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 |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya very much likes computers. Recently he has received a new "Ternatron IV" as a gift from his mother. Unlike other modern computers, "Ternatron IV" operates with ternary and not binary logic. Petya immediately wondered how the xor operation is performed on this computer (and whether there is anything like it).
It turned out that the operation does exist (however, it is called tor) and it works like this. Suppose that we need to calculate the value of the expression a tor b. Both numbers a and b are written in the ternary notation one under the other one (b under a). If they have a different number of digits, then leading zeroes are added to the shorter number until the lengths are the same. Then the numbers are summed together digit by digit. The result of summing each two digits is calculated modulo 3. Note that there is no carry between digits (i. e. during this operation the digits aren't transferred). For example: 1410 tor 5010 = 01123 tor 12123 = 10213 = 3410.
Petya wrote numbers a and c on a piece of paper. Help him find such number b, that a tor b = c. If there are several such numbers, print the smallest one.
Input
The first line contains two integers a and c (0 β€ a, c β€ 109). Both numbers are written in decimal notation.
Output
Print the single integer b, such that a tor b = c. If there are several possible numbers b, print the smallest one. You should print the number in decimal notation.
Examples
Input
14 34
Output
50
Input
50 34
Output
14
Input
387420489 225159023
Output
1000000001
Input
5 5
Output
0
Submitted Solution:
```
def toTernary(n):
i = 0
while(3**i < n):
i+=1
i-=1
ternary = []
while(i>=0):
multi = (n//3**i)
ternary.append(multi)
n -= multi * 3**i
i-=1
return ternary
def sameLen(t1, t2):
l1 = len(t1)
l2 = len(t2)
dif = abs(l1 - l2)
nL = []
for i in range(dif):
nL.append(0)
if l1 == l2:
return
elif l1 < l2:
t1 = nL + t1
return t1, t2
elif l1 > l2:
t2 = nL + t2
return t1, t2
ac = list(map(int, input().split()))
a = ac[0]
c = ac[1]
aT = toTernary(a)
cT = toTernary(c)
print(a, aT, c, cT)
aT, cT = sameLen(aT, cT)
print(aT, cT)
bT = []
for i in range(len(aT)):
b = cT[i] - aT[i]
if(b<0):
b = 3 + b
bT.append(b)
print(bT)
b = 0
for i in range(len(bT)):
b += (3**(len(bT)-i-1))*bT[i]
print(b)
``` | instruction | 0 | 58,197 | 20 | 116,394 |
No | output | 1 | 58,197 | 20 | 116,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya very much likes computers. Recently he has received a new "Ternatron IV" as a gift from his mother. Unlike other modern computers, "Ternatron IV" operates with ternary and not binary logic. Petya immediately wondered how the xor operation is performed on this computer (and whether there is anything like it).
It turned out that the operation does exist (however, it is called tor) and it works like this. Suppose that we need to calculate the value of the expression a tor b. Both numbers a and b are written in the ternary notation one under the other one (b under a). If they have a different number of digits, then leading zeroes are added to the shorter number until the lengths are the same. Then the numbers are summed together digit by digit. The result of summing each two digits is calculated modulo 3. Note that there is no carry between digits (i. e. during this operation the digits aren't transferred). For example: 1410 tor 5010 = 01123 tor 12123 = 10213 = 3410.
Petya wrote numbers a and c on a piece of paper. Help him find such number b, that a tor b = c. If there are several such numbers, print the smallest one.
Input
The first line contains two integers a and c (0 β€ a, c β€ 109). Both numbers are written in decimal notation.
Output
Print the single integer b, such that a tor b = c. If there are several possible numbers b, print the smallest one. You should print the number in decimal notation.
Examples
Input
14 34
Output
50
Input
50 34
Output
14
Input
387420489 225159023
Output
1000000001
Input
5 5
Output
0
Submitted Solution:
```
def toTernary(n):
i = 0
while(3**i < n):
i+=1
i-=1
ternary = []
while(i>=0):
multi = (n//3**i)
ternary.append(multi)
n -= multi * 3**i
i-=1
return ternary
def sameLen(t1, t2):
l1 = len(t1)
l2 = len(t2)
dif = abs(l1 - l2)
nL = []
for i in range(dif):
nL.append(0)
if l1 == l2:
return t1, t2
elif l1 < l2:
t1 = nL + t1
return t1, t2
elif l1 > l2:
t2 = nL + t2
return t1, t2
ac = list(map(int, input().split()))
a = ac[0]
c = ac[1]
aT = toTernary(a)
cT = toTernary(c)
#print(a, aT, c, cT)
aT, cT = sameLen(aT, cT)
#print(aT, cT)
bT = []
for i in range(len(aT)):
b = cT[i] - aT[i]
if(b<0):
b = 3 + b
bT.append(b)
#print(bT)
b = 0
for i in range(len(bT)):
b += (3**(len(bT)-i-1))*bT[i]
print(b)
``` | instruction | 0 | 58,198 | 20 | 116,396 |
No | output | 1 | 58,198 | 20 | 116,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya very much likes computers. Recently he has received a new "Ternatron IV" as a gift from his mother. Unlike other modern computers, "Ternatron IV" operates with ternary and not binary logic. Petya immediately wondered how the xor operation is performed on this computer (and whether there is anything like it).
It turned out that the operation does exist (however, it is called tor) and it works like this. Suppose that we need to calculate the value of the expression a tor b. Both numbers a and b are written in the ternary notation one under the other one (b under a). If they have a different number of digits, then leading zeroes are added to the shorter number until the lengths are the same. Then the numbers are summed together digit by digit. The result of summing each two digits is calculated modulo 3. Note that there is no carry between digits (i. e. during this operation the digits aren't transferred). For example: 1410 tor 5010 = 01123 tor 12123 = 10213 = 3410.
Petya wrote numbers a and c on a piece of paper. Help him find such number b, that a tor b = c. If there are several such numbers, print the smallest one.
Input
The first line contains two integers a and c (0 β€ a, c β€ 109). Both numbers are written in decimal notation.
Output
Print the single integer b, such that a tor b = c. If there are several possible numbers b, print the smallest one. You should print the number in decimal notation.
Examples
Input
14 34
Output
50
Input
50 34
Output
14
Input
387420489 225159023
Output
1000000001
Input
5 5
Output
0
Submitted Solution:
```
a, c = input().split()
a = int(a)
c = int(c)
lc = []
la = []
while a != 0:
la.append(a % 3)
a //= 3
while c != 0:
lc.append(c % 3)
c //= 3
d = max(len(la), len(lc))
d1 = d - len(la)
d2 = d - len(lc)
for i in range(d1):
la.append(0)
for i in range(d2):
lc.append(0)
la = la[-1::-1]
lc = lc[-1::-1]
l = []
sum = 0
print(la)
print(lc)
for i in range(d):
x = la[i]
y = lc[i]
for j in range(100):
if (x + j) % 3 == y:
break
l.append(j)
l = l[-1::-1]
for i in range(len(l)):
sum += (l[i] * (3 ** i))
print(sum)
``` | instruction | 0 | 58,199 | 20 | 116,398 |
No | output | 1 | 58,199 | 20 | 116,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya very much likes computers. Recently he has received a new "Ternatron IV" as a gift from his mother. Unlike other modern computers, "Ternatron IV" operates with ternary and not binary logic. Petya immediately wondered how the xor operation is performed on this computer (and whether there is anything like it).
It turned out that the operation does exist (however, it is called tor) and it works like this. Suppose that we need to calculate the value of the expression a tor b. Both numbers a and b are written in the ternary notation one under the other one (b under a). If they have a different number of digits, then leading zeroes are added to the shorter number until the lengths are the same. Then the numbers are summed together digit by digit. The result of summing each two digits is calculated modulo 3. Note that there is no carry between digits (i. e. during this operation the digits aren't transferred). For example: 1410 tor 5010 = 01123 tor 12123 = 10213 = 3410.
Petya wrote numbers a and c on a piece of paper. Help him find such number b, that a tor b = c. If there are several such numbers, print the smallest one.
Input
The first line contains two integers a and c (0 β€ a, c β€ 109). Both numbers are written in decimal notation.
Output
Print the single integer b, such that a tor b = c. If there are several possible numbers b, print the smallest one. You should print the number in decimal notation.
