message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a grid with A horizontal rows and B vertical columns, with the squares painted white. On this grid, we will repeatedly apply the following operation:
* Assume that the grid currently has a horizontal rows and b vertical columns. Choose "vertical" or "horizontal".
* If we choose "vertical", insert one row at the top of the grid, resulting in an (a+1) \times b grid.
* If we choose "horizontal", insert one column at the right end of the grid, resulting in an a \times (b+1) grid.
* Then, paint one of the added squares black, and the other squares white.
Assume the grid eventually has C horizontal rows and D vertical columns. Find the number of ways in which the squares can be painted in the end, modulo 998244353.
Constraints
* 1 \leq A \leq C \leq 3000
* 1 \leq B \leq D \leq 3000
* A, B, C, and D are integers.
Input
Input is given from Standard Input in the following format:
A B C D
Output
Print the number of ways in which the squares can be painted in the end, modulo 998244353.
Examples
Input
1 1 2 2
Output
3
Input
2 1 3 4
Output
65
Input
31 41 59 265
Output
387222020
Submitted Solution:
```
a, b, c, d = map(int, input().split())
MOD = 998244353
dp = [[0] * (d - b + 1) for _ in range(c - a + 1)]
dp[0][0] = 1
for x in range(1, d - b + 1):
dp[0][x] = dp[0][x - 1] * a % MOD
for y in range(1, c - a + 1):
dp[y][0] = dp[y - 1][0] * b % MOD
for x in range(1, c - a + 1):
for y in range(1, d - b + 1):
dp[x][y] = (dp[x - 1][y] * (b + y) + dp[x][y - 1] * (a + x)) % MOD
print(dp[-1][-1])
``` | instruction | 0 | 73,958 | 7 | 147,916 |
No | output | 1 | 73,958 | 7 | 147,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have a grid with A horizontal rows and B vertical columns, with the squares painted white. On this grid, we will repeatedly apply the following operation:
* Assume that the grid currently has a horizontal rows and b vertical columns. Choose "vertical" or "horizontal".
* If we choose "vertical", insert one row at the top of the grid, resulting in an (a+1) \times b grid.
* If we choose "horizontal", insert one column at the right end of the grid, resulting in an a \times (b+1) grid.
* Then, paint one of the added squares black, and the other squares white.
Assume the grid eventually has C horizontal rows and D vertical columns. Find the number of ways in which the squares can be painted in the end, modulo 998244353.
Constraints
* 1 \leq A \leq C \leq 3000
* 1 \leq B \leq D \leq 3000
* A, B, C, and D are integers.
Input
Input is given from Standard Input in the following format:
A B C D
Output
Print the number of ways in which the squares can be painted in the end, modulo 998244353.
Examples
Input
1 1 2 2
Output
3
Input
2 1 3 4
Output
65
Input
31 41 59 265
Output
387222020
Submitted Solution:
```
a,b,c,d = map(int,input().split())
mod = 998244353
dp = [[0]*(d+1) for _ in range(c+1)]
dp[a][b] = 1
for i in range(a+1,c+1):
dp[i][b] = dp[i-1][b]*b
dp[i][b] %= mod
for i in range(a,c+1):
for j in range(b+1,d+1):
dp[i][j] += dp[i][j-1]*i+dp[i-1][j]*j-dp[i-1][j-1]*(i-1)*(j-1)
dp[i][j] %= mod
print(dp[c][d])
``` | instruction | 0 | 73,959 | 7 | 147,918 |
No | output | 1 | 73,959 | 7 | 147,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya's birthday is approaching and Lena decided to sew a patterned handkerchief to him as a present. Lena chose digits from 0 to n as the pattern. The digits will form a rhombus. The largest digit n should be located in the centre. The digits should decrease as they approach the edges. For example, for n = 5 the handkerchief pattern should look like that:
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 4 5 4 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Your task is to determine the way the handkerchief will look like by the given n.
Input
The first line contains the single integer n (2 ≤ n ≤ 9).
Output
Print a picture for the given n. You should strictly observe the number of spaces before the first digit on each line. Every two adjacent digits in the same line should be separated by exactly one space. There should be no spaces after the last digit at the end of each line.
Examples
Input
2
Output
0
0 1 0
0 1 2 1 0
0 1 0
0
Input
3
Output
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Submitted Solution:
```
n=int(input(""))
for i in range (n+1):
for k in range (n-i):
print(" ",end=" ")
for k in range (i):
print(k,end=" ")
for k in range (i+1):
if k==i:
print(i-k,end="")
else:
print(i-k,end=" ")
print("")
for i in range (n-1,-1,-1):
for k in range (n-i):
print(" ",end=" ")
for k in range (i):
print(k,end=" ")
for k in range (i+1):
if k==i:
print(i-k,end="")
else:
print(i-k,end=" ")
print("")
``` | instruction | 0 | 74,247 | 7 | 148,494 |
Yes | output | 1 | 74,247 | 7 | 148,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya's birthday is approaching and Lena decided to sew a patterned handkerchief to him as a present. Lena chose digits from 0 to n as the pattern. The digits will form a rhombus. The largest digit n should be located in the centre. The digits should decrease as they approach the edges. For example, for n = 5 the handkerchief pattern should look like that:
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 4 5 4 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Your task is to determine the way the handkerchief will look like by the given n.
Input
The first line contains the single integer n (2 ≤ n ≤ 9).
Output
Print a picture for the given n. You should strictly observe the number of spaces before the first digit on each line. Every two adjacent digits in the same line should be separated by exactly one space. There should be no spaces after the last digit at the end of each line.
Examples
Input
2
Output
0
0 1 0
0 1 2 1 0
0 1 0
0
Input
3
Output
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Submitted Solution:
```
n, num = int(input()), '0123456789'
for i in num[:n] + num[n::-1]:
print(' '.join(' '*(n-int(i))+num[:int(i)]+num[int(i)::-1]))
``` | instruction | 0 | 74,248 | 7 | 148,496 |
Yes | output | 1 | 74,248 | 7 | 148,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya's birthday is approaching and Lena decided to sew a patterned handkerchief to him as a present. Lena chose digits from 0 to n as the pattern. The digits will form a rhombus. The largest digit n should be located in the centre. The digits should decrease as they approach the edges. For example, for n = 5 the handkerchief pattern should look like that:
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 4 5 4 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Your task is to determine the way the handkerchief will look like by the given n.
Input
The first line contains the single integer n (2 ≤ n ≤ 9).
Output
Print a picture for the given n. You should strictly observe the number of spaces before the first digit on each line. Every two adjacent digits in the same line should be separated by exactly one space. There should be no spaces after the last digit at the end of each line.
Examples
Input
2
Output
0
0 1 0
0 1 2 1 0
0 1 0
0
Input
3
Output
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Submitted Solution:
```
n=int(input())
count=0
for i in range(n+1):
for j in range(2*(n-i)):
print(end=" ")
for j in range(i):
print(j,end=" ")
for j in range(i,-1,-1):
if j==0:
print(j)
else:
print(j,end=" ")
for i in range(1,n + 2):
for j in range(2*i):
print(end=" ")
for j in range(n-i):
print(j, end=" ")
for j in range(n-i,-1,-1):
if j==0:
print(j)
else:
print(j, end=" ")
``` | instruction | 0 | 74,249 | 7 | 148,498 |
Yes | output | 1 | 74,249 | 7 | 148,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya's birthday is approaching and Lena decided to sew a patterned handkerchief to him as a present. Lena chose digits from 0 to n as the pattern. The digits will form a rhombus. The largest digit n should be located in the centre. The digits should decrease as they approach the edges. For example, for n = 5 the handkerchief pattern should look like that:
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 4 5 4 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Your task is to determine the way the handkerchief will look like by the given n.
Input
The first line contains the single integer n (2 ≤ n ≤ 9).
Output
Print a picture for the given n. You should strictly observe the number of spaces before the first digit on each line. Every two adjacent digits in the same line should be separated by exactly one space. There should be no spaces after the last digit at the end of each line.
