text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n.
Input
The only line contains two integers n and k (1 โค n, k โค 5 โ
10^5).
Output
Print one integer โ the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Tags: combinatorics, math, number theory
Correct Solution:
```
n,k=map(int,input().split())
mod = 998244353
f = [1]
for i in range(n):
f += [f[-1]*(i+1) % mod]
def comb(a, b):
return f[a]*pow(f[b], mod-2, mod)*pow(f[a-b], mod-2, mod) % mod
ans=0
for i in range(1,n+1):
m=(n//i)-1
if m<(k-1):break
ans+=comb(m,k-1)
ans%=mod
print(ans)
```
| 100,500 |
Provide tags and a correct Python 2 solution for this coding contest problem.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n.
Input
The only line contains two integers n and k (1 โค n, k โค 5 โ
10^5).
Output
Print one integer โ the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Tags: combinatorics, math, number theory
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=998244353
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
def inv(x):
return pow(x,mod-2,mod)
n,k=li()
fac=[1]*(n+1)
for i in range(2,n+1):
fac[i]=(fac[i-1]*i)%mod
ans=0
for i in range(1,n+1):
tp=n/i
if tp<k:
break
temp=(fac[tp-1]*inv(fac[k-1]))%mod
temp=(temp*inv(fac[tp-k]))%mod
ans=(ans+temp)%mod
pn(ans)
```
| 100,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n.
Input
The only line contains two integers n and k (1 โค n, k โค 5 โ
10^5).
Output
Print one integer โ the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
def getInts(): return list(map(int,input().split()))
#MOD = 10**9 + 7
MOD = 998244353
def ncr(n,r):
num = denom = 1
for i in range(r):
num = (num*(n-i))%MOD
denom = (denom*(i+1))%MOD
return (num * pow(denom,MOD-2,MOD))%MOD
"""
Chains of multiples
Iterate over first of chain
"""
N, K = map(int,input().split())
if N < K:
print(0)
else:
if K == 1:
print(N)
else:
ans = 0
for i in range(1,N+1):
#how many multiples of i are less than or equal to N?
if N//i > K:
ans += ncr(N//i-1,K-1)
ans %= MOD
elif N//i == K:
ans += 1
ans %= MOD
else:
break
print(ans)
```
Yes
| 100,502 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n.
Input
The only line contains two integers n and k (1 โค n, k โค 5 โ
10^5).
Output
Print one integer โ the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa=ifa[::-1]
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
'''
class SMT:
def __init__(self,arr):
self.n=len(arr)-1
self.arr=[0]*(self.n<<2)
self.lazy=[0]*(self.n<<2)
def Build(l,r,rt):
if l==r:
self.arr[rt]=arr[l]
return
m=(l+r)>>1
Build(l,m,rt<<1)
Build(m+1,r,rt<<1|1)
self.pushup(rt)
Build(1,self.n,1)
def pushup(self,rt):
self.arr[rt]=self.arr[rt<<1]+self.arr[rt<<1|1]
def pushdown(self,rt,ln,rn):#lr,rn่กจๅบ้ดๆฐๅญๆฐ
if self.lazy[rt]:
self.lazy[rt<<1]+=self.lazy[rt]
self.lazy[rt<<1|1]+=self.lazy[rt]
self.arr[rt<<1]+=self.lazy[rt]*ln
self.arr[rt<<1|1]+=self.lazy[rt]*rn
self.lazy[rt]=0
def update(self,L,R,c,l=1,r=None,rt=1):#L,R่กจ็คบๆไฝๅบ้ด
if r==None: r=self.n
if L<=l and r<=R:
self.arr[rt]+=c*(r-l+1)
self.lazy[rt]+=c
return
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
if L<=m: self.update(L,R,c,l,m,rt<<1)
if R>m: self.update(L,R,c,m+1,r,rt<<1|1)
self.pushup(rt)
def query(self,L,R,l=1,r=None,rt=1):
if r==None: r=self.n
#print(L,R,l,r,rt)
if L<=l and R>=r:
return self.arr[rt]
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
ans=0
if L<=m: ans+=self.query(L,R,l,m,rt<<1)
if R>m: ans+=self.query(L,R,m+1,r,rt<<1|1)
return ans
'''
class DSU:#ๅฎน้+่ทฏๅพๅ็ผฉ
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]<self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#็งฉ+่ทฏๅพ
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
#print(flag)
return flag
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
def nb(i,j):
for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
if 0<=ni<n and 0<=nj<m:
yield ni,nj
@bootstrap
def gdfs(r,p):
if len(g[r])==1 and p!=-1:
yield None
for ch in g[r]:
if ch!=p:
yield gdfs(ch,r)
yield None
t=1
for i in range(t):
n,k=RL()
mod=998244353
ans=0
fact(n,mod)
ifact(n,mod)
for i in range(1,n+1):
c=n//i
if c<k:
break
ans+=com(c-1,k-1,mod)
ans%=mod
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
```
Yes
| 100,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n.
Input
The only line contains two integers n and k (1 โค n, k โค 5 โ
10^5).
Output
Print one integer โ the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
n,k=map(int,input().split())
from math import factorial
if n<k:
print(0)
else:
ans=0
m=998244353
def ncr(n, r, p):
# initialize numerator
# and denominator
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
if k==1:
print(n)
else:
for i in range(1,n):
if n//i>k:
ans+=ncr(n//i-1,k-1,m)
ans%=m
elif n//i==k:
ans+=1
ans%=m
else:
break
print(ans)
```
Yes
| 100,504 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n.
Input
The only line contains two integers n and k (1 โค n, k โค 5 โ
10^5).
Output
Print one integer โ the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
#!/usr/bin/env python
# coding:utf-8
# Copyright (C) dirlt
MOD = 998244353
def pow(a, x):
ans = 1
while x > 0:
if x % 2 == 1:
ans = (ans * a) % MOD
a = (a * a) % MOD
x = x // 2
return ans
def run(n, k):
fac = [1] * (n + 1)
for i in range(1, n + 1):
fac[i] = (fac[i - 1] * i) % MOD
ans = 0
for a0 in range(1, n + 1):
nn = n // a0 - 1
if (nn - k + 1) < 0: break
a = fac[nn]
b = fac[k - 1]
c = fac[nn - k + 1]
d = (b * c) % MOD
e = pow(d, MOD - 2)
f = (a * e) % MOD
ans = (ans + f) % MOD
return ans
# this is codeforces main function
def main():
from sys import stdin
def read_int():
return int(stdin.readline())
def read_int_array(sep=None):
return [int(x) for x in stdin.readline().split(sep)]
def read_str_array(sep=None):
return [x.strip() for x in stdin.readline().split(sep)]
import os
if os.path.exists('tmp.in'):
stdin = open('tmp.in')
n, k = read_int_array()
ans = run(n, k)
print(ans)
if __name__ == '__main__':
main()
```
Yes
| 100,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n.
Input
The only line contains two integers n and k (1 โค n, k โค 5 โ
10^5).
Output
Print one integer โ the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
from sys import stdin as f
from math import factorial as fac
def binom_coeff(n, k):
return fac(n) / (fac(k) * fac(n - k))
def find_arr_num(n, k, m):
if k == 1:
return n
result, a = 0, 1
while n // a >= k:
count = n // a
result = result + (binom_coeff(count-1, k-1) % m)
a = a + 1
#print(a)
#print(result)
return int(result % m)
m = 998244353
n, k = [int(i) for i in f.readline().strip().split()]
print(find_arr_num(n, k, m))
```
No
| 100,506 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n.
Input
The only line contains two integers n and k (1 โค n, k โค 5 โ
10^5).
Output
Print one integer โ the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
n,k=map(int,input().split())
from math import factorial
if n<k:
print(0)
else:
ans=0
m=998244353
for i in range(1,n):
if n//i>k:
ans+=factorial(n//i-1)//(factorial(k-1)*factorial(n//i-k))
ans%=m
elif n//i==k:
ans+=1
ans%=m
else:
break
print(ans)
```
No
| 100,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n.
Input
The only line contains two integers n and k (1 โค n, k โค 5 โ
10^5).
Output
Print one integer โ the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
n,k = map(int,input().split())
p = 998244353
def pot(a,d):
res = 1
mno = a
while d > 0:
if d%2 == 1:
res = (res*mno)%p
mno = (mno*mno)%p
d//=2
return res
silnia = [1] * (n+1)
for i in range(2,n+1):
silnia[i] = (i*silnia[i-1])%p
#dla a_1 = i ile mozliwych (k-1)-tek liczb od 2 do n//a_i
wyn = 0
for a1 in range(1,n):
if n//a1 < k:
break
else:
zb = (n//a1)-1
#wyn += zb po k-1
wyn += silnia[zb] * pot(silnia[zb-k+1], p-2) * pot(silnia[k-1], p-2)
print(wyn%p)
```
No
| 100,508 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n.
Input
The only line contains two integers n and k (1 โค n, k โค 5 โ
10^5).
Output
Print one integer โ the number of stable arrays [a_1, a_2, ..., a_k] such that 1 โค a_1 < a_2 < ... < a_k โค n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
#Sorry Pacy
print(16)
```
No
| 100,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Tags: dp, greedy, math, sortings
Correct Solution:
```
t = int(input())
for i in range(t):
a, p = map(int, input().split())
binarr = list(input())
gaps = []
acts = 0
if binarr[0] == '1':
acts += 1
cost = 0
current = 0
for bit in binarr:
if bit == '0':
current += 1
else:
if current:
gaps.append(current)
current = 0
acts += 1
if current:
gaps.append(current)
if binarr[0] == '0':
del gaps[0]
if binarr[-1] == '0' and acts:
del gaps[-1]
gaps.sort()
for gap in gaps:
if gap * p <= a:
cost += gap * p
acts -= 1
else:
break
print(cost + acts * a)
```
| 100,510 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Tags: dp, greedy, math, sortings
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
a,b=mi()
s=si()
n=len(s)
dp=[[0]*2 for i in range(n+1)]
dp[-1][1]=a
for i in range(n):
if s[i]=='0':
dp[i][0]=min(dp[i-1])
dp[i][1]=min(dp[i-1][1]+b,dp[i-1][0]+a+b)
else:
dp[i][0]=10**18
dp[i][1]=min(dp[i-1][1],dp[i-1][0]+a)
#print(dp[i])
print(min(dp[n-1]))
```
| 100,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Tags: dp, greedy, math, sortings
Correct Solution:
```
from math import *
from operator import *
t = int(input())
for _ in range(t):
a, b = map(int, input().split())
s = input()
l = list()
p, q = 0, 0
s = '0' + s + '0'
for i in range(len(s)):
if s[i] == '1' and s[i-1] == '0':
p = i
elif s[i] == '0' and s[i-1] == '1':
q = i - 1
l.append([p, q])
answer = 0
f = False
if len(l) == 0:
print(0)
else:
for i in range(0, len(l)-1):
if (l[i+1][0] - l[i][1] - 1) * b <= a:
answer += (l[i+1][0] - l[i][1] - 1) * b
f = True
elif f:
f = False
answer += a
else:
answer += a
print(answer + a)
```
| 100,512 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Tags: dp, greedy, math, sortings
Correct Solution:
```
def get_min_cost(mines, a, b):
mines = mines.strip('0')
if len(mines) == 0:
return 0
max_zeros_len = a // b
ones_run_count = 0
short_zeros_count = 0
short_zeros_sum_length = 0
for i, ch in enumerate(mines):
if ch == '1':
if i == 0 or mines[i - 1] == '0':
ones_run_count += 1
if i > 0 and mines[i - 1] == '0':
l = i - start_index
if l <= max_zeros_len:
short_zeros_count += 1
short_zeros_sum_length += l
if ch == '0':
if mines[i - 1] == '1':
start_index = i
return a * (ones_run_count - short_zeros_count) + b * short_zeros_sum_length
if __name__ == '__main__':
n_samples = int(input())
for t in range(n_samples):
a, b = map(int, input().split(" "))
mines = input()
print(get_min_cost(mines, a, b))
```
| 100,513 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Tags: dp, greedy, math, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
from collections import deque, defaultdict, namedtuple
import heapq
from math import sqrt, factorial, gcd, ceil, atan, pi
from itertools import permutations
# def input(): return sys.stdin.readline().strip()
# def input(): return sys.stdin.buffer.readline()[:-1] # warning bytes
# def input(): return sys.stdin.buffer.readline().strip() # warning bytes
def input(): return sys.stdin.buffer.readline().decode('utf-8').strip()
import string
import operator
import random
# string.ascii_lowercase
from bisect import bisect_left, bisect_right
from functools import lru_cache, reduce
MOD = int(1e9)+7
INF = float('inf')
# sys.setrecursionlimit(MOD)
def solve():
a, b = [int(x) for x in input().split()]
s = [int(x) for x in input()]
d = deque(s)
while d and d[0] == 0:
d.popleft()
while d and d[-1] == 0:
d.pop()
if not d:
print(0)
return
s = ''.join([str(x) for x in d])
seg = [len(x) for x in s.split('1') if x]
ans = a * (len(seg) + 1)
for s in seg:
if s * b < a:
ans -= a
ans += s * b
print(ans)
T = 1
T = int(input())
for case in range(1,T+1):
ans = solve()
"""
1 1 1 1 2 2 2 3 3 5 6 6 8 8 11
2 2 2 1 2 2 1
1 1 1 1 1 2 1
"""
```
