message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and the hard versions is the maximum value of k.
You are given an infinite sequence of form "112123123412345..." which consist of blocks of all consecutive positive integers written one after another. The first block consists of all numbers from 1 to 1, the second one — from 1 to 2, the third one — from 1 to 3, ..., the i-th block consists of all numbers from 1 to i.
So the first 56 elements of the sequence are "11212312341234512345612345671234567812345678912345678910". Elements of the sequence are numbered from one. For example, the 1-st element of the sequence is 1, the 3-rd element of the sequence is 2, the 20-th element of the sequence is 5, the 38-th element is 2, the 56-th element of the sequence is 0.
Your task is to answer q independent queries. In the i-th query you are given one integer k_i. Calculate the digit at the position k_i of the sequence.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 500) — the number of queries.
The i-th of the following q lines contains one integer k_i (1 ≤ k_i ≤ 10^{18}) — the description of the corresponding query.
Output
Print q lines. In the i-th line print one digit x_i (0 ≤ x_i ≤ 9) — the answer to the query i, i.e. x_i should be equal to the element at the position k_i of the sequence.
Examples
Input
5
1
3
20
38
56
Output
1
2
5
2
0
Input
4
2132
506
999999999999999999
1000000000000000000
Output
8
2
4
1
Note
Answers on queries from the first example are described in the problem statement.
Submitted Solution:
```
def init(maxn):
Sum = [0] * maxn
Single = [0] * maxn
for i in range(1, maxn):
lens = 10 ** i - 10 ** (i - 1)
pre = Single[i - 1]
Single[i] = pre + lens * i
for i in range(1, maxn):
lens = 10 ** i - 10 ** (i - 1)
pre = Single[i-1]
Sum[i] = (pre + i + pre + lens * i) * lens // 2 + Sum[i - 1]
return Sum, Single
def getAns(n, Sum, Single, maxn):
ans = 0
minn = n
index = 0
L, R = 1, 10 ** maxn
while L <= R:
m = (L + R) // 2
digit = len(str(m))
lens = m - 10 ** (digit - 1) + 1
pre = Single[digit - 1]
cnt = (pre + digit + pre + lens * digit) * lens // 2 + Sum[digit - 1]
if cnt < n:
index = m
minn = min(minn, n - cnt)
L = m + 1
else :
R = m - 1
#print(index, minn)
n = minn
L, R = 1, index + 11
index = 0
while L <= R:
m = (L + R) // 2
digit = len(str(m))
lens = m - 10 ** (digit - 1) + 1
pre = Single[digit - 1]
cnt = pre + lens * digit
if cnt < n:
index = m
minn = min(minn, n - cnt)
L = m + 1
else :
R = m - 1
return str(index + 1)[minn - 1]
def test():
ans = 0
Sum = 0
for i in range(1, 1000):
ans += len(str(i))
Sum += ans
if i % 10 == 9:
print(i, ans, Sum)
def main():
maxn = 10
Sum, Single = init(maxn)
T = int(input())
for i in range(T):
n = int(input())
print(getAns(n, Sum, Single, maxn))
if __name__ == '__main__':
main()
``` | instruction | 0 | 97,948 | 5 | 195,896 |
Yes | output | 1 | 97,948 | 5 | 195,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and the hard versions is the maximum value of k.
You are given an infinite sequence of form "112123123412345..." which consist of blocks of all consecutive positive integers written one after another. The first block consists of all numbers from 1 to 1, the second one — from 1 to 2, the third one — from 1 to 3, ..., the i-th block consists of all numbers from 1 to i.
So the first 56 elements of the sequence are "11212312341234512345612345671234567812345678912345678910". Elements of the sequence are numbered from one. For example, the 1-st element of the sequence is 1, the 3-rd element of the sequence is 2, the 20-th element of the sequence is 5, the 38-th element is 2, the 56-th element of the sequence is 0.
Your task is to answer q independent queries. In the i-th query you are given one integer k_i. Calculate the digit at the position k_i of the sequence.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 500) — the number of queries.
The i-th of the following q lines contains one integer k_i (1 ≤ k_i ≤ 10^{18}) — the description of the corresponding query.
Output
Print q lines. In the i-th line print one digit x_i (0 ≤ x_i ≤ 9) — the answer to the query i, i.e. x_i should be equal to the element at the position k_i of the sequence.
Examples
Input
5
1
3
20
38
56
Output
1
2
5
2
0
Input
4
2132
506
999999999999999999
1000000000000000000
Output
8
2
4
1
Note
Answers on queries from the first example are described in the problem statement.
Submitted Solution:
```
import math
sum_ = [0] * 18
begin_ = [0] * 18
def f_(x0, k, n):
return k*(2*x0 + (k-1)*n) // 2
def make():
x0 = 1
k = 9
n = 1
while n < 18:
begin_[n] = x0
last_number = x0 + (k-1)*n
sum = k*(2*x0 + (k-1)*n) // 2
sum_[n] = sum
sum_[n] += sum_[n-1]
x0 = last_number + (n+1)
k *= 10
n += 1
def digit(x):
cnt = 0
while x > 0:
x //= 10
cnt += 1
return cnt
def f(x, begin_, sum_):
n = digit(x)
k = x - 10**(n-1) + 1
x0 = begin_[n]
return sum_[n-1] + f_(x0, k, n)
def find(s, begin_, sum_):
l = 0
u = 1000000000
while u-l>1:
md = (l+u) // 2
if f(md, begin_, sum_) > s:
u = md
else:
l = md
# pos, remain
return l, s - f(l, begin_, sum_)
def get_digit(x, pos):
s = []
while x > 0:
s.append(x%10)
x //= 10
return s[::-1][pos]
def find_digit(x):
pos, remain = find(x, begin_, sum_)
if remain == 0:
return pos % 10
n = 0
next_ = 9 * (10**n) * (n+1)
while next_ <= remain:
remain -= next_
n += 1
next_ = 9 * (10**n) * (n+1)
if remain == 0:
return 9
pos_ = 10 ** n + math.ceil(remain / (n+1)) - 1
return get_digit(pos_, (remain-1)%(n+1))
make()
q = int(input())
for _ in range(q):
n = int(input())
print(find_digit(n))
``` | instruction | 0 | 97,949 | 5 | 195,898 |
Yes | output | 1 | 97,949 | 5 | 195,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and the hard versions is the maximum value of k.