Examples
Input
2
Output
0
0 1 0
0 1 2 1 0
0 1 0
0
Input
3
Output
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Submitted Solution:
```
a=int(input())
x=a
w=[]
while a>=0:
b=[]
b.append(a)
for i in range(a):
b.append(str(a-1-i))
b.insert(0,(a-1-i))
b.insert(0,' '*(2*(x-a)))
w.append(b)
w.insert(0,b)
a=a-1
w.pop(x)
for j in w:
xx=str(j)
xxx=xx.replace(',','')
xxxx=xxx.replace("'",'')
print(xxxx[2:-1])
``` | instruction | 0 | 74,250 | 7 | 148,500 |
Yes | output | 1 | 74,250 | 7 | 148,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya's birthday is approaching and Lena decided to sew a patterned handkerchief to him as a present. Lena chose digits from 0 to n as the pattern. The digits will form a rhombus. The largest digit n should be located in the centre. The digits should decrease as they approach the edges. For example, for n = 5 the handkerchief pattern should look like that:
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 4 5 4 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Your task is to determine the way the handkerchief will look like by the given n.
Input
The first line contains the single integer n (2 ≤ n ≤ 9).
Output
Print a picture for the given n. You should strictly observe the number of spaces before the first digit on each line. Every two adjacent digits in the same line should be separated by exactly one space. There should be no spaces after the last digit at the end of each line.
Examples
Input
2
Output
0
0 1 0
0 1 2 1 0
0 1 0
0
Input
3
Output
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Submitted Solution:
```
n = int(input())
spaces = n*2
for i in range(n):
for s in range(spaces):
print(' ',end='')
for j in range(i+1):
print(j,end=' ')
for j in range(i-1,-1,-1):
print(j,end=' ')
print()
spaces-=2
for i in range(n+1):
print(i,end=' ')
for i in range(n-1,-1,-1):
print(i,end=' ')
print()
spaces=2
for i in range(n-1,-1,-1):
for s in range(spaces):
print(' ',end='')
for j in range(i+1):
print(j,end=' ')
for j in range(i-1,-1,-1):
print(j,end=' ')
print()
spaces+=2
``` | instruction | 0 | 74,251 | 7 | 148,502 |
No | output | 1 | 74,251 | 7 | 148,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya's birthday is approaching and Lena decided to sew a patterned handkerchief to him as a present. Lena chose digits from 0 to n as the pattern. The digits will form a rhombus. The largest digit n should be located in the centre. The digits should decrease as they approach the edges. For example, for n = 5 the handkerchief pattern should look like that:
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 4 5 4 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Your task is to determine the way the handkerchief will look like by the given n.
Input
The first line contains the single integer n (2 ≤ n ≤ 9).
Output
Print a picture for the given n. You should strictly observe the number of spaces before the first digit on each line. Every two adjacent digits in the same line should be separated by exactly one space. There should be no spaces after the last digit at the end of each line.
Examples
Input
2
Output
0
0 1 0
0 1 2 1 0
0 1 0
0
Input
3
Output
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Submitted Solution:
```
n=int(input())
for i in range(n+1):
k=-1
for j in range(n+1):
if(j>=n-i):
k+=1
if(i==0):
print(k,end="")
else:
print(k,end=" ")
else:
print(' ',end=" ")
for j in range(0,i-1):
k=k-1
print(k,end=" ")
if(i>0):
print(k-1)
else:
print()
for i in range(n):
k=-1
for j in range(n+1):
if(j>i):
k=k+1
print(k,end=" ")
else:
print(' ',end=" ")
for j in range(n-i-2):
k=k-1
print(k,end=" ")
if(i<n-1):
print(k-1)
``` | instruction | 0 | 74,252 | 7 | 148,504 |
No | output | 1 | 74,252 | 7 | 148,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya's birthday is approaching and Lena decided to sew a patterned handkerchief to him as a present. Lena chose digits from 0 to n as the pattern. The digits will form a rhombus. The largest digit n should be located in the centre. The digits should decrease as they approach the edges. For example, for n = 5 the handkerchief pattern should look like that:
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 4 5 4 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Your task is to determine the way the handkerchief will look like by the given n.
Input
The first line contains the single integer n (2 ≤ n ≤ 9).
Output
Print a picture for the given n. You should strictly observe the number of spaces before the first digit on each line. Every two adjacent digits in the same line should be separated by exactly one space. There should be no spaces after the last digit at the end of each line.
Examples
Input
2
Output
0
0 1 0
0 1 2 1 0
0 1 0
0
Input
3
Output
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Submitted Solution:
```
def solution():
num = int(input())
out= []
for i in range(num+1):
res = ' ' * ((num) - i)
part = [i for i in range(i+1)]
part2 = part + [i for i in range(i, -1, -1) if i != part[-1]]
for j in part2:
res += str(j) + ' '
out.append(res)
mask = [] + out
mask.pop()
out += list(reversed(mask))
for i in out:
print(i)
solution()
``` | instruction | 0 | 74,253 | 7 | 148,506 |
No | output | 1 | 74,253 | 7 | 148,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya's birthday is approaching and Lena decided to sew a patterned handkerchief to him as a present. Lena chose digits from 0 to n as the pattern. The digits will form a rhombus. The largest digit n should be located in the centre. The digits should decrease as they approach the edges. For example, for n = 5 the handkerchief pattern should look like that:
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 4 5 4 3 2 1 0
0 1 2 3 4 3 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Your task is to determine the way the handkerchief will look like by the given n.
Input
The first line contains the single integer n (2 ≤ n ≤ 9).
Output
Print a picture for the given n. You should strictly observe the number of spaces before the first digit on each line. Every two adjacent digits in the same line should be separated by exactly one space. There should be no spaces after the last digit at the end of each line.
Examples
Input
2
Output
0
0 1 0
0 1 2 1 0
0 1 0
0
Input
3
Output
0
0 1 0
0 1 2 1 0
0 1 2 3 2 1 0
0 1 2 1 0
0 1 0
0
Submitted Solution:
```
n = int(input())
for i in range(n+1):
s = n - i
for t in range(s):
print(" ",end=' ')
if(i==0):
print("0")
else:
print("0",end=' ')
r = 0
for j in range(1,i+1):
r = j
print("{}".format(j),end=' ')
for y in range(r-1,-1,-1):
if(y!=0):
print("{}".format(y),end=' ')
elif(y==0):
print("{}".format(y))
print("\n")
for i in range(n):
s = i + 1
for j in range(s):
print(" ",end=" ")
r = 0
for t in range(n-i):
r = t
print(t,end=" ")
for y in range(r-1,-1,-1):
print(y,end=" ")
print("\n")
``` | instruction | 0 | 74,254 | 7 | 148,508 |
No | output | 1 | 74,254 | 7 | 148,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7. | instruction | 0 | 74,255 | 7 | 148,510 |
Tags: greedy, implementation, math
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat Sep 14 09:52:12 2019
@author: jeff
"""
def paint():
total = input()
unique = [] * int(total)
inputs = list(map(int, input().split()))
counter = 0;
inputs.sort()
for x in inputs:
if x not in unique:
unique.append(x)
val = [False] * len(unique)
for z in range(len(unique)):
for p in range(len(unique)):
if unique[z] < unique[p] and unique[p]%unique[z] == 0 and val[p] != True:
val[p] = True
counter = counter + 1
if(len(unique) == 0):
print(1)
else:
print(len(unique) - counter)
paint()
``` | output | 1 | 74,255 | 7 | 148,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7. | instruction | 0 | 74,256 | 7 | 148,512 |
Tags: greedy, implementation, math
Correct Solution:
```
def main():
n = int(input())
arr = list(map(int,input().split()))
colors = [0]*n
arr.sort()
color = 1
for i in range(n):
if colors[i] == 0:
colors[i] = color
for j in range(i+1,n):
if arr[j]%arr[i] == 0 and colors[j] == 0:
colors[j] = color
color += 1
print(max(colors))
main()
``` | output | 1 | 74,256 | 7 | 148,513 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7. | instruction | 0 | 74,257 | 7 | 148,514 |
Tags: greedy, implementation, math
Correct Solution:
```
import sys
def main():
n = int(sys.stdin.readline())
numbers = list(map(int, sys.stdin.readline().split()))
numbers.sort()
painted = {}; check = []
for i in range(n):
for j in range(i, n):
if numbers[j]%numbers[i] == 0:
if numbers[i] not in painted:
painted[numbers[i]] = []
painted[numbers[i]].append(j)
else:
if j not in painted[numbers[i]]:
painted[numbers[i]].append(j)
count={}
for i in painted:
for j in painted[i]:
if j not in check:
check.append(j)
count[i] = count.get(i, 0) + 1
sys.stdout.write(f'{len(count)}')
if __name__ == '__main__':
main()
``` | output | 1 | 74,257 | 7 | 148,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7. | instruction | 0 | 74,258 | 7 | 148,516 |
Tags: greedy, implementation, math
Correct Solution:
```
i = input()
nums = input().split(' ')
for i in range(len(nums)):
nums[i] = int(nums[i])
nums.sort()
color = 0
while len(nums) > 0:
num1 = nums.pop(0)
color += 1
for i in range(len(nums)-1, -1, -1):
if nums[i] % num1 == 0:
nums.remove(nums[i])
print(str(color))
``` | output | 1 | 74,258 | 7 | 148,517 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7. | instruction | 0 | 74,259 | 7 | 148,518 |
Tags: greedy, implementation, math
Correct Solution:
```
a = int(input())
b = [int(i) for i in input().strip().split()]
b.sort()
ans = []
ans.append(b[0])
for i in range(1,len(b)):
flag = 0
for j in ans:
if b[i]%j==0:
flag=1
break
if(flag == 0):
ans.append(b[i])
print(len(ans))
``` | output | 1 | 74,259 | 7 | 148,519 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7. | instruction | 0 | 74,260 | 7 | 148,520 |
Tags: greedy, implementation, math
Correct Solution:
```
def solve():
global a
a.sort()
k = 0
m = [False for _ in range(n)]
for i in range(n):
if not m[i]:
k += 1
for j in range(i + 1, n):
if not m[j] and a[j] % a[i] == 0:
m[j] = True
return k
def main():
global n, a
n = int(input())
a = list(map(int, input().split()))
print(solve())
main()
``` | output | 1 | 74,260 | 7 | 148,521 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7. | instruction | 0 | 74,261 | 7 | 148,522 |
Tags: greedy, implementation, math
Correct Solution:
```
n=int(input())
a=list(map(int,input().strip().split(" ")))
a.sort()
p=[a[0]]
c=1
for i in range(1,n):
k=1
for j in p:
if a[i]%j==0:
k=0
if k:
p.append(a[i])
c+=1
print(c)
``` | output | 1 | 74,261 | 7 | 148,523 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7. | instruction | 0 | 74,262 | 7 | 148,524 |
Tags: greedy, implementation, math
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
a.sort()
lst = []
ans = 0
for i in range(len(a)):
if (a[i] not in lst):
for j in range(len(lst)):
if (a[i] % lst[j] == 0):
break
else:
lst.append(a[i])
print(len(lst))
``` | output | 1 | 74,262 | 7 | 148,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7.
