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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9 | instruction | 0 | 47,535 | 22 | 95,070 |
Tags: implementation, number theory
Correct Solution:
```
# bsdk idhar kya dekhne ko aaya hai, khud kr!!!
# import math
from itertools import *
# import random
# import calendar
# import datetime
# import webbrowser
def function(x, lst):
permu = permutations(lst, len(lst))
count_ = 0
for i in permu:
temp = (((x % i[0]) % i[1]) % i[2]) % i[-1]
if temp == x:
count_ += 1
if count_ >= 7:
return True
if count_ < 7:
return False
p1, p2, p3, p4, a, b = map(int, input().split())
arr = [p1, p2, p3, p4]
count = 0
for i in range(a, b+1):
if function(i, arr):
count += 1
print(count)
``` | output | 1 | 47,535 | 22 | 95,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9 | instruction | 0 | 47,536 | 22 | 95,072 |
Tags: implementation, number theory
Correct Solution:
```
from itertools import permutations
def compute(p1,p2,p3,p4,x):
return (((x % p1)%p2)%p3)%p4
def solve(p1,p2,p3,p4,a,b):
per = list(permutations([p1,p2,p3,p4]))
count = 0
for i in range(a,b+1):
temp = 0
for j in per:
if i == compute(*j,i):
temp += 1
if temp == 7 :
count += 1
break
return count
def main():
# n = int(input())
a = list(map(int, input().split(" ")))
# b = list(map(int, input().split(" ")))
print(solve(*a))
main()
``` | output | 1 | 47,536 | 22 | 95,073 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9 | instruction | 0 | 47,537 | 22 | 95,074 |
Tags: implementation, number theory
Correct Solution:
```
# cook your dish here
from sys import stdin,stdout
from collections import Counter
from itertools import permutations
I=lambda: map(int,stdin.readline().split())
I1=lambda: stdin.readline()
p1,p2,p3,p4,a,b=I()
l=list(permutations([p1,p2,p3,p4]))
ans=0
for i in range(a,b+1):
c=0
for j in range(24):
a,x=list(l[j]),i
for y in a:
x=x%y
if x==i: c+=1
if c>=7: ans+=1
print(ans)
``` | output | 1 | 47,537 | 22 | 95,075 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9 | instruction | 0 | 47,538 | 22 | 95,076 |
Tags: implementation, number theory
Correct Solution:
```
a,b,c,d,x,y=map(int,input().split())
if x>=min(a,b,c,d):
print(0)
else:
print(min(min(a,b,c,d),y+1)-x)
``` | output | 1 | 47,538 | 22 | 95,077 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9 | instruction | 0 | 47,539 | 22 | 95,078 |
Tags: implementation, number theory
Correct Solution:
```
import math
p1,p2,p3,p4,a,b=map(int,input().split())
c=0
for i in range(a,b+1):
if (((i%p1)%p2)%p3)%p4==i:
c+=1
print (c)
``` | output | 1 | 47,539 | 22 | 95,079 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9 | instruction | 0 | 47,540 | 22 | 95,080 |
Tags: implementation, number theory
Correct Solution:
```
def readln(): return tuple(map(int, input().split()))
inp = readln()
mod = min(inp[:4])
print(len([_ for _ in range(inp[4], inp[5] + 1) if _ < mod]))
``` | output | 1 | 47,540 | 22 | 95,081 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9 | instruction | 0 | 47,541 | 22 | 95,082 |
Tags: implementation, number theory
Correct Solution:
```
l=list(map(int,input().split()))
x=min(l[:4])
if x>l[5]:
print(l[5]-l[4]+1)
elif x>=l[4]:
print(x-l[4])
else:
print("0")
``` | output | 1 | 47,541 | 22 | 95,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9
Submitted Solution:
```
mylist=list(map(int,input().split()))
a=0
b=min(mylist[:4:]) - mylist[4]
if(b>=0):
a=min(b,mylist[5]-mylist[4]+1)
print(a)
``` | instruction | 0 | 47,542 | 22 | 95,084 |
Yes | output | 1 | 47,542 | 22 | 95,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9
Submitted Solution:
```
*l, a, b = map(int, input().split())
print(max(0, min(b + 1, *l) - a))
``` | instruction | 0 | 47,543 | 22 | 95,086 |
Yes | output | 1 | 47,543 | 22 | 95,087 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9
Submitted Solution:
```
p,q,r,s,a,b = input().split()
p,q,r,s,a,b = [int(p),int(q),int(r),int(s),int(a),int(b)]
i = a
answer = 0
for j in range(a,b+1):
if (j<p and j<q and j<r and j<s):
answer = answer+1
else:
j=j+1
print(answer)
``` | instruction | 0 | 47,544 | 22 | 95,088 |
Yes | output | 1 | 47,544 | 22 | 95,089 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9
Submitted Solution:
```
def f(x):
return (((x%p1)%p2)%p3)%p4
p1,p2,p3,p4,a,b = map(int,input().split())
ans = 0
for i in range(a,b+1):
if f(i) == i: ans += 1
print(ans)
``` | instruction | 0 | 47,545 | 22 | 95,090 |
Yes | output | 1 | 47,545 | 22 | 95,091 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9
Submitted Solution:
```
a,b,c,d,e,f = map(int,input().split())
g = min([a,b,c,d])
print(max(0,min(g,f)-e))
``` | instruction | 0 | 47,546 | 22 | 95,092 |
No | output | 1 | 47,546 | 22 | 95,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9
Submitted Solution:
```
import sys
I=sys.stdin.readline
p1,p2,p3,p4,a,b=map(int,I().split())
x=min(p1,p2,p3,p4)
if x>=a:
print(0)
else:
if x>b:
print(b-a+1)
else:
print(x-a)
``` | instruction | 0 | 47,547 | 22 | 95,094 |
No | output | 1 | 47,547 | 22 | 95,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9
Submitted Solution:
```
p1,p2,p3,p4,a,b=[int(x) for x in input().split()]
k=min(p1,p2,p3,p4,b)
t=min(p1,p2,p3,p4)
if(a==b and t>a):
print(1)
elif(k>a and b!=k):
print(k-a+1)
elif(k>a and b==k):
print(k-a)
else:
print(0)
``` | instruction | 0 | 47,548 | 22 | 95,096 |
No | output | 1 | 47,548 | 22 | 95,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Petya was given this problem for homework:
You are given function <image> (here <image> represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.
It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.
Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers <image> with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number x if there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.
Input
First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).
It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.
Output
Output the number of integers in the given range that have the given property.
Examples
Input
2 7 1 8 2 8
Output
0
Input
20 30 40 50 0 100
Output
20
Input
31 41 59 26 17 43
Output
9
Submitted Solution:
```
s = list(map(int, input().split()))
lis = [s[0]]
for i in range(1,4):
if s[i]<s[i-1]:lis.append(s[i])
cnt = 0
for i in range(s[4], min(lis)):
m = i
for j in lis:
m = m%j
if m == i: cnt += 1
print(cnt)
``` | instruction | 0 | 47,549 | 22 | 95,098 |
No | output | 1 | 47,549 | 22 | 95,099 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little penguin Polo adores integer segments, that is, pairs of integers [l; r] (l ≤ r).
He has a set that consists of n integer segments: [l1; r1], [l2; r2], ..., [ln; rn]. We know that no two segments of this set intersect. In one move Polo can either widen any segment of the set 1 unit to the left or 1 unit to the right, that is transform [l; r] to either segment [l - 1; r], or to segment [l; r + 1].
The value of a set of segments that consists of n segments [l1; r1], [l2; r2], ..., [ln; rn] is the number of integers x, such that there is integer j, for which the following inequality holds, lj ≤ x ≤ rj.
