message stringlengths 2 59.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 37 108k | cluster float64 20 20 | __index_level_0__ int64 74 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13 | instruction | 0 | 5,285 | 20 | 10,570 |
Tags: *special, combinatorics
Correct Solution:
```
a=int(input())
print(a*(a-1)*6+1)
``` | output | 1 | 5,285 | 20 | 10,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13 | instruction | 0 | 5,286 | 20 | 10,572 |
Tags: *special, combinatorics
Correct Solution:
```
a = int(input())
res = 1
for i in range(2, a + 1):
res += (i - 1) * 12
print(res)
``` | output | 1 | 5,286 | 20 | 10,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13 | instruction | 0 | 5,287 | 20 | 10,574 |
Tags: *special, combinatorics
Correct Solution:
```
"""====================================================================================
====================================================================================
___ _______ ___ _______ ___ ___
| /\ | | \ | | / | | | | |\ /|
| / \ | | \ | | / | | | | | \ / |
|___ /____\ | | \ | |/ |___| | | | \/ |
| / \ | | / | |\ |\ | | | |
| / \ | | / | | \ | \ | | | |
___|/ \___|___ |___/ ___|___ | \ | \ |___| | |
====================================================================================
====================================================================================
"""
# ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥
n = int(input())
a = 1
s = 12
for i in range(n-1):
a += s
s += 12
print(a)
# ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥
"""====================================================================================
====================================================================================
___ _______ ___ _______ ___ ___
| /\ | | \ | | / | | | | |\ /|
| / \ | | \ | | / | | | | | \ / |
|___ /____\ | | \ | |/ |___| | | | \/ |
| / \ | | / | |\ |\ | | | |
| / \ | | / | | \ | \ | | | |
___|/ \___|___ |___/ ___|___ | \ | \ |___| | |
====================================================================================
====================================================================================
"""
``` | output | 1 | 5,287 | 20 | 10,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13 | instruction | 0 | 5,288 | 20 | 10,576 |
Tags: *special, combinatorics
Correct Solution:
```
def star(n):
f = 1
while n:
n -=1
f += n*12
return f
print(star(int(input())))
``` | output | 1 | 5,288 | 20 | 10,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13 | instruction | 0 | 5,289 | 20 | 10,578 |
Tags: *special, combinatorics
Correct Solution:
```
#In the name of Allah
from sys import stdin, stdout
input = stdin.readline
a = int(input())
ans = 1
for i in range(1, a):
ans += 6 * (2 * i)
stdout.write(str(ans))
``` | output | 1 | 5,289 | 20 | 10,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13 | instruction | 0 | 5,290 | 20 | 10,580 |
Tags: *special, combinatorics
Correct Solution:
```
n = int(input().split()[0])
print((2 * n - 1) * (2 * n - 1) + 4 * (n - 1) * n // 2)
``` | output | 1 | 5,290 | 20 | 10,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13 | instruction | 0 | 5,291 | 20 | 10,582 |
Tags: *special, combinatorics
Correct Solution:
```
n = int(input())
print(n * (n - 1) * 6 + 1)
``` | output | 1 | 5,291 | 20 | 10,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13 | instruction | 0 | 5,292 | 20 | 10,584 |
Tags: *special, combinatorics
Correct Solution:
```
n = int(input())
b = (6*n*(n - 1)) + 1
print(b)
``` | output | 1 | 5,292 | 20 | 10,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change his mind, Roman is ready to solve any mathematical problem. After some thought, Sereja asked Roma to find, how many numbers are close to number n, modulo m.
Number x is considered close to number n modulo m, if:
* it can be obtained by rearranging the digits of number n,
* it doesn't have any leading zeroes,
* the remainder after dividing number x by m equals 0.
Roman is a good mathematician, but the number of such numbers is too huge for him. So he asks you to help him.
Input
The first line contains two integers: n (1 ≤ n < 1018) and m (1 ≤ m ≤ 100).
Output
In a single line print a single integer — the number of numbers close to number n modulo m.
Examples
Input
104 2
Output
3
Input
223 4
Output
1
Input
7067678 8
Output
47
Note
In the first sample the required numbers are: 104, 140, 410.
In the second sample the required number is 232.
Submitted Solution:
```
n, m = input().split()
m = int(m)
def variations(string):
if string:
head = string[0]
tail = string[1:]
for var in variations(tail):
for i in range(len(var)+1):
yield var[:i] + head + var[i:]
else:
yield string
res = 0
for v in variations(n):
if not v.startswith("0"):
if int(v) % m == 0:
res += 1
print(res)
``` | instruction | 0 | 5,371 | 20 | 10,742 |
No | output | 1 | 5,371 | 20 | 10,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change his mind, Roman is ready to solve any mathematical problem. After some thought, Sereja asked Roma to find, how many numbers are close to number n, modulo m.
Number x is considered close to number n modulo m, if:
* it can be obtained by rearranging the digits of number n,
* it doesn't have any leading zeroes,
* the remainder after dividing number x by m equals 0.
Roman is a good mathematician, but the number of such numbers is too huge for him. So he asks you to help him.
Input
The first line contains two integers: n (1 ≤ n < 1018) and m (1 ≤ m ≤ 100).
Output
In a single line print a single integer — the number of numbers close to number n modulo m.
Examples
Input
104 2
Output
3
Input
223 4
Output
1
Input
7067678 8
Output
47
Note
In the first sample the required numbers are: 104, 140, 410.
In the second sample the required number is 232.
Submitted Solution:
```
def fact(n):
j = 1
for i in range(1,n+1):
j*=i
return j
n,m = map(int,input().split())
l = len(str(n))
n = str(n)
mp = dict()
mp2 = dict()
for i in range(l):
if int(n[i]) in mp2.keys():
mp2[int(n[i])]+=1
else :
mp2[int(n[i])] = 1
for i in range(l):
for j in range(l):
for k in range(l):
if i!=j and j!=k and i!=k :
y = int(n[i]+n[j]+n[k])%m
if not y :
mp[str(n[i]+n[j]+n[k])] = y
if l<=3:
if l==1:
if int(n)==0 or int(n)%m!=0:
print(0)
else:
print(1)
elif l==2:
x = n[1]+n[0]
c = 0
if int(n)%m==0:
c+=1
if x[0]!='0' and int(x)%m==0:
c+=1
print(c)
else:
c = 0
for i in mp:
if i[0]!='0':
c+=1
print(c)
exit(0)
a = 0
for i in mp.keys():
mp3 = dict(mp2)
c,c2 = 0,1
for l in i:
mp3[int(l)]-=1
for j in mp3.keys():
c2 *= fact(mp3[j])
c += mp3[j]
if 0 in mp3.keys() and mp3[0]>=1:
a+= fact(c-1)*(c-1)//c2
else:
a += fact(c)//c2
print(a)
``` | instruction | 0 | 5,372 | 20 | 10,744 |
No | output | 1 | 5,372 | 20 | 10,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change his mind, Roman is ready to solve any mathematical problem. After some thought, Sereja asked Roma to find, how many numbers are close to number n, modulo m.
