message stringlengths 2 57.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 61 108k | cluster float64 22 22 | __index_level_0__ int64 122 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n integers a_1, a_2, ..., a_n.
For each a_i find its two divisors d_1 > 1 and d_2 > 1 such that \gcd(d_1 + d_2, a_i) = 1 (where \gcd(a, b) is the greatest common divisor of a and b) or say that there is no such pair.
Input
The first line contains single integer n (1 β€ n β€ 5 β
10^5) β the size of the array a.
The second line contains n integers a_1, a_2, ..., a_n (2 β€ a_i β€ 10^7) β the array a.
Output
To speed up the output, print two lines with n integers in each line.
The i-th integers in the first and second lines should be corresponding divisors d_1 > 1 and d_2 > 1 such that \gcd(d_1 + d_2, a_i) = 1 or -1 and -1 if there is no such pair. If there are multiple answers, print any of them.
Example
Input
10
2 3 4 5 6 7 8 9 10 24
Output
-1 -1 -1 -1 3 -1 -1 -1 2 2
-1 -1 -1 -1 2 -1 -1 -1 5 3
Note
Let's look at a_7 = 8. It has 3 divisors greater than 1: 2, 4, 8. As you can see, the sum of any pair of divisors is divisible by 2 as well as a_7.
There are other valid pairs of d_1 and d_2 for a_{10}=24, like (3, 4) or (8, 3). You can print any of them.
Submitted Solution:
```
import io
import os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
sieve = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011, 2017, 2027, 2029, 2039, 2053, 2063, 2069, 2081, 2083, 2087, 2089, 2099, 2111, 2113, 2129, 2131, 2137, 2141, 2143, 2153, 2161, 2179, 2203, 2207, 2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287, 2293, 2297, 2309, 2311, 2333, 2339, 2341, 2347, 2351, 2357, 2371, 2377, 2381, 2383, 2389, 2393, 2399, 2411, 2417, 2423, 2437, 2441, 2447, 2459, 2467, 2473, 2477, 2503, 2521, 2531, 2539, 2543, 2549, 2551, 2557, 2579, 2591, 2593, 2609, 2617, 2621, 2633, 2647, 2657, 2659, 2663, 2671, 2677, 2683, 2687, 2689, 2693, 2699, 2707, 2711, 2713, 2719, 2729, 2731, 2741, 2749, 2753, 2767, 2777, 2789, 2791, 2797, 2801, 2803, 2819, 2833, 2837, 2843, 2851, 2857, 2861, 2879, 2887, 2897, 2903, 2909, 2917, 2927, 2939, 2953, 2957, 2963, 2969, 2971, 2999, 3001, 3011, 3019, 3023, 3037, 3041, 3049, 3061, 3067, 3079, 3083, 3089, 3109, 3119, 3121, 3137, 3163]
n = int(input())
a = list(map(int, input().split()))
out1 = [int(-1) for _ in range(n)]
out2 = [int(-1) for _ in range(n)]
for i in range(n):
m = a[i]
for p in sieve:
if a[i] % p:
continue
while not m % p:
m //= p
if m != 1:
out1[i] = m
out2[i] = a[i]//m
break
print(' '.join(map(str, out1)))
print(' '.join(map(str, out2)))
``` | instruction | 0 | 103,859 | 22 | 207,718 |
Yes | output | 1 | 103,859 | 22 | 207,719 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n integers a_1, a_2, ..., a_n.
For each a_i find its two divisors d_1 > 1 and d_2 > 1 such that \gcd(d_1 + d_2, a_i) = 1 (where \gcd(a, b) is the greatest common divisor of a and b) or say that there is no such pair.
Input
The first line contains single integer n (1 β€ n β€ 5 β
10^5) β the size of the array a.
The second line contains n integers a_1, a_2, ..., a_n (2 β€ a_i β€ 10^7) β the array a.
Output
To speed up the output, print two lines with n integers in each line.
The i-th integers in the first and second lines should be corresponding divisors d_1 > 1 and d_2 > 1 such that \gcd(d_1 + d_2, a_i) = 1 or -1 and -1 if there is no such pair. If there are multiple answers, print any of them.
Example
Input
10
2 3 4 5 6 7 8 9 10 24
Output
-1 -1 -1 -1 3 -1 -1 -1 2 2
-1 -1 -1 -1 2 -1 -1 -1 5 3
Note
Let's look at a_7 = 8. It has 3 divisors greater than 1: 2, 4, 8. As you can see, the sum of any pair of divisors is divisible by 2 as well as a_7.
There are other valid pairs of d_1 and d_2 for a_{10}=24, like (3, 4) or (8, 3). You can print any of them.
Submitted Solution:
```
from sys import stdin, stdout
# 10
# 2 3 4 5 6 7 8 9 10 24
# 6 = 2,3
# 10 = 2,5
# 24 = 2,3
# 30 = 2,3,5
# a = 30
# 1. get all prime numbers
# 2. put into two groups, {p1}, {p2*p3*p4...pn}
# 3. then p1 and p2*p3*p4...pn are the answer
# proof:
# then (p1 + p2*p3*p4...pn)%p1 != 0
# then (p1 + p2*p3*p4...pn)%p2 != 0
# then (p1 + p2*p3*p4...pn)%p3 != 0
# .....
# then (p1 + p2*p3*p4...pn)%pn != 0
def two_divisors(a, div):
r1 = []
r2 = []
for v in a:
p = getprimelist(v, div)
if len(p) < 2:
r1.append(-1)
r2.append(-1)
else:
r1.append(p[0])
d = 1
for i in range(1, len(p)):
d *= p[i]
r2.append(d)
return [r1, r2]
def getprimelist(a, div):
p = []
while a > 1:
if len(p) == 0 or p[-1] != div[a]:
p.append(div[a])
a //= div[a]
return p
def getmindiv(max):
max += 1
div = [i for i in range(max)]
for k in range(2, max):
if div[k] != k:
continue
curv = k
while curv < max:
div[curv] = k
curv += k
return div
if __name__ == '__main__':
n = int(stdin.readline())
a = list(map(int, stdin.readline().split()))
div = getmindiv(max(a))
r = two_divisors(a, div)
print(' '.join(map(str, r[0])))
print(' '.join(map(str, r[1])))
#stdin.write(' '.join(r[1]) + '\n')
``` | instruction | 0 | 103,860 | 22 | 207,720 |
Yes | output | 1 | 103,860 | 22 | 207,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n integers a_1, a_2, ..., a_n.
For each a_i find its two divisors d_1 > 1 and d_2 > 1 such that \gcd(d_1 + d_2, a_i) = 1 (where \gcd(a, b) is the greatest common divisor of a and b) or say that there is no such pair.
Input
The first line contains single integer n (1 β€ n β€ 5 β
10^5) β the size of the array a.
The second line contains n integers a_1, a_2, ..., a_n (2 β€ a_i β€ 10^7) β the array a.
Output
To speed up the output, print two lines with n integers in each line.
The i-th integers in the first and second lines should be corresponding divisors d_1 > 1 and d_2 > 1 such that \gcd(d_1 + d_2, a_i) = 1 or -1 and -1 if there is no such pair. If there are multiple answers, print any of them.
