message stringlengths 2 44.5k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi wants to print a document with N pages double-sided, where two pages of data can be printed on one sheet of paper.
At least how many sheets of paper does he need?
Constraints
* N is an integer.
* 1 \leq N \leq 100
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer.
Examples
Input
5
Output
3
Input
2
Output
1
Input
100
Output
50
Submitted Solution:
```
n=input()
a=(int(n)+1)//2
print(a)
``` | instruction | 0 | 35,771 | 5 | 71,542 |
Yes | output | 1 | 35,771 | 5 | 71,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi wants to print a document with N pages double-sided, where two pages of data can be printed on one sheet of paper.
At least how many sheets of paper does he need?
Constraints
* N is an integer.
* 1 \leq N \leq 100
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer.
Examples
Input
5
Output
3
Input
2
Output
1
Input
100
Output
50
Submitted Solution:
```
n=int(input())
p=int((n+1)/2)
print(p)
``` | instruction | 0 | 35,772 | 5 | 71,544 |
Yes | output | 1 | 35,772 | 5 | 71,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi wants to print a document with N pages double-sided, where two pages of data can be printed on one sheet of paper.
At least how many sheets of paper does he need?
Constraints
* N is an integer.
* 1 \leq N \leq 100
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer.
Examples
Input
5
Output
3
Input
2
Output
1
Input
100
Output
50
Submitted Solution:
```
bingo = []
for i in range(3):
array = list(map(int, input().strip().split()))
bingo.append(array)
N = int(input())
number = []
for j in range(N):
number.append(int(input()))
result = []
for n in range(N):
for t in range(3):
for y in range(3):
if(bingo[t][y] == number[n]):
result.append([t,y])
bingo_a = 'No'
t_a = 0
for r in range(len(result)):
for t in range(3):
for y in range(3):
if(result[r][0] == t and result[r][1] == y):
t_a += 1
if(t_a == 3):
bingo_a = 'Yes'
y_a = 0
for r in range(len(result)):
for y in range(3):
for t in range(3):
if(result[r][0] == t and result[r][1] == y):
y_a += 1
if(y_a == 3):
bingo_a = 'Yes'
ans = 0
for rr in range(len(result)):
if(result[rr][0] == 0 or result[rr][1] == 0):
ans = ans + 1
if(result[rr][1] == 1 or result[rr][1] == 1):
ans = ans + 1
if(result[rr][0] == 2 or result[rr][1] == 2):
ans = ans + 1
if(ans == 3):
bingo_a = 'Yes'
ans = 0
for rr in range(len(result)):
if(result[rr][0] == 0 or result[rr][1] == 2):
ans = ans + 1
if(result[rr][1] == 1 or result[rr][1] == 1):
ans = ans + 1
if(result[rr][0] == 2 or result[rr][1] == 0):
ans = ans + 1
if(ans == 3):
bingo_a = 'Yes'
print(bingo_a)
``` | instruction | 0 | 35,774 | 5 | 71,548 |
No | output | 1 | 35,774 | 5 | 71,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi wants to print a document with N pages double-sided, where two pages of data can be printed on one sheet of paper.
At least how many sheets of paper does he need?
Constraints
* N is an integer.
* 1 \leq N \leq 100
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer.
Examples
Input
5
Output
3
Input
2
Output
1
Input
100
Output
50
Submitted Solution:
```
A11,A12,A13 = map(int, input().split())
A21,A22,A23 = map(int, input().split())
A31,A32,A33 = map(int, input().split())
N = int(input())
b = [int(input()) for _ in range(N)]
cnt11=0
cnt12=0
cnt13=0
cnt21=0
cnt22=0
cnt23=0
cnt31=0
cnt32=0
cnt33=0
for i in range(N):
if b[i]==A11:
cnt11+=1
elif b[i]==A12:
cnt12+=1
elif b[i]==A13:
cnt13+=1
elif b[i]==A21:
cnt21+=1
elif b[i]==A22:
cnt22+=1
elif b[i]==A23:
cnt23+=1
elif b[i]==A31:
cnt31+=1
elif b[i]==A32:
cnt32+=1
elif b[i]==A33:
cnt33+=1
else:
pass
bingo=0
if cnt11>0:
if((cnt12>0 and cnt13>0)or(cnt22>0 and cnt33>0)or(cnt21>0 and cnt31>0)):
bingo+=1
else:
pass
elif cnt12>0:
if (cnt22>0 and cnt32>0):
bingo+=1
else:
pass
elif cnt13>0:
if((cnt22>0 and cnt31>0)or(cnt23>0 and cnt33>0)):
bingo+=1
else:
pass
elif cnt21>0:
if(cnt22>0 and cnt23>0):
bingo+=1
else:
pass
elif cnt31>0:
if(cnt32>0 and cnt33>0):
bingo+=1
else:
pass
else:
pass
if bingo>0:
print('Yes')
else:
print('No')
``` | instruction | 0 | 35,775 | 5 | 71,550 |
No | output | 1 | 35,775 | 5 | 71,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi wants to print a document with N pages double-sided, where two pages of data can be printed on one sheet of paper.
At least how many sheets of paper does he need?
Constraints
* N is an integer.
* 1 \leq N \leq 100
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer.
Examples
Input
5
Output
3
Input
2
Output
1
Input
100
Output
50
Submitted Solution:
```
N = int(input())
if N % 2 == 0:
N = N / 2
print(N)
else:
N = N // 2 + 1
print(N)
``` | instruction | 0 | 35,776 | 5 | 71,552 |
No | output | 1 | 35,776 | 5 | 71,553 |
Provide a correct Python 3 solution for this coding contest problem.
A: IP Address (Internet Protocol Address)
problem
Since one number string is given, find the number of valid delimiters for IPv4 IP addresses. However, the effective delimiters for IPv4 IP addresses are as follows.
* The sequence of numbers is divided into four, and each of the separated columns satisfies all of the following.
* Not an empty string.
* When read as a decimal number, it is an integer between 0 and 255.
* If the number is 0, then the column is `0` (` 00` etc. are not allowed).
* If the number is non-zero, the first digit is not `0` (`01` etc. are not allowed).
Input format
S
Constraint
* 4 \ leq | S | \ leq 12
* S contains only numbers.
Output format
Output the number on one line according to the valid delimiter.
Input example 1
123456789
Output example 1
1
There is one valid delimiter, `123.45.67.89`.
Input example 2
768426
Output example 2
6
Example
Input
123456789
Output
1 | instruction | 0 | 35,952 | 5 | 71,904 |
"Correct Solution:
```
def check(num):
if num == "":return False
if num[0] == "0":return num == "0"
return 0 <= int(num) <= 255
s = input()
ans = 0
for i in range(1, 4):
for j in range(1, 4):
for k in range(1, 4):
n1 = s[:i]
n2 = s[i:i+j]
n3 = s[i+j:i+j+k]
n4 = s[i+j+k:]
if check(n1) and check(n2) and check(n3) and check(n4):ans += 1
print(ans)
``` | output | 1 | 35,952 | 5 | 71,905 |
Provide a correct Python 3 solution for this coding contest problem.
A: IP Address (Internet Protocol Address)
problem
Since one number string is given, find the number of valid delimiters for IPv4 IP addresses. However, the effective delimiters for IPv4 IP addresses are as follows.
* The sequence of numbers is divided into four, and each of the separated columns satisfies all of the following.
* Not an empty string.
* When read as a decimal number, it is an integer between 0 and 255.
* If the number is 0, then the column is `0` (` 00` etc. are not allowed).
* If the number is non-zero, the first digit is not `0` (`01` etc. are not allowed).
Input format
S
Constraint
* 4 \ leq | S | \ leq 12
* S contains only numbers.
Output format
Output the number on one line according to the valid delimiter.
Input example 1
123456789
Output example 1
1
There is one valid delimiter, `123.45.67.89`.