Number x is considered close to number n modulo m, if:
* it can be obtained by rearranging the digits of number n,
* it doesn't have any leading zeroes,
* the remainder after dividing number x by m equals 0.
Roman is a good mathematician, but the number of such numbers is too huge for him. So he asks you to help him.
Input
The first line contains two integers: n (1 ≤ n < 1018) and m (1 ≤ m ≤ 100).
Output
In a single line print a single integer — the number of numbers close to number n modulo m.
Examples
Input
104 2
Output
3
Input
223 4
Output
1
Input
7067678 8
Output
47
Note
In the first sample the required numbers are: 104, 140, 410.
In the second sample the required number is 232.
Submitted Solution:
```
nums=list(map(int,input().split()))
nums_modified=[]
d1={str(i):0 for i in range(0,10)}
d2={str(i):0 for i in range(0,10)}
for index in range(1,len(nums)):
x=list(str(nums[index]))
x2=list(map(int,x))
small=min(x2);large=max(x2)
n=str(small*7+large*11)
if len(n)>=3:
n=n[1:]
nums_modified.append(n)
if index%2!=0:
d1[n[0]]+=1
else:d2[n[0]]+=1
ans=[0]* 10
for key,value in d1.items():
if value==2:
ans[int(key)]+=1
elif value>2:
ans[int(key)]+=2
for key,value in d2.items():
if value==2:
ans[int(key)]+=1
elif value>2:
ans[int(key)]+=2
c=0
for i in ans:
if i>2:
c+=2
else:
c+=i
print(c)
``` | instruction | 0 | 5,373 | 20 | 10,746 |
No | output | 1 | 5,373 | 20 | 10,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change his mind, Roman is ready to solve any mathematical problem. After some thought, Sereja asked Roma to find, how many numbers are close to number n, modulo m.
Number x is considered close to number n modulo m, if:
* it can be obtained by rearranging the digits of number n,
* it doesn't have any leading zeroes,
* the remainder after dividing number x by m equals 0.
Roman is a good mathematician, but the number of such numbers is too huge for him. So he asks you to help him.
Input
The first line contains two integers: n (1 ≤ n < 1018) and m (1 ≤ m ≤ 100).
Output
In a single line print a single integer — the number of numbers close to number n modulo m.
Examples
Input
104 2
Output
3
Input
223 4
Output
1
Input
7067678 8
Output
47
Note
In the first sample the required numbers are: 104, 140, 410.
In the second sample the required number is 232.
Submitted Solution:
```
def fact(n):
j = 1
for i in range(1,n+1):
j*=i
return j
n,m = map(int,input().split())
l = len(str(n))
n = str(n)
mp = dict()
mp2 = dict()
for i in range(l):
if int(n[i]) in mp2.keys():
mp2[int(n[i])]+=1
else :
mp2[int(n[i])] = 1
for i in range(l):
for j in range(l):
for k in range(l):
if i!=j and j!=k and i!=k :
y = int(n[i]+n[j]+n[k])%m
if not y :
mp[str(n[i]+n[j]+n[k])] = y
if l<=3:
if l==1:
if int(n)==0 or int(n)%m!=0:
print(0)
else:
print(1)
elif l==2:
x = n[1]+n[0]
c = 0
if int(n)%m==0:
c+=1
if x[0]!='0' and int(x)%m==0:
c+=1
print(c)
else:
c = 0
for i in mp:
if i[0]!='0':
c+=1
print(c)
exit(0)
a = 0
print(mp2)
for i in mp.keys():
mp3 = dict(mp2)
c,c2 = 0,1
print(i)
for l in i:
mp3[int(l)]-=1
for j in mp3.keys():
c2 *= fact(mp3[j])
c += mp3[j]
if 0 in mp3.keys() and mp3[0]>=1:
a+= fact(c-1)*(c-1)//c2
else:
a += fact(c)//c2
print(mp3)
print(a)
``` | instruction | 0 | 5,374 | 20 | 10,748 |
No | output | 1 | 5,374 | 20 | 10,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fibonotci sequence is an integer recursive sequence defined by the recurrence relation
Fn = sn - 1·Fn - 1 + sn - 2·Fn - 2 with F0 = 0, F1 = 1
Sequence s is an infinite and almost cyclic sequence with a cycle of length N. A sequence s is called almost cyclic with a cycle of length N if <image>, for i ≥ N, except for a finite number of values si, for which <image> (i ≥ N).
Following is an example of an almost cyclic sequence with a cycle of length 4:
s = (5,3,8,11,5,3,7,11,5,3,8,11,…)
Notice that the only value of s for which the equality <image> does not hold is s6 (s6 = 7 and s2 = 8). You are given s0, s1, ...sN - 1 and all the values of sequence s for which <image> (i ≥ N).
Find <image>.
Input
The first line contains two numbers K and P. The second line contains a single number N. The third line contains N numbers separated by spaces, that represent the first N numbers of the sequence s. The fourth line contains a single number M, the number of values of sequence s for which <image>. Each of the following M lines contains two numbers j and v, indicating that <image> and sj = v. All j-s are distinct.
* 1 ≤ N, M ≤ 50000
* 0 ≤ K ≤ 1018
* 1 ≤ P ≤ 109
* 1 ≤ si ≤ 109, for all i = 0, 1, ...N - 1
* N ≤ j ≤ 1018
* 1 ≤ v ≤ 109
* All values are integers
Output
Output should contain a single integer equal to <image>.
Examples
Input
10 8
3
1 2 1
2
7 3
5 4
Output
4
Submitted Solution:
```
__author__ = 'dwliv_000'
(n,m)=(int(i) for i in input().split())
k=int(input())
g=[int(i) for i in input().split()]
j=int(input())
c={}
q=0
for j in range(j):
(a,b) = (int(i) for i in input().split())
c[a]=b
if(a>q):
q=a
if(q%k)==0:
f=q
else:
f=(q//k+1)*k
d=[]
for e in range(f):
if(e in c):
d.append(c[e])
else:
d.append(g[e%k])
l=0
l2=1
for j in range(2,n+2):
t=l2
l2=l2*d[(j-1)%len(d)]+l*d[(j-2)%len(d)]
l=t
print(l2%m)
``` | instruction | 0 | 5,434 | 20 | 10,868 |
No | output | 1 | 5,434 | 20 | 10,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fibonotci sequence is an integer recursive sequence defined by the recurrence relation
Fn = sn - 1·Fn - 1 + sn - 2·Fn - 2 with F0 = 0, F1 = 1
Sequence s is an infinite and almost cyclic sequence with a cycle of length N. A sequence s is called almost cyclic with a cycle of length N if <image>, for i ≥ N, except for a finite number of values si, for which <image> (i ≥ N).
