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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Boboniu gives you
* r red balls,
* g green balls,
* b blue balls,
* w white balls.
He allows you to do the following operation as many times as you want:
* Pick a red ball, a green ball, and a blue ball and then change their color to white.
You should answer if it's possible to arrange all the balls into a palindrome after several (possibly zero) number of described operations.
Input
The first line contains one integer T (1≤ T≤ 100) denoting the number of test cases.
For each of the next T cases, the first line contains four integers r, g, b and w (0≤ r,g,b,w≤ 10^9).
Output
For each test case, print "Yes" if it's possible to arrange all the balls into a palindrome after doing several (possibly zero) number of described operations. Otherwise, print "No".
Example
Input
4
0 1 1 1
8 1 9 3
0 0 0 0
1000000000 1000000000 1000000000 1000000000
Output
No
Yes
Yes
Yes
Note
In the first test case, you're not able to do any operation and you can never arrange three balls of distinct colors into a palindrome.
In the second test case, after doing one operation, changing (8,1,9,3) to (7,0,8,6), one of those possible palindromes may be "rrrwwwbbbbrbbbbwwwrrr".
A palindrome is a word, phrase, or sequence that reads the same backwards as forwards. For example, "rggbwbggr", "b", "gg" are palindromes while "rgbb", "gbbgr" are not. Notice that an empty word, phrase, or sequence is palindrome. | instruction | 0 | 81,893 | 7 | 163,786 |
Tags: brute force, math
Correct Solution:
```
for i in range(int(input())):
a=list(map(int,input().split()))
b=[a[j]%2 for j in range(len(a))]
if b.count(0)==4:
print("Yes")
elif b.count(0)==3:
print("Yes")
elif min(a[:3])>=1:
a[0]-=1
a[1]-=1
a[2]-=1
a[3]+=3
c=[a[j]%2 for j in range(4)]
if c.count(0)==4:
print("Yes")
elif c.count(0)==3:
print("Yes")
else:
print("No")
else:
print("No")
``` | output | 1 | 81,893 | 7 | 163,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu gives you
* r red balls,
* g green balls,
* b blue balls,
* w white balls.
He allows you to do the following operation as many times as you want:
* Pick a red ball, a green ball, and a blue ball and then change their color to white.
You should answer if it's possible to arrange all the balls into a palindrome after several (possibly zero) number of described operations.
Input
The first line contains one integer T (1≤ T≤ 100) denoting the number of test cases.
For each of the next T cases, the first line contains four integers r, g, b and w (0≤ r,g,b,w≤ 10^9).
Output
For each test case, print "Yes" if it's possible to arrange all the balls into a palindrome after doing several (possibly zero) number of described operations. Otherwise, print "No".
Example
Input
4
0 1 1 1
8 1 9 3
0 0 0 0
1000000000 1000000000 1000000000 1000000000
Output
No
Yes
Yes
Yes
Note
In the first test case, you're not able to do any operation and you can never arrange three balls of distinct colors into a palindrome.
In the second test case, after doing one operation, changing (8,1,9,3) to (7,0,8,6), one of those possible palindromes may be "rrrwwwbbbbrbbbbwwwrrr".
A palindrome is a word, phrase, or sequence that reads the same backwards as forwards. For example, "rggbwbggr", "b", "gg" are palindromes while "rgbb", "gbbgr" are not. Notice that an empty word, phrase, or sequence is palindrome.
Submitted Solution:
```
def check(r,g,b,w):
if r%2+g%2+b%2+w%2>1:
return False
return True
T = int(input())
for ttt in range(T):
r,g,b,w = map(int,input().split())
if check(r,g,b,w):
print("Yes")
elif r>0 and g>0 and b>0 and check(r-1,g-1,b-1,w+1):
print("Yes")
else :
print("No")
``` | instruction | 0 | 81,894 | 7 | 163,788 |
Yes | output | 1 | 81,894 | 7 | 163,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu gives you
* r red balls,
* g green balls,
* b blue balls,
* w white balls.
He allows you to do the following operation as many times as you want:
* Pick a red ball, a green ball, and a blue ball and then change their color to white.
You should answer if it's possible to arrange all the balls into a palindrome after several (possibly zero) number of described operations.
Input
The first line contains one integer T (1≤ T≤ 100) denoting the number of test cases.
For each of the next T cases, the first line contains four integers r, g, b and w (0≤ r,g,b,w≤ 10^9).
Output
For each test case, print "Yes" if it's possible to arrange all the balls into a palindrome after doing several (possibly zero) number of described operations. Otherwise, print "No".
Example
Input
4
0 1 1 1
8 1 9 3
0 0 0 0
1000000000 1000000000 1000000000 1000000000
Output
No
Yes
Yes
Yes
Note
In the first test case, you're not able to do any operation and you can never arrange three balls of distinct colors into a palindrome.
In the second test case, after doing one operation, changing (8,1,9,3) to (7,0,8,6), one of those possible palindromes may be "rrrwwwbbbbrbbbbwwwrrr".
A palindrome is a word, phrase, or sequence that reads the same backwards as forwards. For example, "rggbwbggr", "b", "gg" are palindromes while "rgbb", "gbbgr" are not. Notice that an empty word, phrase, or sequence is palindrome.
Submitted Solution:
```
for _ in range(int(input())):
rgb = list(map(int, input().split()))
a = sum([1 for i in rgb if i%2 == 0])
if a==2:
print('NO')
elif a==1 and (rgb[0] == 0 or rgb[1] == 0 or rgb[2] == 0):
print('NO')
else:
print('YES')
``` | instruction | 0 | 81,895 | 7 | 163,790 |
Yes | output | 1 | 81,895 | 7 | 163,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu gives you
* r red balls,
* g green balls,
* b blue balls,
* w white balls.
He allows you to do the following operation as many times as you want:
* Pick a red ball, a green ball, and a blue ball and then change their color to white.
You should answer if it's possible to arrange all the balls into a palindrome after several (possibly zero) number of described operations.
Input
The first line contains one integer T (1≤ T≤ 100) denoting the number of test cases.
For each of the next T cases, the first line contains four integers r, g, b and w (0≤ r,g,b,w≤ 10^9).
Output
For each test case, print "Yes" if it's possible to arrange all the balls into a palindrome after doing several (possibly zero) number of described operations. Otherwise, print "No".
Example
Input
4
0 1 1 1
8 1 9 3
0 0 0 0
1000000000 1000000000 1000000000 1000000000
Output
No
Yes
Yes
Yes
Note
In the first test case, you're not able to do any operation and you can never arrange three balls of distinct colors into a palindrome.
In the second test case, after doing one operation, changing (8,1,9,3) to (7,0,8,6), one of those possible palindromes may be "rrrwwwbbbbrbbbbwwwrrr".
A palindrome is a word, phrase, or sequence that reads the same backwards as forwards. For example, "rggbwbggr", "b", "gg" are palindromes while "rgbb", "gbbgr" are not. Notice that an empty word, phrase, or sequence is palindrome.
Submitted Solution:
```
q=lambda:int(r&1==1)+int(g&1==1)+int(b&1==1)+int(w&1==1)
for i in range(int(input())):
t=1
r,g,b,w=map(int,input().split())
if q()==2or(q()==3and r*g*b==0):t=0
print("YES"if t else"NO")
``` | instruction | 0 | 81,896 | 7 | 163,792 |
Yes | output | 1 | 81,896 | 7 | 163,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu gives you
* r red balls,
* g green balls,
* b blue balls,
* w white balls.
He allows you to do the following operation as many times as you want:
* Pick a red ball, a green ball, and a blue ball and then change their color to white.
You should answer if it's possible to arrange all the balls into a palindrome after several (possibly zero) number of described operations.
Input
The first line contains one integer T (1≤ T≤ 100) denoting the number of test cases.
For each of the next T cases, the first line contains four integers r, g, b and w (0≤ r,g,b,w≤ 10^9).
Output
For each test case, print "Yes" if it's possible to arrange all the balls into a palindrome after doing several (possibly zero) number of described operations. Otherwise, print "No".
Example
Input
4
0 1 1 1
8 1 9 3
0 0 0 0
1000000000 1000000000 1000000000 1000000000
Output
No
Yes
Yes
Yes
Note
In the first test case, you're not able to do any operation and you can never arrange three balls of distinct colors into a palindrome.
In the second test case, after doing one operation, changing (8,1,9,3) to (7,0,8,6), one of those possible palindromes may be "rrrwwwbbbbrbbbbwwwrrr".
A palindrome is a word, phrase, or sequence that reads the same backwards as forwards. For example, "rggbwbggr", "b", "gg" are palindromes while "rgbb", "gbbgr" are not. Notice that an empty word, phrase, or sequence is palindrome.
Submitted Solution:
```
t=int(input())
for i in range(t):
r,g,b,w=map(int,input().split())
c=0
val=min(r,g,b)
if(r%2==1):
c+=1
if(g%2==1):
c+=1
if(b%2==1):
c+=1
if(w%2==1):
c+=1
if(c<=1):
print('Yes')
elif(val!=0):
r-=1
g-=1
b-=1
w-=1
c=0
if(r%2==1):
c+=1
if(g%2==1):
c+=1
if(b%2==1):
c+=1
if(w%2==1):
c+=1
if(c<=1):
print('Yes')
else:
print('No')
else:
print('No')
``` | instruction | 0 | 81,897 | 7 | 163,794 |
Yes | output | 1 | 81,897 | 7 | 163,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu gives you
* r red balls,
* g green balls,
* b blue balls,
* w white balls.