Examples
Input
14 34
Output
50
Input
50 34
Output
14
Input
387420489 225159023
Output
1000000001
Input
5 5
Output
0
Submitted Solution:
```
a,c=map(int,input().split())
a1="0"
while a>0:
a1=str(a%3)+a1
a//=3
c1="0"
while c>0:
c1=str(c%3)+c1
c//=3
if len(c1)>len(a1):
a1='0'*(len(c1)-len(a1))+a1
else:
c1='0'*(len(a1)-len(c1))+c1
i,b1=len(c1)-1,""
while i>=0:
if c1[i]=='0':
if a1[i]=='0':
b1='0'+b1
elif a1[i]=='1':
b1='2'+b1
else:
b1='1'+b1
elif c1[i]=='1':
if a1[i]=='0':
b1='1'+b1
elif a1[i]=='1':
b1='0'+b1
else:
b1='2'+b1
else:
if a1[i]=='0':
b1='2'+b1
elif a1[i]=='1':
b1='1'+b1
else:
b1='0'+b1
i-=1
print(int(b1,3))
``` | instruction | 0 | 58,200 | 20 | 116,400 |
No | output | 1 | 58,200 | 20 | 116,401 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11 | instruction | 0 | 58,434 | 20 | 116,868 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
import sys,os.path
if __name__ == '__main__':
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
n = input()
ans = []
no = int(n)
while no>0:
# print(no)
temp = ""
for i in range(len(n)):
a = int(n[i])
if a>=1:
temp+="1"
else:
temp+="0"
no = int(n)-int(temp)
n = str(no)
ans.append(temp)
print(len(ans))
print(*ans)
``` | output | 1 | 58,434 | 20 | 116,869 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11 | instruction | 0 | 58,435 | 20 | 116,870 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
strN = input()
l = len(strN)
strN = list(map(int, list(strN)))
k = max(strN)
print(k)
res = [0 for i in range(k)]
for i in range(l):
for j in range(strN[i]):
res[j] += 10**(l-i-1);
print(*res, sep = " ")
``` | output | 1 | 58,435 | 20 | 116,871 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11 | instruction | 0 | 58,436 | 20 | 116,872 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
digits = list(map(int, list(input())))
quasis = list()
for x in range(max(digits)) :
nextQuasi = "";
for y in range(len(digits)) :
if(digits[y] > 0) :
nextQuasi += "1"
digits[y]-=1
else :
nextQuasi += "0"
quasis.append(int(nextQuasi))
print (len(quasis))
print (" ".join(map(str, quasis)))
``` | output | 1 | 58,436 | 20 | 116,873 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11 | instruction | 0 | 58,437 | 20 | 116,874 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
n=int(input())
res=[]
while (n>0):
b=''
for i in str(n):
b+=str(min(int(i),1))
n-=int(b)
res.append(int(b))
print(len(res))
print(*res)
``` | output | 1 | 58,437 | 20 | 116,875 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11 | instruction | 0 | 58,438 | 20 | 116,876 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
n=input()
a=[int(x) for x in n]
ans=[]
mx=max(a)
for i in range(mx):
l=[0 for x in range(len(a))]
ans+=[l]
for j in range(len(a)):
val=a[j]
for i in range(mx):
if(val>0):
ans[i][j]=1
val-=1
else:
ans[i][j]=0
res=[]
for i in range(mx):
v=""
j=0
while(j<len(a) and ans[i][j]==0):
j+=1
while(j<len(a)):
v+=str(ans[i][j])
j+=1
res+=[int("".join(v))]
print(mx)
print(*res)
``` | output | 1 | 58,438 | 20 | 116,877 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11 | instruction | 0 | 58,439 | 20 | 116,878 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
l = list(map(int,list(input())))
print(max(l))
ans = []
while l.count(0) != len(l):
k = ['0' for i in range(len(l))]
for i in range(len(l)):
if l[i] != 0:
l[i] -= 1
k[i] = '1'
ans.append(str(int(''.join(k))))
print(' '.join(ans))
``` | output | 1 | 58,439 | 20 | 116,879 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11 | instruction | 0 | 58,440 | 20 | 116,880 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
s = input()
mat=[]
for i in s:
mat.append(int(i))
sat = [0] * max(mat)
for i in range(len(mat)):
for j in range(mat[i]):
sat[j] +=10**(len(mat) - i - 1)
print(len(sat))
print(*sat)
``` | output | 1 | 58,440 | 20 | 116,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11 | instruction | 0 | 58,441 | 20 | 116,882 |
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
n=str(int(input()))
l=[]
while(1):
s1=''
k=0
for i in range(len(n)):
if(int(n[i])>1):
s1+='1'
k=1
else:
s1+=n[i]
l.append(int(s1))
if(k==0):
break
n=str(int(n)-int(s1))
print(len(l))
print(*l)
``` | output | 1 | 58,441 | 20 | 116,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11
Submitted Solution:
```
n=list(map(int,input()))
m=max(n)
print(m)
for j in range(m):
l=[]
for i in n:
if i>j:l.append(str(1))
else:l.append(str(0))
print(int(''.join(l)))
``` | instruction | 0 | 58,442 | 20 | 116,884 |
Yes | output | 1 | 58,442 | 20 | 116,885 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11
Submitted Solution:
```
n = int(input())
a = []
while n > 0:
d = int(''.join(min(_, '1') for _ in str(n)))
n -= d;
a += [d];
print(len(a))
print(' '.join(map(str, a)))
``` | instruction | 0 | 58,443 | 20 | 116,886 |
Yes | output | 1 | 58,443 | 20 | 116,887 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11
Submitted Solution:
```
n=input()
n1=len(n)
k=0
for i in range(n1):
if(int(n[i])>k):
k=int(n[i])
print(k)
l=[]
for i in range(k):
l.append("")
for i in range(n1):
for j in range(int(n[i])):
l[j]+="1"
for j in range(int(n[i]),k):
l[j]+="0"
def fix(l):
l1=len(l)
p=[]
for i in range(l1):
s=0
if(l[i][0]=="0"):
j=1
r=len(l[i])
s=1
while(j<r):
if(l[i][j]=="0"):
s+=1
else:
break
j+=1
p.append(s)
for i in range(l1):
if(p[i]>0):
m=""
r=len(l[i])
for j in range(p[i],r):
m+=l[i][j]
l[i]=m
return l
fix(l)
for i in range(k):
print(l[i],"",end="")
if(k==0):
print(k)
``` | instruction | 0 | 58,444 | 20 | 116,888 |
Yes | output | 1 | 58,444 | 20 | 116,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11
Submitted Solution:
```
n = int(input())
ans = []
while n:
sol = 0
val = 1
k = n
while k:
if k%10:
sol += val
k = k//10
val *= 10
ans.append(sol)
n -= sol
print(len(ans))
for i in ans:
print(i,end=' ')
``` | instruction | 0 | 58,445 | 20 | 116,890 |
Yes | output | 1 | 58,445 | 20 | 116,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11
Submitted Solution:
```
kw = []
for i in range(63, 0, -1):
kw.append(int(str(bin(i))[2:]))
n = int(input())
res = []
i = 0
while n > 0:
while kw[i] <= n:
n -= kw[i]
res.append(kw[i])
i += 1
print(len(res))
for elem in res:
print(elem, end = ' ')
``` | instruction | 0 | 58,446 | 20 | 116,892 |
No | output | 1 | 58,446 | 20 | 116,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11
Submitted Solution:
```
from bisect import insort,bisect_right,bisect_left
from sys import stdout, stdin, setrecursionlimit
from heapq import heappush, heappop, heapify
from io import BytesIO, IOBase
from collections import *
from itertools import *
from random import *
from string import *
from queue import *
from math import *
from re import *
from os import *
# sqrt,ceil,floor,factorial,gcd,log2,log10,comb
####################################---fast-input-output----#########################################
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 = read(self._fd, max(fstat(self._fd).st_size, 8192))
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 = read(self._fd, max(fstat(self._fd).st_size, 8192))
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:
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")
stdin, stdout = IOWrapper(stdin), IOWrapper(stdout)
graph, mod, szzz = {}, 10**9 + 7, lambda: sorted(zzz())
def getStr(): return input()
def getInt(): return int(input())
def listStr(): return list(input())
def getStrs(): return input().split()
def isInt(s): return '0' <= s[0] <= '9'
def input(): return stdin.readline().strip()
def zzz(): return [int(i) for i in input().split()]
def output(answer, end='\n'): stdout.write(str(answer) + end)
def lcd(xnum1, xnum2): return (xnum1 * xnum2 // gcd(xnum1, xnum2))
def getPrimes(N = 10**5):
SN = int(sqrt(N))
sieve = [i for i in range(N+1)]
sieve[1] = 0
for i in sieve:
if i > SN:
break
if i == 0:
continue
for j in range(2*i, N+1, i):
sieve[j] = 0
prime = [i for i in range(N+1) if sieve[i] != 0]
return prime
def primeFactor(n,prime=getPrimes()):
lst = []
mx=int(sqrt(n))+1
for i in prime:
if i>mx:break
while n%i==0:
lst.append(i)
n//=i
if n>1:
lst.append(n)
return lst
dx = [-1, 1, 0, 0, 1, -1, 1, -1]
dy = [0, 0, 1, -1, 1, -1, -1, 1]
daysInMounth = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
#################################################---Some Rule For Me To Follow---#################################
"""
--instants of Reading problem continuously try to understand them.
--If you Know some-one , Then you probably don't know him !
--Try & again try, maybe you're just one statement away!