Example
Input
10
2 3 4 5 6 7 8 9 10 24
Output
-1 -1 -1 -1 3 -1 -1 -1 2 2
-1 -1 -1 -1 2 -1 -1 -1 5 3
Note
Let's look at a_7 = 8. It has 3 divisors greater than 1: 2, 4, 8. As you can see, the sum of any pair of divisors is divisible by 2 as well as a_7.
There are other valid pairs of d_1 and d_2 for a_{10}=24, like (3, 4) or (8, 3). You can print any of them.
Submitted Solution:
```
import math
def is_power_of_two(ai):
i = 1
while (2 ** i) <= ai:
if ai == 2 ** i:
return True
i += 1
return False
# Β Gotten from https://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n/3035188#3035188
def get_primes(n):
""" Returns a list of primes < n """
sieve = [True] * n
for i in range(3, int(n ** 0.5) + 1, 2):
if sieve[i]:
sieve[i * i :: 2 * i] = [False] * ((n - i * i - 1) // (2 * i) + 1)
return [2] + [i for i in range(3, n, 2) if sieve[i]]
n = int(input())
max_ai = 10 ** 7 + 1
s = int(math.sqrt(max_ai)) + 1
primes = get_primes(s)
np = len(primes)
set_primes = set(primes)
a = list(map(int, input().split()))
d1 = [-1 for i in range(n)]
d2 = [-1 for i in range(n)]
for i in range(n):
ai = a[i]
if ai in set_primes:
r1 = -1
r2 = -1
elif is_power_of_two(ai):
r1 = -1
r2 = -1
else:
# print(ai)
try:
if (ai % 2) == 0:
for i1 in range(1, np):
p1 = primes[i1]
if (ai % p1) == 0:
r1 = 2
r2 = p1
assert False
else:
for i1 in range(1, np - 1):
p1 = primes[i1]
for i2 in range(i1 + 1, np):
p2 = primes[i2]
if (ai % i1) == 0 and (ai % p2) == 0:
r1 = p1
r2 = p2
assert False
except AssertionError:
pass
d1[i] = r1
d2[i] = r2
print(" ".join((map(str, d1))))
print(" ".join((map(str, d2))))
``` | instruction | 0 | 103,861 | 22 | 207,722 |
No | output | 1 | 103,861 | 22 | 207,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n integers a_1, a_2, ..., a_n.
For each a_i find its two divisors d_1 > 1 and d_2 > 1 such that \gcd(d_1 + d_2, a_i) = 1 (where \gcd(a, b) is the greatest common divisor of a and b) or say that there is no such pair.
Input
The first line contains single integer n (1 β€ n β€ 5 β
10^5) β the size of the array a.
The second line contains n integers a_1, a_2, ..., a_n (2 β€ a_i β€ 10^7) β the array a.
Output
To speed up the output, print two lines with n integers in each line.
The i-th integers in the first and second lines should be corresponding divisors d_1 > 1 and d_2 > 1 such that \gcd(d_1 + d_2, a_i) = 1 or -1 and -1 if there is no such pair. If there are multiple answers, print any of them.
Example
Input
10
2 3 4 5 6 7 8 9 10 24
Output
-1 -1 -1 -1 3 -1 -1 -1 2 2
-1 -1 -1 -1 2 -1 -1 -1 5 3
Note
Let's look at a_7 = 8. It has 3 divisors greater than 1: 2, 4, 8. As you can see, the sum of any pair of divisors is divisible by 2 as well as a_7.
There are other valid pairs of d_1 and d_2 for a_{10}=24, like (3, 4) or (8, 3). You can print any of them.
Submitted Solution:
```
from math import log,gcd
def get_divisor(number):
i = 2
divisor = []
while i*i <=number:
if number % i == 0:
if i == number//i:
divisor.append(i)
else:
divisor.append(i)
divisor.append(number//i)
i = i + 1
return divisor
def main():
first = []
second = []
n = int(input())
number = list(map(int,input().split()))
for k in range(n):
divisors = get_divisor(number[k])
if gcd(number[k],sum(divisors)) != 1:
first.append(-1)
second.append(-1)
else:
for i in range(len(divisors)):
for j in range(i,len(divisors)):
if gcd(divisors[i] + divisors[j],number[k]) == 1:
first.append(divisors[i])
second.append(divisors[j])
break;
print(*first)
print(*second)
main()
``` | instruction | 0 | 103,862 | 22 | 207,724 |
No | output | 1 | 103,862 | 22 | 207,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n integers a_1, a_2, ..., a_n.
For each a_i find its two divisors d_1 > 1 and d_2 > 1 such that \gcd(d_1 + d_2, a_i) = 1 (where \gcd(a, b) is the greatest common divisor of a and b) or say that there is no such pair.
Input
The first line contains single integer n (1 β€ n β€ 5 β
10^5) β the size of the array a.
The second line contains n integers a_1, a_2, ..., a_n (2 β€ a_i β€ 10^7) β the array a.
Output
To speed up the output, print two lines with n integers in each line.
The i-th integers in the first and second lines should be corresponding divisors d_1 > 1 and d_2 > 1 such that \gcd(d_1 + d_2, a_i) = 1 or -1 and -1 if there is no such pair. If there are multiple answers, print any of them.
Example
Input
10
2 3 4 5 6 7 8 9 10 24
Output
-1 -1 -1 -1 3 -1 -1 -1 2 2
-1 -1 -1 -1 2 -1 -1 -1 5 3
Note
Let's look at a_7 = 8. It has 3 divisors greater than 1: 2, 4, 8. As you can see, the sum of any pair of divisors is divisible by 2 as well as a_7.
There are other valid pairs of d_1 and d_2 for a_{10}=24, like (3, 4) or (8, 3). You can print any of them.
Submitted Solution:
```
from sys import stdin, stdout
import math
from random import randint
def primes(n):
""" Returns a list of primes < n """
sieve = [True] * (n//2)
for i in range(3,int(n**0.5)+1,2):
if sieve[i//2]:
sieve[i*i//2::i] = [False] * ((n-i*i-1)//(2*i)+1)
return [2] + [2*i+1 for i in range(1,n//2) if sieve[i]]
def solve(n, P):
ori = n
s = set()
for i in P:
if n == 1: break
if i > math.sqrt(n):
i = n
first = True
while n % i == 0:
if first:
first = False
for x in s:
if gcd(x+i, ori) == 1:
return x, i
s.add(i)
n //= i
return -1, -1
def gcd(a, b):
while b:
a, b = b, a%b
return a
n = int(stdin.readline())
l = list(map(int, stdin.readline().strip().split()))
#l = [randint(2, 10**7) for _ in range(n)]
#print(l)
a1, a2 = [-1]*n, [-1]*n
P = primes(max(l))
for i in range(len(l)):
a1[i], a2[i] = solve(l[i], P)
# print("=====================")
# print(l[i]%a1[i], l[i]%a2[i], gcd(a1[i]+a2[i], l[i]))
# primeFactors(l[i])
# if a1[i] > 0:
# primeFactors(a1[i]+a2[i])
stdout.write(" ".join(map(str, a1))+"\n")
stdout.write(" ".join(map(str, a2))+"\n")
``` | instruction | 0 | 103,863 | 22 | 207,726 |
No | output | 1 | 103,863 | 22 | 207,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n integers a_1, a_2, ..., a_n.
