t][i] = (t, i-1, -1)
print(dp[dmax-1][N])
t, i, j = dmax-1, N, -1
res = []
while t > 0 and i > 0:
t, i, j = track[t][i]
if j > 0:
res.append(j)
print(len(res))
print(' '.join([str(x) for x in reversed(res)]))
```
No
| 102,616 | [
0.49072265625,
0.25390625,
0.1114501953125,
0.29248046875,
-0.43017578125,
-0.2357177734375,
-0.382568359375,
0.032196044921875,
0.54248046875,
0.402587890625,
0.8125,
-0.09027099609375,
0.2802734375,
-0.58203125,
-0.426513671875,
-0.178955078125,
-0.46044921875,
-0.33349609375,
... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Submitted Solution:
```
def count_spaces(l, t, x):
# up to ind x.
total = prev = 0; n = len(l)
for i in range(x + 1):
total += l[i][1] - prev
prev = l[i][1] + t[l[i][0]]
return total
def move_blocks(x, dist, l, t):
# blocks up to x will move.
last = -1
gaps = []
for i in range(x, 0, -1):
gap = l[i][1] - (l[i - 1][1] + t[l[i - 1][0]])
gaps.append(gap)
if dist > gap:
dist -= gap
else:
if gap > dist:
gaps[-1] = dist
last = i
dist = 0
break
if dist > 0:
last = 0
gaps.append(dist)
#print(gaps, last, x, "ago")
pref = 0
for j in range(last, x + 1): # == len(gaps), prob
#pref += gaps[j - last] # 0 1 2..
pref += gaps[-(j - last + 1)] # -1 -2 -3..
l[j][1] -= pref
def main():
n = int(input())
t = []; d = []; p = []; v = []
nu = 0
for i in range(n):
ti, di, pi = map(int, input().split())
if ti >= di:
continue
di -= 1
t.append(ti); d.append(di); p.append(pi)
v.append(nu)
nu += 1
v.sort(key = lambda x : [p[x] / d[x], d[x]], reverse = True)
l = []
#print(v)
for idx in v:
#print(l, "hello", idx)
placed = False
for j in range(len(l) - 1, -1, -1):
el = l[j]
#print(el, "jumankjo", d, t, idx)
if d[idx] >= el[1] + t[el[0]]: # back of new after back of old
placed = True
front_diff = el[1] + t[el[0]] - (d[idx] - t[idx]) # overlap btwn old back and new front
if j == len(l) - 1:
back_diff = 0
else:
#back_diff = max(0, d[idx] - (l[j + 1][1] + t[l[j + 1][0]])) # overlap btwn back of new and front of next old
back_diff = max(0, d[idx] - l[j + 1][1])
if abs(front_diff) >= back_diff:
diff = max(0, front_diff) + back_diff
else:
diff = back_diff + front_diff
#print(front_diff, back_diff, diff)
if diff == 0:
#l.append((idx, d[idx] - t[idx]))
l.insert(j + 1, [idx, d[idx] - t[idx]])
else:
spaces = count_spaces(l, t, j)
#print(spaces)
if spaces >= diff:
move_blocks(j, diff, l, t)
#l.insert(j + 1, [idx, d[idx] - t[idx]])
#l.insert(j + 1, [idx, l[j][1] + t[l[j][0]]])
l.insert(j + 1, [idx, l[j][1] + t[l[j][0]] - back_diff])
break
#print("apple", placed)
if not placed:
if len(l) == 0:
l.append([idx, d[idx] - t[idx]])
elif l[0][1] >= t[idx]:
#l.append((idx, l[0][1] - t[idx]))
l.insert(0, [idx, l[0][1] - t[idx]])
o = []; total = 0
#print(l)
for k in range(len(l)):
total += p[l[k][0]]
o.append(l[k][0] + 1)
print(total)
print(len(o))
print(*o)
main()
'''
try:
main()
except Exception as e:
print(e)
'''
```
No
| 102,617 | [
0.49072265625,
0.25390625,
0.1114501953125,
0.29248046875,
-0.43017578125,
-0.2357177734375,
-0.382568359375,
0.032196044921875,
0.54248046875,
0.402587890625,
0.8125,
-0.09027099609375,
0.2802734375,
-0.58203125,
-0.426513671875,
-0.178955078125,
-0.46044921875,
-0.33349609375,
... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is in really serious trouble β his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di β the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti β₯ di, then i-th item cannot be saved.
Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b β in ta + tb seconds after fire started.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of items in Polycarp's house.
Each of the following n lines contains three integers ti, di, pi (1 β€ ti β€ 20, 1 β€ di β€ 2 000, 1 β€ pi β€ 20) β the time needed to save the item i, the time after which the item i will burn completely and the value of item i.
Output
In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m β the number of items in the desired set. In the third line print m distinct integers β numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.
Examples
Input
3
3 7 4
2 6 5
3 7 6
Output
11
2
2 3
Input
2
5 6 1
3 3 5
Output
1
1
1
Note
In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.
In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.
Submitted Solution:
```
n = int(input())
t, d, p = [], [], []
for _ in range(n):
a, b, c = map(int, input().split())
t.append(a); d.append(b); p.append(c)
dp = [[0, -1] for i in range(10)]
for i in range(1, 10):
# dp[i] = dp[i - 1].copy()
for j in range(n):
if d[j] > i >= t[j] and dp[i][0] < p[j] + dp[i - t[j]][0]:
dp[i][0] = p[j] + dp[i - t[j]][0]
dp[i][1] = j
print(max(dp)[0])
i = dp.index(max(dp))
ans = []
while i > 0 and dp[i][1] != -1:
ans.append(dp[i][1] + 1)
i = dp[i][1]
print(len(ans))
print(*ans[::-1])
```
No
| 102,618 | [
0.49072265625,
0.25390625,
0.1114501953125,
0.29248046875,
-0.43017578125,
-0.2357177734375,
-0.382568359375,
0.032196044921875,
0.54248046875,
0.402587890625,
0.8125,
-0.09027099609375,
0.2802734375,
-0.58203125,
-0.426513671875,
-0.178955078125,
-0.46044921875,
-0.33349609375,
... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Tags: dp
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
from collections import deque, defaultdict, namedtuple
from math import sqrt, factorial, gcd, ceil, atan, pi
def input(): return sys.stdin.readline()[:-1] # warning not \n
# def input(): return sys.stdin.buffer.readline().strip() # warning bytes
# def input(): return sys.stdin.buffer.readline().decode('utf-8')
import string
import operator
# string.ascii_lowercase
from bisect import bisect_left, bisect_right
from functools import lru_cache, reduce
MOD = int(1e9)+7
INF = float('inf')
def solve():
a, n, m = [int(x) for x in input().split()]
rain = [0] * (a + 1)
for _ in range(n):
l, r = [int(x) for x in input().split()]
for i in range(l + 1, r + 1):
rain[i] = 1
umb = []
for _ in range(m):
x, p = [int(x) for x in input().split()]
umb.append((x, p))
umb.sort()
dp = [INF for _ in range(a + 1)]
dp[0] = 0
for i in range(1, a + 1):
if rain[i]:
for x, p in umb:
if x >= i: break
dp[i] = min(dp[i], dp[x] + p * (i - x))
else:
dp[i] = dp[i-1]
if dp[a] == INF:
print(-1)
else:
print(dp[a])
t = 1
# t = int(input())
for case in range(1,t+1):
ans = solve()
"""
1 2
dp[x] = min()
"""
```
| 102,673 | [
0.314697265625,
0.44482421875,
0.461181640625,
0.1304931640625,
-0.41064453125,
-0.3134765625,
-0.3310546875,
0.0159454345703125,
0.420166015625,
0.73486328125,
0.75146484375,
0.0311431884765625,
0.0242156982421875,
-0.7080078125,
-0.147705078125,
0.1475830078125,
-0.7705078125,
-0... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Tags: dp
Correct Solution:
```
# Author: S Mahesh Raju
# Username: maheshraju2020
# Date: 03/07/2020
from sys import stdin,stdout
from math import gcd, ceil, sqrt
from collections import Counter
ii1 = lambda: int(stdin.readline().strip())
is1 = lambda: stdin.readline().strip()
iia = lambda: list(map(int, stdin.readline().strip().split()))
isa = lambda: stdin.readline().strip().split()
mod = 1000000007
a, n, m = iia()
rain = []
for _ in range(n):
l, r = iia()
for i in range(l, r):
rain.append(i)
umb = []
for _ in range(m):
umb.append(iia())
rain.sort()
umb.sort()
dp = [0] * (a + 1)
for i in range(a + 1):
if i not in rain:
if i != 0:
dp[i] = dp[i - 1]
else:
for j in umb:
if j[0] <= i:
temp = (i + 1 - j[0]) * j[1]
if j[0] - 1 >= 0:
temp += dp[j[0] - 1]
if dp[i] > 0:
dp[i] = min(dp[i], temp)
else:
dp[i] = temp
else:
break
# print(dp)
if umb[0][0] > rain[0]:
print(-1)
else:
print(dp[-1])
```
| 102,674 | [
0.314697265625,
0.44482421875,
0.461181640625,
0.1304931640625,
-0.41064453125,
-0.3134765625,
-0.3310546875,
0.0159454345703125,
0.420166015625,
0.73486328125,
0.75146484375,
0.0311431884765625,
0.0242156982421875,
-0.7080078125,
-0.147705078125,
0.1475830078125,
-0.7705078125,
-0... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Tags: dp
Correct Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop,heapify
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
from itertools import accumulate
from functools import lru_cache
M = mod = 998244353
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n')]
def li3():return [int(i) for i in input().rstrip('\n')]
sys.setrecursionlimit(10 ** 6)
a, n, m = li()
l = []
rain = defaultdict(int)
for i in range(n):
c, b = li()
for j in range(c, b):
rain[(j, j + 1)] = 1
# print(rain)
umbrellas = [float('inf')] * (a + 5)
for i in range(m):
c, b = li()
umbrellas[c] = min(umbrellas[c], b)
# print(umbrellas[:a + 1])
@lru_cache(None)
def dp(i = 0, umbon = 0):
# print(i, umbon)
if i == a:
if rain[(i - 1, i)]:
if umbon:return umbon
return float('inf')
else:return umbon
else:
ans = float('inf')
last = umbon
if rain[(i - 1, i)]:
umbon = min(umbon, umbrellas[i]) if umbon else umbrellas[i]
if not last:last = float('inf')
ans = min(ans, last + dp(i + 1, umbon))
ans = min(ans, last + dp(i + 1, 0))
else:
ans = min(ans, dp(i + 1, 0) + last)
umbon = min(umbon, umbrellas[i]) if umbon else umbrellas[i]
ans = min(ans, dp(i + 1, umbon) + last)
return ans
print(dp(0, 0) if dp(0, 0) != float('inf') else -1)
```
| 102,675 | [
0.314697265625,
0.44482421875,
0.461181640625,
0.1304931640625,
-0.41064453125,
-0.3134765625,
-0.3310546875,
0.0159454345703125,
0.420166015625,
0.73486328125,
0.75146484375,
0.0311431884765625,
0.0242156982421875,
-0.7080078125,
-0.147705078125,
0.1475830078125,
-0.7705078125,
-0... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Tags: dp
Correct Solution:
```
import io