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
a.sort()
i=0
while i<n:
j=i
k=a[i]
while j<n:
if a[j]%k==0:
a[j]=a[i]
j+=1
i=i+1
print(len(set(a)))
``` | instruction | 0 | 74,263 | 7 | 148,526 |
Yes | output | 1 | 74,263 | 7 | 148,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7.
Submitted Solution:
```
# 1209A
n = int(input())
a = input().split()
a = list(map(int,a))
a.sort()
count = 0
while len(a) > 0:
cur = a[0]
count += 1
i = 0
while i < len(a):
if a[i]%cur == 0:
del a[i]
else:
i += 1
print(count)
``` | instruction | 0 | 74,264 | 7 | 148,528 |
Yes | output | 1 | 74,264 | 7 | 148,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7.
Submitted Solution:
```
x=int(input())
p=list(map(int,input().split()))
t=0
while(p):
m=min(p)
p=[x for x in p if x%m!=0]
t+=1
print(t)
``` | instruction | 0 | 74,265 | 7 | 148,530 |
Yes | output | 1 | 74,265 | 7 | 148,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7.
Submitted Solution:
```
n=int(input())
l=[]
q=[]
w=[]
S=0;
D=0;
p=input().rstrip().split(' ')
p.sort(key=int)
for i in range(0,len(p)):
if(1):
if len(l)==0:
D+=1;
l.append(int(p[i]))
else:
for j in range(0,len(l)):
if int(p[i])%l[j]==0:
break;
else:
D+=1;
l.append(int(p[i]))
print(D)
``` | instruction | 0 | 74,266 | 7 | 148,532 |
Yes | output | 1 | 74,266 | 7 | 148,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7.
Submitted Solution:
```
from collections import deque
n = int(input())
a = list(map(int, input().split()))
cnt = 0
done = deque()
prime = [2,3,5,7,11,13,17,19,23,29,31,33,37,41,43,47,53,59,61,67,71,73,79,83,89,97]
i = 0
while len(done) < n and i < 26:
x = len(done)
for j in range(n):
if j not in done and a[j] % prime[i] == 0:
done.append(j)
i += 1
if x != len(done):
cnt += 1
print(cnt)
``` | instruction | 0 | 74,267 | 7 | 148,534 |
No | output | 1 | 74,267 | 7 | 148,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7.
Submitted Solution:
```
n = int(input())
l = [*map(int,input().split())]
l.sort()
cnt = 0
for i in range(n):
for j in range(i + 1,n):
if(l[j] != -1 and l[j] % l[i] == 0):
l[j] = -1
cnt += 1
print(cnt)
``` | instruction | 0 | 74,268 | 7 | 148,536 |
No | output | 1 | 74,268 | 7 | 148,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
a=sorted(a)
c=0
while(len(set(a))==1):
for j in range(0,n):
if a[j]!=0:
if a[j]%a[c]==0:
a[j]=0
c+=1
print(c)
``` | instruction | 0 | 74,269 | 7 | 148,538 |
No | output | 1 | 74,269 | 7 | 148,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of integers a_1, a_2, ..., a_n. You need to paint elements in colors, so that:
* If we consider any color, all elements of this color must be divisible by the minimal element of this color.
* The number of used colors must be minimized.
For example, it's fine to paint elements [40, 10, 60] in a single color, because they are all divisible by 10. You can use any color an arbitrary amount of times (in particular, it is allowed to use a color only once). The elements painted in one color do not need to be consecutive.
For example, if a=[6, 2, 3, 4, 12] then two colors are required: let's paint 6, 3 and 12 in the first color (6, 3 and 12 are divisible by 3) and paint 2 and 4 in the second color (2 and 4 are divisible by 2). For example, if a=[10, 7, 15] then 3 colors are required (we can simply paint each element in an unique color).
Input
The first line contains an integer n (1 ≤ n ≤ 100), where n is the length of the given sequence.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100). These numbers can contain duplicates.
Output
Print the minimal number of colors to paint all the given numbers in a valid way.
Examples
Input
6
10 2 3 5 4 2
Output
3
Input
4
100 100 100 100
Output
1
Input
8
7 6 5 4 3 2 2 3
Output
4
Note
In the first example, one possible way to paint the elements in 3 colors is:
* paint in the first color the elements: a_1=10 and a_4=5,
* paint in the second color the element a_3=3,
* paint in the third color the elements: a_2=2, a_5=4 and a_6=2.
In the second example, you can use one color to paint all the elements.
In the third example, one possible way to paint the elements in 4 colors is:
* paint in the first color the elements: a_4=4, a_6=2 and a_7=2,
* paint in the second color the elements: a_2=6, a_5=3 and a_8=3,
* paint in the third color the element a_3=5,
* paint in the fourth color the element a_1=7.
Submitted Solution:
```
import math
n = int(input())
arr = list(map(int,input().split()))
def prime(n):
res = []
while n%2==0:
res.append(2)
n = n//2
for i in range(3,int(math.sqrt(n))+1,2):
while n%i==0:
res.append(i)
n = n//i
if n>2:
res.append(n)
return res
new_arr = []
for a in arr:
new_arr.extend(prime(a))
print(len(set(new_arr)))
``` | instruction | 0 | 74,270 | 7 | 148,540 |
No | output | 1 | 74,270 | 7 | 148,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared a lot of food, left some money and washed all Arseniy's clothes.
Ten minutes before her leave she realized that it would be also useful to prepare instruction of which particular clothes to wear on each of the days she will be absent. Arseniy's family is a bit weird so all the clothes is enumerated. For example, each of Arseniy's n socks is assigned a unique integer from 1 to n. Thus, the only thing his mother had to do was to write down two integers li and ri for each of the days — the indices of socks to wear on the day i (obviously, li stands for the left foot and ri for the right). Each sock is painted in one of k colors.
When mother already left Arseniy noticed that according to instruction he would wear the socks of different colors on some days. Of course, that is a terrible mistake cause by a rush. Arseniy is a smart boy, and, by some magical coincidence, he posses k jars with the paint — one for each of k colors.