Find the minimum number of moves needed to make the value of the set of Polo's segments divisible by k.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 105). Each of the following n lines contain a segment as a pair of integers li and ri ( - 105 ≤ li ≤ ri ≤ 105), separated by a space.
It is guaranteed that no two segments intersect. In other words, for any two integers i, j (1 ≤ i < j ≤ n) the following inequality holds, min(ri, rj) < max(li, lj).
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
2 3
1 2
3 4
Output
2
Input
3 7
1 2
3 3
4 7
Output
0 | instruction | 0 | 48,226 | 22 | 96,452 |
Tags: brute force, implementation
Correct Solution:
```
n, k= map(int,input().split())
r = 0
for i in range(n):
num = list(map(int,input().split()))
r += abs(num[0]-num[1])+1
if r % k == 0:
print(0)
else:
print((-r) % k)
``` | output | 1 | 48,226 | 22 | 96,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little penguin Polo adores integer segments, that is, pairs of integers [l; r] (l ≤ r).
He has a set that consists of n integer segments: [l1; r1], [l2; r2], ..., [ln; rn]. We know that no two segments of this set intersect. In one move Polo can either widen any segment of the set 1 unit to the left or 1 unit to the right, that is transform [l; r] to either segment [l - 1; r], or to segment [l; r + 1].
The value of a set of segments that consists of n segments [l1; r1], [l2; r2], ..., [ln; rn] is the number of integers x, such that there is integer j, for which the following inequality holds, lj ≤ x ≤ rj.
Find the minimum number of moves needed to make the value of the set of Polo's segments divisible by k.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 105). Each of the following n lines contain a segment as a pair of integers li and ri ( - 105 ≤ li ≤ ri ≤ 105), separated by a space.
It is guaranteed that no two segments intersect. In other words, for any two integers i, j (1 ≤ i < j ≤ n) the following inequality holds, min(ri, rj) < max(li, lj).
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
2 3
1 2
3 4
Output
2
Input
3 7
1 2
3 3
4 7
Output
0 | instruction | 0 | 48,233 | 22 | 96,466 |
Tags: brute force, implementation
Correct Solution:
```
n,k = [int(i) for i in input().split()]
total = 0
for x in range(n):
l,r = [int(i) for i in input().split()]
total = total+(abs(l-r)+1)
count = 0
while total%k!=0:
total+=1
count+=1
print(count)
``` | output | 1 | 48,233 | 22 | 96,467 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Simon has an array a1, a2, ..., an, consisting of n positive integers. Today Simon asked you to find a pair of integers l, r (1 ≤ l ≤ r ≤ n), such that the following conditions hold:
1. there is integer j (l ≤ j ≤ r), such that all integers al, al + 1, ..., ar are divisible by aj;
2. value r - l takes the maximum value among all pairs for which condition 1 is true;
Help Simon, find the required pair of numbers (l, r). If there are multiple required pairs find all of them.
Input
The first line contains integer n (1 ≤ n ≤ 3·105).
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 106).
Output
Print two integers in the first line — the number of required pairs and the maximum value of r - l. On the following line print all l values from optimal pairs in increasing order.
Examples
Input
5
4 6 9 3 6
Output
1 3
2
Input
5
1 3 5 7 9
Output
1 4
1
Input
5
2 3 5 7 11
Output
5 0
1 2 3 4 5
Note
In the first sample the pair of numbers is right, as numbers 6, 9, 3 are divisible by 3.
In the second sample all numbers are divisible by number 1.
In the third sample all numbers are prime, so conditions 1 and 2 are true only for pairs of numbers (1, 1), (2, 2), (3, 3), (4, 4), (5, 5). | instruction | 0 | 48,258 | 22 | 96,516 |
Tags: binary search, brute force, data structures, math, two pointers
Correct Solution:
```
import math
#import math
#------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now----------------------------------------------------
class SegmentTree:
def __init__(self, data, default=0, func=lambda a,b:math.gcd(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=9999999999999999999999999999999, func=lambda a,b:min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
n=int(input())
l=list(map(int,input().split()))
s2=SegmentTree(l)
s1=SegmentTree1(l)
s=0
e=0
d=dict()
d1=dict()
ed=[]+l[::-1]
s3=SegmentTree(ed)
s4=SegmentTree1(ed)
d2=dict()
while(s<n):
if s in d1:
d1[s]=e-s-1
else:
d1.update({s:e-s-1})
if e==n:
e-=1
s+=1
if s2.query(s,e)==s1.query(s,e):
e+=1
else:
s+=1
if s>e:
e+=1
if e>n-1:
e=n
if s in d1:
d1[s]=e-s-1
else:
d1.update({s:e-s-1})
e=0
s=0
while(s<n):
if s in d2:
d2[n-1-s]=e-s-1
else:
d2.update({n-1-s:e-s-1})
if e==n:
e-=1
s+=1
if s3.query(s,e)==s4.query(s,e):
e+=1
else:
s+=1
if s>e:
e+=1
if e>n-1:
e=n
if s in d2:
d2[n-1-s]=e-s-1
else:
d2.update({n-1-s:e-s-1})
ans=0
#print(d1,d2)
for j in d1:
if j in d2:
if 0<=j+d1[j]<n and 0<=j-d2[j]<n and 0<=j<n and s2.query(j,j+d1[j])==s2.query(j-d2[j],j):
d1[j]+=d2[j]
ans=max(ans,d1[j])
for j in d2:
ans=max(ans,d2[j])
s=0
e=s+ans
w=[]
while(e<n):
if s2.query(s,e)==s1.query(s,e):
w.append(s+1)
s+=1
e+=1
print(len(w),ans)
print(*w,sep=' ')
``` | output | 1 | 48,258 | 22 | 96,517 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Simon has an array a1, a2, ..., an, consisting of n positive integers. Today Simon asked you to find a pair of integers l, r (1 ≤ l ≤ r ≤ n), such that the following conditions hold:
1. there is integer j (l ≤ j ≤ r), such that all integers al, al + 1, ..., ar are divisible by aj;
2. value r - l takes the maximum value among all pairs for which condition 1 is true;
Help Simon, find the required pair of numbers (l, r). If there are multiple required pairs find all of them.
Input
The first line contains integer n (1 ≤ n ≤ 3·105).
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 106).
Output
Print two integers in the first line — the number of required pairs and the maximum value of r - l. On the following line print all l values from optimal pairs in increasing order.
Examples
Input
5
4 6 9 3 6
Output
1 3
2
Input
5
1 3 5 7 9
Output
1 4
1
Input
5
2 3 5 7 11
Output
5 0
1 2 3 4 5
Note
In the first sample the pair of numbers is right, as numbers 6, 9, 3 are divisible by 3.
In the second sample all numbers are divisible by number 1.