For each a_i find its two divisors d_1 > 1 and d_2 > 1 such that \gcd(d_1 + d_2, a_i) = 1 (where \gcd(a, b) is the greatest common divisor of a and b) or say that there is no such pair.
Input
The first line contains single integer n (1 β€ n β€ 5 β
10^5) β the size of the array a.
The second line contains n integers a_1, a_2, ..., a_n (2 β€ a_i β€ 10^7) β the array a.
Output
To speed up the output, print two lines with n integers in each line.
The i-th integers in the first and second lines should be corresponding divisors d_1 > 1 and d_2 > 1 such that \gcd(d_1 + d_2, a_i) = 1 or -1 and -1 if there is no such pair. If there are multiple answers, print any of them.
Example
Input
10
2 3 4 5 6 7 8 9 10 24
Output
-1 -1 -1 -1 3 -1 -1 -1 2 2
-1 -1 -1 -1 2 -1 -1 -1 5 3
Note
Let's look at a_7 = 8. It has 3 divisors greater than 1: 2, 4, 8. As you can see, the sum of any pair of divisors is divisible by 2 as well as a_7.
There are other valid pairs of d_1 and d_2 for a_{10}=24, like (3, 4) or (8, 3). You can print any of them.
Submitted Solution:
```
# Contest No.: Edu89
# Problem No.: D
# Solver: JEMINI
# Date: 20200611
import sys
import heapq
def gcd(a, b):
if a < b:
a, b = b, a
if b == 0:
return a
return gcd(b, a % b)
def main():
temp = [True] * (10 ** 4)
temp[0] = False
temp[1] = False
for i in range(10 ** 4):
if temp[i]:
cnt = 2 * i
while cnt < 10 ** 4:
temp[cnt] = False
cnt += i
prime = set()
for i in range(10 ** 4):
if temp[i]:
prime.add(i)
n = int(input())
nums = list(map(int, sys.stdin.readline().split()))
ans1 = []
ans2 = []
for i in range(n):
flag = False
tempAns = None
num = nums[i]
fprime = None
for j in range(2, int(num ** 0.5) + 2):
#print(num, fprime, j)
if num % j == 0:
if not fprime:
fprime = j
elif j in prime and gcd(j + fprime, num) == 1:
flag = True
tempAns = [j, fprime]
break
if fprime and num // j in prime and gcd(num // j + fprime, num) == 1:
flag = True
tempAns = [num // j, fprime]
break
if flag:
ans1.append(tempAns[0])
ans2.append(tempAns[1])
else:
ans1.append(-1)
ans2.append(-1)
for i in ans1:
print(i, end= " ")
print("")
for i in ans2:
print(i, end= " ")
return
if __name__ == "__main__":
main()
``` | instruction | 0 | 103,864 | 22 | 207,728 |
No | output | 1 | 103,864 | 22 | 207,729 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0. | instruction | 0 | 104,057 | 22 | 208,114 |
Tags: combinatorics, dfs and similar, dsu, math, number theory
Correct Solution:
```
import sys
# sys.stdin = open('data/D.in', 'r')
T = 1
M = 1000000007
def process():
P, K = list(map(int, input().split()))
k = [K * x % P for x in range(P)]
# print(k)
# f(0) = k[f(0)]
# f(1) = k[f(4)]
# f(2) = k[f(3)]
# f(3) = k[f(2)]
# f(4) = k[f(1)]
if not K:
print(pow(P, P - 1, M))
return
f = [0] * P
c = [0] * P
ans = 1
for i in range(0, P):
if f[i]: continue
cnt = 1
u = i
f[u] = 1
while not f[k[u]]:
u = k[u]
f[u] = 1
cnt = cnt + 1
c[cnt] = c[cnt] + 1
# print(c)
for i in range(1, P):
if c[i] != 0:
cnt = i * c[i] + (1 if i > 1 else 0)
ans = ans * pow(cnt, c[i], M) % M
print(ans)
for _ in range(T):
process()
``` | output | 1 | 104,057 | 22 | 208,115 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0. | instruction | 0 | 104,058 | 22 | 208,116 |
Tags: combinatorics, dfs and similar, dsu, math, number theory
Correct Solution:
```
__author__ = 'MoonBall'
import sys
# sys.stdin = open('data/D.in', 'r')
T = 1
M = 1000000007
def process():
P, K = list(map(int, input().split()))
k = [K * x % P for x in range(P)]
# print(k)
# f(0) = k[f(0)]
# f(1) = k[f(4)]
# f(2) = k[f(3)]
# f(3) = k[f(2)]
# f(4) = k[f(1)]
if not K:
print(pow(P, P - 1, M))
return
f = [0] * P
c = [0] * P
ans = 1
for i in range(0, P):
if f[i]: continue
cnt = 1
u = i
f[u] = 1
while not f[k[u]]:
u = k[u]
f[u] = 1
cnt = cnt + 1
c[cnt] = c[cnt] + 1
# print(c)
for i in range(1, P):
if c[i] != 0:
cnt = i * c[i] + (c[1] if i > 1 else 0)
ans = ans * pow(cnt, c[i], M) % M
print(ans)
for _ in range(T):
process()
``` | output | 1 | 104,058 | 22 | 208,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0. | instruction | 0 | 104,059 | 22 | 208,118 |
Tags: combinatorics, dfs and similar, dsu, math, number theory
Correct Solution:
```
import math
def expmod(base, expon, mod):
ans = 1
for i in range(1, expon + 1):
ans = (ans * base) % mod
return ans
p, k = input().split()
s = 10 ** 9 + 7
k = int(k)
p = int(p)
ord = 1
done = 0
if k == 0:
z = p - 1
if k == 1:
z = p
else:
for i in range(2,p + 1):
if done == 0:
if (p-1) % i == 0:
if expmod(k, i, p) == 1:
ord = i
done = 1
z = int((p-1) / ord)
rem = expmod(p, z, s)
print(int(rem))
``` | output | 1 | 104,059 | 22 | 208,119 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0. | instruction | 0 | 104,060 | 22 | 208,120 |
Tags: combinatorics, dfs and similar, dsu, math, number theory
Correct Solution:
```
def main():
t = 1
a = k
if k == 0: return n - 1
if k == 1: return n
while a != 1:
a = a * k % n
t += 1
p = (n - 1) // t
return p
n, k = map(int, input().split())
mod = 10 ** 9 + 7
p = main()
print(pow(n, p, mod))
``` | output | 1 | 104,060 | 22 | 208,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0. | instruction | 0 | 104,061 | 22 | 208,122 |
Tags: combinatorics, dfs and similar, dsu, math, number theory
Correct Solution:
```
MOD = 10**9+7
def f(a,b):
if b == 1:
return a%MOD
elif b % 2 == 0:
return f((a*a)%MOD,b//2)
else:
return (a*f((a*a)%MOD,b//2)) % MOD
p,k = map(int,input().split())
if k == 0:
print(f(p,p-1))
exit()
if k == 1:
print(f(p,p))
exit()
t = 1
a = k
while a != 1:
a = (a*k % p)
t += 1
n = (p-1)//t
print(f(p,n))
``` | output | 1 | 104,061 | 22 | 208,123 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0. | instruction | 0 | 104,062 | 22 | 208,124 |
Tags: combinatorics, dfs and similar, dsu, math, number theory
Correct Solution:
```
M = 10**9 + 7
def period(p, k):
c = 1
ek = k
while ek != 1:
ek = (ek * k) % p
c += 1
return c
def functions(p, k):
if k == 0:
return pow(p, p-1, M)
elif k == 1:
return pow(p, p, M)
else:
c = (p-1)//period(p, k)
return pow(p, c, M)
if __name__ == '__main__':
p, k = map(int, input().split())
print(functions(p, k))
``` | output | 1 | 104,062 | 22 | 208,125 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0. | instruction | 0 | 104,063 | 22 | 208,126 |
Tags: combinatorics, dfs and similar, dsu, math, number theory
Correct Solution:
```
MOD=int(1e9+7)
n,k=map(int,input().split())
if k==0:
p=n-1
elif k==1:
p=n
else:
t=1
a=k
while a!=1:
a=a*k%n
t+=1
p=(n-1)//t
print(pow(n,p,MOD))
``` | output | 1 | 104,063 | 22 | 208,127 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0. | instruction | 0 | 104,064 | 22 | 208,128 |
Tags: combinatorics, dfs and similar, dsu, math, number theory
Correct Solution:
```
__author__ = 'MoonBall'
import sys
# sys.stdin = open('data/D.in', 'r')
T = 1
M = 1000000007
def process():
P, K = list(map(int, input().split()))
k = [K * x % P for x in range(P)]
# print(k)
# f(0) = k[f(0)]
# f(1) = k[f(4)]
# f(2) = k[f(3)]
# f(3) = k[f(2)]
# f(4) = k[f(1)]
if not K:
print(pow(P, P - 1, M))
return
f = [0] * P
c = [0] * P
ans = 1
for i in range(0, P):
if f[i]: continue
cnt = 1
u = i
f[u] = 1
while not f[k[u]]:
u = k[u]
f[u] = 1
cnt = cnt + 1
c[cnt] = c[cnt] + 1
# print(c)
for i in range(1, P):
if c[i] != 0:
cnt = i * c[i] + (1 if i > 1 else 0)
ans = ans * pow(cnt, c[i], M) % M
print(ans)
for _ in range(T):
process()
``` | output | 1 | 104,064 | 22 | 208,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0.
Submitted Solution:
```
c=1000000007
#special cases for k=0, 1
p, k = map(int, input().split(" "))
"""def binary(number):
#turn into string
s=""
while number!=0:
r = number%2
s = str(r) + s
number = number//2
return s
def mod(base, expo, mod):
#write expo as binary and calculate to the power of 2 power
ans=1
s=binary(expo)
l=len(s)
#calculate a list of remainders of length l
mods = []
current=base
for i in range (l):
mods = [current] + mods
current = (current**2) % mod
#print(mods, s)
for i in range (l):
if s[i] == "1":
ans *= mods[i]
#print(ans)
ans = ans%mod
return ans
"""
if k==0:
answer=pow(p, p-1, c)
elif k==1:
answer=pow(p, p, c)
else:
#counted = [False] * (p-1)
size = 1
"""def modify(init):
global counted
new=init*k
new = new%p
counted[init-1]=True
while new!=init:
#print(new)
counted[new-1]=True
new=new*k
new = new%p
#print(new)
for j in range (p-1):
if counted[j]==False:
i=j+1
#print("i", i)
modify(i)
#print(counted)
cycles+=1
"""
new=k
while new!=1:
size+=1
new*=k
new=new%p
cycles = (p-1)//size
#print(cycles)
answer=pow(p, cycles, c)
#too slow, issue is not with mod
#fuck python and fuck c++ dont have time to recode this in c++ why can't python be faster
print(answer)
``` | instruction | 0 | 104,065 | 22 | 208,130 |
Yes | output | 1 | 104,065 | 22 | 208,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0.
Submitted Solution:
```
p,k=map(int,input().split())
res=set()
ex=p-1
if k!=0:
if k==1:
ex=p
else:
c=k
for i in range(p):
res.add(c)
c*=k
c%=p
ex=(p-1)//len(res)
ans=1
for i in range(ex):
ans*=p
ans%=1000000007
print(ans)
``` | instruction | 0 | 104,066 | 22 | 208,132 |
Yes | output | 1 | 104,066 | 22 | 208,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0.
Submitted Solution:
```
__author__ = 'MoonBall'
import sys
# sys.stdin = open('data/D.in', 'r')
T = 1
M = 1000000007
def pow(x, y, m):
if y == 0:
return 1
if (y & 1):
return pow(x, y - 1, m) * x % m
else:
t = pow(x, y >> 1, m)
return t * t % m
def process():
P, K = list(map(int, input().split()))
k = [K * x % P for x in range(P)]
# print(k)
# f(0) = k[f(0)]
# f(1) = k[f(4)]
# f(2) = k[f(3)]
# f(3) = k[f(2)]
# f(4) = k[f(1)]
if not K:
print(pow(P, P - 1, M))
return
if K == 1:
print(pow(P, P, M))
return
f = [0] * P
c = [0] * P
ans = 1
for i in range(P):
if f[i]: continue
cnt = 1
u = i
f[u] = 1
while not f[k[u]]:
u = k[u]
f[u] = 1
cnt = cnt + 1
c[cnt] = c[cnt] + 1
# print(c)
for i in range(2, P):
if c[i] != 0:
cnt = i * c[i] + 1
ans = ans * pow(cnt, c[i], M) % M
print(ans)
for _ in range(T):
process()
``` | instruction | 0 | 104,067 | 22 | 208,134 |
Yes | output | 1 | 104,067 | 22 | 208,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0.
Submitted Solution:
```
def m_pow(x, y, m):
if y == 0:
return 1
if (y & 1):
return m_pow(x, y - 1, m) * x % m
else:
t = m_pow(x, y >> 1, m)
return t * t % m
#
(p, k) = map(int, input().split())
used = [0] * p
if k == 0:
print(m_pow(p, p - 1, 1000000007))
else:
c = 1 if k == 1 else 0
for x in range(1, p):
if not used[x]:
y = x
while not used[y]:
used[y] = True
y = k * y % p
c += 1
print(m_pow(p, c, 1000000007))
``` | instruction | 0 | 104,068 | 22 | 208,136 |
Yes | output | 1 | 104,068 | 22 | 208,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0.
Submitted Solution:
```
MOD = 10**9+7
p,k = map(int,input().split())
bool = False
for i in range(1,p):
if (i**2-k) % p == 0:
bool = True
break
if bool:
print((p*p)%MOD)
else:
print(p)
``` | instruction | 0 | 104,069 | 22 | 208,138 |
No | output | 1 | 104,069 | 22 | 208,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0.