He allows you to do the following operation as many times as you want:
* Pick a red ball, a green ball, and a blue ball and then change their color to white.
You should answer if it's possible to arrange all the balls into a palindrome after several (possibly zero) number of described operations.
Input
The first line contains one integer T (1≤ T≤ 100) denoting the number of test cases.
For each of the next T cases, the first line contains four integers r, g, b and w (0≤ r,g,b,w≤ 10^9).
Output
For each test case, print "Yes" if it's possible to arrange all the balls into a palindrome after doing several (possibly zero) number of described operations. Otherwise, print "No".
Example
Input
4
0 1 1 1
8 1 9 3
0 0 0 0
1000000000 1000000000 1000000000 1000000000
Output
No
Yes
Yes
Yes
Note
In the first test case, you're not able to do any operation and you can never arrange three balls of distinct colors into a palindrome.
In the second test case, after doing one operation, changing (8,1,9,3) to (7,0,8,6), one of those possible palindromes may be "rrrwwwbbbbrbbbbwwwrrr".
A palindrome is a word, phrase, or sequence that reads the same backwards as forwards. For example, "rggbwbggr", "b", "gg" are palindromes while "rgbb", "gbbgr" are not. Notice that an empty word, phrase, or sequence is palindrome.
Submitted Solution:
```
import sys
t=int(sys.stdin.readline())
for _ in range(t):
a=list(map(int,input().split()))
e=0
o=0
ze=-1
for i in range(len(a)):
if a[i]%2==0:
e+=1
else:
o+=1
if a[i]==0 and i!=4:
ze=i
if e>=3 or (o>=3 and ze==-1):
print('YES')
else:
print('NO')
``` | instruction | 0 | 81,898 | 7 | 163,796 |
No | output | 1 | 81,898 | 7 | 163,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu gives you
* r red balls,
* g green balls,
* b blue balls,
* w white balls.
He allows you to do the following operation as many times as you want:
* Pick a red ball, a green ball, and a blue ball and then change their color to white.
You should answer if it's possible to arrange all the balls into a palindrome after several (possibly zero) number of described operations.
Input
The first line contains one integer T (1≤ T≤ 100) denoting the number of test cases.
For each of the next T cases, the first line contains four integers r, g, b and w (0≤ r,g,b,w≤ 10^9).
Output
For each test case, print "Yes" if it's possible to arrange all the balls into a palindrome after doing several (possibly zero) number of described operations. Otherwise, print "No".
Example
Input
4
0 1 1 1
8 1 9 3
0 0 0 0
1000000000 1000000000 1000000000 1000000000
Output
No
Yes
Yes
Yes
Note
In the first test case, you're not able to do any operation and you can never arrange three balls of distinct colors into a palindrome.
In the second test case, after doing one operation, changing (8,1,9,3) to (7,0,8,6), one of those possible palindromes may be "rrrwwwbbbbrbbbbwwwrrr".
A palindrome is a word, phrase, or sequence that reads the same backwards as forwards. For example, "rggbwbggr", "b", "gg" are palindromes while "rgbb", "gbbgr" are not. Notice that an empty word, phrase, or sequence is palindrome.
Submitted Solution:
```
n = int(input())
for i in range(n):
arr = list(map(int,input().split()))
even = 0
for i in arr:
even += i%2
if even >= 3 or even == 0:
print('Yes')
elif even==1 and (arr[0]*arr[1]*arr[2]) > 0:
print('Yes')
else:
print('No')
``` | instruction | 0 | 81,899 | 7 | 163,798 |
No | output | 1 | 81,899 | 7 | 163,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu gives you
* r red balls,
* g green balls,
* b blue balls,
* w white balls.
He allows you to do the following operation as many times as you want:
* Pick a red ball, a green ball, and a blue ball and then change their color to white.
You should answer if it's possible to arrange all the balls into a palindrome after several (possibly zero) number of described operations.
Input
The first line contains one integer T (1≤ T≤ 100) denoting the number of test cases.
For each of the next T cases, the first line contains four integers r, g, b and w (0≤ r,g,b,w≤ 10^9).
Output
For each test case, print "Yes" if it's possible to arrange all the balls into a palindrome after doing several (possibly zero) number of described operations. Otherwise, print "No".
Example
Input
4
0 1 1 1
8 1 9 3
0 0 0 0
1000000000 1000000000 1000000000 1000000000
Output
No
Yes
Yes
Yes
Note
In the first test case, you're not able to do any operation and you can never arrange three balls of distinct colors into a palindrome.
In the second test case, after doing one operation, changing (8,1,9,3) to (7,0,8,6), one of those possible palindromes may be "rrrwwwbbbbrbbbbwwwrrr".
A palindrome is a word, phrase, or sequence that reads the same backwards as forwards. For example, "rggbwbggr", "b", "gg" are palindromes while "rgbb", "gbbgr" are not. Notice that an empty word, phrase, or sequence is palindrome.
Submitted Solution:
```
def mi(): return map(int, input().split())
def ii(): return int(input())
def li(): return list(map(int, input().split()))
def si(): return input().split()
#######################################################
t = ii()
for _ in range(t):
r, g, b, w = mi()
s = r+g+b+w
if s%2==0:
if r%2==0 and b%2==0 and w%2==0 and g%2==0:
print("Yes")
continue
else:
odd = 0
for i in [r,g,w,b]:
if i%2!=0:
odd+=1
if odd==1:
print("Yes")
continue
ma = min(r,g,b)
r-=ma
g-=ma
b-=ma
w += 3*ma
s = w+r+g+b
if s%2==0:
if r%2==0 and b%2==0 and w%2==0 and g%2==0:
print("Yes")
else:
print("No")
else:
odd = 0
for i in [r,g,w,b]:
if i%2!=0:
odd+=1
if odd==1:
print("Yes")
else:
print("No")
``` | instruction | 0 | 81,900 | 7 | 163,800 |
No | output | 1 | 81,900 | 7 | 163,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Boboniu gives you
* r red balls,
* g green balls,
* b blue balls,
* w white balls.
He allows you to do the following operation as many times as you want:
* Pick a red ball, a green ball, and a blue ball and then change their color to white.
You should answer if it's possible to arrange all the balls into a palindrome after several (possibly zero) number of described operations.
Input
The first line contains one integer T (1≤ T≤ 100) denoting the number of test cases.
For each of the next T cases, the first line contains four integers r, g, b and w (0≤ r,g,b,w≤ 10^9).
Output
For each test case, print "Yes" if it's possible to arrange all the balls into a palindrome after doing several (possibly zero) number of described operations. Otherwise, print "No".
Example
Input
4
0 1 1 1
8 1 9 3
0 0 0 0
1000000000 1000000000 1000000000 1000000000
Output
No
Yes
Yes
Yes
Note
In the first test case, you're not able to do any operation and you can never arrange three balls of distinct colors into a palindrome.
In the second test case, after doing one operation, changing (8,1,9,3) to (7,0,8,6), one of those possible palindromes may be "rrrwwwbbbbrbbbbwwwrrr".
A palindrome is a word, phrase, or sequence that reads the same backwards as forwards. For example, "rggbwbggr", "b", "gg" are palindromes while "rgbb", "gbbgr" are not. Notice that an empty word, phrase, or sequence is palindrome.
Submitted Solution:
```
n = int(input())
for i in range(n):
arr = list(map(int,input().split()))
if sum(arr) == 0 or sum(arr) == 1:
print("YES")
else:
white = arr[-1]
minimum = min(arr[:3])
if minimum != 0:
for i in range(3):
arr[i] -= minimum
arr.remove(0)
white += 3*minimum
if white%2!=0:
if arr[0]%2!=0 or arr[1]%2!=0:
print("NO")
else:
print('YES')
else:
if arr[0]%2==0 or arr[1]%2==0:
print('YES')
else:
print('NO')
``` | instruction | 0 | 81,901 | 7 | 163,802 |
No | output | 1 | 81,901 | 7 | 163,803 |
Provide tags and a correct Python 3 solution for this coding contest problem.
John Doe has an n × m table. John Doe can paint points in some table cells, not more than one point in one table cell. John Doe wants to use such operations to make each square subtable of size n × n have exactly k points.
John Doe wondered, how many distinct ways to fill the table with points are there, provided that the condition must hold. As this number can be rather large, John Doe asks to find its remainder after dividing by 1000000007 (109 + 7).
You should assume that John always paints a point exactly in the center of some cell. Two ways to fill a table are considered distinct, if there exists a table cell, that has a point in one way and doesn't have it in the other.
Input
A single line contains space-separated integers n, m, k (1 ≤ n ≤ 100; n ≤ m ≤ 1018; 0 ≤ k ≤ n2) — the number of rows of the table, the number of columns of the table and the number of points each square must contain.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print a single integer — the remainder from dividing the described number of ways by 1000000007 (109 + 7).