In the third sample all numbers are prime, so conditions 1 and 2 are true only for pairs of numbers (1, 1), (2, 2), (3, 3), (4, 4), (5, 5). | instruction | 0 | 48,259 | 22 | 96,518 |
Tags: binary search, brute force, data structures, math, two pointers
Correct Solution:
```
import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a,b:math.gcd(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=9999999999999999999999999999999, func=lambda a,b:min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
n=int(input())
l=list(map(int,input().split()))
s2=SegmentTree(l)
s1=SegmentTree1(l)
s=0
e=0
d=dict()
d1=dict()
ed=[]+l[::-1]
s3=SegmentTree(ed)
s4=SegmentTree1(ed)
d2=dict()
while(s<n):
if s in d1:
d1[s]=e-s-1
else:
d1.update({s:e-s-1})
if e==n:
e-=1
s+=1
if s2.query(s,e)==s1.query(s,e):
e+=1
else:
s+=1
if s>e:
e+=1
if e>n-1:
e=n
if s in d1:
d1[s]=e-s-1
else:
d1.update({s:e-s-1})
e=0
s=0
while(s<n):
if s in d2:
d2[n-1-s]=e-s-1
else:
d2.update({n-1-s:e-s-1})
if e==n:
e-=1
s+=1
if s3.query(s,e)==s4.query(s,e):
e+=1
else:
s+=1
if s>e:
e+=1
if e>n-1:
e=n
if s in d2:
d2[n-1-s]=e-s-1
else:
d2.update({n-1-s:e-s-1})
ans=0
#print(d1,d2)
for j in d1:
if j in d2:
if 0<=j+d1[j]<n and 0<=j-d2[j]<n and 0<=j<n and s2.query(j,j+d1[j])==s2.query(j-d2[j],j):
d1[j]+=d2[j]
ans=max(ans,d1[j])
for j in d2:
ans=max(ans,d2[j])
s=0
e=s+ans
w=[]
while(e<n):
if s2.query(s,e)==s1.query(s,e):
w.append(s+1)
s+=1
e+=1
print(len(w),ans)
print(*w,sep=' ')
``` | output | 1 | 48,259 | 22 | 96,519 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Simon has an array a1, a2, ..., an, consisting of n positive integers. Today Simon asked you to find a pair of integers l, r (1 ≤ l ≤ r ≤ n), such that the following conditions hold:
1. there is integer j (l ≤ j ≤ r), such that all integers al, al + 1, ..., ar are divisible by aj;
2. value r - l takes the maximum value among all pairs for which condition 1 is true;
Help Simon, find the required pair of numbers (l, r). If there are multiple required pairs find all of them.
Input
The first line contains integer n (1 ≤ n ≤ 3·105).
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 106).
Output
Print two integers in the first line — the number of required pairs and the maximum value of r - l. On the following line print all l values from optimal pairs in increasing order.
Examples
Input
5
4 6 9 3 6
Output
1 3
2
Input
5
1 3 5 7 9
Output
1 4
1
Input
5
2 3 5 7 11
Output
5 0
1 2 3 4 5
Note
In the first sample the pair of numbers is right, as numbers 6, 9, 3 are divisible by 3.
In the second sample all numbers are divisible by number 1.
In the third sample all numbers are prime, so conditions 1 and 2 are true only for pairs of numbers (1, 1), (2, 2), (3, 3), (4, 4), (5, 5). | instruction | 0 | 48,260 | 22 | 96,520 |
Tags: binary search, brute force, data structures, math, two pointers
Correct Solution:
```
import math
# import collections
# from itertools import permutations
# from itertools import combinations
# import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
'''def is_prime(n):
j=2
while j*j<=n:
if n%j==0:
return 0
j+=1
return 1'''
'''def gcd(x, y):
while(y):
x, y = y, x % y
return x'''
'''fact=[]
def factors(n) :
i = 1
while i <= math.sqrt(n):
if (n % i == 0) :
if (n / i == i) :
fact.append(i)
else :
fact.append(i)
fact.append(n//i)
i = i + 1'''
def prob():
n = int(input())
n+=1
# s=input()
l=[1]
l+=[int(x) for x in input().split()]
l+=[1]
# a,b = list(map(int , input().split()))
p = [True]*n
s=0
q = [i for i in range(1,n)]
for i in range(1,n):
if p[i]:
a=i
b=i
d=l[i]
if d==1:
s,q = n-2 , [1]
break
while l[a-1] % d == 0:
a-=1
while l[b+1] % d == 0:
b+=1
p[b]=False
d = b-a
if d>s:
s,q = d,[a]
elif d==s and s!=0:
q.append(a)
print(len(q) , s)
print(*q)
t=1
# t=int(input())
for _ in range(0,t):
prob()
``` | output | 1 | 48,260 | 22 | 96,521 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Simon has an array a1, a2, ..., an, consisting of n positive integers. Today Simon asked you to find a pair of integers l, r (1 ≤ l ≤ r ≤ n), such that the following conditions hold:
1. there is integer j (l ≤ j ≤ r), such that all integers al, al + 1, ..., ar are divisible by aj;
2. value r - l takes the maximum value among all pairs for which condition 1 is true;
Help Simon, find the required pair of numbers (l, r). If there are multiple required pairs find all of them.
Input
The first line contains integer n (1 ≤ n ≤ 3·105).
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 106).
Output
Print two integers in the first line — the number of required pairs and the maximum value of r - l. On the following line print all l values from optimal pairs in increasing order.
Examples
Input
5
4 6 9 3 6
Output
1 3
2
Input
5
1 3 5 7 9
Output
1 4
1
Input
5
2 3 5 7 11
Output
5 0
1 2 3 4 5
Note
In the first sample the pair of numbers is right, as numbers 6, 9, 3 are divisible by 3.
In the second sample all numbers are divisible by number 1.
In the third sample all numbers are prime, so conditions 1 and 2 are true only for pairs of numbers (1, 1), (2, 2), (3, 3), (4, 4), (5, 5). | instruction | 0 | 48,261 | 22 | 96,522 |
Tags: binary search, brute force, data structures, math, two pointers
Correct Solution:
```
n = int(input()) + 1
t = [1] + list(map(int, input().split())) + [1]
p = [True] * n
s, q = 0, list(range(1, n))
for i in range(1, n):
if p[i]:
a = b = i
d = t[i]
if d == 1:
s, q = n - 2, [1]
break
while t[a - 1] % d == 0: a -= 1
while t[b + 1] % d == 0:
b += 1
p[b] = False
d = b - a
if d > s: s, q = d, [a]
elif d == s != 0: q.append(a)
print(len(q), s)
print(' '.join(map(str, q)))
``` | output | 1 | 48,261 | 22 | 96,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Simon has an array a1, a2, ..., an, consisting of n positive integers. Today Simon asked you to find a pair of integers l, r (1 ≤ l ≤ r ≤ n), such that the following conditions hold:
1. there is integer j (l ≤ j ≤ r), such that all integers al, al + 1, ..., ar are divisible by aj;
2. value r - l takes the maximum value among all pairs for which condition 1 is true;
Help Simon, find the required pair of numbers (l, r). If there are multiple required pairs find all of them.
Input
The first line contains integer n (1 ≤ n ≤ 3·105).
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 106).
Output
Print two integers in the first line — the number of required pairs and the maximum value of r - l. On the following line print all l values from optimal pairs in increasing order.
Examples
Input
5
4 6 9 3 6
Output
1 3
2
Input
5
1 3 5 7 9
Output
1 4
1
Input
5
2 3 5 7 11
Output
5 0
1 2 3 4 5
Note
In the first sample the pair of numbers is right, as numbers 6, 9, 3 are divisible by 3.
In the second sample all numbers are divisible by number 1.
In the third sample all numbers are prime, so conditions 1 and 2 are true only for pairs of numbers (1, 1), (2, 2), (3, 3), (4, 4), (5, 5).
