text stringlengths 198 433k | conversation_id int64 0 109k |
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Provide tags and a correct Python 3 solution for this coding contest problem.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Tags: brute force, math, number theory
Correct Solution:
```
import math
def pf(n):
fac = {}
while n % 2 == 0:
fac[2] = fac.get(2, 0)+1
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
fac[i] = fac.get(i, 0)+1
n = n // i
if n > 2:
fac[n] = fac.get(n, 0)+1
return fac
def gp(f, p):
s = 1
e = int(math.ceil(math.log(p,f)))
val = 1
while(s<=e):
mid = (s+e)//2
if p%(f**mid)!=0:
e = mid - 1
else:
s = mid+1
val = mid
return val
test = int(input())
for _ in range(test):
p, q = map(int, input().split())
if p%q:
print(p)
else:
pfac = pf(q)
val = float("inf")
for k in pfac:
u = gp(k, p)
val = min(val, (k**(u-pfac[k]+1)))
ans = p//val
print(ans)
```
| 13,500 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Tags: brute force, math, number theory
Correct Solution:
```
from sys import stdin, stdout
from math import sqrt
#stdin = open('Q3.txt', 'r')
def II(): return int(stdin.readline())
def MI(): return map(int, stdin.readline().split())
bigp=10**18+7
primes=[]
def SieveOfEratosthenes(n,primes):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for p in range(2, n):
if prime[p]:
primes.append(p)
def solve():
p,q=MI()
if p%q != 0:
ans=p
else:
x,y=q,p
mind=bigp
sqrtq=int(sqrt(q))
sp=[i for i in primes if i<=sqrtq]+[bigp]
for i in sp:
j=i
if x==1:
break
qe=0
while x%j==0:
qe+=1
x=x//j
if i==bigp:
qe,j=1,x
if qe>0:
pe=qe
y=y//pow(j,qe)
while y%j==0:
pe+=1
y=y//j
mind=min(mind,pow(j,pe-qe+1))
ans=p//mind
stdout.write(str(ans)+"\n")
SieveOfEratosthenes(32000,primes)
t=II()
for _ in range(t):
solve()
```
| 13,501 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Tags: brute force, math, number theory
Correct Solution:
```
import sys
#from collections import deque
#from functools import *
#from fractions import Fraction as f
from copy import *
from bisect import *
#from heapq import *
from math import gcd,ceil,sqrt
from itertools import permutations as prm,product
def eprint(*args):
print(*args, file=sys.stderr)
zz=1
#sys.setrecursionlimit(10**6)
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
di=[[-1,0],[1,0],[0,1],[0,-1]]
def string(s):
return "".join(s)
def fori(n):
return [fi() for i in range(n)]
def inc(d,c,x=1):
d[c]=d[c]+x if c in d else x
def bo(i):
return ord(i)-ord('A')
def li():
return [int(xx) for xx in input().split()]
def fli():
return [float(x) for x in input().split()]
def comp(a,b):
if(a>b):
return 2
return 2 if a==b else 0
def gi():
return [xx for xx in input().split()]
def cil(n,m):
return n//m+int(n%m>0)
def fi():
return int(input())
def pro(a):
return reduce(lambda a,b:a*b,a)
def swap(a,i,j):
a[i],a[j]=a[j],a[i]
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
def gh():
sys.stdout.flush()
def isvalid(i,j,n,m):
return 0<=i<n and 0<=j<m
def bo(i):
return ord(i)-ord('a')
def graph(n,m):
for i in range(m):
x,y=mi()
a[x].append(y)
a[y].append(x)
t=fi()
while t>0:
t-=1
p,q=mi()
d={}
if p%q:
print(p)
else:
for j in range(2,ceil(sqrt(q))+3):
if q%j==0:
d[j]=0
while q%j==0:
q//=j
d[j]+=1
if q>=2:
d[q]=1
mini=10**18
n=p
for i in d:
c=0
if p%i==0:
while p%i==0:
p//=i
c+=1
mini=min(mini,i**(c-d[i]+1))
print(n//mini)
```
| 13,502 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Tags: brute force, math, number theory
Correct Solution:
```
from sys import stdin,stdout
from collections import *
from math import gcd,floor,ceil
st=lambda:list(stdin.readline().strip())
li=lambda:list(map(int,stdin.readline().split()))
mp=lambda:map(int,stdin.readline().split())
inp=lambda:int(stdin.readline())
pr=lambda n: stdout.write(str(n)+"\n")
INF=float('inf')
def factors(n):
l=[]
i=2
while i*i<=n:
if n%i==0:
x=[i,0]
while n%i==0:
n//=i
x[1]+=1
l.append(x)
i+=1
if n>1:
l.append([n,1])
return l
def solve():
a,b=mp()
if a%b:
pr(a)
return
fact=factors(b)
mi=a
for i in fact:
p=a
count=0
while p%i[0]==0:
count+=1
p//=i[0]
cur=1
while count >= i[1]:
cur*=i[0]
count-=1
mi=min(mi,cur)
pr(a//mi)
for _ in range(inp()):
solve()
```
| 13,503 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Tags: brute force, math, number theory
Correct Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop
import math
from collections import *
from functools import reduce,cmp_to_key,lru_cache
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# import sys
# input = sys.stdin.readline
M = mod = 10**9 + 7
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip().split()]
def st():return str(input().rstrip())[2:-1]
def val():return int(input().rstrip())
def li2():return [str(i)[2:-1] for i in input().rstrip().split()]
def li3():return [int(i) for i in st()]
def factorize(n):
d = defaultdict(int)
for i in range(2, int(n ** 0.5) + 1):
while n % i == 0:
d[i] += 1
n = n // i
if n - 1:d[n] = 1
return d
for i in range(val()):
a, b = li()
if a % b:
print(a)
continue
mydict = factorize(b)
l = list(mydict)
ans = a
finans = 1
for i in l:
temp = ans
while temp % i == 0:temp = temp // i
temp *= i ** (mydict[i] - 1)
finans = max(finans, temp)
print(finans)
```
| 13,504 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Tags: brute force, math, number theory
Correct Solution:
```
for _ in range(int(input())):
p, q = map(int,input().split())
c = q
d = p
i = 1
factor = []
while i*i <= q:
if q % i == 0:
factor.append(i)
if q//i != i:
factor.append(q//i)
i += 1
factor.sort(reverse=True)
factor.pop()
m = 1
for i in factor:
d = p
while d % c == 0:
d //= i
m = max(m, d)
print(m)
```
| 13,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Tags: brute force, math, number theory
Correct Solution:
```
T=int(input())
for _ in range(T):
p,q=map(int,input().split())
if p<q:
print(p)
elif p%q!=0:
print(p)
else:
s=set()
i=2
while i*i<=q:
if q%i==0:
s.add(i)
s.add(q//i)
i+=1
s.add(q)
l=list(s)
#print(l)
lis=[1]
for i in l:
ch=p
while ch%i==0:
ch//=i
if ch%q:
lis.append(ch)
break
print(max(lis))
```
| 13,506 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Tags: brute force, math, number theory
Correct Solution:
```
"""T=int(input())
for _ in range(0,T):
n=int(input())
a,b=map(int,input().split())
s=input()
s=[int(x) for x in input().split()]
for i in range(0,len(s)):
a,b=map(int,input().split())"""
import math
T=int(input())
for _ in range(0,T):
p,q=map(int,input().split())
if(p%q!=0):
print(p)
else:
n = q
L=[]
if(n%2==0):
L.append(2)
while(n%2==0):
n=n//2
for i in range(3,int(math.sqrt(n))+1,2):
if(n%i==0):
L.append(i)
while(n%i==0):
n=n//i
if(n>2):
L.append(n)
ans=1
for i in range(0,len(L)):
num=p
ele=L[i]
while(num%q==0 and num%ele==0):
num=num//ele
ans=max(ans, num)
print(ans)
```
| 13,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Submitted Solution:
```
"""
#If FastIO not needed, used this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict as dd, deque as dq
import math, string
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
MOD = 10**9+7
"""
Start with P
P has factors pi**qi
X = P, is X % Q == 0, doesn't work
Need the largest factor of P s.t. factor % Q = 0
30
10
1 2 3 5 6 10 15 30 so 15 is the answer
If P//Q is the smallest factor of P, the answer is the next smallest factor
Otherwise, the answer is either the next factor, or a smaller factor which doesn't divide P//Q
We need to remove a combination of factors from P until it doesn't divide Q any more
50 = 2*2*5
100 = 2*2*5*5
10 = 2*5
30 = 2*3*5
Answer = 5*5
"""
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
def solve():
P, Q = getInts()
if P < Q:
return P
if P % Q > 0:
return P
facs = sorted(list(factors(Q)))
best = 0
for fac in facs:
if fac == 1:
continue
power = 1
while P % fac**power == 0:
#print(P,fac,power,flush=True)
tmp = P//(fac**power)
if tmp % Q > 0:
best = max(best,tmp)
power += 1
return best
for _ in range(getInt()):
print(solve())
```
Yes
| 13,508 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**5)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
def PrimeFactorization(x):
def plist(x):
if x < 2: return []
if x & 1 == 0: return [2] + plist(x >> 1)
for p in range(3, x + 1, 2):
if x % p == 0: return [p] + plist(x // p)
if p ** 2 > x: return [x]
pl = plist(x)
pp, ee = [], []
for p in pl:
if not pp or p != pp[-1]:
pp += [p]
ee += [0]
ee[-1] += 1
return [(p, e) for p, e in zip(pp, ee)]
def solve():
if p%q:return p
be=PrimeFactorization(q)
mn=p
for b,e in be:
c=p
c//=b**e
cur=b
while c%b==0:
cur*=b
c//=b
mn=min(mn,cur)
return p//mn
for _ in range(II()):
p,q=MI()
print(solve())
```
Yes
| 13,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Submitted Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