"""
##################################################---START-CODING---###############################################
n = getInt()
need = int(log2(n))
bank = [ int(bin(i)[2:]) for i in range(need+2) if int(bin(i)[2:]) <= n]
dp =[[]]+[None]*n
for i in range(n+1):
for j in bank:
if i+j>n:continue
if dp[i+j]==None:
dp[i+j]=[]
dp[i+j]=dp[i]+[j]
else:
if len(dp[i+j])>len(dp[i])+1 and dp[i]!=None:
dp[i+j]=dp[i]+[j]
print(len(dp[n]))
print(*dp[n])
``` | instruction | 0 | 58,447 | 20 | 116,894 |
No | output | 1 | 58,447 | 20 | 116,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11
Submitted Solution:
```
import itertools
from functools import reduce
check = list(itertools.product([0,1], repeat=7))
check_int = sorted([reduce(lambda a, x: a * 10 + x, t) for t in check],reverse=True)
n = int(input())
res = []
c = 0
while n>0:
if n<check_int[c]:
c+=1
else:
n-= check_int[c]
res.append(str(check_int[c]))
print (len(res))
print (' '.join(res[::-1]))
``` | instruction | 0 | 58,448 | 20 | 116,896 |
No | output | 1 | 58,448 | 20 | 116,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 β are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer n. Represent it as a sum of minimum number of quasibinary numbers.
Input
The first line contains a single integer n (1 β€ n β€ 106).
Output
In the first line print a single integer k β the minimum number of numbers in the representation of number n as a sum of quasibinary numbers.
In the second line print k numbers β the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal n. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Examples
Input
9
Output
9
1 1 1 1 1 1 1 1 1
Input
32
Output
3
10 11 11
Submitted Solution:
```
s = input()
n = len(s)
ans = "NO"
for i in range(n):
for j in range(i, n):
if s[:i] + s[j:] == "CODEFORCES":
ans = "YES"
print(ans)
``` | instruction | 0 | 58,449 | 20 | 116,898 |
No | output | 1 | 58,449 | 20 | 116,899 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karen has just arrived at school, and she has a math test today!
<image>
The test is about basic addition and subtraction. Unfortunately, the teachers were too busy writing tasks for Codeforces rounds, and had no time to make an actual test. So, they just put one question in the test that is worth all the points.
There are n integers written on a row. Karen must alternately add and subtract each pair of adjacent integers, and write down the sums or differences on the next row. She must repeat this process on the values on the next row, and so on, until only one integer remains. The first operation should be addition.
Note that, if she ended the previous row by adding the integers, she should start the next row by subtracting, and vice versa.
The teachers will simply look at the last integer, and then if it is correct, Karen gets a perfect score, otherwise, she gets a zero for the test.
Karen has studied well for this test, but she is scared that she might make a mistake somewhere and it will cause her final answer to be wrong. If the process is followed, what number can she expect to be written on the last row?
Since this number can be quite large, output only the non-negative remainder after dividing it by 109 + 7.
Input
The first line of input contains a single integer n (1 β€ n β€ 200000), the number of numbers written on the first row.
The next line contains n integers. Specifically, the i-th one among these is ai (1 β€ ai β€ 109), the i-th number on the first row.
Output
Output a single integer on a line by itself, the number on the final row after performing the process above.
Since this number can be quite large, print only the non-negative remainder after dividing it by 109 + 7.
Examples
Input
5
3 6 9 12 15
Output
36
Input
4
3 7 5 2
Output
1000000006
Note
In the first test case, the numbers written on the first row are 3, 6, 9, 12 and 15.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is still 36, so this is the correct output.
In the second test case, the numbers written on the first row are 3, 7, 5 and 2.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is 109 + 6, so this is the correct output. | instruction | 0 | 58,569 | 20 | 117,138 |
Tags: brute force, combinatorics, constructive algorithms, math
Correct Solution:
```
from sys import exit, stdin, stdout
n = int(stdin.readline())
a = [int(i) for i in stdin.readline().split()]
if n == 1:
print(a[0])
exit(0)
mod = 1000000007
f = [0] * (n + 1)
f[0] = 1
for i in range(1, n + 1):
f[i] = (f[i-1] * i) % mod
def f_pow(a, k):
if k == 0:
return 1
if k % 2 == 1:
return f_pow(a, k - 1) * a % mod
else:
return f_pow(a * a % mod, k // 2) % mod
def c(n, k):
d = f[k] * f[n - k] % mod
return f[n] * f_pow(d, mod - 2) % mod
oper = 1
while not (oper and n % 2 == 0):
for i in range(n - 1):
a[i] = a[i] + oper * a[i + 1]
oper *= -1
n -= 1
oper *= 1 if (n//2 % 2) != 0 else -1
sm1 = 0
sm2 = 0
for i in range(n):
if i % 2 == 0:
sm1 = (sm1 + c(n // 2 - 1, i // 2) * a[i]) % mod
else:
sm2 = (sm2 + c(n // 2 - 1, i // 2) * a[i]) % mod
stdout.write(str((sm1 + oper * sm2) % mod))
``` | output | 1 | 58,569 | 20 | 117,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karen has just arrived at school, and she has a math test today!
<image>
The test is about basic addition and subtraction. Unfortunately, the teachers were too busy writing tasks for Codeforces rounds, and had no time to make an actual test. So, they just put one question in the test that is worth all the points.
There are n integers written on a row. Karen must alternately add and subtract each pair of adjacent integers, and write down the sums or differences on the next row. She must repeat this process on the values on the next row, and so on, until only one integer remains. The first operation should be addition.
Note that, if she ended the previous row by adding the integers, she should start the next row by subtracting, and vice versa.
The teachers will simply look at the last integer, and then if it is correct, Karen gets a perfect score, otherwise, she gets a zero for the test.
Karen has studied well for this test, but she is scared that she might make a mistake somewhere and it will cause her final answer to be wrong. If the process is followed, what number can she expect to be written on the last row?
Since this number can be quite large, output only the non-negative remainder after dividing it by 109 + 7.
Input
The first line of input contains a single integer n (1 β€ n β€ 200000), the number of numbers written on the first row.
The next line contains n integers. Specifically, the i-th one among these is ai (1 β€ ai β€ 109), the i-th number on the first row.
Output
Output a single integer on a line by itself, the number on the final row after performing the process above.
Since this number can be quite large, print only the non-negative remainder after dividing it by 109 + 7.
Examples
Input
5
3 6 9 12 15
Output
36
Input
4
3 7 5 2
Output
1000000006
Note
In the first test case, the numbers written on the first row are 3, 6, 9, 12 and 15.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is still 36, so this is the correct output.
In the second test case, the numbers written on the first row are 3, 7, 5 and 2.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is 109 + 6, so this is the correct output. | instruction | 0 | 58,570 | 20 | 117,140 |
Tags: brute force, combinatorics, constructive algorithms, math
Correct Solution:
```
import sys
def CreateFact(n,mod):
cant=n+1
factorials=[1]*cant
for x in range(1,cant):
val=factorials[x-1]*x%mod
factorials[x]=val
inv_factorial=[1]*cant
inv_factorial[n]=pow(factorials[-1], mod - 2, mod)
for x in reversed(range(0,n)):
val=inv_factorial[x+1]*(x+1)%mod
inv_factorial[x]=val
return factorials,inv_factorial
def CalculateNPairsCoef(n,mod):
factorial,inv_factorial=CreateFact(n,mod)
coef=[1]*n
middle=int((n+1)/2)
last=n-1
for x in range(1,middle):
o=factorial[n-1]*inv_factorial[n-1-x]%mod*inv_factorial[x]%mod
coef[x]=o
coef[last-x]=o
return coef
def KarenAdTest():
n=int(sys.stdin.readline())
line =sys.stdin.readline().split()
i=0
while i<n:
x=int(line[i])
line[i]=x
i+=1
mod=1000000007
if n==1:
val=line[0]%mod
print(val)
return
if n==2:
val=(line[0]+line[1])%mod
print(val)
return
if n==3:
val=(line[0]+2*line[1]-line[2])%mod
print(val)
return
if n==4:
val=(line[0]-line[1]+line[2]-line[3])%mod
print(val)
return
if n==5:
val=(line[0]+2*line[2]+line[4])%mod
print(val)
return
#si el numero es mayor que 5 valos a calcular sus coeficientes finales
#Como es multiplo de 2 se calcula directo los coeficientes d la primera fila
#que son los d los n valores iniciles
coefi=[1]*n
if n%2==0:
m=int(n/2)
c=CalculateNPairsCoef(m,mod)
pos=0
if n%4==0:
for x in range(0,m):
coefi[pos]=c[x]
pos+=1
coefi[pos]=-c[x]
pos+=1
else:
for x in range(0,m):
coefi[pos]=c[x]
pos+=1
coefi[pos]=c[x]
pos+=1
#Como no es multiplo de dos se calculan los coeficientes d la 2da fila
else:
sr=n-1
m=int(sr/2)
c=CalculateNPairsCoef(m,mod)
co=[1]*sr
pos=2
for x in range(1,m):
co[pos]=c[x]
pos+=1
co[pos]=c[x]
pos+=1
if n%4==1:
coefi=[0]*n
coefi[0]=coefi[n-1]=1
for x in range(2,sr,2):
coefi[x]=co[x-1]+co[x]
else:
for x in range(2,sr,2):
coefi[x]=-co[x-1]+co[x]
for x in range(1,sr,2):
coefi[x]=co[x-1]+co[x]
coefi[sr]=-1
res=0
for x in range(0,n):
res+=(coefi[x]*line[x])%mod
res=int(res%mod)
print(res)
KarenAdTest()
``` | output | 1 | 58,570 | 20 | 117,141 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karen has just arrived at school, and she has a math test today!