import os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
INF = 10**15
def dp_min(dp, pos, new_val):
if pos not in dp:
dp[pos] = new_val
else:
dp[pos] = min(dp[pos], new_val)
def solve():
a, n, m = list(map(int, input().split()))
has_rain = [False] * a
for _ in range(n):
l, r = map(int, input().split())
for i in range(l, r):
has_rain[i] = True
umbrella = [INF] * a
for _ in range(m):
x, p = map(int, input().split())
if x == a:
continue
umbrella[x] = min(umbrella[x], p)
# dp: best cost so far if I am currently carrying no umbrella (-1), or some umbrella of certain weight
dp = {-1: 0}
for i in range(a):
new_dp = {}
for umbrella_weight, best_cost in dp.items():
if not has_rain[i]:
# carry no umbrella
dp_min(new_dp, -1, best_cost)
# take the new umbrella
dp_min(new_dp, umbrella[i], best_cost + umbrella[i])
if umbrella_weight != -1:
# continue same umbrella
dp_min(new_dp, umbrella_weight, best_cost + umbrella_weight)
dp = new_dp
best = min(dp.values())
if best >= INF:
print(-1)
else:
print(best)
t = 1
for _ in range(t):
solve()
```
| 102,676 | [
0.314697265625,
0.44482421875,
0.461181640625,
0.1304931640625,
-0.41064453125,
-0.3134765625,
-0.3310546875,
0.0159454345703125,
0.420166015625,
0.73486328125,
0.75146484375,
0.0311431884765625,
0.0242156982421875,
-0.7080078125,
-0.147705078125,
0.1475830078125,
-0.7705078125,
-0... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Tags: dp
Correct Solution:
```
import sys
a, n, m = map(int, input().split(' '))
seg = []
for i in range(n):
rained = tuple(map(int, input().split(' ')))
for k in range(rained[0], rained[1]):
seg.append(k+1)
umbrella = []
for j in range(m):
u = tuple(map(int, input().split(' ')))
umbrella.append(u)
memo = [0] * (a+1)
umbrella = sorted(umbrella, key=lambda x: x[0])
if umbrella[0][0] > seg[0] - 1:
print(-1)
sys.exit(0)
for index in range(1, len(memo)):
if index not in seg:
memo[index] = memo[index-1]
continue
for each in umbrella:
if index >= each[0]:
cur = (index - each[0]) * each[1] + memo[each[0]]
if memo[index] > 0:
if cur < memo[index]:
memo[index] = cur
else:
memo[index] = cur
print(memo[-1])
```
| 102,677 | [
0.314697265625,
0.44482421875,
0.461181640625,
0.1304931640625,
-0.41064453125,
-0.3134765625,
-0.3310546875,
0.0159454345703125,
0.420166015625,
0.73486328125,
0.75146484375,
0.0311431884765625,
0.0242156982421875,
-0.7080078125,
-0.147705078125,
0.1475830078125,
-0.7705078125,
-0... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Tags: dp
Correct Solution:
```
# Codeforces Round #486 (Div. 3)
from functools import cmp_to_key
#key=cmp_to_key(lambda x,y: 1 if x not in y else -1 )
import sys
def getIntList():
return list(map(int, input().split()))
a,n,m = getIntList()
rainend = set()
umbr = {}
keyPointSet = set([0,a])
for i in range(n):
t =tuple(getIntList())
for j in range(t[0] + 1, t[1] + 1):
rainend.add(j)
keyPointSet.add(t[1])
for i in range(m):
t =getIntList()
if t[0] not in umbr or t[1]< umbr[t[0]]:
umbr[t[0]] = t[1]
keyPointSet.add(t[0])
keyPoint = list(keyPointSet)
keyPoint.sort()
dp = {}
dp[0] = {}
dp[0] [0] = 0
if 0 in umbr:
dp[0][umbr[0]] = 0
for i in range(1, len(keyPoint)):
x = keyPoint[i]
lx = keyPoint[i-1]
ifrain = x in rainend
dp[x] = {}
nowdp = dp[x]
lastdp = dp[lx]
for z in lastdp:
if z == 0 :
if not ifrain:
nowdp[0] = lastdp[0]
else:
nowdp[z] = lastdp[z] + z * (x-lx)
if len(nowdp) >0:
nowdp[0] = min(nowdp.values())
if x in umbr:
if umbr[x] not in nowdp or nowdp[0] < nowdp[umbr[x]]:
nowdp[umbr[x]] = nowdp[0]
else:
print(-1)
sys.exit()
print( min(dp[a].values()) )
```
| 102,678 | [
0.314697265625,
0.44482421875,
0.461181640625,
0.1304931640625,
-0.41064453125,
-0.3134765625,
-0.3310546875,
0.0159454345703125,
0.420166015625,
0.73486328125,
0.75146484375,
0.0311431884765625,
0.0242156982421875,
-0.7080078125,
-0.147705078125,
0.1475830078125,
-0.7705078125,
-0... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Tags: dp
Correct Solution:
```
import sys
a,m,n=list(map(int,input().split()))
aux=[0]*(a+1)
inf=10**15
dp=[aux.copy() for i in range(n+1)]
m1=10**12
m2=10**12
for i in range(m):
l,r=list(map(int,input().split()))
if l<m1:
m1=l
for j in range(l,r):
dp[0][j+1]=inf
s=[]
for i in range(1,n+1):
x,w=list(map(int,input().split()))
s.append(tuple([x,w]))
if x<m2:
m2=x
if m2>m1:
print(-1)
sys.exit()
s.sort()
for i in range(1,n+1):
x=s[i-1][0]
w=s[i-1][1]
for j in range(x+1):
dp[i][j]=dp[i-1][j]
for j in range(x+1,a+1):
if i!=1:
dp[i][j]=min(dp[0][j]+dp[i][j-1],dp[i-1][j],w*(j-x)+dp[i][x])
else:
dp[i][j]=min(dp[0][j]+dp[i][j-1],w*(j-x)+dp[i][x])
ans=dp[-1][-1]
if ans>=inf:
print(-1)
else:
print(ans)
```
| 102,679 | [
0.314697265625,
0.44482421875,
0.461181640625,
0.1304931640625,
-0.41064453125,
-0.3134765625,
-0.3310546875,
0.0159454345703125,
0.420166015625,
0.73486328125,
0.75146484375,
0.0311431884765625,
0.0242156982421875,
-0.7080078125,
-0.147705078125,
0.1475830078125,
-0.7705078125,
-0... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Tags: dp
Correct Solution:
```
import sys
def input():
return sys.stdin.readline().strip()
def input_l():
return map(int, input().split())
def input_t():
return tuple(input_l())
def main():
a, s, d = input_l()
q = []
e = []
z = [0] * (a + 1)
for i in range(s):
w = input_t()
for k in range(w[0], w[1]):
q.append(k + 1)
for j in range(d):
e.append(input_t())
e = sorted(e, key = lambda x: x[0])
if e[0][0] > q[0] - 1:
print(-1)
sys.exit(0)
for i in range(1, len(z)):
if i not in q:
z[i] = z[i-1]
continue
for j in e:
if i >= j[0]:
c = (i - j[0]) * j[1] + z[j[0]]
if z[i] > 0:
if c < z[i]:
z[i] = c
else:
z[i] = c
print(z[-1])
main()
```
| 102,680 | [
0.314697265625,
0.44482421875,
0.461181640625,
0.1304931640625,
-0.41064453125,
-0.3134765625,
-0.3310546875,
0.0159454345703125,
0.420166015625,
0.73486328125,
0.75146484375,
0.0311431884765625,
0.0242156982421875,
-0.7080078125,
-0.147705078125,
0.1475830078125,
-0.7705078125,
-0... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
import math
def main():
a, n, m = map(int, input().split())
rain = []
u = []
w = []
raining = [False] * (a+1)
for i in range(n):
l, r = map(int, input().split())
rain.append((l, r))
for j in range(l, r):
raining[j] = True
for i in range(m):
x, y = map(int, input().split())
u.append(x)
w.append(y)
rain_int = [0] * a
for i in range(n):
rain_int[rain[i][0]-1] = 1
rain_int[rain[i][1]-1] = -1
umbrellas = [-1 for _ in range(a+1)]
for i, x in enumerate(u):
if umbrellas[x] == -1 or w[umbrellas[x]] > w[i]:
umbrellas[x] = i
dp = [[math.inf for _ in range(m+1)] for _ in range(a+1)]
dp[0][m] = 0
for i in range(a):
for j in range(m+1):
if dp[i][j] == math.inf:
continue
if not raining[i]:
dp[i+1][m] = min(dp[i+1][m], dp[i][j])
if j < m:
dp[i+1][j] = min(dp[i+1][j], dp[i][j] + w[j])
if umbrellas[i] != -1:
dp[i+1][umbrellas[i]] = min(dp[i+1][umbrellas[i]], dp[i][j]+w[umbrellas[i]])
ans = math.inf
for i in range(m+1):
ans = min(ans, dp[a][i])
print(-1 if ans == math.inf else ans)
if __name__ == '__main__':
main()
```
Yes
| 102,681 | [
0.404296875,
0.43896484375,
0.374267578125,
0.08331298828125,
-0.388427734375,
-0.1744384765625,
-0.33251953125,
0.0889892578125,
0.405029296875,
0.7373046875,
0.72119140625,
0.10003662109375,
0.0450439453125,
-0.69287109375,
-0.185546875,
0.0791015625,
-0.71533203125,
-0.619140625... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
#!/usr/bin/env pypy
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
MOD = 10**9 + 7
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
a, s, d = map(int, input().split())
q = []
e = []
z = [0] * (a + 1)
for i in range(s):
w = tuple(map(int, input().split()))
for k in range(w[0], w[1]):
q.append(k + 1)
for j in range(d):
e.append(tuple(map(int, input().split())))
e = sorted(e, key = lambda x: x[0])
# ~ print(e)
if e[0][0] > q[0] - 1:
print(-1)
exit()
for i in range(1, a +1):
if i not in q:
z[i] = z[i-1]
continue
for j in e:
if i >= j[0]:
c = (i - j[0]) * j[1] + z[j[0]]
if z[i] > 0:
if c < z[i]:
z[i] = c
else:
z[i] = c
print(z[-1])
```
Yes
| 102,682 | [
0.404296875,
0.43896484375,
0.374267578125,
0.08331298828125,
-0.388427734375,
-0.1744384765625,
-0.33251953125,
0.0889892578125,
0.405029296875,
0.7373046875,
0.72119140625,
0.10003662109375,
0.0450439453125,
-0.69287109375,
-0.185546875,
0.0791015625,
-0.71533203125,
-0.619140625... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
rd = lambda: map(int, input().split())
a, n, m = rd()
s = set()
u = {}
k = set([0, a])
for _ in range(n):
l, r = rd()
for x in range(l + 1, r + 1):
s.add(x)
k.add(r)
for _ in range(m):
x, p = rd()
u[x] = min(p, u.get(x, 1e9))
k.add(x)
k = sorted(list(k))
dp = {}
dp[0] = {}
dp[0][0] = 0
if 0 in u:
dp[0][u[0]] = 0
for i in range(1, len(k)):
x = k[i]
y = k[i - 1]
dp[x] = {}
for z in dp[y]:
if z:
dp[x][z] = dp[y][z] + z * (x - y)
else:
if x not in s:
dp[x][0] = dp[y][0]
if len(dp[x]):
dp[x][0] = min(dp[x].values())
if x in u:
if u[x] not in dp[x] or dp[x][0] < dp[x][u[x]]:
dp[x][u[x]] = dp[x][0]
else:
print(-1)
exit()
print(dp[a][0])
```
Yes
| 102,683 | [
0.404296875,
0.43896484375,
0.374267578125,
0.08331298828125,
-0.388427734375,
-0.1744384765625,
-0.33251953125,
0.0889892578125,
0.405029296875,
0.7373046875,
0.72119140625,
0.10003662109375,
0.0450439453125,
-0.69287109375,
-0.185546875,
0.0791015625,
-0.71533203125,
-0.619140625... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
a,m,n=list(map(int,input().split()))
aux=[0]*(a+1)
inf=10**15
dp=[aux.copy() for i in range(n+1)]
for i in range(m):
l,r=list(map(int,input().split()))