Input example 2
768426
Output example 2
6
Example
Input
123456789
Output
1 | instruction | 0 | 35,953 | 5 | 71,906 |
"Correct Solution:
```
import itertools
S = input()
ans = 0
for ptn in itertools.combinations(range(1,len(S)), 3):
ai,bi,ci = ptn
a = S[:ai]
b = S[ai:bi]
c = S[bi:ci]
d = S[ci:]
if a[0] == '0' and a != '0': continue
if b[0] == '0' and b != '0': continue
if c[0] == '0' and c != '0': continue
if d[0] == '0' and d != '0': continue
if not 0 <= int(a) <= 255: continue
if not 0 <= int(b) <= 255: continue
if not 0 <= int(c) <= 255: continue
if not 0 <= int(d) <= 255: continue
ans += 1
print(ans)
``` | output | 1 | 35,953 | 5 | 71,907 |
Provide a correct Python 3 solution for this coding contest problem.
Given a non-negative decimal integer $x$, convert it to binary representation $b$ of 32 bits. Then, print the result of the following operations to $b$ respecitvely.
* Inversion: change the state of each bit to the opposite state
* Logical left shift: shift left by 1
* Logical right shift: shift right by 1
Constraints
* $0 \leq x \leq 2^{32} - 1$
Input
The input is given in the following format.
$x$
Output
Print the given bits, results of inversion, left shift and right shift in a line respectively.
Examples
Input
8
Output
00000000000000000000000000001000
11111111111111111111111111110111
00000000000000000000000000010000
00000000000000000000000000000100
Input
13
Output
00000000000000000000000000001101
11111111111111111111111111110010
00000000000000000000000000011010
00000000000000000000000000000110 | instruction | 0 | 35,970 | 5 | 71,940 |
"Correct Solution:
```
x = int(input())
BITMASK = (1 << 32) - 1
print("{:032b}".format(x))
print("{:032b}".format(~x & BITMASK))
print("{:032b}".format((x << 1) & BITMASK))
print("{:032b}".format((x >> 1) & BITMASK))
``` | output | 1 | 35,970 | 5 | 71,941 |
Provide a correct Python 3 solution for this coding contest problem.
Given a non-negative decimal integer $x$, convert it to binary representation $b$ of 32 bits. Then, print the result of the following operations to $b$ respecitvely.
* Inversion: change the state of each bit to the opposite state
* Logical left shift: shift left by 1
* Logical right shift: shift right by 1
Constraints
* $0 \leq x \leq 2^{32} - 1$
Input
The input is given in the following format.
$x$
Output
Print the given bits, results of inversion, left shift and right shift in a line respectively.
Examples
Input
8
Output
00000000000000000000000000001000
11111111111111111111111111110111
00000000000000000000000000010000
00000000000000000000000000000100
Input
13
Output
00000000000000000000000000001101
11111111111111111111111111110010
00000000000000000000000000011010
00000000000000000000000000000110 | instruction | 0 | 35,971 | 5 | 71,942 |
"Correct Solution:
```
x = int(input())
MASK = (1 << 32) - 1
print("{:032b}".format(x))
print("{:032b}".format(~x & MASK))
print("{:032b}".format((x << 1) & MASK))
print("{:032b}".format((x >> 1) & MASK))
``` | output | 1 | 35,971 | 5 | 71,943 |
Provide a correct Python 3 solution for this coding contest problem.
Given a non-negative decimal integer $x$, convert it to binary representation $b$ of 32 bits. Then, print the result of the following operations to $b$ respecitvely.
* Inversion: change the state of each bit to the opposite state
* Logical left shift: shift left by 1
* Logical right shift: shift right by 1
Constraints
* $0 \leq x \leq 2^{32} - 1$
Input
The input is given in the following format.
$x$
Output
Print the given bits, results of inversion, left shift and right shift in a line respectively.
Examples
Input
8
Output
00000000000000000000000000001000
11111111111111111111111111110111
00000000000000000000000000010000
00000000000000000000000000000100
Input
13
Output
00000000000000000000000000001101
11111111111111111111111111110010
00000000000000000000000000011010
00000000000000000000000000000110 | instruction | 0 | 35,972 | 5 | 71,944 |
"Correct Solution:
```
n=int(input())
m=(1<<32)-1
print("{:032b}".format(n))
print("{:032b}".format(~n&m))
print("{:032b}".format(n<<1&m))
print("{:032b}".format(n>>1))
``` | output | 1 | 35,972 | 5 | 71,945 |
Provide a correct Python 3 solution for this coding contest problem.
Given a non-negative decimal integer $x$, convert it to binary representation $b$ of 32 bits. Then, print the result of the following operations to $b$ respecitvely.
* Inversion: change the state of each bit to the opposite state
* Logical left shift: shift left by 1
* Logical right shift: shift right by 1
Constraints
* $0 \leq x \leq 2^{32} - 1$
Input
The input is given in the following format.
$x$
Output
Print the given bits, results of inversion, left shift and right shift in a line respectively.
Examples
Input
8
Output
00000000000000000000000000001000
11111111111111111111111111110111
00000000000000000000000000010000
00000000000000000000000000000100
Input
13
Output
00000000000000000000000000001101
11111111111111111111111111110010
00000000000000000000000000011010
00000000000000000000000000000110 | instruction | 0 | 35,973 | 5 | 71,946 |
"Correct Solution:
```
n=format(int(input()),'032b')
n_li=list(n)
inversion_n=["1" if n_li[i]=="0" else "0" for i in range(len(n_li))]
inversion_n= "".join(inversion_n)
left=n[1:len(n)]+"0"
right="0"+n[0:len(n)-1]
print(n)
print(inversion_n)
print(left)
print(right)
``` | output | 1 | 35,973 | 5 | 71,947 |
Provide a correct Python 3 solution for this coding contest problem.
Given a non-negative decimal integer $x$, convert it to binary representation $b$ of 32 bits. Then, print the result of the following operations to $b$ respecitvely.
* Inversion: change the state of each bit to the opposite state
* Logical left shift: shift left by 1
* Logical right shift: shift right by 1
Constraints
* $0 \leq x \leq 2^{32} - 1$
Input
The input is given in the following format.
$x$
Output
Print the given bits, results of inversion, left shift and right shift in a line respectively.
Examples
Input
8
Output
00000000000000000000000000001000
11111111111111111111111111110111
00000000000000000000000000010000
00000000000000000000000000000100
Input
13
Output
00000000000000000000000000001101
11111111111111111111111111110010
00000000000000000000000000011010
00000000000000000000000000000110 | instruction | 0 | 35,974 | 5 | 71,948 |
"Correct Solution:
```
n = int(input())
MASK = (1 << 32) -1
print("{:032b}".format(n))
print("{:032b}".format(~n & MASK))
print("{0:032b}".format(n << 1 & MASK))
print("{0:032b}".format(n >> 1 & MASK))
``` | output | 1 | 35,974 | 5 | 71,949 |
Provide a correct Python 3 solution for this coding contest problem.
Given a non-negative decimal integer $x$, convert it to binary representation $b$ of 32 bits. Then, print the result of the following operations to $b$ respecitvely.
* Inversion: change the state of each bit to the opposite state
* Logical left shift: shift left by 1
* Logical right shift: shift right by 1
Constraints
* $0 \leq x \leq 2^{32} - 1$
Input
The input is given in the following format.
$x$
Output
Print the given bits, results of inversion, left shift and right shift in a line respectively.
Examples
Input
8
Output
00000000000000000000000000001000
11111111111111111111111111110111
00000000000000000000000000010000
00000000000000000000000000000100
Input
13
Output
00000000000000000000000000001101
11111111111111111111111111110010
00000000000000000000000000011010
00000000000000000000000000000110 | instruction | 0 | 35,975 | 5 | 71,950 |
"Correct Solution:
```
x=int(input())
b=~x&0b11111111111111111111111111111111
c=(x<<1)&0b11111111111111111111111111111111
d=(x>>1)&0b11111111111111111111111111111111
print(format(x,'032b'))
print(format(b,'032b'))
print(format(c,'032b'))
print(format(d,'032b'))
``` | output | 1 | 35,975 | 5 | 71,951 |
Provide a correct Python 3 solution for this coding contest problem.
Given a non-negative decimal integer $x$, convert it to binary representation $b$ of 32 bits. Then, print the result of the following operations to $b$ respecitvely.