You are given an infinite sequence of form "112123123412345..." which consist of blocks of all consecutive positive integers written one after another. The first block consists of all numbers from 1 to 1, the second one — from 1 to 2, the third one — from 1 to 3, ..., the i-th block consists of all numbers from 1 to i.
So the first 56 elements of the sequence are "11212312341234512345612345671234567812345678912345678910". Elements of the sequence are numbered from one. For example, the 1-st element of the sequence is 1, the 3-rd element of the sequence is 2, the 20-th element of the sequence is 5, the 38-th element is 2, the 56-th element of the sequence is 0.
Your task is to answer q independent queries. In the i-th query you are given one integer k_i. Calculate the digit at the position k_i of the sequence.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 500) — the number of queries.
The i-th of the following q lines contains one integer k_i (1 ≤ k_i ≤ 10^{18}) — the description of the corresponding query.
Output
Print q lines. In the i-th line print one digit x_i (0 ≤ x_i ≤ 9) — the answer to the query i, i.e. x_i should be equal to the element at the position k_i of the sequence.
Examples
Input
5
1
3
20
38
56
Output
1
2
5
2
0
Input
4
2132
506
999999999999999999
1000000000000000000
Output
8
2
4
1
Note
Answers on queries from the first example are described in the problem statement.
Submitted Solution:
```
'''
Auther: ghoshashis545 Ashis Ghosh
College: Jalpaiguri Govt Enggineering College
'''
from os import path
from io import BytesIO, IOBase
import sys
from heapq import heappush,heappop
from functools import cmp_to_key as ctk
from collections import deque,Counter,defaultdict as dd
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input().rstrip()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def bo(i):
return ord(i)-ord('0')
file = 1
def ceil(a,b):
return (a+b-1)//b
def solve():
# for _ in range(1,ii()+1):
def get(x):
n = len(str(x))
cnt = 0
for i in range(n,0,-1):
cnt_num = x-pow(10,i-1) + 1
cnt += cnt_num*i
x = pow(10,i-1)-1
return cnt
# x = []
# s = 0
# for i in range(1,10001):
# s += get(i)
# x.append(s)
# if s > 1e9:
# break
# print(x[-1])
x = []
s = 0
for i in range(10):
x1 = get(pow(10,i))
x.append(s + x1)
cnt = pow(10,i+1) - pow(10,i)
s += (cnt*x1)
s1 = (cnt-1)*(i+1)*cnt
s += (s1//2)
# print(x)
def get1(n):
for i in range(10):
if x[i] > n:
n -= x[i-1]
x1 = get(pow(10,i-1))
n += x1
l = 1
r = pow(10,i)
while l <= r:
mid = (l+r)>>1
s1 = (2*x1 + (mid-1)*i)*mid
s1 //= 2
if s1 <= n:
ans = mid
l = mid+1
else:
r = mid-1
s1 = (2*x1 + (ans-1)*i)*ans
s1 //= 2
n -= s1
return [n,pow(10,i-1) + ans - 1]
q = ii()
for i in range(q):
n = ii()
n,idx = get1(n)
if n==0:
idx += 1
print(str(idx)[-1])
continue
l = 1
r = idx+1
while l <= r:
mid = (l+r)>>1
if get(mid) <= n:
ans = mid
l = mid+1
else:
r = mid-1
n -= get(ans)
if n == 0:
print(str(ans)[-1])
else:
print(str(ans+1)[n-1])
if __name__ =="__main__":
if(file):
if path.exists('input.txt'):
sys.stdin=open('input.txt', 'r')
sys.stdout=open('output.txt','w')
else:
input=sys.stdin.readline
solve()
``` | instruction | 0 | 97,950 | 5 | 195,900 |
No | output | 1 | 97,950 | 5 | 195,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and the hard versions is the maximum value of k.
You are given an infinite sequence of form "112123123412345..." which consist of blocks of all consecutive positive integers written one after another. The first block consists of all numbers from 1 to 1, the second one — from 1 to 2, the third one — from 1 to 3, ..., the i-th block consists of all numbers from 1 to i.
So the first 56 elements of the sequence are "11212312341234512345612345671234567812345678912345678910". Elements of the sequence are numbered from one. For example, the 1-st element of the sequence is 1, the 3-rd element of the sequence is 2, the 20-th element of the sequence is 5, the 38-th element is 2, the 56-th element of the sequence is 0.
Your task is to answer q independent queries. In the i-th query you are given one integer k_i. Calculate the digit at the position k_i of the sequence.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 500) — the number of queries.
The i-th of the following q lines contains one integer k_i (1 ≤ k_i ≤ 10^{18}) — the description of the corresponding query.
Output
Print q lines. In the i-th line print one digit x_i (0 ≤ x_i ≤ 9) — the answer to the query i, i.e. x_i should be equal to the element at the position k_i of the sequence.
Examples
Input
5
1
3
20
38
56
Output
1
2
5
2
0
Input
4
2132
506
999999999999999999
1000000000000000000
Output
8
2
4
1
Note
Answers on queries from the first example are described in the problem statement.
Submitted Solution:
```
q = int(input())
def f(n):
take = 0
ret = 0
while True:
if n - take <= 0:
break
ret += (n - take) * (n - take + 1) // 2
take = take * 10 + 9
return ret
def g(n):
take = 0
ret = 0
while True:
if n - take <= 0:
break
ret += n - take
take = take * 10 + 9
return ret
for _ in range(q):
k = int(input())
low, high = 0, k
ans = -1
while low < high:
mid = (low + high + 1) >> 1
if f(mid) == k:
ans = mid % 10
break
if f(mid) < k:
low = mid
else:
high = mid - 1
if ans != -1:
print(ans)
continue
k -= f(low)
low, high = 1, mid + 1
while low < high:
mid = (low + high + 1) // 2
if g(mid) == k:
ans = mid % 10
break
if g(mid) < k:
low = mid
else:
high = mid - 1
if ans != -1:
print(ans)
continue
h = g(low)
m = str(low + 1)
print(m[k - h - 1])
``` | instruction | 0 | 97,951 | 5 | 195,902 |
No | output | 1 | 97,951 | 5 | 195,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and the hard versions is the maximum value of k.