| 100,514 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Tags: dp, greedy, math, sortings
Correct Solution:
```
from sys import stdin
input = stdin.readline
t=int(input())
for i in range(t):
a,b=map(int,input().split())
s=input()
n=len(s)-1
i=0
arr=[]
while i<n:
if s[i]=="0":
i=i+1
continue
j=i
while j<n:
if s[j]=="0":
break
j=j+1
arr.append((i,j-1))
i=j
l=len(arr)
res=0
i=0
while i<l:
ans=a
j=i+1
while j<l:
if b*(arr[j][0]-arr[j-1][1]-1)>a:
break
ans=ans+b*(arr[j][0]-arr[j-1][1]-1)
j=j+1
i=j
res=res+ans
print(res)
```
| 100,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Tags: dp, greedy, math, sortings
Correct Solution:
```
t = int(input().strip())
for case in range(t):
a, b = map(int, input().split(' '))
m_line = input().strip()
if len(m_line) == 0:
print(0)
continue
onepart_cnt = 0
pre_m = ''
for m in m_line:
if pre_m == '1' and m == '0':
onepart_cnt = onepart_cnt + 1
pre_m = m
if pre_m == '1':
onepart_cnt = onepart_cnt + 1
min_c = onepart_cnt * a
pre_zero_cnt = 1
flag = False
pre_m = ''
for m in m_line:
if pre_m == '1' and m == '0':
pre_zero_cnt = 0
flag = True
if flag is True and m == '0':
pre_zero_cnt = pre_zero_cnt + 1
if flag is True and m == '1':
flag = False
tmp_c = pre_zero_cnt * b - a
if tmp_c < 0:
min_c = min_c + tmp_c
pre_m = m
print(min_c)
```
| 100,516 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Tags: dp, greedy, math, sortings
Correct Solution:
```
for _ in range(int(input())):
a,b=map(int,input().split())
c,d=-100000,a
for i in input():
if i=='1':
if c>0: d+=min(a,c*b)
c=0
else: c+=1
print(0 if c<0 else d)
```
| 100,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
t = int(input())
for _ in range(t):
a,b = map(int,input().split())
s = input()
res = []
i = 0
n = len(s)
ans = 0
while(i<n):
if s[i] == '1':
temp = [i]
while(i < n and s[i] == '1'):
i += 1
temp.append(i-1)
if len(res):
if (b * (temp[0]-res[-1][1] -1) + a) <= 2*a:
ans += (b * (temp[0]-res[-1][-1] -1))
res[-1][-1] = temp[1]
else:
res.append(temp)
else:
res.append(temp)
else:
i += 1
print(len(res) * a + ans)
```
Yes
| 100,518 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
noOfTestCases = int(input())
for testCase in range(noOfTestCases):
activationCost, placeMine = [int(i) for i in input().split()]
buildings = input()
noOfBuildings = len(buildings)
totalCost = 0
atIndex = 0
previousMine = None
while atIndex<noOfBuildings:
if buildings[atIndex] == '0':
pass
else:
if previousMine is None:
pass
elif (atIndex - previousMine - 1) * placeMine < activationCost:
totalCost += (atIndex - previousMine - 1) * placeMine
else:
totalCost += activationCost
previousMine = atIndex
atIndex += 1
if previousMine is not None:
totalCost += activationCost
print(totalCost)
```
Yes
| 100,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
from sys import stdin, stdout
from math import factorial as f
from random import randint, shuffle
input = stdin.readline
for _ in range(int(input())):
a, b = map(int, input().split())
s = input()[:-1]
s = s.lstrip('0')
if not s:
print(0)
continue
ans = a
cnt = 0
for i in s:
if i == '0':
cnt += 1
elif cnt:
ans += min(a, cnt * b)
cnt = 0
print(ans)
```
Yes
| 100,520 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
# ------------------- fast io --------------------
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")
# ------------------- fast io --------------------
from math import ceil,floor,sqrt
from collections import Counter,defaultdict,deque
def int1():
return int(input())
def map1():
return map(int,input().split())
def list1():
return list(map(int,input().split()))
mod=pow(10,9)+7
def solve():
a,b=map1()
l1=list(input().strip("0"))
n=len(l1)
l2=[]
c,flag,sum1=0,0,0
for i in range(n):
if(l1[i]=="0"):
c=c+1
else:
if(flag==0):
v=c*b+a
v1=2*a
flag=1
else:
v=c*b
v1=a
sum1+=min(v,v1)
#print(sum1)
c=0
print(sum1)
for _ in range(int(input())):
solve()
```
Yes
| 100,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
''' https://codeforces.com/problemset/problem/1443/B '''
from typing import List
def solv(city: List[int], a: int, b: int) -> int:
n_elem = len(city)
if n_elem == 0:
return 0
expl_c = [0] * n_elem
mine_c = [0] * n_elem
if city[0] == 0:
mine_c[0] = b
else:
expl_c[0] = a
for i in range(1, n_elem):
if city[i] == 0:
expl_c[i] = min(expl_c[i-1], mine_c[i-1] + a)
mine_c[i] = mine_c[i-1] + b
else:
expl_c[i] = min(expl_c[i-1] + a, mine_c[i-1] + a)
mine_c[i] = mine_c[i-1]
return expl_c[-1]
TESTS = int(input())
try:
for _ in range(TESTS):
ab = input()
a, b = [int(v) for v in ab.split()]
strarr = input()
arr = [int(v) for v in strarr.strip('0')]
print(solv(arr, a, b))
except Exception as e:
print(strarr)
'''
strarr = "01101110"
arr = [int(v) for v in strarr.strip('0')]
a, b = 5, 1
print(solv(arr, a, b))
'''
```
No
| 100,522 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
for _ in range(int(input())):
[x,y] = list(map(int,input().split()))
n = str(input())
output = 0
gear = 0
temp = 0
flag = 0
go =0
for i in range(len(n)):
if n[i]=='1':
go = 1
break
if go == 1:
while n[0]=='0' and len(n)!=0:
n = n[1:]
while n[-1]=='0' and len(n)!=0:
n = n[:(len(n)-1)]
for i in range(len(n)):
if n[i]=='1':
if temp<=x and gear != 0:
output+=temp
temp = 0
flag = 1
if flag == 0:
flag = 1
output+=x
if n[i] == '0':
flag = 0
temp += y
if temp<=x:
gear = 1
else:
output= 0
print(output)
```
No
| 100,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
for _ in range(int(input())):
a,b=map(int,input().split())
mines=input()
lent=len(mines)
if mines.count('1')==0:
print(0)
continue
for i in range(lent):
if mines[i]=='1':
start=i
break
for i in range(lent-1,-1,-1):
if mines[i]=='1':
end=i
break
mod=mines[start:end+1]
new=''
found=False
for i in range(len(mod)):
if mod[i]=='1' and not found:
new+=mod[i]
found=True
if mod[i]=='0':
new+=mod[i]
found=False
zeros=new.count('0')
if zeros==0:
print(a)
continue
ones=new.count('1')
zeros_cost=(zeros*b)+a
ones_cost=ones*a
print(min(zeros_cost,ones_cost))
```
No
| 100,524 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Bertown is a city with n buildings in a straight line.
The city's security service discovered that some buildings were mined. A map was compiled, which is a string of length n, where the i-th character is "1" if there is a mine under the building number i and "0" otherwise.
Bertown's best sapper knows how to activate mines so that the buildings above them are not damaged. When a mine under the building numbered x is activated, it explodes and activates two adjacent mines under the buildings numbered x-1 and x+1 (if there were no mines under the building, then nothing happens). Thus, it is enough to activate any one mine on a continuous segment of mines to activate all the mines of this segment. For manual activation of one mine, the sapper takes a coins. He can repeat this operation as many times as you want.
Also, a sapper can place a mine under a building if it wasn't there. For such an operation, he takes b coins. He can also repeat this operation as many times as you want.
The sapper can carry out operations in any order.
You want to blow up all the mines in the city to make it safe. Find the minimum number of coins that the sapper will have to pay so that after his actions there are no mines left in the city.
Input
The first line contains one positive integer t (1 โค t โค 10^5) โ the number of test cases. Then t test cases follow.
Each test case begins with a line containing two integers a and b (1 โค a, b โค 1000) โ the cost of activating and placing one mine, respectively.
The next line contains a map of mines in the city โ a string consisting of zeros and ones.
The sum of the string lengths for all test cases does not exceed 10^5.
Output
For each test case, output one integer โ the minimum number of coins that the sapper will have to pay.
Example
Input
2
1 1
01000010
5 1
01101110
Output
2
6
Note
In the second test case, if we place a mine under the fourth building and then activate it, then all mines on the field are activated. The cost of such operations is six, b=1 coin for placing a mine and a=5 coins for activating.
Submitted Solution:
```
t = int(input())
for _ in range(t):
a, b = map(int, input().split())
mines = input()
l = -1
mines_s = list()
for i in range(len(mines)):
if l == -1 and mines[i] == "1":
l = i
elif l != -1 and mines[i] == "0":
mines_s.append((l, i - 1))
l = -1
if l != -1:
mines_s.append((l, len(mines) - 1))
s = 0
i = 0
while i < len(mines_s) - 1:
l1, r1 = mines_s.pop(0)
l2, r2 = mines_s.pop(0)
if b * (l2 - r1 - 1) <= a:
s += b * (l2 - r1 - 1)
mines_s = [(l1, r2)] + mines_s
else:
mines_s = [(l1, r1), (l2, r2)] + mines_s
i += 1
s += a * len(mines_s)
print(s)
```
No
| 100,525 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Tags: brute force, implementation
Correct Solution:
```
# from __future__ import print_function,division
# range = xrange
import sys
input = sys.stdin.readline
# sys.setrecursionlimit(10**9)
from sys import stdin, stdout
from collections import defaultdict, Counter
M = 10**9+7
def main():
for _ in range(int(input())):
s = input().strip()
n = len(s)
pos = [[0,0]]
curr = [0,0]
for i in range(n):
if s[i]=="L":
curr = [curr[0]-1,curr[1]]
elif s[i]=="R":
curr = [curr[0]+1,curr[1]]
elif s[i]=="D":
curr = [curr[0],curr[1]-1]
else:
curr = [curr[0],curr[1]+1]
pos.append(curr)
ans = [0,0]
for i in range(n):
ob = pos[i+1]
curr = [0,0]
# print("ob:",ob)
for j in range(n):
# print(curr)
if s[j]=="L" and (curr[0]-1!=ob[0] or curr[1]!=ob[1]):
curr = [curr[0]-1,curr[1]]
elif s[j]=="R" and (curr[0]+1!=ob[0] or curr[1]!=ob[1]):
curr = [curr[0]+1,curr[1]]
elif s[j]=="D" and (curr[0]!=ob[0] or curr[1]-1!=ob[1]):
curr = [curr[0],curr[1]-1]
elif s[j]=="U" and (curr[0]!=ob[0] or curr[1]+1!=ob[1]):
curr = [curr[0],curr[1]+1]
if curr==[0,0]:
ans = ob
break
print(*ans)
if __name__== '__main__':
main()
```
| 100,526 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Tags: brute force, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
def move(x, d):
m = {'L': (-1, 0), 'R': (1, 0), 'D': (0, -1), 'U': (0, 1)}
return tuple([x[i] + m[d][i] for i in range(2)])
def solve(s):
p = (0,0)
p_set = {p}
for char in s:
z = move(p, char)
p_set.add(z)
p = z
ans = (0, 0)
for x in p_set:
q = (0, 0)
for char in s:
z = move(q, char)
if z != x:
q = z
if q == (0,0):
ans = x
break
return ans
t = int(input())
for case in range(t):
ans = solve(input().strip())
print(*ans)
```
| 100,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Tags: brute force, implementation
Correct Solution:
```
T = int(input())
r = 1
while r<=T:
s = input()
n = len(s)
direc = {'U':[0,1],'D':[0,-1],'L':[-1,0],'R':[1,0]}
x = 0
y = 0
dic = {}
for i in range(n):
x += direc[s[i]][0]
y += direc[s[i]][1]
if(x,y) not in dic:
dic[(x,y)] = i
ans = [0,0]
for ele in dic:
already = dic[ele]
restpath = s[already:]
x = ele[0]
y = ele[1]
x -= direc[s[already]][0]
y -= direc[s[already]][1]
for c in restpath:
if x+direc[c][0] == ele[0] and y+direc[c][1] == ele[1]: continue
x += direc[c][0]
y += direc[c][1]
if x==0 and y==0:
ans = ele
break
print(ans[0],ans[1])
r += 1
```
| 100,528 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Tags: brute force, implementation
Correct Solution:
```
d = {'L': [-1, 0], 'R': [1, 0], 'U': [0, 1], 'D': [0, -1]}
for _ in range(int(input())):
s = input()
x, y = 0, 0
h = dict()
for i in range(len(s)):
dir = s[i]
x += d[dir][0]
y += d[dir][1]
if x not in h:
h[x] = dict()
if y not in h[x]:
h[x][y] = i
blocks = []
for x in h:
for y in h[x]:
blocks += [[x, y]]
for block in blocks:
x, y = 0, 0
for dir in s:
x += d[dir][0]
y += d[dir][1]
if [x, y] == block:
x -= d[dir][0]
y -= d[dir][1]
if [x, y] == [0, 0]:
break
if [x, y] == [0, 0]:
print(block[0], block[1])
else:
print("0 0")
```
| 100,529 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Tags: brute force, implementation
Correct Solution:
```
import sys
import os
from sys import stdin, stdout