for j in range(l,r):
dp[0][j+1]=inf
s=[]
for i in range(1,n+1):
x,w=list(map(int,input().split()))
s.append(tuple([x,w]))
s.sort()
for i in range(1,n+1):
x=s[i-1][0]
w=s[i-1][1]
for j in range(x+1):
dp[i][j]=dp[i-1][j]
for j in range(x+1,a+1):
if i!=1:
dp[i][j]=min(dp[0][j]+dp[i][j-1],dp[i-1][j],w*(j-x)+dp[i][x])
else:
dp[i][j]=min(dp[0][j]+dp[i][j-1],w*(j-x)+dp[i][x])
ans=dp[-1][-1]
if ans>=inf:
print(-1)
else:
print(ans)
```
No
| 102,684 | [
0.404296875,
0.43896484375,
0.374267578125,
0.08331298828125,
-0.388427734375,
-0.1744384765625,
-0.33251953125,
0.0889892578125,
0.405029296875,
0.7373046875,
0.72119140625,
0.10003662109375,
0.0450439453125,
-0.69287109375,
-0.185546875,
0.0791015625,
-0.71533203125,
-0.619140625... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
#!python3
# http://codeforces.com/contest/988/problem/F
from bisect import bisect_left as bl
a, n, m = [int(i) for i in input().strip().split(' ')]
R = []
for i in range(n):
l, r = [int(j) for j in input().strip().split(' ')]
inx = bl(R, l)
R.insert(inx, l)
R.insert(inx+1, r)
A = []
for i in range(m):
A += [tuple(int(j) for j in input().strip().split(' '))]
A = sorted(A)
class S: # Solution
def __init__(self, a, t):
self.t = t # tiredness
self.a = a # ambrella
def __repr__(self):
return "({} {})".format(self.a, self.t)
O = [S(0,0)] # no ambrellas, not tired
x = 0 # starts at 0
x_prev = 0
is_rain = False
optimal = None
while x <= a:
# check if rain:
rain_inx = bl(R, x)
if rain_inx < len(R):
rain_coord = R[rain_inx]
if x == rain_coord:
is_rain = not is_rain
# update tiredness in solutions:
distance = x - x_prev
optimal = 9*999
for o in O:
amb = o.a
if amb > 0:
p = A[amb-1][1]
o.t += p*distance
if o.t < optimal:
optimal = o.t
# pick up new ambrellas:
k = bl(A, (x, 0)) # inx of next ambrellas or one at x
if k < len(A): # have some here or next
while k < len(A) and A[k][0] == x: # every ambrella at x
amb = k + 1
O += [S(amb, optimal)]
k += 1
assert k >= len(A) or A[k][0] != x, "k should be next amb or none"
# drop solution without ambrella if rains:
index_list = [i for i, el in enumerate(O) if el.a == 0]
s_inx = index_list[0] if len(index_list) > 0 else None
if is_rain and s_inx is not None: # can not go without ambrella
assert O[s_inx].a == 0
del O[s_inx]
elif not is_rain and s_inx is None: # can go without one
O.append(S(0, optimal))
if not any(O) and x != a:
optimal = -1
break
# get next x:
next_x = a
if rain_inx < len(R):
if x < R[rain_inx]:
next_x = R[rain_inx]
elif rain_inx + 1 < len(R):
next_x = R[rain_inx+1]
if k < len(A) and A[k][0] < next_x:
next_x = A[k][0]
if x == next_x:
break
x_prev = x
x = next_x
print(optimal)
```
No
| 102,685 | [
0.404296875,
0.43896484375,
0.374267578125,
0.08331298828125,
-0.388427734375,
-0.1744384765625,
-0.33251953125,
0.0889892578125,
0.405029296875,
0.7373046875,
0.72119140625,
0.10003662109375,
0.0450439453125,
-0.69287109375,
-0.185546875,
0.0791015625,
-0.71533203125,
-0.619140625... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
def main():
a, n, m = map(int, input().split())
rain = [False] * (a + 1)
for _ in range(n):
x, y = map(int, input().split())
for i in range(x, y + 1):
rain[i] = True
umrella = [0] * (a + 1)
for _ in range(m):
x, y = map(int, input().split())
if umrella[x] == 0:
umrella[x] = y
else:
umrella[x] = min(umrella[x], y)
max_value = 10 ** 18
dp = [max_value] * (a + 1)
ok = True
if not rain[0] or umrella[0] > 0:
dp[0] = 0
else:
ok = False
if ok:
for i in range(1, a + 1):
if not rain[i]:
dp[i] = dp[i - 1]
else:
for j in range(0, i + 1):
if umrella[j] > 0:
if j >= 1:
dp[i] = min(dp[i], dp[j - 1] + umrella[j] * (i - j))
else:
dp[i] = min(dp[i], umrella[j] * (i - j))
if dp[a] == max_value:
print(-1)
else:
print(dp[a])
else:
print(-1)
if __name__ == "__main__":
main()
```
No
| 102,686 | [
0.404296875,
0.43896484375,
0.374267578125,
0.08331298828125,
-0.388427734375,
-0.1744384765625,
-0.33251953125,
0.0889892578125,
0.405029296875,
0.7373046875,
0.72119140625,
0.10003662109375,
0.0450439453125,
-0.69287109375,
-0.185546875,
0.0791015625,
-0.71533203125,
-0.619140625... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
s=input()[::-1]
m=I=41
f=s.find('5')+1
i=s.find('0')+1
t=len(s)
if i:
j=min(s.find('0',i)+1or I,f or I)
if j<I:
m=i+j-3
if j<i:m+=1
if f:
j=min(s.find('2')+1or I,s.find('7')+1or I)
if j<I:
l=f+j-3
if j<f:l+=1
if f==t:
i=t-1
while i==j or s[i-1]=='0':
if i!=j:l+=1
i-=1
m=min(m,l)
print((-1,m)[m<I])
```
No
| 102,687 | [
0.404296875,
0.43896484375,
0.374267578125,
0.08331298828125,
-0.388427734375,
-0.1744384765625,
-0.33251953125,
0.0889892578125,
0.405029296875,
0.7373046875,
0.72119140625,
0.10003662109375,
0.0450439453125,
-0.69287109375,
-0.185546875,
0.0791015625,
-0.71533203125,
-0.619140625... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
cycle = False
def DFS(s):
global cycle
visited[s] = 1
stack = [s]
while len(stack) != 0:
u = stack[-1]
visited[u] = 1
if len(graph[u]) != 0:
v = graph[u].pop()
if visited[v] == 1:
cycle = True
return
if visited[v] == 2:
continue
stack.append(v)
else:
result.append(u)
stack.pop()
visited[u] = 2
def TopoSort(graph, result, visited):
for i in range (m):
if not visited[requiredCourse[i]]:
DFS(requiredCourse[i])
return result
n, m = map(int, input().split())
requiredCourse = list(map(int, input().split()))
visited = [0 for i in range (n + 1)]
graph = [[] for i in range (n + 1)]
result = []
for i in range (1, n + 1):
tmp = list(map(int, input().split()))
if tmp[0] == 0:
continue
for j in range (1, tmp[0] + 1):
graph[i].append(tmp[j])
res = TopoSort(graph, result, visited)
if cycle:
print(-1)
else:
print(len(res))
for i in res:
print(i, end = " ")
```
| 103,362 | [
0.5400390625,
0.05194091796875,
0.038330078125,
0.1500244140625,
-0.28369140625,
0.1053466796875,
-0.0635986328125,
0.1451416015625,
0.059326171875,
0.92822265625,
1.0390625,
-0.0243072509765625,
0.60888671875,
-0.80419921875,
-0.5087890625,
0.222900390625,
-0.73681640625,
-1.15429... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
# https://codeforces.com/problemset/problem/770/C
n, k = map(int, input().split())
K = set(list(map(int, input().split())))
g = {}
rg = {}
deg = {}
def push_d(deg, u, val):
if u not in deg:
deg[u] = 0
deg[u] += val
def push_g(g, u, v):
if u not in g:
g[u] = []
g[u].append(v)
for u in range(1, n+1):
list_v = list(map(int, input().split()))[1:]
deg[u] = 0
for v in list_v:
push_d(deg, u, 1)
push_g(g, v, u)
push_g(rg, u, v)
S = [x for x in K]
used = [0] * (n+1)
i = 0
while i<len(S):
u = S[i]
if u in rg:
for v in rg[u]:
if used[v] == 0:
used[v] = 1
S.append(v)
i+=1
S = {x:1 for x in S}
deg0 = [x for x in S if deg[x]==0]
ans = []
def process(g, deg, deg0, u):
if u in g:
for v in g[u]:
if v in S:
push_d(deg, v, -1)
if deg[v] == 0:
deg0.append(v)
while len(deg0) > 0 and len(K) > 0:
u = deg0.pop()
ans.append(u)
if u in K:
K.remove(u)
process(g, deg, deg0, u)
if len(K) > 0:
print(-1)
else:
print(len(ans))
print(' '.join([str(x) for x in ans]))
#6 2
#5 6
#0
#1 1
#1 4 5
#2 2 1
#1 4
#2 5 3
```
| 103,363 | [
0.5400390625,
0.05194091796875,
0.038330078125,
0.1500244140625,
-0.28369140625,
0.1053466796875,
-0.0635986328125,
0.1451416015625,
0.059326171875,
0.92822265625,
1.0390625,
-0.0243072509765625,
0.60888671875,
-0.80419921875,
-0.5087890625,
0.222900390625,
-0.73681640625,
-1.15429... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
def dfs(start_node, edges, colors, result):
stack = [start_node]
while stack:
current_node = stack[-1]
if colors[current_node] == 2:
stack.pop()
continue
colors[current_node] = 1
children = edges[current_node]
if not children:
colors[current_node] = 2
result.append(stack.pop())
else:
child = children.pop()
if colors[child] == 1:
return False
stack.append(child)
return True
def find_courses_sequence(member_of_node, find_nodes, edges):
colors = [0] * member_of_node
result = []
for node in find_nodes:
if not dfs(node, edges, colors, result):
return []
return result
if __name__ == '__main__':
n, k = map(int, input().split())
main_courses = [int(c)-1 for c in input().split()]
courses = dict()
for index in range(n):
courses[index] = [int(d)-1 for d in input().split()[1:]]
result = find_courses_sequence(n, main_courses, courses)
if result:
print(len(result))
for v in result:
print(v+1, end=" ")
else:
print(-1)
```
| 103,364 | [
0.5400390625,
0.05194091796875,
0.038330078125,
0.1500244140625,
-0.28369140625,
0.1053466796875,
-0.0635986328125,
0.1451416015625,
0.059326171875,
0.92822265625,
1.0390625,
-0.0243072509765625,
0.60888671875,
-0.80419921875,
-0.5087890625,
0.222900390625,
-0.73681640625,
-1.15429... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
n,k=list(map(lambda x: int(x), input().split()))
m=list(map(lambda x: int(x), input().split()))
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
class Graph:
def __init__(self, V):
self.V = V
self.adj = [[] for i in range(V)]
@bootstrap
def DFSUtil(self, temp, v, visited):
visited[v] = True
for i in self.adj[v]:
if visited[i] == False:
yield self.DFSUtil(temp, i, visited)
temp.append(v)
yield temp
def addEdge(self, v, w):
self.adj[v].append(w)
# self.adj[w].append(v)
@bootstrap
def isCyclicUtil(self, v, visited, recStack):
# Mark current node as visited and
# adds to recursion stack
visited[v] = True
recStack[v] = True
# Recur for all neighbours
# if any neighbour is visited and in
# recStack then graph is cyclic
for neighbour in self.adj[v]:
if visited[neighbour] == False:
ans =yield self.isCyclicUtil(neighbour, visited, recStack)