You are given an infinite sequence of form "112123123412345..." which consist of blocks of all consecutive positive integers written one after another. The first block consists of all numbers from 1 to 1, the second one — from 1 to 2, the third one — from 1 to 3, ..., the i-th block consists of all numbers from 1 to i.
So the first 56 elements of the sequence are "11212312341234512345612345671234567812345678912345678910". Elements of the sequence are numbered from one. For example, the 1-st element of the sequence is 1, the 3-rd element of the sequence is 2, the 20-th element of the sequence is 5, the 38-th element is 2, the 56-th element of the sequence is 0.
Your task is to answer q independent queries. In the i-th query you are given one integer k_i. Calculate the digit at the position k_i of the sequence.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 500) — the number of queries.
The i-th of the following q lines contains one integer k_i (1 ≤ k_i ≤ 10^{18}) — the description of the corresponding query.
Output
Print q lines. In the i-th line print one digit x_i (0 ≤ x_i ≤ 9) — the answer to the query i, i.e. x_i should be equal to the element at the position k_i of the sequence.
Examples
Input
5
1
3
20
38
56
Output
1
2
5
2
0
Input
4
2132
506
999999999999999999
1000000000000000000
Output
8
2
4
1
Note
Answers on queries from the first example are described in the problem statement.
Submitted Solution:
```
def digits_until_block_(n):
result = 0
for bas in range(1, 20): # bas for basamak
minimum = int(10 ** (bas - 1))
maximum = int((10 ** bas) - 1)
if n < maximum:
maximum = n
if maximum < minimum:
break
result += sum_between(n - maximum + 1, n - minimum + 1) * bas
return result
def digits_until_(n):
if n == 0:
return 0
if n == 1:
return 1
return digits_until_block_(n) - digits_until_block_(n - 1)
def sum_between(x, y):
return int((x + y) * (y - x + 1) / 2)
def solve(q):
left = 1
right = 1000000000
while left < right:
mid = (left + right) // 2
if digits_until_block_(mid) < q:
left = mid + 1
else:
right = mid
q = q - digits_until_block_(left - 1)
if q == 0:
return str(left - 1)[-1]
left = 1
right = 1000000000
while left < right:
mid = (left + right) // 2
if digits_until_(mid) < q:
left = mid + 1
else:
right = mid
q = q - digits_until_(left - 1)
if q == 0:
return str(left - 1)[-1]
return str(left)[q - 1]
q = int(input(""))
q_list = []
for _ in range(q):
q_list.append(int(input("")))
for query in q_list:
print(solve(query))
``` | instruction | 0 | 97,952 | 5 | 195,904 |
No | output | 1 | 97,952 | 5 | 195,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between the easy and the hard versions is the maximum value of k.
You are given an infinite sequence of form "112123123412345..." which consist of blocks of all consecutive positive integers written one after another. The first block consists of all numbers from 1 to 1, the second one — from 1 to 2, the third one — from 1 to 3, ..., the i-th block consists of all numbers from 1 to i.
So the first 56 elements of the sequence are "11212312341234512345612345671234567812345678912345678910". Elements of the sequence are numbered from one. For example, the 1-st element of the sequence is 1, the 3-rd element of the sequence is 2, the 20-th element of the sequence is 5, the 38-th element is 2, the 56-th element of the sequence is 0.
Your task is to answer q independent queries. In the i-th query you are given one integer k_i. Calculate the digit at the position k_i of the sequence.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 500) — the number of queries.
The i-th of the following q lines contains one integer k_i (1 ≤ k_i ≤ 10^{18}) — the description of the corresponding query.
Output
Print q lines. In the i-th line print one digit x_i (0 ≤ x_i ≤ 9) — the answer to the query i, i.e. x_i should be equal to the element at the position k_i of the sequence.
Examples
Input
5
1
3
20
38
56
Output
1
2
5
2
0
Input
4
2132
506
999999999999999999
1000000000000000000
Output
8
2
4
1
Note
Answers on queries from the first example are described in the problem statement.
Submitted Solution:
```
dp, cnt = [0], 1
dp2 = [0]
while dp[-1] <= int(1e18):
ans = dp2[-1] + (10 ** cnt - 10 ** (cnt - 1)) * cnt
dp2.append(ans)
ans = dp[-1] + dp2[-2] * (10 ** cnt - 10 ** (cnt - 1)) + cnt * int((10 ** cnt - 10 ** (cnt - 1) + 1) * (10 ** cnt - 10 ** (cnt - 1)) / 2)
cnt += 1
dp.append(ans)
def Cal(a, b):
return dp2[b - 1] * a + b * int(a * (a + 1) / 2)
q = int(input())
for _ in range(q):
k = int(input())
i = 0
while k > dp[i]:
i += 1
k -= dp[i - 1]
l, r = 0, 10 ** i - 10 ** (i - 1)
last = int((l + r) / 2)
while not(Cal(last, i) < k and Cal(last + 1, i) >= k):
if(Cal(last, i) < k):
l = last
last = int((l + r) / 2 + 1)
else:
r = last
last = int((l + r) / 2)
k -= Cal(last, i)
j = 0
while dp2[j] < k:
j += 1
k -= dp2[j - 1]
a = int((k - 1) / j)
k -= a * j
Long = str(10 ** (j - 1) + a)
print(Long[k - 1])
``` | instruction | 0 | 97,953 | 5 | 195,906 |
No | output | 1 | 97,953 | 5 | 195,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given 2 integers u and v, find the shortest array such that [bitwise-xor](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of its elements is u, and the sum of its elements is v.
Input
The only line contains 2 integers u and v (0 ≤ u,v ≤ 10^{18}).
Output
If there's no array that satisfies the condition, print "-1". Otherwise:
The first line should contain one integer, n, representing the length of the desired array. The next line should contain n positive integers, the array itself. If there are multiple possible answers, print any.
Examples
Input
2 4
Output
2
3 1
Input
1 3
Output
3
1 1 1
Input
8 5
Output
-1
Input
0 0
Output
0
Note
In the first sample, 3⊕ 1 = 2 and 3 + 1 = 4. There is no valid array of smaller length.