import math
def main():
t = int(stdin.readline())
for _ in range(t):
st = stdin.readline().strip()
px, py = 0,0
target = set()
for s in st:
if s == "R":
px += 1
elif s == "L":
px -= 1
elif s == "U":
py += 1
else:
py -= 1
target.add((px,py))
if (0, 0) in target:
target.remove((0,0))
is_found = False
for coor in target:
X,Y = coor
px, py = 0,0
for s in st:
tx, ty = px, py
if s == "R":
tx = px +1
elif s == "L":
tx = px - 1
elif s == "U":
ty = py + 1
else:
ty = py - 1
if tx == X and ty == Y:
continue
px, py = tx, ty
if (px, py) == (0, 0):
is_found = True
print(X,Y)
break
if not is_found:
print("0 0")
main()
```
| 100,530 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Tags: brute force, implementation
Correct Solution:
```
from sys import stdin as inp
from sys import stdout as out
def calcvalue(s):
x,y=0,0
for i in s:
if i=="L":
x-=1
elif i=="R":
x+=1
elif i=="U":
y+=1
elif i=="D":
y-=1
return x,y
for _ in range(int(inp.readline())):
s=input()
x,y=0,0
flag=True
for i in range(len(s)):
if s[i]=="L":
x-=1
elif s[i]=="R":
x+=1
elif s[i]=="U":
y+=1
elif s[i]=="D":
y-=1
obs=[x,y]
a,b=0,0
for j in range(len(s)):
if s[j]=="L":
a-=1
if [a,b]==obs:
a+=1
elif s[j]=="R":
a+=1
if [a,b]==obs:a-=1
elif s[j]=="U":
b+=1
if [a,b]==obs:b-=1
elif s[j]=="D":
b-=1
if [a,b]==obs:b+=1
if a==0 and b==0:
print(*obs)
flag=False
break
if flag:
print("0 0")
```
| 100,531 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Tags: brute force, implementation
Correct Solution:
```
def upper(s,k,l,u):
x=-1
if u==-1:
return -1
while(l<=u):
mid=(l+u)//2
if s[mid]<=k:
x=max(x,mid)
l=mid+1
else:
u=mid-1
return x
for i in range(int(input())):
a=input()
n=len(a)
x=0
y=0
d=-1
for i in range(n):
if a[i]=="L":x-=1
if a[i]=="R":x+=1
if a[i]=="U":y+=1
if a[i]=="D":y-=1
p=0
q=0
for j in range(n):
u=0
v=0
if a[j]=="L":u-=1
if a[j]=="R":u+=1
if a[j]=="U":v+=1
if a[j]=="D":v-=1
if p+u==x and q+v==y:
continue
else:
p+=u
q+=v
if p==0 and q==0:
d=1
an=[x,y]
break
if d==1:print(an[0],an[1])
else:
print(0,0)
```
| 100,532 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Tags: brute force, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
from itertools import groupby
for _ in range(int(input())):
s = input()[:-1]
n = len(s)
x, y = [0] * (n + 1), [0] * (n + 1)
for i, c in enumerate(s):
x[i + 1] = x[i] + (1 if c == 'R' else -1 if c == 'L' else 0)
y[i + 1] = y[i] + (1 if c == 'U' else -1 if c == 'D' else 0)
ansx = ansy = 0
for i in range(1, n + 1):
x0, y0 = 0, 0
for c in s:
if c == 'R': x0 += 1
elif c == 'L': x0 -= 1
elif c == 'U': y0 += 1
else: y0 -= 1
if x0 == x[i] and y0 == y[i]:
if c == 'R': x0 -= 1
elif c == 'L': x0 += 1
elif c == 'U': y0 -= 1
else: y0 += 1
if not x0 and not y0:
ansx, ansy = x[i], y[i]
break
print("{} {}".format(ansx, ansy))
```
| 100,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Submitted Solution:
```
test=int(input())
for i1 in range(test):
start_x=0
start_y=0
s=input()
s=s.replace('\r','')
l=[]
for i in s:
if i=='L':
start_x-=1
if i=='R':
start_x+=1
if i=='U':
start_y+=1
if i=='D':
start_y-=1
l.append((start_x,start_y))
#print(l)
orig_x=0
orig_u=0
flag=0
for i in l:
start_x = 0
start_y = 0
for j in s:
if j=='L' and (start_x-1,start_y)!=i:
start_x-=1
if j=='R' and (start_x+1,start_y)!=i:
start_x+=1
if j=='U' and (start_x,start_y+1)!=i:
start_y+=1
if j=='D' and (start_x,start_y-1)!=i:
start_y-=1
#print('Here',start_x,start_y)
if start_x == 0 and start_y == 0:
a,b=i
print(a,b)
flag=1
break
if flag==0:
print(0,0)
```
Yes
| 100,534 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Submitted Solution:
```
#------------------Important Modules------------------#
from sys import stdin,stdout
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import *
input=stdin.readline
prin=stdout.write
from random import sample
from collections import Counter,deque
from math import sqrt,ceil,log2,gcd
#dist=[0]*(n+1)
mod=10**9+7
"""
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)]
# 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:
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
# Driver code
"""
def f(arr,i,j,d,dist):
if i==j:
return
nn=max(arr[i:j])
for tl in range(i,j):
if arr[tl]==nn:
dist[tl]=d
#print(tl,dist[tl])
f(arr,i,tl,d+1,dist)
f(arr,tl+1,j,d+1,dist)
#return dist
def ps(n):
cp=0;lk=0;arr={}
#print(n)
while n%2==0:
n=n//2
#arr.append(n);arr.append(2**(lk+1))
lk+=1
arr[2]=lk;
for ps in range(3,ceil(sqrt(n))+1,2):
lk=0
while n%ps==0:
n=n//ps
arr.append(n);arr.append(ps**(lk+1))
lk+=1
arr[ps]=lk;
if n!=1:
lk+=1
arr[n]=lk
#return arr
#print(arr)
return arr
#count=0
#dp=[[0 for i in range(m)] for j in range(n)]
#[int(x) for x in input().strip().split()]
def gcd(x, y):
while(y):
x, y = y, x % y
return x
# Driver Code
def factorials(n,r):
#This calculates ncr mod 10**9+7
slr=n;dpr=r
qlr=1;qs=1
mod=10**9+7
for ip in range(n-r+1,n+1):
qlr=(qlr*ip)%mod
for ij in range(1,r+1):
qs=(qs*ij)%mod
#print(qlr,qs)
ans=(qlr*modInverse(qs))%mod
return ans
def modInverse(b):
qr=10**9+7
return pow(b, qr - 2,qr)
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
def power(arr):
listrep = arr
subsets = []
for i in range(2**len(listrep)):
subset = []
for k in range(len(listrep)):
if i & 1<<k:
subset.append(listrep[k])
subsets.append(subset)
return subsets
def pda(n) :
list = []
for i in range(1, int(sqrt(n) + 1)) :
if (n % i == 0) :
if (n // i == i) :
list.append(i)
else :
list.append(n//i);list.append(i)
# The list will be printed in reverse
return list
def dis(xa,ya,xb,yb):
return sqrt((xa-xb)**2+(ya-yb)**2)
#### END ITERATE RECURSION ####
#===============================================================================================
#----------Input functions--------------------#
def ii():
return int(input())
def ilist():
return [int(x) for x in input().strip().split()]
def out(array:list)->str:
array=[str(x) for x in array]
return ' '.join(array);
def islist():
return list(map(str,input().split().rstrip()))
def outfast(arr:list)->str:
ss=''
for ip in arr:
ss+=str(ip)+' '
return prin(ss);
def kk(n):
kp=0
while n%2==0:
n=n//2
kp+=1
if n==1:
return [True,kp]
else:
return [False,0]
###-------------------------CODE STARTS HERE--------------------------------###########
#########################################################################################
t=int(input())
#t=1
for jj in range(t):
dicts={'R':'L','L':'R','U':'D','D':'U'}
ds={'R':1,"L":-1,'U':1,'D':-1}
kk=input().strip();n=len(kk)
stack=[];x=0;y=0;maxx=0;maxy=0;minx=0;miny=0;
arr=[];
for i in range(n):
if kk[i]=='R' or kk[i]=='L':
x+=ds[kk[i]]
arr.append([x,y])
#maxx=max(maxx,x);minx=min(minx,x)
else:
y+=ds[kk[i]]
#arr.append([x,y])
#maxy=max(maxy,y);miny=min(miny,y)
arr.append([x,y])
#print(x,y,maxy,miny,"ji")
kpk=len(arr)
for jj in range(kpk):
a,b=arr[jj]
s,r=0,0
for i in range(n):
if kk[i]=='R' or kk[i]=='L':
s+=ds[kk[i]]
if s==a and r==b:
s-=ds[kk[i]]
else:
r+=ds[kk[i]]
if s==a and r==b:
r-=ds[kk[i]]
if s==0 and r==0:
print(a,b)
break
elif jj==kpk-1:
print(0,0)
#break
else:
continue
```
Yes
| 100,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Submitted Solution:
```
t = int(input())
for _ in range(t):
s = input()
c = set()
x = y = 0
for i in s:
if i == "U":
y += 1
elif i == "D":
y -= 1
elif i == "L":
x -= 1
else:
x += 1
c.add((x, y))
for ox, oy in c:
x = y = 0
for i in s:
if i == "U" and (x, y + 1) != (ox, oy):
y += 1
elif i == "D" and (x, y - 1) != (ox, oy):
y -= 1
elif i == "L" and (x - 1, y) != (ox, oy):
x -= 1
elif i == "R" and (x + 1, y) != (ox, oy):
x += 1
if x == y == 0:
print(ox, oy)
break
else:
print(0, 0)
```
Yes
| 100,536 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Submitted Solution:
```
def find(s, n, c):
pos = [0, 0]
for i in range(n):
if s[i] == 'L':
if (pos[0] - 1, pos[1]) == c:
continue
pos[0] -= 1
elif s[i] == 'R':
if (pos[0] + 1, pos[1]) == c:
continue
pos[0] += 1
elif s[i] == 'D':
if (pos[0], pos[1] - 1) == c:
continue
pos[1] -= 1
else:
if (pos[0], pos[1] + 1) == c:
continue
pos[1] += 1
return pos
for _ in range(int(input())):
s = input().strip()
d = {}
n = len(s)
pos = [0, 0]
for i in range(n):
if s[i] == 'L':
pos[0] -= 1
elif s[i] == 'R':
pos[0] += 1
elif s[i] == 'D':
pos[1] -= 1
else:
pos[1] += 1
d[(pos[0], pos[1])] = 1
res = (0, 0)
#print(d)
for i in d:
if i != (0, 0):
ans = find(s, n, i)
#print("@", ans)
if ans == [0, 0]:
res = i
break
print(res[0], res[1])
```
Yes
| 100,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Submitted Solution:
```
def maker(l,b,u):
e = b.count(l)
c = 0
for i in range(len(b)):
if b[i] == u:
c += 1
if c == e + 1:
return i
a = int(input())
for i in range(a):
b = input()
k = []
x = 0
y = 0
c = 0
for j in b:
if j == 'R':
x += 1
elif j == 'L':
x -= 1
elif j == 'U':
y += 1
elif j == 'D':
y -= 1
k.append([x,y])
if x!= 0 and y !=0:
print(0,0)
else:
if x >= 1:
print(*k[maker('L', b, 'R')])
if x < 0:
print(*k[maker('R',b,'L')])
if y >= 1:
print(*k[maker('D',b,'U')])
if y < 0:
print(*k[maker('U',b ,'D')])
```
No
| 100,538 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Submitted Solution:
```
n = int(input())
for i in range(n):
sec = input()
x = 0
y = 0
for i in range(len(sec)):
if (sec[i] == 'U'):
y += 1
elif sec[i] == 'D':
y -= 1
elif sec[i] == 'L':
x -= 1
else:
x += 1
curr_x = 0
curr_y = 0
here = False
for j in range(len(sec)):
if (i == j):
here = True
elif (not here or sec[i] != sec[j]):
here = False
if (sec[j] == 'U'):
curr_y += 1
elif sec[j] == 'D':
curr_y -= 1
elif sec[j] == 'L':
curr_x -= 1
else:
curr_x += 1
if curr_x == 0 and curr_y == 0:
break
if curr_x == 0 and curr_y == 0:
print(x,y)
else:
print(0,0)
```
No
| 100,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Submitted Solution:
```
s = ''
def takeStep(curX,curY,canObstruct,stepNum,obsX,obsY,space):
# print(' '*space,curX,curY)
if stepNum==len(s):
return (curX==0 and curY==0), obsX , obsY, canObstruct
step = s[stepNum]
if canObstruct:
if step=='R':
i = takeStep(curX,curY,0,stepNum+1,curX+1,curY,space+1)
if i[0]:
return i
elif step=='L':
i = takeStep(curX,curY,0,stepNum+1,curX-1,curY,space+1)
if i[0]:
return i
elif step=='U':
i = takeStep(curX,curY,0,stepNum+1,curX,curY+1,space+1)
if i[0]:
return i
elif step=='D':
i = takeStep(curX,curY,0,stepNum+1,curX,curY-1,space+1)
if i[0]:
return i
if step=='R':
if curX+1==obsX and curY==obsY and canObstruct==0:
return takeStep(curX,curY,canObstruct,stepNum+1,obsX,obsY,space+1)
return takeStep(curX+1,curY,canObstruct,stepNum+1,obsX,obsY,space+1)
elif step=='L':
if curX-1==obsX and curY==obsY and canObstruct==0:
return takeStep(curX,curY,canObstruct,stepNum+1,obsX,obsY,space+1)
return takeStep(curX-1,curY,canObstruct,stepNum+1,obsX,obsY,space+1)
elif step=='U':
if curX==obsX and curY+1==obsY and canObstruct==0:
return takeStep(curX,curY,canObstruct,stepNum+1,obsX,obsY,space+1)
return takeStep(curX,curY+1,canObstruct,stepNum+1,obsX,obsY,space+1)
elif step=='D':
if curX==obsX and curY-1==obsY and canObstruct==0:
return takeStep(curX,curY,canObstruct,stepNum+1,obsX,obsY,space+1)
return takeStep(curX,curY-1,canObstruct,stepNum+1,obsX,obsY,space+1)
t = int(input())
for _ in range(t):
s= input()
i = takeStep(0,0,1,0,0,0,0)
if i[0]:
if i[-1]==0:
print(i[1],i[2])
else:
print(4999,4999)
else:
print(0,0)
```
No
| 100,540 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot on a checkered field that is endless in all directions. Initially, the robot is located in the cell with coordinates (0, 0). He will execute commands which are described by a string of capital Latin letters 'L', 'R', 'D', 'U'. When a command is executed, the robot simply moves in the corresponding direction:
* 'L': one cell to the left (the x-coordinate of the current cell decreases by 1);
* 'R': one cell to the right (the x-coordinate of the current cell is increased by 1);
* 'D': one cell down (the y-coordinate of the current cell decreases by 1);
* 'U': one cell up (the y-coordinate of the current cell is increased by 1).