Submitted Solution:
```
p,k = map(int,input().split())
m = 10**9+7
if k == 0:
print(pow(p,p-1,m))
else:
used = set()
res = 0
for i in range(1, p):
a = i
if a not in used:
res += 1
while a not in used:
used.add(a)
a = k * a % p
print(pow(p, res, m))
``` | instruction | 0 | 104,070 | 22 | 208,140 |
No | output | 1 | 104,070 | 22 | 208,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0.
Submitted Solution:
```
__author__ = 'MoonBall'
import sys
# sys.stdin = open('data/D.in', 'r')
T = 1
M = 1000000007
def process():
P, K = list(map(int, input().split()))
k = [K * x % P for x in range(P)]
# print(k)
# f(0) = k[f(0)]
# f(1) = k[f(4)]
# f(2) = k[f(3)]
# f(3) = k[f(2)]
# f(4) = k[f(1)]
if not K:
print(pow(P, P - 1, M))
return
f = [0] * P
c = [0] * P
ans = 1
for i in range(1, P):
if f[i]: continue
cnt = 1
u = i
f[u] = 1
while not f[k[u]]:
u = k[u]
f[u] = 1
cnt = cnt + 1
c[cnt] = c[cnt] + 1
# print(c)
for i in range(1, P):
if c[i] != 0:
cnt = i * c[i] + 1
ans = ans * pow(cnt, c[i], M) % M
print(ans)
for _ in range(T):
process()
``` | instruction | 0 | 104,071 | 22 | 208,142 |
No | output | 1 | 104,071 | 22 | 208,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As behooves any intelligent schoolboy, Kevin Sun is studying psycowlogy, cowculus, and cryptcowgraphy at the Bovinia State University (BGU) under Farmer Ivan. During his Mathematics of Olympiads (MoO) class, Kevin was confronted with a weird functional equation and needs your help. For two fixed integers k and p, where p is an odd prime number, the functional equation states that
<image>
for some function <image>. (This equation should hold for any integer x in the range 0 to p - 1, inclusive.)
It turns out that f can actually be many different functions. Instead of finding a solution, Kevin wants you to count the number of distinct functions f that satisfy this equation. Since the answer may be very large, you should print your result modulo 109 + 7.
Input
The input consists of two space-separated integers p and k (3 β€ p β€ 1 000 000, 0 β€ k β€ p - 1) on a single line. It is guaranteed that p is an odd prime number.
Output
Print a single integer, the number of distinct functions f modulo 109 + 7.
Examples
Input
3 2
Output
3
Input
5 4
Output
25
Note
In the first sample, p = 3 and k = 2. The following functions work:
1. f(0) = 0, f(1) = 1, f(2) = 2.
2. f(0) = 0, f(1) = 2, f(2) = 1.
3. f(0) = f(1) = f(2) = 0.
Submitted Solution:
```
import math
def expmod(base, expon, mod):
ans = 1
for i in range(1, expon + 1):
ans = (ans * base) % mod
return ans
p, k = input().split()
k = int(k)
p = int(p)
ord = 1
done = 0
if k == 0:
z = p - 1
else:
for i in range(1,int((p-1) ** (0.5))+2):
if done == 0:
if (p-1) % i == 0:
if expmod(k, i, p) == 1:
ord = i
done = 1
z = int((p-1) / ord)
rem = expmod(p, z, 1000000007)
print(int(rem))
``` | instruction | 0 | 104,072 | 22 | 208,144 |
No | output | 1 | 104,072 | 22 | 208,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
If you have gone that far, you'll probably skip unnecessary legends anyway...
You are given a binary string <image> and an integer <image>. Find the number of integers k, 0 β€ k < N, such that for all i = 0, 1, ..., m - 1
<image> Print the answer modulo 109 + 7.
Input
In the first line of input there is a string s consisting of 0's and 1's (1 β€ |s| β€ 40).
In the next line of input there is an integer n (1 β€ n β€ 5Β·105).
Each of the next n lines contains two space-separated integers pi, Ξ±i (1 β€ pi, Ξ±i β€ 109, pi is prime). All pi are distinct.
Output
A single integer β the answer to the problem.
Examples
Input
1
2
2 1
3 1
Output
2
Input
01
2
3 2
5 1
Output
15
Input
1011
1
3 1000000000
Output
411979884
Submitted Solution:
```
def gcd(x, y):
if y == 0:
return x
else:
return gcd(y, x%y)
s = input()
n = int(input())
N = 1
for i in range(n):
l = input().split(" ")
N *= int(l[0])**int(l[1])
k = []
for char in range(len(s)):
if s[char] == '1':
k.append(char)
count = 0
for j in range(N):
for i in range(len(k)):
if gcd(k[i] + j, N) == 1:
count += 1
print(count)
``` | instruction | 0 | 104,119 | 22 | 208,238 |
No | output | 1 | 104,119 | 22 | 208,239 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence a0a1...aN-1 digits and a prime number Q. For each i β€ j with ai β 0, the subsequence aiai+1...aj can be read as a decimal representation of a positive integer. Subsequences with leading zeros are not considered. Your task is to count the number of pairs (i, j) such that the corresponding subsequence is a multiple of Q.
Input
The input consists of at most 50 datasets. Each dataset is represented by a line containing four integers N, S, W, and Q, separated by spaces, where 1 β€ N β€ 105, 1 β€ S β€ 109, 1 β€ W β€ 109, and Q is a prime number less than 108. The sequence a0...aN-1 of length N is generated by the following code, in which ai is written as a[i].
int g = S;
for(int i=0; i<N; i++) {
a[i] = (g/7) % 10;
if( g%2 == 0 ) { g = (g/2); }
else { g = (g/2) ^ W; }
}
Note: the operators /, %, and ^ are the integer division, the modulo, and the bitwise exclusiveor, respectively. The above code is meant to be a random number generator. The intended solution does not rely on the way how the sequence is generated.
The end of the input is indicated by a line containing four zeros separated by spaces.
Output
For each dataset, output the answer in a line. You may assume that the answer is less than 230.
Example
Input
3 32 64 7
4 35 89 5
5 555 442 3
5 777 465 11
100000 666 701622763 65537
0 0 0 0
Output
2
4
6
3
68530 | instruction | 0 | 104,366 | 22 | 208,732 |
"Correct Solution:
```
from collections import defaultdict
def gen(N, S, W):
g = S
for i in range(N):
yield (g//7) % 10
if g % 2 == 0:
g //= 2
else:
g = (g//2) ^ W
def solve():
N, S, W, Q = map(int, input().split())
if N == 0:
return False
bs = list(gen(N, S, W))
ans = 0
if Q == 2 or Q == 5:
cnt = 0
for i in range(N):
b = bs[i]
if b != 0:
cnt += 1
if b % Q == 0:
ans += cnt
else:
rev10 = pow(10, Q-2, Q)
D = defaultdict(int)
D[0] = 1
s = 0; v = 1
first = 1
for i in range(N):
b = bs[i]
if first and b == 0:
continue
s = (s + v * b) % Q
v = v * rev10 % Q
ans += D[s]
if i < N-1 and bs[i+1] != 0:
D[s] += 1
first = 0
print(ans)
return True
while solve():
...