Submitted Solution:
```
import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a,b:math.gcd(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=9999999999999999999999999999999, func=lambda a,b:min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
n=int(input())
l=list(map(int,input().split()))
s2=SegmentTree(l)
s1=SegmentTree1(l)
s=0
e=0
d=dict()
d1=dict()
ed=[]+l[::-1]
s3=SegmentTree(ed)
s4=SegmentTree1(ed)
d2=dict()
while(e<n):
if s in d1:
d1[s]=e-s-1
else:
d1.update({s:e-s-1})
if s2.query(s,e)==s1.query(s,e):
e+=1
else:
s+=1
if s>e:
e+=1
if s in d1:
d1[s]=e-s-1
else:
d1.update({s:e-s-1})
e=0
s=0
while(e<n):
if s in d2:
d2[n-1-s]=e-s-1
else:
d2.update({n-1-s:e-s-1})
if s3.query(s,e)==s4.query(s,e):
e+=1
else:
s+=1
if s>e:
e+=1
if s in d2:
d2[n-1-s]=e-s-1
else:
d2.update({n-1-s:e-s-1})
ans=0
for j in d1:
if j in d2:
d1[j]+=d2[j]
ans=max(ans,d1[j])
for j in d2:
ans=max(ans,d2[j])
s=0
e=s+ans
w=[]
while(e<n):
if s2.query(s,e)==s1.query(s,e):
w.append(s+1)
s+=1
e+=1
if n==100:
print(l[-1])
print(len(w),ans)
print(*w,sep=' ')
``` | instruction | 0 | 48,262 | 22 | 96,524 |
No | output | 1 | 48,262 | 22 | 96,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Simon has an array a1, a2, ..., an, consisting of n positive integers. Today Simon asked you to find a pair of integers l, r (1 ≤ l ≤ r ≤ n), such that the following conditions hold:
1. there is integer j (l ≤ j ≤ r), such that all integers al, al + 1, ..., ar are divisible by aj;
2. value r - l takes the maximum value among all pairs for which condition 1 is true;
Help Simon, find the required pair of numbers (l, r). If there are multiple required pairs find all of them.
Input
The first line contains integer n (1 ≤ n ≤ 3·105).
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 106).
Output
Print two integers in the first line — the number of required pairs and the maximum value of r - l. On the following line print all l values from optimal pairs in increasing order.
Examples
Input
5
4 6 9 3 6
Output
1 3
2
Input
5
1 3 5 7 9
Output
1 4
1
Input
5
2 3 5 7 11
Output
5 0
1 2 3 4 5
Note
In the first sample the pair of numbers is right, as numbers 6, 9, 3 are divisible by 3.
In the second sample all numbers are divisible by number 1.
In the third sample all numbers are prime, so conditions 1 and 2 are true only for pairs of numbers (1, 1), (2, 2), (3, 3), (4, 4), (5, 5).
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
left = [0] * n
right = [0] * n
max_length = - 1
left_idx = []
hasOne = False
for i in range(n):
val = a[i]
if val == 1:
hasOne = True
break
l = i - 1
while l >= 0 and a[l] % val == 0:
l -= 1
l += 1
r= i + 1
while r < n and a[r] % val == 0:
r += 1
r -= 1
length = r - l
if length > max_length:
left_idx.clear()
max_length = length
left_idx.append(l+1)
elif length == max_length:
left_idx.append(l + 1)
if hasOne:
print(str(1) +" " + str(n-1))
print(1)
else:
print(str(len(left_idx)) + " " + str(max_length) )
re = ''
for i in left_idx:
re += str(i) + " "
print(re)
``` | instruction | 0 | 48,263 | 22 | 96,526 |
No | output | 1 | 48,263 | 22 | 96,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Simon has an array a1, a2, ..., an, consisting of n positive integers. Today Simon asked you to find a pair of integers l, r (1 ≤ l ≤ r ≤ n), such that the following conditions hold:
1. there is integer j (l ≤ j ≤ r), such that all integers al, al + 1, ..., ar are divisible by aj;
2. value r - l takes the maximum value among all pairs for which condition 1 is true;
Help Simon, find the required pair of numbers (l, r). If there are multiple required pairs find all of them.
Input
The first line contains integer n (1 ≤ n ≤ 3·105).
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 106).
Output
Print two integers in the first line — the number of required pairs and the maximum value of r - l. On the following line print all l values from optimal pairs in increasing order.
Examples
Input
5
4 6 9 3 6
Output
1 3
2
Input
5
1 3 5 7 9
Output
1 4
1
Input
5
2 3 5 7 11
Output
5 0
1 2 3 4 5
Note
In the first sample the pair of numbers is right, as numbers 6, 9, 3 are divisible by 3.
In the second sample all numbers are divisible by number 1.
In the third sample all numbers are prime, so conditions 1 and 2 are true only for pairs of numbers (1, 1), (2, 2), (3, 3), (4, 4), (5, 5).
Submitted Solution:
```
#!/usr/bin/python3
def readln(): return tuple(map(int, input().split()))
n, = readln()
a = readln()
cnt1 = [0] * n
j = 0
for i in range(n):
if a[i] % a[j]:
j += 1
while j < i:
cnt1[j] = cnt1[j - 1] - 1
j += 1
cnt1[j] += 1
cnt2 = [0] * n
j = n - 1
for i in range(n - 1, -1, -1):
if a[i] % a[j]:
j -= 1
while j > i:
cnt2[j] = cnt2[j + 1] - 1
j -= 1
cnt2[j] += 1
cnt = [u + v - 1 for u, v in zip(cnt1, cnt2)]
m = max(cnt)
ans = set()
for i in range(n):
if cnt[i] == m:
ans.add(i - cnt2[i] + 2)
print(len(ans), m - 1)
print(*sorted(ans))
``` | instruction | 0 | 48,264 | 22 | 96,528 |
No | output | 1 | 48,264 | 22 | 96,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Simon has an array a1, a2, ..., an, consisting of n positive integers. Today Simon asked you to find a pair of integers l, r (1 ≤ l ≤ r ≤ n), such that the following conditions hold:
1. there is integer j (l ≤ j ≤ r), such that all integers al, al + 1, ..., ar are divisible by aj;
2. value r - l takes the maximum value among all pairs for which condition 1 is true;
Help Simon, find the required pair of numbers (l, r). If there are multiple required pairs find all of them.
Input
The first line contains integer n (1 ≤ n ≤ 3·105).
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 106).
Output
Print two integers in the first line — the number of required pairs and the maximum value of r - l. On the following line print all l values from optimal pairs in increasing order.
Examples
Input
5
4 6 9 3 6
Output
1 3
2
Input
5
1 3 5 7 9
Output
1 4
1
Input
5
2 3 5 7 11
Output
5 0
1 2 3 4 5
Note
In the first sample the pair of numbers is right, as numbers 6, 9, 3 are divisible by 3.
In the second sample all numbers are divisible by number 1.
In the third sample all numbers are prime, so conditions 1 and 2 are true only for pairs of numbers (1, 1), (2, 2), (3, 3), (4, 4), (5, 5).
Submitted Solution:
```
import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a,b:math.gcd(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=9999999999999999999999999999999, func=lambda a,b:min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
n=int(input())
l=list(map(int,input().split()))
s2=SegmentTree(l)
s1=SegmentTree1(l)
s=0
e=0
d=dict()
d1=dict()
ed=[]+l[::-1]
s3=SegmentTree(ed)
s4=SegmentTree1(ed)
d2=dict()
while(s<n):
if s in d1:
d1[s]=e-s-1
else:
d1.update({s:e-s-1})
if e==n:
e-=1
s+=1
if s2.query(s,e)==s1.query(s,e):
e+=1
else:
s+=1
if s>e:
e+=1
if e>n-1:
e=n
if s in d1:
d1[s]=e-s-1
else:
d1.update({s:e-s-1})
e=0
s=0
while(s<n):
if s in d2:
d2[n-1-s]=e-s-1
else:
d2.update({n-1-s:e-s-1})
if e==n:
e-=1
s+=1
if s3.query(s,e)==s4.query(s,e):
e+=1
else:
s+=1
if s>e:
e+=1
if e>n-1:
e=n
if s in d2:
d2[n-1-s]=e-s-1
else:
d2.update({n-1-s:e-s-1})
ans=0
for j in d1:
if j in d2:
d1[j]+=d2[j]
ans=max(ans,d1[j])
for j in d2:
ans=max(ans,d2[j])
s=0
e=s+ans
w=[]
while(e<n):
if s2.query(s,e)==s1.query(s,e):
w.append(s+1)
s+=1
e+=1
if n==100:
print(l[-1])
print(len(w),ans)
print(*w,sep=' ')
``` | instruction | 0 | 48,265 | 22 | 96,530 |
No | output | 1 | 48,265 | 22 | 96,531 |
Provide tags and a correct Python 2 solution for this coding contest problem.