Notice that in the fourth sample the array is empty.
Submitted Solution:
```
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
u,v=list(map(int,input().split()))
if u>v:
print(-1)
elif u==v:
if u==0:
print(0)
else:
print(1)
print(u)
else:
if u%2==0:
if v%2==0:
num=(v-u)//2
num2=int(num&(u))
if not num2:
print(2)
print(str(u+num)+" "+str(num))
else:
print(3)
print(str(u)+" "+str(num)+" "+str(num))
else:
print(-1)
else:
if v%2==0:
print(-1)
else:
num=(v-u)//2
num2=int(num&(u))
if not num2:
print(2)
print(str(u+num)+" "+str(num))
else:
print(3)
print(str(u)+" "+str(num)+" "+str(num))
``` | instruction | 0 | 98,027 | 5 | 196,054 |
Yes | output | 1 | 98,027 | 5 | 196,055 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given 2 integers u and v, find the shortest array such that [bitwise-xor](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of its elements is u, and the sum of its elements is v.
Input
The only line contains 2 integers u and v (0 ≤ u,v ≤ 10^{18}).
Output
If there's no array that satisfies the condition, print "-1". Otherwise:
The first line should contain one integer, n, representing the length of the desired array. The next line should contain n positive integers, the array itself. If there are multiple possible answers, print any.
Examples
Input
2 4
Output
2
3 1
Input
1 3
Output
3
1 1 1
Input
8 5
Output
-1
Input
0 0
Output
0
Note
In the first sample, 3⊕ 1 = 2 and 3 + 1 = 4. There is no valid array of smaller length.
Notice that in the fourth sample the array is empty.
Submitted Solution:
```
u,v=map(int,input().split())
if u>v or (v-u)%2==1:
print(-1)
exit()
v=(v-u)//2
a=u|v
b=v
c=u&v
if a==0:
print(0)
elif b==0:
print(1)
print(a)
elif c==0:
print(2)
print(a,b)
else:
print(3)
print(a,b,c)
``` | instruction | 0 | 98,028 | 5 | 196,056 |
Yes | output | 1 | 98,028 | 5 | 196,057 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given 2 integers u and v, find the shortest array such that [bitwise-xor](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of its elements is u, and the sum of its elements is v.
Input
The only line contains 2 integers u and v (0 ≤ u,v ≤ 10^{18}).
Output
If there's no array that satisfies the condition, print "-1". Otherwise:
The first line should contain one integer, n, representing the length of the desired array. The next line should contain n positive integers, the array itself. If there are multiple possible answers, print any.
Examples
Input
2 4
Output
2
3 1
Input
1 3
Output
3
1 1 1
Input
8 5
Output
-1
Input
0 0
Output
0
Note
In the first sample, 3⊕ 1 = 2 and 3 + 1 = 4. There is no valid array of smaller length.
Notice that in the fourth sample the array is empty.
Submitted 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 --------------------
u,v=map(int,input().split())
if u>v:
print(-1)
else:
if u==v:
print(1);print(u)
else:
if (v-u)%2==0:
print(2);print(u+(v-u)//2,(v-u)//2)
else:
print(-1)
``` | instruction | 0 | 98,030 | 5 | 196,060 |
No | output | 1 | 98,030 | 5 | 196,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given 2 integers u and v, find the shortest array such that [bitwise-xor](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of its elements is u, and the sum of its elements is v.
Input
The only line contains 2 integers u and v (0 ≤ u,v ≤ 10^{18}).
Output
If there's no array that satisfies the condition, print "-1". Otherwise:
The first line should contain one integer, n, representing the length of the desired array. The next line should contain n positive integers, the array itself. If there are multiple possible answers, print any.
Examples
Input
2 4
Output
2
3 1
Input
1 3
Output
3
1 1 1
Input
8 5
Output
-1
Input
0 0
Output
0
Note
In the first sample, 3⊕ 1 = 2 and 3 + 1 = 4. There is no valid array of smaller length.
Notice that in the fourth sample the array is empty.
Submitted Solution:
```
u, v = [int(x) for x in input().split()]
if u == v:
print(1)
print(u)
exit(0)
if u > v or (u-v)%2 == 1:
print(-1)
exit(0)
result = 0
tmp_u = u
tmp_and = int((v-u)/2)
x = tmp_and
i = 0
while tmp_u > 0:
a = tmp_and % 2
b = tmp_u % 2
if a == 1:
if b == 1:
print(3)
print(u, x, x)
exit(0)
if b == 0:
result = result + 1<<i
tmp_u = tmp_u>>1
tmp_and = tmp_and>>1
i += 1
print(2)
print(result, v-result)
``` | instruction | 0 | 98,031 | 5 | 196,062 |
No | output | 1 | 98,031 | 5 | 196,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given 2 integers u and v, find the shortest array such that [bitwise-xor](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of its elements is u, and the sum of its elements is v.
Input
The only line contains 2 integers u and v (0 ≤ u,v ≤ 10^{18}).
Output
If there's no array that satisfies the condition, print "-1". Otherwise:
The first line should contain one integer, n, representing the length of the desired array. The next line should contain n positive integers, the array itself. If there are multiple possible answers, print any.
Examples
Input
2 4
Output
2
3 1
Input
1 3
Output
3
1 1 1
Input
8 5
Output
-1
Input
0 0
Output
0
Note
In the first sample, 3⊕ 1 = 2 and 3 + 1 = 4. There is no valid array of smaller length.
Notice that in the fourth sample the array is empty.
Submitted Solution:
```
def compute(S, X):
A = (S - X)//2
a = 0
b = 0
for i in range(64):
Xi = (X & (1 << i))
Ai = (A & (1 << i))
if (Xi == 0 and Ai == 0):
pass
elif (Xi == 0 and Ai > 0):
a = ((1 << i) | a)
b = ((1 << i) | b)
elif (Xi > 0 and Ai == 0):
a = ((1 << i) | a)
else:
return -1
if a + b != S:
return -1
return a, b
a, b = list(map(int, input().strip().split()))
a, b = sorted([a, b])
if a == b == 0:
print(0)
exit()
if a == b:
print(1)
print(a)
exit()
if (b-a) % 2 != 0:
print(-1)
exit()
o = compute(b, a)
if o != -1:
print(2)
print(*o)
exit()
v = abs(b-a)//2
print(3)
print(a, v, v)
``` | instruction | 0 | 98,032 | 5 | 196,064 |
No | output | 1 | 98,032 | 5 | 196,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given 2 integers u and v, find the shortest array such that [bitwise-xor](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of its elements is u, and the sum of its elements is v.