* Inversion: change the state of each bit to the opposite state
* Logical left shift: shift left by 1
* Logical right shift: shift right by 1
Constraints
* $0 \leq x \leq 2^{32} - 1$
Input
The input is given in the following format.
$x$
Output
Print the given bits, results of inversion, left shift and right shift in a line respectively.
Examples
Input
8
Output
00000000000000000000000000001000
11111111111111111111111111110111
00000000000000000000000000010000
00000000000000000000000000000100
Input
13
Output
00000000000000000000000000001101
11111111111111111111111111110010
00000000000000000000000000011010
00000000000000000000000000000110 | instruction | 0 | 35,976 | 5 | 71,952 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Bitset I - Bit Operation I
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ITP2_10_A&lang=jp
"""
x = int(input())
print(f'{x:032b}')
print(f'{x:032b}'.translate(str.maketrans('01', '10')))
print(f'{x<<1:032b}'[-32:])
print(f'{x>>1:032b}')
``` | output | 1 | 35,976 | 5 | 71,953 |
Provide a correct Python 3 solution for this coding contest problem.
Given a non-negative decimal integer $x$, convert it to binary representation $b$ of 32 bits. Then, print the result of the following operations to $b$ respecitvely.
* Inversion: change the state of each bit to the opposite state
* Logical left shift: shift left by 1
* Logical right shift: shift right by 1
Constraints
* $0 \leq x \leq 2^{32} - 1$
Input
The input is given in the following format.
$x$
Output
Print the given bits, results of inversion, left shift and right shift in a line respectively.
Examples
Input
8
Output
00000000000000000000000000001000
11111111111111111111111111110111
00000000000000000000000000010000
00000000000000000000000000000100
Input
13
Output
00000000000000000000000000001101
11111111111111111111111111110010
00000000000000000000000000011010
00000000000000000000000000000110 | instruction | 0 | 35,977 | 5 | 71,954 |
"Correct Solution:
```
x = int(input())
print('{:032b}'.format(x))
print('{:032b}'.format(~x & (2 ** 32 - 1)))
print('{:032b}'.format(x << 1 & (2 ** 32 - 1)))
print('{:032b}'.format(x >> 1))
``` | output | 1 | 35,977 | 5 | 71,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a non-negative decimal integer $x$, convert it to binary representation $b$ of 32 bits. Then, print the result of the following operations to $b$ respecitvely.
* Inversion: change the state of each bit to the opposite state
* Logical left shift: shift left by 1
* Logical right shift: shift right by 1
Constraints
* $0 \leq x \leq 2^{32} - 1$
Input
The input is given in the following format.
$x$
Output
Print the given bits, results of inversion, left shift and right shift in a line respectively.
Examples
Input
8
Output
00000000000000000000000000001000
11111111111111111111111111110111
00000000000000000000000000010000
00000000000000000000000000000100
Input
13
Output
00000000000000000000000000001101
11111111111111111111111111110010
00000000000000000000000000011010
00000000000000000000000000000110
Submitted Solution:
```
def out(n):
mask = 2**32 -1
print(format(n&mask,"032b"))
n = int(input())
out(n)
out(~n)
out(n<<1)
out(n>>1)
``` | instruction | 0 | 35,978 | 5 | 71,956 |
Yes | output | 1 | 35,978 | 5 | 71,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a non-negative decimal integer $x$, convert it to binary representation $b$ of 32 bits. Then, print the result of the following operations to $b$ respecitvely.
* Inversion: change the state of each bit to the opposite state
* Logical left shift: shift left by 1
* Logical right shift: shift right by 1
Constraints
* $0 \leq x \leq 2^{32} - 1$
Input
The input is given in the following format.
$x$
Output
Print the given bits, results of inversion, left shift and right shift in a line respectively.
Examples
Input
8
Output
00000000000000000000000000001000
11111111111111111111111111110111
00000000000000000000000000010000
00000000000000000000000000000100
Input
13
Output
00000000000000000000000000001101
11111111111111111111111111110010
00000000000000000000000000011010
00000000000000000000000000000110
Submitted Solution:
```
#!/usr/bin/env python3
# Bitset 1 - Bit Operation 1
def run():
n = int(input())
fmt = "{:032b}"
print(fmt.format(n))
print(fmt.format(~n + 2**32))
print(fmt.format(n << 1 & ~(1 << 32)))
print(fmt.format(n >> 1))
if __name__ == '__main__':
run()
``` | instruction | 0 | 35,979 | 5 | 71,958 |
Yes | output | 1 | 35,979 | 5 | 71,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a non-negative decimal integer $x$, convert it to binary representation $b$ of 32 bits. Then, print the result of the following operations to $b$ respecitvely.
* Inversion: change the state of each bit to the opposite state
* Logical left shift: shift left by 1
* Logical right shift: shift right by 1
Constraints
* $0 \leq x \leq 2^{32} - 1$
Input
The input is given in the following format.
$x$
Output
Print the given bits, results of inversion, left shift and right shift in a line respectively.
Examples
Input
8
Output
00000000000000000000000000001000
11111111111111111111111111110111
00000000000000000000000000010000
00000000000000000000000000000100
Input
13
Output
00000000000000000000000000001101
11111111111111111111111111110010
00000000000000000000000000011010
00000000000000000000000000000110
Submitted Solution:
```
x = int(input())
s = "{:032b}".format(x)
print(s)
print(s.replace("0","2").replace("1","0").replace("2","1"))
print(s[1:] + "0")
print("0" + s[:-1])
``` | instruction | 0 | 35,980 | 5 | 71,960 |
Yes | output | 1 | 35,980 | 5 | 71,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a non-negative decimal integer $x$, convert it to binary representation $b$ of 32 bits. Then, print the result of the following operations to $b$ respecitvely.
* Inversion: change the state of each bit to the opposite state
* Logical left shift: shift left by 1
* Logical right shift: shift right by 1
Constraints
* $0 \leq x \leq 2^{32} - 1$
Input
The input is given in the following format.
$x$
Output
Print the given bits, results of inversion, left shift and right shift in a line respectively.
Examples
Input
8
Output
00000000000000000000000000001000
11111111111111111111111111110111
00000000000000000000000000010000
00000000000000000000000000000100
Input
13
Output
00000000000000000000000000001101
11111111111111111111111111110010
00000000000000000000000000011010
00000000000000000000000000000110
Submitted Solution:
```
x = int(input())
bin_x = format(x, '032b')
bin_x1 = format(~x & 0b11111111111111111111111111111111, '032b')
bin_x2 = format(x << 1 & 0b11111111111111111111111111111111, '032b')
bin_x3 = format(x >> 1 & 0b11111111111111111111111111111111, '032b')
print(bin_x)
print(bin_x1)
print(bin_x2)
print(bin_x3)
``` | instruction | 0 | 35,981 | 5 | 71,962 |
Yes | output | 1 | 35,981 | 5 | 71,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Not long ago Billy came across such a problem, where there were given three natural numbers A, B and C from the range [1, N], and it was asked to check whether the equation AB = C is correct. Recently Billy studied the concept of a digital root of a number. We should remind you that a digital root d(x) of the number x is the sum s(x) of all the digits of this number, if s(x) ≤ 9, otherwise it is d(s(x)). For example, a digital root of the number 6543 is calculated as follows: d(6543) = d(6 + 5 + 4 + 3) = d(18) = 9. Billy has counted that the digital root of a product of numbers is equal to the digital root of the product of the factors' digital roots, i.e. d(xy) = d(d(x)d(y)). And the following solution to the problem came to his mind: to calculate the digital roots and check if this condition is met. However, Billy has doubts that this condition is sufficient. That's why he asks you to find out the amount of test examples for the given problem such that the algorithm proposed by Billy makes mistakes.
Input
The first line contains the only number N (1 ≤ N ≤ 106).
Output
Output one number — the amount of required A, B and C from the range [1, N].
Examples
Input
4
Output
2
Input
5
Output
6
Note
For the first sample the required triples are (3, 4, 3) and (4, 3, 3).