Following is an example of an almost cyclic sequence with a cycle of length 4:
s = (5,3,8,11,5,3,7,11,5,3,8,11,…)
Notice that the only value of s for which the equality <image> does not hold is s6 (s6 = 7 and s2 = 8). You are given s0, s1, ...sN - 1 and all the values of sequence s for which <image> (i ≥ N).
Find <image>.
Input
The first line contains two numbers K and P. The second line contains a single number N. The third line contains N numbers separated by spaces, that represent the first N numbers of the sequence s. The fourth line contains a single number M, the number of values of sequence s for which <image>. Each of the following M lines contains two numbers j and v, indicating that <image> and sj = v. All j-s are distinct.
* 1 ≤ N, M ≤ 50000
* 0 ≤ K ≤ 1018
* 1 ≤ P ≤ 109
* 1 ≤ si ≤ 109, for all i = 0, 1, ...N - 1
* N ≤ j ≤ 1018
* 1 ≤ v ≤ 109
* All values are integers
Output
Output should contain a single integer equal to <image>.
Examples
Input
10 8
3
1 2 1
2
7 3
5 4
Output
4
Submitted Solution:
```
__author__ = 'dwliv_000'
(n,m)=(int(i) for i in input().split())
k=int(input())
g=[int(i) for i in input().split()]
j=int(input())
c={}
q=0
for j in range(j):
(a,b) = (int(i) for i in input().split())
c[a]=b
if(a>q):
q=a
if(q%k)==0:
f=q
else:
f=(q//k+1)*k
d=[]
for e in range(f):
if(e in c):
d.append(c[e])
else:
d.append(g[e%k])
l=0
l2=1
for j in range(2,n+2):
t=l2
l2=l2*d[(j-1)%len(d)]+l*d[(j-2)%len(d)]
l=t
if(n==1):
print(d[0])
else:
if(n==0):
print('0')
else:
print(l2%m)
``` | instruction | 0 | 5,435 | 20 | 10,870 |
No | output | 1 | 5,435 | 20 | 10,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fibonotci sequence is an integer recursive sequence defined by the recurrence relation
Fn = sn - 1·Fn - 1 + sn - 2·Fn - 2 with F0 = 0, F1 = 1
Sequence s is an infinite and almost cyclic sequence with a cycle of length N. A sequence s is called almost cyclic with a cycle of length N if <image>, for i ≥ N, except for a finite number of values si, for which <image> (i ≥ N).
Following is an example of an almost cyclic sequence with a cycle of length 4:
s = (5,3,8,11,5,3,7,11,5,3,8,11,…)
Notice that the only value of s for which the equality <image> does not hold is s6 (s6 = 7 and s2 = 8). You are given s0, s1, ...sN - 1 and all the values of sequence s for which <image> (i ≥ N).
Find <image>.
Input
The first line contains two numbers K and P. The second line contains a single number N. The third line contains N numbers separated by spaces, that represent the first N numbers of the sequence s. The fourth line contains a single number M, the number of values of sequence s for which <image>. Each of the following M lines contains two numbers j and v, indicating that <image> and sj = v. All j-s are distinct.
* 1 ≤ N, M ≤ 50000
* 0 ≤ K ≤ 1018
* 1 ≤ P ≤ 109
* 1 ≤ si ≤ 109, for all i = 0, 1, ...N - 1
* N ≤ j ≤ 1018
* 1 ≤ v ≤ 109
* All values are integers
Output
Output should contain a single integer equal to <image>.
Examples
Input
10 8
3
1 2 1
2
7 3
5 4
Output
4
Submitted Solution:
```
__author__ = 'dwliv_000'
(n,m)=(int(i) for i in input().split())
k=int(input())
g=[int(i) for i in input().split()]
j=int(input())
c={}
q=0
for j in range(j):
(a,b) = (int(i) for i in input().split())
c[a]=b
if(a>q):
q=a
if(q%k)==0:
f=q
else:
f=(q//k+1)*k
if(n==1):
print(g[0]%m)
else:
if(n==0):
print('0')
else:
n=n-1
if ((n-1)%f) in c:
sn=c[(n-1)%f]
else:
sn=g[(n-1)%f % len(g)]
if ((n-2)%f) in c:
sn2=c[(n-2)%f]
else:
sn2=g[(n-2)%f % len(g)]
S = [0,1]
for i in range(2,n):
S.append((S[i-1]+S[i-2])%m)
k = k+1
if (S[i]==1) and (S[i-1]==0):
break
f1=S[(n-1)%k]
f2=S[(n-2)%k]
l2=((sn % m) * f1 + (sn2 % m)*f2)
print(l2)
``` | instruction | 0 | 5,436 | 20 | 10,872 |
No | output | 1 | 5,436 | 20 | 10,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Fibonotci sequence is an integer recursive sequence defined by the recurrence relation
Fn = sn - 1·Fn - 1 + sn - 2·Fn - 2 with F0 = 0, F1 = 1
Sequence s is an infinite and almost cyclic sequence with a cycle of length N. A sequence s is called almost cyclic with a cycle of length N if <image>, for i ≥ N, except for a finite number of values si, for which <image> (i ≥ N).
Following is an example of an almost cyclic sequence with a cycle of length 4:
s = (5,3,8,11,5,3,7,11,5,3,8,11,…)
Notice that the only value of s for which the equality <image> does not hold is s6 (s6 = 7 and s2 = 8). You are given s0, s1, ...sN - 1 and all the values of sequence s for which <image> (i ≥ N).
Find <image>.
Input
The first line contains two numbers K and P. The second line contains a single number N. The third line contains N numbers separated by spaces, that represent the first N numbers of the sequence s. The fourth line contains a single number M, the number of values of sequence s for which <image>. Each of the following M lines contains two numbers j and v, indicating that <image> and sj = v. All j-s are distinct.