<image>
The test is about basic addition and subtraction. Unfortunately, the teachers were too busy writing tasks for Codeforces rounds, and had no time to make an actual test. So, they just put one question in the test that is worth all the points.
There are n integers written on a row. Karen must alternately add and subtract each pair of adjacent integers, and write down the sums or differences on the next row. She must repeat this process on the values on the next row, and so on, until only one integer remains. The first operation should be addition.
Note that, if she ended the previous row by adding the integers, she should start the next row by subtracting, and vice versa.
The teachers will simply look at the last integer, and then if it is correct, Karen gets a perfect score, otherwise, she gets a zero for the test.
Karen has studied well for this test, but she is scared that she might make a mistake somewhere and it will cause her final answer to be wrong. If the process is followed, what number can she expect to be written on the last row?
Since this number can be quite large, output only the non-negative remainder after dividing it by 109 + 7.
Input
The first line of input contains a single integer n (1 β€ n β€ 200000), the number of numbers written on the first row.
The next line contains n integers. Specifically, the i-th one among these is ai (1 β€ ai β€ 109), the i-th number on the first row.
Output
Output a single integer on a line by itself, the number on the final row after performing the process above.
Since this number can be quite large, print only the non-negative remainder after dividing it by 109 + 7.
Examples
Input
5
3 6 9 12 15
Output
36
Input
4
3 7 5 2
Output
1000000006
Note
In the first test case, the numbers written on the first row are 3, 6, 9, 12 and 15.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is still 36, so this is the correct output.
In the second test case, the numbers written on the first row are 3, 7, 5 and 2.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is 109 + 6, so this is the correct output. | instruction | 0 | 58,571 | 20 | 117,142 |
Tags: brute force, combinatorics, constructive algorithms, math
Correct Solution:
```
from sys import exit
n = int(input())
a = [int(i) for i in input().split()]
if n == 1:
print(a[0])
exit(0)
mod = 1000000007
f = [0] * (n + 1)
f[0] = 1
for i in range(1, n + 1):
f[i] = (f[i-1] * i) % mod
def f_pow(a, k):
if k == 0:
return 1
if k % 2 == 1:
return f_pow(a, k - 1) * a % mod
else:
return f_pow(a * a % mod, k // 2) % mod
def c(n, k):
d = f[k] * f[n - k] % mod
return f[n] * f_pow(d, mod - 2) % mod
oper = 1
while not (oper and n % 2 == 0):
for i in range(n - 1):
a[i] = a[i] + oper * a[i + 1]
oper *= -1
n -= 1
oper *= 1 if (n//2 % 2) != 0 else -1
sm1 = 0
sm2 = 0
for i in range(n):
if i % 2 == 0:
sm1 = (sm1 + c(n // 2 - 1, i // 2) * a[i]) % mod
else:
sm2 = (sm2 + c(n // 2 - 1, i // 2) * a[i]) % mod
print((sm1 + oper * sm2) % mod)
``` | output | 1 | 58,571 | 20 | 117,143 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karen has just arrived at school, and she has a math test today!
<image>
The test is about basic addition and subtraction. Unfortunately, the teachers were too busy writing tasks for Codeforces rounds, and had no time to make an actual test. So, they just put one question in the test that is worth all the points.
There are n integers written on a row. Karen must alternately add and subtract each pair of adjacent integers, and write down the sums or differences on the next row. She must repeat this process on the values on the next row, and so on, until only one integer remains. The first operation should be addition.
Note that, if she ended the previous row by adding the integers, she should start the next row by subtracting, and vice versa.
The teachers will simply look at the last integer, and then if it is correct, Karen gets a perfect score, otherwise, she gets a zero for the test.
Karen has studied well for this test, but she is scared that she might make a mistake somewhere and it will cause her final answer to be wrong. If the process is followed, what number can she expect to be written on the last row?
Since this number can be quite large, output only the non-negative remainder after dividing it by 109 + 7.
Input
The first line of input contains a single integer n (1 β€ n β€ 200000), the number of numbers written on the first row.
The next line contains n integers. Specifically, the i-th one among these is ai (1 β€ ai β€ 109), the i-th number on the first row.
Output
Output a single integer on a line by itself, the number on the final row after performing the process above.
Since this number can be quite large, print only the non-negative remainder after dividing it by 109 + 7.
Examples
Input
5
3 6 9 12 15
Output
36
Input
4
3 7 5 2
Output
1000000006
Note
In the first test case, the numbers written on the first row are 3, 6, 9, 12 and 15.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is still 36, so this is the correct output.
In the second test case, the numbers written on the first row are 3, 7, 5 and 2.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is 109 + 6, so this is the correct output. | instruction | 0 | 58,572 | 20 | 117,144 |
Tags: brute force, combinatorics, constructive algorithms, math
Correct Solution:
```
import sys
#Este metodo calcula los coeficientes resultantes para n%2==0 y para los restantes calcula los coeficientes de su segunda fila
def RestWith4(c,m,n,v):
pos=0
coefi=[1]*n
if v:
for x in range(0,m):
coefi[pos]=c[x]
pos+=1
coefi[pos]=-c[x]
pos+=1
else:
for x in range(0,m):
coefi[pos]=c[x]
pos+=1
coefi[pos]=c[x]
pos+=1
return coefi
#Con este metodo se obtienen los arreglos de los valores factoriales de los numeros desde 0-n asi como su inverso en orden lineal
#Para luego calcular la combinatoria segun la posicion
def CreateFact(n,mod):
cant=n+1
factorials=[1]*cant
for x in range(1,cant):
val=factorials[x-1]*x%mod
factorials[x]=val
inv_factorial=[1]*cant
inv_factorial[n]=pow(factorials[-1], mod - 2, mod)
for x in reversed(range(0,n)):
val=inv_factorial[x+1]*(x+1)%mod
inv_factorial[x]=val
return factorials,inv_factorial
#Calcula los coeficientes de las posiciones impares que vimos que cumplen los mismo q las pares solo pueden variar el signo
def CalculateNPairsCoef(n,mod):
factorial,inv_factorial=CreateFact(n,mod)
coef=[1]*n
middle=int((n+1)/2)
last=n-1
for x in range(1,middle):
o=factorial[n-1]*inv_factorial[n-1-x]%mod*inv_factorial[x]%mod
coef[x]=o
coef[last-x]=o
return coef
def KarenAdTest():
#recibiendo y parseando los valores de entrada
n=int(sys.stdin.readline())
line =sys.stdin.readline().split()
i=0
while i<n:
x=int(line[i])
line[i]=x
i+=1
mod=1000000007
#Casos bases 1-5
if n==1:
val=line[0]%mod
print(val)
return
if n==2:
val=(line[0]+line[1])%mod
print(val)
return
if n==3:
val=(line[0]+2*line[1]-line[2])%mod
print(val)
return
if n==4:
val=(line[0]-line[1]+line[2]-line[3])%mod
print(val)
return
if n==5:
val=(line[0]+2*line[2]+line[4])%mod
print(val)
return
#si el numero es mayor que 5 valos a calcular sus coeficientes finales
#Como es multiplo de 2 se calcula directo los coeficientes d la primera fila
#que son los d los n valores iniciales
coefi=[]
if n%2==0:
m=int(n/2)
c=CalculateNPairsCoef(m,mod)
if n%4==0:
coefi=RestWith4(c,m,n,1)
else:
coefi=RestWith4(c,m,n,0)
#Como no es multiplo de dos se calculan los coeficientes d la 2da fila
else:
sr=n-1
m=int(sr/2)
c=CalculateNPairsCoef(m,mod)
co=RestWith4(c,m,sr,0)
#Si deja resto 1 con n entonces las posiciones pares se anulan
# y las pares son iguales a las suma de las dos anteriores
if n%4==1:
coefi=[0]*n
coefi[0]=coefi[n-1]=1
for x in range(2,sr,2):
coefi[x]=co[x-1]+co[x]
#Si deja resto 3 con n entonces las posiciones pares son la suma de las anterior y de ella en co
# y las impares son la diferencia entre ella y su anterior
else:
coefi=[1]*n
for x in range(2,sr,2):
coefi[x]=-co[x-1]+co[x]
for x in range(1,sr,2):
coefi[x]=co[x-1]+co[x]
coefi[sr]=-1
#Una vez calculados los coeficientes estos se multiplican por los valores de entrada
res=0
for x in range(0,n):
res+=(coefi[x]*line[x])%mod
res=int(res%mod)
print(res)
KarenAdTest()
``` | output | 1 | 58,572 | 20 | 117,145 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karen has just arrived at school, and she has a math test today!
<image>
The test is about basic addition and subtraction. Unfortunately, the teachers were too busy writing tasks for Codeforces rounds, and had no time to make an actual test. So, they just put one question in the test that is worth all the points.
There are n integers written on a row. Karen must alternately add and subtract each pair of adjacent integers, and write down the sums or differences on the next row. She must repeat this process on the values on the next row, and so on, until only one integer remains. The first operation should be addition.