Submitted Solution:
```
n = int(input())
corr = 0
for i in range(1, n + 1):
corr += (n // i)
total = 0
count = [0] * 9
for i in range(1, n + 1):
count[i % 9] += 1
mult = [0]*9
for i in range(9):
for j in range(9):
mult[(i * j) % 9] += count[i] * count[j]
for i in range(9):
total += mult[i] * count[i]
print(total - corr)
``` | instruction | 0 | 36,054 | 5 | 72,108 |
Yes | output | 1 | 36,054 | 5 | 72,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Not long ago Billy came across such a problem, where there were given three natural numbers A, B and C from the range [1, N], and it was asked to check whether the equation AB = C is correct. Recently Billy studied the concept of a digital root of a number. We should remind you that a digital root d(x) of the number x is the sum s(x) of all the digits of this number, if s(x) ≤ 9, otherwise it is d(s(x)). For example, a digital root of the number 6543 is calculated as follows: d(6543) = d(6 + 5 + 4 + 3) = d(18) = 9. Billy has counted that the digital root of a product of numbers is equal to the digital root of the product of the factors' digital roots, i.e. d(xy) = d(d(x)d(y)). And the following solution to the problem came to his mind: to calculate the digital roots and check if this condition is met. However, Billy has doubts that this condition is sufficient. That's why he asks you to find out the amount of test examples for the given problem such that the algorithm proposed by Billy makes mistakes.
Input
The first line contains the only number N (1 ≤ N ≤ 106).
Output
Output one number — the amount of required A, B and C from the range [1, N].
Examples
Input
4
Output
2
Input
5
Output
6
Note
For the first sample the required triples are (3, 4, 3) and (4, 3, 3).
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
ans = 0
def cnt(rem):
if not rem:
return n//9
return (n-rem)//9 + 1 if n >= rem else 0
for i in range(9):
for j in range(9):
#number of k's -> 9*k+wanted <= n
wanted = i*j%9
cnta = cnt(i)
cntb = cnt(j)
cntc = cnt(wanted)
ans += cnta * cntb * cntc
for i in range(1, n+1):
ans -= n//i
print(ans)
``` | instruction | 0 | 36,055 | 5 | 72,110 |
Yes | output | 1 | 36,055 | 5 | 72,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Not long ago Billy came across such a problem, where there were given three natural numbers A, B and C from the range [1, N], and it was asked to check whether the equation AB = C is correct. Recently Billy studied the concept of a digital root of a number. We should remind you that a digital root d(x) of the number x is the sum s(x) of all the digits of this number, if s(x) ≤ 9, otherwise it is d(s(x)). For example, a digital root of the number 6543 is calculated as follows: d(6543) = d(6 + 5 + 4 + 3) = d(18) = 9. Billy has counted that the digital root of a product of numbers is equal to the digital root of the product of the factors' digital roots, i.e. d(xy) = d(d(x)d(y)). And the following solution to the problem came to his mind: to calculate the digital roots and check if this condition is met. However, Billy has doubts that this condition is sufficient. That's why he asks you to find out the amount of test examples for the given problem such that the algorithm proposed by Billy makes mistakes.
Input
The first line contains the only number N (1 ≤ N ≤ 106).
Output
Output one number — the amount of required A, B and C from the range [1, N].
Examples
Input
4
Output
2
Input
5
Output
6
Note
For the first sample the required triples are (3, 4, 3) and (4, 3, 3).
Submitted Solution:
```
def dig_root(num):
tot=0
if(num<10):
if(num%2==0):
return(num//2)
else:
return(num+1)
else:
while(num>0):
rem=num%10
tot+=rem
num=num//10
if(tot%2==0):
return(tot//2)
else:
return(tot+1)
num=int(input())
print(dig_root(num))
``` | instruction | 0 | 36,057 | 5 | 72,114 |
No | output | 1 | 36,057 | 5 | 72,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Not long ago Billy came across such a problem, where there were given three natural numbers A, B and C from the range [1, N], and it was asked to check whether the equation AB = C is correct. Recently Billy studied the concept of a digital root of a number. We should remind you that a digital root d(x) of the number x is the sum s(x) of all the digits of this number, if s(x) ≤ 9, otherwise it is d(s(x)). For example, a digital root of the number 6543 is calculated as follows: d(6543) = d(6 + 5 + 4 + 3) = d(18) = 9. Billy has counted that the digital root of a product of numbers is equal to the digital root of the product of the factors' digital roots, i.e. d(xy) = d(d(x)d(y)). And the following solution to the problem came to his mind: to calculate the digital roots and check if this condition is met. However, Billy has doubts that this condition is sufficient. That's why he asks you to find out the amount of test examples for the given problem such that the algorithm proposed by Billy makes mistakes.
Input
The first line contains the only number N (1 ≤ N ≤ 106).
Output
Output one number — the amount of required A, B and C from the range [1, N].
Examples
Input
4
Output
2
Input
5
Output
6
Note
For the first sample the required triples are (3, 4, 3) and (4, 3, 3).
Submitted Solution:
```
def digr(n):
x = sum(int(dig) for dig in str(n))
if x<10:
return x
else:
return digr(x)
n=int(input())
d=0
for i in range(1,n+1):
for j in range(i,n+1):
c=digr(digr(i)*digr(j))
if c<n+1 and c!=i*j:
if i!=j:
d=d+2
else:
d=d+1
print(d)
``` | instruction | 0 | 36,058 | 5 | 72,116 |
No | output | 1 | 36,058 | 5 | 72,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Not long ago Billy came across such a problem, where there were given three natural numbers A, B and C from the range [1, N], and it was asked to check whether the equation AB = C is correct. Recently Billy studied the concept of a digital root of a number. We should remind you that a digital root d(x) of the number x is the sum s(x) of all the digits of this number, if s(x) ≤ 9, otherwise it is d(s(x)). For example, a digital root of the number 6543 is calculated as follows: d(6543) = d(6 + 5 + 4 + 3) = d(18) = 9. Billy has counted that the digital root of a product of numbers is equal to the digital root of the product of the factors' digital roots, i.e. d(xy) = d(d(x)d(y)). And the following solution to the problem came to his mind: to calculate the digital roots and check if this condition is met. However, Billy has doubts that this condition is sufficient. That's why he asks you to find out the amount of test examples for the given problem such that the algorithm proposed by Billy makes mistakes.
Input
The first line contains the only number N (1 ≤ N ≤ 106).
Output
Output one number — the amount of required A, B and C from the range [1, N].
Examples
Input
4
Output
2
Input
5
Output
6
Note
For the first sample the required triples are (3, 4, 3) and (4, 3, 3).
Submitted Solution:
```
def main():
ans = 0
a = [0] * 10
n = int(input())
for i in range (1, 9):
a[i] = n // 9 + int(n % 9 >= i)
a[9] = n // 9
for i in range(1, 10):
for j in range(1, 10):
k = i * j % 9
if k == 0:
k = 9
ans += a[i] * a[j] * a[k]
for i in range(1, n):
ans -= n // i
print(ans)
return
if __name__ == "__main__":
main()
``` | instruction | 0 | 36,059 | 5 | 72,118 |
No | output | 1 | 36,059 | 5 | 72,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Not long ago Billy came across such a problem, where there were given three natural numbers A, B and C from the range [1, N], and it was asked to check whether the equation AB = C is correct. Recently Billy studied the concept of a digital root of a number. We should remind you that a digital root d(x) of the number x is the sum s(x) of all the digits of this number, if s(x) ≤ 9, otherwise it is d(s(x)). For example, a digital root of the number 6543 is calculated as follows: d(6543) = d(6 + 5 + 4 + 3) = d(18) = 9. Billy has counted that the digital root of a product of numbers is equal to the digital root of the product of the factors' digital roots, i.e. d(xy) = d(d(x)d(y)). And the following solution to the problem came to his mind: to calculate the digital roots and check if this condition is met. However, Billy has doubts that this condition is sufficient. That's why he asks you to find out the amount of test examples for the given problem such that the algorithm proposed by Billy makes mistakes.
Input
The first line contains the only number N (1 ≤ N ≤ 106).