# biesect_left is essentially an equivalent of lower_bound function in
# cpp and returns the first index not smaller than x.
from bisect import bisect_left;
from bisect import bisect_right;
from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1;
return x;
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## nCr function efficient using Binomial Cofficient
def nCr(n, k, modulus = 1):
if(k > n - k):
k = n - k;
res = 1;
for i in range(k):
res = res * (n - i);
res = res / (i + 1);
res %= modulus;
return int(res);
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret;
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
for _ in range(int(input())):
p, q = int_array();
if q > p or p % q != 0:
print(p);
continue;
fact = primefs(q);
ans = NINF;
for i in fact:
x = p;
while x % q == 0:
x //= i;
ans = max(ans, x);
print(ans);
```
Yes
| 13,510 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Submitted Solution:
```
import math
for i in range(int(input())):
p,q=map(int,input().split())
l=p
if p%q!=0:
print(p)
else:
g=[]
c=1
if q%2==0:
while q%2==0:
q=q//2
p=p//2
while p%2==0:
c=c*2
p=p//2
g.append(c*2)
for j in range(3,int(math.sqrt(q))+2,2):
if q%j==0:
c=1
while q%j==0:
q=q//j
p=p//j
while p%j==0:
c=c*j
p=p//j
g.append(c*j)
if q>2:
c=1
while p%q==0:
c=c*q
p=p//q
g.append(c)
print(l//(min(g)))
```
Yes
| 13,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Submitted Solution:
```
import math
def smallestDivisor(n):
if (n % 2 == 0):
return 2;
i = 3;
while(i * i <= n):
if (n % i == 0):
return i;
i += 2;
return n;
def prevPowerofK(n, k):
p = int(math.log(n) / math.log(k))
return int(math.pow(k, p))
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(i)
n = n // i
if n > 2:
l.append(n)
return l
t = int(input())
while t:
n,x= map(int,input().split())
if n>x:
if n%x!=0:
print(n)
else:
l2 = primeFactors(x)
d2 = {}
for j in l2:
if j in d2:
d2[j]+=1
else:
d2[j]=1
ans = 0
for i in d2:
if n%i==0:
power = 1
number = i
while n%number==0:
power+=1
number *=i
xxx = n// i**(power-1)
xxx *= i**(d2[i]-1)
ans = max(ans,xxx)
print(ans)
else:
print(n)
t-=1
```
No
| 13,512 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Submitted Solution:
```
import math
''' Get all the prime factors '''
def prime_factorization(x):
result = []
for i in range(2, int(math.sqrt(x)) + 1):
# If 'i' is a divisor of 'x',
if x % i == 0:
# Count how many times 'i' can divide 'x' consecutively.
count = 0
while x % i == 0:
count += 1
x //= i
result.append((i, count))
if x > 1:
result.append((x, 1))
return result
''' Find all divisors of a number '''
def find_all_divisors_of_a_number(x):
result = []
for i in range(1, int(math.sqrt(x)) + 1):
if x % i == 0:
result.append(i)
if i * i != x:
result.append(x // i)
return result
for _ in range(int(input())):
p, q = map(int, input().split())
if p < q:
print(p)
else:
if p % q != 0:
print(p)
else:
now = prime_factorization(q)
max_value = 0
for i in now:
cur = p
prime = i[0]
cnt = i[1]
while cur % (prime ** cnt) == 0:
cur /= prime
max_value = max(max_value, int(cur))
print(max_value)
```
No
| 13,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Submitted Solution:
```
import sys
import math
def II():
return int(sys.stdin.readline())
def LI():
return list(map(int, sys.stdin.readline().split()))
def MI():
return map(int, sys.stdin.readline().split())
def SI():
return sys.stdin.readline().strip()
t = II()
for q in range(t):
p,q = MI()
q2 = q
a = []
while True:
count = 0
for i in range(2,int(q**0.5)+1):
if q%i == 0:
count+=1
a.append(i)
q//=i
break
if count == 0:
a.append(q)
break
ans = []
while p%q2 == 0:
p//=q2
ans+=a
if ans:
ans.sort(reverse=True)
while True:
count = 0
for i in range(len(ans)):
if p*ans[i]%q2!=0:
p*=ans[i]
count = 1
ans.pop(i)
break
if count == 0:
break
print(p)
```
No
| 13,514 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.
To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that:
* p_i is divisible by x_i;
* x_i is not divisible by q_i.
Oleg is really good at division and managed to find all the answers quickly, how about you?
Input
The first line contains an integer t (1 β€ t β€ 50) β the number of pairs.
Each of the following t lines contains two integers p_i and q_i (1 β€ p_i β€ 10^{18}; 2 β€ q_i β€ 10^{9}) β the i-th pair of integers.
Output
Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.
One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.
Example
Input
3
10 4
12 6
179 822
Output
10
4
179
Note
For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.
For the second pair, where p_2 = 12 and q_2 = 6, note that
* 12 is not a valid x_2, since 12 is divisible by q_2 = 6;
* 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6.
The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
Submitted Solution:
```
from sys import stdin, gettrace
if gettrace():
inputi = input
else:
def input():
return next(stdin)[:-1]
def inputi():
return stdin.buffer.readline()
def solve():
p, q = map(int, inputi().split())
if p%q != 0:
print(p)
return
qf = {}
qr = q
i = 2
while i*i < qr:
while qr%i == 0:
qf[i] = qf.get(i, 0) + 1
qr //= i
i+=1
if qr != 1:
qf[qr] = 1
pf = {}
pr = p
for f in qf.keys():
pf[f] = 0
while pr%f == 0:
pf[f] += 1
pr //= f
best = 0
for f in qf.keys():
best = max(p//(f**(pf[f] -qf[f]+1)), best)
print(best)
def main():
t = int(inputi())
for _ in range(t):
solve()
if __name__ == "__main__":
main()
```
No
| 13,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Tags: greedy, implementation
Correct Solution:
```
a=input()
b=input()
sum1=0
sum2=0
for i in range(len(a)):
if(a[i]!=b[i]):
if(a[i]=='4'):
sum1=sum1+1
else:
sum2=sum2+1
print(max(sum1,sum2))
```
| 13,516 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Tags: greedy, implementation
Correct Solution:
```
def question3():
A = list(input())
B = list(input())
count_7_4 = 0
count_4_7 = 0
for i in range(len(A)):
if A[i] != B[i]:
if A[i] == "4":
count_4_7 += 1
else:
count_7_4 += 1
count = min(count_7_4,count_4_7) + max(count_7_4,count_4_7) - min(count_4_7,count_7_4)
return count
remained_test_cases = 1
while remained_test_cases > 0:
print(question3())
remained_test_cases -= 1
```
| 13,517 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Tags: greedy, implementation
Correct Solution:
```
def luckyConversion():
count1 = 0
count2 = 0
for i in range(len(a)):
if a[i] == b[i]:
pass
else:
if a[i] == '4':
count1 += 1
if a[i] == '7':
count2 += 1
#return(count2,count1)
return max(count2,count1)
a = input()
b = input()
x = luckyConversion()
print(x)
```
| 13,518 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Tags: greedy, implementation
Correct Solution:
```
def main():
l = [x == '7' for x, y in zip(input(), input()) if x != y]
x = sum(l)
print(max(x, len(l) - x))
if __name__ == '__main__':
main()
# Made By Mostafa_Khaled
```
| 13,519 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Tags: greedy, implementation
Correct Solution:
```
a = input()
b = input()
n = len(a)
x = 0
y = 0
for i in range(n):
if a[i] == '4' and b[i] == '7':
x += 1
if a[i] == '7' and b[i] == '4':
y += 1
print(max(x, y))
```
| 13,520 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Tags: greedy, implementation
Correct Solution:
```
c47,c74=0,0
a,b=input(),input()
for i in range(len(a)):
if a[i]==b[i]:continue
if a[i]=='4':
c47+=1
else:
c74+=1
print(max(c47,c74))
```
| 13,521 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Tags: greedy, implementation
Correct Solution:
```
a=input()
b=input()
k1=0
k2=0
for i in range (len(a)):
if a[i]!=b[i]:
if a[i]=='4':
k1+=1
else:
k2+=1
d=max(k1,k2)
print (d)
```
| 13,522 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Tags: greedy, implementation
Correct Solution:
```
import sys
read=sys.stdin.buffer.readline
mi=lambda:map(int,read().split())
li=lambda:list(mi())
cin=lambda:int(read())
a=input()
b=input()
d={'4':0,'7':0}
for i in range(len(a)):
if a[i]!=b[i]:
d[b[i]]+=1
print(max(d['4'],d['7']))
```
| 13,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Submitted Solution:
```
def solve(s1,s2):
x,y = 0,0
for i in range(len(s1)):
if s1[i] == '4' and s2[i] == '7':
x += 1
elif s1[i] == '7' and s2[i] == '4':
y += 1
print(max(x,y))
s1,s2 = input(),input()
if __name__ == '__main__':
solve(s1,s2)
```
Yes
| 13,524 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Submitted Solution:
```
a=input()
b=input()
da={'4':0,'7':0}
db={'4':0,'7':0}
for i in a:
da[i]+=1
for i in b:
db[i]+=1
dif=0
for i in range(len(a)):
if(a[i]!=b[i]):
dif+=1
ans=0
if(da==db):
ans=dif//2
else:
x=abs(da['4']-db['4'])
ans+=x
dif-=x
ans+=(dif//2)
print(ans)
```
Yes
| 13,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Submitted Solution:
```
a=input()
b=input()
count4=0
count7=0
for i in range(0,len(a)):
if a[i]!=b[i]:
if ord(a[i])-48 == 4:
count4+=1
else:
count7+=1
print(abs(count4-count7)+min(count4,count7))
```
Yes
| 13,526 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Submitted Solution:
```
a = input()
b = input()
l = len(a)
d = {'4':0,'7':0}
for i in range(l):
if a[i] != b[i]:
d[a[i]] += 1
vals = d.values()
print(min(vals) + (max(vals)-min(vals)))
```
Yes
| 13,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Submitted Solution:
```
a=input()
b=input()
l1=list(a)
l2=list(b)
l3=[]
for i in range(len(l1)):
if(l1[i]!=l2[i]):
l3.append(i)
i=0
sum1=0
while(i<len(l3)-1):
if((l3[i+1]-l3[i])==1):
if(l1[l3[i]]==l1[l3[i+1]]):
l1[l3[i]]=l2[l3[i]]
sum1=sum1+1
i=i+1
else:
l1[l3[i]]=l2[l3[i]]
l1[l3[i+1]]=l2[l3[i+1]]
sum1=sum1+1
i=i+2
else:
l1[l3[i]]=l2[l3[i]]
sum1=sum1+1
i=i+1
if(l1[l3[len(l3)-1]] != l2[l3[len(l3)-1]]):
sum1=sum1+1
print(sum1)
```
No
| 13,528 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Submitted Solution:
```
def question3():
A = list(input())
B = list(input())
count = 0
for i in range(len(A)):
if i != len(A) - 1 :
if A[i] != B[i]:
if A[i+1] != B[i+1] and A[i] != B[i]:
count += 1
A[i],A[i+1] = A[i+1],A[i]
else:
count += 1
A[i] = B[i]
else:
if A[i] != B[i]:
count += 1
A[i] = B[i]
return count
remained_test_cases = 1
while remained_test_cases > 0:
print(question3())
remained_test_cases -= 1
```
No
| 13,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Submitted Solution:
```
def question3():
A = list(input())
B = list(input())
count = 0
for i in range(len(A)):
if i != len(A) - 1 :
if A[i] != B[i]:
if A[i+1] != B[i+1] and A[i] != A[i+1]:
count += 1
A[i],A[i+1] = A[i+1],A[i]
else:
count += 1
A[i] = B[i]
else:
if A[i] != B[i]:
count += 1
A[i] = B[i]
# print("a ",A)
return count
remained_test_cases = 1
while remained_test_cases > 0:
print(question3())
remained_test_cases -= 1
```
No
| 13,530 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:
* replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
* swap any pair of digits in string a.
Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.
Input
The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output
Print on the single line the single number β the minimum number of operations needed to convert string a into string b.
Examples
Input
47
74
Output
1
Input
774
744
Output
1
Input
777
444
Output
3
Note
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites.
Submitted Solution:
```
from collections import *
a, b = [list(input()) for i in range(2)]
ans, d4, d7 = 0, deque([]), deque([])
for i in range(len(a)):
if a[i] != b[i]:
if a[i] == '4':
d4.append(i)
else:
d7.append(i)
for i in range(len(a)):
if a[i] != b[i]:
if a[i] == '4' and d7:
a[i], a[d7[0]] = a[d7[0]], a[i]
d7.popleft()
elif a[i] == '7' and d4:
a[i], a[d4[0]] = a[d4[0]], a[i]
d4.popleft()
ans += 1
print(ans)
```
No
| 13,531 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Tags: dp, math
Correct Solution:
```
for i in range(int(input())):
n,m,k=map(int,input().split())
print('YES' if n*m-1==k else 'NO')
```
| 13,532 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Tags: dp, math
Correct Solution:
```
t = int(input())
for i in range(t):
n,m,k = map(int,input().split())
cost = n*m-1
if cost==k:
print('YES')
else:
print('NO')
```
| 13,533 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Tags: dp, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n,m,k = map(int, input().split())
c = (m-1) + (m*(n-1))
if c==k:
print("YES")
else:
print("NO")
```
| 13,534 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Tags: dp, math
Correct Solution:
```
for i in range(int(input())):
x, y, d = [int(x) for x in input().split()]
if d == x-1 + x * (y - 1):
print("Yes")
else:
print("No")
```
| 13,535 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Tags: dp, math
Correct Solution:
```
t=int(input())
while(t):
t=t-1
n, m, k = map(int, input().split(' '))
i,j = 1, 1
b = True
c=0
while(i!=n or j!=m):
if b:
if j!=m:
j=j+1
c+=i
b = False
if not b:
if i!=n:
i=i+1
c+=j
b=True
if c==k:
print("YES")
else:
print("NO")
```
| 13,536 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Tags: dp, math
Correct Solution:
```
from collections import defaultdict
d = defaultdict(lambda: 0)
def check(a,b):
ans=a
ans+=(b*(a+1))
return ans
for i in range(int(input())):
n,m,k = map(int, input().split())
rx=n-1
ry=m-1
a=check(rx,ry)
b=check(ry,rx)
if a==k or b==k:
print("YES")
else:
print("NO")
```
| 13,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Tags: dp, math
Correct Solution:
```
def arrIn(): return list(map(int,input().split()))
def mapIn(): return map(int,input().split())
for ii in range(int(input())):
n,m,k=mapIn()
ans=n-1+n*(m-1)
if(ans==k):print("YES")
else : print("NO")
```
| 13,538 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Tags: dp, math
Correct Solution:
```
for i in range(int(input())):
x,y,z=map(int,input().split())
if x*y-1==z:
print('YES')
else:
print('NO')
```
| 13,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Submitted Solution:
```
from sys import stdin
input = stdin.readline
from heapq import heapify,heappush,heappop,heappushpop
from collections import defaultdict as dd, deque as dq,Counter as C
from math import factorial as f ,ceil,gcd,sqrt,log
from bisect import bisect_left as bl ,bisect_right as br
from itertools import combinations as c,permutations as p
from math import factorial as f ,ceil,gcd,sqrt,log
mi = lambda : map(int,input().split())
ii = lambda: int(input())
li = lambda : list(map(int,input().split()))
mati = lambda r : [ li() for _ in range(r)]
lcm = lambda a,b : (a*b)//gcd(a,b)
def solve():
n,m,k=mi()
s=min(n-1+n*(m-1),m-1+m*n)
#print(s)
if s==k:
print("YES")
else:
print("NO")
for _ in range(ii()):
solve()
```
Yes
| 13,540 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Submitted Solution:
```
t=int(input())
while t>0:
lst=input()
n, m, k=lst.split(" ")
n=int(n)
m=int(m)
k=int(k)
if n*m-1==k:
print("YES")
else:
print("NO")
t-=1
```
Yes
| 13,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Submitted Solution:
```
T = int(input())
for _ in range(T):
N,M,K = map(int,input().split())
k = N-1 + (M-1)*N
print('YES' if k==K else 'NO')
```
Yes
| 13,542 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Submitted Solution:
```
a=int(input())
for i in range(0,a):
r,b,d=map(int, input().split())
if(r*b)-1==d:
print("YES")
else:
print("NO")
```
Yes
| 13,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Submitted Solution:
```
"""
-> Author : Om Anshuman #
-> About : IT Fresher #
-> Insti : VIT, Vellore #
-> Created : 28.04.2021 #
"""
"""
from sys import stdin, stdout
import math
from functools import reduce
import statistics
import numpy as np
import itertools
import sys
"""
def prog_name():
n,m,k = map(int,input().split())
if n==1 and m==1 :
print("YES")
else:
cost = min(n,m)-1
if cost+((max(n,m)-1)*(cost+1))==k:
print("YES")
else:
print("NO")
t = int(input())
for unique in range(t):
prog_name()
```
No
| 13,544 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Submitted Solution:
```
for i in range(int(input())) :
a = input().split(" ")
m = int(a[0])
n = int(a[1])
k = int(a[2])
if k>=m+n :
print("YES")
else :
print("NO")
```
No
| 13,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Submitted Solution:
```
a = int(input())
for i in range(6):
n,m,k = [int(i) for i in input().split()]
if (n*m-1) == k:print('Yes')
else:print('No')
#(1,1)
#(2,2) => (1,1+1) -> 1 burles, (1+1,1) -> 1 burles 2 burles
```
No
| 13,546 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a n Γ m grid. You are standing at cell (1, 1) and your goal is to finish at cell (n, m).
You can move to the neighboring cells to the right or down. In other words, suppose you are standing at cell (x, y). You can:
* move right to the cell (x, y + 1) β it costs x burles;
* move down to the cell (x + 1, y) β it costs y burles.
Can you reach cell (n, m) spending exactly k burles?
Input
The first line contains the single integer t (1 β€ t β€ 100) β the number of test cases.
The first and only line of each test case contains three integers n, m, and k (1 β€ n, m β€ 100; 0 β€ k β€ 10^4) β the sizes of grid and the exact amount of money you need to spend.
Output
For each test case, if you can reach cell (n, m) spending exactly k burles, print YES. Otherwise, print NO.
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 0
2 2 2
2 2 3
2 2 4
1 4 3
100 100 10000
Output
YES
NO
YES
NO
YES
NO
Note
In the first test case, you are already in the final cell, so you spend 0 burles.
In the second, third and fourth test cases, there are two paths from (1, 1) to (2, 2): (1, 1) β (1, 2) β (2, 2) or (1, 1) β (2, 1) β (2, 2). Both costs 1 + 2 = 3 burles, so it's the only amount of money you can spend.
In the fifth test case, there is the only way from (1, 1) to (1, 4) and it costs 1 + 1 + 1 = 3 burles.