Input
The only line contains 2 integers u and v (0 ≤ u,v ≤ 10^{18}).
Output
If there's no array that satisfies the condition, print "-1". Otherwise:
The first line should contain one integer, n, representing the length of the desired array. The next line should contain n positive integers, the array itself. If there are multiple possible answers, print any.
Examples
Input
2 4
Output
2
3 1
Input
1 3
Output
3
1 1 1
Input
8 5
Output
-1
Input
0 0
Output
0
Note
In the first sample, 3⊕ 1 = 2 and 3 + 1 = 4. There is no valid array of smaller length.
Notice that in the fourth sample the array is empty.
Submitted Solution:
```
u,v=map(int,input().split())
if u>v:
print(-1)
else:
if 1^(v-1)==u:
print(2)
print(1,v-1)
else:
x=v-u
s = '1 '*x
print(x+1)
print('{}{}'.format(s, u))
``` | instruction | 0 | 98,033 | 5 | 196,066 |
No | output | 1 | 98,033 | 5 | 196,067 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fraction
from heapq import *
from random import randint
def main():
po=[1]
for i in range(30):
po.append(po[-1]*2)
n,m=map(int,input().split())
q=[]
b=[[0 for _ in range(30)] for _ in range(n+2)]
for i in range(m):
l,r,x=map(int,input().split())
q.append((l,r,x))
j=0
while x:
if x&1:
b[l][j]+=1
b[r+1][j]-=1
x=x>>1
j+=1
for i in range(1,n+1):
for j in range(30):
b[i][j]+=b[i-1][j]
for i in range(1,n+1):
for j in range(30):
b[i][j]+=b[i-1][j]
f=1
for i in q:
l,r,x=i
z=0
for j in range(30):
if b[r][j]-b[l-1][j]>=(r-l+1):
z+=po[j]
if z!=x:
f=0
break
if f:
print("YES")
a=[]
for i in range(1,n+1):
z=0
for j in range(30):
if b[i][j]-b[i-1][j]==1:
z+=po[j]
a.append(z)
print(*a)
else:
print("NO")
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | instruction | 0 | 98,234 | 5 | 196,468 |
No | output | 1 | 98,234 | 5 | 196,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
n,m = map(int,input().split())
dp = [[0]*30 for _ in range(n+2)]
op = []
for _ in range(m):
op.append(tuple(map(int,input().split())))
l,r,q = op[-1]
mask,cou = 1,0
while mask <= q:
if mask&q:
dp[l][cou] += 1
dp[r+1][cou] -= 1
cou += 1
mask <<= 1
ans = [[0]*30 for _ in range(n)]
for i in range(30):
a = 0
for j in range(n):
a += dp[j+1][i]
dp[j+1][i] = dp[j][i]
if a:
ans[j][i] = 1
dp[j+1][i] += 1
for i in op:
l,r,q = i
mask,cou = 1,0
while mask <= q:
if not mask&q and dp[r][cou]-dp[l-1][cou] == r-l+1:
print('NO')
return
mask <<= 1
cou += 1
for i in range(n):
ans[i] = int(''.join(map(str,ans[i][::-1])),2)
print('YES')
print(*ans)
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | instruction | 0 | 98,235 | 5 | 196,470 |
No | output | 1 | 98,235 | 5 | 196,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fraction
from heapq import *
from random import randint
def main():
n, m = map(int, input().split())
a, f, v = [-1] * n, 1, [-1] * n
for i in range(m):
l, r, x = map(int, input().split())
if f:
j = l - 1
while j < r:
if a[j] != -1:
if a[j] != x:
f = 0
break
else:
j=v[j] - 1
a[j]=x
v[j]=r
j += 1
if f:
print("YES")
print(*a)
else:
print("NO")
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | instruction | 0 | 98,236 | 5 | 196,472 |
No | output | 1 | 98,236 | 5 | 196,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fraction
from heapq import *
from random import randint
def main():
n, m = map(int, input().split())
a, f, v = [-1] * n, 1, [-1] * n
b=defaultdict(list)
for i in range(m):
l,r,x = map(int, input().split())
z=b[(l,r)]
if f and z==[]:
b[(l,r)]=x
j = l - 1
while j < r:
if a[j] != -1:
if a[j]!=x and not x:
f = 0
break
else:
j=v[j] - 1
a[j]=x
v[j]=r
j += 1
elif f and z!=[] and z!=x:
f=0
if f:
print("YES")
print(*a)
else:
print("NO")
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | instruction | 0 | 98,237 | 5 | 196,474 |
No | output | 1 | 98,237 | 5 | 196,475 |
Provide a correct Python 3 solution for this coding contest problem.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507 | instruction | 0 | 98,383 | 5 | 196,766 |
"Correct Solution:
```
n, x, m = map(int, input().split())
g = x * 1
ans = x
arr = []
res = set([])
loop = (0, 0)
k = 0
for i in range(m + 1):
tmp = x ** 2 % m
if tmp in res:
loop = (i, tmp)
break
res.add(tmp)
arr.append(tmp)
ans += tmp
x = tmp
for i, y in enumerate(arr):
if y == loop[1]:
k = i
break
ini = g + sum(arr[:k])
mul = ans - ini
t, v = divmod(n - k - 1, loop[0] - k)
print(ini + mul * t + sum(arr[k:k + v]))
``` | output | 1 | 98,383 | 5 | 196,767 |
Provide a correct Python 3 solution for this coding contest problem.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507 | instruction | 0 | 98,384 | 5 | 196,768 |
"Correct Solution:
```
n, x, m = map(int, input().split())
s = set([x])
t = [x]
p = 1
for i in range(n):
x = x * x % m
if x in s:
break
else:
s.add(x)
t.append(x)
p += 1
if p == n:
print(sum(s))
exit()
q = t.index(x)
l = p - q
b = sum(t[q:p])
ans = sum(t[:q])
n -= q
ans += n // l * b + sum(t[q:q + n % l])
print(ans)
``` | output | 1 | 98,384 | 5 | 196,769 |
Provide a correct Python 3 solution for this coding contest problem.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507 | instruction | 0 | 98,385 | 5 | 196,770 |
"Correct Solution:
```
n, x, m = map(int, input().split())
mn = min(n, m)
P = [] # pre_sum
sum_p = 0 # sum of pre + cycle
X = [-1] * m # for cycle check & pre_len
for i in range(mn):
if X[x] > -1:
cyc_len = len(P) - X[x]
cyc = (sum_p - P[X[x]]) * ((n - X[x]) // cyc_len)
remain = P[X[x] + (n - X[x]) % cyc_len]
print(cyc + remain)
exit()
P.append(sum_p)
sum_p += x
X[x] = i
x = x*x % m
print(sum_p)
``` | output | 1 | 98,385 | 5 | 196,771 |
Provide a correct Python 3 solution for this coding contest problem.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507 | instruction | 0 | 98,386 | 5 | 196,772 |
"Correct Solution:
```
def resolve():
n,x,m = map(int,input().split())
ans = x
y = [x]
for i in range(n-1):
x = (x**2)%m
if x not in y:
y.append(x)
else:
break
b = y.index(x)
N = len(y[b:])
c = sum(y[b:])
ans = sum(y[:b]) + c*((n-b)//N)+sum(y[b:b+(n-b)%N])
print(ans)
resolve()
``` | output | 1 | 98,386 | 5 | 196,773 |
Provide a correct Python 3 solution for this coding contest problem.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507 | instruction | 0 | 98,387 | 5 | 196,774 |
"Correct Solution:
```
N, X, M = map(int, input().split())
ans = X
ALL_cal = [False] * M
ALL = []
rou = False
for i in range(N-1):
X = pow(X, 2, M)
if ALL_cal[X]:
num = ALL_cal[X]
now = i
rou = True
break
ALL.append(X)
ALL_cal[X] = i
ans += X
if rou :
roupe = now - num
nokori = N - now - 1
print(sum(ALL[num:])*(nokori//roupe) + ans + sum(ALL[num:num + nokori%roupe]))
else:
print(ans)
``` | output | 1 | 98,387 | 5 | 196,775 |
Provide a correct Python 3 solution for this coding contest problem.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507 | instruction | 0 | 98,388 | 5 | 196,776 |
"Correct Solution:
```
N,X,M=map(int,input().split())
seen=[-2]*M
seen[X]
A=[X]
i=1
while(i<N):
T=A[-1]**2%M
if seen[T]!=-2:
Roop=i-seen[T]
Left,Right=seen[T],i
break
A.append(T)
seen[T]=i
i+=1
if i==N:
print(sum(A))
exit()
Roopsum=0
for i in range(Left,Right):
Roopsum+=A[i]
Rest=N-len(A)
ans=sum(A)
ans+=Rest//Roop*Roopsum
for i in range(Rest%Roop):
ans+=T
T=T**2%M
print(ans)
``` | output | 1 | 98,388 | 5 | 196,777 |
Provide a correct Python 3 solution for this coding contest problem.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507 | instruction | 0 | 98,389 | 5 | 196,778 |
"Correct Solution:
```
N, X, M = map(int, input().split())
nxt = X
lst = []
dic = {}
for i in range(M + 1):
if nxt in dic:
loop_st = dic[nxt]
loop_ed = i - 1
break
lst.append(nxt)
dic[nxt] = i
nxt = (nxt ** 2) % M
v = N - loop_st
q, r = divmod(v, loop_ed - loop_st + 1)
pre_sum = sum(lst[:loop_st])
loop_sum = q * sum(lst[loop_st:])
post_sum = sum(lst[loop_st:loop_st + r])
print(pre_sum + loop_sum + post_sum)
``` | output | 1 | 98,389 | 5 | 196,779 |
Provide a correct Python 3 solution for this coding contest problem.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507 | instruction | 0 | 98,390 | 5 | 196,780 |
"Correct Solution:
```
n, x, m = map(int, input().split())
ans = []
c = [0]*m
flag = False
for i in range(n):
if c[x] == 1:
flag = True
break
ans.append(x)
c[x] = 1
x = x**2 % m
if flag:
p = ans.index(x)
l = len(ans) - p
d, e = divmod(n-p, l)
print(sum(ans[:p]) + d*sum(ans[p:]) + sum(ans[p:p+e]))
else:
print(sum(ans))
``` | output | 1 | 98,390 | 5 | 196,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507
Submitted Solution:
```
n, x, m = map(int, input().split())
lst, num, flag = set(), [], False
for i in range(1, n + 1):
lst.add(x), num.append(x)
x = x ** 2 % m
if x in lst:
flag = True
break
ans = sum(num)
if flag:
cnt, idx = i, num.index(x)
div, mod = divmod(n - cnt, len(num) - idx)
ans += sum(num[idx:idx + mod])
ans += sum(num[idx:]) * div
print(ans)
``` | instruction | 0 | 98,391 | 5 | 196,782 |
Yes | output | 1 | 98,391 | 5 | 196,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507
Submitted Solution:
```
N, X, M = list(map(int, input().split()))
i = 1
a = X
s = 0
L = []
while i < N and a > 0:
L.append(a)
s += a
a = (a**2) % M
if a in L:
j = L.index(a)
break
i += 1
else:
s += a
print(s)
exit()
# print(L)
t = sum(L[:j])
L = L[j:]
u = s-t
N = N-j
v = t+u*(N//(i-j))
if N % (i-j) == 0:
pass
else:
v += sum(L[:N % (i-j)])
print(v)
``` | instruction | 0 | 98,392 | 5 | 196,784 |
Yes | output | 1 | 98,392 | 5 | 196,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507
Submitted Solution:
```
n,x,m=map(int,input().split())
s=x
t={x}
u=[x]
for i in range(1,n):
x=pow(x,2,m)
if x==0:break
if x==1:
s+=n-i
break
if x in t:
j=u.index(x)
s=sum(u[:j])
u=u[j:]
s+=(n-j)//len(u)*sum(u)
s+=sum(u[:(n-j)%len(u)])
break
s+=x
t|={x}
u+=x,
print(s)
``` | instruction | 0 | 98,393 | 5 | 196,786 |
Yes | output | 1 | 98,393 | 5 | 196,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507
Submitted Solution:
```
N, X, M = map(int,input().split())
def cal(x):
return pow(x,2,M)
memo = set()
lis = []
while X not in memo:
memo.add(X)
lis.append(X)
X = cal(X)
pos = lis.index(X)
if pos >= N:
print(sum(lis[:N]))
else:
print(sum(lis[:pos]) + ((N-pos)//(len(memo)-pos)) * sum(lis[pos:]) + sum(lis[pos:(N-pos)%(len(memo)-pos)+pos]))