Examples
Input
5 6 1
Output
45
Note
Let's consider the first test case:
<image> The gray area belongs to both 5 × 5 squares. So, if it has one point, then there shouldn't be points in any other place. If one of the white areas has a point, then the other one also must have a point. Thus, there are about 20 variants, where the point lies in the gray area and 25 variants, where each of the white areas contains a point. Overall there are 45 variants. | instruction | 0 | 82,014 | 7 | 164,028 |
Tags: bitmasks, combinatorics, dp, math
Correct Solution:
```
n,m,k=map(int,input().split())
M=int(1e9+7)
N=n*n
iv=[0]*(N+1)
iv[1]=1
for i in range(2, N+1):
iv[i]=M-M//i*iv[M%i]%M
f1=[1]*(N+1)
for i in range(1, N+1):
f1[i]=f1[i-1]*i%M
f2=[1]*(N+1)
for i in range(1, N+1):
f2[i]=f2[i-1]*iv[i]%M
left=m%n
#m/n+1, m/n
def powM(b, p):
r=1
while p>0:
if p%2>0:
r=r*b%M
b=b*b%M
p//=2
return r
c=[[powM(f1[n]*f2[j]%M*f2[n-j]%M, m//n+i) for j in range(n+1)] for i in range(2)]
#print(c)
dp=[[0]*(k+1) for i in range(n+1)]
dp[0][0]=1
for i in range(n):
for j in range(k+1):
#prune
if j>i*n or j<k-(n-i)*n:
continue
for l in range(min(n, k-j)+1):
# i,j -> i+1,j+l
dp[i+1][j+l]=(dp[i+1][j+l]+c[i<left][l]*dp[i][j])%M
print(dp[n][k])
``` | output | 1 | 82,014 | 7 | 164,029 |
Provide a correct Python 3 solution for this coding contest problem.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3 | instruction | 0 | 82,347 | 7 | 164,694 |
"Correct Solution:
```
N=int(input())
c=str(input())
R=c.count('R')
W=c[:R].count('W')
print(W)
``` | output | 1 | 82,347 | 7 | 164,695 |
Provide a correct Python 3 solution for this coding contest problem.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3 | instruction | 0 | 82,348 | 7 | 164,696 |
"Correct Solution:
```
N = int(input())
C = input()
r = C.count('R')
ans = C[:r].count('W')
print(ans)
``` | output | 1 | 82,348 | 7 | 164,697 |
Provide a correct Python 3 solution for this coding contest problem.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3 | instruction | 0 | 82,349 | 7 | 164,698 |
"Correct Solution:
```
n=int(input())
s=input()
nr=s.count("R")
print(s[:nr].count("W"))
``` | output | 1 | 82,349 | 7 | 164,699 |
Provide a correct Python 3 solution for this coding contest problem.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3 | instruction | 0 | 82,350 | 7 | 164,700 |
"Correct Solution:
```
_ = input()
c = input()
r_num = c.count("R")
print(c[:r_num].count("W"))
``` | output | 1 | 82,350 | 7 | 164,701 |
Provide a correct Python 3 solution for this coding contest problem.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3 | instruction | 0 | 82,351 | 7 | 164,702 |
"Correct Solution:
```
N = int(input())
C = input()
Nw = C.count('W')
Nr = N - Nw
print(Nw - C[(-1*Nw):].count('W'))
``` | output | 1 | 82,351 | 7 | 164,703 |
Provide a correct Python 3 solution for this coding contest problem.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3 | instruction | 0 | 82,352 | 7 | 164,704 |
"Correct Solution:
```
N,c=input(),input()
print(c[:c.count('R')].count('W'))
``` | output | 1 | 82,352 | 7 | 164,705 |
Provide a correct Python 3 solution for this coding contest problem.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3 | instruction | 0 | 82,353 | 7 | 164,706 |
"Correct Solution:
```
n=int(input())
c=list(input())
x=c.count("R")
m=c[:x]
y=m.count("W")
print(y)
``` | output | 1 | 82,353 | 7 | 164,707 |
Provide a correct Python 3 solution for this coding contest problem.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3 | instruction | 0 | 82,354 | 7 | 164,708 |
"Correct Solution:
```
N=int(input())
S=input()
le=S.count('R')
P=S[le:]
print(P.count('R'))
``` | output | 1 | 82,354 | 7 | 164,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3
Submitted Solution:
```
n = int(input())
C = list(input())
Nr = C.count("R")
ans = C[:Nr].count("W")
print(ans)
``` | instruction | 0 | 82,355 | 7 | 164,710 |
Yes | output | 1 | 82,355 | 7 | 164,711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3
Submitted Solution:
```
n = int(input())
c = input()
dv = c.count('R')
print(c[:dv].count('W'))
``` | instruction | 0 | 82,356 | 7 | 164,712 |
Yes | output | 1 | 82,356 | 7 | 164,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3
Submitted Solution:
```
n = int(input())
c = list(input())
n = c.count("R")
m = c[:n]
print(n-m.count("R"))
``` | instruction | 0 | 82,357 | 7 | 164,714 |
Yes | output | 1 | 82,357 | 7 | 164,715 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3
Submitted Solution:
```
n = int(input())
wr = input()
R_cnt = wr.count('R')
ans = wr.count('W',0,R_cnt)
print(ans)
``` | instruction | 0 | 82,358 | 7 | 164,716 |
Yes | output | 1 | 82,358 | 7 | 164,717 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3
Submitted Solution:
```
import datetime
import string
import re
import math
N=int(input())
input = input()
str = []
for i in range(N):
str.append(input[i])
count = 0
begin = 0
end = len(str) - 1
while True:
if begin < len(str):
while str[begin] == 'W':
begin += 1
if begin == len(str):
break
if end >= 0:
while str[end] == 'R':
end -= 1
if end == -1:
break
if begin < end:
str[begin] = 'R'
str[end] = 'W'
begin += 1
end -= 1
count += 1
elif begin == len(str) and end != -1:
if str[end + 1] == 'R':
str[end] = 'W'
end -= 1
count += 1
else:
end -= 1
elif begin != len(str) and end == -1:
if str[begin - 1] == 'W':
str[begin] = 'R'
begin += 1
count += 1
else:
begin += 1
else:
print(count)
break
``` | instruction | 0 | 82,359 | 7 | 164,718 |
No | output | 1 | 82,359 | 7 | 164,719 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3
Submitted Solution:
```
N = int(input())
c = input()
d = list(c)
i = 0
j = N - 1
result = 0
while i != j and i < j:
if d[i] == 'W':
while d[j] != 'R' and i < j:
j -= 1
d[i] = 'R'
d[j] = 'W'
result += 1
i += 1
print(result)
``` | instruction | 0 | 82,360 | 7 | 164,720 |
No | output | 1 | 82,360 | 7 | 164,721 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3
Submitted Solution:
```
# -*- coding: utf-8 -*-
N=int(input())
C=input()
J=0
ans=0
for i in range(N):
if C[i]=="W":
for j in range(J,N):
if i+j>N:break
if C[N-1-j]=="R":
ans+=1
#print(i,N-1-j)
J=j+1
break
if i+J+1>N:break
print(ans)
``` | instruction | 0 | 82,361 | 7 | 164,722 |
No | output | 1 | 82,361 | 7 | 164,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.
You can do the following two kinds of operations any number of times in any order:
* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).
According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?
Constraints
* 2 \leq N \leq 200000
* c_i is `R` or `W`.
Input
Input is given from Standard Input in the following format:
N
c_{1}c_{2}...c_{N}
Output
Print an integer representing the minimum number of operations needed.
Examples
Input
4
WWRR
Output
2
Input
2
RR
Output
0
Input
8
WRWWRWRR
Output
3
Submitted Solution:
```
N = int(input())
c = list(input())
ans = 0
FLG = False
for i in range(N):
if c[i] == "W":
j = N - 1
while c[j] == "W":
j -= 1
if j == i:
FLG = True
break
if FLG:
break
ans += 1
c[i], c[j] = c[j], c[i]
print(ans)
``` | instruction | 0 | 82,362 | 7 | 164,724 |
No | output | 1 | 82,362 | 7 | 164,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo got interested in modern art. He decided to draw a big picture on the board with N+2 rows and M+2 columns of squares constructed in the venue of CODE FESTIVAL 2017, using some people.
The square at the (i+1)-th row and (j+1)-th column in the board is represented by the pair of integers (i,j). That is, the top-left square is (0,0), and the bottom-right square is (N+1,M+1). Initially, the squares (x,y) satisfying 1 \leq x \leq N and 1 \leq y \leq M are painted white, and the other (outermost) squares are painted black.
Ringo arranged people at some of the outermost squares, facing inward. More specifically, the arrangement of people is represented by four strings A, B, C and D, as follows:
* For each row except the top and bottom, if the i-th character (1 \leq i \leq N) in A is `1`, place a person facing right at the square (i,0); otherwise, do nothing.
* For each row except the top and bottom, if the i-th character (1 \leq i \leq N) in B is `1`, place a person facing left at the square (i,M+1); otherwise, do nothing.
* For each column except the leftmost and rightmost, if the i-th character (1 \leq i \leq M) in C is `1`, place a person facing down at the square (0,i); otherwise, do nothing.
* For each column except the leftmost and rightmost, if the i-th character (1 \leq i \leq M) in D is `1`, place a person facing up at the square (N+1,i); otherwise, do nothing.
Each person has a sufficient amount of non-white paint. No two people have paint of the same color.
<image>
An example of an arrangement of people (For convenience, black squares are displayed in gray)
Ringo repeats the following sequence of operations until all people are dismissed from the venue.