if ans == True:
yield True
elif recStack[neighbour] == True:
yield True
# The node needs to be poped from
# recursion stack before function ends
recStack[v] = False
yield False
# Returns true if graph is cyclic else false
def isCyclic(self,nodes):
visited = [False] * self.V
recStack = [False] * self.V
for node in nodes:
if visited[node] == False:
if self.isCyclicUtil(node, visited, recStack) == True:
return True
return False
G=Graph(n)
for i in range(0,n):
x=list(map(lambda x: int(x), input().split()))
if x[0]==0:
continue
else:
for k in range(1,x[0]+1):
G.addEdge(i,x[k]-1)
visited=[False for _ in range(n)]
path=[]
# print(G.adj)
for subj in m:
temp = []
if visited[subj-1]==False:
G.DFSUtil(temp,subj-1,visited)
path.extend(temp)
if G.isCyclic([x-1 for x in m]):
print(-1)
else:
print(len(path))
for p in path:
print(p+1,end=" ")
print()
```
| 103,365 | [
0.5400390625,
0.05194091796875,
0.038330078125,
0.1500244140625,
-0.28369140625,
0.1053466796875,
-0.0635986328125,
0.1451416015625,
0.059326171875,
0.92822265625,
1.0390625,
-0.0243072509765625,
0.60888671875,
-0.80419921875,
-0.5087890625,
0.222900390625,
-0.73681640625,
-1.15429... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
import sys
def main():
n,k = map(int,sys.stdin.readline().split())
courses = list(map(int,sys.stdin.readline().split()))
courses = [x-1 for x in courses]
visited = [False]*n
used = [False]*n
ans = []
t = []
for i in range(n):
temp = list(map(int,sys.stdin.readline().split()))
temp = [x-1 for x in temp]
t.append(temp[1:])
for i in range(k):
c = courses[i]
if used[c]:
continue
q = [c]
visited[c]=True
while len(q)>0:
cur = q[-1]
if len(t[cur])!=0:
s = t[cur].pop()
if visited[s] and not used[s]:
print(-1)
return
if used[s]:
continue
q.append(s)
visited[s]=True
else:
ans.append(cur)
q.pop()
used[cur] = True
ans = [str(x+1) for x in ans]
print(len(ans))
print(" ".join(ans))
main()
```
| 103,366 | [
0.5400390625,
0.05194091796875,
0.038330078125,
0.1500244140625,
-0.28369140625,
0.1053466796875,
-0.0635986328125,
0.1451416015625,
0.059326171875,
0.92822265625,
1.0390625,
-0.0243072509765625,
0.60888671875,
-0.80419921875,
-0.5087890625,
0.222900390625,
-0.73681640625,
-1.15429... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
import collections as col
import itertools as its
import sys
import operator
from copy import copy, deepcopy
class Solver:
def __init__(self):
pass
def solve(self):
n, k = map(int, input().split())
q = list(map(lambda x: int(x) - 1, input().split()))
used = [False] * n
for e in q:
used[e] = True
edges = [[] for _ in range(n)]
redges = [[] for _ in range(n)]
for i in range(n):
l = list(map(lambda x: int(x) - 1, input().split()))[1:]
edges[i] = l
for e in l:
redges[e].append(i)
degs = [len(edges[i]) for i in range(n)]
d = 0
while d < len(q):
v = q[d]
d += 1
for e in edges[v]:
if not used[e]:
used[e] = True
q.append(e)
q = q[::-1]
nq = []
for v in q:
if degs[v] == 0:
nq.append(v)
d = 0
while d < len(nq):
v = nq[d]
d += 1
for e in redges[v]:
if not used[e]:
continue
degs[e] -= 1
if degs[e] == 0:
nq.append(e)
#print(nq)
if len(q) != len(nq):
print(-1)
return
print(len(nq))
print(' '.join(map(lambda x: str(x + 1), nq)))
if __name__ == '__main__':
s = Solver()
s.solve()
```
| 103,367 | [
0.5400390625,
0.05194091796875,
0.038330078125,
0.1500244140625,
-0.28369140625,
0.1053466796875,
-0.0635986328125,
0.1451416015625,
0.059326171875,
0.92822265625,
1.0390625,
-0.0243072509765625,
0.60888671875,
-0.80419921875,
-0.5087890625,
0.222900390625,
-0.73681640625,
-1.15429... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
f = lambda: map(int, input().split())
g = lambda: [int(q) - 1 for q in f()]
class T:
def __init__(s, i):
s.i, s.t = i, g()[1:]
s.a = s.q = 0
n, k = f()
d = g()
p = [T(i) for i in range(n)]
s = []
while d:
x = p[d.pop()]
if x.a: continue
d.append(x.i)
q = 1
for i in x.t:
y = p[i]
if y.a: continue
d.append(y.i)
q = 0
if q:
s.append(x.i + 1)
x.a = 1
elif x.q:
print(-1)
exit()
else: x.q = 1
print(len(s), *s)
```
| 103,368 | [
0.5400390625,
0.05194091796875,
0.038330078125,
0.1500244140625,
-0.28369140625,
0.1053466796875,
-0.0635986328125,
0.1451416015625,
0.059326171875,
0.92822265625,
1.0390625,
-0.0243072509765625,
0.60888671875,
-0.80419921875,
-0.5087890625,
0.222900390625,
-0.73681640625,
-1.15429... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
from sys import *
f = lambda: list(map(int, stdin.readline().split()))
class T:
def __init__(self, i):
self.i, self.t = i, f()[1:]
self.a = self.q = 0
n, k = f()
d = f()
p = [None] + [T(i + 1) for i in range(n)]
s = []
while d:
x = p[d.pop()]
if x.a: continue
d.append(x.i)
q = 1
for i in x.t:
y = p[i]
if y.a: continue
d.append(y.i)
q = 0
if q:
s.append(x.i)
x.a = 1
elif x.q:
print(-1)
exit()
else: x.q = 1
print(len(s), *s)
```
| 103,369 | [
0.5400390625,
0.05194091796875,
0.038330078125,
0.1500244140625,
-0.28369140625,
0.1053466796875,
-0.0635986328125,
0.1451416015625,
0.059326171875,
0.92822265625,
1.0390625,
-0.0243072509765625,
0.60888671875,
-0.80419921875,
-0.5087890625,
0.222900390625,
-0.73681640625,
-1.15429... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 8)
cycle = False
def DFS(i, graph, result, visited):
global cycle
stack = [i]
while len(stack) != 0:
u = stack[-1]
visited[u] = 1
if len(graph[u]) != 0:
v = graph[u].pop()
if visited[v] == 1:
cycle = True
return
if visited[v] == 2:
continue
stack.append(v)
visited[u] = 1
else:
result.append(u)
stack.pop()
visited[u] = 2
def TopoSort(graph, result):
for i in range(m):
if not visited[requiredCourse[i]]:
DFS(requiredCourse[i], graph, result, visited)
return result
n, m = map(int, input().split())
requiredCourse = list(map(int, input().split()))
graph = [[] for i in range(n + 1)]
result = []
visited = [False for i in range(n + 1)]
for i in range(1, n + 1):
tmp = list(map(int, input().split()))
if tmp[0] == 0:
continue
for j in range(1, tmp[0] + 1):
graph[i].append(tmp[j])
res = TopoSort(graph, result)
if cycle == True:
print(-1)
else:
print(len(res))
for i in res:
print(i, end=" ")
```
Yes
| 103,370 | [
0.5517578125,
0.114990234375,
0.020294189453125,
0.11083984375,
-0.3349609375,
0.1529541015625,
-0.09344482421875,
0.2010498046875,
0.081298828125,
0.9287109375,
1.046875,
0.0196990966796875,
0.572265625,
-0.81494140625,
-0.458740234375,
0.164306640625,
-0.7314453125,
-1.1943359375... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
'''import sys
flag=True
sys.setrecursionlimit(2000000)
c=[];st=[];
def topo(s):#Traversing the array and storing the vertices
global c,st,flag;
c[s]=1; #Being Visited
for i in adjli[s]:#visiting neighbors
if c[i]==0:
topo(i)
if c[i]==1:
flag=False# If Back Edge , Then Not Possible
st.append(str(s))
c[s]=2 # Visited
try:
n,k=map(int,input().split(' '))#Number Of Courses,Dependencies
main=list(map(int,input().split(' ')))#Main Dependencies
depen=[]#Dependencies List
for i in range(n):
depen.append(list(map(int,input().split(' ')))[1:]);c.append(0)#Append Input To Dependencies List, Marking Visited as 0(False)
c.append(0)
adjli=[]
adjli.append(main)#Assuming Main Course at index 0 with dependencies as Main Dependency(main)
for i in range(len(depen)):
adjli.append(depen[i])#Appending Other Dependencies
topo(0)#TopoLogical Sort Order
st.pop(-1)#popping the assumed Main Couse
if flag:# IF possible then print
print(len(st))
print(' '.join(st))
else:
print(-1)
except Exception as e:
print(e,"error")'''
import sys
flag=True
sys.setrecursionlimit(2000000000)
c=[];st=[];
cur_adj=[]
def topo(s):#Traversing the array and storing the vertices
global c,st,flag;
stack = [s]
while(stack):
s = stack[-1]
c[s]=1; #Being Visited
if(cur_adj[s] < len(adjli[s])):
cur = adjli[s][cur_adj[s]]
if(c[cur]==0):
stack.append(cur)
if(c[cur]==1):
flag=False# If Back Edge , Then Not Possible
cur_adj[s]+=1
else:
c[s]=2
st.append(str(s))
del stack[-1]
try:
n,k=map(int,input().split(' '))
main=list(map(int,input().split(' ')))
depen=[]
for i in range(n):
depen.append(list(map(int,input().split(' ')))[1:]);c.append(0)
cur_adj.append(0)
c.append(0)
cur_adj.append(0)
adjli=[]
adjli.append(main)#Assuming Main Course at index 0 with dependencies as Main Dependency(main)
for i in range(len(depen)):
adjli.append(depen[i])#Appending Other Dependencies
topo(0)#TopoLogical Sort Order
st.pop(-1)#popping the assumed Main Couse
if flag:# IF possible then print
print(len(st))
print(' '.join(st))
else:
print(-1)
except Exception as e:
print(e,"error")
```
Yes
| 103,371 | [
0.5517578125,
0.114990234375,
0.020294189453125,
0.11083984375,
-0.3349609375,
0.1529541015625,
-0.09344482421875,
0.2010498046875,
0.081298828125,
0.9287109375,
1.046875,
0.0196990966796875,
0.572265625,
-0.81494140625,
-0.458740234375,
0.164306640625,
-0.7314453125,
-1.1943359375... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
import sys
flag=True
sys.setrecursionlimit(2000000000)
c=[];st=[];
cur_adj=[]
def topo(s):#Traversing the array and storing the vertices
global c,st,flag;
stack = [s]
while(stack):
s = stack[-1]
c[s]=1; #Being Visited
if(cur_adj[s] < len(adjli[s])):
cur = adjli[s][cur_adj[s]]
if(c[cur]==0):
stack.append(cur)
if(c[cur]==1):
flag=False# If Back Edge , Then Not Possible
cur_adj[s]+=1
else:
c[s]=2
st.append(str(s))
del stack[-1]
try:
n,k=map(int,input().split(' '))
main=list(map(int,input().split(' ')))
depen=[]
for i in range(n):
depen.append(list(map(int,input().split(' ')))[1:]);c.append(0)
cur_adj.append(0)
c.append(0)
cur_adj.append(0)
adjli=[]