Output
Output one number — the amount of required A, B and C from the range [1, N].
Examples
Input
4
Output
2
Input
5
Output
6
Note
For the first sample the required triples are (3, 4, 3) and (4, 3, 3).
Submitted Solution:
```
n=int(input())
dct={}
m=[]
def root(num):
if(num<=9):
return num
if(num in dct.keys()):
m.append(0)
return dct[num]
s=str(num)
sm=0
for i in s:
sm=sm+int(i)
dct[num]=root(sm)
return dct[num]
dct1={}
dctn={}
for i in range(1,n+1):
for j in range(1,n+1):
x=root(i*j)
if(x in dct1.keys()):
if(i==1):
dctn[x].append(j)
dct1[x].append(i*j)
else:
if(i==1):
dctn[x]=[j]
dct1[x]=[i*j]
s=0
# print(len(m))
# print(dctn)
# print(dct1)
for i in dctn.keys():
l=len(dctn[i])
for j in dct1[i]:
if(j not in dctn[i]):
s=s+l
print(s)
``` | instruction | 0 | 36,060 | 5 | 72,120 |
No | output | 1 | 36,060 | 5 | 72,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer x of n digits a_1, a_2, …, a_n, which make up its decimal notation in order from left to right.
Also, you are given a positive integer k < n.
Let's call integer b_1, b_2, …, b_m beautiful if b_i = b_{i+k} for each i, such that 1 ≤ i ≤ m - k.
You need to find the smallest beautiful integer y, such that y ≥ x.
Input
The first line of input contains two integers n, k (2 ≤ n ≤ 200 000, 1 ≤ k < n): the number of digits in x and k.
The next line of input contains n digits a_1, a_2, …, a_n (a_1 ≠ 0, 0 ≤ a_i ≤ 9): digits of x.
Output
In the first line print one integer m: the number of digits in y.
In the next line print m digits b_1, b_2, …, b_m (b_1 ≠ 0, 0 ≤ b_i ≤ 9): digits of y.
Examples
Input
3 2
353
Output
3
353
Input
4 2
1234
Output
4
1313
Submitted Solution:
```
n,m = map(int,input().split())
s = input()
s1 = s[:m]
s2 = ""
for i in range(n):
s2+= s1[i%m]
print(n)
if s2>=s:
print(s2)
else:
j = m-1
while s1[j]=='9':
j-=1
s1= s1[:j]+chr(ord(s1[j])+1)
s1+= '0'*(m-len(s1))
s2=""
for i in range(n):
s2+= s1[i%m]
print(s2)
``` | instruction | 0 | 36,183 | 5 | 72,366 |
Yes | output | 1 | 36,183 | 5 | 72,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer x of n digits a_1, a_2, …, a_n, which make up its decimal notation in order from left to right.
Also, you are given a positive integer k < n.
Let's call integer b_1, b_2, …, b_m beautiful if b_i = b_{i+k} for each i, such that 1 ≤ i ≤ m - k.
You need to find the smallest beautiful integer y, such that y ≥ x.
Input
The first line of input contains two integers n, k (2 ≤ n ≤ 200 000, 1 ≤ k < n): the number of digits in x and k.
The next line of input contains n digits a_1, a_2, …, a_n (a_1 ≠ 0, 0 ≤ a_i ≤ 9): digits of x.
Output
In the first line print one integer m: the number of digits in y.
In the next line print m digits b_1, b_2, …, b_m (b_1 ≠ 0, 0 ≤ b_i ≤ 9): digits of y.
Examples
Input
3 2
353
Output
3
353
Input
4 2
1234
Output
4
1313
Submitted Solution:
```
n, k = map(int, input().split())
s = input()
x=s[:k]*(n//k)+s[:n%k]
print(n)
if s<=x:
print(x)
else:
v=str(int(s[:k])+1)
x=v*(n//k)+v[:n%k]
print(x)
``` | instruction | 0 | 36,185 | 5 | 72,370 |
Yes | output | 1 | 36,185 | 5 | 72,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer x of n digits a_1, a_2, …, a_n, which make up its decimal notation in order from left to right.
Also, you are given a positive integer k < n.
Let's call integer b_1, b_2, …, b_m beautiful if b_i = b_{i+k} for each i, such that 1 ≤ i ≤ m - k.
You need to find the smallest beautiful integer y, such that y ≥ x.
Input
The first line of input contains two integers n, k (2 ≤ n ≤ 200 000, 1 ≤ k < n): the number of digits in x and k.
The next line of input contains n digits a_1, a_2, …, a_n (a_1 ≠ 0, 0 ≤ a_i ≤ 9): digits of x.
Output
In the first line print one integer m: the number of digits in y.
In the next line print m digits b_1, b_2, …, b_m (b_1 ≠ 0, 0 ≤ b_i ≤ 9): digits of y.
Examples
Input
3 2
353
Output
3
353
Input
4 2
1234
Output
4
1313
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
n,k=map(int,input().split())
a=list(map(int,input().rstrip()))
f=0
for i in range(k, n):
if a[i] != a[i - k]:
if a[i]>a[i-k]:
f=1
break
if f:
for i in range(k-1,-1,-1):
if a[i]!=9:
a[i]+=1
break
for i in range(k, n):
if a[i] != a[i - k]:
a[i]=a[i-k]
print(n)
print(*a,sep="")
# 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()
``` | instruction | 0 | 36,186 | 5 | 72,372 |
No | output | 1 | 36,186 | 5 | 72,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer x of n digits a_1, a_2, …, a_n, which make up its decimal notation in order from left to right.
Also, you are given a positive integer k < n.
Let's call integer b_1, b_2, …, b_m beautiful if b_i = b_{i+k} for each i, such that 1 ≤ i ≤ m - k.
You need to find the smallest beautiful integer y, such that y ≥ x.
Input
The first line of input contains two integers n, k (2 ≤ n ≤ 200 000, 1 ≤ k < n): the number of digits in x and k.
The next line of input contains n digits a_1, a_2, …, a_n (a_1 ≠ 0, 0 ≤ a_i ≤ 9): digits of x.
Output
In the first line print one integer m: the number of digits in y.
In the next line print m digits b_1, b_2, …, b_m (b_1 ≠ 0, 0 ≤ b_i ≤ 9): digits of y.
Examples
Input
3 2
353
Output
3
353
Input
4 2
1234
Output
4
1313
Submitted Solution:
```
nk=input().split()
n=int(nk[0])
k=int(nk[1])
a1=input()
a=[x for x in a1]
y=[0 for x in range(n)]
firstindex=[]
count=0
change=0
done={}
for x in range(len(a)):
try:
if done[x]==0:
continue
else:
y[x]=a[x]
if x+k<len(a):
y[x+k]=a[x]
except:
y[x]=a[x]
if x+k<len(a):
y[x+k]=a[x]
done[x]=0
done[x+k]=0
if y==a:
print(n)
print(a1)
elif k==n-1:
count=1
for x in range(n-2,0,-1):
if int(y[x])<9:
count+=1
change=int(y[x])+1
break
else:
count+=1
y[-count]=change
print(n)
for x in y:
print(x,end='')
else:
for x in y[::-1]:
if int(x)<9:
count+=1
change=int(x)+1
break
else:
count+=1
y[-count]=change
y[-count-k]=change
print(n)
for x in y:
print(x,end='')
``` | instruction | 0 | 36,187 | 5 | 72,374 |
No | output | 1 | 36,187 | 5 | 72,375 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer x of n digits a_1, a_2, …, a_n, which make up its decimal notation in order from left to right.
Also, you are given a positive integer k < n.
Let's call integer b_1, b_2, …, b_m beautiful if b_i = b_{i+k} for each i, such that 1 ≤ i ≤ m - k.
You need to find the smallest beautiful integer y, such that y ≥ x.
Input
The first line of input contains two integers n, k (2 ≤ n ≤ 200 000, 1 ≤ k < n): the number of digits in x and k.
The next line of input contains n digits a_1, a_2, …, a_n (a_1 ≠ 0, 0 ≤ a_i ≤ 9): digits of x.