* Select a person who is still in the venue.
* The selected person repeats the following action while the square in front of him/her is white: move one square forward, and paint the square he/she enters with his/her paint. When the square in front of him/her is not white, he/she stops doing the action.
* The person is now dismissed from the venue.
<image>
An example of a way the board is painted
How many different states of the board can Ringo obtain at the end of the process? Find the count modulo 998244353.
Two states of the board are considered different when there is a square painted in different colors.
Constraints
* 1 \leq N,M \leq 10^5
* |A|=|B|=N
* |C|=|D|=M
* A, B, C and D consist of `0` and `1`.
Input
Input is given from Standard Input in the following format:
N M
A
B
C
D
Output
Print the number of the different states of the board Ringo can obtain at the end of the process, modulo 998244353.
Examples
Input
2 2
10
01
10
01
Output
6
Input
2 2
11
11
11
11
Output
32
Input
3 4
111
111
1111
1111
Output
1276
Input
17 21
11001010101011101
11001010011010111
111010101110101111100
011010110110101000111
Output
548356548
Input
3 4
000
101
1111
0010
Output
21
Input
9 13
111100001
010101011
0000000000000
1010111111101
Output
177856
Input
23 30
01010010101010010001110
11010100100100101010101
000101001001010010101010101101
101001000100101001010010101000
Output
734524988
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
import numpy as np
MOD = 998244353
N,M = map(int,readline().split())
A = np.array(list(readline().rstrip()),dtype=np.int64) - 48
B = np.array(list(readline().rstrip()),dtype=np.int64) - 48
C = np.array(list(readline().rstrip()),dtype=np.int64) - 48
D = np.array(list(readline().rstrip()),dtype=np.int64) - 48
def cumprod(arr,MOD):
L = len(arr); Lsq = int(L**.5+1)
arr = np.resize(arr,Lsq**2).reshape(Lsq,Lsq)
for n in range(1,Lsq):
arr[:,n] *= arr[:,n-1]; arr[:,n] %= MOD
for n in range(1,Lsq):
arr[n] *= arr[n-1,-1]; arr[n] %= MOD
return arr.ravel()[:L]
def make_fact(U,MOD):
x = np.arange(U,dtype=np.int64); x[0] = 1
fact = cumprod(x,MOD)
x = np.arange(U,0,-1,dtype=np.int64); x[0] = pow(int(fact[-1]),MOD-2,MOD)
fact_inv = cumprod(x,MOD)[::-1]
return fact,fact_inv
U = 10**6
fact, fact_inv = make_fact(U,MOD)
def cnt_vertical(N,M,A,B,C,D):
"""
・ABが長さN、CDが長さM。
・初手に縦線を入れる場合。
・縦線の右端の位置に応じて場合分けして集計
・右端の位置を決めると、（左端の位置ごとの左側の入れ方の累積和） x （右側の入れ方の累積和）、適当に重みをつける必要あり
"""
AB = A+B
# ある場所を初めての縦線としたときに、左側にできる部分。上にある点の個数と下にある点の個数
L = C.sum(); LU = A.cumsum() - A; LD = B.cumsum() - B
if L >= 1:
NL = fact[LU+LD+L-1] * fact_inv[LU+LD] % MOD * fact_inv[L-1] % MOD
else:
NL = np.zeros(N,dtype=np.int64)
i = np.where(AB>0)[0][0]
NL[i] = 1
R = D.sum(); RU = A[::-1].cumsum() - A[::-1]; RD = B[::-1].cumsum() - B[::-1]
NR = fact[RU+RD+R-1] * fact_inv[RU+RD] % MOD * fact_inv[R-1] % MOD; NR = NR[::-1]
x = np.full(N,1,np.int64); x[AB==2] = 2; coef = cumprod(x,MOD)
x = np.full(N,1,np.int64); x[AB==2] = (MOD+1)//2; coef_inv = cumprod(x,MOD)
NL *= coef_inv; NR *= coef; NL %= MOD; NR %= MOD
NL[AB==2] *= 2; NL[NL>=MOD] -= MOD
NL[AB==0] = 0; NR[AB==0] = 0
# sum(l<=r) NL[l] NR[r] を求めればよい。
NL_cum = NL.cumsum() % MOD
return (NL_cum * NR % MOD).sum() % MOD
def AB_only(N,A,B):
x = np.ones(N,dtype=np.int64)
x[A+B==2] = 2
return cumprod(x,MOD)[-1]
def ABC_only(N,M,A,B,C):
AB = A+B
R = C.sum(); RU = A[::-1].cumsum() - A[::-1]; RD = B[::-1].cumsum() - B[::-1]
NR = fact[RU+RD+R-1] * fact_inv[RU+RD] % MOD * fact_inv[R-1] % MOD; NR = NR[::-1]
NR[AB==0] = 0
x = np.full(N,1,np.int64); x[AB==2] = 2; coef = cumprod(x,MOD)
x=(coef * NR % MOD).sum()%MOD
assert x == cnt_vertical(N,M,A,B,np.zeros_like(C),C)
n = A.sum()+B.sum()+C.sum()-1; m = C.sum()-1
y = fact[n] * fact_inv[m] % MOD * fact_inv[n-m] % MOD
return (x+y)%MOD
def F(N,M,A,B,C,D):
# 問題の答えを出します。
NA,NB,NC,ND = [np.count_nonzero(x) for x in [A,B,C,D]]
if all(x != 0 for x in [NA,NB,NC,ND]):
return (cnt_vertical(N,M,A,B,C,D) + cnt_vertical(M,N,C,D,A,B)) % MOD
# A,Cに0列を寄せる
if NA != 0:
NA,NB = NB,NA; A,B = B,A
if NC != 0:
NC,ND = ND,NC; C,D = D,C
if NB == 0:
return AB_only(M,C,D)
if ND == 0:
return AB_only(N,A,B)
# B,Dは0ではない
if NA == 0 and NC == 0:
# 2面B,Dのみ -> binom(B+D,B)
return fact[NB+ND] * fact_inv[NB] % MOD * fact_inv[ND] % MOD
if NA == 0:
return ABC_only(M,N,C,D,B)
if NC == 0:
return ABC_only(N,M,A,B,D)
raise exception
answer = F(N,M,A,B,C,D)
print(answer)
``` | instruction | 0 | 82,459 | 7 | 164,918 |
No | output | 1 | 82,459 | 7 | 164,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo got interested in modern art. He decided to draw a big picture on the board with N+2 rows and M+2 columns of squares constructed in the venue of CODE FESTIVAL 2017, using some people.
The square at the (i+1)-th row and (j+1)-th column in the board is represented by the pair of integers (i,j). That is, the top-left square is (0,0), and the bottom-right square is (N+1,M+1). Initially, the squares (x,y) satisfying 1 \leq x \leq N and 1 \leq y \leq M are painted white, and the other (outermost) squares are painted black.
Ringo arranged people at some of the outermost squares, facing inward. More specifically, the arrangement of people is represented by four strings A, B, C and D, as follows:
* For each row except the top and bottom, if the i-th character (1 \leq i \leq N) in A is `1`, place a person facing right at the square (i,0); otherwise, do nothing.
* For each row except the top and bottom, if the i-th character (1 \leq i \leq N) in B is `1`, place a person facing left at the square (i,M+1); otherwise, do nothing.
* For each column except the leftmost and rightmost, if the i-th character (1 \leq i \leq M) in C is `1`, place a person facing down at the square (0,i); otherwise, do nothing.
* For each column except the leftmost and rightmost, if the i-th character (1 \leq i \leq M) in D is `1`, place a person facing up at the square (N+1,i); otherwise, do nothing.
Each person has a sufficient amount of non-white paint. No two people have paint of the same color.
<image>
An example of an arrangement of people (For convenience, black squares are displayed in gray)
Ringo repeats the following sequence of operations until all people are dismissed from the venue.
* Select a person who is still in the venue.
* The selected person repeats the following action while the square in front of him/her is white: move one square forward, and paint the square he/she enters with his/her paint. When the square in front of him/her is not white, he/she stops doing the action.
* The person is now dismissed from the venue.
<image>
An example of a way the board is painted
How many different states of the board can Ringo obtain at the end of the process? Find the count modulo 998244353.
Two states of the board are considered different when there is a square painted in different colors.
Constraints
* 1 \leq N,M \leq 10^5
* |A|=|B|=N
* |C|=|D|=M
* A, B, C and D consist of `0` and `1`.
Input
Input is given from Standard Input in the following format:
N M
A
B
C
D
Output
Print the number of the different states of the board Ringo can obtain at the end of the process, modulo 998244353.