Submitted Solution:
```
#!/usr/bin/env python3
import os
from sys import stdin
def solve(tc):
n, m, k = map(int, stdin.readline().split())
if n == 1 or m == 1:
if k == (n-1)+(m-1):
print("YES")
else:
print("NO")
return
check = 0
for i in range(1, n):
check += i
for j in range(1, m):
check += n
if k == check:
print("YES")
else:
print("NO")
tcs = 1
tcs = int(stdin.readline().strip())
tc = 1
while tc <= tcs:
solve(tc)
tc += 1
```
No
| 13,547 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Tags: interactive, math
Correct Solution:
```
alpha = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
def stringsum(a):
ans = 0
for x in a:
ans+=alpha.index(x)
return ans
t = int(input())
for i in range(t):
n,m = list(map(int,input().strip().split()))
ostrings = []
rstrings = []
bsum1=[]
bsum2=[]
bsum3 = []
for i in range(n):
ostrings.append(input())
for i in range(n-1):
rstrings.append(input())
for i in range(m):
s=0
for j in range(n):
s+= alpha.index(ostrings[j][i])
bsum1.append(s)
for i in range(m):
s = 0
for j in range(n-1):
s+=alpha.index(rstrings[j][i])
bsum2.append(s)
for i in range(m):
bsum3.append(bsum1[i]-bsum2[i])
ans = ''
for x in bsum3:
ans = ans+alpha[x]
print(ans)
```
| 13,548 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Tags: interactive, math
Correct Solution:
```
t = int(input())
# testcases
for tests in range(t):
# plz don't copy the code, to avoid plagiarism issue
n,m = [int(i) for i in input().split(' ')]
hash_chrs = [{} for i in range(m)]
for i in range(n):
inp = input()
# plz don't copy the code, to avoid plagiarism issue
for j in range(m):
if inp[j] in hash_chrs[j].keys():
hash_chrs[j][inp[j]] = hash_chrs[j][inp[j]]+1
else:
hash_chrs[j][inp[j]] = 1
# plz don't copy the code, to avoid plagiarism issue
for i in range(n-1):
inp = input()
for j in range(m):
hash_chrs[j][inp[j]] = hash_chrs[j][inp[j]]-1
# plz don't copy the code, to avoid plagiarism issue
sol = ''
for i in range(m):
for key in hash_chrs[i].keys():
if hash_chrs[i][key]>0:
sol+=key
print(sol)
# plz don't copy the code, to avoid plagiarism issue
########################################################################################
```
| 13,549 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Tags: interactive, math
Correct Solution:
```
import sys
input = sys.stdin.readline
def solve():
n, m = map(int, input().split())
original = [input() for _ in range(n)]
modified = [input() for _ in range(n-1)]
stolen_chars = []
for j in range(m):
chars_available = 0
for i in range(n):
chars_available += ord(original[i][j])
for i in range(n-1):
chars_available -= ord(modified[i][j])
stolen_chars.append(chr(chars_available))
return ''.join(stolen_chars)
t = int(input())
output = []
for _ in range(t):
print(solve())
sys.stdout.flush()
```
| 13,550 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Tags: interactive, math
Correct Solution:
```
import sys
from bisect import bisect
from math import sqrt, ceil, floor
def input(): return sys.stdin.readline().strip()
def iinput(): return int(input())
def rinput(): return map(int, sys.stdin.readline().strip().split())
def get_list(): return list(map(int, sys.stdin.readline().strip().split()))
mod = int(1e9)+7
for _ in range(iinput()):
n, m = rinput()
l = [{} for _ in range(m)]
ans = ""
for _ in range(n):
s = input()
for i in range(m):
l[i][s[i]] = l[i].get(s[i], 0) + 1
for _ in range(n-1):
s = input()
for i in range(m):
l[i][s[i]] = l[i].get(s[i], 0) + 1
for i in range(m):
for key in l[i]:
if l[i][key]%2 == 1:
ans += key
break
print(ans)
```
| 13,551 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Tags: interactive, math
Correct Solution:
```
import sys
from string import ascii_lowercase as alph
t = int(input())
for _ in range(t):
n, m = map(int, input().split())
cnt = [0] * m
for i in range(n):
ind = 0
for x in input():
cnt[ind] += alph.index(x) + 1
ind += 1
for i in range(n - 1):
ind = 0
for x in input():
cnt[ind] -= alph.index(x) + 1
ind += 1
for i in range(m):
ind = cnt[i]
if ind != 0:
print(alph[ind - 1], end='')
print()
sys.stdout.flush()
```
| 13,552 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Tags: interactive, math
Correct Solution:
```
import sys
from copy import deepcopy
input = sys.stdin.readline
for _ in range(int(input())):
n, m = map(int, input().split())
total = [0 for _ in range(m)]
for i in range(n):
tmp = list(input().rstrip())
for j in range(len(tmp)):
total[j] += ord(tmp[j])
for i in range(n - 1):
tmp = list(input().rstrip())
for j in range(len(tmp)):
total[j] -= ord(tmp[j])
for i in range(m):
print(chr(total[i]), end = '', flush=True)
print()
```
| 13,553 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Tags: interactive, math
Correct Solution:
```
#!/usr/bin/python3
# -*- coding: utf-8 -*-
'''
f = open('test.txt')
byt = f.readlines()
for x in byt:
print(x)
print(byt)
'''
t = int(input())
while t > 0:
t -= 1
s = input().split()
n = int(s[0])
m = int(s[1])
#ε»Ίη«ζ―εηεε
Έηε葨
a = []
for i in range(n):
s = input()
if i == 0:
for x in s:
ai = dict()
ai[x] = 1
a.append(ai)
else:
for j in range(m):
ai = a[j]
if ai.get(s[j]) == None:
ai[s[j]] = 1
else:
ai[s[j]] += 1
a[j] = ai
#δ»ζ―εηεε
ΈδΈε ι€εΊη°θΏη
for i in range(n-1):
s = input()
for j in range(m):
ai = a[j]
if ai[s[j]] == 1:
del ai[s[j]]
else:
ai[s[j]] -= 1
#ε°ζ―εε©δ½ηδΈδΈͺζΌζ₯ε¨δΈεεΏ
ans = ''
for j in range(m):
ai = a[j]
key = list(ai.keys())[0]
ans += key
print(ans)
```
| 13,554 |
Provide tags and a correct Python 3 solution for this coding contest problem.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Tags: interactive, math
Correct Solution:
```
from collections import Counter
for _ in range(int(input())):
n, m = map(int, input().split())
f = [];l = []
f1={};l1={}
for i in range(n):
f += [input()]
for j in range(n - 1):
l += [input()]
for i in range(m):
s="";t=""
for j in range(n):
s+=f[j][i]
if j <len(l):
t+=l[j][i]
f1[i]=list(s)
l1[i]=list(t)
res=[]
for i in range(m):
w=list(f1[i])
t=list(l1[i])
res+= list((Counter(w) - Counter(t)).elements())
print("".join(res))
```
| 13,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#######################################
from collections import *
from collections import deque
from operator import itemgetter , attrgetter
from decimal import *
import bisect
import math
import heapq as hq
#import sympy
MOD=10**9 +7
def is_prime(n):
if n == 2 or n == 3: return True
if n < 2 or n%2 == 0: return False
if n < 9: return True
if n%3 == 0: return False
r = int(n**0.5)
# since all primes > 3 are of the form 6n Β± 1
# start with f=5 (which is prime)
# and test f, f+2 for being prime
# then loop by 6.
f = 5
while f <= r:
if n % f == 0: return False
if n % (f+2) == 0: return False
f += 6
return True
def pow(a,b,m):
ans=1
while b:
if b&1:
ans=(ans*a)%m
b//=2
a=(a*a)%m
return ans
#vis=[]
#graph=[]
def ispalindrome(s):
if s[:]==s[::-1]:
return 1
return 0
dp=[]
limit=[]
v=[]
def dpdfs(u,t=-1):
dp[0][u]=0
dp[1][u]=0
for i in v[u]:
if i==t:
continue
if dp[1][i]==-1:
dpdfs(i,u)
dp[0][u]+=max(abs(limit[0][u]-limit[1][i])+dp[1][i],abs(limit[0][u]-limit[0][i])+dp[0][i])
dp[1][u] += max(abs(limit[1][u] - limit[1][i]) + dp[1][i], abs(limit[1][u] - limit[0][i]) + dp[0][i])
vis=[]
f=0
def dfs(i):
vis[i]=1
act[i]=1
for j in v[i]:
if act[j]:
f=1
#print(-1)
return -1
if vis[j]==0:
if dfs(j)==-1:
return -1
act[i]=0
ans.append(i)
return 0
from queue import PriorityQueue
def z_algorithm(s):
res = [0] * len(s)
res[0] = len(s)
i, j = 1, 0
while i < len(s):
while i + j < len(s) and s[j] == s[i + j]:
j += 1
res[i] = j
if j == 0:
i += 1
continue
k = 1
while i + k < len(s) and k + res[k] < j:
res[i + k] = res[k]
k += 1
i, j = i + k, j - k
return res
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
# Function to return LCM of two numbers
def lcm(a, b):
return (a / gcd(a, b)) * b
for _ in range(int(input())):
n,m=map(int,input().split())
l=[]
for i in range(2*n-1):
l.append(input())
ma={}
ans=[]
for i in range(m):
ma={}
for j in range(2*n-1):
if l[j][i] in ma:
ma.pop(l[j][i])
else:
ma[l[j][i]]=1
t=list(ma.keys())
ans.append(t[0])
print("".join(ans))
```
Yes
| 13,556 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Submitted Solution:
```
######code with keroo##############
####################################
t = int(input())
for tests in range(t):
n,m = [int(i) for i in input().split(' ')]
strs = list()
hashchar=[{}for i in range(m)]
for i in range(n):
inputt = input()
strs.append(inputt)
for j in range(m):
if inputt[j] in hashchar[j].keys():
hashchar[j][inputt[j]] = hashchar[j][inputt[j]]+1
else:
hashchar[j][inputt[j]] = 1
for i in range(n-1):
inputt = input()
for j in range(m):
hashchar[j][inputt[j]] = hashchar[j][inputt[j]]-1
ans=""
for i in range(m):
for k in hashchar[i].keys():
if hashchar[i][k]>0:
ans+=k
print(ans)
```
Yes
| 13,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Submitted Solution:
```
#DaRk DeveLopeR
import sys
#taking input as string
input = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda: list(map(int,sys.stdin.readline().rstrip("\r\n").split()))
mod = 10**9+7; Mod = 998244353; INF = float('inf')
#______________________________________________________________________________________________________
import math
from bisect import *
from heapq import *
from collections import defaultdict as dd
from collections import OrderedDict as odict
from collections import Counter as cc
from collections import deque
from itertools import groupby
sys.setrecursionlimit(20*20*20*20+10) #this is must for dfs
def solve():
n,m=takeivr()
arr=[]
for i in range(2*n-1):
string=takesr()
arr.append(string)
ans=""
for i in range(m):
xor=0
for j in range(2*n-1):
# print(ord(arr[j][i]))
xor^=ord(arr[j][i])
# print(xor)
ans+=chr(xor)
# print()
print(ans)
def main():
global tt
if not ONLINE_JUDGE:
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
t = 1
t = takein()
#t = 1
for tt in range(1,t + 1):
solve()
if not ONLINE_JUDGE:
print("Time Elapsed :",time.time() - start_time,"seconds")
sys.stdout.close()
#---------------------- USER DEFINED INPUT FUNCTIONS ----------------------#
def takein():
return (int(sys.stdin.readline().rstrip("\r\n")))
# input the string
def takesr():
return (sys.stdin.readline().rstrip("\r\n"))
# input int array
def takeiar():
return (list(map(int, sys.stdin.readline().rstrip("\r\n").split())))
# input string array
def takesar():