Your task is to put an obstacle in one cell of the field so that after executing the commands, the robot will return to the original cell of its path (0, 0). Of course, an obstacle cannot be placed in the starting cell (0, 0). It is guaranteed that if the obstacle is not placed, then the robot will not return to the starting cell.
An obstacle affects the movement of the robot in the following way: if it tries to go in a certain direction, and there is an obstacle, then it simply remains in place (the obstacle also remains, that is, it does not disappear).
Find any such cell of the field (other than (0, 0)) that if you put an obstacle there, the robot will return to the cell (0, 0) after the execution of all commands. If there is no solution, then report it.
Input
The first line contains one integer t (1 โค t โค 500) โ the number of test cases.
Each test case consists of a single line containing s โ the sequence of commands, which are uppercase Latin letters 'L', 'R', 'D', 'U' only. The length of s is between 1 and 5000, inclusive. Additional constraint on s: executing this sequence of commands leads the robot to some cell other than (0, 0), if there are no obstacles.
The sum of lengths of all s in a test doesn't exceed 5000.
Output
For each test case print a single line:
* if there is a solution, print two integers x and y (-10^9 โค x,y โค 10^9) such that an obstacle in (x, y) will force the robot to return back to the cell (0, 0);
* otherwise, print two zeroes (i. e. 0 0).
If there are multiple answers, you can print any of them.
Example
Input
4
L
RUUDL
LLUU
DDDUUUUU
Output
-1 0
1 2
0 0
0 1
Submitted Solution:
```
for _ in range(int(input())):
s=list(input())
up=s.count('U')
do=s.count('D')
r=s.count('R')
le=s.count('L')
if r!=le and up!=do:
print(0,0)
else:
if r==le:
x=0
y=0
c=0
if up>do:
for i in s:
if i=='U':
c+=1
y+=1
elif i=='D':
y-=1
elif i=='L':
x-=1
elif i=='R':
x+=1
if c==do+1:
print(x,y)
break
else:
for i in s:
if i=='U':
y+=1
elif i=='D':
c+=1
y-=1
elif i=='L':
x-=1
elif i=='R':
x+=1
if c==up+1:
print(x,y)
break
elif up==do:
x=0
y=0
c=0
if r>le:
for i in s:
if i=='U':
y+=1
elif i=='D':
y-=1
elif i=='L':
x-=1
elif i=='R':
c+=1
x+=1
if c==le+1:
print(x,y)
break
else:
for i in s:
if i=='U':
y+=1
elif i=='D':
y-=1
elif i=='L':
c+=1
x-=1
elif i=='R':
x+=1
if c==r+1:
print(x,y)
break
```
No
| 100,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
i=int
p=input
N=i(p());L=p();R=p()
l=i(L,2);r=i(R,2);a=R
if L[0]<R[0]:a='1'*N
elif L==R:a=L
elif L[-1]=='1'and l+1==r:a=R
elif l//2<r//2:a=R[:-1]+'1'
print(a)
```
| 100,542 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
n = int(input())
l = input(); L = int(l,2)
r = input(); R = int(r,2)
if (l[0] != r[0]):
print('1'*n)
exit()
if (R - L < 2 or r[-1] == 1):
print(r)
else:
print(r[:-1] + '1')
```
| 100,543 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
n = int(input())
l = input()
r = input()
if n == 1:
print(r)
elif l[0] == '0':
print('1'*n)
elif r[-1] == '0' and int(l,2)+1 < int(r,2):
print(r[:-1] + "1")
else:
print(r)
```
| 100,544 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
i=int;p=input
N=i(p());L=p();R=p()
l=i(L,2);r=i(R,2);a=R
if l-l%2<r:a=R[:-1]+'1'
if i(L[-1])and l+1==r:a=R
if L==R:a=L
if L[0]<R[0]:a='1'*N
print(a)
```
| 100,545 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
n = int(input());l = input();r = input()
if n == 1: print(r)
elif l[0] == '0': print('1'*n)
elif r[-1] == '0' and int(l,2)+1 < int(r,2): print(r[:-1] + "1")
else: print(r)
```
| 100,546 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
def solve2(l, r):
n = len(l)
if l[0] != r[0]:
return [1]*n
if r[-1] == 1:
return r
x = r.copy()
for j in range(2):
if x == l:
return r
for k in range(n-1,-1,-1):
if x[k] == 1:
x[k] = 0
break
else:
x[k] = 1
x = r.copy()
r[-1] = 1
return r
def solve():
n = int(input())
l = list(map(int,input().strip()))
r = list(map(int,input().strip()))
ans = solve2(l, r)
print(''.join(map(str,ans)))
solve()
```
| 100,547 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
i=int
p=input
N=i(p());L=p();R=p();a=R
if L[0]<R[0]:a='1'*N
elif L==R:a=L
elif L[-1]=='1'and i(L,2)+1==i(R,2):a=R
elif i(L,2)//2<i(R,2)//2:a=''.join(R[:-1])+'1'
print(a)
```
| 100,548 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Tags: bitmasks, constructive algorithms, greedy, math, strings, two pointers
Correct Solution:
```
i=int
p=input
N=i(p());L=p();R=p()
l=i(L,2);r=i(R,2);a=R
if l//2<r//2:a=R[:-1]+'1'
if i(L[-1])and l+1==r:a=R
if L==R:a=L
if L[0]<R[0]:a='1'*N
print(a)
```
| 100,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Submitted Solution:
```
N = int(input())
L = input(); l = int(L, 2)
R = input(); r = int(R, 2)
if L[0] != R[0]:
print('1' * N)
exit()
print(R if R[-1] == '1' or r - l < 2 else R[:-1] + '1')
```
Yes
| 100,550 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def solve2(l, r):
n = len(l)
if l[0] != r[0]:
return [1]*n
rr = r.copy()
ans = r.copy()
for i in range(2):
ll = rr.copy()
if r[-1] == 0:
a = rr.copy()
for j in range(4):
if ll == l:
break
for k in range(n-1,-1,-1):
if ll[k] == 1:
ll[k] = 0
break
else:
ll[k] = 1
for k in range(n):
a[k] ^= ll[k]
if ans < a:
ans = a.copy()
break
if rr == l:
break
for k in range(n-1,-1,-1):
if rr[k] == 1:
rr[k] = 0
break
else:
rr[k] = 1
return ans
def solve():
n = int(input())
l = list(map(int,input().strip()))
r = list(map(int,input().strip()))
ans = solve2(l, r)
print(''.join(map(str,ans)))
solve()
```
Yes
| 100,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Submitted Solution:
```
i=int
p=input
def solve(N,L,R):
if L[0]<R[0]:return'1'*N
if L==R:return L
if L[-1]=='1'and i(L,2)+1==i(R,2):return R
if i(L,2)//2<i(R,2)//2:return''.join(R[:-1])+'1'
return R
print(solve(i(p()),p(),p()))
```
Yes
| 100,552 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Submitted Solution:
```
n = input()
l = input()
r = input()
ans = ''
if l[0] == '0' and r[0] == '1':
for _ in range(0, int(n)):
ans = ans + '1'
else:
numl = int(l, 2)
numr = int(r, 2)
if r[int(n) - 1] == '0' and numl + 1 < numr:
ans = bin(numr + 1)[2:]
else:
ans = r
print(ans)
```
Yes
| 100,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Submitted Solution:
```
n = int(input())
l = input()
r = input()
if n == 1:
print(r)
if l[0] == '0':
print('1'*n)
elif r[-1] == '0' and int(l,2)+1 < int(r,2):
print(r[:-1] + "1")
else:
print(r)
```
No
| 100,554 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Submitted Solution:
```
n = int(input())
l = input()
r = input()
n = len(r)
if l[0] == '0':
print('1'*n)
elif r[-1] == '0' and int(l,2)+1 < int(r):
print(r[:-1] + "1")
else:
print(r)
```
No
| 100,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Submitted Solution:
```
n= int (input())
l= int (input())
r= int (input())
if (r==0):
print(0);
exit(0);
st2=pow(10,n-1);
def f(a,b):
A = str(a);
B = str(b);
res=[]
for i in range(len(A)):
if (A[i]==B[i]):
res.append('0');
else:
res.append('1');
s="";
for i in res:
s+=i;
return int(s);
if (l<st2):
for i in range(n):
print(1,end='');
exit(0);
res=r;
for i in range(r-3,r+1):
x=i;
if (i<l):
continue;
for j in range(i+1,r+1):
x=f(x,j);
res=max(res,x);
print(i);
print(res);
```
No
| 100,556 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers l and r in binary representation. Let g(x, y) be equal to the [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all integers from x to y inclusive (that is x โ (x+1) โ ... โ (y-1) โ y). Let's define f(l, r) as the maximum of all values of g(x, y) satisfying l โค x โค y โค r.
Output f(l, r).
Input
The first line contains a single integer n (1 โค n โค 10^6) โ the length of the binary representation of r.
The second line contains the binary representation of l โ a string of length n consisting of digits 0 and 1 (0 โค l < 2^n).
The third line contains the binary representation of r โ a string of length n consisting of digits 0 and 1 (0 โค r < 2^n).
It is guaranteed that l โค r. The binary representation of r does not contain any extra leading zeros (if r=0, the binary representation of it consists of a single zero). The binary representation of l is preceded with leading zeros so that its length is equal to n.
Output
In a single line output the value of f(l, r) for the given pair of l and r in binary representation without extra leading zeros.
Examples
Input
7
0010011
1111010
Output
1111111
Input
4
1010
1101
Output
1101
Note
In sample test case l=19, r=122. f(x,y) is maximal and is equal to 127, with x=27, y=100, for example.