Arseniy wants to repaint some of the socks in such a way, that for each of m days he can follow the mother's instructions and wear the socks of the same color. As he is going to be very busy these days he will have no time to change the colors of any socks so he has to finalize the colors now.
The new computer game Bota-3 was just realised and Arseniy can't wait to play it. What is the minimum number of socks that need their color to be changed in order to make it possible to follow mother's instructions and wear the socks of the same color during each of m days.
Input
The first line of input contains three integers n, m and k (2 ≤ n ≤ 200 000, 0 ≤ m ≤ 200 000, 1 ≤ k ≤ 200 000) — the number of socks, the number of days and the number of available colors respectively.
The second line contain n integers c1, c2, ..., cn (1 ≤ ci ≤ k) — current colors of Arseniy's socks.
Each of the following m lines contains two integers li and ri (1 ≤ li, ri ≤ n, li ≠ ri) — indices of socks which Arseniy should wear during the i-th day.
Output
Print one integer — the minimum number of socks that should have their colors changed in order to be able to obey the instructions and not make people laugh from watching the socks of different colors.
Examples
Input
3 2 3
1 2 3
1 2
2 3
Output
2
Input
3 2 2
1 1 2
1 2
2 1
Output
0
Note
In the first sample, Arseniy can repaint the first and the third socks to the second color.
In the second sample, there is no need to change any colors.
Submitted Solution:
```
def main():
from collections import Counter
n, m, _ = map(int, input().split())
cc = list(map(int, input().split()))
pairs, avail = [[] for _ in range(n)], [True] * n
for _ in range(m):
a, b = map(int, input().split())
pairs[a - 1].append(b - 1)
pairs[b - 1].append(a - 1)
for a, f in enumerate(avail):
if f:
stack, cnt, avail[a] = [a], Counter(), False
while stack:
a = stack.pop()
cnt[cc[a]] += 1
for b in pairs[a]:
if avail[b]:
avail[b] = False
stack.append(b)
n -= cnt.most_common(1)[0][1]
print(n)
if __name__ == '__main__':
main()
# Made By Mostafa_Khaled
``` | instruction | 0 | 74,650 | 7 | 149,300 |
Yes | output | 1 | 74,650 | 7 | 149,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared a lot of food, left some money and washed all Arseniy's clothes.
Ten minutes before her leave she realized that it would be also useful to prepare instruction of which particular clothes to wear on each of the days she will be absent. Arseniy's family is a bit weird so all the clothes is enumerated. For example, each of Arseniy's n socks is assigned a unique integer from 1 to n. Thus, the only thing his mother had to do was to write down two integers li and ri for each of the days — the indices of socks to wear on the day i (obviously, li stands for the left foot and ri for the right). Each sock is painted in one of k colors.
When mother already left Arseniy noticed that according to instruction he would wear the socks of different colors on some days. Of course, that is a terrible mistake cause by a rush. Arseniy is a smart boy, and, by some magical coincidence, he posses k jars with the paint — one for each of k colors.
Arseniy wants to repaint some of the socks in such a way, that for each of m days he can follow the mother's instructions and wear the socks of the same color. As he is going to be very busy these days he will have no time to change the colors of any socks so he has to finalize the colors now.
The new computer game Bota-3 was just realised and Arseniy can't wait to play it. What is the minimum number of socks that need their color to be changed in order to make it possible to follow mother's instructions and wear the socks of the same color during each of m days.
Input
The first line of input contains three integers n, m and k (2 ≤ n ≤ 200 000, 0 ≤ m ≤ 200 000, 1 ≤ k ≤ 200 000) — the number of socks, the number of days and the number of available colors respectively.
The second line contain n integers c1, c2, ..., cn (1 ≤ ci ≤ k) — current colors of Arseniy's socks.
Each of the following m lines contains two integers li and ri (1 ≤ li, ri ≤ n, li ≠ ri) — indices of socks which Arseniy should wear during the i-th day.
Output
Print one integer — the minimum number of socks that should have their colors changed in order to be able to obey the instructions and not make people laugh from watching the socks of different colors.
Examples
Input
3 2 3
1 2 3
1 2
2 3
Output
2
Input
3 2 2
1 1 2
1 2
2 1
Output
0
Note
In the first sample, Arseniy can repaint the first and the third socks to the second color.
In the second sample, there is no need to change any colors.
Submitted Solution:
```
import sys
n, m, k = [int(x) for x in sys.stdin.readline().replace('\n', '').split(' ')]
c = [int(x) for x in sys.stdin.readline().replace('\n', '').split(' ')]
# print ((n,m,k))
# Graphs with python
socks = [[] for _ in range(n)]
# populate a Graph
for i in range(m):
l, r = [int(x)-1 for x in sys.stdin.readline().replace('\n', '').split(' ')]
socks[r] += [l]
socks[l] += [r]
# search a Graph
visited = [False for _ in range(n)]
forest = {}
for i, v in enumerate(visited):
if v:
continue
visited[i] = True
queue = [(i, i)]
forest[i] = {'nodes': 1, 'colours': {c[i]: 1}}
while len(queue) != 0:
# print(queue)
representant, current = queue.pop()
for node in socks[current]:
if not visited[node]:
queue += [(representant, node)]
forest[representant]['nodes'] += 1
if c[node] in forest[representant]['colours']:
forest[representant]['colours'][c[node]] += 1
else:
forest[representant]['colours'][c[node]] = 1
visited[node] = True
# print(forest)
total = 0
for key in forest:
maximun = 0
for i in forest[key]['colours']:
if forest[key]['colours'][i] > maximun:
maximun = forest[key]['colours'][i]
total += forest[key]['nodes'] - maximun
sys.stdout.write(str(total))
``` | instruction | 0 | 74,651 | 7 | 149,302 |
Yes | output | 1 | 74,651 | 7 | 149,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared a lot of food, left some money and washed all Arseniy's clothes.
Ten minutes before her leave she realized that it would be also useful to prepare instruction of which particular clothes to wear on each of the days she will be absent. Arseniy's family is a bit weird so all the clothes is enumerated. For example, each of Arseniy's n socks is assigned a unique integer from 1 to n. Thus, the only thing his mother had to do was to write down two integers li and ri for each of the days — the indices of socks to wear on the day i (obviously, li stands for the left foot and ri for the right). Each sock is painted in one of k colors.
When mother already left Arseniy noticed that according to instruction he would wear the socks of different colors on some days. Of course, that is a terrible mistake cause by a rush. Arseniy is a smart boy, and, by some magical coincidence, he posses k jars with the paint — one for each of k colors.
Arseniy wants to repaint some of the socks in such a way, that for each of m days he can follow the mother's instructions and wear the socks of the same color. As he is going to be very busy these days he will have no time to change the colors of any socks so he has to finalize the colors now.
The new computer game Bota-3 was just realised and Arseniy can't wait to play it. What is the minimum number of socks that need their color to be changed in order to make it possible to follow mother's instructions and wear the socks of the same color during each of m days.
Input
The first line of input contains three integers n, m and k (2 ≤ n ≤ 200 000, 0 ≤ m ≤ 200 000, 1 ≤ k ≤ 200 000) — the number of socks, the number of days and the number of available colors respectively.
The second line contain n integers c1, c2, ..., cn (1 ≤ ci ≤ k) — current colors of Arseniy's socks.
Each of the following m lines contains two integers li and ri (1 ≤ li, ri ≤ n, li ≠ ri) — indices of socks which Arseniy should wear during the i-th day.
Output
Print one integer — the minimum number of socks that should have their colors changed in order to be able to obey the instructions and not make people laugh from watching the socks of different colors.
Examples
Input
3 2 3
1 2 3
1 2
2 3
Output
2
Input
3 2 2
1 1 2
1 2
2 1
Output
0
Note
In the first sample, Arseniy can repaint the first and the third socks to the second color.
In the second sample, there is no need to change any colors.