return (list(map(str, sys.stdin.readline().rstrip("\r\n").split())))
# innut values for the diffrent variables
def takeivr():
return (map(int, sys.stdin.readline().rstrip("\r\n").split()))
def takesvr():
return (map(str, sys.stdin.readline().rstrip("\r\n").split()))
#------------------ USER DEFINED PROGRAMMING FUNCTIONS ------------------#
def ispalindrome(s):
return s==s[::-1]
def invert(bit_s):
# convert binary string
# into integer
temp = int(bit_s, 2)
# applying Ex-or operator
# b/w 10 and 31
inverse_s = temp ^ (2 ** (len(bit_s) + 1) - 1)
# convert the integer result
# into binary result and then
# slicing of the '0b1'
# binary indicator
rslt = bin(inverse_s)[3 : ]
return str(rslt)
def counter(a):
q = [0] * max(a)
for i in range(len(a)):
q[a[i] - 1] = q[a[i] - 1] + 1
return(q)
def counter_elements(a):
q = dict()
for i in range(len(a)):
if a[i] not in q:
q[a[i]] = 0
q[a[i]] = q[a[i]] + 1
return(q)
def string_counter(a):
q = [0] * 26
for i in range(len(a)):
q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
return(q)
def factorial(n,m = 1000000007):
q = 1
for i in range(n):
q = (q * (i + 1)) % m
return(q)
def factors(n):
q = []
for i in range(1,int(n ** 0.5) + 1):
if n % i == 0: q.append(i); q.append(n // i)
return(list(sorted(list(set(q)))))
def prime_factors(n):
q = []
while n % 2 == 0: q.append(2); n = n // 2
for i in range(3,int(n ** 0.5) + 1,2):
while n % i == 0: q.append(i); n = n // i
if n > 2: q.append(n)
return(list(sorted(q)))
def transpose(a):
n,m = len(a),len(a[0])
b = [[0] * n for i in range(m)]
for i in range(m):
for j in range(n):
b[i][j] = a[j][i]
return(b)
def power_two(x):
return (x and (not(x & (x - 1))))
def ceil(a, b):
return -(-a // b)
def seive(n):
a = [1]
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p ** 2,n + 1, p):
prime[i] = False
p = p + 1
for p in range(2,n + 1):
if prime[p]:
a.append(p)
return(a)
def pref(li):
pref_sum = [0]
for i in li:
pref_sum.append(pref_sum[-1]+i)
return pref_sum
def kadane(x): # maximum sum contiguous subarray
sum_so_far = 0
current_sum = 0
for i in x:
current_sum += i
if current_sum < 0:
current_sum = 0
else:
sum_so_far = max(sum_so_far, current_sum)
return sum_so_far
def binary_search(li, val):
# print(lb, ub, li)
ans = -1
lb = 0
ub = len(li)-1
while (lb <= ub):
mid = (lb+ub) // 2
# print('mid is',mid, li[mid])
if li[mid] > val:
ub = mid-1
elif val > li[mid]:
lb = mid+1
else:
ans = mid # return index
break
return ans
def upper_bound(li, num):
answer = -1
start = 0
end = len(li)-1
while (start <= end):
middle = (end+start) // 2
if li[middle] <= num:
answer = middle
start = middle+1
else:
end = middle-1
return answer # max index where x is not greater than num
def lower_bound(li, num):
answer = -1
start = 0
end = len(li)-1
while (start <= end):
middle = (end+start) // 2
if li[middle] >= num:
answer = middle
end = middle-1
else:
start = middle+1
return answer # min index where x is not less than num
#-----------------------------------------------------------------------#
ONLINE_JUDGE = __debug__
if ONLINE_JUDGE:
input = sys.stdin.readline
main()
```
Yes
| 13,558 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Submitted Solution:
```
from collections import Counter
from collections import defaultdict
t = 0
for i in range(int(input())):
n,m = map(int,input().split())
l = []
ll = []
ans = ""
for i in range(n):
l.append(str(input()))
for i in range(n-1):
ll.append(str(input()))
t = 0
tt = 0
for i in range(m):
d = defaultdict(int)
dd = defaultdict(int)
for j in range(n):
d[l[j][i]]+=1
for j in range(n-1):
dd[ll[j][i]]+=1
for j in d:
if d[j]-dd[j]>0:
ans+=j
print(ans)
```
Yes
| 13,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Submitted Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for i in range(t):
n, m = [int(x) for x in input().split()]
A = []
B = []
d = {I: {} for I in range(m)}
for j in range(n):
S = input()
print(S, len(S))
for I in range(m):
c = S[I]
if c not in d[I]:
d[I][c] = 0
d[I][c]+=1
for j in range(n-1):
S = input()
for I in range(m):
c = S[I]
d[I][c]-=1
if d[I][c]==0:
d[I].pop(c)
answer = ''
for I in range(m):
for c in d[I]:
answer+=chr(c)
break
print(answer)
sys.stdout.flush()
```
No
| 13,560 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Submitted Solution:
```
T = int(input())
def task():
n, m = [int(x) for x in input().split()]
start, end = [], []
for _ in range(n):
start.append(input())
for _ in range(n-1):
end.append(input())
for i in range(m):
d = {}
for x in end:
v = x[i]
if v not in d:
d[v] = 1
else:
d[v] += 1
for s in start:
v = s[i]
if v in d:
if d[v] == 0:
return s
d[v] -= 1
else:
return s
return -1
for _ in range(T):
print(task())
```
No
| 13,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Submitted Solution:
```
import sys
for _ in range(int(input())):
n,m=map(int,input().split())
gd = [0]*26
wd = [0]*26
arr1=[input() for i in range(n)]
arr2=[input() for i in range(n-1)]
for i in arr1:
for j in i:
gd[ord(j)-97] += 1
for i in arr2:
for j in i:
wd[ord(j)-97] += 1
ans = ''
# print(gd[:5])
# print(wd[:5])
for i in arr1:
cp = gd[:]
for j in i:
cp[ord(j)-97] -= 1
# print(cp[:5])
# print(wd)
if cp == wd:
ans = i
break
print(ans)
sys.stdout.flush()
```
No
| 13,562 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AquaMoon had n strings of length m each. n is an odd number.
When AquaMoon was gone, Cirno tried to pair these n strings together. After making (n-1)/(2) pairs, she found out that there was exactly one string without the pair!
In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 1 and at most m) and swapped the letters in the two strings of this pair at the selected positions.
For example, if m = 6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 2, 3 and 6 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".
Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.
AquaMoon found the remaining n-1 strings in complete disarray. Also, she remembers the initial n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?
Input
This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the [interactive problems guide](https://codeforces.com/blog/entry/45307) for the detailed information about flushing the output buffer.
The input consists of multiple test cases. The first line contains a single integer t (1 β€ t β€ 100) β the number of test cases.
The first line of each test case contains two integers n, m (1 β€ n β€ 10^5, 1 β€ m β€ 10^5) β the number of strings and the length of each string, respectively.
The next n lines each contain a string with length m, describing the original n strings. All string consists of lowercase Latin letters.
The next n-1 lines each contain a string with length m, describing the strings after Cirno exchanged and reordered them.
It is guaranteed that n is odd and that the sum of n β
m over all test cases does not exceed 10^5.
Hack format:
The first line should contain a single integer t. After that t test cases should follow in the following format:
The first line should contain two integers n and m.
The following n lines should contain n strings of length m, describing the original strings.
The following (n-1)/(2) lines should describe the pairs. They should contain, in the following order: the index of the first string i (1 β€ i β€ n), the index of the second string j (1 β€ j β€ n, i β j), the number of exchanged positions k (1 β€ k β€ m), and the list of k positions that are exchanged (k distinct indices from 1 to m in any order).
The final line should contain a permutation of integers from 1 to n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.
Output
For each test case print a single line with the stolen string.
Example
Input
3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk
Output
ccccc
cccc
kekeke
Note
In the first test case, "aaaaa" and "bbbbb" exchanged all positions, and "ccccc" is the stolen string.
In the second test case, "aaaa" and "bbbb" exchanged two first positions, and "cccc" is the stolen string.
This is the first test in the hack format:
3
3 5
aaaaa
bbbbb
ccccc
1 2 5 1 2 3 4 5
2 1 3
3 4
aaaa
bbbb
cccc
1 2 2 1 2
2 1 3
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
1 5 3 2 3 6
2 4 3 2 4 6
5 4 1 2 3
Submitted Solution:
```
alpha = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
def stringsum(a):
ans = 0
for x in a:
ans+=alpha.index(x)
return ans
t = int(input())
for i in range(t):
n,m = list(map(int,input().strip().split()))
ostrings = []
rstrings = []
osord=[]
rsord=[]
for i in range(n):
ostrings.append(input())
for i in range(n-1):
rstrings.append(input())
for x in ostrings:
osord.append(stringsum(x))
for x in rstrings:
rsord.append(stringsum(x))
print(ostrings[osord.index(sum(osord)-sum(rsord))])
```
No
| 13,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick has some permutation consisting of p integers from 1 to n. A segment [l, r] (l β€ r) is a set of elements pi satisfying l β€ i β€ r.
Nick calls a pair of segments [a0, a1] and [b0, b1] (1 β€ a0 β€ a1 < b0 β€ b1 β€ n) good if all their (a1 - a0 + b1 - b0 + 2) elements, when sorted in ascending order, form an arithmetic progression with a difference of 1. That is, when they sorted in ascending order, the elements are in the form {x, x + 1, x + 2, ..., x + m - 1}, for some x and m.
Your task is to find the number of distinct pairs of good segments in the given permutation. Two pairs of segments are considered distinct if the sets of elements contained in these pairs of segments are distinct. For example, any segment [l, r] (l < r) can be represented as a pair of segments, as [l, i] and [i + 1, r] (l β€ i β€ r). As all these pairs consist of the same set of elements, they are considered identical.
See the notes accompanying the sample tests for clarification.
Input
The first line contains integer n (1 β€ n β€ 3Β·105) β the permutation size. The second line contains n space-separated distinct integers pi, (1 β€ pi β€ n).
Output
Print a single integer β the number of good pairs of segments of permutation p.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
3
1 2 3
Output
3
Input
5
1 4 5 3 2
Output
10
Input
5
5 4 3 1 2
Output
10
Note
In the first sample the following pairs of segments are good: ([1, 1], [2, 2]); ([2, 2], [3, 3]); ([1, 2], [3, 3]). Pair of segments ([1, 1], [2, 3]) is by definition equivalent to pair ([1, 2], [3, 3]), since both of them covers the same set of elements, namely {1, 2, 3}.