Note that, if she ended the previous row by adding the integers, she should start the next row by subtracting, and vice versa.
The teachers will simply look at the last integer, and then if it is correct, Karen gets a perfect score, otherwise, she gets a zero for the test.
Karen has studied well for this test, but she is scared that she might make a mistake somewhere and it will cause her final answer to be wrong. If the process is followed, what number can she expect to be written on the last row?
Since this number can be quite large, output only the non-negative remainder after dividing it by 109 + 7.
Input
The first line of input contains a single integer n (1 β€ n β€ 200000), the number of numbers written on the first row.
The next line contains n integers. Specifically, the i-th one among these is ai (1 β€ ai β€ 109), the i-th number on the first row.
Output
Output a single integer on a line by itself, the number on the final row after performing the process above.
Since this number can be quite large, print only the non-negative remainder after dividing it by 109 + 7.
Examples
Input
5
3 6 9 12 15
Output
36
Input
4
3 7 5 2
Output
1000000006
Note
In the first test case, the numbers written on the first row are 3, 6, 9, 12 and 15.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is still 36, so this is the correct output.
In the second test case, the numbers written on the first row are 3, 7, 5 and 2.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is 109 + 6, so this is the correct output. | instruction | 0 | 58,573 | 20 | 117,146 |
Tags: brute force, combinatorics, constructive algorithms, math
Correct Solution:
```
#!/usr/bin/env python3
from sys import stdin, stdout
def main():
N = 400005
MOD = 10 ** 9 + 7
fact = [1] + [i+1 for i in range(N)]
for i in range(1, N + 1):
fact[i] *= fact[i - 1]
fact[i] %= MOD
def inv(n): return pow(n, MOD - 2, MOD)
def simplex(n, k):
if n == 0: return 1
if k == 0: return 0
return fact[n + k - 1] * inv(fact[n]) * inv(fact[k - 1]) % MOD
def F1(z, i):
if z == 1:
return 1
if z % 4 == 2:
x = z // 4 * 2
j = i % 4
if j == 0:
return simplex(x, i // 4 * 2 + 1)
elif j == 1:
return simplex(x, i // 4 * 2 + 1) * 2 % MOD
elif j == 2:
return MOD - simplex(x, i // 4 * 2 + 2)
else:
return 0
elif z % 4 == 0:
if i == 0:
return MOD - 1
if i == 1:
return 0
i -= 2
x = z // 4 * 2 - 1
j = i % 4
if j == 0:
return simplex(x, i // 4 * 2 + 2)
elif j == 1:
return simplex(x, i // 4 * 2 + 2) * 2 % MOD
elif j == 2:
return MOD - simplex(x, i // 4 * 2 + 3)
else:
return 0
elif z % 4 == 3:
if i == 0:
return MOD - 1
i -= 1
x = z // 4 * 2 + 1
y = z // 4 * 2 - 1
j = i % 4
if j % 4 == 0:
return simplex(x, i // 4 * 2 + 1)
elif j % 4 == 1 or j % 4 == 2:
return simplex(x, i // 4 * 2 + 2)
else:
z1 = 0 if y < 0 else simplex(y, i // 4 * 2 + 3)
ans = simplex(x, i // 4 * 2 + 1) - z1
if ans < 0: ans += MOD
return ans
else:
if i < 2:
return 1
if i == 2:
return MOD - (z // 4 * 2 - 1)
i -= 3
x = z // 4 * 2
y = z // 4 * 2 - 2
j = i % 4
if j % 4 == 0:
return simplex(x, i // 4 * 2 + 2)
elif j % 4 == 1 or j % 4 == 2:
return simplex(x, i // 4 * 2 + 3)
else:
z1 = 0 if y < 0 else simplex(y, i // 4 * 2 + 4)
ans = simplex(x, i // 4 * 2 + 2) - z1
if ans < 0: ans += MOD
return ans
def F2(n):
if n == 1: return [1]
return [F1(i, n - i) for i in range(1, n + 1)]
n = int(stdin.readline())
a = [int(x) for x in stdin.readline().split()]
coeffs = F2(n)
ans = 0
for c, x in zip(coeffs, a):
ans += c * x % MOD
if ans >= MOD:
ans -= MOD
stdout.write('{}\n'.format(ans))
if __name__ == "__main__":
main()
``` | output | 1 | 58,573 | 20 | 117,147 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karen has just arrived at school, and she has a math test today!
<image>
The test is about basic addition and subtraction. Unfortunately, the teachers were too busy writing tasks for Codeforces rounds, and had no time to make an actual test. So, they just put one question in the test that is worth all the points.
There are n integers written on a row. Karen must alternately add and subtract each pair of adjacent integers, and write down the sums or differences on the next row. She must repeat this process on the values on the next row, and so on, until only one integer remains. The first operation should be addition.
Note that, if she ended the previous row by adding the integers, she should start the next row by subtracting, and vice versa.
The teachers will simply look at the last integer, and then if it is correct, Karen gets a perfect score, otherwise, she gets a zero for the test.
Karen has studied well for this test, but she is scared that she might make a mistake somewhere and it will cause her final answer to be wrong. If the process is followed, what number can she expect to be written on the last row?
Since this number can be quite large, output only the non-negative remainder after dividing it by 109 + 7.
Input
The first line of input contains a single integer n (1 β€ n β€ 200000), the number of numbers written on the first row.
The next line contains n integers. Specifically, the i-th one among these is ai (1 β€ ai β€ 109), the i-th number on the first row.
Output
Output a single integer on a line by itself, the number on the final row after performing the process above.
Since this number can be quite large, print only the non-negative remainder after dividing it by 109 + 7.
Examples
Input
5
3 6 9 12 15
Output
36
Input
4
3 7 5 2
Output
1000000006
Note
In the first test case, the numbers written on the first row are 3, 6, 9, 12 and 15.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is still 36, so this is the correct output.
In the second test case, the numbers written on the first row are 3, 7, 5 and 2.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is 109 + 6, so this is the correct output. | instruction | 0 | 58,574 | 20 | 117,148 |
Tags: brute force, combinatorics, constructive algorithms, math
Correct Solution:
```
n = int(input())
p = list(map(int,input().split()))
MOD = 10**9+7
mode = 0
if n%4 == 3:
n-= 1
new = []
for i in range(n):
if mode == 0: new.append(p[i]+p[i+1])
else: new.append(p[i]-p[i+1])
mode = 1-mode
p = new
def calc0(p):
res = 0
ncr = 1
n = len(p)//2-1
for i in range(n+1):
res = (res+ncr*(p[i*2]-p[i*2+1])) % MOD
ncr = (ncr*(n-i)*pow(i+1,MOD-2,MOD)) % MOD
return res
def calc1(p):
res = 0
ncr = 1
n = len(p)//2
for i in range(n+1):
res = (res+ncr*(p[i*2])) % MOD
ncr = (ncr*(n-i)*pow(i+1,MOD-2,MOD)) % MOD
return res
def calc2(p):
res = 0
ncr = 1
n = len(p)//2-1
for i in range(n+1):
res = (res+ncr*(p[i*2]+p[i*2+1])) % MOD
ncr = (ncr*(n-i)*pow(i+1,MOD-2,MOD)) % MOD
return res
print([calc0,calc1,calc2,-1][n%4](p))
``` | output | 1 | 58,574 | 20 | 117,149 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karen has just arrived at school, and she has a math test today!
<image>
The test is about basic addition and subtraction. Unfortunately, the teachers were too busy writing tasks for Codeforces rounds, and had no time to make an actual test. So, they just put one question in the test that is worth all the points.
There are n integers written on a row. Karen must alternately add and subtract each pair of adjacent integers, and write down the sums or differences on the next row. She must repeat this process on the values on the next row, and so on, until only one integer remains. The first operation should be addition.
Note that, if she ended the previous row by adding the integers, she should start the next row by subtracting, and vice versa.
The teachers will simply look at the last integer, and then if it is correct, Karen gets a perfect score, otherwise, she gets a zero for the test.
Karen has studied well for this test, but she is scared that she might make a mistake somewhere and it will cause her final answer to be wrong. If the process is followed, what number can she expect to be written on the last row?
Since this number can be quite large, output only the non-negative remainder after dividing it by 109 + 7.
Input
The first line of input contains a single integer n (1 β€ n β€ 200000), the number of numbers written on the first row.
The next line contains n integers. Specifically, the i-th one among these is ai (1 β€ ai β€ 109), the i-th number on the first row.
Output
Output a single integer on a line by itself, the number on the final row after performing the process above.
Since this number can be quite large, print only the non-negative remainder after dividing it by 109 + 7.
Examples
Input
5
3 6 9 12 15
Output
36
Input
4
3 7 5 2
Output
1000000006
Note
In the first test case, the numbers written on the first row are 3, 6, 9, 12 and 15.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is still 36, so this is the correct output.