Submitted Solution:
```
from collections import deque
import sys
input = sys.stdin.buffer.readline
n,m,k = map(int,input().split())
colors = list(map(int,input().split()))
adj = [[] for i in range(n)]
for i in range(m):
u,v = map(int,input().split())
u -= 1
v -= 1
adj[u].append(v)
adj[v].append(u)
total_change = 0
visited = [0]*n
for i in range(n):
if not visited[i]:
curr_comp = []
s = deque([i])
while s:
curr = s[-1]
if not visited[curr]:
for neighbor in adj[curr]:
s.append(neighbor)
visited[curr] = 1
curr_comp.append(curr)
else:
s.pop()
total = len(curr_comp)
most = {}
for a in curr_comp:
if colors[a] in most:
most[colors[a]] += 1
else:
most[colors[a]] = 1
total_change += total - max(most.values())
print(total_change)
``` | instruction | 0 | 74,652 | 7 | 149,304 |
Yes | output | 1 | 74,652 | 7 | 149,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared a lot of food, left some money and washed all Arseniy's clothes.
Ten minutes before her leave she realized that it would be also useful to prepare instruction of which particular clothes to wear on each of the days she will be absent. Arseniy's family is a bit weird so all the clothes is enumerated. For example, each of Arseniy's n socks is assigned a unique integer from 1 to n. Thus, the only thing his mother had to do was to write down two integers li and ri for each of the days — the indices of socks to wear on the day i (obviously, li stands for the left foot and ri for the right). Each sock is painted in one of k colors.
When mother already left Arseniy noticed that according to instruction he would wear the socks of different colors on some days. Of course, that is a terrible mistake cause by a rush. Arseniy is a smart boy, and, by some magical coincidence, he posses k jars with the paint — one for each of k colors.
Arseniy wants to repaint some of the socks in such a way, that for each of m days he can follow the mother's instructions and wear the socks of the same color. As he is going to be very busy these days he will have no time to change the colors of any socks so he has to finalize the colors now.
The new computer game Bota-3 was just realised and Arseniy can't wait to play it. What is the minimum number of socks that need their color to be changed in order to make it possible to follow mother's instructions and wear the socks of the same color during each of m days.
Input
The first line of input contains three integers n, m and k (2 ≤ n ≤ 200 000, 0 ≤ m ≤ 200 000, 1 ≤ k ≤ 200 000) — the number of socks, the number of days and the number of available colors respectively.
The second line contain n integers c1, c2, ..., cn (1 ≤ ci ≤ k) — current colors of Arseniy's socks.
Each of the following m lines contains two integers li and ri (1 ≤ li, ri ≤ n, li ≠ ri) — indices of socks which Arseniy should wear during the i-th day.
Output
Print one integer — the minimum number of socks that should have their colors changed in order to be able to obey the instructions and not make people laugh from watching the socks of different colors.
Examples
Input
3 2 3
1 2 3
1 2
2 3
Output
2
Input
3 2 2
1 1 2
1 2
2 1
Output
0
Note
In the first sample, Arseniy can repaint the first and the third socks to the second color.
In the second sample, there is no need to change any colors.
Submitted Solution:
```
def main():
from collections import Counter
n, m, _ = map(int, input().split())
cc = list(map(int, input().split()))
pairs, avail = [[] for _ in range(n)], [True] * n
for _ in range(m):
a, b = map(int, input().split())
pairs[a - 1].append(b - 1)
pairs[b - 1].append(a - 1)
for a, f in enumerate(avail):
if f:
stack, cnt, avail[a] = [a], Counter(), False
while stack:
a = stack.pop()
cnt[cc[a]] += 1
for b in pairs[a]:
if avail[b]:
avail[b] = False
stack.append(b)
n -= cnt.most_common(1)[0][1]
print(n)
if __name__ == '__main__':
main()
``` | instruction | 0 | 74,653 | 7 | 149,306 |
Yes | output | 1 | 74,653 | 7 | 149,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared a lot of food, left some money and washed all Arseniy's clothes.
Ten minutes before her leave she realized that it would be also useful to prepare instruction of which particular clothes to wear on each of the days she will be absent. Arseniy's family is a bit weird so all the clothes is enumerated. For example, each of Arseniy's n socks is assigned a unique integer from 1 to n. Thus, the only thing his mother had to do was to write down two integers li and ri for each of the days — the indices of socks to wear on the day i (obviously, li stands for the left foot and ri for the right). Each sock is painted in one of k colors.
When mother already left Arseniy noticed that according to instruction he would wear the socks of different colors on some days. Of course, that is a terrible mistake cause by a rush. Arseniy is a smart boy, and, by some magical coincidence, he posses k jars with the paint — one for each of k colors.
Arseniy wants to repaint some of the socks in such a way, that for each of m days he can follow the mother's instructions and wear the socks of the same color. As he is going to be very busy these days he will have no time to change the colors of any socks so he has to finalize the colors now.
The new computer game Bota-3 was just realised and Arseniy can't wait to play it. What is the minimum number of socks that need their color to be changed in order to make it possible to follow mother's instructions and wear the socks of the same color during each of m days.
Input
The first line of input contains three integers n, m and k (2 ≤ n ≤ 200 000, 0 ≤ m ≤ 200 000, 1 ≤ k ≤ 200 000) — the number of socks, the number of days and the number of available colors respectively.
The second line contain n integers c1, c2, ..., cn (1 ≤ ci ≤ k) — current colors of Arseniy's socks.
Each of the following m lines contains two integers li and ri (1 ≤ li, ri ≤ n, li ≠ ri) — indices of socks which Arseniy should wear during the i-th day.
Output
Print one integer — the minimum number of socks that should have their colors changed in order to be able to obey the instructions and not make people laugh from watching the socks of different colors.
Examples
Input
3 2 3
1 2 3
1 2
2 3
Output
2
Input
3 2 2
1 1 2
1 2
2 1
Output
0
Note
In the first sample, Arseniy can repaint the first and the third socks to the second color.
In the second sample, there is no need to change any colors.
Submitted Solution:
```
import math,sys,bisect,heapq,os
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
from functools import lru_cache
#sys.setrecursionlimit(200000000)
int1 = lambda x: -int(x)
def input(): return sys.stdin.readline().rstrip('\r\n')
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
aj = lambda: list(map(int, input().split()))
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
def solve():
aj2 = lambda: list(map(int1, input().split()))
def find(a):
if par[a] < 0:
return a
par[a] = find(par[a])
return par[a]
def merge(a,b):
if a!=b:
#par[a] += par[b]
par[b] = a
n,m,k = aj()
par = [0] + aj2()
ans = 0
s = set()
for ii,i in enumerate(par):
if i in s:
par[ii] = abs(i)
else:
s.add(i)
# print(par)
for i in range(m):
a,b = aj()
aa = find(a)
bb = find(b)
if aa!=bb:
merge(aa,bb)
ans += 1
print(ans)
try:
#os.system("online_judge.py")
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
solve()
``` | instruction | 0 | 74,654 | 7 | 149,308 |
No | output | 1 | 74,654 | 7 | 149,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared a lot of food, left some money and washed all Arseniy's clothes.
Ten minutes before her leave she realized that it would be also useful to prepare instruction of which particular clothes to wear on each of the days she will be absent. Arseniy's family is a bit weird so all the clothes is enumerated. For example, each of Arseniy's n socks is assigned a unique integer from 1 to n. Thus, the only thing his mother had to do was to write down two integers li and ri for each of the days — the indices of socks to wear on the day i (obviously, li stands for the left foot and ri for the right). Each sock is painted in one of k colors.
When mother already left Arseniy noticed that according to instruction he would wear the socks of different colors on some days. Of course, that is a terrible mistake cause by a rush. Arseniy is a smart boy, and, by some magical coincidence, he posses k jars with the paint — one for each of k colors.
Arseniy wants to repaint some of the socks in such a way, that for each of m days he can follow the mother's instructions and wear the socks of the same color. As he is going to be very busy these days he will have no time to change the colors of any socks so he has to finalize the colors now.
The new computer game Bota-3 was just realised and Arseniy can't wait to play it. What is the minimum number of socks that need their color to be changed in order to make it possible to follow mother's instructions and wear the socks of the same color during each of m days.
Input
The first line of input contains three integers n, m and k (2 ≤ n ≤ 200 000, 0 ≤ m ≤ 200 000, 1 ≤ k ≤ 200 000) — the number of socks, the number of days and the number of available colors respectively.
The second line contain n integers c1, c2, ..., cn (1 ≤ ci ≤ k) — current colors of Arseniy's socks.
Each of the following m lines contains two integers li and ri (1 ≤ li, ri ≤ n, li ≠ ri) — indices of socks which Arseniy should wear during the i-th day.
Output
Print one integer — the minimum number of socks that should have their colors changed in order to be able to obey the instructions and not make people laugh from watching the socks of different colors.