You have an array of positive integers a[1], a[2], ..., a[n] and a set of bad prime numbers b1, b2, ..., bm. The prime numbers that do not occur in the set b are considered good. The beauty of array a is the sum <image>, where function f(s) is determined as follows:
* f(1) = 0;
* Let's assume that p is the minimum prime divisor of s. If p is a good prime, then <image>, otherwise <image>.
You are allowed to perform an arbitrary (probably zero) number of operations to improve array a. The operation of improvement is the following sequence of actions:
* Choose some number r (1 ≤ r ≤ n) and calculate the value g = GCD(a[1], a[2], ..., a[r]).
* Apply the assignments: <image>, <image>, ..., <image>.
What is the maximum beauty of the array you can get?
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 5000) showing how many numbers are in the array and how many bad prime numbers there are.
The second line contains n space-separated integers a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — array a. The third line contains m space-separated integers b1, b2, ..., bm (2 ≤ b1 < b2 < ... < bm ≤ 109) — the set of bad prime numbers.
Output
Print a single integer — the answer to the problem.
Examples
Input
5 2
4 20 34 10 10
2 5
Output
-2
Input
4 5
2 4 8 16
3 5 7 11 17
Output
10
Note
Note that the answer to the problem can be negative.
The GCD(x1, x2, ..., xk) is the maximum positive integer that divides each xi. | instruction | 0 | 48,282 | 22 | 96,564 |
Tags: dp, greedy, math, number theory
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
from fractions import gcd
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
n,m=li()
l=li()
d=Counter(li())
mp=Counter()
def f(x):
ret=0
i=2
tp=x
while i*i<=x:
if tp%i==0:
if d[i]:
while tp%i==0:
ret-=1
tp/=i
else:
while tp%i==0:
ret+=1
tp/=i
i+=1
if tp>1:
if d[tp]:
ret-=1
else:
ret+=1
return ret
ans=0
g=[0]*n
g1=0
for i in range(n):
g1=gcd(g1,l[i])
if not mp[l[i]]:
mp[l[i]]=f(l[i])
ans+=mp[l[i]]
g[i]=g1
p=1
for i in range(n-1,-1,-1):
g[i]/=p
if not mp[g[i]]:
mp[g[i]]=f(g[i])
temp=mp[g[i]]
if temp<0:
ans-=(i+1)*temp
p*=g[i]
pn(ans)
``` | output | 1 | 48,282 | 22 | 96,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1 | instruction | 0 | 48,303 | 22 | 96,606 |
Tags: brute force, data structures, math
Correct Solution:
```
from sys import stdin
import math
from collections import defaultdict
input = stdin.readline
# n, k = map(int, input().split(" "))
# l = list(map(int, input().split(" ")))
n = int(input())
l = list(map(int, input().split(" ")))
d = defaultdict(int)
g = defaultdict(int)
d[l[0]] = 1
g[l[0]] = 1
for i in range(1, n):
t = defaultdict(int)
t[l[i]] = 1
g[l[i]]+=1
for k, v in d.items():
gcd = math.gcd(k, l[i])
t[gcd] += v
g[gcd] += v
d = t
for i in range(int(input())):
a = g[int(input())]
print(a)
``` | output | 1 | 48,303 | 22 | 96,607 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1 | instruction | 0 | 48,304 | 22 | 96,608 |
Tags: brute force, data structures, math
Correct Solution:
```
import sys
from collections import defaultdict
from math import gcd, sqrt
MAX = pow(10, 5)
# stdin = open("testdata.txt", "r")
ip = sys.stdin
n = int(ip.readline())
a = list(map(int, ip.readline().split()))
gcd_count = defaultdict(int)
main_gcd = defaultdict(int)
main_gcd[a[0]] = 1
gcd_count[a[0]] = 1
for i in range(1, n):
ele = a[i]
temp_gcd = defaultdict(int)
temp_gcd[ele] = 1
gcd_count[ele] += 1
for k, v in main_gcd.items():
temp = gcd(ele, k)
temp_gcd[temp] += v
gcd_count[temp] += v
main_gcd = temp_gcd
q = int(ip.readline())
for _ in range(q):
k = int(ip.readline())
print(gcd_count[k])
``` | output | 1 | 48,304 | 22 | 96,609 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1 | instruction | 0 | 48,305 | 22 | 96,610 |
Tags: brute force, data structures, math
Correct Solution:
```
from sys import stdin
import math
from collections import defaultdict
input = stdin.readline
n = int(input())
arr = list(map(int, input().rstrip().split(" ")))
q = int(input())
d = defaultdict(lambda : 0)
current = defaultdict(lambda : 0)
# totalCount1 = 0
# count1 = 0
for i in range(n):
newCurrent = defaultdict(lambda : 0)
newCurrent[arr[i]] += 1
for key, value in current.items():
g = math.gcd(arr[i], key)
if g > 1:
newCurrent[g] += value
for key, value in newCurrent.items():
d[key] += value
current = newCurrent
totalCombi = (n * (n + 1)) // 2
#print(d)
d[1] = totalCombi - sum(d.values()) + arr.count(1)
#print(d)
for _ in range(q):
x = int(input())
print(d[x])
``` | output | 1 | 48,305 | 22 | 96,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1 | instruction | 0 | 48,306 | 22 | 96,612 |
Tags: brute force, data structures, math
Correct Solution:
```
from math import gcd
from collections import defaultdict
from sys import stdin, stdout
##This method is better cause for all the same results we only calculate once
def main():
GCD_count = defaultdict(int)
GCD_map = defaultdict(int)
arr_len = int(stdin.readline())
arr = [int(x) for x in stdin.readline().split()]
for start in range(arr_len):
temp = defaultdict(int)
GCD_count[arr[start]] += 1
temp[arr[start]] += 1
for gcd_now, occurence in GCD_map.items():
res = gcd(gcd_now, arr[start])
temp[res] += occurence
GCD_count[res] += occurence
GCD_map = temp
num_queries = int(stdin.readline())
for _ in range(num_queries):
print(GCD_count[int(stdin.readline())])
main()
``` | output | 1 | 48,306 | 22 | 96,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1 | instruction | 0 | 48,307 | 22 | 96,614 |
Tags: brute force, data structures, math
Correct Solution:
```
import os, sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
mod = 998244353
from math import ceil, gcd,log2
n=int(int(input()))
a=list(map(int,input().split()))
d=dict()
ck=[int(input()) for _ in range(int(input()))]
curr=dict()
for i in a:
curr1=dict()
for j in curr:
t=gcd(i,j)
d[t]=d.get(t,0)+curr[j]
curr1[t]=curr1.get(t,0)+curr[j]
curr1[i]=curr1.get(i,0)+1
d[i]=d.get(i,0)+1
curr=curr1
for i in ck:
if i in d:
print(d[i])
else:
print(0)
``` | output | 1 | 48,307 | 22 | 96,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1 | instruction | 0 | 48,308 | 22 | 96,616 |
Tags: brute force, data structures, math
Correct Solution:
```
from sys import stdin
import math
from collections import defaultdict
input = stdin.readline
n = int(input())
arr = list(map(int, input().rstrip().split(" ")))
q = int(input())
d = defaultdict(lambda : 0)
current = defaultdict(lambda : 0)
for i in range(n):
newCurrent = defaultdict(lambda : 0)
newCurrent[arr[i]] += 1
for key, value in current.items():
g = math.gcd(arr[i], key)
newCurrent[g] += value
for key, value in newCurrent.items():
d[key] += value
current = newCurrent
for _ in range(q):
x = int(input())
print(d[x])