adjli.append(main)#Assuming Main Course at index 0 with dependencies as Main Dependency(main)
for i in range(len(depen)):
adjli.append(depen[i])#Appending Other Dependencies
topo(0)#TopoLogical Sort Order
st.pop(-1)#popping the assumed Main Couse
if flag:# IF possible then print
print(len(st))
print(' '.join(st))
else:
print(-1)
except Exception as e:
print(e,"error")
```
Yes
| 103,372 | [
0.5517578125,
0.114990234375,
0.020294189453125,
0.11083984375,
-0.3349609375,
0.1529541015625,
-0.09344482421875,
0.2010498046875,
0.081298828125,
0.9287109375,
1.046875,
0.0196990966796875,
0.572265625,
-0.81494140625,
-0.458740234375,
0.164306640625,
-0.7314453125,
-1.1943359375... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
#This code is dedicated to Vlada S.
class Course:
def __init__(self, reqs, number):
self.reqs = list(map(int, reqs.split()[1:]))
self.available = False
self.in_stack = False
self.number = number
n, k = list(map(int, input().split()))
requirements = list(map(int, input().split()))
courses = {}
answer = ""
for i in range(n):
courses[i + 1]= Course(input(), i + 1)
for i in range(len(requirements)):
requirements[i] = courses[requirements[i]]
while requirements:
data = {}
course = requirements.pop()
if not course.available:
requirements.append(course)
done = True
for c in course.reqs:
c = courses[c]
if not c.available:
requirements.append(c)
done = False
if done:
answer += " " + str(course.number)
course.available = True
else:
if course.in_stack:
print(-1)
break
course.in_stack = True
else:
print(answer.count(" "))
print(answer[1:])
# Made By Mostafa_Khaled
```
Yes
| 103,373 | [
0.5517578125,
0.114990234375,
0.020294189453125,
0.11083984375,
-0.3349609375,
0.1529541015625,
-0.09344482421875,
0.2010498046875,
0.081298828125,
0.9287109375,
1.046875,
0.0196990966796875,
0.572265625,
-0.81494140625,
-0.458740234375,
0.164306640625,
-0.7314453125,
-1.1943359375... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
putninja=[]
courses=[]
B=True
def mff(A, num):
global putninja
global courses
if A[0]=='0':
putninja.append(num)
courses[int(num)-1][0]='stop'
elif A[0]=='stop':
pass
else:
for j in range(int(A[0])):
mff(courses[int(A[j+1])-1],A[j+1])
putninja.append(num)
courses[int(num)-1][0]='stop'
kn=input().split()
k=int(kn[0])
n=int(kn[1])
mk=input().split()
for i in range(k):
s=input()
courses.append(s.split(' '))
for i in range(n):
try:
mff(courses[int(mk[i])-1],mk[i])
except:
B=False
break
if B:
print(len(putninja))
print(' '.join(putninja))
else:
print(-1)
```
No
| 103,374 | [
0.5517578125,
0.114990234375,
0.020294189453125,
0.11083984375,
-0.3349609375,
0.1529541015625,
-0.09344482421875,
0.2010498046875,
0.081298828125,
0.9287109375,
1.046875,
0.0196990966796875,
0.572265625,
-0.81494140625,
-0.458740234375,
0.164306640625,
-0.7314453125,
-1.1943359375... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
from sys import stdin, stdout
n,k = stdin.readline().split()
n = int(n)
k = int(k)
toLearn = [-1] * n
kList = input().split()
for i in range(len(kList)):
kList[i] = int(kList[i])
nList = []
flag = True
for i in range(n):
nL = stdin.readline().split()
nL.pop(0)
for j in range(len(nL)):
nL[j] = int(nL[j])
if len(nL) == 0:
flag = False
nList.append(nL)
# n = 100000
# k = 1
# kList = [100000]
# nList =[[]]
# flag = False
# for i in range(1, n):
# nList.append([i])
res = []
if k == 0:
print(0)
print("")
elif flag:
print(-1)
else:
res = []
notCircle = True
current = 0
while len(kList) > 0 and notCircle:
temp = []
for i in kList:
res.append(i)
toLearn[i - 1] = current
for j in nList[i - 1]:
if toLearn[j - 1] >= 0:
notCircle = False
break
break
if toLearn[j - 1] == -1:
temp.append(j)
kList = temp
current += 1
if notCircle:
res.reverse()
s = ""
for i in res:
s += str(i) + " "
stdout.write(str(len(res)) + '\n')
stdout.write(s[: -1] + '\n')
else:
stdout.write('-1\n')
```
No
| 103,375 | [
0.5517578125,
0.114990234375,
0.020294189453125,
0.11083984375,
-0.3349609375,
0.1529541015625,
-0.09344482421875,
0.2010498046875,
0.081298828125,
0.9287109375,
1.046875,
0.0196990966796875,
0.572265625,
-0.81494140625,
-0.458740234375,
0.164306640625,
-0.7314453125,
-1.1943359375... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
itog = ''
result = []
iskl = 0
control = 0
maxway = -1
wayStolb = 0
table = []
fs = input()
inp = fs.split(' ')
nk = [int(i) for i in inp]
ss = input()
inp = ss.split(' ')
maincour = [int(i) for i in inp]
for i in range(0,nk[0]):
table.append([])
s = input()
inp = s.split(' ')
t = [int(i) for i in inp]
for j in range(0,t[0]+1):
table[i].append(t[j])
for i in range (0,len(maincour)):
#print(table[maincour[i]-1])
if table[maincour[i]-1][0] > maxway:
wayStolb = maincour[i] - 1
maxway = table[maincour[i]-1][0]
#print(wayStolb)
#print(maxway)
e = maxway
stolb = wayStolb #4
startStolb = stolb
while control != len(maincour) :
while table[wayStolb][0] != 0:
maxway = -1
e = table[stolb][0]
for i in range(1,e+1):
if table[table[stolb][i]-1][0] > maxway:
wayStolb = table[stolb][i] - 1
maxway = table[table[stolb][i]-1][0]
e = maxway
if e > 0:
stolb = wayStolb
iskl = iskl + 1;
if iskl > 100000:
iskl = -1
break
#print('WayStolb = '+str(wayStolb))
#print('ΠΡΠ» Π·Π΄Π΅ΡΡ'+str(stolb))
if iskl == -1:
result = '-1'
break
#print('ΠΡΡΠ°Π½ΠΎΠ²ΠΊΠ° ' +str(wayStolb))
result.append(wayStolb+1)
for i in range(0,len(maincour)):
if wayStolb == maincour[i] - 1:
control = control + 1;
if control != len(maincour):
table[stolb][0] = table[stolb][0] - 1
try:
table[stolb].remove(wayStolb+1)
except ValueError:
for i in range(0,len(maincour)):
if table[maincour[i]-1][0] == 0 and result.count(maincour[i]) == 0:
result.append(maincour[i])
control = control + 1
if control != len(maincour):
for i in range (0,len(maincour)):
#print(table[maincour[i]-1])
if table[maincour[i]-1][0] > maxway:
wayStolb = maincour[i] - 1
maxway = table[maincour[i]-1][0]
#print(wayStolb)
#print(maxway)
e = maxway
stolb = wayStolb #4
startStolb = stolb
#print(table[stolb])
#print(table[stolb][1])
wayStolb = startStolb
#print('wayStolb = '+str(wayStolb))
stolb = wayStolb
if result != '-1':
itog = itog + str(result[0])
for i in range(1,len(result)):
itog = itog + ' ' + str(result[i])
print(len(result))
print(itog)
else:
print(result)
```
No
| 103,376 | [
0.5517578125,
0.114990234375,
0.020294189453125,
0.11083984375,
-0.3349609375,
0.1529541015625,
-0.09344482421875,
0.2010498046875,
0.081298828125,
0.9287109375,
1.046875,
0.0196990966796875,
0.572265625,
-0.81494140625,
-0.458740234375,
0.164306640625,
-0.7314453125,
-1.1943359375... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
import collections as col
import itertools as its
import sys
import operator
from copy import copy, deepcopy
class Solver:
def __init__(self):
pass
def solve(self):
n, k = map(int, input().split())
q = list(map(lambda x: int(x) - 1, input().split()))
used = [False] * n
for e in q:
used[e] = True
edges = [[] for _ in range(n)]
redges = [[] for _ in range(n)]
for i in range(n):
l = list(map(lambda x: int(x) - 1, input().split()))[1:]
edges[i] = l
for e in l:
redges[e].append(i)
degs = [len(edges[i]) for i in range(n)]
d = 0
while d < len(q):
v = q[d]
d += 1
for e in edges[v]:
if not used[e]:
used[e] = True
q.append(e)
q = q[::-1]
nq = []
for v in q:
if degs[v] == 0:
nq.append(v)
d = 0
while d < len(nq):
v = nq[d]
d += 1
for e in redges[v]:
degs[e] -= 1
if degs[e] == 0:
nq.append(e)
if len(q) != len(nq):
print(-1)
return
print(len(nq))
print(' '.join(map(lambda x: str(x + 1), nq)))
if __name__ == '__main__':
s = Solver()
s.solve()
```
No
| 103,377 | [
0.5517578125,
0.114990234375,
0.020294189453125,
0.11083984375,
-0.3349609375,
0.1529541015625,
-0.09344482421875,
0.2010498046875,
0.081298828125,
0.9287109375,
1.046875,
0.0196990966796875,
0.572265625,
-0.81494140625,
-0.458740234375,
0.164306640625,
-0.7314453125,
-1.1943359375... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Tags: dp, math, number theory, sortings
Correct Solution:
```
for _ in range(int(input())):
length=int(input())
array=list(map(int,input().split()))
array.sort()
cnt=[0]*(max(array)+1)
dp=[0]*(max(array)+1)
for x in array:
cnt[x]+=1
"""
[3, 7, 9, 14, 63]
"""
for i in range(1,len(dp)):
dp[i]+=cnt[i]
for j in range(i,len(dp),i):
dp[j]=max(dp[j],dp[i])
print(length-max(dp))
```
| 103,903 | [
0.3251953125,
0.01552581787109375,
0.498779296875,
0.12744140625,
-0.517578125,
-0.55712890625,
-0.2139892578125,
0.0394287109375,
0.34228515625,
0.6083984375,
0.87841796875,
-0.1024169921875,
0.025665283203125,
-1.0341796875,
-0.65234375,
-0.283447265625,
-0.3916015625,
-0.4079589... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Tags: dp, math, number theory, sortings
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fraction
from heapq import *
from random import randint
def main():
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
ma=max(a)
b=Counter(a)
c=sorted(set(a))
d=[0]*(ma+1)
d[c[-1]] =b[c[-1]]
ans = b[c[-1]]
c.pop()
c.reverse()
for i in range(len(c)):
j=2*c[i]
zz=0
while j<=ma:
zz=max(zz,d[j])
j+=c[i]
d[c[i]]=b[c[i]]+zz
ans=max(ans,d[c[i]])
print(n-ans)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
| 103,904 | [
0.337890625,
0.021636962890625,
0.48291015625,
0.126953125,
-0.5224609375,
-0.56298828125,
-0.22021484375,
0.035064697265625,
0.338623046875,
0.6201171875,
0.8798828125,
-0.1417236328125,
0.043060302734375,
-1.05078125,
-0.64990234375,
-0.29296875,
-0.397705078125,
-0.3828125,
-0... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Tags: dp, math, number theory, sortings