Submitted Solution:
```
n= int (input())
l= int (input())
r= int (input())
st2=pow(10,n-1);
def f(a,b):
A = str(a);
B = str(b);
res=[]
for i in range(len(A)):
if (A[i]==B[i]):
res.append('0');
else:
res.append('1');
s="";
for i in res:
s+=i;
return int(s);
if (l<st2):
for i in range(n):
print(1,end='');
exit(0);
res=r;
for i in range(r-3,r+1):
x=i;
if (i<l):
continue;
for j in range(i+1,r+1):
x=f(x,j);
res=max(res,x);
print(res);
```
No
| 100,557 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
import math, sys
from collections import defaultdict, Counter, deque
from heapq import heapify, heappush, heappop
MOD = int(1e9) + 7
def main():
n = int(input())
arr = list(map(int, input().split()))
pre = [0 for i in range(n)]
suf = [0 for i in range(n)]
x = 0
for i in range(n):
x ^= arr[i]
pre[i] = x
if x == 0:
print("YES")
return
s = 0
for i in range(n - 1, -1, -1):
s ^= arr[i]
suf[i] = s
for i in range(n - 2):
p = pre[i]
m = arr[i + 1]
for j in range(i + 2, n):
s = suf[j]
# print(p, s, m)
if p == m == s:
print("YES")
return
m ^= arr[j]
print("NO")
t = 1
t = int(input())
for _ in range(t):
main()
```
| 100,558 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from bisect import bisect_right
from math import gcd,log
from collections import Counter,defaultdict,deque
from pprint import pprint
from itertools import permutations
from bisect import bisect_right
from random import randint as rti
# import deque
MOD=10**9+7
def main():
n=int(input())
arr=list(map(int,input().split()))
total=0
for i in arr:
total^=i
til=0
for i in range(n-1):
til^=arr[i]
if til == total^til:
print('YES')
return
pf=0
for i in range(n):
pf^=arr[i]
till=0
for j in range(i+1,n-1):
till^=arr[j]
if pf==till==total^till^pf:
print('YES')
return
print('NO')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
for _ in range(int(input())):
main()
```
| 100,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
from collections import Counter
t = int(input())
for _ in range(t):
n = int(input())
lst = list(map(int,input().split()))
pre = []
pos = [-1]
x = 0
for i in lst:
x = x^i
pre.append(x)
x = 0
for i in lst[::-1]:
x = x^i
pos.append(x)
pos = pos[::-1]
pos = pos[1:]
# print(pre)
flag = True
for i in range(n):
if pre[i] == pos[i]:
print("YES")
flag = False
break
elif pre[i] == 0:
z = pos[i]
if z in pre[:i+1]:
print("YES")
flag = False
break
elif pos[i] == 0:
z = pre[i]
if z in pos[i:]:
print("YES")
flag = False
break
if flag:
print("NO")
```
| 100,560 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
import math as mt
# from collections import defaultdict
# from collections import Counter, deque
# from itertools import permutations
# from functools import reduce
# from heapq import heapify, heappop, heappush, heapreplace
def getInput(): return sys.stdin.readline().strip()
def getInt(): return int(getInput())
def getInts(): return map(int, getInput().split())
def getArray(): return list(getInts())
# sys.setrecursionlimit(10**7)
# INF = float('inf')
# MOD1, MOD2 = 10**9+7, 998244353
# def def_value():
# return 0
# Defining the dict
def solve(li,n):
xor = 0
for i in li:
xor ^= i
if(xor == 0 ):
print("YES")
return
pre = li[0]
for i in range(1,n):
mid = li[i]
for j in range(i+1,n):
if( pre == mid and mid == xor):
print("YES")
return
mid ^= li[j]
pre ^= li[i]
print("NO")
for _ in range(int(input())):
n = int(input())
li = list(map(int,input().split()))
solve(li,n)
```
| 100,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
from itertools import accumulate
tests = int(input())
for t in range(tests):
n = int(input())
arr = map(int, input().split(' '))
pref = list(accumulate(arr, lambda a, b: a ^ b))
if pref[-1] == 0:
# can split into 2
print("YES")
else:
if pref[-1] in pref[:n-1] and 0 in pref[pref[:n-1].index(pref[-1]):]:
print("YES")
else:
print("NO")
```
| 100,562 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
input()
l = list(map(int, input().split()))
v = []
for x in l:
if not v: v.append(x)
else: v.append(x ^ v[-1])
if v[-1] == 0:
print("YES")
else:
ok = False
for i in range(1, len(l)):
for j in range(i+1, len(l)):
v1 = v[i-1]
v2 = v[j-1] ^ v[i-1]
v3 = v[-1] ^ v[j-1]
if v1 == v2 == v3:
print("YES")
ok = True
break
if ok:
break
if not ok:
print("NO")
```
| 100,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
x = 0
for v in a:
x ^= v
if x == 0:
print("YES")
else:
y = a[0]
si = 0
while y != x:
si += 1
if si >= n:
break
y ^= a[si]
z = a[n - 1]
ei = n - 1
while z != x:
ei -= 1
if ei <= si:
break
z ^= a[ei]
if si < n - 1 and ei > 0 and (ei - si - 1) > 0:
print("YES")
else:
print("NO")
```
| 100,564 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import math
import os
import sys
from fractions import *
from sys import *
from decimal import *
from io import BytesIO, IOBase
from itertools import *
from collections import *
# sys.setrecursionlimit(10**5)
M = 10 ** 9 + 7
# print(math.factorial(5))
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
# sys.setrecursionlimit(10**6)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def inpu(): return int(inp())
# -----------------------------------------------------------------
def regularbracket(t):
p = 0
for i in t:
if i == "(":
p += 1
else:
p -= 1
if p < 0:
return False
else:
if p > 0:
return False
else:
return True
# -------------------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <= key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# ------------------------------reverse string(pallindrome)
def reverse1(string):
pp = ""
for i in string[::-1]:
pp += i
if pp == string:
return True
return False
# --------------------------------reverse list(paindrome)
def reverse2(list1):
l = []
for i in list1[::-1]:
l.append(i)
if l == list1:
return True
return False
def mex(list1):
# list1 = sorted(list1)
p = max(list1) + 1
for i in range(len(list1)):
if list1[i] != i:
p = i
break
return p
def sumofdigits(n):
n = str(n)
s1 = 0
for i in n:
s1 += int(i)
return s1
def perfect_square(n):
s = math.sqrt(n)
if s == int(s):
return True
return False
# -----------------------------roman
def roman_number(x):
if x > 15999:
return
value = [5000, 4000, 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
symbol = ["F", "MF", "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
roman = ""
i = 0
while x > 0:
div = x // value[i]
x = x % value[i]
while div:
roman += symbol[i]
div -= 1
i += 1
return roman
def soretd(s):
for i in range(1, len(s)):
if s[i - 1] > s[i]:
return False
return True
# print(soretd("1"))
# ---------------------------
def countRhombi(h, w):
ct = 0
for i in range(2, h + 1, 2):
for j in range(2, w + 1, 2):
ct += (h - i + 1) * (w - j + 1)
return ct
def countrhombi2(h, w):
return ((h * h) // 4) * ((w * w) // 4)
# ---------------------------------
def binpow(a, b):
if b == 0:
return 1
else:
res = binpow(a, b // 2)
if b % 2 != 0:
return res * res * a
else:
return res * res
# -------------------------------------------------------
def binpowmodulus(a, b, m):
a %= m
res = 1
while (b > 0):
if (b & 1):
res = res * a % m
a = a * a % m
b >>= 1
return res
# -------------------------------------------------------------
def coprime_to_n(n):
result = n
i = 2
while (i * i <= n):
if (n % i == 0):
while (n % i == 0):
n //= i
result -= result // i
i += 1
if (n > 1):
result -= result // n
return result
# -------------------prime
def prime(x):
if x == 1:
return False
else:
for i in range(2, int(math.sqrt(x)) + 1):
# print(x)
if (x % i == 0):
return False
else:
return True
def luckynumwithequalnumberoffourandseven(x,n,a):
if x >= n and str(x).count("4") == str(x).count("7"):
a.append(x)
else:
if x < 1e12:
luckynumwithequalnumberoffourandseven(x * 10 + 4,n,a)
luckynumwithequalnumberoffourandseven(x * 10 + 7,n,a)
return a
#----------------------
def luckynum(x,l,r,a):
if x>=l and x<=r:
a.append(x)
if x>r:
a.append(x)
return a
if x < 1e10:
luckynum(x * 10 + 4, l,r,a)
luckynum(x * 10 + 7, l,r,a)
return a
def luckynuber(x, n, a):
p = set(str(x))
if len(p) <= 2:
a.append(x)
if x < n:
luckynuber(x + 1, n, a)
return a
# ------------------------------------------------------interactive problems
def interact(type, x):
if type == "r":
inp = input()
return inp.strip()
else:
print(x, flush=True)
# ------------------------------------------------------------------zero at end of factorial of a number
def findTrailingZeros(n):
# Initialize result
count = 0
# Keep dividing n by
# 5 & update Count
while (n >= 5):
n //= 5
count += n
return count
# -----------------------------------------------merge sort
# Python program for implementation of MergeSort
def mergeSort(arr):
if len(arr) > 1:
# Finding the mid of the array
mid = len(arr) // 2
# Dividing the array elements
L = arr[:mid]
# into 2 halves
R = arr[mid:]
# Sorting the first half
mergeSort(L)
# Sorting the second half
mergeSort(R)
i = j = k = 0
# Copy data to temp arrays L[] and R[]
while i < len(L) and j < len(R):
if L[i] < R[j]:
arr[k] = L[i]
i += 1
else:
arr[k] = R[j]
j += 1
k += 1
# Checking if any element was left
while i < len(L):
arr[k] = L[i]
i += 1
k += 1
while j < len(R):
arr[k] = R[j]
j += 1
k += 1
# -----------------------------------------------lucky number with two lucky any digits
res = set()
def solven(p, l, a, b, n): # given number
if p > n or l > 10:
return
if p > 0:
res.add(p)
solven(p * 10 + a, l + 1, a, b, n)
solven(p * 10 + b, l + 1, a, b, n)
# problem
"""
n = int(input())
for a in range(0, 10):
for b in range(0, a):
solve(0, 0)
print(len(res))
"""
# Python3 program to find all subsets
# by backtracking.
# In the array A at every step we have two
# choices for each element either we can
# ignore the element or we can include the
# element in our subset
def subsetsUtil(A, subset, index, d):
print(*subset)
s = sum(subset)
d.append(s)
for i in range(index, len(A)):
# include the A[i] in subset.
subset.append(A[i])
# move onto the next element.
subsetsUtil(A, subset, i + 1, d)
# exclude the A[i] from subset and
# triggers backtracking.
subset.pop(-1)
return d
def subsetSums(arr, l, r, d, sum=0):
if l > r:
d.append(sum)
return
subsetSums(arr, l + 1, r, d, sum + arr[l])
# Subset excluding arr[l]
subsetSums(arr, l + 1, r, d, sum)
return d
def print_factors(x):
factors = []
for i in range(1, x + 1):
if x % i == 0:
factors.append(i)
return (factors)
# -----------------------------------------------
def calc(X, d, ans, D):
# print(X,d)
if len(X) == 0:
return
i = X.index(max(X))
ans[D[max(X)]] = d
Y = X[:i]
Z = X[i + 1:]
calc(Y, d + 1, ans, D)
calc(Z, d + 1, ans, D)
# ---------------------------------------
def factorization(n, l):
c = n
if prime(n) == True:
l.append(n)
return l
for i in range(2, c):
if n == 1:
break
while n % i == 0:
l.append(i)
n = n // i
return l
# endregion------------------------------
def good(b):
l = []
i = 0
while (len(b) != 0):
if b[i] < b[len(b) - 1 - i]:
l.append(b[i])
b.remove(b[i])
else:
l.append(b[len(b) - 1 - i])
b.remove(b[len(b) - 1 - i])
if l == sorted(l):
# print(l)
return True
return False
# arr=[]
# print(good(arr))
def generate(st, s):
if len(s) == 0:
return
# If current string is not already present.
if s not in st:
st.add(s)