``` | output | 1 | 48,308 | 22 | 96,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1 | instruction | 0 | 48,309 | 22 | 96,618 |
Tags: brute force, data structures, math
Correct Solution:
```
from os import path;import sys,time
mod = int(1e9 + 7)
from math import ceil, floor,gcd,log,log2 ,factorial,sqrt
from collections import defaultdict ,Counter , OrderedDict , deque;from itertools import combinations,permutations
# from string import ascii_lowercase ,ascii_uppercase
from bisect import *;from functools import reduce;from operator import mul;maxx = float('inf')
I = lambda :int(sys.stdin.buffer.readline())
lint = lambda :[int(x) for x in sys.stdin.buffer.readline().split()]
S = lambda: sys.stdin.readline().strip('\n')
grid = lambda r :[lint() for i in range(r)]
localsys = 0
start_time = time.time()
#left shift --- num*(2**k) --(k - shift)
nCr = lambda n, r: reduce(mul, range(n - r + 1, n + 1), 1) // factorial(r)
def ceill(n,x):
return (n+x -1 )//x
T =0
def solve():
n =I()
a = lint()
ans = defaultdict(int) ; d = defaultdict(int)
for i in a:
t = defaultdict(int )
t[i] =1
ans[i]+=1
for j in d:
gc = gcd(i, j)
ans[gc]+=d[j]
t[gc]+=d[j]
d = t
for _ in range(I()):
print(ans[I()])
def run():
if (path.exists('input.txt')):
sys.stdin=open('input.txt','r')
sys.stdout=open('output.txt','w')
run()
T = I() if T else 1
for _ in range(T):
solve()
if localsys:
print("\n\nTime Elased :",time.time() - start_time,"seconds")
``` | output | 1 | 48,309 | 22 | 96,619 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1 | instruction | 0 | 48,310 | 22 | 96,620 |
Tags: brute force, data structures, math
Correct Solution:
```
from bisect import insort,bisect_right,bisect_left
from sys import stdout, stdin, setrecursionlimit
from math import sqrt,ceil,floor,factorial,gcd
from io import BytesIO, IOBase
from collections import *
from itertools import *
from random import *
from string import *
from queue import *
from heapq import *
from re import *
from os import *
####################################---fast-input-output----#########################################
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = read(self._fd, max(fstat(self._fd).st_size, 8192))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = read(self._fd, max(fstat(self._fd).st_size, 8192))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
stdin, stdout = IOWrapper(stdin), IOWrapper(stdout)
graph, mod, szzz = {}, 10**9 + 7, lambda: sorted(zzz())
def getStr(): return input()
def getInt(): return int(input())
def listStr(): return list(input())
def getStrs(): return input().split()
def isInt(s): return '0' <= s[0] <= '9'
def input(): return stdin.readline().strip()
def zzz(): return [int(i) for i in input().split()]
def output(answer, end='\n'): stdout.write(str(answer) + end)
def lcd(xnum1, xnum2): return (xnum1 * xnum2 // gcd(xnum1, xnum2))
def getPrimes(N = 10**5):
SN = int(sqrt(N))
sieve = [i for i in range(N+1)]
sieve[1] = 0
for i in sieve:
if i > SN:
break
if i == 0:
continue
for j in range(2*i, N+1, i):
sieve[j] = 0
prime = [i for i in range(N+1) if sieve[i] != 0]
return prime
def primeFactor(n,prime=getPrimes()):
lst = []
mx=int(sqrt(n))+1
for i in prime:
if i>mx:break
while n%i==0:
lst.append(i)
n//=i
if n>1:
lst.append(n)
return lst
dx = [-1, 1, 0, 0, 1, -1, 1, -1]
dy = [0, 0, 1, -1, 1, -1, -1, 1]
daysInMounth = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
#################################################---Some Rule For Me To Follow---#################################
"""
--instants of Reading problem continuously try to understand them.
--If you Know some-one , Then you probably don't know him !
--Try & again try, maybe you're just one statement away!
"""
##################################################---START-CODING---###############################################
n = getInt()
l = zzz()
d = defaultdict(int)
g = defaultdict(int)
d[l[0]] = 1
g[l[0]] = 1
for i in range(1, n):
t = defaultdict(int)
t[l[i]] = 1
g[l[i]]+=1
for k, v in d.items():
gg = gcd(k, l[i])
t[gg] += v
g[gg] += v
d = t
for i in range(int(input())):
a = g[getInt()]
print(a)
``` | output | 1 | 48,310 | 22 | 96,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1
Submitted Solution:
```
from sys import stdin
n = int(stdin.readline())
a = [int(x) for x in stdin.readline().split()]
q = int(stdin.readline())
def gcd(a,b):
while a != 0:
a,b = b%a, a
return b
totals = {}
new = {}
for x in a[::-1]:
old = new
new = {}
for y in old:
g = gcd(x,y)
if g in new:
new[g] += old[y]
else:
new[g] = old[y]
if x in new:
new[x] += 1
else:
new[x] = 1
for y in new:
if y in totals:
totals[y] += new[y]
else:
totals[y] = new[y]
for x in range(q):
c = int(stdin.readline())
if c in totals:
print(totals[c])
else:
print(0)
``` | instruction | 0 | 48,311 | 22 | 96,622 |
Yes | output | 1 | 48,311 | 22 | 96,623 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1
Submitted Solution:
```
from sys import stdin
import math
from collections import defaultdict
input = stdin.readline
# n, k = map(int, input().split(" "))
# l = list(map(int, input().split(" ")))
n = int(input())
l = list(map(int, input().split(" ")))
k = math.ceil(math.log2(n))
dp = [[0] * k for i in range(2 ** k)]
# def make_sparse(l, n, k):
# """Making sparse table, replace max with needed function like[GCD, Min, max, sum]"""
# for i in range(n):
# dp[i][0] = l[i]
#
# for j in range(1, k + 1):
# i = 0
# while i + (1 << j) <= n:
# dp[i][j] = math.gcd(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1])
# i += 1
#
#
# def querys(l, r):
# j = int(math.log2(r - l + 1))
# return math.gcd(dp[l][j], dp[r - (1 << j) + 1][j])
# make_sparse(l, n, k)
d = defaultdict(int)
g = defaultdict(int)
d[l[0]] = 1
g[l[0]] = 1
for i in range(1, n):
t = defaultdict(int)
t[l[i]] = 1
g[l[i]]+=1
for k, v in d.items():
gcd = math.gcd(k, l[i])
t[gcd] += v
g[gcd] += v
d = t
for i in range(int(input())):
a = g[int(input())]
if not a:
print(0)
else:
print(a)
``` | instruction | 0 | 48,312 | 22 | 96,624 |
Yes | output | 1 | 48,312 | 22 | 96,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1
Submitted Solution:
```
from sys import stdin
import math
from collections import defaultdict
input = stdin.readline
# n, k = map(int, input().split(" "))
# l = list(map(int, input().split(" ")))
n = int(input())
l = list(map(int, input().split(" ")))
k = math.ceil(math.log2(n))
dp = [[0] * k for i in range(2 ** k)]
def make_sparse(l, n, k):
"""Making sparse table, replace max with needed function like[GCD, Min, max, sum]"""
for i in range(n):
dp[i][0] = l[i]
for j in range(1, k + 1):
i = 0
while i + (1 << j) <= n:
dp[i][j] = math.gcd(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1])
i += 1
def querys(l, r):
j = int(math.log2(r - l + 1))
return math.gcd(dp[l][j], dp[r - (1 << j) + 1][j])
make_sparse(l, n, k)
d = defaultdict(int)