* 1 ≤ N, M ≤ 50000
* 0 ≤ K ≤ 1018
* 1 ≤ P ≤ 109
* 1 ≤ si ≤ 109, for all i = 0, 1, ...N - 1
* N ≤ j ≤ 1018
* 1 ≤ v ≤ 109
* All values are integers
Output
Output should contain a single integer equal to <image>.
Examples
Input
10 8
3
1 2 1
2
7 3
5 4
Output
4
Submitted Solution:
```
def solve(K,P,N,s,M,V):
q = 0
memo = {}
# Fn = Sn-1 * Fn-1 + Sn-2 * Fn-2
if K == 0:
return 0
elif K == 1:
return 1
memo_pp = 0
memo_p = 1
k = 2
while k <= K:
Fn1 = memo_p
Fn2 = memo_pp
Sn1 = s[(k-1)%N]
Sn2 = s[(k-2)%N]
if (k-1) in V:
Sn1 = V[k-1]
if (k-2) in V:
Sn2 = V[k-2]
memo_pp = memo_p
memo_p = Sn1*Fn1 + Sn2*Fn2
k += 1
return memo_p % P
def main(infile, outfile):
K,P = map(int,infile.readline().split())
N = int(infile.readline())
s = list(map(int,infile.readline().split()))
M = int(infile.readline())
V = {}
for i in range(M):
j,v = map(int,infile.readline().split())
V[j] = v
outfile.write(str(solve(K,P,N,s,M,V)) + '\n')
if __name__ == '__main__':
from sys import stdin, stdout
main(stdin, stdout)
``` | instruction | 0 | 5,437 | 20 | 10,874 |
No | output | 1 | 5,437 | 20 | 10,875 |
Provide a correct Python 3 solution for this coding contest problem.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja | instruction | 0 | 5,581 | 20 | 11,162 |
"Correct Solution:
```
n = int(input())
ans = ""
while n != 0:
n -= 1
ans += chr(ord('a') + n%26)
n //= 26
print(ans[::-1])
``` | output | 1 | 5,581 | 20 | 11,163 |
Provide a correct Python 3 solution for this coding contest problem.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja | instruction | 0 | 5,582 | 20 | 11,164 |
"Correct Solution:
```
N = int(input())
ans = ""
while N > 0:
N -=1
ans = chr(ord("a") + (N%26)) + ans
N //= 26
print(ans)
``` | output | 1 | 5,582 | 20 | 11,165 |
Provide a correct Python 3 solution for this coding contest problem.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja | instruction | 0 | 5,583 | 20 | 11,166 |
"Correct Solution:
```
n=int(input())
ans = ""
while(n > 0):
n -= 1
ans += chr(ord("a") + n%26)
n = n//26
print(ans[::-1])
``` | output | 1 | 5,583 | 20 | 11,167 |
Provide a correct Python 3 solution for this coding contest problem.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja | instruction | 0 | 5,584 | 20 | 11,168 |
"Correct Solution:
```
N = int(input())
l = []
while N:
N -= 1
l.append(chr(97+(N % 26)))
N //= 26
print("".join(l[::-1]))
``` | output | 1 | 5,584 | 20 | 11,169 |
Provide a correct Python 3 solution for this coding contest problem.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja | instruction | 0 | 5,585 | 20 | 11,170 |
"Correct Solution:
```
n = int(input())
ans = ''
while n:
n -= 1
ans = chr(n%26 + ord('a')) + ans
n //= 26
print(ans)
``` | output | 1 | 5,585 | 20 | 11,171 |
Provide a correct Python 3 solution for this coding contest problem.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja | instruction | 0 | 5,586 | 20 | 11,172 |
"Correct Solution:
```
n = int(input())
s = ''
while n > 0:
n -= 1
s = chr(ord('a') + n % 26) + s
n //= 26
print(s)
``` | output | 1 | 5,586 | 20 | 11,173 |
Provide a correct Python 3 solution for this coding contest problem.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja | instruction | 0 | 5,587 | 20 | 11,174 |
"Correct Solution:
```
n=int(input())
ans=""
while 0 < n:
d=n%26
if d==0:
d=26
ans=chr(int(d)+96) + ans
n=(n-d)/26
print(ans)
``` | output | 1 | 5,587 | 20 | 11,175 |
Provide a correct Python 3 solution for this coding contest problem.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja | instruction | 0 | 5,588 | 20 | 11,176 |
"Correct Solution:
```
n = int(input())
al = 'abcdefghijklmnopqrstuvwxyz'
s = ''
while n:
n -= 1
e = n % 26
s = al[e] + s
n //= 26
print(s)
``` | output | 1 | 5,588 | 20 | 11,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja
Submitted Solution:
```
n = int(input())
ans = str()
while n != 0:
n -= 1
ans += chr(ord('a') + n % 26)
n //= 26
print(ans[::-1])
``` | instruction | 0 | 5,589 | 20 | 11,178 |
Yes | output | 1 | 5,589 | 20 | 11,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja
Submitted Solution:
```
N = int(input())
name = ''
while N > 0:
name += chr(ord('a') + (N - 1) % 26)
N = (N - 1) // 26
print(name[::-1])
``` | instruction | 0 | 5,590 | 20 | 11,180 |
Yes | output | 1 | 5,590 | 20 | 11,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja
Submitted Solution:
```
n=int(input())
a=[]
while n>0:
n-=1
a.append(chr(ord("a")+n%26))
n//=26
ans="".join(a)
print(ans[::-1])
``` | instruction | 0 | 5,591 | 20 | 11,182 |
Yes | output | 1 | 5,591 | 20 | 11,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja
Submitted Solution:
```
MOD=10**9+1
n=int(input())
ans=''
while n:
ans=chr(ord('a')+(n+25)%26)+ans
n=(n-1)//26
print(ans)
``` | instruction | 0 | 5,592 | 20 | 11,184 |
Yes | output | 1 | 5,592 | 20 | 11,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja
Submitted Solution:
```
n = int(input())
ans = ""
while n:
n -= 1
ans = chr(96 + n % 26) + ans
n //= 26
print(ans)
``` | instruction | 0 | 5,593 | 20 | 11,186 |
No | output | 1 | 5,593 | 20 | 11,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja
Submitted Solution:
```
def run(n):
'''
'''
_base_array = string.ascii_lowercase
results = []
while n > 0:
results.append(_base_array[n%26 - 1])
if n%26 == 0:
n = n//26 - 1
continue
n = n//26
print(''.join(results[::-1]))
if __name__ == '__main__':
n = int(input())
run(n)
``` | instruction | 0 | 5,594 | 20 | 11,188 |
No | output | 1 | 5,594 | 20 | 11,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja
Submitted Solution:
```
N = int(input())
def alf(n):
if n == 0:
return 'a'
return chr(ord('a') + n -1)
ans = alf(N % 26)
cnt = 1
while True:
if N <= 26 ** cnt:
break
ans = alf((N // (26 ** cnt)) % 26) + ans
cnt += 1
print(ans)
``` | instruction | 0 | 5,595 | 20 | 11,190 |
No | output | 1 | 5,595 | 20 | 11,191 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:
* the dogs numbered 1,2,\cdots,26 were respectively given the names `a`, `b`, ..., `z`;
* the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names `aa`, `ab`, `ac`, ..., `zy`, `zz`;
* the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names `aaa`, `aab`, `aac`, ..., `zzy`, `zzz`;
* the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names `aaaa`, `aaab`, `aaac`, ..., `zzzy`, `zzzz`;
* the dogs numbered 475255,475256,\cdots were respectively given the names `aaaaa`, `aaaab`, ...;
* and so on.