# Traverse current string, one by one
# remove every character and recur.
for i in range(len(s)):
t = list(s).copy()
t.remove(s[i])
t = ''.join(t)
generate(st, t)
return
#=--------------------------------------------longest increasing subsequence
def largestincreasingsubsequence(A):
l = [1]*len(A)
sub=[]
for i in range(1,len(l)):
for k in range(i):
if A[k]<A[i]:
sub.append(l[k])
l[i]=1+max(sub,default=0)
return max(l,default=0)
#----------------------------------longest palindromic substring
# Python3 program for the
# above approach
# Function to calculate
# Bitwise OR of sums of
# all subsequences
def findOR(nums, N):
# Stores the prefix
# sum of nums[]
prefix_sum = 0
# Stores the bitwise OR of
# sum of each subsequence
result = 0
# Iterate through array nums[]
for i in range(N):
# Bits set in nums[i] are
# also set in result
result |= nums[i]
# Calculate prefix_sum
prefix_sum += nums[i]
# Bits set in prefix_sum
# are also set in result
result |= prefix_sum
# Return the result
return result
#l=[]
def OR(a, n):
ans = a[0]
for i in range(1, n):
ans |= a[i]
#l.append(ans)
return ans
#print(prime(12345678987766))
"""
def main():
q=inpu()
x = q
v1 = 0
v2 = 0
i = 2
while i * i <= q:
while q % i == 0:
if v1!=0:
v2 = i
else:
v1 = i
q //= i
i += 1
if q - 1!=0:
v2 = q
if v1 * v2 - x!=0:
print(1)
print(v1 * v2)
else:
print(2)
if __name__ == '__main__':
main()
"""
"""
def main():
l,r = sep()
a=[]
luckynum(0,l,r,a)
a.sort()
#print(a)
i=0
ans=0
l-=1
#print(a)
while(True):
if r>a[i]:
ans+=(a[i]*(a[i]-l))
l=a[i]
else:
ans+=(a[i]*(r-l))
break
i+=1
print(ans)
if __name__ == '__main__':
main()
"""
"""
def main():
sqrt = {i * i: i for i in range(1, 1000)}
#print(sqrt)
a, b = sep()
for y in range(1, a):
x2 = a * a - y * y
if x2 in sqrt:
x = sqrt[x2]
if b * y % a == 0 and b * x % a == 0 and b * x // a != y:
print('YES')
print(-x, y)
print(0, 0)
print(b * y // a, b * x // a)
exit()
print('NO')
if __name__ == '__main__':
main()
"""
"""
def main():
m=inpu()
q=lis()
n=inpu()
arr=lis()
q=min(q)
arr.sort(reverse=True)
s=0
cnt=0
i=0
while(i<n):
cnt+=1
s+=arr[i]
#print(cnt,q)
if cnt==q:
i+=2
cnt=0
i+=1
print(s)
if __name__ == '__main__':
main()
"""
"""
def main():
n,k = sep()
if k * 2 >= (n - 1) * n:
print('no solution')
else:
for i in range(n):
print(0,i)
if __name__ == '__main__':
main()
"""
"""
def main():
t = inpu()
for _ in range(t):
n,k = sep()
arr = lis()
i=0
j=0
while(k!=0):
if i==n-1:
break
if arr[i]!=0:
arr[i]-=1
arr[n-1]+=1
k-=1
else:
i+=1
print(*arr)
if __name__ == '__main__':
main()
"""
def main():
t = int(input())
for _ in range(t):
n=int(input())
arr = lis()
s = 0
for x in arr:
s ^= x
if s == 0:
print('YES')
else:
c = 0
y=0
for i in range(n):
y ^= arr[i]
if y == s:
c += 1
y=0
if c >= 2:
print("YES")
else:
print("NO")
if __name__ == '__main__':
main()
```
| 100,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Submitted Solution:
```
import sys
from collections import defaultdict
input = sys.stdin.readline
def solve(n):
arr = list(map(int,input().split()))
dp = defaultdict(int)
for i in range(n):
dp[i] = dp[i-1] ^ arr[i]
temp = 0
if dp[n-1]==0:
for i in range(n-1):
if dp[i]==dp[n-1]^dp[i]:
temp = 1
if temp:
print("YES")
else:
print("NO")
else:
temp = 0
for i in range(n-2):
if dp[i]==dp[n-1]:
for j in range(i+1,n-1):
if dp[j]==0:
temp = 1
break
if temp:
print("YES")
else:
print("NO")
t = int(input())
for _ in range(t):
n = int(input())
solve(n)
```
Yes
| 100,566 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=[int(x) for x in input().split()]
x=0
for i in a:
x=x^i
if x==0:
print("YES")
else:
arr=[0 for i in range(31)]
y=0
f=0
for i in range(n):
y=y^a[i]
if y==x:
f+=1
y=0
if f<3:
print("NO")
else:
print("YES")
```
Yes
| 100,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
res = 0
for i in range(n):
res ^= arr[i]
if res == 0:
print("YES")
else:
count = 0
xor = 0
for i in range(n):
xor ^= arr[i]
if xor == res:
count += 1
xor = 0
if count >= 2:
print("YES")
else:
print("NO")
```
Yes
| 100,568 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Submitted Solution:
```
#######puzzleVerma#######
import sys
import math
LI=lambda:[int(k) for k in input().split()]
input = lambda: sys.stdin.readline().rstrip()
IN=lambda:int(input())
S=lambda:input()
for t in range(IN()):
n=IN()
a=LI()
pre=0
for i in range(n):
pre^=a[i]
prea=0
flag=0
for j in range(i+1,n):
prea^=a[j]
if prea==pre:
prea=0
flag=1
if (flag and (not prea)):
print("YES")
break
if (not flag):
print("NO")
```
Yes
| 100,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Submitted Solution:
```
tt = int ( input ())
for _ in range (tt):
n = int(input ())
l = []
l = list (map(int , input().split()))
if l.count(l[0])== n :
print ("YES")
continue
ans = 1
bit = [0 for i in range (30)]
for el in l:
for i in range (30) :
if ((1<<i)&el)!=0:
bit[i]+=1
for i in range (30):
if (bit[i]%2 == 1) and (bit[i]!=n):
ans = 0
break
if ans == 1:
print ("YES")
else :
print ("NO")
#jflsdjlsdfjlsdjla;jfdsj;af
#jfd;sajfl;jdsa;fjsda
#da;sfj;sdjf;
```
No
| 100,570 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Submitted Solution:
```
#!/usr/bin/env python3
# created : 2021
from sys import stdin, stdout
from random import randint
def solve(tc):
n = int(stdin.readline().strip())
seq = list(map(int, stdin.readline().split()))
allSame = True
for i in range(1, n):
if seq[i] != seq[0]:
allSame = False
break
if allSame:
print("YES")
return
bits = [0 for i in range(30)]
for i in range(n):
for j in range(30):
if (seq[i] >> j) & 1:
bits[j] += 1
for j in range(30):
if bits[j] & 1:
print("NO")
return
print("YES")
tcs = 1
tcs = int(stdin.readline().strip())
tc = 1
while tc <= tcs:
solve(tc)
tc += 1
```
No
| 100,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
arr= list(map(int, input().split()))
pre=arr[0]
for x in arr[1:-1]:
pre = x^pre
if pre == arr[-1]:
print("YES")
else:
print("NO")
```
No
| 100,572 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 โค t โค 15) โ the number of test cases you need to solve.
The first line of each test case contains an integers n (2 โค n โค 2000) โ the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, โฆ, a_{n} (0 โค a_i < 2^{30}) โ the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 โ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal.
Submitted Solution:
```
from collections import defaultdict
for _ in range(int(input())):
n = map(int, input().split(" "))
l = list(map(int, input().split(" ")))
if len(set(l)) == 1:
print('YES')
else:
a = 0
for i in l:
a^=i
if a==0:
print("YES")
else:
if l.count(a)>1:
print("YES")
else:
print("NO")
```
No
| 100,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Tags: bitmasks, brute force, dfs and similar, implementation, math, probabilities
Correct Solution:
```
scale = 10 ** 6
def calculate_expectation(c, m, p, v):
expectation = 1
if c != 0:
if c <= v:
if m != 0:
expectation += c/scale * calculate_expectation(0, m + c/2, p + c/2, v)
else:
expectation += c/scale
else:
if m != 0:
expectation += c/scale * calculate_expectation(c - v, m + v/2, p + v/2, v)
else:
expectation += c/scale * calculate_expectation(c - v, 0, p + v, v)
if m != 0:
if m <= v:
if c != 0:
expectation += m/scale * calculate_expectation(c + m/2, 0, p + m/2, v)
else:
expectation += m/scale
else:
if c != 0:
expectation += m/scale * calculate_expectation(c + v/2, m - v, p + v/2, v)
else:
expectation += m/scale * calculate_expectation(0, m - v, p + v, v)
return expectation
t = int(input())
results = []
for i in range(t):
c, m, p, v = map(float, input().split())
c = c * scale
m = m * scale
p = p * scale
v = v * scale
results.append(calculate_expectation(c, m, p, v))
for i in range(t):
print(results[i])
```
| 100,574 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Tags: bitmasks, brute force, dfs and similar, implementation, math, probabilities
Correct Solution:
```
from math import *
dic = {}
def brt(lst):
if(tuple(lst) in dic):
return dic[tuple(lst)]
prob1 = lst[0]/(lst[0]+lst[1]+lst[2])
prob2 = lst[1]/(lst[0]+lst[1]+lst[2])
prob3 = lst[2]/(lst[0]+lst[1]+lst[2])
val = prob3
if(lst[0]>0):
nxt1 = list(lst)
dff = min(v,nxt1[0])
nxt1[0]-=v
if(nxt1[0]<0):
nxt1[0]=0
ctt = 1 + (nxt1[1] > 0)
nxt1[2]+=dff/ctt
if(nxt1[1]>0):
nxt1[1]+=dff/ctt
#print(lst,':',nxt1)
if(prob1>0):
val = val + prob1*(1 + brt(nxt1))
if(lst[1] > 0):
nxt2 = list(lst)
dff = min(v,nxt2[1])
nxt2[1]-=v
if(nxt2[1]<0):
nxt2[1]=0
ctt = 1 + (nxt2[0] > 0)
nxt2[2]+=dff/ctt
if(nxt2[0]>0):
nxt2[0]+=dff/ctt
#print(lst,':',nxt2)
if(prob2>0):
val = val + prob2*(1 + brt(nxt2))
dic[tuple(lst)] = val
return val
n = int(input())
for i in range(n):
dic.clear()
sc,sm,sp,sv = input().split(' ')
c = float(sc)*10000
m = float(sm)*10000
p = float(sp)*10000
v = float(sv)*10000
print(brt([c,m,p]))
```
| 100,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Tags: bitmasks, brute force, dfs and similar, implementation, math, probabilities
Correct Solution:
```
from collections import deque
import sys
sys.setrecursionlimit(10**6)
scale=10**6
def rec(c,m,p,v):
res=p/scale
if c>0:
if c>v:
if m>0:
res+=c/scale*(1+rec(c-v,m+v/2,p+v/2,v))
else:
res+=c/scale*(1+rec(c-v,0,p+v,v))
else:
if m>0:
res+=c/scale*(1+rec(0,m+c/2,p+c/2,v))
else:
res+=c/scale*(1+rec(0,0,p+c,v))
if m>0:
if m>v:
if c>0:
res+=m/scale*(1+rec(c+v/2,m-v,p+v/2,v))
else:
res+=m/scale*(1+rec(0,m-v,p+v,v))
else:
if c>0:
res+=m/scale*(1+rec(c+m/2,0,p+m/2,v))
else:
res+=m/scale*(1+rec(0,0,p+m,v))
return res
t=int(input())
for i in range(t):
cm,mm,pm,vv=[float(i) for i in input().split()]
print(rec(cm*scale,mm*scale,pm*scale,vv*scale))
```
| 100,576 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Tags: bitmasks, brute force, dfs and similar, implementation, math, probabilities
Correct Solution:
```
def solve():
c, m, p, v = map(float, input().split())
def rec1(a, p, cur, r):
cur += 1
ans = cur * p * r
if (a > 1e-8):
if (a >= v):
ans += rec1(a - v, p + v, cur, r * a)
else:
ans += rec1(0, p + a, cur, r * a)
return ans
def rec(c, m, p, cur, r):
cur += 1
ans = cur * p * r
if (c > 1e-8):
if (c > v + 1e-9):
ans += rec(c - v, m + v/2, p+v/2, cur, r * c)
else:
ans += rec1( m + c/2, p + c/2, cur, r * c)
if (m > 1e-8):
if (m > v + 1e-9):
ans += rec(c + v/2, m - v, p + v/2, cur, r * m)
else:
ans += rec1(c + m/2, p + m/2, cur, r * m)
return ans
print(f'{rec(c, m, p, 0, 1):.{9}f}')
k = int(input())
for i in range(k):
solve()
```
| 100,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Tags: bitmasks, brute force, dfs and similar, implementation, math, probabilities
Correct Solution:
```
import sys
from itertools import zip_longest
EPS = 0.00000001
def read_floats():
return [float(i) for i in sys.stdin.readline().strip().split()]
def read_int():
return int(sys.stdin.readline().strip())
def probs(c, m, p, v):
#print(f"calculating p({c, m, p, v})")
if c > m:
return probs(m, c, p, v)
# so c <= m
if c < EPS:
if v >= m:
return [p, 1 - p]
else:
return [p] + [(1 - p) * a for a in probs(0, m - v, p + v, v)]
if c < EPS:
pc = []
elif v > c:
pc = probs(0, m + c / 2, p + c / 2, v)
else:
pc = probs(c - v, m + v / 2, p + v / 2, v)
if m < EPS:
pm = []
elif v > m:
pm = probs(c + m / 2, 0, p + m / 2, v)
else:
pm = probs(c + v / 2, m - v, p + v / 2, v)
return [p] + [m * a + c * b for a, b in zip_longest(pm, pc, fillvalue=0)]
t = read_int()
for i in range(t):
c, m, p, v = read_floats()
pb = probs(c, m, p, v)
print(sum(a * b for a, b in zip(range(1, 20), pb)))
```
| 100,578 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Tags: bitmasks, brute force, dfs and similar, implementation, math, probabilities
Correct Solution:
```
def process(c, m, p, v):
d = {'': [1, (c, m, p), (True, True, True)]}
I = 1
answer = 0
while len(d) > 0:
d2 = {}
for x in d:
prob, t, R = d[x]
c2, m2, p2 = t
c_r, m_r, p_r = R
for i in range(3):
if i==0 and R[0]:
step = 'C'
if c2 > v and (c2-v) > 0.00001:
if R[1]:
t2 = [c2-v, m2+v/2, p2+v/2]
d2[x+step] = [prob*c2, t2, R]
else:
t2 = [c2-v, 0, p2+v]
d2[x+step] = [prob*c2, t2, R]
else:
if R[1]:
t2 = [0, m2+c2/2, p2+c2/2]
d2[x+step] = [prob*c2, t2, (False, m_r, p_r)]
else:
t2 = [0, 0, 1]
d2[x+step] = [prob*c2, t2, (False, m_r, p_r)]
elif i==1 and R[1]:
step = 'M'
if m2 > v and (m2-v) > 0.00001:
if R[0]:
t2 = [c2+v/2, m2-v, p2+v/2]
d2[x+step] = [prob*m2, t2, R]
else:
t2 = [0, m2-v, p2+v]
d2[x+step] = [prob*m2, t2, R]
else:
if R[0]:
t2 = [c2+m2/2, 0, p2+m2/2]
d2[x+step] = [prob*m2, t2, (c_r, False, p_r)]
else:
t2 = [0, 0, 1]
d2[x+step] = [prob*m2, t2, (c_r, False, p_r)]
elif i==2:
answer+=(I*prob*p2)
I+=1
d = d2
return answer
t= int(input())
for i in range(t):
c, m, p, v = [float(x) for x in input().split()]
print(process(c, m, p, v))
```
| 100,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Tags: bitmasks, brute force, dfs and similar, implementation, math, probabilities
Correct Solution:
```
import sys, os
from io import BytesIO, IOBase
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
stdin, stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
mod = pow(10, 9) + 7
mod2 = 998244353
def inp(): return stdin.readline().strip()
def iinp(): return int(inp())
def out(var, end="\n"): stdout.write(str(var)+"\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def ceil(a, b): return (a+b-1)//b
S1 = 'abcdefghijklmnopqrstuvwxyz'
S2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
def isprime(x):
if x<=1: return False
if x in (2, 3): return True
if x%2 == 0: return False
for i in range(3, int(sqrt(x))+1, 2):
if x%i == 0: return False
return True
mndif = 10**-6
def solve(curr, currp, c, m, p, v):
ans = (curr+1)*currp*p
if c<mndif and m<mndif:
return ans
if c<mndif:
if m > v:
ans += solve(curr+1, currp*m, 0, m-v, p+v, v)
else:
ans += solve(curr+1, currp*m, 0, 0, p+m, v)
return ans
if m<mndif:
if c > v:
ans += solve(curr+1, currp*c, c-v, 0, p+v, v)
else:
ans += solve(curr+1, currp*c, 0, 0, p+c, v)
return ans
if c > v:
ans += solve(curr+1, currp*c, c-v, m+v/2, p+v/2, v)
else:
ans += solve(curr+1, currp*c, 0, m+c/2, p+c/2, v)
if m > v:
ans += solve(curr+1, currp*m, c+v/2, m-v, p+v/2, v)
else:
ans += solve(curr+1, currp*m, c+m/2, 0, p+m/2, v)
return ans
for _ in range(int(inp())):
c, m, p, v = map(float, inp().split())
ans = solve(0, 1, c, m, p, v)
print(ans)
```
| 100,580 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Tags: bitmasks, brute force, dfs and similar, implementation, math, probabilities
Correct Solution:
```
import math;import heapq;import sys;input=sys.stdin.readline;S=lambda:input();I=lambda:int(S());M=lambda:map(int,S().split());L=lambda:list(M());H=1000000000+7
def solve(c,m,p,v,s,res,a):
if len(s)!=0 and s[-1]=="P":
res[0]+=len(s)*a
return
solve(c,m,p,v,s+"P",res,a*p)
if c>pow(10,-6):
if c>v:
k=1
if m>pow(10,-6):
k+=1
x=v/k
if m>pow(10,-6):
t=m+x
else:
t=m
solve(c-v,t,p+x,v,s+"C",res,a*c)
else:
k=1
if m>pow(10,-6):
k+=1
x=c/k
if m>pow(10,-6):
t=m+x
else:
t=m
solve(0,t,p+x,v,s+"C",res,a*c)
if m>pow(10,-6):
if m>v:
k=1
if c>pow(10,-6):
k+=1
x=v/k
if c>pow(10,-6):
t=c+x
else:
t=c
solve(t,m-v,p+x,v,s+"M",res,a*m)
else:
k=1
if c>pow(10,-6):
k+=1
x=m/k
if c>pow(10,-6):
t=c+x
else:
t=c
solve(t,0,p+x,v,s+"M",res,a*m)
for _ in range(I()):
c,m,p,v=map(float,input().split())
res=[0]
solve(c,m,p,v,"",res,1)
print(res[0])
```
| 100,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Submitted Solution:
```
def E(c, m, p, v):
ans = 0
if p > 0.000001:
ans += p
if c > 0.000001:
m2 = m
p2 = p
if m > 0.000001 and p > 0.000001:
m2 += min(c, v) / 2
p2 += min(c, v) / 2
elif p > 0.000001:
p2 += min(c, v)
ans += c * (1 + E(max(c - v, 0), m2, p2, v))
if m > 0.000001:
c1 = c
p1 = p
if c > 0.000001 and p > 0.000001:
c1 += min(m, v) / 2
p1 += min(m, v) / 2
elif p > 0.000001:
p1 += min(m, v)
ans += m * (1 + E(c1, max(m - v, 0), p1, v))
return ans
for _ in range(int(input())):
c, m, p, v = list(map(float, input().split()))
print(E(c, m, p, v))
```
Yes
| 100,582 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Submitted Solution:
```
"""
ID: happyn61
LANG: PYTHON3
PROB: loan
"""
from itertools import product
import itertools
#from collections import defaultdict
import sys
import heapq
from collections import deque
MOD=1000000000007
#fin = open ('loan.in', 'r')
#fout = open ('loan.out', 'w')
#print(dic["4734"])
def find(parent,i):
if parent[i] != i:
parent[i]=find(parent,parent[i])
return parent[i]
# A utility function to do union of two subsets
def union(parent,rank,xx,yy):
x=find(parent,xx)
y=find(parent,yy)
if rank[x]>rank[y]:
parent[y]=x
elif rank[y]>rank[x]:
parent[x]=y
else:
parent[y]=x
rank[x]+=1
ans=0
#NK=sys.stdin.readline().strip().split()
K=int(sys.stdin.readline().strip())
#N=int(NK[0])
#K=int(NK[1])
#M=int(NK[2])
#ol=list(map(int,sys.stdin.readline().strip().split()))
#d={0:0,1:0}
x=0
y=0
#d={"N":(0,1),"S":(0,-1),"W":(-1,0),"E":(1,0)}
for _ in range(K):
#a=int(sys.stdin.readline().strip())
c,m,p,v=list(map(float,sys.stdin.readline().strip().split()))
#print(c,m,p,v)
stack=deque([(c,m,p,1,1)])
ans=0
while stack:
c,m,p,kk,w=stack.popleft()
#print(c,m,p,w)
ans+=p*w*kk
if (c-0.0000001)>0:
k=c*kk
if c<=v:
#pp+=(c/2)
#m+=(c/2)
#c=0
if (m-0.0000001)>0:
stack.append((0,m+c/2,p+c/2,k,w+1))
else:
stack.append((0,0,p+c,k,w+1))
else:
if (m-0.0000001)>0:
stack.append((c-v,m+v/2,p+v/2,k,w+1))
else:
stack.append((c-v,m,p+v,k,w+1))
if (m-0.0000001)>0:
k=m*kk
if m<=v:
if (c-0.0000001)>0:
stack.append((c+m/2,0,p+m/2,k,w+1))
else:
stack.append((c,0,p+m,k,w+1))
else:
if (c-0.0000001)>0:
stack.append((c+v/2,m-v,p+v/2,k,w+1))
else:
stack.append((c,m-v,p+v,k,w+1))
#print(stack)
print(ans)
```
Yes
| 100,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Submitted Solution:
```
from decimal import *
getcontext().prec = 20
input = __import__('sys').stdin.readline
mis = lambda: map(int, input().split())
ii = lambda: int(input())
dp = {}
v = None
def solve(s) :
if s in dp : return dp[s]
a, b, c = s
ret = (c, c)
if a > 0 :
na = max(Decimal(0), a-v)
x = a-na
nb = b
nc = c
if b > 0 :
nb += x/2
nc += x/2
else : nc += x
nret = solve((na, nb, nc))
ret = (ret[0] + a*(nret[0]+nret[1]), ret[1] + a*nret[1])
if b > 0 :
nb = max(Decimal(0), b-v)
x = b-nb
na = a
nc = c
if a > 0 :
na += x/2
nc += x/2
else : nc += x
nret = solve((na, nb, nc))
ret = (ret[0] + b*(nret[0]+nret[1]), ret[1] + b*nret[1])
dp[s] = ret
return ret
for tc in range(ii()) :
dp = {}
a,b,c,_v = map(Decimal, input().split())
v = _v
print(solve((a,b,c))[0])
```
Yes
| 100,584 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Submitted Solution:
```
def dfs(c,m,p,v,steps,prob):
ans =0
if c!=-1:
if m!=-1:
if c<=v:
ans+=dfs(-1,m+c/2,p+c/2,v,steps+1,prob*c/r)
else:
ans+=dfs(c-v,m+v/2,p+v/2,v,steps+1,prob*c/r)
else:
if c<=v:
ans+=dfs(-1,-1,p+c,v,steps+1,prob*c/r)
else:
ans+=dfs(c-v,-1,p+v,v,steps+1,prob*c/r)
if m!=-1:
if c!=-1:
if m<=v:
ans+=dfs(c+m/2,-1,p+m/2,v,steps+1,prob*m/r)
else:
ans+=dfs(c+v/2,m-v,p+v/2,v,steps+1,prob*m/r)
else:
if m<=v:
ans+=dfs(-1,-1,p+m,v,steps+1,prob*m/r)
else:
ans+=dfs(-1,m-v,p+v,v,steps+1,prob*m/r)
if p:
ans+=steps*prob*p/r
return ans
r = 10**9
T = int(input())
for case in range(T):
c,m,p,v = list(map(float,input().split()))
c*=r
m*=r
p*=r
v*=r
answer = dfs(c,m,p,v,1,1)
print(answer)
```
Yes
| 100,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Submitted Solution:
```
#Fast I/O
import sys,os
#User Imports
from math import *
from bisect import *
from heapq import *
from collections import *
# To enable the file I/O i the below 2 lines are uncommented.
# read from in.txt if uncommented
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
# will print on Console if file I/O is not activated
#if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
# inputs template
from io import BytesIO, IOBase
#Main Logic
def main():
for _ in range(int(input())):
scale=10**6
c,m,p,v=map(float,input().split())
c,m,p,v=int(c*scale),int(m*scale),int(p*scale),int(v*scale)
def rec(c,m,p):
out=p/scale
if c:
if c>v:
if m:
out+=(c/scale)*(1+rec(c-v,m+v//2,p+v//2))
else:
out+=(c/scale)*(1+rec(c-v,0,p+v))
else:
if m:
out+=(c/scale)*(1+rec(0,m+c//2,p+c//2))
else:
out+=(c/scale)*(1+rec(0,0,p+c))
if m:
if m>v:
if c:
out+=(m/scale)*(1+rec(c+v//2,m-v,p+v//2))
else:
out+=(m/scale)*(1+rec(c,m-v,p+v))
else:
if c:
out+=(m/scale)*(1+rec(c+m//2,0,p+m//2))
else:
out+=(m/scale)*(1+rec(c,0,p+m))
return out
print(rec(c,m,p))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#for array of integers
def MI():return (map(int,input().split()))
# endregion
#for fast output, always take string
def outP(var): sys.stdout.write(str(var)+'\n')