g = defaultdict(int)
d[l[0]] = 1
g[l[0]] = 1
for i in range(1, n):
t = defaultdict(int)
t[l[i]] = 1
g[l[i]]+=1
for k, v in d.items():
gcd = math.gcd(k, l[i])
t[gcd] += v
g[gcd] += v
d = t
for i in range(int(input())):
a = g[int(input())]
if not a:
print(0)
else:
print(a)
``` | instruction | 0 | 48,313 | 22 | 96,626 |
Yes | output | 1 | 48,313 | 22 | 96,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1
Submitted Solution:
```
import math
def gcd(a, b):
c = a % b
while c:
a = b
b = c
c = a % b
return b
n = int(input())
a = list(map(int, input().split()))
k = math.floor(math.log(n, 2))
rangeGcd = [[0]*n for i in range(k+1)]
for i in range(n):
rangeGcd[0][i] = a[i]
for i in range(n):
for j in range(1, k+1):
step = 2**(j-1)
if i+step < n:
rangeGcd[j][i] = gcd(rangeGcd[j-1][i], rangeGcd[j-1][i+step])
else:
rangeGcd[j][i] = rangeGcd[j-1][i]
q = int(input())
for i in range(q):
count = 0
d = int(input())
for x in range(n):
for y in range(k+1):
if rangeGcd[y][x] < d:
break
while rangeGcd[y][x] == d:
count += 1
if y == 0:
break
if x+2**y >= n:
break
x += 2**y
y -= 1
print(count)
``` | instruction | 0 | 48,314 | 22 | 96,628 |
No | output | 1 | 48,314 | 22 | 96,629 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1
Submitted Solution:
```
from sys import stdin
import math
from collections import defaultdict
input = stdin.readline
n = int(input())
arr = list(map(int, input().rstrip().split(" ")))
q = int(input())
d = defaultdict(lambda : 0)
current = []
# totalCount1 = 0
# count1 = 0
for i in range(n):
newCurrent = []
newCurrent.append(arr[i])
toAdd = 0
for j in range(len(current)):
g = math.gcd(arr[i], current[j])
if g > 1:
newCurrent.append(g)
else:
#toAdd = len(current) - j
break
# count1 += toAdd
# totalCount1 += count1
#print(newCurrent)
for ele in newCurrent:
d[ele] += 1
current = newCurrent
totalCombi = (n * (n + 1)) // 2
if totalCombi - sum(d.values()) != 1196:
d[1] = totalCombi - sum(d.values())
else:
d[1] = 1197
#print(d)
for _ in range(q):
x = int(input())
print(d[x])
``` | instruction | 0 | 48,315 | 22 | 96,630 |
No | output | 1 | 48,315 | 22 | 96,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1
Submitted Solution:
```
from sys import stdin
import math
input = stdin.readline
# n, k = map(int, input().split(" "))
# l = list(map(int, input().split(" ")))
n = int(input())
l = list(map(int, input().split(" ")))
k = math.ceil(math.log2(n))
dp = [[0] * k for i in range(2 ** k)]
test = {}
def make_sparse(l, n, k):
"""Making sparse table, replace max with needed function like[GCD, Min, max, sum]"""
for i in range(n):
dp[i][0] = l[i]
for j in range(1, k + 1):
i = 0
while i + (1 << j) <= n:
dp[i][j] = math.gcd(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1])
i += 1
def querys(l, r):
j = int(math.log2(r - l + 1))
return math.gcd(dp[l][j], dp[r - (1 << j) + 1][j])
make_sparse(l, n, k)
d = {}
c = i = 0
while i < n:
j = i
while j < n and l[j] == l[i]:
j += 1
c += 1
d[l[i]] = d.get(l[i], 0) + (c * (c + 1)) // 2
for k in range(j, n):
a = str([l[i], l[k]])
g = test.get(a, 0)
if not g:
g = querys(i, k)
test[a] = g
if g == 1:
d[1] = d.get(1, 0) + c * (n - k)
break
else:
d[g] = d.get(g, 0) + c
i += c
c = 0
for i in range(int(input())):
a = d.get(int(input(), 0))
if not a:
print(0)
else:
print(a)
``` | instruction | 0 | 48,316 | 22 | 96,632 |
No | output | 1 | 48,316 | 22 | 96,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given a sequence of integers a1, ..., an and q queries x1, ..., xq on it. For each query xi you have to count the number of pairs (l, r) such that 1 ≤ l ≤ r ≤ n and gcd(al, al + 1, ..., ar) = xi.
<image> is a greatest common divisor of v1, v2, ..., vn, that is equal to a largest positive integer that divides all vi.
Input
The first line of the input contains integer n, (1 ≤ n ≤ 105), denoting the length of the sequence. The next line contains n space separated integers a1, ..., an, (1 ≤ ai ≤ 109).
The third line of the input contains integer q, (1 ≤ q ≤ 3 × 105), denoting the number of queries. Then follows q lines, each contain an integer xi, (1 ≤ xi ≤ 109).
Output
For each query print the result in a separate line.
Examples
Input
3
2 6 3
5
1
2
3
4
6
Output
1
2
2
0
1
Input
7
10 20 3 15 1000 60 16
10
1
2
3
4
5
6
10
20
60
1000
Output
14
0
2
2
2
0
2
2
1
1
Submitted Solution:
```
from sys import stdin
import math
input = stdin.readline
# n, k = map(int, input().split(" "))
# l = list(map(int, input().split(" ")))
n = int(input())
l = list(map(int, input().split(" ")))
k = math.ceil(math.log2(n))
dp = [[0] * k for i in range(2 ** k)]
test = {}
def make_sparse(l, n, k):
"""Making sparse table, replace max with needed function like[GCD, Min, max, sum]"""
for i in range(n):
dp[i][0] = l[i]
for j in range(1, k + 1):
i = 0
while i + (1 << j) <= n:
dp[i][j] = math.gcd(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1])
i += 1
def querys(l, r):
j = int(math.log2(r - l + 1))
return math.gcd(dp[l][j], dp[r - (1 << j) + 1][j])
make_sparse(l, n, k)
d = {}
c = i = 0
while i < n:
j = i
while j < n and l[j] == l[i]:
j += 1
c += 1
d[l[i]] = d.get(l[i], 0) + (c * (c + 1)) // 2
for k in range(j, n):
a = str([l[j-1], l[k]])
g = test.get(a, 0)
if not g:
g = querys(j-1, k)
test[a] = g
if g == 1:
d[1] = d.get(1, 0) + c * (n - k)
break
else:
d[g] = d.get(g, 0) + c
i += c
c = 0
for i in range(int(input())):
a = d.get(int(input(), 0))
if not a:
print(0)
else:
print(a)
``` | instruction | 0 | 48,317 | 22 | 96,634 |
No | output | 1 | 48,317 | 22 | 96,635 |
Provide a correct Python 3 solution for this coding contest problem.
problem
One day, Sosusa, who loves prime numbers, was playing with the pair $ (p, q) $, where $ p + q $ is a prime number. Suddenly, Sosusa wondered how many of these pairs were prime numbers with $ p $ and $ q $ both less than or equal to $ N $. Find the number on your behalf.
output
Output the number of pairs. Also, output a line break at the end.
Example
Input
3
Output
2 | instruction | 0 | 48,781 | 22 | 97,562 |
"Correct Solution:
```
def main():
n = int(input())
root = int(n ** (1 / 2)) + 1
isPrime = [True] * (n + 3)
for i in range(4, n + 3, 2):
isPrime[i] = False
ans = 0
prePrime = -1
for i in range(3, n + 3, 2):
if isPrime[i]:
if prePrime + 2 == i:
ans += 2
prePrime = i
if i > root:
continue
for j in range(i * i, n + 3, i):
isPrime[j] = False
print(ans)
main()
``` | output | 1 | 48,781 | 22 | 97,563 |
Provide a correct Python 3 solution for this coding contest problem.
problem
One day, Sosusa, who loves prime numbers, was playing with the pair $ (p, q) $, where $ p + q $ is a prime number. Suddenly, Sosusa wondered how many of these pairs were prime numbers with $ p $ and $ q $ both less than or equal to $ N $. Find the number on your behalf.
output
Output the number of pairs. Also, output a line break at the end.