In the second test case, the numbers written on the first row are 3, 7, 5 and 2.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is 109 + 6, so this is the correct output. | instruction | 0 | 58,575 | 20 | 117,150 |
Tags: brute force, combinatorics, constructive algorithms, math
Correct Solution:
```
#!/usr/bin/env pypy3
import math
def make_nCr_mod(max_n=2*10**5 + 100, mod=10**9 + 7):
fact, inv_fact = [0] * (max_n + 1), [0] * (max_n + 1)
fact[0] = 1
for i in range(max_n):
fact[i + 1] = fact[i] * (i + 1) % mod
inv_fact[-1] = pow(fact[-1], mod - 2, mod)
for i in reversed(range(max_n)):
inv_fact[i] = inv_fact[i + 1] * (i + 1) % mod
def nCr_mod(n, r):
res = 1
while n or r:
a, b = n % mod, r % mod
if a < b:
return 0
res = res * fact[a] % mod * inv_fact[b] % mod * inv_fact[a - b] % mod
n //= mod
r //= mod
return res
return nCr_mod
nCr_mod = make_nCr_mod()
MODULUS = 10**9+7
input()
A = input().split(' ')
A = list(map(int, A))
if len(A) == 1:
print(A[0])
exit(0)
if len(A) % 2 == 1:
new_A = []
next_plus = True
for i in range(len(A) - 1):
if next_plus:
new_A += [A[i] + A[i+1]]
else:
new_A += [A[i] - A[i+1]]
next_plus = not next_plus
A = new_A
if len(A) % 4 == 2:
new_A = []
for i in range(len(A) // 2):
new_A += [A[2*i] + A[2*i+1]]
A = new_A
else:
new_A = []
for i in range(len(A) // 2):
new_A += [A[2*i] - A[2*i+1]]
A = new_A
# binomial sum
N = len(A)-1
ret = 0
for i in range(N+1):
ret += A[i]*nCr_mod(N, i)
ret = ret % MODULUS
print(ret)
``` | output | 1 | 58,575 | 20 | 117,151 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karen has just arrived at school, and she has a math test today!
<image>
The test is about basic addition and subtraction. Unfortunately, the teachers were too busy writing tasks for Codeforces rounds, and had no time to make an actual test. So, they just put one question in the test that is worth all the points.
There are n integers written on a row. Karen must alternately add and subtract each pair of adjacent integers, and write down the sums or differences on the next row. She must repeat this process on the values on the next row, and so on, until only one integer remains. The first operation should be addition.
Note that, if she ended the previous row by adding the integers, she should start the next row by subtracting, and vice versa.
The teachers will simply look at the last integer, and then if it is correct, Karen gets a perfect score, otherwise, she gets a zero for the test.
Karen has studied well for this test, but she is scared that she might make a mistake somewhere and it will cause her final answer to be wrong. If the process is followed, what number can she expect to be written on the last row?
Since this number can be quite large, output only the non-negative remainder after dividing it by 109 + 7.
Input
The first line of input contains a single integer n (1 β€ n β€ 200000), the number of numbers written on the first row.
The next line contains n integers. Specifically, the i-th one among these is ai (1 β€ ai β€ 109), the i-th number on the first row.
Output
Output a single integer on a line by itself, the number on the final row after performing the process above.
Since this number can be quite large, print only the non-negative remainder after dividing it by 109 + 7.
Examples
Input
5
3 6 9 12 15
Output
36
Input
4
3 7 5 2
Output
1000000006
Note
In the first test case, the numbers written on the first row are 3, 6, 9, 12 and 15.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is still 36, so this is the correct output.
In the second test case, the numbers written on the first row are 3, 7, 5 and 2.
Karen performs the operations as follows:
<image>
The non-negative remainder after dividing the final number by 109 + 7 is 109 + 6, so this is the correct output. | instruction | 0 | 58,576 | 20 | 117,152 |
Tags: brute force, combinatorics, constructive algorithms, math
Correct Solution:
```
"""
5
3 6 9 12 15 should output 36
4
3 7 5 2 should output 1000000006
"""
from sys import exit
from typing import List
MOD = 10 ** 9 + 7;
def mod_inv(n: int) -> int:
return pow(n, MOD - 2, MOD)
def tri_sum(row: List[int]) -> int:
pascal_row = [1]
for i in range(1, len(row)):
last = pascal_row[-1]
curr = (last * (len(row) - i) * mod_inv(i)) % MOD
pascal_row.append(curr)
total = 0
for contrib, val in zip(pascal_row, row):
total = (total + contrib * val) % MOD
return total
row_len = int(input())
row = [int(i) for i in input().split()]
if len(row) != row_len:
raise ValueError(f"the lengths {row_len} and {len(row)} are different so uh yeah")
if len(row) == 1: # stupid edge cases
print(row[0])
exit()
if row_len % 2 == 1:
row_len -= 1
new_row = []
for i in range(row_len):
new_row.append(row[i] + row[i + 1] if i % 2 == 0 else row[i] - row[i + 1])
row = new_row
alt_rows = [[], []]
for i, v in enumerate(row):
alt_rows[i % 2].append(v)
first = tri_sum(alt_rows[0])
second = tri_sum(alt_rows[1])
ans = (first + second if ((row_len - 1) * row_len // 2) % 2 == 1 else first - second) % MOD
print(ans)
``` | output | 1 | 58,576 | 20 | 117,153 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Absent-minded Masha got set of n cubes for her birthday.
At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural x such she can make using her new cubes all integers from 1 to x.
To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number.
The number can't contain leading zeros. It's not required to use all cubes to build a number.
Pay attention: Masha can't make digit 6 from digit 9 and vice-versa using cube rotations.
Input
In first line integer n is given (1 β€ n β€ 3) β the number of cubes, Masha got for her birthday.
Each of next n lines contains 6 integers aij (0 β€ aij β€ 9) β number on j-th face of i-th cube.
Output
Print single integer β maximum number x such Masha can make any integers from 1 to x using her cubes or 0 if Masha can't make even 1.
Examples
Input
3
0 1 2 3 4 5
6 7 8 9 0 1
2 3 4 5 6 7
Output
87
Input
3
0 1 3 5 6 8
1 2 4 5 7 8
2 3 4 6 7 9
Output
98
Note
In the first test case, Masha can build all numbers from 1 to 87, but she can't make 88 because there are no two cubes with digit 8. | instruction | 0 | 58,598 | 20 | 117,196 |
Tags: brute force, implementation
Correct Solution:
```
order=["012","021","102","120","201","210"]
cube=["","",""]
for _ in range(int(input())):
cube[_]=input()
for num in range(1,999):
num=str(num)
flag=0
for o in range(6):
i=0
for ele in num:
if ele not in cube[int(order[o][i])]:
break
i+=1
else:
flag=1
if flag==1:
break
if not flag:
break
print(int(num)-1)
``` | output | 1 | 58,598 | 20 | 117,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Absent-minded Masha got set of n cubes for her birthday.
At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural x such she can make using her new cubes all integers from 1 to x.
To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number.
The number can't contain leading zeros. It's not required to use all cubes to build a number.
Pay attention: Masha can't make digit 6 from digit 9 and vice-versa using cube rotations.
Input
In first line integer n is given (1 β€ n β€ 3) β the number of cubes, Masha got for her birthday.
Each of next n lines contains 6 integers aij (0 β€ aij β€ 9) β number on j-th face of i-th cube.
Output
Print single integer β maximum number x such Masha can make any integers from 1 to x using her cubes or 0 if Masha can't make even 1.
Examples
Input
3
0 1 2 3 4 5
6 7 8 9 0 1
2 3 4 5 6 7
Output
87
Input
3
0 1 3 5 6 8
1 2 4 5 7 8
2 3 4 6 7 9
Output
98
Note
In the first test case, Masha can build all numbers from 1 to 87, but she can't make 88 because there are no two cubes with digit 8. | instruction | 0 | 58,599 | 20 | 117,198 |
Tags: brute force, implementation
Correct Solution:
```
def check(s, a, used):
if len(used) == len(s):
return True
n = len(a)
index = len(used)
for i in range(n):
if i not in used and s[index] in a[i]:
result = check(s, a, used.union(set([i])))
if result:
return True
return False
def valid(s, a):
if len(s) > len(a):
return False
return check(s, a, set())
n = int(input())
a = []
for _ in range(n):
a.append(list(input().split()))
count = 1
while True:
if not valid(str(count), a):
print(count - 1)
break
count += 1
``` | output | 1 | 58,599 | 20 | 117,199 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Absent-minded Masha got set of n cubes for her birthday.
At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural x such she can make using her new cubes all integers from 1 to x.
To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number.
The number can't contain leading zeros. It's not required to use all cubes to build a number.
Pay attention: Masha can't make digit 6 from digit 9 and vice-versa using cube rotations.
Input
In first line integer n is given (1 β€ n β€ 3) β the number of cubes, Masha got for her birthday.
Each of next n lines contains 6 integers aij (0 β€ aij β€ 9) β number on j-th face of i-th cube.
Output
Print single integer β maximum number x such Masha can make any integers from 1 to x using her cubes or 0 if Masha can't make even 1.