Output
In the first line print one integer m: the number of digits in y.
In the next line print m digits b_1, b_2, …, b_m (b_1 ≠ 0, 0 ≤ b_i ≤ 9): digits of y.
Examples
Input
3 2
353
Output
3
353
Input
4 2
1234
Output
4
1313
Submitted Solution:
```
n,k=map(int,input().split())
s=list(input().strip())
i=0
l=[-1]*n
while i<k-1:
j=i
while j<n:
l[j]=s[i]
j+=k
i+=1
j=i
ll=set()
while j<n:
ll.add(int(s[j]))
j+=k
#print(ll,i)
ll=list(ll)
ll.sort()
if(len(ll)==1):
x=ll[0]
else:
mx=max(ll)
if(mx==int(s[i])):
x=mx
else:
x=int(s[i])+1
j=i
while j<n:
l[j]=str(x)
j+=k
#print(l)
for i in range(n):
if(l[i]==-1):
l[i]=s[i]
s="".join(l)
print(len(s))
print(s)
``` | instruction | 0 | 36,188 | 5 | 72,376 |
No | output | 1 | 36,188 | 5 | 72,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A continued fraction of height n is a fraction of form <image>. You are given two rational numbers, one is represented as <image> and the other one is represented as a finite fraction of height n. Check if they are equal.
Input
The first line contains two space-separated integers p, q (1 ≤ q ≤ p ≤ 1018) — the numerator and the denominator of the first fraction.
The second line contains integer n (1 ≤ n ≤ 90) — the height of the second fraction. The third line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1018) — the continued fraction.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
Print "YES" if these fractions are equal and "NO" otherwise.
Examples
Input
9 4
2
2 4
Output
YES
Input
9 4
3
2 3 1
Output
YES
Input
9 4
3
1 2 4
Output
NO
Note
In the first sample <image>.
In the second sample <image>.
In the third sample <image>.
Submitted Solution:
```
# some math vlad
p,q=map(int,input().split())
n=int(input())
a=list(map(int,input().split()))
x,y=a[n-1],1
for i in range(n-1)[::-1]:x,y=a[i]*x+y,x
print (['NO','YES'][p*y==q*x])
``` | instruction | 0 | 36,420 | 5 | 72,840 |
Yes | output | 1 | 36,420 | 5 | 72,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A continued fraction of height n is a fraction of form <image>. You are given two rational numbers, one is represented as <image> and the other one is represented as a finite fraction of height n. Check if they are equal.
Input
The first line contains two space-separated integers p, q (1 ≤ q ≤ p ≤ 1018) — the numerator and the denominator of the first fraction.
The second line contains integer n (1 ≤ n ≤ 90) — the height of the second fraction. The third line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1018) — the continued fraction.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
Print "YES" if these fractions are equal and "NO" otherwise.
Examples
Input
9 4
2
2 4
Output
YES
Input
9 4
3
2 3 1
Output
YES
Input
9 4
3
1 2 4
Output
NO
Note
In the first sample <image>.
In the second sample <image>.
In the third sample <image>.
Submitted Solution:
```
#by: Liger Zero
p,q = input().split()
p = int(p)
q = int(q)
n = int(input())
A = input().split()
a = 1
b = int(A[n-1])
i = n-2
while( i >= 0 ):
aux = b
b = a + b*int(A[i])
a = aux
i = i-1
print ("YES" if p*a == q*b else "NO" )
``` | instruction | 0 | 36,421 | 5 | 72,842 |
Yes | output | 1 | 36,421 | 5 | 72,843 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A continued fraction of height n is a fraction of form <image>. You are given two rational numbers, one is represented as <image> and the other one is represented as a finite fraction of height n. Check if they are equal.
Input
The first line contains two space-separated integers p, q (1 ≤ q ≤ p ≤ 1018) — the numerator and the denominator of the first fraction.
The second line contains integer n (1 ≤ n ≤ 90) — the height of the second fraction. The third line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1018) — the continued fraction.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
Print "YES" if these fractions are equal and "NO" otherwise.
Examples
Input
9 4
2
2 4
Output
YES
Input
9 4
3
2 3 1
Output
YES
Input
9 4
3
1 2 4
Output
NO
Note
In the first sample <image>.
In the second sample <image>.
In the third sample <image>.
Submitted Solution:
```
p,q = list(map(float, input().split()))
input()
l = list(map(float, input().split()))
val = p/q
sum = 0
for n in l[::-1]:
sum += n
sum = 1/sum
if 1/sum == val:
print("YES")
else:
print("NO")
``` | instruction | 0 | 36,422 | 5 | 72,844 |
No | output | 1 | 36,422 | 5 | 72,845 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A continued fraction of height n is a fraction of form <image>. You are given two rational numbers, one is represented as <image> and the other one is represented as a finite fraction of height n. Check if they are equal.
Input
The first line contains two space-separated integers p, q (1 ≤ q ≤ p ≤ 1018) — the numerator and the denominator of the first fraction.
The second line contains integer n (1 ≤ n ≤ 90) — the height of the second fraction. The third line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1018) — the continued fraction.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
Print "YES" if these fractions are equal and "NO" otherwise.
Examples
Input
9 4
2
2 4
Output
YES
Input
9 4
3
2 3 1
Output
YES
Input
9 4
3
1 2 4
Output
NO
Note
In the first sample <image>.
In the second sample <image>.
In the third sample <image>.
Submitted Solution:
```
p, q = map(int, input().split())
n, t = int(input()), list(map(int, input().split()))
t.reverse()
a, b = 1, t[0]
for c in t[1: ]:
a, b = b, c * b + a
print(a, b)
print('YNEOS'[a * p != b * q :: 2])
``` | instruction | 0 | 36,423 | 5 | 72,846 |
No | output | 1 | 36,423 | 5 | 72,847 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A continued fraction of height n is a fraction of form <image>. You are given two rational numbers, one is represented as <image> and the other one is represented as a finite fraction of height n. Check if they are equal.
Input
The first line contains two space-separated integers p, q (1 ≤ q ≤ p ≤ 1018) — the numerator and the denominator of the first fraction.
The second line contains integer n (1 ≤ n ≤ 90) — the height of the second fraction. The third line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1018) — the continued fraction.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
Print "YES" if these fractions are equal and "NO" otherwise.
Examples
Input
9 4
2
2 4
Output
YES
Input
9 4
3
2 3 1
Output
YES
Input
9 4
3
1 2 4
Output
NO
Note
In the first sample <image>.
In the second sample <image>.
In the third sample <image>.
Submitted Solution:
```
p,q=list(map(int,input().split()))
n=int(input())
num=list(map(int,input().split()))
frac=1
for i in reversed(range(n)):
if i==len(num)-1:
frac=num[i]
else:
frac=num[i]+1/(frac)
if abs(p/q-frac)<=1e-10:
print("YES")
else:
print("NO")
``` | instruction | 0 | 36,424 | 5 | 72,848 |
No | output | 1 | 36,424 | 5 | 72,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A continued fraction of height n is a fraction of form <image>. You are given two rational numbers, one is represented as <image> and the other one is represented as a finite fraction of height n. Check if they are equal.
Input
The first line contains two space-separated integers p, q (1 ≤ q ≤ p ≤ 1018) — the numerator and the denominator of the first fraction.
The second line contains integer n (1 ≤ n ≤ 90) — the height of the second fraction. The third line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1018) — the continued fraction.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
Print "YES" if these fractions are equal and "NO" otherwise.
Examples
Input
9 4
2
2 4
Output
YES
Input
9 4
3
2 3 1
Output
YES
Input
9 4
3
1 2 4
Output
NO
Note
In the first sample <image>.
In the second sample <image>.
In the third sample <image>.