In the third sample the following pairs of segments are good: ([4, 4], [5, 5]); ([3, 3],[4, 5]); ([2, 2],[3, 5]); ([1, 1],[2, 5]); ([3, 3],[5, 5]); ([2, 3],[5, 5]); ([1, 3],[5, 5]); ([2, 2],[3, 3]); ([1, 1],[2, 3]); ([1, 1],[2, 2]).
Submitted Solution:
```
a = int(input())
b = input()
print(a*(a-1)/2)
```
No
| 13,564 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick has some permutation consisting of p integers from 1 to n. A segment [l, r] (l β€ r) is a set of elements pi satisfying l β€ i β€ r.
Nick calls a pair of segments [a0, a1] and [b0, b1] (1 β€ a0 β€ a1 < b0 β€ b1 β€ n) good if all their (a1 - a0 + b1 - b0 + 2) elements, when sorted in ascending order, form an arithmetic progression with a difference of 1. That is, when they sorted in ascending order, the elements are in the form {x, x + 1, x + 2, ..., x + m - 1}, for some x and m.
Your task is to find the number of distinct pairs of good segments in the given permutation. Two pairs of segments are considered distinct if the sets of elements contained in these pairs of segments are distinct. For example, any segment [l, r] (l < r) can be represented as a pair of segments, as [l, i] and [i + 1, r] (l β€ i β€ r). As all these pairs consist of the same set of elements, they are considered identical.
See the notes accompanying the sample tests for clarification.
Input
The first line contains integer n (1 β€ n β€ 3Β·105) β the permutation size. The second line contains n space-separated distinct integers pi, (1 β€ pi β€ n).
Output
Print a single integer β the number of good pairs of segments of permutation p.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
3
1 2 3
Output
3
Input
5
1 4 5 3 2
Output
10
Input
5
5 4 3 1 2
Output
10
Note
In the first sample the following pairs of segments are good: ([1, 1], [2, 2]); ([2, 2], [3, 3]); ([1, 2], [3, 3]). Pair of segments ([1, 1], [2, 3]) is by definition equivalent to pair ([1, 2], [3, 3]), since both of them covers the same set of elements, namely {1, 2, 3}.
In the third sample the following pairs of segments are good: ([4, 4], [5, 5]); ([3, 3],[4, 5]); ([2, 2],[3, 5]); ([1, 1],[2, 5]); ([3, 3],[5, 5]); ([2, 3],[5, 5]); ([1, 3],[5, 5]); ([2, 2],[3, 3]); ([1, 1],[2, 3]); ([1, 1],[2, 2]).
Submitted Solution:
```
a = int(input())
b = input()
print(int(a*(a-1)/2))
```
No
| 13,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mirror Box is a name of a popular game in the Iranian National Amusement Park (INAP). There is a wooden box, 105 cm long and 100 cm high in this game. Some parts of the box's ceiling and floor are covered by mirrors. There are two negligibly small holes in the opposite sides of the box at heights hl and hr centimeters above the floor. The picture below shows what the box looks like.
<image>
In the game, you will be given a laser gun to shoot once. The laser beam must enter from one hole and exit from the other one. Each mirror has a preset number vi, which shows the number of points players gain if their laser beam hits that mirror. Also β to make things even funnier β the beam must not hit any mirror more than once.
Given the information about the box, your task is to find the maximum score a player may gain. Please note that the reflection obeys the law "the angle of incidence equals the angle of reflection".
Input
The first line of the input contains three space-separated integers hl, hr, n (0 < hl, hr < 100, 0 β€ n β€ 100) β the heights of the holes and the number of the mirrors.
Next n lines contain the descriptions of the mirrors. The i-th line contains space-separated vi, ci, ai, bi; the integer vi (1 β€ vi β€ 1000) is the score for the i-th mirror; the character ci denotes i-th mirror's position β the mirror is on the ceiling if ci equals "T" and on the floor if ci equals "F"; integers ai and bi (0 β€ ai < bi β€ 105) represent the x-coordinates of the beginning and the end of the mirror.
No two mirrors will share a common point. Consider that the x coordinate increases in the direction from left to right, so the border with the hole at height hl has the x coordinate equal to 0 and the border with the hole at height hr has the x coordinate equal to 105.
Output
The only line of output should contain a single integer β the maximum possible score a player could gain.
Examples
Input
50 50 7
10 F 1 80000
20 T 1 80000
30 T 81000 82000
40 T 83000 84000
50 T 85000 86000
60 T 87000 88000
70 F 81000 89000
Output
100
Input
80 72 9
15 T 8210 15679
10 F 11940 22399
50 T 30600 44789
50 F 32090 36579
5 F 45520 48519
120 F 49250 55229
8 F 59700 80609
35 T 61940 64939
2 T 92540 97769
Output
120
Note
The second sample is depicted above. The red beam gets 10 + 50 + 5 + 35 + 8 + 2 = 110 points and the blue one gets 120.
The red beam on the picture given in the statement shows how the laser beam can go approximately, this is just illustration how the laser beam can gain score. So for the second sample there is no such beam that gain score 110.
Submitted Solution:
```
print(302)
print("0 1000000")
coord, rad = 2000, 300
for i in range(300):
print(coord, rad)
coord, rad = coord+600, rad-1
print("1000000 1000000")
```
No
| 13,566 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Tags: constructive algorithms, graph matchings, implementation, shortest paths
Correct Solution:
```
n,t = map(int,input().split())
seq = input().rstrip().lstrip()
main = [i for i in seq]
def find_boys(s):
d = []
for i in range(len(s)) :
if s[i] == "B":
d.append(i)
return d
for i in range(t):
li = find_boys(main)
for j in li:
if j != len(main) -1 and main[j+1] == "G":
main[j] , main[j+1] = main[j+1] , main[j]
print("".join(main))
```
| 13,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Tags: constructive algorithms, graph matchings, implementation, shortest paths
Correct Solution:
```
n,p=map(int,input().split())
queue=input()
queue=list(queue)
i=0
for t in range(p):
while i<n-1:
if queue[i]=="B" and queue[i+1]=="G":
queue[i]="G"
queue[i+1]="B"
i+=2
else:
i+=1
i=0
print("".join(queue))
```
| 13,568 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Tags: constructive algorithms, graph matchings, implementation, shortest paths
Correct Solution:
```
n,t = map(int,input().split(" "))
q = input()
klist = list(i for i in q)
for _ in range(t):
i = 0
while i < n-1:
if klist[i] == "B" and klist [i+1] == "G":
klist[i],klist[i+1] = klist[i+1],klist[i]
i += 1
i += 1
print("".join(klist))
```
| 13,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Tags: constructive algorithms, graph matchings, implementation, shortest paths
Correct Solution:
```
import sys
u = sys.stdin.read().split("\n")
t = int(u[0].split()[1])
s = u[1]
newS = ""
for i in range(t):
if newS != "":
s = newS
newS = ""
switched = 0
for j in range(len(s)):
if s[j] == 'G':
if switched == 1:
newS += 'B'
switched = 0
else:
newS += s[j]
else:
if(j+1 == len(s)):
newS += s[j]
else:
if s[j+1] == 'G':
newS += 'G'
switched = 1
else:
newS += s[j]
print(newS)
```
| 13,570 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Tags: constructive algorithms, graph matchings, implementation, shortest paths
Correct Solution:
```
from sys import stdin,stdout
from collections import Counter
def ai(): return list(map(int, stdin.readline().split()))
def ei(): return map(int, stdin.readline().split())
def ip(): return int(stdin.readline().strip())
def op(ans): return stdout.write(str(ans) + '\n')
from math import ceil
n,t = ei()
s = [i for i in input()]
for i in range(t):
x = set()
for j in range(n-1):
if s[j] == 'B' and s[j+1] == 'G' and j not in x and j+1 not in x:
s[j],s[j+1] = s[j+1],s[j]
x.add(j);x.add(j+1)
print("".join(s))
```
| 13,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Tags: constructive algorithms, graph matchings, implementation, shortest paths
Correct Solution:
```
n,k = map(int,input().split())
q = input()
for i in range(k):
q = q.replace('BG','GB')
print(q)
```
| 13,572 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Tags: constructive algorithms, graph matchings, implementation, shortest paths
Correct Solution:
```
b, c = map(int, input().split())
# a = []
# x = ['B']
y = []
# g = ""
e = input()
def swap(s, i, j):
lst = list(s)
# print("length of lst ", len(lst))
if i > len(lst)-1:
return
elif i < len(lst)-1:
lst[i], lst[j] = lst[j], lst[i]
elif i == len(lst)-1:
lst = list(s)
aa = ''.join(lst)
# return ''.join(lst)
return aa
# def callit(e):
# for n in range(0,len(y)):
# e = swap(e,int(y[n]),int(y[n])+1)
# return e
#
# p = int(0)
# if len(e) > 0:
# for j in e :
# if j == 'B':
# print(e.index(j)+p)
# p = e.index(j)
# print("VALUE OF p",p)
# e = e[e.index(j):]
for j in range(0,len(e)):
if e[j] == 'B':
y.append(j)
# print(y)
# for n in range(0, len(y)):
# e = swap(e, int(y[n]), int(y[n]) + 1)
# print(e)
count = 0
while (count <c):
for n in range(0, len(y)):
e = swap(e, int(y[n]), int(y[n]) + 1)
e = e
y.clear()
Y=[]
for j in range(0, len(e)):
if e[j] == 'B':
y.append(j)
# print(y)
count = count + 1
print(e)
```
| 13,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Tags: constructive algorithms, graph matchings, implementation, shortest paths
Correct Solution:
```
seconds = [int(i) for i in input().split()][1]
queue = [i for i in input()]
for i in range(seconds):
pointer = 0
while pointer < len(queue):
try:
if queue[pointer] == 'B' and queue[pointer + 1] == 'G':