Examples
Input
3
0 1 2 3 4 5
6 7 8 9 0 1
2 3 4 5 6 7
Output
87
Input
3
0 1 3 5 6 8
1 2 4 5 7 8
2 3 4 6 7 9
Output
98
Note
In the first test case, Masha can build all numbers from 1 to 87, but she can't make 88 because there are no two cubes with digit 8. | instruction | 0 | 58,603 | 20 | 117,206 |
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
a = [list(map(int, input().split())) for _ in range(n)]
if n == 1:
avail = set(a[0])
elif n == 2:
avail = set(a[0] + a[1])
avail = avail.union({i * 10 + j for i in a[0] for j in a[1]})
avail = avail.union({i * 10 + j for i in a[1] for j in a[0]})
else:
avail = set(a[0] + a[1] + a[2])
avail = avail.union({i * 10 + j for i in a[0] for j in a[1]})
avail = avail.union({i * 10 + j for i in a[1] for j in a[0]})
avail = avail.union({i * 10 + j for i in a[0] for j in a[2]})
avail = avail.union({i * 10 + j for i in a[2] for j in a[0]})
avail = avail.union({i * 10 + j for i in a[2] for j in a[1]})
avail = avail.union({i * 10 + j for i in a[1] for j in a[2]})
for c in [[0, 1, 2], [0, 2, 1], [1, 0, 2], [1, 2, 0], [2, 0, 1], [2, 1, 0]]:
avail = avail.union({i * 100 + j * 10 + k for i in a[c[0]] for j in a[c[1]] for k in a[c[2]]})
for i in range(1, 1000):
if i not in avail:
print(i - 1)
exit()
print(999)
``` | output | 1 | 58,603 | 20 | 117,207 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...
Your task is to print the k-th digit of this sequence.
Input
The first and only line contains integer k (1 β€ k β€ 10^{12}) β the position to process (1-based index).
Output
Print the k-th digit of the resulting infinite sequence.
Examples
Input
7
Output
7
Input
21
Output
5 | instruction | 0 | 58,896 | 20 | 117,792 |
Tags: binary search, divide and conquer, implementation
Correct Solution:
```
n = int(input())
strt = 1
n -= 1
cnt = 9
while n > strt * cnt:
n -= cnt * strt
cnt *= 10
strt += 1
ans = (8 + 2) ** (strt - 1)
ans += n // strt
print(str(ans)[n % strt])
``` | output | 1 | 58,896 | 20 | 117,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...
Your task is to print the k-th digit of this sequence.
Input
The first and only line contains integer k (1 β€ k β€ 10^{12}) β the position to process (1-based index).
Output
Print the k-th digit of the resulting infinite sequence.
Examples
Input
7
Output
7
Input
21
Output
5 | instruction | 0 | 58,897 | 20 | 117,794 |
Tags: binary search, divide and conquer, implementation
Correct Solution:
```
k=int(input())
i=0
r=1
while(k>=r):
r+=9*(i+1)*10**i
i+=1
# print(i,r)
r=r-(9*i*10**(i-1))
# print(r)
ans=str(((k-r)//i)+10**(i-1))
# print/(ans)
ans=ans[(k-r)%i]
print(ans)
``` | output | 1 | 58,897 | 20 | 117,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...
Your task is to print the k-th digit of this sequence.
Input
The first and only line contains integer k (1 β€ k β€ 10^{12}) β the position to process (1-based index).
Output
Print the k-th digit of the resulting infinite sequence.
Examples
Input
7
Output
7
Input
21
Output
5 | instruction | 0 | 58,898 | 20 | 117,796 |
Tags: binary search, divide and conquer, implementation
Correct Solution:
```
k = int(input())
n = 1
up_bnd = 9
while(k > up_bnd):
n += 1
up_bnd += (9*n)*(10**(n-1))
low_bnd = 0
for i in range(1, n):
low_bnd += (9*i)*(10**(i-1))
num = int((k-low_bnd)/n)
lb_val = 0
for i in range(n-1):
lb_val = (lb_val*10)+9
num += lb_val
rm = (k-low_bnd) % n
if(rm != 0):
num += 1
ans = 0
if(rm == 0):
ans = num % 10
else:
for i in range(n-rm+1):
j = (num % 10)
num = int(num/10)
ans = j
print(int(ans))
``` | output | 1 | 58,898 | 20 | 117,797 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...
Your task is to print the k-th digit of this sequence.
Input
The first and only line contains integer k (1 β€ k β€ 10^{12}) β the position to process (1-based index).
Output
Print the k-th digit of the resulting infinite sequence.
Examples
Input
7
Output
7
Input
21
Output
5 | instruction | 0 | 58,899 | 20 | 117,798 |
Tags: binary search, divide and conquer, implementation
Correct Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from bisect import bisect_left as bsl
def main():
cur=9;count=1;tot=0
num=[];cc=[]
for s in range(11):
num.append(cur*count)
tot+=cur
cc.append(tot)
cur*=10;count+=1
ans=[num[0]]
for s in range(1,11):
ans.append(ans[-1]+num[s])
k=int(input())
ind=min(bsl(ans,k),10)
left=k
if ind>0:
left-=ans[ind-1]
#sort out this bit below, might be ceil instead of //
nums=left//(ind+1);rem=left%(ind+1)
if left%(ind+1)!=0:
nums+=1
if ind>0:
nums+=cc[ind-1]
answer=[int(k) for k in str(nums)]
print(answer[rem-1])
main()
``` | output | 1 | 58,899 | 20 | 117,799 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...
Your task is to print the k-th digit of this sequence.
Input
The first and only line contains integer k (1 β€ k β€ 10^{12}) β the position to process (1-based index).
Output
Print the k-th digit of the resulting infinite sequence.
Examples
Input
7
Output
7
Input
21
Output
5 | instruction | 0 | 58,900 | 20 | 117,800 |
Tags: binary search, divide and conquer, implementation
Correct Solution:
```
#Bhargey Mehta (Junior)
#DA-IICT, Gandhinagar
import sys, math, queue, bisect
#sys.stdin = open('input.txt', 'r')
MOD = 998244353
sys.setrecursionlimit(1000000)
n = int(input())
if n < 10:
print(n)
exit()
d = 1
while n > 9*d*pow(10, d-1):
n -= 9*d*pow(10, d-1)
d += 1
x = pow(10, d-1) + (n-1)//d
p = n % d
x = str(x).zfill(d)
print(x[p-1])
``` | output | 1 | 58,900 | 20 | 117,801 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...
Your task is to print the k-th digit of this sequence.
Input
The first and only line contains integer k (1 β€ k β€ 10^{12}) β the position to process (1-based index).
Output
Print the k-th digit of the resulting infinite sequence.
Examples
Input
7
Output
7
Input
21
Output
5 | instruction | 0 | 58,901 | 20 | 117,802 |
Tags: binary search, divide and conquer, implementation
Correct Solution:
```
k = int(input())
k_=k
ca = 9
di = 1
tem = 9
while k_>0:
k_-=ca*di
ca*=10
di+=1
tem += ca*di
if k_==0:
break
tem -= ca*di
ca=int(ca/10)
di-=1
tem -= ca*di
ca=int(ca/10)
ca_=0
while ca>0:
ca_+=ca
ca= int(ca/10)
k -= tem
re = int((k+di-1)//di)+ca_
re_= k%di
if re_==0:
l=1
else:
l = 10**(di-re_)
re = int(re//l)
print(re%10)
``` | output | 1 | 58,901 | 20 | 117,803 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...
Your task is to print the k-th digit of this sequence.
Input
The first and only line contains integer k (1 β€ k β€ 10^{12}) β the position to process (1-based index).
Output
Print the k-th digit of the resulting infinite sequence.
Examples
Input
7
Output
7
Input
21
Output
5 | instruction | 0 | 58,902 | 20 | 117,804 |
Tags: binary search, divide and conquer, implementation
Correct Solution:
```
N = int(input())
beg = 0
end = 9
i = 0 # Π² ΡΠΊΠΎΠ»ΡΠΊΠΈ Π·Π½Π°ΡΠ½ΡΡ
ΡΠΈΡΠ»Π°Ρ
ΠΌΡ Π½Π°Ρ
ΠΎΠ΄ΠΈΠΌΡΡ - 1
while N > end:
i += 1
beg, end = end, end + (i + 1) * 9 * 10**i
n = N - beg - 1 # ΡΡΠΎ N ΠΎΡΠ½ΠΎΡΠΈΡΠ΅Π»ΡΠ½ΠΎ Π½Π°ΡΠ°Π»Π° ΡΠΈΡΠ΅Π» Ρ Π΄Π»ΠΈΠ½Π½ΠΎΠΉ i, Π½Π°ΡΠΈΠ½Π°Ρ Ρ 0
lvl = i - n % (i + 1) # Π½ΠΎΠΌΠ΅Ρ ΡΠΈΠΌΠ²Π°Π»Π°, Π½Π°ΡΠΈΠ½Π°Ρ ΠΎΡ ΠΊΠΎΠ½ΡΠ° ΡΠΈΡΠ»Π°
period = (i + 1) * 10**lvl # ΠΏΠ΅ΡΠΈΠΎΠ΄ ΡΠΌΠ΅Π½Ρ ΡΠΈΡΠ»Π°
res = n//period % 10
if lvl == i:
res += 1
print(res)
``` | output | 1 | 58,902 | 20 | 117,805 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...