Submitted Solution:
```
p, q = map(int, input().split())
n, t = int(input()), list(map(int, input().split()))
t.reverse()
a, b = 1, t[0]
for c in t[1: ]:
a, b = b, c * b + a
print('YNEOS'[(a, b) != (q, p) :: 2])
``` | instruction | 0 | 36,425 | 5 | 72,850 |
No | output | 1 | 36,425 | 5 | 72,851 |
Provide a correct Python 3 solution for this coding contest problem.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1 | instruction | 0 | 36,667 | 5 | 73,334 |
"Correct Solution:
```
def gn(n,k):
k -= 1
r = k+2*n+n//2 - (1 + 2*k + 2*n)
if r < 0:
print(-1)
return
print(k+1, 2*n+k, 3*n+k)
n-=1
k+=1
for i in range(1, n+1):
x = i
y = n+(i-((n+1)//2)+n)%n+1
print(k+x, k+y, 2*k+x+y+r)
def kn(n,k):
k -= 1
r = k+2*n+(n+1)//2 - (1 + 2*k + 2*n)
if r < 0:
print(-1)
return
for i in range(1, n+1):
x = i
y = n+(i-((n+1)//2)+n)%n+1
print(k+x, k+y, 2*k+x+y+r)
N, K = map(int, input().split())
if N%2:
kn(N,K)
else:
gn(N,K)
``` | output | 1 | 36,667 | 5 | 73,335 |
Provide a correct Python 3 solution for this coding contest problem.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1 | instruction | 0 | 36,668 | 5 | 73,336 |
"Correct Solution:
```
import math
N, K = map(int,input().split())
if (2*N)*(K+ K+2*N-1) > N*(K+2*N + K+3*N-1):
print(-1)
exit()
flag = [1]*(2*N)
for i in range(math.floor(N/2),N):
a = K + i
b = K + 2*N -1 -(i-math.floor(N/2))*2
c = K + 3*N -1 -(i-math.floor(N/2))
print(a,b,c)
for i in range(math.floor(N/2)):
a = K + i
b = K + 2*N -2 -i*2
c = K + 3*N -1 -i-math.ceil(N/2)
print(a,b,c)
``` | output | 1 | 36,668 | 5 | 73,337 |
Provide a correct Python 3 solution for this coding contest problem.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1 | instruction | 0 | 36,669 | 5 | 73,338 |
"Correct Solution:
```
n,k=map(int,input().split())
sum1=sum(range(k,k+2*n))
sum2=sum(range(k+2*n,k+3*n))
if sum1<=sum2 and k-1<=n//2:
arr1=set([k+i for i in range(n)])
arr2=set([k+n+i for i in range(n)])
ans1=[]
for i in range(k-1):
ans1.append([k+i,k+2*n-2-2*i,2*k+2*n-2-i])
arr1.discard(k+i)
arr2.discard(k+2*n-2-2*i)
arr1=list(arr1)
arr2=list(arr2)
arr2=sorted(arr2,reverse=True)
ans2=[]
for i in range(n-(k-1)):
ans2.append([arr1[i],arr2[i],arr1[i]+arr2[i]])
ans2=sorted(ans2,key=lambda x:x[2])
for i in range(1,n-(k-1)):
if ans2[i][2]<=ans2[i-1][2]:
ans2[i][2]=ans2[i-1][2]+1
ans=ans1+ans2
for a,b,c in ans:
print(a,b,c)
else:
print(-1)
``` | output | 1 | 36,669 | 5 | 73,339 |
Provide a correct Python 3 solution for this coding contest problem.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1 | instruction | 0 | 36,670 | 5 | 73,340 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
from itertools import accumulate
from collections import Counter
from bisect import bisect as br, bisect_left as bl
class PMS:
#1-indexed
def __init__(self, A, B, issum = False):
#Aに初期状態の要素をすべて入れる,Bは値域のリスト
self.X, self.comp = self.compress(B)
self.size = len(self.X)
self.tree = [0] * (self.size + 1)
self.p = 2**(self.size.bit_length() - 1)
self.dep = self.size.bit_length()
CA = Counter(A)
S = [0] + list(accumulate([CA[self.X[i]] for i in range(self.size)]))
for i in range(1, 1+self.size):
self.tree[i] = S[i] - S[i - (i&-i)]
if issum:
self.sumtree = [0] * (self.size + 1)
Ssum = [0] + list(accumulate([CA[self.X[i]]*self.X[i] for i in range(self.size)]))
for i in range(1, 1+self.size):
self.sumtree[i] = Ssum[i] - Ssum[i - (i&-i)]
def compress(self, L):
#座圧
L2 = list(set(L))
L2.sort()
C = {v : k for k, v in enumerate(L2, 1)}
# 1-indexed
return L2, C
def leng(self):
#今入っている個数を取得
return self.count(self.X[-1])
def count(self, v):
#v(Bの元)以下の個数を取得
i = self.comp[v]
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def less(self, v):
#v(Bの元である必要はない)未満の個数を取得
i = bl(self.X, v)
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def leq(self, v):
#v(Bの元である必要はない)以下の個数を取得
i = br(self.X, v)
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, v, x):
#vをx個入れる,負のxで取り出す,iの個数以上取り出すとエラーを出さずにバグる
i = self.comp[v]
while i <= self.size:
self.tree[i] += x
i += i & -i
def get(self, i):
# i番目の値を取得
if i <= 0:
return -1
s = 0
k = self.p
for _ in range(self.dep):
if s + k <= self.size and self.tree[s+k] < i:
s += k
i -= self.tree[s]
k //= 2
return self.X[s]
def gets(self, v):
#累積和がv以下となる最大のindexを返す
v1 = v
s = 0
k = self.p
for _ in range(self.dep):
if s + k <= self.size and self.sumtree[s+k] < v:
s += k
v -= self.sumtree[s]
k //= 2
if s == self.size:
return self.leng()
return self.count(self.X[s]) + (v1 - self.countsum(self.X[s]))//self.X[s]
def addsum(self, i, x):
#sumを扱いたいときにaddの代わりに使う
self.add(i, x)
x *= i
i = self.comp[i]
while i <= self.size:
self.sumtree[i] += x
i += i & -i
def countsum(self, v):
#v(Bの元)以下のsumを取得
i = self.comp[v]
s = 0
while i > 0:
s += self.sumtree[i]
i -= i & -i
return s
def getsum(self, i):
#i番目までのsumを取得
x = self.get(i)
return self.countsum(x) - x*(self.count(x) - i)
N, K = map(int, readline().split())
C = list(range(K+2*N, K+3*N))
A = list(range(K, K+N))
B = list(range(K+N, K+2*N))
Ans = []
ans = 1
BB = PMS(B, B)
if sum(A) + sum(B) > sum(C):
ans = -1
else:
Ao = [A[i] for i in range(N) if A[i]%2]
Ae = [A[i] for i in range(N) if not A[i]%2]
if len(Ao) < len(Ae):
AA = Ae + Ao
else:
AA = Ao + Ae
for a, c in zip(AA, C):
bx = c - a
k = BB.leq(bx)
if k == 0:
ans = -1
break
b = BB.get(k)
Ans.append((a, b, c))
BB.add(b, -1)
if ans != -1:
print('\n'.join('{} {} {}'.format(*an) for an in Ans))
else:
print(ans)
``` | output | 1 | 36,670 | 5 | 73,341 |
Provide a correct Python 3 solution for this coding contest problem.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1 | instruction | 0 | 36,671 | 5 | 73,342 |
"Correct Solution:
```
import sys
n,k=map(int,input().split())
if 2*k-1>n:
print(-1)
sys.exit()
a=[i for i in range(k,k+n)]
A=[]
for i in range(len(a)):
if i%2==0:
A.append(a[i])
for i in range(len(a)):
if i%2==1:
A.append(a[i])
b1=[i for i in range(k+n,k+n+(n+1)//2)]
b2=[i for i in range(k+n+(n+1)//2,k+2*n)]
b1.reverse()
b2.reverse()
B=b1+b2
C=[i for i in range(k+2*n,k+3*n)]
for i in range(n):
print(A[i],B[i],C[i])
``` | output | 1 | 36,671 | 5 | 73,343 |
Provide a correct Python 3 solution for this coding contest problem.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1 | instruction | 0 | 36,672 | 5 | 73,344 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
def solve():