``` | output | 1 | 104,366 | 22 | 208,733 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence a0a1...aN-1 digits and a prime number Q. For each i β€ j with ai β 0, the subsequence aiai+1...aj can be read as a decimal representation of a positive integer. Subsequences with leading zeros are not considered. Your task is to count the number of pairs (i, j) such that the corresponding subsequence is a multiple of Q.
Input
The input consists of at most 50 datasets. Each dataset is represented by a line containing four integers N, S, W, and Q, separated by spaces, where 1 β€ N β€ 105, 1 β€ S β€ 109, 1 β€ W β€ 109, and Q is a prime number less than 108. The sequence a0...aN-1 of length N is generated by the following code, in which ai is written as a[i].
int g = S;
for(int i=0; i<N; i++) {
a[i] = (g/7) % 10;
if( g%2 == 0 ) { g = (g/2); }
else { g = (g/2) ^ W; }
}
Note: the operators /, %, and ^ are the integer division, the modulo, and the bitwise exclusiveor, respectively. The above code is meant to be a random number generator. The intended solution does not rely on the way how the sequence is generated.
The end of the input is indicated by a line containing four zeros separated by spaces.
Output
For each dataset, output the answer in a line. You may assume that the answer is less than 230.
Example
Input
3 32 64 7
4 35 89 5
5 555 442 3
5 777 465 11
100000 666 701622763 65537
0 0 0 0
Output
2
4
6
3
68530 | instruction | 0 | 104,367 | 22 | 208,734 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+9
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
def f(n,s,w,q):
g = s
a = []
for i in range(n):
a.append(g // 7 % 10)
if g % 2 == 1:
g = (g // 2) ^ w
else:
g //= 2
if q % 2 == 0 or q % 5 == 0:
r = 0
t = 0
for c in a:
if c > 0:
t += 1
if c % q == 0:
r += t
return r
b = [0] * (n+1)
k = 1
for i in range(n-1,-1,-1):
b[i] = (b[i+1] + a[i] * k) % q
k = k * 10 % q
d = collections.defaultdict(int)
r = 0
for i in range(n):
if a[i] > 0:
d[b[i]] += 1
r += d[b[i+1]]
return r
while 1:
n,m,l,o = LI()
if n == 0:
break
rr.append(f(n,m,l,o))
return '\n'.join(map(str, rr))
print(main())
``` | output | 1 | 104,367 | 22 | 208,735 |
Provide a correct Python 3 solution for this coding contest problem.
Division of Big Integers
Given two integers $A$ and $B$, compute the quotient, $\frac{A}{B}$. Round down to the nearest decimal.
Input
Two integers $A$ and $B$ separated by a space character are given in a line.
Output
Print the quotient in a line.
Constraints
* $-1 \times 10^{1000} \leq A, B \leq 10^{1000}$
* $B \ne 0$
Sample Input 1
5 8
Sample Output 1
0
Sample Input 2
100 25
Sample Output 2
4
Sample Input 3
-1 3
Sample Output 3
0
Sample Input 4
12 -3
Sample Output 4
-4
Example
Input
5 8
Output
0 | instruction | 0 | 104,412 | 22 | 208,824 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Big Integers - Division of Big Integers
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=NTL_2_D&lang=jp
"""
import sys
def main(args):
A, B = map(int, input().split())
sign = 1 if A*B >= 0 else -1
print(abs(A) // abs(B) * sign)
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 104,412 | 22 | 208,825 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72 | instruction | 0 | 105,321 | 22 | 210,642 |
Tags: brute force, greedy, implementation
Correct Solution:
```
import sys
q=int(sys.stdin.readline())
for _ in range(q):
n= int(sys.stdin.readline())
count =0
while n>1:
if n%2==0:
n//=2
count+=1
elif n%3==0:
n//=3
n*=2
count+=1
elif n%5==0:
n//=5
n*=4
count+=1
else:
count=-1
break
print(count)
``` | output | 1 | 105,321 | 22 | 210,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72 | instruction | 0 | 105,322 | 22 | 210,644 |
Tags: brute force, greedy, implementation
Correct Solution:
```
import sys
q = int(input())
for i in range(q):
#print ('i:',i)
n = int(input())
cnt = 0
while n > 1:
if n%2==0:
n = n // 2
cnt += 1
elif n%3 ==0:
n = (2*n) // 3
cnt += 1
elif n%5 ==0:
n = (n*4) // 5
cnt += 1
else:
cnt = -1
break
print (cnt)
``` | output | 1 | 105,322 | 22 | 210,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72 | instruction | 0 | 105,323 | 22 | 210,646 |
Tags: brute force, greedy, implementation
Correct Solution:
```
q = int(input())
for _ in range(q):
n = int(input())
count = 0
while n!=1:
if n%5 == 0:
n//=5
count += 3
elif n%3 == 0:
n//=3
count+=2
elif n%2 == 0:
n//= 2
count +=1
else:
print (-1)
break
else:
print (count)
``` | output | 1 | 105,323 | 22 | 210,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72 | instruction | 0 | 105,324 | 22 | 210,648 |
Tags: brute force, greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
cnt = 0
while n > 1:
if n%2 == 0:
cnt += 1
n //= 2
elif n%3 == 0:
cnt += 1
n = (2 * n // 3)
elif n%5 == 0:
cnt += 1
n = (4 * n // 5)
else:
cnt = -1
break
print(cnt)
``` | output | 1 | 105,324 | 22 | 210,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72 | instruction | 0 | 105,325 | 22 | 210,650 |
Tags: brute force, greedy, implementation
Correct Solution:
```
q=int(input())
for query in range(q):
n=int(input())
ANS=0
while n>1:
if n%3==0:
ANS+=1
n=n//3*2
elif n%5==0:
ANS+=1
n=n//5*4
elif n%2==0:
ANS+=1
n//=2
else:
print(-1)
break
else:
print(ANS)
``` | output | 1 | 105,325 | 22 | 210,651 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72 | instruction | 0 | 105,326 | 22 | 210,652 |
Tags: brute force, greedy, implementation
Correct Solution:
```
def to(n):
if n==1:return 0
c=0
while n>1:
if n%2==0:
n=n//2;c+=1
elif n%3==0:
n=(n//3)*2;c+=1
elif n%5==0:
n=(n//5)*4;c+=1
else:
return -1
break
return c
for _ in range(int(input())):
n=int(input())
print(to(n))
``` | output | 1 | 105,326 | 22 | 210,653 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72 | instruction | 0 | 105,327 | 22 | 210,654 |
Tags: brute force, greedy, implementation
Correct Solution:
```
# import sys
# sys.stdin=open("input.in","r")
for i in range(int(input())):
a=int(input())
c=0
while a>1:
if a%2==0:
a=a//2
c+=1
elif a%3==0:
a=2*(a//3)
c+=1
elif a%5==0:
a=4*(a//5)
c+=1
else:
c=-1
break
print(c)
``` | output | 1 | 105,327 | 22 | 210,655 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer n.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72 | instruction | 0 | 105,328 | 22 | 210,656 |
Tags: brute force, greedy, implementation
Correct Solution:
```
def mul5(num):
if num%5==0:
return 4*(num//5)
def mul2(num):
if num%2 ==0:
return num//2
def mul3(num):
if num%3 == 0:
return 2*(num//3)
#check 2 then 3 then 5
while True:
try:
val = int(input( ))