To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:
`a`, `b`, ..., `z`, `aa`, `ab`, ..., `az`, `ba`, `bb`, ..., `bz`, ..., `za`, `zb`, ..., `zz`, `aaa`, `aab`, ..., `aaz`, `aba`, `abb`, ..., `abz`, ..., `zzz`, `aaaa`, ...
Now, Roger asks you:
"What is the name for the dog numbered N?"
Constraints
* N is an integer.
* 1 \leq N \leq 1000000000000001
Input
Input is given from Standard Input in the following format:
N
Output
Print the answer to Roger's question as a string consisting of lowercase English letters.
Examples
Input
2
Output
b
Input
27
Output
aa
Input
123456789
Output
jjddja
Submitted Solution:
```
import string
n = int(input())
alpha = string.ascii_lowercase
p = []
while True:
x, y = divmod(n, 26)
p.append((x, y))
if x == 0 or (x, y) == (1, 0):
break
n = x
p = p[::-1]
print("".join([alpha[b-1] for a, b in p]))
``` | instruction | 0 | 5,596 | 20 | 11,192 |
No | output | 1 | 5,596 | 20 | 11,193 |
Provide a correct Python 3 solution for this coding contest problem.
FizzBuzz is a game in which integers of 1 or more are spoken in order according to the following rules.
* "Fizz" when divisible by 3
* "Buzz" when divisible by 5
* "FizzBuzz" when divisible by both 3 and 5
* At other times, that number
An example of the progress of the game is shown below.
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16,…
The character string obtained by combining the obtained remarks into one character string is called FizzBuzz String. Since the index s is given, output 20 characters from the s character of the FizzBuzz String. However, the index may start from 1, and the length of the obtained character string may be sufficiently large (s + 20 or more).
Constraints
* s is an integer
* 1 ≤ s ≤ 1018
Input
Input is given in the following format
> s
>
Output
Output 20 characters from the s character of FizzBuzz String on one line
Examples
Input
1
Output
12Fizz4BuzzFizz78Fiz
Input
20
Output
zzBuzz11Fizz1314Fizz
Input
10000000000
Output
93FizzBuzz1418650796 | instruction | 0 | 5,770 | 20 | 11,540 |
"Correct Solution:
```
def count_div(start, end, div):
# [start, end)
return (end - 1) // div - (start - 1) // div
def calc_start(mid):
cnt, i = 0, 1
while 10 ** i < mid:
d, p = 10 ** i, 10 ** (i - 1)
fif, three, five = count_div(p, d, 15), count_div(p, d, 3), count_div(p, d, 5)
cnt += i * (d - p) + (three + five) * 4 - (three + five - fif) * i
i += 1
d = 10 ** (i - 1)
fif, three, five = count_div(d, mid, 15), count_div(d, mid, 3), count_div(d, mid, 5)
cnt += i * (mid - d) + (three + five) * 4 - (three + five - fif) * i
return cnt
N = int(input()) - 1
left, right = 1, 10 ** 18
while left + 1 < right:
mid = (left + right) // 2
start = calc_start(mid)
if start <= N:
left = mid
else:
right = mid
ans = ''
for i in range(left, left + 30):
tmp = ''
if i % 3 == 0:
tmp += "Fizz"
if i % 5 == 0:
tmp += "Buzz"
if not tmp:
tmp = str(i)
ans += tmp
start = calc_start(left)
print(ans[N - start:N - start + 20])
``` | output | 1 | 5,770 | 20 | 11,541 |
Provide a correct Python 3 solution for this coding contest problem.
FizzBuzz is a game in which integers of 1 or more are spoken in order according to the following rules.
* "Fizz" when divisible by 3
* "Buzz" when divisible by 5
* "FizzBuzz" when divisible by both 3 and 5
* At other times, that number
An example of the progress of the game is shown below.
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16,…
The character string obtained by combining the obtained remarks into one character string is called FizzBuzz String. Since the index s is given, output 20 characters from the s character of the FizzBuzz String. However, the index may start from 1, and the length of the obtained character string may be sufficiently large (s + 20 or more).
Constraints
* s is an integer
* 1 ≤ s ≤ 1018
Input
Input is given in the following format
> s
>
Output
Output 20 characters from the s character of FizzBuzz String on one line
Examples
Input
1
Output
12Fizz4BuzzFizz78Fiz
Input
20
Output
zzBuzz11Fizz1314Fizz
Input
10000000000
Output
93FizzBuzz1418650796 | instruction | 0 | 5,771 | 20 | 11,542 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
n = I()
if n < 30:
s = ''
for i in range(1, 100):
if i % 15 == 0:
s += 'FizzBuzz'
elif i % 5 == 0:
s += 'Buzz'
elif i % 3 == 0:
s += 'Fizz'
else:
s += str(i)
# print(s)
return s[n-1:n+19]
i = 2
n -= 4 * 4 + 5
while 1:
t = 10**i - (10**(i-1))
s3 = t // 3
s5 = t // 5
s15 = t // 15
sl = s3 + s5 - s15
tl = (s3 + s5) * 4 + (t - sl) * i
if tl >= n:
break
n -= tl
i += 1
si = 10**(i-1) - 1
for j in range(i-1,0,-1):
t = 10**j
for k in range(1,10):
s3 = (si+t) // 3 - si // 3
s5 = (si+t) // 5 - si // 5
s15 = (si+t) // 15 - si // 15
sl = s3 + s5 - s15
tl = (s3 + s5) * 4 + (t - sl) * i
if tl >= n:
break
n -= tl
si += t
# print('si,n',si,n)
s = ''
for i in range(si+1,si+100):
if i % 15 == 0:
s += 'FizzBuzz'
elif i % 5 == 0:
s += 'Buzz'
elif i % 3 == 0:
s += 'Fizz'
else:
s += str(i)
# print(s)
return s[n-1:n+19]
print(main())
``` | output | 1 | 5,771 | 20 | 11,543 |
Provide a correct Python 3 solution for this coding contest problem.