Correct Solution:
```
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from collections import deque
for i in range(int(input())):
n=int(input())
alist=list(map(int,input().split()))
maxN=int(2e5+1)
anslist=[0]*maxN
blist=[0]*maxN
for i,val in enumerate(alist):blist[val]+=1
for i in range(1,maxN):
if blist[i]==0:
continue
anslist[i]+=blist[i]
for j in range(2*i,maxN,i):
anslist[j]=max(anslist[i],anslist[j])
print(n-max(anslist))
```
| 103,905 | [
0.34375,
0.033905029296875,
0.49853515625,
0.1181640625,
-0.53076171875,
-0.5703125,
-0.228759765625,
0.045318603515625,
0.3525390625,
0.63232421875,
0.853515625,
-0.1414794921875,
0.044189453125,
-1.0419921875,
-0.63671875,
-0.282470703125,
-0.392578125,
-0.413330078125,
-0.8828... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Tags: dp, math, number theory, sortings
Correct Solution:
```
import sys
import math
def II():
return int(sys.stdin.readline())
def LI():
return list(map(int, sys.stdin.readline().split()))
def MI():
return map(int, sys.stdin.readline().split())
def SI():
return sys.stdin.readline().strip()
def FACT(n, mod):
s = 1
facts = [1]
for i in range(1,n+1):
s*=i
s%=mod
facts.append(s)
return facts[n]
def C(n, k, mod):
return (FACT(n,mod) * pow((FACT(k,mod)*FACT(n-k,mod))%mod,mod-2, mod))%mod
t = II()
for q in range(t):
n = II()
a = LI()
inf = max(a)+1
cnt = [0 for i in range(inf)]
for i in a:
cnt[i]+=1
dp = [0 for i in range(inf)]
for i in range(1,inf):
dp[i]+=cnt[i]
for j in range(i*2,inf, i):
dp[j] = max(dp[j], dp[i])
print(n-max(dp))
```
| 103,906 | [
0.32568359375,
0.049835205078125,
0.494873046875,
0.1212158203125,
-0.548828125,
-0.55029296875,
-0.2340087890625,
0.05499267578125,
0.34326171875,
0.63330078125,
0.86767578125,
-0.1146240234375,
0.04034423828125,
-1.05859375,
-0.6376953125,
-0.280517578125,
-0.3974609375,
-0.41674... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Tags: dp, math, number theory, sortings
Correct Solution:
```
import sys
import copy
input=sys.stdin.readline
for z in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
cnt=[0]*(2*10**5+10)
cnt2=[0]*(2*10**5+10)
a.sort()
num10=-1
cnt10=0
a2=[]
for i in range(n):
if num10!=a[i]:
if cnt10!=0:
a2.append(num10)
cnt[num10]=cnt10
cnt10=1
num10=a[i]
else:
cnt10+=1
if cnt10!=0:
a2.append(num10)
cnt[num10]=cnt10
cnt2=copy.copy(cnt)
a2.sort()
for i in range(len(a2)):
b=a2[i]
num11=cnt2[b]
for j in range(b*2,2*10**5+10,b):
cnt2[j]=max(cnt2[j],num11+cnt[j])
ans=n+10
for i in range(len(a2)):
ans=min(ans,n-cnt2[a2[i]])
print(ans)
```
| 103,907 | [
0.343994140625,
0.0302734375,
0.49951171875,
0.1273193359375,
-0.53955078125,
-0.5576171875,
-0.2379150390625,
0.040771484375,
0.33837890625,
0.62353515625,
0.84619140625,
-0.1072998046875,
0.04608154296875,
-1.056640625,
-0.6474609375,
-0.28369140625,
-0.383056640625,
-0.419677734... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Tags: dp, math, number theory, sortings
Correct Solution:
```
from sys import stdin
from collections import Counter
input = stdin.readline
for test in range(int(input())):
n = int(input())
lst = list(map(int, input().strip().split()))
N = max(lst) + 1
po = [0] * N
cnt = Counter(lst)
for k in range(1, N):
po[k] += cnt.get(k, 0)
for i in range(2 * k, N, k):
po[i] = max(po[i], po[k])
print(n - max(po))
```
| 103,908 | [
0.322998046875,
0.037139892578125,
0.50927734375,
0.126220703125,
-0.5556640625,
-0.55419921875,
-0.2376708984375,
0.036224365234375,
0.3369140625,
0.625,
0.84423828125,
-0.1171875,
0.047332763671875,
-1.068359375,
-0.65771484375,
-0.29345703125,
-0.40869140625,
-0.40673828125,
-... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Tags: dp, math, number theory, sortings
Correct Solution:
```
from bisect import bisect_left
for _ in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
l.sort()
m = max(l)
z = [0] * (m+1)
f = [0] * (m+1)
for i in range(n):
z[l[i]] += 1
for i in sorted(set(l)):
b = i+i
f[i] += z[i]
while b <= m:
f[b] = max(f[b], f[i])
b += i
print(n-max(f))
```
| 103,909 | [
0.265869140625,
0.062744140625,
0.489013671875,
0.136474609375,
-0.59375,
-0.5673828125,
-0.1788330078125,
-0.0245361328125,
0.337158203125,
0.66796875,
0.8837890625,
-0.06781005859375,
0.09027099609375,
-1.1064453125,
-0.67138671875,
-0.30908203125,
-0.406005859375,
-0.45776367187... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Tags: dp, math, number theory, sortings
Correct Solution:
```
from collections import Counter
#fk this shit
def f(arr):
cnt=Counter(arr)
mxn=max(arr)+100
factor=[0]*(mxn)
for i in range(1,len(factor)):
factor[i]+=cnt.get(i,0)
for j in range(2*i,mxn,i):
factor[j]=max(factor[i],factor[j])
return len(arr)-max(factor)
for i in range(int(input())):
s=input()
lst=list(map(int,input().strip().split()))
print(f(lst))
```
| 103,910 | [
0.3955078125,
0.0115966796875,
0.420166015625,
0.1820068359375,
-0.5380859375,
-0.57666015625,
-0.20947265625,
0.056915283203125,
0.378173828125,
0.6162109375,
0.86083984375,
-0.07135009765625,
0.04791259765625,
-1.0830078125,
-0.650390625,
-0.29833984375,
-0.387451171875,
-0.39453... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
from collections import Counter
MX = 200001
def solve(arr):
total = len(arr)
cnt = [0] * MX
for a in arr: cnt[a] += 1
u = sorted(set(arr))
loc = [-1] * MX
for i,a in enumerate(u): loc[a] = i
N = len(u)
dp = [0] * N
for i,a in enumerate(u):
dp[i] = max(dp[i], cnt[a])
for b in range(2*a,MX,a):
j = loc[b]
if j >= 0:
dp[j] = max(dp[j], dp[i] + cnt[b])
return total - max(dp)
for _ in range(int(input())):
input()
arr = list(map(int,input().split()))
print(solve(arr))
```
Yes
| 103,911 | [
0.367919921875,
0.06451416015625,
0.364990234375,
0.111572265625,
-0.63916015625,
-0.5087890625,
-0.256103515625,
0.089111328125,
0.285888671875,
0.56494140625,
0.76220703125,
-0.045318603515625,
0.02227783203125,
-1.0595703125,
-0.70654296875,
-0.2958984375,
-0.34326171875,
-0.404... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
def __init__(self, file):
self.newlines = 0
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline()
# --------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def S(): return input().strip()
def print_list(l): print(' '.join(map(str, l)))
# sys.setrecursionlimit(200000)
# import random
# from functools import reduce
# from functools import lru_cache
# from heapq import *
# from collections import deque as dq
# import math
# import bisect as bs
from collections import Counter
# from collections import defaultdict as dc
for t in range(N()):
n = N()
a = RLL()
count = Counter(a)
dp = [0] * 200001
for i in range(1, 200001):
if count[i] == 0: continue
dp[i] += count[i]
for j in range(2 * i, 200001, i):
dp[j] = max(dp[i], dp[j])
print(n - max(dp))
```
Yes
| 103,912 | [
0.3583984375,
0.0275115966796875,
0.369384765625,
0.1343994140625,
-0.67431640625,
-0.50390625,
-0.234375,
0.09619140625,
0.277587890625,
0.60009765625,
0.7666015625,
-0.049835205078125,
0.01364898681640625,
-1.0400390625,
-0.69873046875,
-0.28369140625,
-0.317138671875,
-0.4509277... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
import sys,math
def main():
for _ in range(int(sys.stdin.readline().strip())):
n=int(sys.stdin.readline().strip())
arr=list(map(int,sys.stdin.readline().strip().split()))
table=[0]*(200001)
dp=[0]*(200001)
for i in range(n):
table[arr[i]]+=1
for i in range(1,200001):
dp[i]+=table[i]
if dp[i]>0:
for j in range(2*i,200001,i):
dp[j]=max(dp[j],dp[i])
ans=n-max(dp)
sys.stdout.write(str(ans)+'\n')
main()
```
Yes
| 103,913 | [
0.340576171875,
0.06109619140625,
0.389404296875,
0.1475830078125,
-0.63818359375,
-0.4853515625,
-0.252197265625,
0.09002685546875,
0.251708984375,
0.5693359375,
0.79052734375,
-0.08367919921875,
-0.0006961822509765625,
-1.078125,
-0.71142578125,
-0.292236328125,
-0.32275390625,
-... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
factorsMemo=[-1 for _ in range(200001)]
def getAllFactors(x): #returns a sorted list of all the factors or divisors of x (including 1 and x)
if factorsMemo[x]==-1:
factors=[]
for i in range(1,int(x**0.5)+1):
if x%i==0:
factors.append(i)
if x//i!=i:
factors.append(x//i)
factorsMemo[x]=factors
return factorsMemo[x]
def main():
t=int(input())
allAns=[]
for _ in range(t):
n=int(input())
a=readIntArr()
a.sort()
dp=[-inf for _ in range(200001)] #dp[previous number]max number of elements before and including it to be beautiful}
maxSizeOfBeautifulArray=0
for x in a:
maxPrevCnts=0
for f in getAllFactors(x):
maxPrevCnts=max(maxPrevCnts,dp[f])
dp[x]=1+maxPrevCnts
maxSizeOfBeautifulArray=max(maxSizeOfBeautifulArray,dp[x])
allAns.append(n-maxSizeOfBeautifulArray)
multiLineArrayPrint(allAns)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
#import sys
#input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
inf=float('inf')
MOD=10**9+7
main()
```
Yes
| 103,914 | [
0.373779296875,
0.058074951171875,
0.391845703125,
0.133056640625,
-0.6298828125,
-0.51904296875,
-0.266357421875,
0.07781982421875,
0.1917724609375,
0.5517578125,
0.80224609375,
-0.06939697265625,
0.034515380859375,
-1.0927734375,
-0.693359375,
-0.2900390625,
-0.30517578125,
-0.45... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
import sys,math
# FASTEST IO
from io import BytesIO, IOBase
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# End of FASTEST IO
def main():
for _ in range(int(sys.stdin.readline().strip())):
n=int(sys.stdin.readline().strip())
arr=list(map(int,sys.stdin.readline().strip().split()))
table=[0]*(200001)
dp=[0]*(200001)
for i in range(n):
table[arr[i]]+=1