queue[pointer] = 'G'
queue[pointer + 1] = 'B'
pointer += 1
except:
pass
pointer += 1
print(''.join(queue))
```
| 13,574 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Submitted Solution:
```
# cook your dish here
n, t = tuple(map(int,input().split()))
s1=input()
s=[]
for tn in s1:
s.append(tn)
for lgc in range(t):
i=0
while(i<n-1):
if(s[i]=='B' and s[i+1]=='G' ):
s[i]='G'
s[i+1]='B'
i+=1
i+=1
for i in s:
print(i,end='')
```
Yes
| 13,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Submitted Solution:
```
n,t = map(int,input().split())
li = str(input())
l=list(li)
for n in range(t):
i=1
while(i<len(l)):
if(l[i]=='G' and l[i-1]=='B'):
l[i-1]='G'
l[i]='B'
i=i+1
i=i+1
print("".join(l))
```
Yes
| 13,576 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Submitted Solution:
```
n,b=map(int,input().split())
a=input()
for i in range(b):
a=a.replace("BG","GB")
print(a)
```
Yes
| 13,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Submitted Solution:
```
N=[int(x) for x in input().split()]
queue=input()
a=[]
b=[]
for i in range(0,N[0]):
a.append(queue[i])
b.append(queue[i])
for i in range(0,N[1]):
for j in range(0,N[0]-1):
if a[j]=='B' and a[j+1]=='G':
b[j]='G'
b[j+1]='B'
for j in range(0,N[0]):
a[j]=b[j]
print(''.join(a))
```
Yes
| 13,578 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Submitted Solution:
```
array = list(map(int, input().split(' ')))
queue = input()
n = array[0]
t = array[1]
item = 0
i = 0
while item != t:
pos = queue.find('B', i, n)
i = pos+2
queue = [i for i in queue]
if pos == n-1:
queue = ''.join(queue)
break
elif queue[pos+1] == 'G' and pos >= 0:
queue[pos],queue[pos+1] = queue[pos+1],queue[pos]
if i >= n:
i = 0
item += 1
queue = ''.join(queue)
print(queue)
```
No
| 13,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Submitted Solution:
```
x=list(map(int,input().split()))
s=list(input())
print(s)
for i in range(x[1]):
for j in range(len(s)-1):
if s[j]=='B' and s[j+1]=='G':
s[j]='G'
s[j+1]='B'
print(*s)
```
No
| 13,580 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Submitted Solution:
```
size,time = map(int,input().split(" "))
string = input()
length = len(string)
my_list = [a for a in string]
while time > 0:
i = 0
while i < size-1:
if my_list[i] == "B" and my_list[i+1] == "G":
my_list[i+1] = "B"
my_list[i] = "G"
i = i + 2
else:
i = i + 1
time-=1
print(my_list)
```
No
| 13,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the (i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 β€ n, t β€ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
Input
5 1
BGGBG
Output
GBGGB
Input
5 2
BGGBG
Output
GGBGB
Input
4 1
GGGB
Output
GGGB
Submitted Solution:
```
lis = list(map(int, input().split()))
st = list(input())
for j in range(lis[1]):
for i in range(lis[0] -1):
if st[i] == 'B' and st[i+1] == 'G':
st[i], st[i+1] = st[i+1], st[i]
stc = ("".join(st))
print(stc)
```
No
| 13,582 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
import math,sys,bisect,heapq,os
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
from functools import lru_cache
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
def input(): return sys.stdin.readline().rstrip('\r\n')
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
aj = lambda: list(map(int, input().split()))
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
def solve():
G = defaultdict(list)
def addEdge(a,b):
G[a].append(b)
G[b].append(a)
def dfs(node):
d = deque()
d.append(node)
vis[node] = True
ind = [];t = []
while d:
x = d.pop()
ind.append(x)
t.append(A[x])
for i in G.get(x,[]):
if not vis[i]:
vis[i] = True
d.append(i)
ind.sort(reverse = True)
t.sort(reverse = True)
while t:
Ans[ind.pop()] = t.pop()
n, =aj()
vis = [False]*(n+1)
A = aj()
B = aj()
for i in range(n):
p = B[i]
if i - p >= 0:
addEdge(i-p,i)
if i + p < n:
addEdge(i,i+p)
Ans = [-1]*n
for i in range(n):
if not vis[i]:
dfs(i)
Y(Ans == sorted(Ans))
try:
#os.system("online_judge.py")
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
solve()
```
| 13,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
n = int(input())
data = list(map(int, input().split()))
spans = list(map(int, input().split()))
connect = [[False for c in range(n)] for r in range(n)]
for p, span in enumerate(spans):
for r in [p-span, p+span]:
if r >= 0 and r < n:
connect[p][r] = connect[r][p] = True
def visit(data, connect, seen, group, i):
if not seen[i]:
seen[i] = True
group.append((i, data[i]))
for j in range(n):
if connect[i][j]:
visit(data, connect, seen, group, j)
seen = [False for i in range(n)]
for i in range(n):
group = []
visit(data, connect, seen, group, i)
group.sort()
#print()
#print(group)
values = sorted([value for (index, value) in group])
#print(values)
for i, value in enumerate(values):
data[group[i][0]] = value
#print(data)
if data == list(range(1, n+1)):
print('YES')
else:
print('NO')
```
| 13,584 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
n=int(input())
a=list(map(lambda x:int(x),input().split()))
fav=list(map(lambda x:int(x),input().split()))
class DisjSet:
def __init__(self, n):
# Constructor to create and
# initialize sets of n items
self.rank = [1] * n
self.parent = [i for i in range(n)]
self.extra = []
# Finds set of given item x
def find(self, x):
# Finds the representative of the set
# that x is an element of
if (self.parent[x] != x):
# if x is not the parent of itself
# Then x is not the representative of
# its set,
self.parent[x] = self.find(self.parent[x])
# so we recursively call Find on its parent
# and move i's node directly under the
# representative of this set
return self.parent[x]
# Do union of two sets represented
# by x and y.
def Union(self, x, y):
# Find current sets of x and y
xset = self.find(x)
yset = self.find(y)
# If they are already in same set
if xset == yset:
self.extra.append((x, y))
return
# Put smaller ranked item under
# bigger ranked item if ranks are
# different
if self.rank[xset] < self.rank[yset]:
self.parent[xset] = yset
elif self.rank[xset] > self.rank[yset]:
self.parent[yset] = xset
# If ranks are same, then move y under
# x (doesn't matter which one goes where)
# and increment rank of x's tree
else:
self.parent[yset] = xset
self.rank[xset] = self.rank[xset] + 1
obj=DisjSet(n)
for i in range(0,n):
d=fav[i]
if i-d>=0:
obj.Union(i,i-d)
if i+d<=n-1:
obj.Union(i, i + d)
hashmap={}
for i in range(n):
temp=obj.find(i)
if temp not in hashmap:
hashmap[temp]=set()
hashmap[temp].add(i)
flag=1
for i in range(n):
temp=obj.find(i)
if a[i]-1 not in hashmap[temp]:
flag=0
break
# print(hashmap)
if flag==0:
print("NO")
else:
print("YES")
```
| 13,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
n=int(input())
b=list(map(int,input().split()))
c=list(map(int,input().split()))
d=[[] for i in range(n+1)]
for i in range(n):
p=(i+1)-c[i]
q=(i+1)+c[i]
if p<=0 and q<=n:
d[i+1].append(q)
d[q].append(i+1)
elif p > 0 and q > n:
d[i + 1].append(p)
d[p].append(i + 1)
elif p>0 and q<=n:
d[i + 1].append(p)
d[p].append(i + 1)
d[i + 1].append(q)
d[q].append(i + 1)
v=[0]*(n+1)
j=1
k=0
while(j<=n):
if v[j]!=1:
f=[0]*(n+1)
s=[j]
r=[]
while(s):
m=s.pop()
r.append(m)
v[m]=1
f[b[m-1]]=1
for i in d[m]:
if v[i]==0:
s.append(i)
for p in r:
if f[p]==0:
k=1
break
if k==1:
break
j+=1
if k==0:
print("YES")
else:
print('NO')
```
| 13,586 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
n=int(input())
orden= input().split()
fav= input().split()
arr=[]
for i in range(n):
orden[i]=int(orden[i])
fav[i]=int(fav[i])
arr.append(i)
def union(el):
if arr[el] != el:
arr[el] = union(arr[el])
return arr[el]
cont=0
for i in fav:
if cont >= i:
arr[union(cont)] = union(cont - i)
if cont < n - i:
arr[union(cont)] = union(cont + i)
cont += 1
flag = True
for i in range(n):
if union(arr[i])==union(orden[i]-1):
continue
else:
flag = False
if flag:
print("YES")
else:
print("NO")
```
| 13,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
def dsu(nodo):
if nueva[nodo] != nodo:
nueva[nodo] = dsu(nueva[nodo])
return nueva[nodo]
else:
return nodo
def main():
n = int(input())
permutaciones = list(map(int, input().split()))
favoritos = list(map(int, input().split()))
l = []
for i in range(n):
l.append(i)
#lista sol
global nueva
nueva = l.copy()
#print(permutaciones)
#print(favoritos)
#print(l)
#print(nueva)
for i in l:
#nueva[dsu(i)] = dsu(abs(i - favoritos[i]))
if (i - favoritos[i] >= 0):
nueva[dsu(i)] = dsu(abs(i - favoritos[i]))
if (i + favoritos[i] < n):
nueva[dsu(i)] = dsu(i + favoritos[i])
for i in l:
if dsu(i) != dsu(permutaciones[i]-1):
print("NO")
return
print("YES")
return
main()
```
| 13,588 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
arr_size = int( input())
arr_fin = input().split(" ")
arr_fin = map(int, arr_fin)
cell_fav = input().split(" ")
cell_fav = map(int, cell_fav)
graph = {}
# A este array le voy a aplicar DSU
arr_ini = [i for i in range(1,arr_size+1)]
for fav, i in zip(cell_fav, arr_ini):
graph[i] = [i]
if(i-fav > 0):
graph[i].append(abs(fav-i))
if(fav+i <= arr_size):
graph[i].append(fav+i)
dsu_arr = arr_ini[::]
for i in range(0,10):
for key in graph:
min_node = [key]
for node in graph[key]:
min_node.append(dsu_arr[node-1])
min_node = min(min_node)
for node in graph[key]:
dsu_arr[node-1] = min_node
# if(arr_size == 71):
# #print(graph)
# #print(dsu_arr)
cont = 0
flag = True
for cell in arr_fin:
if(dsu_arr[cell-1] != dsu_arr[cont]):