FizzBuzz is a game in which integers of 1 or more are spoken in order according to the following rules.
* "Fizz" when divisible by 3
* "Buzz" when divisible by 5
* "FizzBuzz" when divisible by both 3 and 5
* At other times, that number
An example of the progress of the game is shown below.
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16,…
The character string obtained by combining the obtained remarks into one character string is called FizzBuzz String. Since the index s is given, output 20 characters from the s character of the FizzBuzz String. However, the index may start from 1, and the length of the obtained character string may be sufficiently large (s + 20 or more).
Constraints
* s is an integer
* 1 ≤ s ≤ 1018
Input
Input is given in the following format
> s
>
Output
Output 20 characters from the s character of FizzBuzz String on one line
Examples
Input
1
Output
12Fizz4BuzzFizz78Fiz
Input
20
Output
zzBuzz11Fizz1314Fizz
Input
10000000000
Output
93FizzBuzz1418650796 | instruction | 0 | 5,772 | 20 | 11,544 |
"Correct Solution:
```
# coding: utf-8
def f(n):
return (n*8+32)*6*10**(n-2)-5
i=2
count=0
s=int(input())
while 1:
if count+f(i)<s:
count+=f(i)
else:
break
i+=1
j=0
x=10**(i-1)-9
tmp=(s-count)//(i*8+32)
count-=(5 if tmp!=0 else 0)
count+=tmp*(i*8+32)
x+=tmp*15
fzbz=''
while len(fzbz)<=s-count+100:
fzbz+=('FizzBuzz' if x%15==0 else 'Fizz' if x%3==0 else 'Buzz' if x%5==0 else str(x))
x+=1
print(fzbz[s-count-1:s-count-1+20])
``` | output | 1 | 5,772 | 20 | 11,545 |
Provide a correct Python 3 solution for this coding contest problem.
FizzBuzz is a game in which integers of 1 or more are spoken in order according to the following rules.
* "Fizz" when divisible by 3
* "Buzz" when divisible by 5
* "FizzBuzz" when divisible by both 3 and 5
* At other times, that number
An example of the progress of the game is shown below.
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16,…
The character string obtained by combining the obtained remarks into one character string is called FizzBuzz String. Since the index s is given, output 20 characters from the s character of the FizzBuzz String. However, the index may start from 1, and the length of the obtained character string may be sufficiently large (s + 20 or more).
Constraints
* s is an integer
* 1 ≤ s ≤ 1018
Input
Input is given in the following format
> s
>
Output
Output 20 characters from the s character of FizzBuzz String on one line
Examples
Input
1
Output
12Fizz4BuzzFizz78Fiz
Input
20
Output
zzBuzz11Fizz1314Fizz
Input
10000000000
Output
93FizzBuzz1418650796 | instruction | 0 | 5,773 | 20 | 11,546 |
"Correct Solution:
```
def f(n):
i = 0
while (10**i <= n):
i += 1
i -= 1
return 8 if n % 15 == 0 else 4 if n % 3 == 0 or n % 5 == 0 else i
def fizzbuzzlen(n):
if n == 0:
return 0
i = 0
ans = 0
while (10**i <= n):
start = 10**i
end = min(10**(i+1)-1, n)
numbase = (end-start+1)
num3 = end//3-(start-1)//3
num5 = end//5-(start-1)//5
num15 = end//15-(start-1)//15
numbase = numbase-num3-num5+num15
ans += (i+1)*numbase+4*num3+4*num5 # +8*num15
i += 1
return ans
n = int(input())
# print("".join(["FizzBuzz" if i % 15 == 0 else "Buzz" if i % 5 ==
# 0 else "Fizz" if i % 3 == 0 else str(i) for i in range(1, n+20)])[n-1:n-1+20])
l = 0
r = 10**18+1
while (l+1 != r):
med = (l+r)//2
if (fizzbuzzlen(med) < n):
l = med
else:
r = med
s = "".join(["0"if i == 0 else "FizzBuzz" if i % 15 == 0 else "Buzz" if i % 5 ==
0 else "Fizz" if i % 3 == 0 else str(i) for i in range(l+1, l+21)])
print(s[n-fizzbuzzlen(l)-1:n-fizzbuzzlen(l)-1+20])
``` | output | 1 | 5,773 | 20 | 11,547 |
Provide a correct Python 3 solution for this coding contest problem.
FizzBuzz is a game in which integers of 1 or more are spoken in order according to the following rules.
* "Fizz" when divisible by 3
* "Buzz" when divisible by 5
* "FizzBuzz" when divisible by both 3 and 5
* At other times, that number
An example of the progress of the game is shown below.
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16,…
The character string obtained by combining the obtained remarks into one character string is called FizzBuzz String. Since the index s is given, output 20 characters from the s character of the FizzBuzz String. However, the index may start from 1, and the length of the obtained character string may be sufficiently large (s + 20 or more).
Constraints
* s is an integer
* 1 ≤ s ≤ 1018
Input
Input is given in the following format
> s
>
Output
Output 20 characters from the s character of FizzBuzz String on one line
Examples
Input
1
Output
12Fizz4BuzzFizz78Fiz
Input
20
Output
zzBuzz11Fizz1314Fizz
Input
10000000000
Output
93FizzBuzz1418650796 | instruction | 0 | 5,774 | 20 | 11,548 |
"Correct Solution:
```
def count_div(start, end, div):
# [start, end)
return (end - 1) // div - (start - 1) // div
def calc_start(mid):
cnt, i = 0, 1
while 10 ** i < mid:
d, p = 10 ** i, 10 ** (i - 1)
fif, three, five = count_div(p, d, 15), count_div(p, d, 3), count_div(p, d, 5)
cnt += i * (d - p) + (three + five) * 4 - (three + five - fif) * i
i += 1
d = 10 ** (i - 1)
fif, three, five = count_div(d, mid, 15), count_div(d, mid, 3), count_div(d, mid, 5)
cnt += i * (mid - d) + (three + five) * 4 - (three + five - fif) * i
return cnt
N = int(input()) - 1
left, right = 1, 10 ** 18
while left + 1 < right:
mid = (left + right) // 2
start = calc_start(mid)
if start <= N:
left = mid
else:
right = mid
ans = ''
for i in range(left, left + 9):
tmp = ''
if i % 3 == 0:
tmp += "Fizz"
if i % 5 == 0:
tmp += "Buzz"
if not tmp:
tmp = str(i)
ans += tmp
start = calc_start(left)
print(ans[N - start:N - start + 20])
``` | output | 1 | 5,774 | 20 | 11,549 |
Provide a correct Python 3 solution for this coding contest problem.