N, K = map(int,input().split())
if K > (N+1)/2:
return '-1'
L = [[K,K+N,K+2*N] for i in range(N)]
for i in range(N):
if N%2:
j = (N-1-2*i)%N
k = (2*i)%N
else:
j = (N-1-2*i)%N-i//(N//2)
k = (1+2*i)%N-i//(N//2)
L[i][0] += i
L[j][1] += i
L[k][2] += i
res = '\n'.join(' '.join(map(str, l)) for l in L)
return res
if __name__ == '__main__':
print(solve())
``` | output | 1 | 36,672 | 5 | 73,345 |
Provide a correct Python 3 solution for this coding contest problem.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1 | instruction | 0 | 36,673 | 5 | 73,346 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
N,K=map(int,input().split())
ANS=[]
if N%2==1:
for i in range(N//2+1):
ANS.append((K+i,K+2*N-1-2*i,K+3*N-1-i-N//2))
for i in range(N//2):
ANS.append((K+N//2+1+i,K+2*N-2-2*i,K+3*N-1-i))
for x,y,z in ANS:
if x+y>z:
print(-1)
sys.exit()
for x,y,z in ANS:
print(x,y,z)
else:
for i in range(N//2):
ANS.append((K+i,K+2*N-1-2*i,K+3*N-1-i-N//2))
for i in range(N//2):
ANS.append((K+N//2+i,K+2*N-2-2*i,K+3*N-1-i))
for x,y,z in ANS:
if x+y>z:
print(-1)
sys.exit()
for x,y,z in ANS:
print(x,y,z)
``` | output | 1 | 36,673 | 5 | 73,347 |
Provide a correct Python 3 solution for this coding contest problem.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1 | instruction | 0 | 36,674 | 5 | 73,348 |
"Correct Solution:
```
n, k = map(int, input().split())
if 2*k-1 > n:
print(-1)
exit()
l = (n+1)//2
if n % 2 != 0:
a = []
b = []
c = []
for x in range(k, k+2*l-2+1,2):
a.append(x)
for x in range(k+1, k+2*l-2, 2):
a.append(x)
for x in range(k+3*l-2, k+2*l-2, -1):
b.append(x)
for x in range(k+4*l-3, k+3*l-2, -1):
b.append(x)
for x in range(k+4*l-2, k+6*l-3):
c.append(x)
for i in range(n):
print(a[i], b[i], c[i])
else:
a = []
b = []
c = []
for x in range(k, k+2*l-2+1, 2):
a.append(x)
for x in range(k+1, k+2*l, 2):
a.append(x)
for x in range(k+3*l-1, k+2*l-1, -1):
b.append(x)
for x in range(k+4*l-1, k+3*l-1, -1):
b.append(x)
for x in range(k+4*l, k+6*l):
c.append(x)
for i in range(n):
print(a[i], b[i], c[i])
``` | output | 1 | 36,674 | 5 | 73,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1
Submitted Solution:
```
import sys
N,K = map(int,input().split())
if 2*K-1 > N:
print (-1)
sys.exit()
ans = []
nc = K+2*N + (N+1) // 2
na = K+1
nb = K + 2*N -1
for i in range(N // 2):
ans.append([na,nb,nc])
na += 2
nb -= 1
nc += 1
na = K
nc = K + 2*N
for i in range((N+1) // 2):
ans.append([na,nb,nc])
na += 2
nb -= 1
nc += 1
for i in ans:
print (i[0],i[1],i[2])
``` | instruction | 0 | 36,675 | 5 | 73,350 |
Yes | output | 1 | 36,675 | 5 | 73,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1
Submitted Solution:
```
import sys
sys.setrecursionlimit(2147483647)
INF=float("inf")
MOD=10**9+7
input=lambda :sys.stdin.readline().rstrip()
def resolve():
n,k=map(int,input().split())
if(2*k>n+1):
print(-1)
return
h=(2*k+3*n-1)//2
for i in range(0,(n+1)//2):
print(k+2*i,h-i,k+2*n+i)
for i in range(0,n//2):
print(k+1+i*2,k+2*n-1-i,k+2*n+((n-1)//2)+i+1)
resolve()
``` | instruction | 0 | 36,676 | 5 | 73,352 |
Yes | output | 1 | 36,676 | 5 | 73,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1
Submitted Solution:
```
N,K=map(int,input().split());print(-1 if 2*K-1>N else"\n".join("%d %d %d"%(((N+1)//2-k-1)%N+K,N+2*k%N+K+(2*k>=N)*(N+1)%2,2*N+k+K)for k in range(N)))
``` | instruction | 0 | 36,677 | 5 | 73,354 |
Yes | output | 1 | 36,677 | 5 | 73,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1
Submitted Solution:
```
def solve(n, k):
s1 = n * (k + (k + 2 * n - 1))
s2 = n * ((k + 2 * n) + (k + 3 * n - 1)) // 2
res = []
if s1 > s2:
return None
if n % 2 == 1:
m = ((n + 1) // 4) * 2
for i in range(m):
a = k + n - 1 - i
b = k + n + i + (m - 1 - 2 * i)
c = k + 2 * n + (n - 1) // 2 + (m - 1 - 2 * i)
res.append([a, b, c])
# print(1, a, b, c)
for i in range(n - m):
a = k + n - 1 - m - i
b = k + 2 * n - 1 - i
c = k + 2 * n + (n - 1) // 2 + 2 * ((n - m) // 2 - i)
res.append([a, b, c])
# print(2, a, b, c)
return res
else:
m = n // 2
for i in range(m):
a = k + n - 1 - i
b = k + n + i + (m - 1 - 2 * i)
c = k + 2 * n + m + (m - 2 - 2 * i)
res.append([a, b, c])
# print(1, a, b, c)
for i in range(n - m):
a = k + n - 1 - m - i
b = k + 2 * n - 1 - i
c = k + 3 * n - 1 - 2 * i
res.append([a, b, c])
# print(2, a, b, c)
return res
def main():
n, k = map(int, input().split())
res = solve(n, k)
if res is None:
print(-1)
else:
for r in res:
print(*r)
if __name__ == "__main__":
main()
``` | instruction | 0 | 36,678 | 5 | 73,356 |
Yes | output | 1 | 36,678 | 5 | 73,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1
Submitted Solution:
```
import bisect
n,k=map(int,input().split())
if n*(k-1)>n*(n-1)//2:
print(-1)
exit()
a=k
c=k+2*n
b=c-a
z=set(range(k,k+2*n))
ans=[]
for i in range(n):
bi=bisect.bisect_right(list(z),b)
if bi>0:
if list(z)[bi-1] in z:
print(a,list(z)[bi-1],c)
z.remove(list(z)[bi-1])
else:
a+=1
continue
c+=1
a+=1
b=c-a
``` | instruction | 0 | 36,679 | 5 | 73,358 |
No | output | 1 | 36,679 | 5 | 73,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given are positive integers N and K.
Determine if the 3N integers K, K+1, ..., K+3N-1 can be partitioned into N triples (a_1,b_1,c_1), ..., (a_N,b_N,c_N) so that the condition below is satisfied. Any of the integers K, K+1, ..., K+3N-1 must appear in exactly one of those triples.
* For every integer i from 1 to N, a_i + b_i \leq c_i holds.
If the answer is yes, construct one such partition.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq K \leq 10^9
Input
Input is given from Standard Input in the following format:
N K
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Output
If it is impossible to partition the integers satisfying the condition, print `-1`. If it is possible, print N triples in the following format:
a_1 b_1 c_1
:
a_N b_N c_N
Examples
Input
1 1
Output
1 2 3
Input
3 3
Output
-1
Submitted Solution:
```
N, K = map(int, input().split())
if N < 2*K - 1:
print(-1)
else:
A = [a for a in range(K, K+N, 2)]
A += [a for a in range(K+1, K+N, 2)]
B = [b for b in range(K+2*N-N//2-1, K+N-1, -1)]
B += [b for b in range(K+2*N-N//2, K+2*N)]
C = [c for c in range(K+2*N, K+3*N)]
[print(*p) for p in list(map(list, zip(A, B, C)))]
``` | instruction | 0 | 36,680 | 5 | 73,360 |
No | output | 1 | 36,680 | 5 | 73,361 |
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