for i in range(1,200001):
for j in range(2*i,200001,i):
dp[j]=max(dp[j],table[i]+dp[i])
ans=n-max(dp)
sys.stdout.write(str(ans)+'\n')
main()
```
No
| 103,915 | [
0.36572265625,
0.033905029296875,
0.3564453125,
0.11541748046875,
-0.67822265625,
-0.5107421875,
-0.242431640625,
0.097412109375,
0.274658203125,
0.5908203125,
0.7451171875,
-0.075927734375,
0.051727294921875,
-1.060546875,
-0.69482421875,
-0.331298828125,
-0.30517578125,
-0.424072... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
import heapq
__all__ = ['MappedQueue']
class MappedQueue(object):
"""The MappedQueue class implements an efficient minimum heap. The
smallest element can be popped in O(1) time, new elements can be pushed
in O(log n) time, and any element can be removed or updated in O(log n)
time. The queue cannot contain duplicate elements and an attempt to push an
element already in the queue will have no effect.
MappedQueue complements the heapq package from the python standard
library. While MappedQueue is designed for maximum compatibility with
heapq, it has slightly different functionality.
Examples
--------
A `MappedQueue` can be created empty or optionally given an array of
initial elements. Calling `push()` will add an element and calling `pop()`
will remove and return the smallest element.
>>> q = MappedQueue([916, 50, 4609, 493, 237])
>>> q.push(1310)
True
>>> x = [q.pop() for i in range(len(q.h))]
>>> x
[50, 237, 493, 916, 1310, 4609]
Elements can also be updated or removed from anywhere in the queue.
>>> q = MappedQueue([916, 50, 4609, 493, 237])
>>> q.remove(493)
>>> q.update(237, 1117)
>>> x = [q.pop() for i in range(len(q.h))]
>>> x
[50, 916, 1117, 4609]
References
----------
.. [1] Cormen, T. H., Leiserson, C. E., Rivest, R. L., & Stein, C. (2001).
Introduction to algorithms second edition.
.. [2] Knuth, D. E. (1997). The art of computer programming (Vol. 3).
Pearson Education.
"""
def __init__(self, data=[]):
"""Priority queue class with updatable priorities.
"""
self.h = list(data)
self.d = dict()
self._heapify()
def __len__(self):
return len(self.h)
def _heapify(self):
"""Restore heap invariant and recalculate map."""
heapq.heapify(self.h)
self.d = dict([(elt, pos) for pos, elt in enumerate(self.h)])
if len(self.h) != len(self.d):
raise AssertionError("Heap contains duplicate elements")
def push(self, elt):
"""Add an element to the queue."""
# If element is already in queue, do nothing
if elt in self.d:
return False
# Add element to heap and dict
pos = len(self.h)
self.h.append(elt)
self.d[elt] = pos
# Restore invariant by sifting down
self._siftdown(pos)
return True
def pop(self):
"""Remove and return the smallest element in the queue."""
# Remove smallest element
elt = self.h[0]
del self.d[elt]
# If elt is last item, remove and return
if len(self.h) == 1:
self.h.pop()
return elt
# Replace root with last element
last = self.h.pop()
self.h[0] = last
self.d[last] = 0
# Restore invariant by sifting up, then down
pos = self._siftup(0)
self._siftdown(pos)
# Return smallest element
return elt
def update(self, elt, new):
"""Replace an element in the queue with a new one."""
# Replace
pos = self.d[elt]
self.h[pos] = new
del self.d[elt]
self.d[new] = pos
# Restore invariant by sifting up, then down
pos = self._siftup(pos)
self._siftdown(pos)
def remove(self, elt):
"""Remove an element from the queue."""
# Find and remove element
try:
pos = self.d[elt]
del self.d[elt]
except KeyError:
# Not in queue
raise
# If elt is last item, remove and return
if pos == len(self.h) - 1:
self.h.pop()
return
# Replace elt with last element
last = self.h.pop()
self.h[pos] = last
self.d[last] = pos
# Restore invariant by sifting up, then down
pos = self._siftup(pos)
self._siftdown(pos)
def _siftup(self, pos):
"""Move element at pos down to a leaf by repeatedly moving the smaller
child up."""
h, d = self.h, self.d
elt = h[pos]
# Continue until element is in a leaf
end_pos = len(h)
left_pos = (pos << 1) + 1
while left_pos < end_pos:
# Left child is guaranteed to exist by loop predicate
left = h[left_pos]
try:
right_pos = left_pos + 1
right = h[right_pos]
# Out-of-place, swap with left unless right is smaller
if right < left:
h[pos], h[right_pos] = right, elt
pos, right_pos = right_pos, pos
d[elt], d[right] = pos, right_pos
else:
h[pos], h[left_pos] = left, elt
pos, left_pos = left_pos, pos
d[elt], d[left] = pos, left_pos
except IndexError:
# Left leaf is the end of the heap, swap
h[pos], h[left_pos] = left, elt
pos, left_pos = left_pos, pos
d[elt], d[left] = pos, left_pos
# Update left_pos
left_pos = (pos << 1) + 1
return pos
def _siftdown(self, pos):
"""Restore invariant by repeatedly replacing out-of-place element with
its parent."""
h, d = self.h, self.d
elt = h[pos]
# Continue until element is at root
while pos > 0:
parent_pos = (pos - 1) >> 1
parent = h[parent_pos]
if parent > elt:
# Swap out-of-place element with parent
h[parent_pos], h[pos] = elt, parent
parent_pos, pos = pos, parent_pos
d[elt] = pos
d[parent] = parent_pos
else:
# Invariant is satisfied
break
return pos
##########################################################
##########################################################
##########################################################
from collections import Counter
# from sortedcontainers import SortedList
t = int(input())
DEBUG = False
for asdasd in range(t):
# with open('temp.txt', 'w') as file:
# for i in range(1, 200001):
# file.write(str(i) + ' ')
# break
n = int(input())
arr = [int(i) for i in input().split(" ")]
# if n==200000:
# DEBUG = True
# if DEBUG and asdasd == 3:
# arr.insert(0,n)
# line = str(arr)
# n = 400
# x = [line[i:i+n] for i in range(0, len(line), n)]
# for asd in x:
# print(asd)
arr.sort(reverse = True)
# print(arr)
count = dict(Counter(arr))
edges = {}
##########################################
# DOESNT REALLY WORK
##########################################
# heads = set()
# for x in count:
# edges[x] = set()
# heads_to_remove = set()
# for head in heads:
# if head % x == 0:
# edges[x].update(edges[head])
# edges[x].add(head)
# heads_to_remove.add(head)
# heads -= heads_to_remove
# heads.add(x)
# # print(edges)
# edges_copy = edges.copy()
# for x, edge_set in edges.items():
# for e in edge_set:
# edges_copy[e].add(x)
# edges = edges_copy
############################################
print('x1')
##########################################
# TRYING N^2 EDGEs
##########################################
for x in count:
edges[x] = set()
count_list = list(count.items())
for i in range(len(count_list)):
if i % 10 == 0:
print(i/len(count_list))
for j in range(i+1, len(count_list)):
if count_list[i][0] % count_list[j][0] == 0 or count_list[j][0] % count_list[i][0] == 0:
edges[count_list[i][0]].add(count_list[j][0])
edges[count_list[j][0]].add(count_list[i][0])
############################################
print('x2')
edge_count = {}
total_edges = 0
for x, edge_set in edges.items():
edge_count[x] = count[x]-1
for e in edge_set:
edge_count[x] += count[e]
total_edges += count[x] * edge_count[x]
total_edges //= 2
NODE_VALUE = 1
N_EDGES = 0
# queue = SortedList(list(edge_count.items()), key=lambda x: -x[N_EDGES])
# for key, value in edge_count.items():
# edge_count[key] = value * count[key]
queue = MappedQueue([(n_edges, node_value) for node_value, n_edges in edge_count.items()])
# print('edges:', edges)
# print('edge_count:', edge_count)
# print('total_edges', total_edges)
# print()
# print('queue:', queue)
# print()
remaining_nodes = n
# print('total_edges:', total_edges, 'remaining_nodes:', remaining_nodes)
# print()
# print(count)
while total_edges != remaining_nodes*(remaining_nodes-1)//2:
small = queue.pop()
# remaining_nodes -= count[small[NODE_VALUE]]
remaining_nodes -= 1
# print('popping node:', small[NODE_VALUE], 'edges:', small[N_EDGES])
# print(queue)
for e in edges[small[NODE_VALUE]]:
# print('updating edge', e)
# queue.remove((e, edge_count[e]))
queue.remove((edge_count[e], e))
# print('rem2', count[e] * count[small[NODE_VALUE]])
# edge_count[e] -= count[e] * count[small[NODE_VALUE]]
# total_edges -= count[e] * count[small[NODE_VALUE]]
# print('decreasing edge', e, ' edges by', 1)
edge_count[e] -= 1
# print('decreasing total edges by', count[e])
total_edges -= count[e]
if count[small[NODE_VALUE]] == 1:
edges[e].remove(small[NODE_VALUE])
# queue.add((e, edge_count[e]))
queue.push((edge_count[e], e))
# Edges between nodes of same value
t = count[small[NODE_VALUE]]
# print('rem1', t*(t-1)//2)
t2 = t-1
total_edges -= t2
# print(t, '(*) decreasing total edges by', t2)
count[small[NODE_VALUE]] -= 1
if count[small[NODE_VALUE]] != 0:
queue.push((small[N_EDGES]-1, small[NODE_VALUE]))
# print(queue)
# print('total_edges:', total_edges, 'remaining_nodes:', remaining_nodes)
# print()
print(n - remaining_nodes)
```
No
| 103,916 | [
0.31640625,
0.12176513671875,
0.26416015625,
0.2032470703125,
-0.60498046875,
-0.455810546875,
-0.267822265625,
0.06500244140625,
0.247802734375,
0.63427734375,
0.7265625,
-0.018768310546875,
0.049041748046875,
-1.0654296875,
-0.6943359375,
-0.294677734375,
-0.317138671875,
-0.4838... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
def main():
pass
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict,Counter
m=2*10**5+5
t=int(input())
for _ in range(t):
n=int(input())
a=list(map(int,input().split()))
c=Counter(a)
greater_count=defaultdict(lambda: 0) #Stores number of numbers a given number(not necessarily in a) is greater than or equal to the numbers in a.