Examples
Input
3 2 3
1 2 3
1 2
2 3
Output
2
Input
3 2 2
1 1 2
1 2
2 1
Output
0
Note
In the first sample, Arseniy can repaint the first and the third socks to the second color.
In the second sample, there is no need to change any colors.
Submitted Solution:
```
n, m, k = map(int, input().split())
c = [0] + list(map(int, input().split()))
#socks = [list(map(int, input().split())) for i in range(m)]
ans = 0
for i in range(m):
l, r = map(int, input().split())
if c[l] != c[r]:
ans += 1
print(ans)
``` | instruction | 0 | 74,655 | 7 | 149,310 |
No | output | 1 | 74,655 | 7 | 149,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared a lot of food, left some money and washed all Arseniy's clothes.
Ten minutes before her leave she realized that it would be also useful to prepare instruction of which particular clothes to wear on each of the days she will be absent. Arseniy's family is a bit weird so all the clothes is enumerated. For example, each of Arseniy's n socks is assigned a unique integer from 1 to n. Thus, the only thing his mother had to do was to write down two integers li and ri for each of the days — the indices of socks to wear on the day i (obviously, li stands for the left foot and ri for the right). Each sock is painted in one of k colors.
When mother already left Arseniy noticed that according to instruction he would wear the socks of different colors on some days. Of course, that is a terrible mistake cause by a rush. Arseniy is a smart boy, and, by some magical coincidence, he posses k jars with the paint — one for each of k colors.
Arseniy wants to repaint some of the socks in such a way, that for each of m days he can follow the mother's instructions and wear the socks of the same color. As he is going to be very busy these days he will have no time to change the colors of any socks so he has to finalize the colors now.
The new computer game Bota-3 was just realised and Arseniy can't wait to play it. What is the minimum number of socks that need their color to be changed in order to make it possible to follow mother's instructions and wear the socks of the same color during each of m days.
Input
The first line of input contains three integers n, m and k (2 ≤ n ≤ 200 000, 0 ≤ m ≤ 200 000, 1 ≤ k ≤ 200 000) — the number of socks, the number of days and the number of available colors respectively.
The second line contain n integers c1, c2, ..., cn (1 ≤ ci ≤ k) — current colors of Arseniy's socks.
Each of the following m lines contains two integers li and ri (1 ≤ li, ri ≤ n, li ≠ ri) — indices of socks which Arseniy should wear during the i-th day.
Output
Print one integer — the minimum number of socks that should have their colors changed in order to be able to obey the instructions and not make people laugh from watching the socks of different colors.
Examples
Input
3 2 3
1 2 3
1 2
2 3
Output
2
Input
3 2 2
1 1 2
1 2
2 1
Output
0
Note
In the first sample, Arseniy can repaint the first and the third socks to the second color.
In the second sample, there is no need to change any colors.
Submitted Solution:
```
n, m, k = [int(x) for x in input().split()]
ks = [int(x) for x in input().split()]
lrs = []
res = 0
#[{socks: {s1,s2,s3,..}, colors:{k1: counter, k2: counter,}}...]
clusters = []
for i in range(m):
lr = [int(x) - 1 for x in input().split()]
find_c = False
for c in clusters:
if lr[0] in c['socks'] or lr[1] in c['socks']:
find_c = True
c['socks'].add(lr[0])
c['socks'].add(lr[1])
if ks[lr[0]] not in c['colors']:
c['colors'][ks[lr[0]]] = 1
else:
c['colors'][ks[lr[0]]] += 1
if ks[lr[1]] not in c['colors']:
c['colors'][ks[lr[1]]] = 1
else:
c['colors'][ks[lr[1]]] += 1
if not find_c:
clusters.append({'socks': set(lr), 'colors': {ks[lr[0]]: 1, ks[lr[1]]: 1}})
for c in clusters:
summ = 0
max_counter = 0
for k in c['colors']:
summ += c['colors'][k]
if c['colors'][k] > max_counter:
kk = k
max_counter = c['colors'][k]
res += summ - max_counter
print(res)
``` | instruction | 0 | 74,656 | 7 | 149,312 |
No | output | 1 | 74,656 | 7 | 149,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared a lot of food, left some money and washed all Arseniy's clothes.
Ten minutes before her leave she realized that it would be also useful to prepare instruction of which particular clothes to wear on each of the days she will be absent. Arseniy's family is a bit weird so all the clothes is enumerated. For example, each of Arseniy's n socks is assigned a unique integer from 1 to n. Thus, the only thing his mother had to do was to write down two integers li and ri for each of the days — the indices of socks to wear on the day i (obviously, li stands for the left foot and ri for the right). Each sock is painted in one of k colors.
When mother already left Arseniy noticed that according to instruction he would wear the socks of different colors on some days. Of course, that is a terrible mistake cause by a rush. Arseniy is a smart boy, and, by some magical coincidence, he posses k jars with the paint — one for each of k colors.
Arseniy wants to repaint some of the socks in such a way, that for each of m days he can follow the mother's instructions and wear the socks of the same color. As he is going to be very busy these days he will have no time to change the colors of any socks so he has to finalize the colors now.
The new computer game Bota-3 was just realised and Arseniy can't wait to play it. What is the minimum number of socks that need their color to be changed in order to make it possible to follow mother's instructions and wear the socks of the same color during each of m days.
Input
The first line of input contains three integers n, m and k (2 ≤ n ≤ 200 000, 0 ≤ m ≤ 200 000, 1 ≤ k ≤ 200 000) — the number of socks, the number of days and the number of available colors respectively.
The second line contain n integers c1, c2, ..., cn (1 ≤ ci ≤ k) — current colors of Arseniy's socks.
Each of the following m lines contains two integers li and ri (1 ≤ li, ri ≤ n, li ≠ ri) — indices of socks which Arseniy should wear during the i-th day.
Output
Print one integer — the minimum number of socks that should have their colors changed in order to be able to obey the instructions and not make people laugh from watching the socks of different colors.
Examples
Input
3 2 3
1 2 3
1 2
2 3
Output
2
Input
3 2 2
1 1 2
1 2
2 1
Output
0
Note
In the first sample, Arseniy can repaint the first and the third socks to the second color.
In the second sample, there is no need to change any colors.
Submitted Solution:
```
import logging
import copy
import sys
import math
logging.basicConfig(stream=sys.stderr, level=logging.DEBUG)
#def solve(firstLine):
def solve(colors, days):
union_set = list(range(0,len(colors)))
change_num = [0] * len(colors)
def getHead(n):
while n != union_set[n]:
n = union_set[n]
return n
def getColor(n):
while n != union_set[n]:
n = union_set[n]
return colors[n]
count = 0
for d in days:
c1, c2 = d[0]-1, d[1]-1
if getColor(c1) != getColor(c2):
h1, h2 = getHead(c1), getHead(c2)
if h1 < h2:
union_set[c2] = h1
change_num[c2] += 1
else:
union_set[c1] = h2
change_num[c1] += 1
head_map = {}
for i,n in enumerate(union_set):
if i == n:
continue
head = n
while head != union_set[head]:
head = union_set[head]
if head not in head_map:
head_map[head] = [colors[head]]
else:
head_map[head].append(colors[i])
for h, unions in head_map.items():
unions.append(h)
color_count = {}
for n in unions:
if n not in color_count:
color_count[n] = 1
else:
color_count[n] += 1
max_n = max(list(color_count.values()))
count += (len(unions) - max_n)
count -= sum(list(filter(lambda x: x>1, change_num)))
return count
def main():
firstLine = input().split()
firstLine = list(map(int, firstLine))
colors = input().split()
colors = list(map(int, colors))
lines = []
for i in range(firstLine[1]):
l = input().split()
l = list(map(int, l))
lines.append(l)
print(solve(colors, lines))
def log(*message):
logging.debug(message)
if __name__ == "__main__":
main()
``` | instruction | 0 | 74,657 | 7 | 149,314 |
No | output | 1 | 74,657 | 7 | 149,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A brick is defined as a rectangle with integer side lengths with either width 1 or height 1 (or both).
There is an n× m grid, and each cell is colored either black or white. A tiling is a way to place bricks onto the grid such that each black cell is covered by exactly one brick, and each white cell is not covered by any brick. In other words, bricks are placed on black cells only, cover all black cells, and no two bricks overlap.
<image> An example tiling of the first test case using 5 bricks. It is possible to do better, using only 4 bricks.