num=[0]*val
break
except:
print("this is not a valid input")
for i in range(0, val):
num[i]= int(input())
for x in range(0, val):
count=0
if num[x]==1:
print(0)
else:
while (num[x])/5 >=1 and num[x]%5==0:
num[x] = mul5(num[x])
count +=1
while num[x]/3>=1 and num[x]%3==0:
num[x] = mul3(num[x])
count +=1
while num[x]/2>=1 and num[x]%2==0:
num[x] = mul2(num[x])
count +=1
if count>0 and num[x] ==1:
count= count
else:
count=-1
print(count)
``` | output | 1 | 105,328 | 22 | 210,657 |
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.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72
Submitted Solution:
```
# cook your dish here
from math import *
t=int(input())
for _ in range(t):
n=int(input())
if n==1:
print('0')
else:
c=0
f=0
while n>1:
if n%5==0:
c=c+3
n=n//5
elif n%3==0:
c=c+2
n=n//3
elif n%2==0:
c=c+1
n=n//2
else:
f=1
break
if f==1:
print('-1')
else:
print(c)
``` | instruction | 0 | 105,329 | 22 | 210,658 |
Yes | output | 1 | 105,329 | 22 | 210,659 |
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.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72
Submitted Solution:
```
a=int(input())
i=0
while i<a:
op=0
n=int(input())
if n<=0:
op=-1
else:
while n!=1:
if n%5==0:
op+=1
n=(4*n)//5
elif n%3==0:
op+=1
n=(2*n)//3
elif n%2==0:
op+=1
n=n//2
else:
op=-1
break
print(op)
i+=1
``` | instruction | 0 | 105,330 | 22 | 210,660 |
Yes | output | 1 | 105,330 | 22 | 210,661 |
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.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72
Submitted Solution:
```
def divide(n):
iter = 0
while n != 1:
if n % 2 == 0:
n //= 2
elif n % 3 == 0:
n = n * 2//3
elif n % 5 == 0:
n = n * 4//5
else:
return -1
iter += 1
return iter
q = int(input())
for _ in range(q):
n = int(input())
print(divide(n))
``` | instruction | 0 | 105,331 | 22 | 210,662 |
Yes | output | 1 | 105,331 | 22 | 210,663 |
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.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72
Submitted Solution:
```
q=int(input())
for i in range(q):
n=int(input())
t2,t3,t5=0,0,0
while((n%2)==0):
n=n//2
t2+=1
while((n%3)==0):
n=n//3
t3+=1
while((n%5)==0):
n=n//5
t5+=1
if(n!=1):
print(-1)
else:
print(t2+t3*2+t5*3)
``` | instruction | 0 | 105,332 | 22 | 210,664 |
Yes | output | 1 | 105,332 | 22 | 210,665 |
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.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72
Submitted Solution:
```
a=int(input())
while(a>0):
count=0
flag=0
q=int(input())
while(q!=1):
if(q%5==0):
count+=3
q=q/5
elif(q%3==0):
count+=2
q=q/3
elif(q%2==0):
count+=1
q=q/2
else:
print(-1)
flag=1
break
if(flag!=1):
print(count)
a-=1
``` | instruction | 0 | 105,333 | 22 | 210,666 |
No | output | 1 | 105,333 | 22 | 210,667 |
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.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72
Submitted Solution:
```
l = int(input())
for i in range(l):
count = 0
n = int(input())
if n==1:
print(count)
elif n%2!=0 and n%3!=0 and n%5 !=0:
print(-1)
else:
while n!=1:
charas=0
if n%2!=0 and n%3!=0 and n%5 !=0:
if n!=1:
print(-1)
n=1
charas=1
else:
if n%2==0:
n = n/2
count = count+1
elif n%3==0:
n = 2*n/3
count = count + 1
elif n%5==0:
n = 4*n/5
count = count + 1
if charas==0:
print(count)
``` | instruction | 0 | 105,334 | 22 | 210,668 |
No | output | 1 | 105,334 | 22 | 210,669 |
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.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72
Submitted Solution:
```
t=int(input())
s=[]
for hfvjv in range(0,t):
n=int(input())
g=0
k=[0,0,0]
jj=[1,1,1]
while k!=jj:
if n%5==0:
n=n/5
g+=3
else:
k[2]=1
if n%3==0:
n=n/3
g+=2
else:
k[1]=1
if n%2==0:
n=n/2
g+=1
else:
k[0]=1
if n==1:
s.append(g)
else:
s.append(-1)
for i in s:
print(i)
``` | instruction | 0 | 105,335 | 22 | 210,670 |
No | output | 1 | 105,335 | 22 | 210,671 |
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.
You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:
1. Replace n with n/2 if n is divisible by 2;
2. Replace n with 2n/3 if n is divisible by 3;
3. Replace n with 4n/5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.
Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries.
The next q lines contain the queries. For each query you are given the integer number n (1 β€ n β€ 10^{18}).
Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.
Example
Input
7
1
10
25
30
14
27
1000000000000000000
Output
0
4
6
6
-1
6
72
Submitted Solution:
```
from sys import stdin, stdout
q = int(stdin.readline())
while q != 0:
n = int(stdin.readline())
ans2, ans3, ans5 = 0, 0, 0
while n % 5 == 0:
n /= 5
ans5 += 1
ans2 += 2
while n % 3 == 0:
n /= 3
ans2 += 1
ans3 += 1
while n % 2 == 0:
n /= 2
ans2 += 1
if n > 1:
print(-1)
else:
print(int(ans2 + ans3 + ans5))
q -= 1
``` | instruction | 0 | 105,336 | 22 | 210,672 |
No | output | 1 | 105,336 | 22 | 210,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a fraction x/y good if there exists at least one another fraction (x')/(y') such that x/y = (x')/(y'), 1 β€ x', y' β€ 9, the digit denoting x' is contained in the decimal representation of x, and the digit denoting y' is contained in the decimal representation of y. For example, 26/13 is a good fraction, because 26/13 = 2/1.
You are given an integer number n. Please calculate the number of good fractions x/y such that 1 β€ x β€ n and 1 β€ y β€ n. The answer may be really large, so print it modulo 998244353.
Input
The only line of the input contains one integer n (1 β€ n < 10^{100}).
Output
Print the number of good fractions x/y such that 1 β€ x β€ n and 1 β€ y β€ n. The answer may be really large, so print it modulo 998244353.
Examples
Input
42
Output
150
Input
3141592653589793238462643383279
Output
459925407
Submitted Solution:
```
n=int(input("Enter the number:"))
c=0
for x in range(n+1):
for y in range(n+1):
for k in range(1,x+1):
if (x%k==0 and y%k==0):
q=x/k
w=y/k
if (q<10 and w<10):
m = list(map(int, str(x)))
l = list(map(int, str(y)))
if q in m:
if w in l:
c=c+1
print(c%998244353)
``` | instruction | 0 | 105,337 | 22 | 210,674 |
No | output | 1 | 105,337 | 22 | 210,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call a fraction x/y good if there exists at least one another fraction (x')/(y') such that x/y = (x')/(y'), 1 β€ x', y' β€ 9, the digit denoting x' is contained in the decimal representation of x, and the digit denoting y' is contained in the decimal representation of y. For example, 26/13 is a good fraction, because 26/13 = 2/1.