Your task is to print the k-th digit of this sequence.
Input
The first and only line contains integer k (1 β€ k β€ 10^{12}) β the position to process (1-based index).
Output
Print the k-th digit of the resulting infinite sequence.
Examples
Input
7
Output
7
Input
21
Output
5 | instruction | 0 | 58,903 | 20 | 117,806 |
Tags: binary search, divide and conquer, implementation
Correct Solution:
```
from collections import deque, defaultdict, Counter
from itertools import product, groupby, permutations, combinations
from math import gcd, floor, inf, log2, sqrt, log10
from bisect import bisect_right, bisect_left
from statistics import mode
from string import ascii_uppercase
k = int(input()) -1
y = 9
x = 1
while k > x*y:
k -= x*y
y *= 10
x += 1
start = 10**(x-1)
start += k//x
print(str(start)[k%x])
``` | output | 1 | 58,903 | 20 | 117,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...
Your task is to print the k-th digit of this sequence.
Input
The first and only line contains integer k (1 β€ k β€ 10^{12}) β the position to process (1-based index).
Output
Print the k-th digit of the resulting infinite sequence.
Examples
Input
7
Output
7
Input
21
Output
5
Submitted Solution:
```
n=int(input())
s=0; pred=0
for i in range(1,20):
m=9*pow(10,i-1)*i
s+=m
if n<=s:
nd=pow(10,i-1)
sme=n-pred
num=sme//i
ost=sme%i
if ost==0:
dig=nd+num-1
else:
dig=nd+num
d=i
rez=[]
ddig=dig
while d>0:
o=ddig%10
a=ddig//10
rez.append(o)
d-=1
ddig=a
break
pred=s
print(str(rez[-ost]))
``` | instruction | 0 | 58,904 | 20 | 117,808 |
Yes | output | 1 | 58,904 | 20 | 117,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...
Your task is to print the k-th digit of this sequence.
Input
The first and only line contains integer k (1 β€ k β€ 10^{12}) β the position to process (1-based index).
Output
Print the k-th digit of the resulting infinite sequence.
Examples
Input
7
Output
7
Input
21
Output
5
Submitted Solution:
```
digit=1
x=int(input())
start_range=1
end_range=9
nine=9
total=0
while(x>end_range):
start_range=end_range+1
digit+=1
end_range+=(digit*nine*pow(10,digit-1))
total=total+start_range
start=pow(10,digit-1)
number=int((x-start_range)/digit)
index=int((x-start_range)%digit)
start=start+number
string=str(start)
print(string[index])
``` | instruction | 0 | 58,906 | 20 | 117,812 |
Yes | output | 1 | 58,906 | 20 | 117,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...
Your task is to print the k-th digit of this sequence.
Input
The first and only line contains integer k (1 β€ k β€ 10^{12}) β the position to process (1-based index).
Output
Print the k-th digit of the resulting infinite sequence.
Examples
Input
7
Output
7
Input
21
Output
5
Submitted Solution:
```
k=int(input(''))
ch=''
for i in range (1, k + 1):
while (len(ch)<k):
ch+=str(i)
print(ch[k-1])
``` | instruction | 0 | 58,909 | 20 | 117,818 |
No | output | 1 | 58,909 | 20 | 117,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's write all the positive integer numbers one after another from 1 without any delimiters (i.e. as a single string). It will be the infinite sequence starting with 123456789101112131415161718192021222324252627282930313233343536...
Your task is to print the k-th digit of this sequence.
Input
The first and only line contains integer k (1 β€ k β€ 10^{12}) β the position to process (1-based index).
Output
Print the k-th digit of the resulting infinite sequence.
Examples
Input
7
Output
7
Input
21
Output
5
Submitted Solution:
```
n=int(input())
x=1
while n>((len(str(x)))**10*9):
n-=len(str(x))**10*9
x*=10
while n>len(str(x)):
n-=len(str(x))
x+=1
for i in range(len(str(x))):
if n!=0:
s=str(x)[i]
n-=1
print(s)
``` | instruction | 0 | 58,911 | 20 | 117,822 |
No | output | 1 | 58,911 | 20 | 117,823 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the n-th position of the sequence.
Input
The only line contains integer n (1 β€ n β€ 1014) β the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output
Print the element in the n-th position of the sequence (the elements are numerated from one).
Examples
Input
3
Output
2
Input
5
Output
2
Input
10
Output
4
Input
55
Output
10
Input
56
Output
1 | instruction | 0 | 59,282 | 20 | 118,564 |
Tags: implementation, math
Correct Solution:
```
def checker(n, m):
return n - (m * (m + 1) / 2)
n = int(input())
if n == 1:
print(1)
exit(0)
l, r = 0, n + 1
while r - l > 1:
m = (l + r) // 2
if checker(n, m) >= 0:
l = m
else:
r = m
#print(l, r)
if checker(n, l) == 0:
print(l)
else:
print(int(checker(n, l)))
``` | output | 1 | 59,282 | 20 | 118,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the n-th position of the sequence.
Input
The only line contains integer n (1 β€ n β€ 1014) β the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output
Print the element in the n-th position of the sequence (the elements are numerated from one).
Examples
Input
3
Output
2
Input
5
Output
2
Input
10
Output
4
Input
55
Output
10
Input
56
Output
1 | instruction | 0 | 59,283 | 20 | 118,566 |
Tags: implementation, math
Correct Solution:
```
import math
n = int(input())
# 1 positions
# 0 1 3 6 10 15 21 28
# 1 2 3 4 5 6 7
# n = (block*(block+1))/2
block = (1/2)*(math.sqrt(8*n-1)-1)
block = math.floor(block)
print(n - (block*(block+1))//2)
``` | output | 1 | 59,283 | 20 | 118,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the n-th position of the sequence.
Input
The only line contains integer n (1 β€ n β€ 1014) β the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output
Print the element in the n-th position of the sequence (the elements are numerated from one).
Examples
Input
3
Output
2
Input
5
Output
2
Input
10
Output
4
Input
55
Output
10
Input
56
Output
1 | instruction | 0 | 59,284 | 20 | 118,568 |
Tags: implementation, math
Correct Solution:
```
import math
n = int(input())
e = (-1 + int(math.sqrt(1+8*n)))//2
z = e*(e+1)//2
r = n - z
if(r==0):
print(e)
else:
print(r)
``` | output | 1 | 59,284 | 20 | 118,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the n-th position of the sequence.
Input
The only line contains integer n (1 β€ n β€ 1014) β the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output
Print the element in the n-th position of the sequence (the elements are numerated from one).
Examples
Input
3
Output
2
Input
5
Output
2
Input
10
Output
4
Input
55
Output
10
Input
56
Output
1 | instruction | 0 | 59,285 | 20 | 118,570 |
Tags: implementation, math
Correct Solution:
```
n=int(input())
low=1
high=n
p=0
while low<=high:
mid=(low+high)//2
s=(mid+1)*mid//2
if s>=n and mid*(mid-1)//2<n:
p=mid
break
if s<n:
low=mid+1
else:
high=mid-1
n-=p*(p-1)//2
print(n)
``` | output | 1 | 59,285 | 20 | 118,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the n-th position of the sequence.
Input
The only line contains integer n (1 β€ n β€ 1014) β the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output
Print the element in the n-th position of the sequence (the elements are numerated from one).
Examples
Input
3
Output
2
Input
5
Output
2
Input
10
Output
4
Input
55
Output
10
Input
56
Output
1 | instruction | 0 | 59,286 | 20 | 118,572 |
Tags: implementation, math
Correct Solution:
```
#!/usr/bin/env python3
def main(args):
n = int(input())
current = 1
prev = 0
while n > current + prev:
prev += current
current += 1
print(n-prev)
if __name__ == '__main__':
import sys
sys.exit(main(sys.argv))
``` | output | 1 | 59,286 | 20 | 118,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the n-th position of the sequence.
Input
The only line contains integer n (1 β€ n β€ 1014) β the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output
Print the element in the n-th position of the sequence (the elements are numerated from one).
Examples
Input
3
Output
2
Input
5
Output
2
Input
10
Output
4
Input
55
Output
10
Input
56
Output
1 | instruction | 0 | 59,287 | 20 | 118,574 |
Tags: implementation, math
Correct Solution:
```
a=int(input())
b=int((2*a)**.5)
c=a-(b**2+b)//2
if c==0:print(b)
else:print(c%b)
``` | output | 1 | 59,287 | 20 | 118,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the n-th position of the sequence.
Input
The only line contains integer n (1 β€ n β€ 1014) β the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output
Print the element in the n-th position of the sequence (the elements are numerated from one).
Examples
Input
3
Output
2
Input
5
Output
2
Input
10
Output
4
Input
55
Output
10
Input
56
Output
1 | instruction | 0 | 59,288 | 20 | 118,576 |
Tags: implementation, math
Correct Solution:
```
n = int(input())
a, b = 0, 10**14
while a < b:
c = (a + b + 1) // 2
if c * (c + 1) // 2 >= n:
b = c - 1
else:
a = c
print(n - (a * (a + 1) // 2))
``` | output | 1 | 59,288 | 20 | 118,577 |
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