FizzBuzz is a game in which integers of 1 or more are spoken in order according to the following rules.
* "Fizz" when divisible by 3
* "Buzz" when divisible by 5
* "FizzBuzz" when divisible by both 3 and 5
* At other times, that number
An example of the progress of the game is shown below.
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16,…
The character string obtained by combining the obtained remarks into one character string is called FizzBuzz String. Since the index s is given, output 20 characters from the s character of the FizzBuzz String. However, the index may start from 1, and the length of the obtained character string may be sufficiently large (s + 20 or more).
Constraints
* s is an integer
* 1 ≤ s ≤ 1018
Input
Input is given in the following format
> s
>
Output
Output 20 characters from the s character of FizzBuzz String on one line
Examples
Input
1
Output
12Fizz4BuzzFizz78Fiz
Input
20
Output
zzBuzz11Fizz1314Fizz
Input
10000000000
Output
93FizzBuzz1418650796 | instruction | 0 | 5,775 | 20 | 11,550 |
"Correct Solution:
```
def calc_start(mid):
cnt = -1
i = 1
while 10 ** i < mid:
cnt += i * (10 ** i - 10 ** (i - 1))
fif = (10 ** i - 1) // 15 - (10 ** (i - 1) - 1) // 15
three = (10 ** i - 1) // 3 - (10 ** (i - 1) - 1) // 3
five = (10 ** i - 1) // 5 - (10 ** (i - 1) - 1) // 5
cnt += (three + five) * 4 - (three + five - fif) * i
i += 1
cnt += i * (mid - 10 ** (i - 1))
fif = (mid - 1) // 15 - (10 ** (i - 1) - 1) // 15
three = (mid - 1) // 3 - (10 ** (i - 1) - 1) // 3
five = (mid - 1) // 5 - (10 ** (i - 1) - 1) // 5
cnt += (three + five) * 4 - (three + five - fif) * i
return cnt + 1
N = int(input()) - 1
left, right = 1, 10 ** 18
while left + 1 < right:
mid = (left + right) // 2
start = calc_start(mid)
if start <= N:
left = mid
else:
right = mid
ans = ''
for i in range(left, left + 30):
tmp = ''
if i % 3 == 0:
tmp += "Fizz"
if i % 5 == 0:
tmp += "Buzz"
if not tmp:
tmp = str(i)
ans += tmp
start = calc_start(left)
print(ans[N - start:N - start + 20])
``` | output | 1 | 5,775 | 20 | 11,551 |
Provide a correct Python 3 solution for this coding contest problem.
FizzBuzz is a game in which integers of 1 or more are spoken in order according to the following rules.
* "Fizz" when divisible by 3
* "Buzz" when divisible by 5
* "FizzBuzz" when divisible by both 3 and 5
* At other times, that number
An example of the progress of the game is shown below.
1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16,…
The character string obtained by combining the obtained remarks into one character string is called FizzBuzz String. Since the index s is given, output 20 characters from the s character of the FizzBuzz String. However, the index may start from 1, and the length of the obtained character string may be sufficiently large (s + 20 or more).
Constraints
* s is an integer
* 1 ≤ s ≤ 1018
Input
Input is given in the following format
> s
>
Output
Output 20 characters from the s character of FizzBuzz String on one line
Examples
Input
1
Output
12Fizz4BuzzFizz78Fiz
Input
20
Output
zzBuzz11Fizz1314Fizz
Input
10000000000
Output
93FizzBuzz1418650796 | instruction | 0 | 5,776 | 20 | 11,552 |
"Correct Solution:
```
def fizzbuzz_length(n):
f3 = 0
f5 = 0
f15 = 0
folen = 0
keta = 1
while True:
ub = 10**keta - 1
lim = min(n, ub)
folen += keta * (lim-10**(keta-1)+1 - (lim // 3 - f3) - (lim // 5 - f5) + (lim//15 - f15))
f3 = lim // 3
f5 = lim // 5
f15 = lim // 15
if n <= ub:
break
keta += 1
return folen + 4 * (f3 + f5)
def fizzbuzz(i):
if i % 15 == 0:
return "FizzBuzz"
elif i % 3 == 0:
return "Fizz"
elif i % 5 == 0:
return "Buzz"
else:
return str(i)
s = int(input())
lb = 0
ub = s
while ub - lb > 1:
m = (ub + lb) // 2
if s <= fizzbuzz_length(m):
ub = m
else:
lb = m
ans = ""
for i in range(ub, ub + 20):
ans += fizzbuzz(i)
pos = s - fizzbuzz_length(ub-1) - 1
print(ans[pos:pos+20])
``` | output | 1 | 5,776 | 20 | 11,553 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero. | instruction | 0 | 5,808 | 20 | 11,616 |
Tags: number theory
Correct Solution:
```
from math import gcd
n, k = map(int, input().split())
a = list(map(int, input().split()))
g = 0
for i in a:
g = gcd(g, i)
ans = set()
ans.add(g % k)
for i in range(k):
ans.add((i * g) % k)
print(len(ans))
print(*sorted(list(ans)))
``` | output | 1 | 5,808 | 20 | 11,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero. | instruction | 0 | 5,809 | 20 | 11,618 |
Tags: number theory
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
import threading
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-----------------------------------------------------
n,k=map(int,input().split())
a=list(map(int,input().split()))
s=set()
g=a[0]
for i in a:
g=math.gcd(g,i)
#print(g)
for i in range (k):
s.add((g*i)%k)
l=list(s)
l.sort()
print(len(s))
print(*l)
``` | output | 1 | 5,809 | 20 | 11,619 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero. | instruction | 0 | 5,810 | 20 | 11,620 |
Tags: number theory
Correct Solution:
```
def gcd(a,b):
if b == 0:
return a
if b > a:
return gcd(b,a)
return gcd(b,a%b)
n,k = list(map(int,input().split()))
l = list(map(int,input().split()))
out = k
for i in l:
out = gcd(i,out)
print(k//out)
print(' '.join(list(map(str,range(0,k,out)))))
``` | output | 1 | 5,810 | 20 | 11,621 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero. | instruction | 0 | 5,811 | 20 | 11,622 |
Tags: number theory
Correct Solution:
```
def nod(a,b):
while a!=0 and b!=0:
if a>b:
a,b=b,a%b
else:
b,a=a,b%a
return a+b
n ,k = map(int, input().split())
a = [int(j) for j in input().split()]
c = a[0]
for i in range(1,n):
c = nod(c,a[i])
if c==1:
break
e = nod(c,k)
if c==1 or e==1:
print(k)
for i in range(k):
print(i, end=" ")
if e>1:
c = k//e
print(c)
for i in range(c):
print(i*e, end=' ')