You are given an integer number n. Please calculate the number of good fractions x/y such that 1 β€ x β€ n and 1 β€ y β€ n. The answer may be really large, so print it modulo 998244353.
Input
The only line of the input contains one integer n (1 β€ n < 10^{100}).
Output
Print the number of good fractions x/y such that 1 β€ x β€ n and 1 β€ y β€ n. The answer may be really large, so print it modulo 998244353.
Examples
Input
42
Output
150
Input
3141592653589793238462643383279
Output
459925407
Submitted Solution:
```
n = int(input())
count = 0
for i in range (1, n +1):
for x in range (1, n + 1):
for y in range (1, n +1):
if i % x == 0 and i % y == 0 and i // x < 10 and i // y < 10:
count += 1
print(count % 998244353)
``` | instruction | 0 | 105,338 | 22 | 210,676 |
No | output | 1 | 105,338 | 22 | 210,677 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 β€ t β€ 5β
10^5) - number of test cases.
On next line there are t numbers, m_i (1 β€ m_i β€ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4. | instruction | 0 | 105,469 | 22 | 210,938 |
Tags: bitmasks, constructive algorithms, dp, math
Correct Solution:
```
T = input()
mod = int(1e9 + 7)
a = map(int, input().split())
c = []
for n in a:
b = (n // 2 + 2)
b = b * b
b //= 4
c.append(str(b % mod))
print(' '.join(c))
``` | output | 1 | 105,469 | 22 | 210,939 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 β€ t β€ 5β
10^5) - number of test cases.
On next line there are t numbers, m_i (1 β€ m_i β€ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4. | instruction | 0 | 105,470 | 22 | 210,940 |
Tags: bitmasks, constructive algorithms, dp, math
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def readvars():
k = map(int,input().split())
return(k)
def readlist():
li = list(map(int,input().split()))
return(li)
def anslist(li):
ans = " ".join([str(v) for v in li])
return(ans)
def main():
t = 1
# t = int(input())
for xx in range(t):
t = int(input())
n = readlist()
for m in n: print(((((m//2) - (m//4) + 1)%(1000000007))*((m//4) + 1))%1000000007)
# 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")
# endregion
if __name__ == "__main__":
main()
``` | output | 1 | 105,470 | 22 | 210,941 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 β€ t β€ 5β
10^5) - number of test cases.
On next line there are t numbers, m_i (1 β€ m_i β€ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4. | instruction | 0 | 105,471 | 22 | 210,942 |
Tags: bitmasks, constructive algorithms, dp, math
Correct Solution:
```
# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
# ===============================================================================================
# some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def pow(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1): # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# discrete binary search
# minimise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# if isvalid(l):
# return l
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m - 1):
# return m
# if isvalid(m):
# r = m + 1
# else:
# l = m
# return m
# maximise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# # print(l,r)
# if isvalid(r):
# return r
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m + 1):
# return m
# if isvalid(m):
# l = m
# else:
# r = m - 1
# return m
##to find factorial and ncr
# N=100000
# mod = 10**9 +7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, N + 1):
# fac.append((fac[-1] * i) % mod)
# inv.append(mod - (inv[mod % i] * (mod // i) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
#
#
# def comb(n, r):
# if n < r:
# return 0
# else:
# return fac[n] * (finv[r] * finv[n - r] % mod) % mod
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def N():
return int(inp())
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(i):
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
def solve():
n=N()
ar=lis()
for i in range(len(ar)):
m=ar[i]
v = m // 2
u = v // 2
w = (v - u)
print((u * w + u + w + 1) % mod)
solve()
#testcase(int(inp()))
``` | output | 1 | 105,471 | 22 | 210,943 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 β€ t β€ 5β
10^5) - number of test cases.
On next line there are t numbers, m_i (1 β€ m_i β€ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4. | instruction | 0 | 105,472 | 22 | 210,944 |
Tags: bitmasks, constructive algorithms, dp, math
Correct Solution:
```
t = int(input())
a = list(map(int, input().split()))
out = []
for n in a:
ans = (n//2 + 2)
ans = ans*ans
ans //= 4
out.append(ans%1000000007)
print(' '.join(str(x) for x in out))
``` | output | 1 | 105,472 | 22 | 210,945 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dreamoon loves summing up something for no reason. One day he obtains two integers a and b occasionally. He wants to calculate the sum of all nice integers. Positive integer x is called nice if <image> and <image>, where k is some integer number in range [1, a].
By <image> we denote the quotient of integer division of x and y. By <image> we denote the remainder of integer division of x and y. You can read more about these operations here: http://goo.gl/AcsXhT.
The answer may be large, so please print its remainder modulo 1 000 000 007 (109 + 7). Can you compute it faster than Dreamoon?
Input
The single line of the input contains two integers a, b (1 β€ a, b β€ 107).
Output
Print a single integer representing the answer modulo 1 000 000 007 (109 + 7).
Examples
Input
1 1
Output
0
Input
2 2
Output
8
Note
For the first sample, there are no nice integers because <image> is always zero.
For the second sample, the set of nice integers is {3, 5}. | instruction | 0 | 105,632 | 22 | 211,264 |
Tags: math
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
import heapq
#range = xrange # not for python 3.0+
# main code
mod=10**9+7
ans=0
a,b=map(int,stdin.read().split())
for i in range(1,b):
#print val
val=a
t1=(val*(val+1))//2
t1%=mod
t1=(t1*b)%mod
temp=(t1*i)%mod
temp=(temp+((i*val)%mod))%mod
ans=(ans+temp)%mod
print(ans)
``` | output | 1 | 105,632 | 22 | 211,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dreamoon loves summing up something for no reason. One day he obtains two integers a and b occasionally. He wants to calculate the sum of all nice integers. Positive integer x is called nice if <image> and <image>, where k is some integer number in range [1, a].
By <image> we denote the quotient of integer division of x and y. By <image> we denote the remainder of integer division of x and y. You can read more about these operations here: http://goo.gl/AcsXhT.
The answer may be large, so please print its remainder modulo 1 000 000 007 (109 + 7). Can you compute it faster than Dreamoon?
Input
The single line of the input contains two integers a, b (1 β€ a, b β€ 107).
Output
Print a single integer representing the answer modulo 1 000 000 007 (109 + 7).
Examples
Input
1 1
Output
0
Input
2 2
Output
8
Note
For the first sample, there are no nice integers because <image> is always zero.
For the second sample, the set of nice integers is {3, 5}. | instruction | 0 | 105,633 | 22 | 211,266 |
Tags: math
Correct Solution:
```
MOD = int(1e9+7)
a, b = list(map(int, input().split()))
res = b * (b - 1) // 2 * (a + (1 + a) * a * b // 2)
print(res % MOD)
``` | output | 1 | 105,633 | 22 | 211,267 |
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