Example
Input
3
Output
2 | instruction | 0 | 48,782 | 22 | 97,564 |
"Correct Solution:
```
import math
N = int(input())
P = [True for x in range(N + 3)]
P[0] = False
P[1] = False
for i in range(2, int(math.sqrt(N + 3)) + 1):
if P[i]:
for j in range(i * 2, N + 3, i):
P[j] = False
count = 0
for q in range(3, N + 1):
if P[q] and P[2 + q]:
count += 2
print(count)
``` | output | 1 | 48,782 | 22 | 97,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an infinite set generated as follows:
* 1 is in this set.
* If x is in this set, x ⋅ a and x+b both are in this set.
For example, when a=3 and b=6, the five smallest elements of the set are:
* 1,
* 3 (1 is in this set, so 1⋅ a=3 is in this set),
* 7 (1 is in this set, so 1+b=7 is in this set),
* 9 (3 is in this set, so 3⋅ a=9 is in this set),
* 13 (7 is in this set, so 7+b=13 is in this set).
Given positive integers a, b, n, determine if n is in this set.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.
Output
For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.
Example
Input
5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264
Output
Yes
No
Yes
No
Yes
Note
In the first test case, 24 is generated as follows:
* 1 is in this set, so 3 and 6 are in this set;
* 3 is in this set, so 9 and 8 are in this set;
* 8 is in this set, so 24 and 13 are in this set.
Thus we can see 24 is in this set.
The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them. | instruction | 0 | 49,090 | 22 | 98,180 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
from math import gcd
def solve(a, b, n):
cur = 1
if a==1 and b==1:
return "Yes"
if n%b==1:
return "Yes"
if a==1:
return "No"
while cur <= n:
rem = n - cur
if rem % b == 0:
return "Yes"
cur*=a
return "No"
for case in range(int(input())):
#n = int(input())
n, a, b = list(map(int, input().split()))
ans = solve(a, b, n)
print(ans)
``` | output | 1 | 49,090 | 22 | 98,181 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an infinite set generated as follows:
* 1 is in this set.
* If x is in this set, x ⋅ a and x+b both are in this set.
For example, when a=3 and b=6, the five smallest elements of the set are:
* 1,
* 3 (1 is in this set, so 1⋅ a=3 is in this set),
* 7 (1 is in this set, so 1+b=7 is in this set),
* 9 (3 is in this set, so 3⋅ a=9 is in this set),
* 13 (7 is in this set, so 7+b=13 is in this set).
Given positive integers a, b, n, determine if n is in this set.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.
Output
For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.
Example
Input
5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264
Output
Yes
No
Yes
No
Yes
Note
In the first test case, 24 is generated as follows:
* 1 is in this set, so 3 and 6 are in this set;
* 3 is in this set, so 9 and 8 are in this set;
* 8 is in this set, so 24 and 13 are in this set.
Thus we can see 24 is in this set.
The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them. | instruction | 0 | 49,095 | 22 | 98,190 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
from math import *
from collections import deque
from copy import deepcopy
import sys
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def multi(): return map(int, input().split())
def strmulti(): return map(str, inp().split())
def lis(): return list(map(int, inp().split()))
def lcm(a, b): return (a * b) // gcd(a, b)
def ncr(n, r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def stringlis(): return list(map(str, inp().split()))
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def printlist(a):
print(' '.join(str(a[i]) for i in range(len(a))))
def isPrime(n):
if (n <= 1): return False
if (n <= 3): return True
if (n % 2 == 0 or n % 3 == 0): return False
i = 5
while (i * i <= n):
if (n % i == 0 or n % (i + 2) == 0):
return False
i = i + 6
return True
def google():
t = int(inp())
for T in range(t):
print("Case #" + str(T + 1) + ": ", end="")
solve()
def normal(x):
if(x==0):
t = int(inp())
else:
t=1
for T in range(t):
solve()
# copied functions end
# start coding
def solve():
n,a,b=multi()
if(b==1):
print("YES")
return
if(a==1):
if(floor((n-1)/b)==ceil((n-1)/b)):
print("YES")
else:
print("NO")
return
i=0
while(a**i<=n):
if(n%b==a**i%b):
print("YES")
return
i+=1
print("NO")
return
normal(0)
``` | output | 1 | 49,095 | 22 | 98,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an infinite set generated as follows:
* 1 is in this set.
* If x is in this set, x ⋅ a and x+b both are in this set.
For example, when a=3 and b=6, the five smallest elements of the set are:
* 1,
* 3 (1 is in this set, so 1⋅ a=3 is in this set),
* 7 (1 is in this set, so 1+b=7 is in this set),
* 9 (3 is in this set, so 3⋅ a=9 is in this set),
* 13 (7 is in this set, so 7+b=13 is in this set).
Given positive integers a, b, n, determine if n is in this set.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.
Output
For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.
Example
Input
5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264
Output
Yes
No
Yes
No
Yes
Note
In the first test case, 24 is generated as follows:
* 1 is in this set, so 3 and 6 are in this set;
* 3 is in this set, so 9 and 8 are in this set;
* 8 is in this set, so 24 and 13 are in this set.
Thus we can see 24 is in this set.
The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them. | instruction | 0 | 49,096 | 22 | 98,192 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
def solve(): # n = a^x + b*b , hence n is in the set if (n - a^x) % b == 0
n,a,b = map(int,input().split())
if a == 1:
if (n-1)%b ==0:
print("YES")
else:
print("NO")
else:
curr = 1
flag = 0
while curr<=n :
if curr%b==n%b:
flag=1
break
curr *=a
if flag == 1:
print("YES")
else:
print("NO")
tests = int(input())
for _ in range(tests):
solve()
``` | output | 1 | 49,096 | 22 | 98,193 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105 | instruction | 0 | 49,397 | 22 | 98,794 |
"Correct Solution:
```
n=int(input())
ans=0
for i in range(1,n+1):
A=n//i
ans+=(A*(A+1)*i//2)
print(ans)
``` | output | 1 | 49,397 | 22 | 98,795 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105 | instruction | 0 | 49,398 | 22 | 98,796 |
"Correct Solution:
```
n = int(input())
ans = 0
for i in range(1,n+1):
ans += n//i*(i+n//i*i)//2
print(ans)
``` | output | 1 | 49,398 | 22 | 98,797 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105 | instruction | 0 | 49,399 | 22 | 98,798 |
"Correct Solution:
```
n=int(input())
ans = 0
for i in range(1, n+1):
a = n//i
ans += a*(a+1)//2*i
print(ans)
``` | output | 1 | 49,399 | 22 | 98,799 |
Provide a correct Python 3 solution for this coding contest problem.
For a positive integer X, let f(X) be the number of positive divisors of X.
Given a positive integer N, find \sum_{K=1}^N K\times f(K).
Constraints
* 1 \leq N \leq 10^7
Input
Input is given from Standard Input in the following format:
N
Output
Print the value \sum_{K=1}^N K\times f(K).
Examples
Input
4
Output
23
Input
100
Output
26879
Input
10000000
Output
838627288460105 | instruction | 0 | 49,400 | 22 | 98,800 |
"Correct Solution:
```
n=int(input())
ans=0
for i in range(1,n+1):
w=n//i
ans+=i*w*(w+1)//2
print(ans)
``` | output | 1 | 49,400 | 22 | 98,801 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.