What is the minimum number of bricks required to make a valid tiling?
Input
The first line contains two integers n, m (1≤ n, m≤ 200) — the number of rows and columns, respectively.
The next n lines describe the grid. The i-th of these lines contains a string of length m, where the j-th character denotes the color of the cell in row i, column j. A black cell is given by "#", and a white cell is given by ".".
It is guaranteed that there is at least one black cell.
Output
Output a single integer, the minimum number of bricks required.
Examples
Input
3 4
#.##
####
##..
Output
4
Input
6 6
######
##....
######
##...#
##...#
######
Output
6
Input
10 8
####..##
#..#.##.
#..#.###
####.#.#
....####
.###.###
###.#..#
########
###..###
.##.###.
Output
18
Note
The first test case can be tiled with 4 bricks placed vertically.
The third test case can be tiled with 18 bricks like this:
<image>
Submitted Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
INF = float("inf")
class Dinic:
def __init__(self, n):
self.lvl = [0] * n
self.ptr = [0] * n
self.q = [0] * n
self.adj = [[] for _ in range(n)]
def add_edge(self, a, b, c, rcap=0):
self.adj[a].append([b, len(self.adj[b]), c, 0])
self.adj[b].append([a, len(self.adj[a]) - 1, rcap, 0])
def dfs(self, v, t, f):
if v == t or not f:
return f
for i in range(self.ptr[v], len(self.adj[v])):
e = self.adj[v][i]
if self.lvl[e[0]] == self.lvl[v] + 1:
p = self.dfs(e[0], t, min(f, e[2] - e[3]))
if p:
self.adj[v][i][3] += p
self.adj[e[0]][e[1]][3] -= p
return p
self.ptr[v] += 1
return 0
def calc(self, s, t):
flow, self.q[0] = 0, s
for l in range(31): # l = 30 maybe faster for random data
while True:
self.lvl, self.ptr = [0] * len(self.q), [0] * len(self.q)
qi, qe, self.lvl[s] = 0, 1, 1
while qi < qe and not self.lvl[t]:
v = self.q[qi]
qi += 1
for e in self.adj[v]:
if not self.lvl[e[0]] and (e[2] - e[3]) >> (30 - l):
self.q[qe] = e[0]
qe += 1
self.lvl[e[0]] = self.lvl[v] + 1
p = self.dfs(s, t, INF)
while p:
flow += p
p = self.dfs(s, t, INF)
if not self.lvl[t]:
break
return flow
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def main():
n,m = map(int,input().split())
grid = []
for _ in range(n):
grid.append(input())
indexToPair = [] # x,y,dir tuple, always take left one's or upper one's index and dir = 0 horizontal 1 vetrical
pairToIndex = {}
cellCount = 0
for i in range(n):
for j in range(m):
if i + 1 < n and grid[i][j] == '#' and grid[i+1][j] == '#':
indexToPair.append((i,j,1))
pairToIndex[(i,j,1)] = len(indexToPair) - 1
if j + 1 < m and grid[i][j] == '#' and grid[i][j+1] == '#':
indexToPair.append((i,j,0))
pairToIndex[(i,j,0)] = len(indexToPair) - 1
if grid[i][j] == '#':
cellCount += 1
flowGraph = Dinic(len(indexToPair) + 2)
for i in range(n):
for j in range(m):
if i + 1 < n and grid[i][j] == '#' and grid[i+1][j] == '#':
flowGraph.add_edge(pairToIndex[(i,j,1)],len(indexToPair),1,1)
if j + 1 < m and grid[i][j] == '#' and grid[i][j+1] == '#':
flowGraph.add_edge(pairToIndex[(i,j,0)],len(indexToPair) + 1,1,1)
for i in range(n - 1):
for j in range(m - 1):
if grid[i][j] == '#' and grid[i+1][j] == '#' and grid[i+1][j+1] == '#':
flowGraph.add_edge(pairToIndex[(i,j,1)],pairToIndex[(i+1,j,0)],1,1)
if grid[i][j] == '#' and grid[i+1][j] == '#' and grid[i][j+1] == '#':
flowGraph.add_edge(pairToIndex[(i,j,1)],pairToIndex[(i,j,0)],1,1)
if grid[i][j] == '#' and grid[i][j+1] == '#' and grid[i+1][j+1] == '#':
flowGraph.add_edge(pairToIndex[(i,j,0)],pairToIndex[(i,j+1,1)],1,1)
if grid[i+1][j] == '#' and grid[i][j+1] == '#' and grid[i+1][j+1] == '#':
flowGraph.add_edge(pairToIndex[(i+1,j,0)],pairToIndex[(i,j+1,1)],1,1)
print(cellCount - len(indexToPair) + flowGraph.calc(len(indexToPair),len(indexToPair) + 1))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 75,205 | 7 | 150,410 |
No | output | 1 | 75,205 | 7 | 150,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A brick is defined as a rectangle with integer side lengths with either width 1 or height 1 (or both).
There is an n× m grid, and each cell is colored either black or white. A tiling is a way to place bricks onto the grid such that each black cell is covered by exactly one brick, and each white cell is not covered by any brick. In other words, bricks are placed on black cells only, cover all black cells, and no two bricks overlap.
<image> An example tiling of the first test case using 5 bricks. It is possible to do better, using only 4 bricks.
What is the minimum number of bricks required to make a valid tiling?
Input
The first line contains two integers n, m (1≤ n, m≤ 200) — the number of rows and columns, respectively.
The next n lines describe the grid. The i-th of these lines contains a string of length m, where the j-th character denotes the color of the cell in row i, column j. A black cell is given by "#", and a white cell is given by ".".
It is guaranteed that there is at least one black cell.
Output
Output a single integer, the minimum number of bricks required.
Examples
Input
3 4
#.##
####
##..
Output
4
Input
6 6
######
##....
######
##...#
##...#
######
Output
6
Input
10 8
####..##
#..#.##.
#..#.###
####.#.#
....####
.###.###
###.#..#
########
###..###
.##.###.
Output
18
Note
The first test case can be tiled with 4 bricks placed vertically.
The third test case can be tiled with 18 bricks like this:
<image>
Submitted Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
INF = float("inf")
class Dinic:
def __init__(self, n):
self.lvl = [0] * n
self.ptr = [0] * n
self.q = [0] * n
self.adj = [[] for _ in range(n)]
def add_edge(self, a, b, c, rcap=0):
self.adj[a].append([b, len(self.adj[b]), c, 0])
self.adj[b].append([a, len(self.adj[a]) - 1, rcap, 0])
def dfs(self, v, t, f):
if v == t or not f:
return f
for i in range(self.ptr[v], len(self.adj[v])):
e = self.adj[v][i]
if self.lvl[e[0]] == self.lvl[v] + 1:
p = self.dfs(e[0], t, min(f, e[2] - e[3]))
if p:
self.adj[v][i][3] += p
self.adj[e[0]][e[1]][3] -= p
return p
self.ptr[v] += 1
return 0
def calc(self, s, t):
flow, self.q[0] = 0, s
for l in range(31): # l = 30 maybe faster for random data
while True:
self.lvl, self.ptr = [0] * len(self.q), [0] * len(self.q)
qi, qe, self.lvl[s] = 0, 1, 1
while qi < qe and not self.lvl[t]:
v = self.q[qi]
qi += 1
for e in self.adj[v]:
if not self.lvl[e[0]] and (e[2] - e[3]) >> (30 - l):
self.q[qe] = e[0]
qe += 1
self.lvl[e[0]] = self.lvl[v] + 1
p = self.dfs(s, t, INF)
while p:
flow += p
p = self.dfs(s, t, INF)
if not self.lvl[t]:
break
return flow
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def main():
import random
n,m = map(int,input().split())
grid = []
for _ in range(n):
grid.append("")
for i in range(m):
if random.random() < 0.9:
grid[_] += "#"
else:
grid[_] += "."