Examples
Input
2 2
10
01
10
01
Output
6
Input
2 2
11
11
11
11
Output
32
Input
3 4
111
111
1111
1111
Output
1276
Input
17 21
11001010101011101
11001010011010111
111010101110101111100
011010110110101000111
Output
548356548
Input
3 4
000
101
1111
0010
Output
21
Input
9 13
111100001
010101011
0000000000000
1010111111101
Output
177856
Input
23 30
01010010101010010001110
11010100100100101010101
000101001001010010101010101101
101001000100101001010010101000
Output
734524988
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
import numpy as np
MOD = 998244353
N,M = map(int,readline().split())
A = np.array(list(readline().rstrip())) - 48
B = np.array(list(readline().rstrip())) - 48
C = np.array(list(readline().rstrip())) - 48
D = np.array(list(readline().rstrip())) - 48
A,B,C,D
def cumprod(arr,MOD):
L = len(arr); Lsq = int(L**.5+1)
arr = np.resize(arr,Lsq**2).reshape(Lsq,Lsq)
for n in range(1,Lsq):
arr[:,n] *= arr[:,n-1]; arr[:,n] %= MOD
for n in range(1,Lsq):
arr[n] *= arr[n-1,-1]; arr[n] %= MOD
return arr.ravel()[:L]
def make_fact(U,MOD):
x = np.arange(U,dtype=np.int64); x[0] = 1
fact = cumprod(x,MOD)
x = np.arange(U,0,-1,dtype=np.int64); x[0] = pow(int(fact[-1]),MOD-2,MOD)
fact_inv = cumprod(x,MOD)[::-1]
return fact,fact_inv
U = 10**6
fact, fact_inv = make_fact(U,MOD)
def cnt_vertical(N,M,A,B,C,D):
"""
・ABが長さN、CDが長さM。
・初手に縦線を入れる場合。
・縦線の右端の位置に応じて場合分けして集計
・右端の位置を決めると、（左端の位置ごとの左側の入れ方の累積和） x （右側の入れ方の累積和）、適当に重みをつける必要あり
"""
AB = A+B
# ある場所を初めての縦線としたときに、左側にできる部分。上にある点の個数と下にある点の個数
L = C.sum(); LU = A.cumsum() - A; LD = B.cumsum() - B
if L >= 1:
NL = fact[LU+LD+L-1] * fact_inv[LU+LD] % MOD * fact_inv[L-1] % MOD
else:
NL = np.zeros(N,dtype=np.int64)
i = np.where(AB>0)[0][0]
NL[i] = 1
R = D.sum(); RU = A[::-1].cumsum() - A[::-1]; RD = B[::-1].cumsum() - B[::-1]
NR = fact[RU+RD+R-1] * fact_inv[RU+RD] % MOD * fact_inv[R-1] % MOD; NR = NR[::-1]
# とりあえず
x = np.full(N,1,np.int64); x[AB==2] = 2; coef = cumprod(x,MOD)
x = np.full(N,1,np.int64); x[AB==2] = (MOD+1)//2; coef_inv = cumprod(x,MOD)
NL *= coef_inv; NR *= coef; NL %= MOD; NR %= MOD
NL[AB==2] *= 2; NL[NL>=MOD] -= MOD
NL[AB==0] = 0; NR[AB==0] = 0
# sum(l<=r) NL[l] NR[r] を求めればよい。
NL_cum = NL.cumsum() % MOD
return (NL_cum * NR % MOD).sum() % MOD
def AB_only(N,A,B):
x = np.ones(N,dtype=np.int64)
x[A+B==2] = 2
return cumprod(x,MOD)[-1]
def ABC_only(N,M,A,B,C):
x = cnt_vertical(N,M,A,B,np.zeros_like(C),C)
x += cnt_vertical(M,N,C,np.zeros_like(C),A,B)
return x%MOD
def F(N,M,A,B,C,D):
# 問題の答えを出します。
NA,NB,NC,ND = [np.count_nonzero(x) for x in [A,B,C,D]]
if all(x != 0 for x in [NA,NB,NC,ND]):
return (cnt_vertical(N,M,A,B,C,D) + cnt_vertical(M,N,C,D,A,B)) % MOD
# A,Cに0列を寄せる
if NA != 0:
NA,NB = NB,NA; A,B = B,A
if NC != 0:
NC,ND = ND,NC; C,D = D,C
if NB == 0:
return AB_only(M,C,D)
if ND == 0:
return AB_only(N,A,B)
# B,Dは0ではない
if NA == 0 and NC == 0:
# 2面B,Dのみ -> binom(B+D,B)
return fact[NB+ND] * fact_inv[NB] % MOD * fact_inv[ND] % MOD
if NA == 0:
return ABC_only(M,N,C,D,B)
if NC == 0:
return ABC_only(N,M,A,B,D)
answer = F(N,M,A,B,C,D)
print(answer)
``` | instruction | 0 | 82,460 | 7 | 164,920 |
No | output | 1 | 82,460 | 7 | 164,921 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo got interested in modern art. He decided to draw a big picture on the board with N+2 rows and M+2 columns of squares constructed in the venue of CODE FESTIVAL 2017, using some people.
The square at the (i+1)-th row and (j+1)-th column in the board is represented by the pair of integers (i,j). That is, the top-left square is (0,0), and the bottom-right square is (N+1,M+1). Initially, the squares (x,y) satisfying 1 \leq x \leq N and 1 \leq y \leq M are painted white, and the other (outermost) squares are painted black.
Ringo arranged people at some of the outermost squares, facing inward. More specifically, the arrangement of people is represented by four strings A, B, C and D, as follows:
* For each row except the top and bottom, if the i-th character (1 \leq i \leq N) in A is `1`, place a person facing right at the square (i,0); otherwise, do nothing.
* For each row except the top and bottom, if the i-th character (1 \leq i \leq N) in B is `1`, place a person facing left at the square (i,M+1); otherwise, do nothing.
* For each column except the leftmost and rightmost, if the i-th character (1 \leq i \leq M) in C is `1`, place a person facing down at the square (0,i); otherwise, do nothing.
* For each column except the leftmost and rightmost, if the i-th character (1 \leq i \leq M) in D is `1`, place a person facing up at the square (N+1,i); otherwise, do nothing.
Each person has a sufficient amount of non-white paint. No two people have paint of the same color.
<image>
An example of an arrangement of people (For convenience, black squares are displayed in gray)
Ringo repeats the following sequence of operations until all people are dismissed from the venue.
* Select a person who is still in the venue.
* The selected person repeats the following action while the square in front of him/her is white: move one square forward, and paint the square he/she enters with his/her paint. When the square in front of him/her is not white, he/she stops doing the action.
* The person is now dismissed from the venue.
<image>
An example of a way the board is painted
How many different states of the board can Ringo obtain at the end of the process? Find the count modulo 998244353.
Two states of the board are considered different when there is a square painted in different colors.
Constraints
* 1 \leq N,M \leq 10^5
* |A|=|B|=N
* |C|=|D|=M
* A, B, C and D consist of `0` and `1`.
Input
Input is given from Standard Input in the following format:
N M
A
B
C
D
Output
Print the number of the different states of the board Ringo can obtain at the end of the process, modulo 998244353.
Examples
Input
2 2
10
01
10
01
Output
6
Input
2 2
11
11
11
11
Output
32
Input
3 4
111
111
1111
1111
Output
1276
Input
17 21
11001010101011101
11001010011010111
111010101110101111100
011010110110101000111
Output
548356548
Input
3 4
000
101
1111
0010
Output
21
Input
9 13
111100001
010101011
0000000000000
1010111111101
Output
177856
Input
23 30
01010010101010010001110
11010100100100101010101
000101001001010010101010101101
101001000100101001010010101000
Output
734524988
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
import numpy as np
MOD = 998244353
N,M = map(int,readline().split())
A = np.array(list(readline().rstrip())) - 48
B = np.array(list(readline().rstrip())) - 48
C = np.array(list(readline().rstrip())) - 48
D = np.array(list(readline().rstrip())) - 48
A,B,C,D
def cumprod(arr,MOD):
L = len(arr); Lsq = int(L**.5+1)
arr = np.resize(arr,Lsq**2).reshape(Lsq,Lsq)
for n in range(1,Lsq):
arr[:,n] *= arr[:,n-1]; arr[:,n] %= MOD
for n in range(1,Lsq):
arr[n] *= arr[n-1,-1]; arr[n] %= MOD
return arr.ravel()[:L]
def make_fact(U,MOD):
x = np.arange(U,dtype=np.int64); x[0] = 1
fact = cumprod(x,MOD)
x = np.arange(U,0,-1,dtype=np.int64); x[0] = pow(int(fact[-1]),MOD-2,MOD)
fact_inv = cumprod(x,MOD)[::-1]
return fact,fact_inv
U = 10**6
fact, fact_inv = make_fact(U,MOD)
def cnt_vertical(N,M,A,B,C,D):
"""
・ABが長さN、CDが長さM。
・初手に縦線を入れる場合。
・縦線の右端の位置に応じて場合分けして集計
・右端の位置を決めると、（左端の位置ごとの左側の入れ方の累積和） x （右側の入れ方の累積和）、適当に重みをつける必要あり
"""
AB = A+B
# ある場所を初めての縦線としたときに、左側にできる部分。上にある点の個数と下にある点の個数
L = C.sum(); LU = A.cumsum() - A; LD = B.cumsum() - B
NL = fact[LU+LD+L-1] * fact_inv[LU+LD] % MOD * fact_inv[L-1] % MOD
R = D.sum(); RU = A[::-1].cumsum() - A[::-1]; RD = B[::-1].cumsum() - B[::-1]
NR = fact[RU+RD+R-1] * fact_inv[RU+RD] % MOD * fact_inv[R-1] % MOD; NR = NR[::-1]
# とりあえず
x = np.full(N,1,np.int64); x[AB==2] = 2; coef = cumprod(x,MOD)
x = np.full(N,1,np.int64); x[AB==2] = (MOD+1)//2; coef_inv = cumprod(x,MOD)
NL *= coef_inv; NR *= coef; NL %= MOD; NR %= MOD
NL[AB==2] *= 2; NL[NL>=MOD] -= MOD
NL[AB==0] = 0; NR[AB==0] = 0
# sum(l<=r) NL[l] NR[r] を求めればよい。
NL_cum = NL.cumsum() % MOD
return (NL_cum * NR % MOD).sum() % MOD
def LR_only(N,A,B):
raise Exception
def LRU_only(N,M,A,B,C):
raise Exception
def F(N,M,A,B,C,D):
# 問題の答えを出します。
NA,NB,NC,ND = [np.count_nonzero(x) for x in [A,B,C,D]]
if all(x != 0 for x in [NA,NB,NC,ND]):
return (cnt_vertical(N,M,A,B,C,D) + cnt_vertical(M,N,C,D,A,B)) % MOD
# A,Cに0列を寄せる
if NA != 0:
NA,NB = NB,NA; A,B = B,A
if NC != 0:
NC,ND = ND,NC; C,D = D,C
if NB == 0:
return LR_only(M,C,D)
if ND == 0:
return LR_only(N,A,B)
# B,Dは0ではない
if NA == 0 and NC == 0:
# 2面B,Dのみ -> binom(B+D,B)
return fact[NB+ND] * fact_inv[NB] % MOD * fact_inv[ND] % MOD
if NA == 0:
return LRU_only(M,N,C,D,B)
if NC == 0:
return LRU_only(N,M,A,B,D)
answer = F(N,M,A,B,C,D)
print(answer)
``` | instruction | 0 | 82,461 | 7 | 164,922 |
No | output | 1 | 82,461 | 7 | 164,923 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo got interested in modern art. He decided to draw a big picture on the board with N+2 rows and M+2 columns of squares constructed in the venue of CODE FESTIVAL 2017, using some people.