print("NO")
flag = False
break
cont += 1
if(flag):
print("YES")
```
| 13,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
n, p, d = int(input()), [x - 1 for x in list(map(int, input().split()))], list(map(int, input().split()))
g = [[False] * n for _ in range(n)]
for i in range(n):
g[i][i] = True
if i - d[i] >= 0:
g[i][i - d[i]] = g[i - d[i]][i] = True
if i + d[i] < n:
g[i][i + d[i]] = g[i + d[i]][i] = True
for k in range(n):
for i in range(n):
for j in range(n):
g[i][j] |= g[i][k] & g[k][j]
ans = True
for i in range(n):
ans &= g[i][p[i]]
print("YES" if ans else "NO")
# Made By Mostafa_Khaled
```
| 13,590 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Submitted Solution:
```
n = int(input())
l = list(map(lambda x: int(x) - 1, input().split()))
f = list(map(int, input().split()))
class UnionFind:
def __init__(self, n):
self.parent = list(range(n))
def find(self, a):
acopy = a
while a != self.parent[a]:
a = self.parent[a]
while acopy != a:
self.parent[acopy], acopy = a, self.parent[acopy]
return a
def union(self, a, b):
self.parent[self.find(b)] = self.find(a)
UF = UnionFind(n)
for i in range(n):
if 0 <= i - f[i]:
UF.union(i, i - f[i])
if i + f[i] < n:
UF.union(i, i + f[i])
works = True
for i in range(n):
if UF.find(i) == UF.find(l[i]):
pass
else:
works = False
break
if works:
print('YES')
else:
print('NO')
```
Yes
| 13,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Submitted Solution:
```
n=int(input())
perm=list(map(int,input().split()))
fav=list(map(int,input().split()))
adj=[[] for i in range(n+1)]
for i in range(1,n+1):
if (i-fav[i-1]) in range(1,n+1):
adj[i].append(i-fav[i-1])
adj[i-fav[i-1]].append(i)
if (i+fav[i-1]) in range(1,n+1):
adj[i].append(i+fav[i-1])
adj[i+fav[i-1]].append(i)
for i in range(n):
q=[perm[i]];vis=[True]*(n+1)
vis[perm[i]]=False;flag=0
if perm[i]==i+1:flag=1
while len(q)!=0:
r=q.pop()
for j in adj[r]:
if vis[j]:
vis[j]=False
q.append(j)
if i+1==j:
flag=1;break
if flag==1:break
if flag==0:
exit(print('NO'))
print('YES')
```
Yes
| 13,592 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Submitted Solution:
```
f = lambda: list(map(int, input().split()))
n, p, d = f()[0], f(), f()
c = list(range(n))
def g(x):
if c[x] != x: c[x] = g(c[x])
return c[x]
for x, k in enumerate(d):
if x >= k: c[g(x)] = g(x - k)
if x < n - k: c[g(x)] = g(x + k)
print('YES' if all(g(x) == g(y - 1) for x, y in zip(c, p)) else 'NO')
```
Yes
| 13,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
def prime_factors(x):
s=set()
n=x
i=2
while i*i<=n:
if n%i==0:
while n%i==0:
n//=i
s.add(i)
i+=1
if n>1:
s.add(n)
return s
from collections import Counter
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
import heapq
# c=sorted((i,int(val))for i,val in enumerate(input().split()))
# n = int(input())
#g=[[] for i in range(n+1)]
#arr=list(map(int, input().split()))
# ls = list(map(int, input().split()))
# n, k = map(int, input().split())
# n =int(input())
#arr=[(i,x) for i,x in enum]
#arr.sort(key=lambda x:x[0])
#print(arr)
#import math
# e=list(map(int, input().split()))
from collections import Counter
#print("\n".join(ls))
#print(os.path.commonprefix(ls[0:2]))
#n=int(input())
from bisect import bisect_right
#d=sorted(d,key=lambda x:(len(d[x]),-x)) d=dictionary d={x:set() for x in arr}
#n=int(input())
#n,m,k= map(int, input().split())
import heapq
#for _ in range(int(input())):
#n,k=map(int, input().split())
#input=sys.stdin.buffer.readline
#for _ in range(int(input())):
#arr = list(map(int, input().split()))import bisect
def par(x):
if root[x]!=x:
return par(root[x])
else:
return x
def max_edge(n):
return n*(n-1)//2
def union(u,v):
x=par(u)
y=par(v)
if x!=y:
if rank[x]>=rank[y]:
root[y]=root[x]
rank[x]+=rank[y]
elif rank[y]>rank[x]:
root[x]=root[y]
rank[y]+=rank[x]
#n,m= map(int, input().split())
n = int(input())
root=[i for i in range(n+1)]
rank=[1]*(n+1)
arr = list(map(int, input().split()))
move= list(map(int, input().split()))
p=[i+1 for i in range(n)]
for i in range(n):
if i+move[i]<n:
union(i,i+move[i])
if i-move[i]>=0:
union(i,i-move[i])
for i in range(n):
if par(arr[i]-1)!=par(i):
print("NO")
break
else:
print("YES")
```
Yes
| 13,594 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Submitted Solution:
```
n=int(input())
from collections import defaultdict
d=defaultdict(list)
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
for i in range(n):
x=b[i]
if i-x>=0:
d[i].append(i-x)
d[i-x].append(i)
if i+x<n:
d[i].append(i+x)
d[i+x].append(i)
#print(d)
f=1
for i in range(0,n):
loc=0
q=[i]
vis=[0]*n
vis[i]=1
while q:
t=q.pop()
# print(t)
if a[t]==a[i]:
loc=1
for z in d[t]:
if not vis[z]:
vis[z]=1
q.append(z)
if loc==0:
# print(i)
f=0
if f:
print('YES')
else:
print('NO')
```
No
| 13,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Submitted Solution:
```
arr_size = int( input())
arr_fin = input().split(" ")
arr_fin = map(int, arr_fin)
cell_fav = input().split(" ")
cell_fav = map(int, cell_fav)
graph = {}
# A este array le voy a aplicar DSU
arr_ini = [i for i in range(1,arr_size+1)]
for fav, i in zip(cell_fav, arr_ini):
graph[i] = [i]
if(i-fav > 0):
graph[i].append(abs(fav-i))
if(fav+i <= arr_size):
graph[i].append(fav+i)
#print(graph)
dsu_arr = arr_ini[::]
for key in graph:
min_node = [key]
for node in graph[key]:
min_node.append(dsu_arr[node-1])
min_node = min(min_node)
for node in graph[key]:
dsu_arr[node-1] = min_node
#print(dsu_arr)
cont = 0
flag = True
for cell in arr_fin:
if(dsu_arr[cell-1] != dsu_arr[cont]):
print("NO")
flag = False
break
cont += 1
if(flag):
print("YES")
```
No
| 13,596 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Submitted Solution:
```
n=int(input())
from collections import defaultdict
d=defaultdict(list)
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
for i in range(n):
x=b[i]
if i-x>=0:
d[i].append(i-x)
d[i-x].append(i)
if i+x<n:
d[i].append(i+x)
d[i+x].append(i)
for i in range(n):
d[i].append(i)
#print(d)
f=1
for i in range(0,n):
loc=0
q=[i]
vis=[0]*n
vis[i]=1
while q:
t=q.pop()
# print(t)
if a[t]==b[i]:
loc=1
for i in d[t]:
if not vis[i]:
vis[i]=1
q.append(i)
if loc==0:
# print(i)
f=0
if f:
print('YES')
else:
print('NO')
```
No
| 13,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day n cells of some array decided to play the following game. Initially each cell contains a number which is equal to it's ordinal number (starting from 1). Also each cell determined it's favourite number. On it's move i-th cell can exchange it's value with the value of some other j-th cell, if |i - j| = di, where di is a favourite number of i-th cell. Cells make moves in any order, the number of moves is unlimited.
The favourite number of each cell will be given to you. You will also be given a permutation of numbers from 1 to n. You are to determine whether the game could move to this state.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of cells in the array. The second line contains n distinct integers from 1 to n β permutation. The last line contains n integers from 1 to n β favourite numbers of the cells.
Output
If the given state is reachable in the described game, output YES, otherwise NO.
Examples
Input
5
5 4 3 2 1
1 1 1 1 1
Output
YES
Input
7
4 3 5 1 2 7 6
4 6 6 1 6 6 1
Output
NO
Input
7
4 2 5 1 3 7 6
4 6 6 1 6 6 1
Output
YES
Submitted Solution:
```
f = lambda: list(map(int, input().split()))
n, p, d = f()[0], f(), f()
c = list(range(n))
def g(x):
if c[x] != x: c[x] = g(c[x])
return c[x]
for x, k in zip(c, d):
if x >= k: c[g(x)] = g(x - k)
if x < n - k: c[g(x)] = g(x + k)
print('YES' if all(g(x) == g(y - 1) for x, y in zip(c, p)) else 'NO')
```
No
| 13,598 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sereja has got an array, consisting of n integers, a1, a2, ..., an. Sereja is an active boy, so he is now going to complete m operations. Each operation will have one of the three forms:
1. Make vi-th array element equal to xi. In other words, perform the assignment avi = xi.
2. Increase each array element by yi. In other words, perform n assignments ai = ai + yi (1 β€ i β€ n).
3. Take a piece of paper and write out the qi-th array element. That is, the element aqi.
Help Sereja, complete all his operations.
Input
The first line contains integers n, m (1 β€ n, m β€ 105). The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the original array.
Next m lines describe operations, the i-th line describes the i-th operation. The first number in the i-th line is integer ti (1 β€ ti β€ 3) that represents the operation type. If ti = 1, then it is followed by two integers vi and xi, (1 β€ vi β€ n, 1 β€ xi β€ 109). If ti = 2, then it is followed by integer yi (1 β€ yi β€ 104). And if ti = 3, then it is followed by integer qi (1 β€ qi β€ n).
Output
For each third type operation print value aqi. Print the values in the order, in which the corresponding queries follow in the input.
Examples
Input
10 11
1 2 3 4 5 6 7 8 9 10
3 2
3 9
2 10
3 1
3 10
1 1 10
2 10
2 10
3 1
3 10
3 9
Output
2
9
11
20
30
40
39
Tags: implementation
Correct Solution:
```
n,q=map(int,input().split())
a=list(map(int, input().split()))
b=0
s=''
for i in range(q):
l=list(map(int, input().split()))
if l[0]==3:
s+=str(a[l[1]-1]+b)+'\n'
elif l[0]==2:
b+=l[1]
elif l[0]==1:
a[l[1]-1]= l[2]-b
print(s)
```
| 13,599 |
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