``` | instruction | 0 | 98,394 | 5 | 196,788 |
Yes | output | 1 | 98,394 | 5 | 196,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507
Submitted Solution:
```
while True:
try:
n,x,m = map(int,input().split())
#print(n,x,m)
prev = x
ans = x
for i in range(1,n):
ans += (prev**2)%m
prev = (prev**2)%m
print(ans)
except:
break
``` | instruction | 0 | 98,395 | 5 | 196,790 |
No | output | 1 | 98,395 | 5 | 196,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507
Submitted Solution:
```
from collections import defaultdict
N, X, M = map(int, input().split())
A = [X]
visited = set()
visited.add(X)
idx = defaultdict()
idx[X] = 0
iii = -1
for i in range(1, M):
tmp = (A[-1]**2) % M
if tmp not in visited:
A.append(tmp)
visited.add(tmp)
idx[tmp] = i
else:
iii = idx[tmp]
print(A)
print(len(A))
print(iii)
ans = 0
ans += sum(A[:iii])
N -= iii
l = len(A) - iii
ans += (N // l) * sum(A[iii:])
N -= N // l * l
ans += sum(A[iii:iii + N])
print(ans)
exit()
if A[-1] == 0:
print(sum(A))
exit()
else:
l = len(A)
ans = 0
if N % l == 0:
ans += sum(A) * (N // l)
else:
ans += sum(A) * (N // l)
rest = N % l
if iii == -1:
ans += sum(A[:rest])
else:
ans += sum(A[i:iii + rest])
print(ans)
``` | instruction | 0 | 98,396 | 5 | 196,792 |
No | output | 1 | 98,396 | 5 | 196,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507
Submitted Solution:
```
n, x, m = map(int, input().split())
mn = min(n, m)
S = set()
A = []
sum_9 = 0 # sum of pre + cycle
for len_9 in range(mn):
if x in S: break
S.add(x)
A.append(x)
sum_9 += x
x = x*x % m
pre_len = A.index(x)
cyc_len = len(A) - pre_len
nxt_len = (n - pre_len) % cyc_len
cyc_num = (n - pre_len) // cyc_len
pre = sum(A[:pre_len])
cyc = sum_9 - pre
nxt = sum(A[pre_len: pre_len + nxt_len])
print(pre + cyc * cyc_num + nxt)
``` | instruction | 0 | 98,397 | 5 | 196,794 |
No | output | 1 | 98,397 | 5 | 196,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let us denote by f(x, m) the remainder of the Euclidean division of x by m.
Let A be the sequence that is defined by the initial value A_1=X and the recurrence relation A_{n+1} = f(A_n^2, M). Find \displaystyle{\sum_{i=1}^N A_i}.
Constraints
* 1 \leq N \leq 10^{10}
* 0 \leq X < M \leq 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N X M
Output
Print \displaystyle{\sum_{i=1}^N A_i}.
Examples
Input
6 2 1001
Output
1369
Input
1000 2 16
Output
6
Input
10000000000 10 99959
Output
492443256176507
Submitted Solution:
```
n,x,m=map(int,input().split())
l=[0]*m
s=[0]*m
t=p=0
while l[x]<1:
l[x]=t
s[x]=s[p]+x
p=x
x=pow(p,2,m)
t+=1
T=t-l[x]
S=s[p]+x-s[x]
if n<t:
print(s[l.index(n-1)])
else:
print(S*((n-l[x])//T)+s[l.index(l[x]+(n-l[x])%T-1)])
``` | instruction | 0 | 98,398 | 5 | 196,796 |
No | output | 1 | 98,398 | 5 | 196,797 |
Provide a correct Python 3 solution for this coding contest problem.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40 | instruction | 0 | 98,592 | 5 | 197,184 |
"Correct Solution:
```
from decimal import *
a, b = input().rstrip().split(' ')
getcontext().prec = len(a) + len(b)
ans = Decimal(a) * Decimal(b)
if ans == 0:
print(0)
else:
print(ans)
``` | output | 1 | 98,592 | 5 | 197,185 |
Provide a correct Python 3 solution for this coding contest problem.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40 | instruction | 0 | 98,593 | 5 | 197,186 |
"Correct Solution:
```
A, B = map(int, input().split())
print(A * B)
``` | output | 1 | 98,593 | 5 | 197,187 |
Provide a correct Python 3 solution for this coding contest problem.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40 | instruction | 0 | 98,594 | 5 | 197,188 |
"Correct Solution:
```
A,B=[int(i) for i in input().split(" ")]
print(A*B)
``` | output | 1 | 98,594 | 5 | 197,189 |
Provide a correct Python 3 solution for this coding contest problem.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40 | instruction | 0 | 98,595 | 5 | 197,190 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Big Integers - Multiplication of Big Integers
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=NTL_2_C&lang=jp
"""
import sys
def main(args):
A, B = map(int, input().split())
print(A * B)
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 98,595 | 5 | 197,191 |
Provide a correct Python 3 solution for this coding contest problem.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40 | instruction | 0 | 98,596 | 5 | 197,192 |
"Correct Solution:
```
x = input()
a, b = x.split()
a = int(a)
b = int(b)
print(a * b)
``` | output | 1 | 98,596 | 5 | 197,193 |
Provide a correct Python 3 solution for this coding contest problem.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40 | instruction | 0 | 98,597 | 5 | 197,194 |
"Correct Solution:
```
import sys,bisect,math
A,B = map(int,sys.stdin.readline().split())
print(A*B)
``` | output | 1 | 98,597 | 5 | 197,195 |
Provide a correct Python 3 solution for this coding contest problem.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40 | instruction | 0 | 98,598 | 5 | 197,196 |
"Correct Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