The square at the (i+1)-th row and (j+1)-th column in the board is represented by the pair of integers (i,j). That is, the top-left square is (0,0), and the bottom-right square is (N+1,M+1). Initially, the squares (x,y) satisfying 1 \leq x \leq N and 1 \leq y \leq M are painted white, and the other (outermost) squares are painted black.
Ringo arranged people at some of the outermost squares, facing inward. More specifically, the arrangement of people is represented by four strings A, B, C and D, as follows:
* For each row except the top and bottom, if the i-th character (1 \leq i \leq N) in A is `1`, place a person facing right at the square (i,0); otherwise, do nothing.
* For each row except the top and bottom, if the i-th character (1 \leq i \leq N) in B is `1`, place a person facing left at the square (i,M+1); otherwise, do nothing.
* For each column except the leftmost and rightmost, if the i-th character (1 \leq i \leq M) in C is `1`, place a person facing down at the square (0,i); otherwise, do nothing.
* For each column except the leftmost and rightmost, if the i-th character (1 \leq i \leq M) in D is `1`, place a person facing up at the square (N+1,i); otherwise, do nothing.
Each person has a sufficient amount of non-white paint. No two people have paint of the same color.
<image>
An example of an arrangement of people (For convenience, black squares are displayed in gray)
Ringo repeats the following sequence of operations until all people are dismissed from the venue.
* Select a person who is still in the venue.
* The selected person repeats the following action while the square in front of him/her is white: move one square forward, and paint the square he/she enters with his/her paint. When the square in front of him/her is not white, he/she stops doing the action.
* The person is now dismissed from the venue.
<image>
An example of a way the board is painted
How many different states of the board can Ringo obtain at the end of the process? Find the count modulo 998244353.
Two states of the board are considered different when there is a square painted in different colors.
Constraints
* 1 \leq N,M \leq 10^5
* |A|=|B|=N
* |C|=|D|=M
* A, B, C and D consist of `0` and `1`.
Input
Input is given from Standard Input in the following format:
N M
A
B
C
D
Output
Print the number of the different states of the board Ringo can obtain at the end of the process, modulo 998244353.
Examples
Input
2 2
10
01
10
01
Output
6
Input
2 2
11
11
11
11
Output
32
Input
3 4
111
111
1111
1111
Output
1276
Input
17 21
11001010101011101
11001010011010111
111010101110101111100
011010110110101000111
Output
548356548
Input
3 4
000
101
1111
0010
Output
21
Input
9 13
111100001
010101011
0000000000000
1010111111101
Output
177856
Input
23 30
01010010101010010001110
11010100100100101010101
000101001001010010101010101101
101001000100101001010010101000
Output
734524988
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
import numpy as np
MOD = 998244353
N,M = map(int,readline().split())
A = np.array(list(readline().rstrip())) - 48
B = np.array(list(readline().rstrip())) - 48
C = np.array(list(readline().rstrip())) - 48
D = np.array(list(readline().rstrip())) - 48
A,B,C,D
def cumprod(arr,MOD):
L = len(arr); Lsq = int(L**.5+1)
arr = np.resize(arr,Lsq**2).reshape(Lsq,Lsq)
for n in range(1,Lsq):
arr[:,n] *= arr[:,n-1]; arr[:,n] %= MOD
for n in range(1,Lsq):
arr[n] *= arr[n-1,-1]; arr[n] %= MOD
return arr.ravel()[:L]
def make_fact(U,MOD):
x = np.arange(U,dtype=np.int64); x[0] = 1
fact = cumprod(x,MOD)
x = np.arange(U,0,-1,dtype=np.int64); x[0] = pow(int(fact[-1]),MOD-2,MOD)
fact_inv = cumprod(x,MOD)[::-1]
return fact,fact_inv
U = 10**6
fact, fact_inv = make_fact(U,MOD)
def cnt_vertical(N,M,A,B,C,D):
"""
・ABが長さN、CDが長さM。
・初手に縦線を入れる場合。
・縦線の右端の位置に応じて場合分けして集計
・右端の位置を決めると、（左端の位置ごとの左側の入れ方の累積和） x （右側の入れ方の累積和）、適当に重みをつける必要あり
"""
AB = A+B
# ある場所を初めての縦線としたときに、左側にできる部分。上にある点の個数と下にある点の個数
L = C.sum(); LU = A.cumsum() - A; LD = B.cumsum() - B
NL = fact[LU+LD+L-1] * fact_inv[LU+LD] % MOD * fact_inv[L-1] % MOD
R = D.sum(); RU = A[::-1].cumsum() - A[::-1]; RD = B[::-1].cumsum() - B[::-1]
NR = fact[RU+RD+R-1] * fact_inv[RU+RD] % MOD * fact_inv[R-1] % MOD; NR = NR[::-1]
# とりあえず
x = np.full(N,1,np.int64); x[AB==2] = 2; coef = cumprod(x,MOD)
x = np.full(N,1,np.int64); x[AB==2] = (MOD+1)//2; coef_inv = cumprod(x,MOD)
NL *= coef_inv; NR *= coef; NL %= MOD; NR %= MOD
NL[AB==2] *= 2; NL[NL>=MOD] -= MOD
NL[AB==0] = 0; NR[AB==0] = 0
# sum(l<=r) NL[l] NR[r] を求めればよい。
NL_cum = NL.cumsum() % MOD
return (NL_cum * NR % MOD).sum() % MOD
def F(N,M,A,B,C,D):
# 問題の答えを出します。
NA,NB,NC,ND = [np.count_nonzero(x) for x in [A,B,C,D]]
if all(x != 0 for x in [NA,NB,NC,ND]):
return (cnt_vertical(N,M,A,B,C,D) + cnt_vertical(M,N,C,D,A,B)) % MOD
raise Exception
answer = F(N,M,A,B,C,D)
print(answer)
``` | instruction | 0 | 82,462 | 7 | 164,924 |
No | output | 1 | 82,462 | 7 | 164,925 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite | instruction | 0 | 82,692 | 7 | 165,384 |
Tags: math, number theory
Correct Solution:
```
import math
for i in range(int(input())):
a,b= map(int,input().split())
if math.gcd(a,b) ==1 :
print('Finite')
else:
print('Infinite')
``` | output | 1 | 82,692 | 7 | 165,385 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite | instruction | 0 | 82,693 | 7 | 165,386 |
Tags: math, number theory
Correct Solution:
```
from math import gcd
import sys
Input = sys.stdin.readline
n = int(Input())
for i in range(n):
a, b = list(map(int, Input().split()))
print("Finite" if gcd(a, b) == 1 else "Infinite")
# Codeforcesian
``` | output | 1 | 82,693 | 7 | 165,387 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite | instruction | 0 | 82,694 | 7 | 165,388 |
Tags: math, number theory
Correct Solution:
```
def gcd(a, b):
while b > 0:
a, b = b, a % b
# print('a', a)
return a
for z in range(int(input())):
a, b = map(int, input().split())
if gcd(a,b) > 1:
print('infinite')
else:
print('finite')
``` | output | 1 | 82,694 | 7 | 165,389 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite | instruction | 0 | 82,695 | 7 | 165,390 |
Tags: math, number theory
Correct Solution:
```
from collections import Counter,defaultdict,deque
from heapq import heapify,heappop,heappush
import sys,bisect,math,itertools,string,queue
sys.setrecursionlimit(10**8)
mod = 10**9+7
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
def inpl_str(): return list(sys.stdin.readline().split())
def inpln(n): return list(int(sys.stdin.readline()) for i in range(n))
n = inp()
ab = [0] * n
for i in range(n):
ab[i] = inpl()
for i in range(n):
a,b = ab[i][0], ab[i][1]
a,b = max(a,b),min(a,b)
if math.gcd(a,b) != 1 and b != 1:
print('inFinite')
else:
print('finite')
``` | output | 1 | 82,695 | 7 | 165,391 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite | instruction | 0 | 82,696 | 7 | 165,392 |
Tags: math, number theory
Correct Solution:
```
def gcd(a,b):
while b:
a,b=b,a%b
return a
t=int(input())
for i in range(t):
a,b=map(int,input().split())
if gcd(a,b)==1:
print("Finite")
else:
print("Infinite")
``` | output | 1 | 82,696 | 7 | 165,393 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite | instruction | 0 | 82,697 | 7 | 165,394 |
Tags: math, number theory
Correct Solution:
```
import math
for _ in range(int(input())):
a,b = map(int,input().split())
re = math.gcd(a,b)
if(re == 1):
print("Finite")
else:
print("Infinite")
``` | output | 1 | 82,697 | 7 | 165,395 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite | instruction | 0 | 82,698 | 7 | 165,396 |