greater_count[0]=c[0]
for i in range(1,m+1):
greater_count[i]=greater_count[i-1]+c[i]
dp=[float('inf') for i in range(m+1)]
for i in range(m,0,-1):
dp[i]=n-greater_count[i-1] #number of numbers in a, strictly greater than i (this is worst case answer for the sublist consisting of all numbers in a >=i)
for j in range(2*i,m,i):
dp[i]=min(dp[i],dp[j]+greater_count[j-1]-greater_count[i]) #number of numbers in a, between (i,j)
print(dp[1])
```
No
| 103,917 | [
0.351806640625,
0.03912353515625,
0.341796875,
0.12225341796875,
-0.6611328125,
-0.51416015625,
-0.2440185546875,
0.08929443359375,
0.2939453125,
0.5947265625,
0.744140625,
-0.036285400390625,
0.0164642333984375,
-1.041015625,
-0.70361328125,
-0.31201171875,
-0.327880859375,
-0.467... | 24 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fraction
from heapq import *
from random import randint
def main():
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
ma=max(a)
b=[0]*(ma+1)
for i in a:
b[i]+=1
c=[]
for i in range(1,ma+1):
if b[i]:
c.append(i)
d = [0] * (ma + 1)
d[c[-1]] =b[c[-1]]
c.pop()
c.reverse()
ans=1
for i in range(len(c)):
j=2*c[i]
zz=0
while j<=ma:
zz=max(zz,d[j])
j+=c[i]
d[c[i]]=b[c[i]]+zz
ans=max(ans,d[c[i]])
print(n-ans)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
No
| 103,918 | [
0.361328125,
0.03729248046875,
0.353759765625,
0.1251220703125,
-0.63818359375,
-0.51806640625,
-0.241943359375,
0.09027099609375,
0.253662109375,
0.58349609375,
0.7744140625,
-0.060150146484375,
0.00506591796875,
-1.0341796875,
-0.70703125,
-0.323974609375,
-0.32421875,
-0.4003906... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 β€ a, b, c β€ 7β
10^8) β the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday β rabbit stew and during Wednesday β chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.
Tags: implementation, math
Correct Solution:
```
a,b,c=map(int,input().split())
fu=0
fu=min(a//3,b//2,c//2)
a-=3*fu
b-=2*fu
c-=2*fu
x=[1,2,3,1,3,2,1]
k=0
m=0
for i in range(len(x)):
d=0
f=0
j=i
a1=a
b1=b
c1=c
z=1
while z==1:
k=x[j]
if (k==1)&(a1>0):
a1-=1
d+=1
if (k==2)&(b1>0):
b1-=1
d+=1
if (k==3)&(c1>0):
c1-=1
d+=1
if d==f:
z=0
else:
f=d
j+=1
j=j%7
if d>m:
m=d
print(7*fu+m)
```
| 105,305 | [
0.488525390625,
0.026641845703125,
0.1583251953125,
0.05682373046875,
-0.58740234375,
-0.0675048828125,
-0.3359375,
0.5205078125,
0.05413818359375,
0.84521484375,
0.55908203125,
-0.2039794921875,
0.304931640625,
-0.69287109375,
-0.6806640625,
0.0122833251953125,
-0.57763671875,
-0.... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 β€ a, b, c β€ 7β
10^8) β the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday β rabbit stew and during Wednesday β chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.
Tags: implementation, math
Correct Solution:
```
def dfs(day_l,day_r,a,b,c):
global span
if day_l<1 or day_r>14:
return
if a<0 or b<0 or c<0:
return
#print(day_l,day_r,a,b,c)
span=max(span,day_r-day_l+1)
if day_l-1 in [1,4,7,8,11,14]:
dfs(day_l-1,day_r,a-1,b,c)
elif day_l-1 in [2,6,9,13]:
dfs(day_l-1,day_r,a,b-1,c)
elif day_l-1 in [3,5,10,12]:
dfs(day_l-1,day_r,a,b,c-1)
if day_r+1 in [1,4,7,8,11,14]:
dfs(day_l,day_r+1,a-1,b,c)
elif day_r+1 in [2,6,9,13]:
dfs(day_l,day_r+1,a,b-1,c)
elif day_r+1 in [3,5,10,12]:
dfs(day_l,day_r+1,a,b,c-1)
a,b,c=list(map(int,input().split()))
weeks=min([a//3,b//2,c//2])
a-=weeks*3
b-=weeks*2
c-=weeks*2
if weeks>0:
span=0
if a>0:
day_min,day_max=7,7
dfs(7,7,a-1,b,c)
ans1=7*weeks+span
span=0
day_min,day_max=7,7
dfs(7,7,a-1+3,b+2,c+2)
ans2=7*(weeks-1)+span
print(max(ans1,ans2))
else:
span=0
for i in [4,5,6,7]:
day_min,day_max=i,i
if i in [4,7]:
dfs(i,i,a-1,b,c)
elif i==6:
dfs(i,i,a,b-1,c)
else:
dfs(i,i,a,b,c-1)
print(span)
```
| 105,306 | [
0.46826171875,
0.007076263427734375,
0.2012939453125,
0.09881591796875,
-0.55810546875,
-0.0200042724609375,
-0.3193359375,
0.52978515625,
0.053558349609375,
0.83740234375,
0.56103515625,
-0.254150390625,
0.304931640625,
-0.64111328125,
-0.67919921875,
0.0208587646484375,
-0.59912109... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 β€ a, b, c β€ 7β
10^8) β the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday β rabbit stew and during Wednesday β chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.
Tags: implementation, math
Correct Solution:
```
a, b, c = map(int, input().split())
x = min(a//3, b//2, c//2)
a,b,c = a-3*x, b-2*x, c-2*x
z = [1,1,2,3,1,3,2]
m = 0
for i in range(7):
lis = [0,a,b,c]
j = i
cnt = 0
while(lis[1]>=0 and lis[2]>=0 and lis[3]>=0):
lis[z[j]] -= 1
j = (j+1)%7
cnt += 1
m = max(m,cnt-1)
print(x*7+m)
```
| 105,307 | [
0.473876953125,
0.03607177734375,
0.14697265625,
0.06829833984375,
-0.568359375,
-0.056243896484375,
-0.3388671875,
0.52294921875,
0.05780029296875,
0.849609375,
0.5517578125,
-0.2105712890625,
0.290771484375,
-0.7041015625,
-0.6767578125,
0.019683837890625,
-0.583984375,
-0.624511... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 β€ a, b, c β€ 7β
10^8) β the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday β rabbit stew and during Wednesday β chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.
Tags: implementation, math
Correct Solution:
```
a = [int(i) for i in input().split()]
days = [0, 1, 2, 0, 2, 1, 0]
week = min(a[0] // 3, a[1] // 2, a[2] // 2)
a[0], a[1], a[2] = a[0] - week * 3, a[1] - week * 2, a[2] - week * 2
m = 0
for day in range(7):
b = a.copy()
c = 0
for this_day in range(7):
if b[days[(day + this_day) % 7]] > 0:
b[days[(day + this_day) % 7]] -= 1
c += 1
else:
break
if c > m:
m = c
print(m + week * 7)
```
| 105,308 | [
0.4990234375,
0.03656005859375,
0.156982421875,
0.0379638671875,
-0.59521484375,
-0.040618896484375,
-0.3134765625,
0.5146484375,
0.07281494140625,
0.8466796875,
0.55224609375,
-0.2156982421875,
0.28271484375,
-0.705078125,
-0.6884765625,
0.0151519775390625,
-0.57763671875,
-0.6284... | 24 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 β€ a, b, c β€ 7β
10^8) β the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday β rabbit stew and during Wednesday β chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.
Tags: implementation, math
Correct Solution:
```
a,b,c = map(int,input().strip().split())
ans = 0
basi = min(c//2,min(b//2,a//3))
#print(basi)
aa = a- 3*basi
bb = b - 2*basi
cc = c - 2*basi
mp = {0:'aa',1:'aa',2:'bb',3:'cc',4:'aa',5:'cc',6:'bb'}
#print(aa,bb,cc)
for i in range(7):
aaa = aa
bbb = bb
ccc = cc
an = 0
for j in range(i,i+15):
if mp[j%7]=='aa':
aaa-=1
if mp[j%7]=='bb':
bbb-=1
if mp[j%7]=='cc':
ccc-=1
if aaa<0 or bbb<0 or ccc<0:
break
else:
an+=1
ans = max(ans,basi*7+an)
print(ans)
```
| 105,309 | [
0.488037109375,
0.0313720703125,
0.153564453125,
0.036224365234375,
-0.56787109375,
-0.05145263671875,
-0.36474609375,
0.513671875,
0.0682373046875,
0.82666015625,
0.5361328125,
-0.2208251953125,
0.292236328125,
-0.68896484375,
-0.6689453125,
-0.007472991943359375,
-0.58349609375,
... | 24 |
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