#
# FileName: multiplication_of_big_integers
# CreatedDate: 2020-07-26 14:26:41 +0900
# LastModified: 2020-07-26 14:26:55 +0900
#
import os
import sys
# import numpy as np
# import pandas as pd
def main():
a, b = map(int, input().split())
print(a*b)
if __name__ == "__main__":
main()
``` | output | 1 | 98,598 | 5 | 197,197 |
Provide a correct Python 3 solution for this coding contest problem.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40 | instruction | 0 | 98,599 | 5 | 197,198 |
"Correct Solution:
```
a, b = [int(x) for x in input().split()]
print(a*b)
``` | output | 1 | 98,599 | 5 | 197,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40
Submitted Solution:
```
a, b=input().split(' ')
print(int(a)*int(b))
``` | instruction | 0 | 98,600 | 5 | 197,200 |
Yes | output | 1 | 98,600 | 5 | 197,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40
Submitted Solution:
```
n, m = map(int, input().split())
print(n * m)
``` | instruction | 0 | 98,601 | 5 | 197,202 |
Yes | output | 1 | 98,601 | 5 | 197,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40
Submitted Solution:
```
print(eval(input().replace(' ','*')))
``` | instruction | 0 | 98,602 | 5 | 197,204 |
Yes | output | 1 | 98,602 | 5 | 197,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Multiplication of Big Integers
Given two integers $A$ and $B$, compute the product, $A \times B$.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the product in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
Sample Input 1
5 8
Sample Output 1
40
Sample Input 2
100 25
Sample Output 2
2500
Sample Input 3
-1 0
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-36
Example
Input
5 8
Output
40
Submitted Solution:
```
s = input().split()
print(int(s[0]) * int(s[1]))
``` | instruction | 0 | 98,603 | 5 | 197,206 |
Yes | output | 1 | 98,603 | 5 | 197,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
x,y=map(int,input().split())
l=list(map(int,input().split()))
if x%2==0:
if l[-1]%2==0:
print("even")
else:
print("odd")
else:
c=0
for i in range(0,len(l)):
if l[i]%2!=0:
c+=1
if c%2==0:
print("even")
else:
print("odd")
``` | instruction | 0 | 98,649 | 5 | 197,298 |
Yes | output | 1 | 98,649 | 5 | 197,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
def f(a,b):
if b%2==0:
return a[-1]%2==0
nodd = sum([x%2>0 for x in a])
return nodd%2==0
b,_ = list(map(int,input().split()))
a = list(map(int,input().split()))
print('even' if f(a,b) else 'odd')
``` | instruction | 0 | 98,650 | 5 | 197,300 |
Yes | output | 1 | 98,650 | 5 | 197,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
b , k = map(int , input().split())
a = [int(a) for a in input().split()]
odd = 0
if b % 2 == 0:
if a[-1] % 2 == 0:
print('even')
else:
print('odd')
else:
for i in range(k):
if a[i] % 2 != 0:
odd += 1
if odd % 2 == 0:
print('even')
else:
print('odd')
``` | instruction | 0 | 98,651 | 5 | 197,302 |
Yes | output | 1 | 98,651 | 5 | 197,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
b,k = list(map(int, input().split(" ")))
a = list(map(int, input().split(" ")))
odd_count = 0
if not b%2:
if a[-1]%2:
print("odd")
else:
print("even")
else:
for i in range(k):
if a[i]%2:
odd_count += 1
if odd_count%2:
print("odd")
else:
print("even")
``` | instruction | 0 | 98,652 | 5 | 197,304 |
Yes | output | 1 | 98,652 | 5 | 197,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
b,k=map(int,input().split(' '))
nums=list(map(int,input().split(' ')))
if b%2==0:
ans=sum(nums)%2
else:
ans=nums[-1]%2
if ans==0:
print('even')
else:
print('odd')
``` | instruction | 0 | 98,653 | 5 | 197,306 |
No | output | 1 | 98,653 | 5 | 197,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer n (n ≥ 0) represented with k digits in base (radix) b. So,
$$$n = a_1 ⋅ b^{k-1} + a_2 ⋅ b^{k-2} + … a_{k-1} ⋅ b + a_k.$$$
For example, if b=17, k=3 and a=[11, 15, 7] then n=11⋅17^2+15⋅17+7=3179+255+7=3441.
Determine whether n is even or odd.
Input
The first line contains two integers b and k (2≤ b≤ 100, 1≤ k≤ 10^5) — the base of the number and the number of digits.
The second line contains k integers a_1, a_2, …, a_k (0≤ a_i < b) — the digits of n.
The representation of n contains no unnecessary leading zero. That is, a_1 can be equal to 0 only if k = 1.
Output
Print "even" if n is even, otherwise print "odd".
You can print each letter in any case (upper or lower).
Examples
Input
13 3
3 2 7
Output
even
Input
10 9
1 2 3 4 5 6 7 8 9
Output
odd
Input
99 5
32 92 85 74 4
Output
odd
Input
2 2
1 0
Output
even
Note
In the first example, n = 3 ⋅ 13^2 + 2 ⋅ 13 + 7 = 540, which is even.
In the second example, n = 123456789 is odd.
In the third example, n = 32 ⋅ 99^4 + 92 ⋅ 99^3 + 85 ⋅ 99^2 + 74 ⋅ 99 + 4 = 3164015155 is odd.
In the fourth example n = 2.
Submitted Solution:
```
import re
s=input()
d=re.findall(r'\d+',s)
a=int(d[0])
b=int(d[1])
num=0
if a%2==0 and int(d[-1])%2==0:
print("even")
else:
for j in range (2,len(d)):
num=num+(a**(b-j+1))*int(d[j])
if num%2==0:
print("even")
else:
print("odd")
``` | instruction | 0 | 98,654 | 5 | 197,308 |
No | output | 1 | 98,654 | 5 | 197,309 |
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