Tags: math, number theory
Correct Solution:
```
from math import gcd
t = int(input())
for _ in range(t):
a, b = [int(_) for _ in input().split()]
if gcd(a, b) == 1:
print('Finite')
else:
print('Infinite')
``` | output | 1 | 82,698 | 7 | 165,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite | instruction | 0 | 82,699 | 7 | 165,398 |
Tags: math, number theory
Correct Solution:
```
def nod(a, b):
if a == 0 or b == 0:
return max(a, b)
if a == 1 or b == 1:
return 1
return nod(b % a, a)
n = int(input())
for i in range(n):
a, b = map(int, input().split())
if nod(min(a, b), max(a, b)) == 1:
print('Finite')
else:
print('Infinite')
``` | output | 1 | 82,699 | 7 | 165,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite
Submitted Solution:
```
from math import gcd
t = int(input())
for i in range(t):
a,b = map(int, input().split())
if gcd(a,b) == 1:
print('finite')
else:
print('infinite')
``` | instruction | 0 | 82,700 | 7 | 165,400 |
Yes | output | 1 | 82,700 | 7 | 165,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite
Submitted Solution:
```
def gcd(a,b):
if b==0:
return a
return gcd(b,a%b)
t=int(input())
for _ in range(t):
a,b=map(int,input().split())
c=gcd(a,b)
if c==1:
print("Finite")
else:
print("Infinite")
``` | instruction | 0 | 82,701 | 7 | 165,402 |
Yes | output | 1 | 82,701 | 7 | 165,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite
Submitted Solution:
```
def gcd(a,b):
if b==0:
return a
return gcd(b,a%b)
tt=int(input())
for t in range(tt):
a,b=map(int,input().split())
if gcd(a,b)!=1:
print("Infinite")
else:
print("Finite")
``` | instruction | 0 | 82,702 | 7 | 165,404 |
Yes | output | 1 | 82,702 | 7 | 165,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite
Submitted Solution:
```
from sys import stdin
from collections import deque
from math import sqrt, floor, ceil, log, log2, log10, pi, gcd
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def si(): return stdin.readline()
for _ in range(ii()):
a, b=mi()
print("Infinite" if gcd(a, b)!=1 else "Finite")
``` | instruction | 0 | 82,703 | 7 | 165,406 |
Yes | output | 1 | 82,703 | 7 | 165,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite
Submitted Solution:
```
t=int(input().strip())
for i in range(t):
a,b=map(int,input().split(' '))
if(min(a,b)-1<=abs(a-b)):
print("Finite")
else:
print("Infinite")
``` | instruction | 0 | 82,704 | 7 | 165,408 |
No | output | 1 | 82,704 | 7 | 165,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite
Submitted Solution:
```
cases = int(input())
for _ in range(cases):
a, b = map(int, input().split())
if a == 1 or a > b or a*2 < b:
print("finite")
else:
print("infinite")
# ans = ["W"]
#
# for i in range(1, 101):
# if i >= a and ans[i-a] == "W" or i >= b and ans[i-b] == "W":
# ans.append("W")
# else:
# ans.append("B")
# print(ans)
``` | instruction | 0 | 82,705 | 7 | 165,410 |
No | output | 1 | 82,705 | 7 | 165,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite
Submitted Solution:
```
t = int(input())
def prime_judge(n):
for i in range(2,int(n**0.5+1)):
if n % i == 0:
return False
else:
return True
for _ in range(t):
a,b = map(int,input().split())
if a==1 or b ==1:
print("Finite")
elif prime_judge(a) and prime_judge(b):
print('Finite')
else:
print('Infinite')
``` | instruction | 0 | 82,706 | 7 | 165,412 |
No | output | 1 | 82,706 | 7 | 165,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Consider the set of all nonnegative integers: {0, 1, 2, ...}. Given two integers a and b (1 ≤ a, b ≤ 10^4). We paint all the numbers in increasing number first we paint 0, then we paint 1, then 2 and so on.
Each number is painted white or black. We paint a number i according to the following rules:
* if i = 0, it is colored white;
* if i ≥ a and i - a is colored white, i is also colored white;
* if i ≥ b and i - b is colored white, i is also colored white;
* if i is still not colored white, it is colored black.
In this way, each nonnegative integer gets one of two colors.
For example, if a=3, b=5, then the colors of the numbers (in the order from 0) are: white (0), black (1), black (2), white (3), black (4), white (5), white (6), black (7), white (8), white (9), ...
Note that:
* It is possible that there are infinitely many nonnegative integers colored black. For example, if a = 10 and b = 10, then only 0, 10, 20, 30 and any other nonnegative integers that end in 0 when written in base 10 are white. The other integers are colored black.
* It is also possible that there are only finitely many nonnegative integers colored black. For example, when a = 1 and b = 10, then there is no nonnegative integer colored black at all.
Your task is to determine whether or not the number of nonnegative integers colored black is infinite.
If there are infinitely many nonnegative integers colored black, simply print a line containing "Infinite" (without the quotes). Otherwise, print "Finite" (without the quotes).
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t lines follow, each line contains two space-separated integers a and b (1 ≤ a, b ≤ 10^4).
Output
For each test case, print one line containing either "Infinite" or "Finite" (without the quotes). Output is case-insensitive (i.e. "infinite", "inFiNite" or "finiTE" are all valid answers).
Example
Input
4
10 10
1 10
6 9
7 3
Output
Infinite
Finite
Infinite
Finite
Submitted Solution:
```
t = int(input())
for _ in range(t):
sieve = ["w"]+["b"]*(10**4)
a,b = map(int,input().split())
if a < b :
a,b = b,a
for i in range(a,(10**4)+1):
if sieve[i-a] == "w":
sieve[i] = "w"
for i in range(b,(10**4)+1):
if sieve[i-b] == "w":
sieve[i] = "w"
count_w = sieve.count("w")
count_b = sieve.count("b")
if count_b > count_w:
print("Infinite")
else:
print("Finite")
``` | instruction | 0 | 82,707 | 7 | 165,414 |
No | output | 1 | 82,707 | 7 | 165,415 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You finally woke up after this crazy dream and decided to walk around to clear your head. Outside you saw your house's fence — so plain and boring, that you'd like to repaint it.
<image>
You have a fence consisting of n planks, where the i-th plank has the color a_i. You want to repaint the fence in such a way that the i-th plank has the color b_i.
You've invited m painters for this purpose. The j-th painter will arrive at the moment j and will recolor exactly one plank to color c_j. For each painter you can choose which plank to recolor, but you can't turn them down, i. e. each painter has to color exactly one plank.
Can you get the coloring b you want? If it's possible, print for each painter which plank he must paint.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of planks in the fence and the number of painters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the initial colors of the fence.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ n) — the desired colors of the fence.
The fourth line of each test case contains m integers c_1, c_2, ..., c_m (1 ≤ c_j ≤ n) — the colors painters have.
It's guaranteed that the sum of n doesn't exceed 10^5 and the sum of m doesn't exceed 10^5 over all test cases.
Output
For each test case, output "NO" if it is impossible to achieve the coloring b.
Otherwise, print "YES" and m integers x_1, x_2, ..., x_m, where x_j is the index of plank the j-th painter should paint.
You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" are all recognized as positive answer).