cnt = 0
pairToIndex = [0] * 130000
cellCount = 0
for i in range(n):
for j in range(m):
if i + 1 < n and grid[i][j] == '#' and grid[i+1][j] == '#':
cnt += 1
pairToIndex[2 * 300 * i + 2 * j + 1] = cnt - 1
if j + 1 < m and grid[i][j] == '#' and grid[i][j+1] == '#':
cnt += 1
pairToIndex[2 * 300 * i + 2 * j] = cnt - 1
if grid[i][j] == '#':
cellCount += 1
flowGraph = Dinic(cnt + 2)
for i in range(n):
for j in range(m):
if i + 1 < n and grid[i][j] == '#' and grid[i+1][j] == '#':
flowGraph.add_edge(cnt,pairToIndex[2 * 300 * i + 2 * j + 1],1)
if j + 1 < m and grid[i][j] == '#' and grid[i][j+1] == '#':
flowGraph.add_edge(pairToIndex[2 * 300 * i + 2 * j],cnt + 1,1)
for i in range(n - 1):
for j in range(m - 1):
if grid[i][j] == '#' and grid[i+1][j] == '#' and grid[i+1][j+1] == '#':
flowGraph.add_edge(pairToIndex[2 * 300 * i + 2 * j + 1],pairToIndex[2 * 300 * (i+1) + 2 * j],1)
if grid[i][j] == '#' and grid[i+1][j] == '#' and grid[i][j+1] == '#':
flowGraph.add_edge(pairToIndex[2 * 300 * i + 2 * j + 1],pairToIndex[2 * 300 * i + 2 * j],1)
if grid[i][j] == '#' and grid[i][j+1] == '#' and grid[i+1][j+1] == '#':
flowGraph.add_edge(pairToIndex[2 * 300 * i + 2 * (j + 1) + 1],pairToIndex[2 * 300 * i + 2 * j],1)
if grid[i+1][j] == '#' and grid[i][j+1] == '#' and grid[i+1][j+1] == '#':
flowGraph.add_edge(pairToIndex[2 * 300 * i + 2 * (j + 1) + 1],pairToIndex[2 * 300 * (i+1) + 2 * j],1)
print(cellCount - cnt + flowGraph.calc(cnt,cnt + 1))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 75,206 | 7 | 150,412 |
No | output | 1 | 75,206 | 7 | 150,413 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | instruction | 0 | 75,353 | 7 | 150,706 |
Tags: math
Correct Solution:
```
def gcd(x, y):
return (x if y == 0 else gcd(y, x % y))
x, y, a, b = map(int, input().split())
g = x * y // gcd(x, y)
print(b // g - (a - 1) // g)
``` | output | 1 | 75,353 | 7 | 150,707 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | instruction | 0 | 75,354 | 7 | 150,708 |
Tags: math
Correct Solution:
```
x, y, a, b = (int(i) for i in input().split())
p = 1
for i in range(2, 1001):
if x % i == 0 and y % i == 0:
x = x // i
y = y // i
p *= i
print(b // (x * y * p) - (a - 1) // (x * y * p))
``` | output | 1 | 75,354 | 7 | 150,709 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | instruction | 0 | 75,355 | 7 | 150,710 |
Tags: math
Correct Solution:
```
def gcd(x, y):
while y:
x, y = y, x % y
return x
x, y, a, b = map(int, input().split())
n = x * y // gcd(x, y)
print(b // n - (a - 1) // n)
``` | output | 1 | 75,355 | 7 | 150,711 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | instruction | 0 | 75,356 | 7 | 150,712 |
Tags: math
Correct Solution:
```
from math import*
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
x, y, a, b = list(map(int, input().split()))
nok = (x * y) // gcd(x, y)
a += (nok - a % nok) % nok
b -= b % nok
print((b - a) // nok + 1)
``` | output | 1 | 75,356 | 7 | 150,713 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | instruction | 0 | 75,357 | 7 | 150,714 |
Tags: math
Correct Solution:
```
def gcd(a,b):
if b==a:
return a
else:
if a>b:
return gcd(a-b,b)
else:
return gcd(a,b-a)
inpList = input()
inp = inpList.split()
x = int(inp[0])
y = int(inp[1])
a = int(inp[2])
b = int(inp[3])
g = gcd(x,y)
l = (x*y)/g
count = int(b/l) - int((a-1)/l)
print(count)
``` | output | 1 | 75,357 | 7 | 150,715 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | instruction | 0 | 75,358 | 7 | 150,716 |
Tags: math
Correct Solution:
```
# coding: utf-8
def gcd(a,b):
if a > b:
a, b = b, a
for i in range(a,0,-1):
if a%i==0 and b%i==0:
return i;
def lcm(a,b):
return a*b//gcd(a,b)
x, y, a, b = [int(i) for i in input().split()]
common = lcm(x,y)
if a%common !=0 :
a = a+common-a%common
b = b-b%common
print((b-a)//common+1)
``` | output | 1 | 75,358 | 7 | 150,717 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | instruction | 0 | 75,359 | 7 | 150,718 |
Tags: math
Correct Solution:
```
def gcd(a,b):
a,b=min(a,b),max(a,b)
while a>0:
a,b=b%a,a
return b
a,b,c,d=map(int,input().split())
t=(a*b//gcd(a,b))
x=c//t
y=d//t
if c%t==0 :
print(y-x+1)
else:
print(y-x)
``` | output | 1 | 75,359 | 7 | 150,719 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18. | instruction | 0 | 75,360 | 7 | 150,720 |
Tags: math
Correct Solution:
```
from fractions import gcd
def main():
x, y, a, b = map(int, input().split())
xy = x * y // gcd(x, y)
print(b // xy - (a - 1) // xy)
if __name__ == '__main__':
main()
``` | output | 1 | 75,360 | 7 | 150,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18.
Submitted Solution:
```
#Time:2014/08/29
x,y,a,b = input().split()
a=int(a)
b=int(b)
x=int(x)
y=int(y)
if x<y :
r=x
x=y
y=r
g=x*y
r=x%y
while r!=0 :
x=y
y=r
r=x%y
g=int(g/y)
if g>=a :
num=int((b-g)/g)+1
if g>b :
num=0
else :
if a%g==0 :
k=a
else :
k=int(a/g+1)*g
num=int((b-k)/g)+1
if k>b :
num=0
print(num)
``` | instruction | 0 | 75,361 | 7 | 150,722 |
Yes | output | 1 | 75,361 | 7 | 150,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18.
Submitted Solution:
```
def GCD(a,b):
while True:
s = a%b
a = b
b = s
# print(a,b)
if b == 0:
return a
x,y,a,b = map(int,input().split())
# print(GCD(2,3))
LCM = int(x*y//GCD(x,y))
print(int(b//LCM-(a-1)//LCM))
# print(LCM)
``` | instruction | 0 | 75,362 | 7 | 150,724 |
Yes | output | 1 | 75,362 | 7 | 150,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18.
Submitted Solution:
```
import math
x, y, a, b = map(int, input().split())
l = max(x, y) * min(x, y) // math.gcd(x, y)
print(b//l - (a-1)//l)
``` | instruction | 0 | 75,363 | 7 | 150,726 |
Yes | output | 1 | 75,363 | 7 | 150,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18.
Submitted Solution:
```
from fractions import gcd
x, y, a, b = [int(x) for x in input().split()]
d = x * y // gcd(x, y)
if a % d == 0:
lowerFactor = a // d
else:
lowerFactor = a // d + 1
higherFactor = b // d
print(higherFactor - lowerFactor + 1)
``` | instruction | 0 | 75,364 | 7 | 150,728 |
Yes | output | 1 | 75,364 | 7 | 150,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18.
Submitted Solution:
```
x, y, a, b = [int(x) for x in input().split()]
g = x*y
for i in range(a, a+g):
if i%g == 0:
m = i
break
print((b-m)//g+1)
``` | instruction | 0 | 75,365 | 7 | 150,730 |
No | output | 1 | 75,365 | 7 | 150,731 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x - 1 consecutive bricks, then he paints the x-th one. That is, he'll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y - 1 consecutive bricks, then he paints the y-th one. Hence he'll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
Input
The input will have a single line containing four integers in this order: x, y, a, b. (1 ≤ x, y ≤ 1000, 1 ≤ a, b ≤ 2·109, a ≤ b).
Output
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
Examples
Input
2 3 6 18
Output
3
Note
Let's look at the bricks from a to b (a = 6, b = 18). The bricks colored in red are numbered 6, 8, 10, 12, 14, 16, 18. The bricks colored in pink are numbered 6, 9, 12, 15, 18. The bricks colored in both red and pink are numbered with 6, 12 and 18.
Submitted Solution:
```
import fractions
x, y, a, b = [int(_) for _ in input().split()]
print(int((b-a+1)/(x*y)/fractions.gcd(x, y))+1)
``` | instruction | 0 | 75,366 | 7 | 150,732 |
No | output | 1 | 75,366 | 7 | 150,733 |
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