``` | output | 1 | 5,811 | 20 | 11,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero. | instruction | 0 | 5,812 | 20 | 11,624 |
Tags: number theory
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=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,m=map(int,input().split())
k=m
def convert(no):
return no%k
l=list(map(int,input().split()))
l=[convert(l[i]) for i in range(n)]
g=0
for i in range(n):
g=math.gcd(g,l[i])
if g==0:
print(1)
print(0)
sys.exit(0)
g=math.gcd(g,k)
t=0
l=[]
while(t<k):
l.append(t%k)
t+=g
l=set(l)
print(len(l))
print(*sorted(list(l)))
``` | output | 1 | 5,812 | 20 | 11,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero. | instruction | 0 | 5,813 | 20 | 11,626 |
Tags: number theory
Correct Solution:
```
from math import gcd
n, k = list(map(int, input().split()))
a = list(map(int, input().split()))[:n]
a = set([b%k for b in a])
a = set ([gcd(k, b) for b in a if b>0])
if len(a) == 0:
print(1)
print(0)
exit()
mg = min(a)
for d in a:
mg = gcd(mg,d)
if mg == 1:
break
res = [mg * j for j in range(k//mg)]
print(len(res))
print(*res)
``` | output | 1 | 5,813 | 20 | 11,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero. | instruction | 0 | 5,814 | 20 | 11,628 |
Tags: number theory
Correct Solution:
```
from functools import reduce
from math import ceil, floor
def gcd(x, y):
"""greatest common divisor of x and y"""
while y:
x, y = y, x % y
return x
def extended_gcd(a, b):
"""returns gcd(a, b), s, r s.t. a * s + b * r == gcd(a, b)"""
s, old_s = 0, 1
r, old_r = b, a
while r:
q = old_r // r
old_r, r = r, old_r - q * r
old_s, s = s, old_s - q * s
return old_r, old_s, (old_r - old_s * a) // b if b else 0
gcdm = lambda args: reduce(gcd, args, 0)
lcm = lambda a, b: a * b // gcd(a, b)
lcmm = lambda *args: reduce(lcm, args, 1)
def main():
n, k = map(int, input().split())
a = list(map(int, input().split()))
g = gcdm(a + [k])
print(k // g)
print(" ".join([str(x * g) for x in range(k // g)]))
main()
# sum(ai xi) - kl = d
# ž g <= k
``` | output | 1 | 5,814 | 20 | 11,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero. | instruction | 0 | 5,815 | 20 | 11,630 |
Tags: number theory
Correct Solution:
```
n, k = map(int,input().split())
v = list(map(int,input().split()))
def gcd(a,b):
if a < b:
return gcd(b,a)
if b == 0:
return a
else:
return gcd(b, a%b)
g = v[0]
for i in range(1,n):
g = gcd(g, v[i])
lst = set()
for i in range(k):
lst.add(g*i % k)
lst = sorted(list(lst))
print(len(lst))
print(' '.join(map(str,lst)))
``` | output | 1 | 5,815 | 20 | 11,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Submitted Solution:
```
import os,io,math
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n,k=map(int,input().split())
a=list(map(int,input().split()))
r=k
for i in range(n):
r=math.gcd(r,a[i])
ans=[]
for i in range(0,k,r):
ans.append(i)
print(len(ans))
print(' '.join(map(str,ans)))
``` | instruction | 0 | 5,816 | 20 | 11,632 |
Yes | output | 1 | 5,816 | 20 | 11,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Submitted Solution:
```
from math import gcd
n,k=map(int,input().split())
arr=list(map(int,input().split()))
gd=0
for i in range(n):
gd=gcd(gd,arr[i])
count=gd
s=set()
for i in range(k):
s.add(count %k)
count +=gd
s=list(s)
s.sort()
print(len(s))
print(*s)
``` | instruction | 0 | 5,817 | 20 | 11,634 |
Yes | output | 1 | 5,817 | 20 | 11,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.
There are n banknote denominations on Mars: the value of i-th banknote is a_i. Natasha has an infinite number of banknotes of each denomination.
Martians have k fingers on their hands, so they use a number system with base k. In addition, the Martians consider the digit d (in the number system with base k) divine. Thus, if the last digit in Natasha's tax amount written in the number system with the base k is d, the Martians will be happy. Unfortunately, Natasha does not know the Martians' divine digit yet.
Determine for which values d Natasha can make the Martians happy.
Natasha can use only her banknotes. Martians don't give her change.
Input
The first line contains two integers n and k (1 ≤ n ≤ 100 000, 2 ≤ k ≤ 100 000) — the number of denominations of banknotes and the base of the number system on Mars.
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — denominations of banknotes on Mars.
All numbers are given in decimal notation.
Output
On the first line output the number of values d for which Natasha can make the Martians happy.
In the second line, output all these values in increasing order.
Print all numbers in decimal notation.
Examples
Input
2 8
12 20
Output
2
0 4
Input
3 10
10 20 30
Output
1
0
Note
Consider the first test case. It uses the octal number system.
If you take one banknote with the value of 12, you will get 14_8 in octal system. The last digit is 4_8.
If you take one banknote with the value of 12 and one banknote with the value of 20, the total value will be 32. In the octal system, it is 40_8. The last digit is 0_8.
If you take two banknotes with the value of 20, the total value will be 40, this is 50_8 in the octal system. The last digit is 0_8.
No other digits other than 0_8 and 4_8 can be obtained. Digits 0_8 and 4_8 could also be obtained in other ways.
The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.
Submitted Solution:
```
from math import *
from collections import *
import sys
sys.setrecursionlimit(10**9)
n,k = map(int,input().split())
a = list(map(int,input().split()))
g = a[0]%k
for i in range(1,n):
g = gcd(g,a[i]%k)
ans = [g]
x = (g+g)%k
while(x != g):
ans.append(x)
x += g
x %= k
print(len(ans))
ans.sort()
for i in ans:
print(i,end = " ")
print()
``` | instruction | 0 | 5,818 | 20 | 11,636 |
Yes | output | 1 | 5,818 | 20 | 11,637 |
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