Example
Input
6
1 1
1
1
1
5 2
1 2 2 1 1
1 2 2 1 1
1 2
3 3
2 2 2
2 2 2
2 3 2
10 5
7 3 2 1 7 9 4 2 7 9
9 9 2 1 4 9 4 2 3 9
9 9 7 4 3
5 2
1 2 2 1 1
1 2 2 1 1
3 3
6 4
3 4 2 4 1 2
2 3 1 3 1 1
2 2 3 4
Output
YES
1
YES
2 2
YES
1 1 1
YES
2 1 9 5 9
NO
NO | instruction | 0 | 82,800 | 7 | 165,600 |
Tags: brute force, constructive algorithms, greedy
Correct Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import os
import sys
from collections import defaultdict
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def main():
t = int(input())
for _ in range(t):
n, m = map(int, input().split())
a = [int(ai) for ai in input().split()]
b = [int(bi) for bi in input().split()]
c = [int(ci) for ci in input().split()]
changes = defaultdict(list)
for i, (ai, bi) in enumerate(zip(a, b)):
if ai != bi:
changes[bi].append(i)
last = -1
sol = [-1] * m
for i, ci in enumerate(c):
if changes[ci]:
last = sol[i] = changes[ci].pop()
if sol[-1] == -1 and c[-1] in b:
last = sol[-1] = b.index(c[-1])
poss = sol[-1] != -1 and all(idx == [] for _, idx in changes.items())
sol = [last if i == -1 else i for i in sol]
if poss:
print("YES")
print(*[i + 1 for i in sol])
else:
print("NO")
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
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()
``` | output | 1 | 82,800 | 7 | 165,601 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You finally woke up after this crazy dream and decided to walk around to clear your head. Outside you saw your house's fence — so plain and boring, that you'd like to repaint it.
<image>
You have a fence consisting of n planks, where the i-th plank has the color a_i. You want to repaint the fence in such a way that the i-th plank has the color b_i.
You've invited m painters for this purpose. The j-th painter will arrive at the moment j and will recolor exactly one plank to color c_j. For each painter you can choose which plank to recolor, but you can't turn them down, i. e. each painter has to color exactly one plank.
Can you get the coloring b you want? If it's possible, print for each painter which plank he must paint.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of planks in the fence and the number of painters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the initial colors of the fence.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ n) — the desired colors of the fence.
The fourth line of each test case contains m integers c_1, c_2, ..., c_m (1 ≤ c_j ≤ n) — the colors painters have.
It's guaranteed that the sum of n doesn't exceed 10^5 and the sum of m doesn't exceed 10^5 over all test cases.
Output
For each test case, output "NO" if it is impossible to achieve the coloring b.
Otherwise, print "YES" and m integers x_1, x_2, ..., x_m, where x_j is the index of plank the j-th painter should paint.
You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" are all recognized as positive answer).
Example
Input
6
1 1
1
1
1
5 2
1 2 2 1 1
1 2 2 1 1
1 2
3 3
2 2 2
2 2 2
2 3 2
10 5
7 3 2 1 7 9 4 2 7 9
9 9 2 1 4 9 4 2 3 9
9 9 7 4 3
5 2
1 2 2 1 1
1 2 2 1 1
3 3
6 4
3 4 2 4 1 2
2 3 1 3 1 1
2 2 3 4
Output
YES
1
YES
2 2
YES
1 1 1
YES
2 1 9 5 9
NO
NO | instruction | 0 | 82,801 | 7 | 165,602 |
Tags: brute force, constructive algorithms, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
N, M = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
t = [[] for _ in range(N + 1)]
for i in range(N):
if a[i] != b[i]: t[b[i]].append(i + 1)
res = [0] * M
same = [0] * (N + 1)
for i in range(M):
x = c[i]
if len(t[x]):
res[i] = t[x].pop()
if len(t[x]) == 0: same[x] = res[i]
for i in range(N):
if a[i] == b[i]: same[a[i]] = i + 1
for i in range(M):
if res[i]: continue
x = c[i]
if same[x]: res[i] = same[x]
for i in range(M):
if res[i]: continue
res[i] = res[-1]
#print(res)
if 0 in res:
print("NO")
continue
for i in range(M):
x = c[i]
j = res[i] - 1
a[j] = x
if a == b:
print("YES")
print(*res)
else: print("NO")
``` | output | 1 | 82,801 | 7 | 165,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You finally woke up after this crazy dream and decided to walk around to clear your head. Outside you saw your house's fence — so plain and boring, that you'd like to repaint it.
<image>
You have a fence consisting of n planks, where the i-th plank has the color a_i. You want to repaint the fence in such a way that the i-th plank has the color b_i.
You've invited m painters for this purpose. The j-th painter will arrive at the moment j and will recolor exactly one plank to color c_j. For each painter you can choose which plank to recolor, but you can't turn them down, i. e. each painter has to color exactly one plank.
Can you get the coloring b you want? If it's possible, print for each painter which plank he must paint.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of planks in the fence and the number of painters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the initial colors of the fence.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ n) — the desired colors of the fence.
The fourth line of each test case contains m integers c_1, c_2, ..., c_m (1 ≤ c_j ≤ n) — the colors painters have.
It's guaranteed that the sum of n doesn't exceed 10^5 and the sum of m doesn't exceed 10^5 over all test cases.
Output
For each test case, output "NO" if it is impossible to achieve the coloring b.
Otherwise, print "YES" and m integers x_1, x_2, ..., x_m, where x_j is the index of plank the j-th painter should paint.
You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" are all recognized as positive answer).
Example
Input
6
1 1
1
1
1
5 2
1 2 2 1 1
1 2 2 1 1
1 2
3 3
2 2 2
2 2 2
2 3 2
10 5
7 3 2 1 7 9 4 2 7 9
9 9 2 1 4 9 4 2 3 9
9 9 7 4 3
5 2
1 2 2 1 1
1 2 2 1 1
3 3
6 4
3 4 2 4 1 2
2 3 1 3 1 1
2 2 3 4
Output
YES
1
YES
2 2
YES
1 1 1
YES
2 1 9 5 9
NO
NO | instruction | 0 | 82,802 | 7 | 165,604 |
Tags: brute force, constructive algorithms, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n, m = map(int, input().split())
now = list(map(int, input().split()))
goal = list(map(int, input().split()))
color = list(map(int, input().split()))
same = {}
dif = {}
for i in range(n):
if now[i] != goal[i]:
if goal[i] not in dif:
dif[goal[i]] = [i]
else:
dif[goal[i]].append(i)
else:
if now[i] not in same:
same[now[i]] = i
ans = []
count = 0
for paint in color:
if paint in dif:
k = dif[paint].pop()
if len(dif[paint]) == 0:
dif.pop(paint)
now[k] = paint
if paint not in same:
same[paint] = k
for _ in range(count+1):
ans.append(k + 1)
count = 0
else:
if paint in same:
for _ in range(count+1):
ans.append(same[paint] + 1)
count = 0
else:
count += 1
count += len(dif)
if count:
print("NO")
else:
print("YES")
ans = map(str, ans)
print(" ".join(ans))
``` | output | 1 | 82,802 | 7 | 165,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You finally woke up after this crazy dream and decided to walk around to clear your head. Outside you saw your house's fence — so plain and boring, that you'd like to repaint it.
<image>
You have a fence consisting of n planks, where the i-th plank has the color a_i. You want to repaint the fence in such a way that the i-th plank has the color b_i.
You've invited m painters for this purpose. The j-th painter will arrive at the moment j and will recolor exactly one plank to color c_j. For each painter you can choose which plank to recolor, but you can't turn them down, i. e. each painter has to color exactly one plank.
Can you get the coloring b you want? If it's possible, print for each painter which plank he must paint.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of planks in the fence and the number of painters.
The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — the initial colors of the fence.
The third line of each test case contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ n) — the desired colors of the fence.
The fourth line of each test case contains m integers c_1, c_2, ..., c_m (1 ≤ c_j ≤ n) — the colors painters have.
It's guaranteed that the sum of n doesn't exceed 10^5 and the sum of m doesn't exceed 10^5 over all test cases.
Output
For each test case, output "NO" if it is impossible to achieve the coloring b.
Otherwise, print "YES" and m integers x_1, x_2, ..., x_m, where x_j is the index of plank the j-th painter should paint.
You may print every letter in any case you want (so, for example, the strings "yEs", "yes", "Yes" and "YES" are all recognized as positive answer).
Example
Input
6
1 1
1
1
1
5 2
1 2 2 1 1
1 2 2 1 1
1 2
3 3
2 2 2
2 2 2
2 3 2
10 5
7 3 2 1 7 9 4 2 7 9
9 9 2 1 4 9 4 2 3 9
9 9 7 4 3
5 2
1 2 2 1 1
1 2 2 1 1
3 3
6 4
3 4 2 4 1 2
2 3 1 3 1 1
2 2 3 4
Output
YES
1
YES
2 2
YES
1 1 1
YES
2 1 9 5 9
NO
NO | instruction | 0 | 82,803 | 7 | 165,606 |
Tags: brute force, constructive algorithms, greedy
Correct Solution:
```
#fi = open("task3.txt")
#def input():
# return fi.readline()
T = int(input())
for _ in range(T):
n,m = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
c = [int(x) for x in input().split()]
needed = [[] for _ in range(n)]
cols_at_end = [-1 for _ in range(n)]
total_needed = 0
for i in range(n):
if a[i] != b[i]:
needed[b[i]-1] += [i]
total_needed += 1
cols_at_end[b[i]-1] = i
plan = []
fail = False
for col in reversed(c):
if len(needed[col-1]) > 0:
where = needed[col-1].pop()
plan += [where]
total_needed -= 1
else:
if cols_at_end[col-1] != -1:
plan += [cols_at_end[col-1]]
elif len(plan)==0:
fail = True
break
else:
plan += [plan[-1]]
if fail or total_needed > 0:
print("NO")
else:
print("YES")
print(' '.join([str(x+1) for x in reversed(plan)]))
``` | output | 1 | 82,803 | 7 | 165,607 |
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