# end of any user-defined functions
MOD=10**9+7
mod=998244353
# main functions for execution of the program.
if __name__ == '__main__':
#This doesn't works here but works wonders when submitted on CodeChef or CodeForces
main()
```
No
| 100,586 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Submitted Solution:
```
import sys
import math
import random
from queue import PriorityQueue as PQ
from bisect import bisect_left as BSL
from bisect import bisect_right as BSR
from collections import OrderedDict as OD
from collections import Counter
from itertools import permutations
# mod = 998244353
mod = 1000000007
sys.setrecursionlimit(1000000)
try:
sys.stdin = open("actext.txt", "r")
OPENFILE = 1
except:
pass
def get_ints():
return map(int,input().split())
def palindrome(s):
mid = len(s)//2
for i in range(mid):
if(s[i]!=s[len(s)-i-1]):
return False
return True
def check(i,n):
if(0<=i<n):
return True
else:
return False
# --------------------------------------------------------------------------
def dpsolve(i,c,m,p,v):
if(c==0 and m==0):
return p*(i)
if(c==0):
if(m>v):
newm = m-v
newp = p+v
else:
newm = 0
newp = 1
return m*dpsolve(i+1,0,newm,newp,v) + p*i
if(m==0):
if(c>v):
newc = c-v
newp = p+v
else:
newc = 0
newp = 1
return c*dpsolve(i+1,newc,0,newp,v) + p*i
first = 0
if(c>v):
newc = c-v
newm = m+v/2
newp = p+v/2
else:
newc = 0
newm = m+c/2
newp = p+c/2
first = c*dpsolve(i+1,newc,newm,newp,v)
second = 0
if(m>v):
newm = m-v
newc = c+v/2
newp = p+v/2
else:
newm = 0
newc = c+m/2
newp = p+m/2
second = m*dpsolve(i+1,newc,newm,newp,v)
return (first+second)+p*i
def solve(c,m,p,v):
ans = dpsolve(1,c,m,p,v)
print(ans)
t = int(input())
for tt in range(t):
c,m,p,v = map(float,input().split())
solve(c,m,p,v)
```
No
| 100,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Submitted Solution:
```
import math
scale = 1e+6
def eq(a, b):
return abs(a - b) < 1e-17
def recur(c, m, p, v, d):
ca, cb, cc = 0, 0, 0
# print(c, m, p, v, d)
if c != None and c > v and not eq(c, v):
if m == None and p:
ca = recur(c - v, m, p + v, v, d + 1)
elif p and m:
ca = recur(c - v, m + v / 2, p + v / 2, v, d + 1)
else:
ca = 0
if c != None and c <= v:
if m == None and p:
ca = recur(None, m, p + c, v, d + 1)
elif p and m:
ca = recur(None, m + c / 2, p + c / 2, v, d + 1)
else:
ca = 0
if m != None and m > v and not eq(m, v):
if c == None and p != None:
cb = recur(c, m - v, p + v, v, d + 1)
elif p != None and c != None:
cb = recur(c + v / 2, m - v, p + v / 2, v, d + 1)
else:
cb = 0
if m != None and m <= v:
if c == None and p != None:
cb = recur(c, None, p + m, v, d + 1)
elif p != None and c != None:
cb = recur(c + m / 2, None, p + m / 2, v, d + 1)
else:
cb = 0
if p:
cc = p
c1, c2, c3 = c if c else 0, m if m else 0, p if p else 0
c1 /= scale
c2 /= scale
# print(c1, c2, c3, ca, cb, cc)
return (c1 * ca + c2 * cb) + p / scale * float(d)
def solve():
c, m, p, v = tuple(float(x) * scale for x in input().split())
# print(c, m, p, v)
print(recur(int(c), int(m), int(p), int(v), 1))
return
for _ in range(int(input())):
solve()
```
No
| 100,588 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After defeating a Blacklist Rival, you get a chance to draw 1 reward slip out of x hidden valid slips. Initially, x=3 and these hidden valid slips are Cash Slip, Impound Strike Release Marker and Pink Slip of Rival's Car. Initially, the probability of drawing these in a random guess are c, m, and p, respectively. There is also a volatility factor v. You can play any number of Rival Races as long as you don't draw a Pink Slip. Assume that you win each race and get a chance to draw a reward slip. In each draw, you draw one of the x valid items with their respective probabilities. Suppose you draw a particular item and its probability of drawing before the draw was a. Then,
* If the item was a Pink Slip, the quest is over, and you will not play any more races.
* Otherwise,
1. If aโค v, the probability of the item drawn becomes 0 and the item is no longer a valid item for all the further draws, reducing x by 1. Moreover, the reduced probability a is distributed equally among the other remaining valid items.
2. If a > v, the probability of the item drawn reduces by v and the reduced probability is distributed equally among the other valid items.
For example,
* If (c,m,p)=(0.2,0.1,0.7) and v=0.1, after drawing Cash, the new probabilities will be (0.1,0.15,0.75).
* If (c,m,p)=(0.1,0.2,0.7) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,0.25,0.75).
* If (c,m,p)=(0.2,Invalid,0.8) and v=0.1, after drawing Cash, the new probabilities will be (0.1,Invalid,0.9).
* If (c,m,p)=(0.1,Invalid,0.9) and v=0.2, after drawing Cash, the new probabilities will be (Invalid,Invalid,1.0).
You need the cars of Rivals. So, you need to find the expected number of races that you must play in order to draw a pink slip.
Input
The first line of input contains a single integer t (1โค tโค 10) โ the number of test cases.
The first and the only line of each test case contains four real numbers c, m, p and v (0 < c,m,p < 1, c+m+p=1, 0.1โค vโค 0.9).
Additionally, it is guaranteed that each of c, m, p and v have at most 4 decimal places.
Output
For each test case, output a single line containing a single real number โ the expected number of races that you must play in order to draw a Pink Slip.
Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.
Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} โค 10^{-6}.
Example
Input
4
0.2 0.2 0.6 0.2
0.4 0.2 0.4 0.8
0.4998 0.4998 0.0004 0.1666
0.3125 0.6561 0.0314 0.2048
Output
1.532000000000
1.860000000000
5.005050776521
4.260163673896
Note
For the first test case, the possible drawing sequences are:
* P with a probability of 0.6;
* CP with a probability of 0.2โ
0.7 = 0.14;
* CMP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* CMMP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006;
* MP with a probability of 0.2โ
0.7 = 0.14;
* MCP with a probability of 0.2โ
0.3โ
0.9 = 0.054;
* MCCP with a probability of 0.2โ
0.3โ
0.1โ
1 = 0.006.
So, the expected number of races is equal to 1โ
0.6 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 + 2โ
0.14 + 3โ
0.054 + 4โ
0.006 = 1.532.
For the second test case, the possible drawing sequences are:
* P with a probability of 0.4;
* CP with a probability of 0.4โ
0.6 = 0.24;
* CMP with a probability of 0.4โ
0.4โ
1 = 0.16;
* MP with a probability of 0.2โ
0.5 = 0.1;
* MCP with a probability of 0.2โ
0.5โ
1 = 0.1.
So, the expected number of races is equal to 1โ
0.4 + 2โ
0.24 + 3โ
0.16 + 2โ
0.1 + 3โ
0.1 = 1.86.
Submitted Solution:
```
blanck=[]
from decimal import *
def rec(a,b,c,v,p,card):
# c=round(c,6)
if (a==float("inf") and b==float("inf")) or c>=1 :
blanck.append(p*(card+1))
return
# print(a,b,c)
l=[a,b,c]
for i in range(3):
if i!=2 and i!=float("inf") and l[i]!=1:
if i==0:
ta=l[i]
if ta<=v:
ta=float("inf")
else:
ta-=v
cnt=2
# if b!=float("inf"):
# cnt+=1
dd=min(l[i],v)
if b!=float("inf"):
rec(ta,b+(dd/cnt),c+(dd/cnt),v,p*l[i],card+1)
else:
rec(ta,b,c+(dd/cnt),v,p*l[i],card+1)
if i==1:
tb=l[i]
if tb<=v:
tb=float("inf")
else:
tb-=v
cnt=1
# if a!=float("inf"):
# cnt+=1
cnt=2
dd=min(l[i],v)
if a!=float("inf"):
rec(a+(dd/cnt),tb,c+(dd/cnt),v,p*l[i],card+1)
else:
rec(a,tb,min(c+(dd/cnt),1.0),v,p*l[i],card+1)
if i==2:
blanck.append(p*c*(card+1))
for t in range(int(input())):
blanck=[]
a,b,c,d=map(float,input().strip().split())
rec(a,b,c,d,1.0,0.0)
ans=0
# blanck=[]
# print(blanck)
# cnt=Counter(blanck)
# print(cnt)
for i in blanck:
if i!=float("inf"):
ans+=i
# print(ans)
print(ans)
```
No
| 100,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"This problem is rubbish! There is not statement, and there are only 5 test cases. The problemsetter took liberties with this problem!" โ people complained in the comments to one round on Codeforces. And even more... No, wait, the checker for the problem was alright, that's a mercy.
Input
The only line of the input contains an integer between 1 and 5, inclusive. All tests for this problem are different. The contents of the test case doesn't need to be equal to its index.
Output
The only line of the output contains an integer between 1 and 3, inclusive.
Examples
Note
This problem has no samples, since there so few test cases.
Tags: *special, brute force
Correct Solution:
```
n = int(input())
print(1 if n == 5 else n%3 + 1)
```
| 100,590 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"This problem is rubbish! There is not statement, and there are only 5 test cases. The problemsetter took liberties with this problem!" โ people complained in the comments to one round on Codeforces. And even more... No, wait, the checker for the problem was alright, that's a mercy.
Input
The only line of the input contains an integer between 1 and 5, inclusive. All tests for this problem are different. The contents of the test case doesn't need to be equal to its index.
Output
The only line of the output contains an integer between 1 and 3, inclusive.
Examples
Note
This problem has no samples, since there so few test cases.
Tags: *special, brute force
Correct Solution:
```
import random
x = int(input())
print(1 if x == 3 or x==5 else 2 if x==1 or x==4 else 3)
```
| 100,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"This problem is rubbish! There is not statement, and there are only 5 test cases. The problemsetter took liberties with this problem!" โ people complained in the comments to one round on Codeforces. And even more... No, wait, the checker for the problem was alright, that's a mercy.
Input
The only line of the input contains an integer between 1 and 5, inclusive. All tests for this problem are different. The contents of the test case doesn't need to be equal to its index.
Output
The only line of the output contains an integer between 1 and 3, inclusive.
Examples
Note
This problem has no samples, since there so few test cases.
Tags: *special, brute force
Correct Solution:
```
a=int(input())
if a==1:
print("2")
if a==2:
print("3")
if a==3:
print("1")
if a==4:
print("2")
if a==5:
print("1")
```
| 100,592 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"This problem is rubbish! There is not statement, and there are only 5 test cases. The problemsetter took liberties with this problem!" โ people complained in the comments to one round on Codeforces. And even more... No, wait, the checker for the problem was alright, that's a mercy.
Input
The only line of the input contains an integer between 1 and 5, inclusive. All tests for this problem are different. The contents of the test case doesn't need to be equal to its index.
Output
The only line of the output contains an integer between 1 and 3, inclusive.
Examples
Note
This problem has no samples, since there so few test cases.
Tags: *special, brute force
Correct Solution:
```
x = int(input())
if x == 1:
print(2)
elif x == 2:
print(3)
elif x == 3:
print(1)
elif x == 4:
print(2)
elif x == 5:
print(1)
```
| 100,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"This problem is rubbish! There is not statement, and there are only 5 test cases. The problemsetter took liberties with this problem!" โ people complained in the comments to one round on Codeforces. And even more... No, wait, the checker for the problem was alright, that's a mercy.
Input
The only line of the input contains an integer between 1 and 5, inclusive. All tests for this problem are different. The contents of the test case doesn't need to be equal to its index.
Output
The only line of the output contains an integer between 1 and 3, inclusive.
Examples
Note
This problem has no samples, since there so few test cases.
Tags: *special, brute force
Correct Solution:
```
N = int(input())
ans = [1, 2, 3, 1, 2, 1]
print(ans[N])
```
| 100,594 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"This problem is rubbish! There is not statement, and there are only 5 test cases. The problemsetter took liberties with this problem!" โ people complained in the comments to one round on Codeforces. And even more... No, wait, the checker for the problem was alright, that's a mercy.
Input
The only line of the input contains an integer between 1 and 5, inclusive. All tests for this problem are different. The contents of the test case doesn't need to be equal to its index.
Output
The only line of the output contains an integer between 1 and 3, inclusive.
Examples
Note
This problem has no samples, since there so few test cases.
Tags: *special, brute force
Correct Solution:
```
print([1,2,3,1,2,1][int(input())])
#
#
#
#
#
#
#
#
#
##############################
```
| 100,595 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"This problem is rubbish! There is not statement, and there are only 5 test cases. The problemsetter took liberties with this problem!" โ people complained in the comments to one round on Codeforces. And even more... No, wait, the checker for the problem was alright, that's a mercy.
Input
The only line of the input contains an integer between 1 and 5, inclusive. All tests for this problem are different. The contents of the test case doesn't need to be equal to its index.
Output
The only line of the output contains an integer between 1 and 3, inclusive.
Examples
Note
This problem has no samples, since there so few test cases.
Tags: *special, brute force
Correct Solution:
```
a=int(input())
if a==5:
print(1)
else:
print(a%3+1)
```
| 100,596 |
Provide tags and a correct Python 3 solution for this coding contest problem.
"This problem is rubbish! There is not statement, and there are only 5 test cases. The problemsetter took liberties with this problem!" โ people complained in the comments to one round on Codeforces. And even more... No, wait, the checker for the problem was alright, that's a mercy.
Input
The only line of the input contains an integer between 1 and 5, inclusive. All tests for this problem are different. The contents of the test case doesn't need to be equal to its index.
Output
The only line of the output contains an integer between 1 and 3, inclusive.
Examples
Note
This problem has no samples, since there so few test cases.
Tags: *special, brute force
Correct Solution:
```
"""====================================================================================
====================================================================================
___ _______ ___ _______ ___ ___
| /\ | | \ | | / | | | | |\ /|
| / \ | | \ | | / | | | | | \ / |
|___ /____\ | | \ | |/ |___| | | | \/ |
| / \ | | / | |\ |\ | | | |
| / \ | | / | | \ | \ | | | |
___|/ \___|___ |___/ ___|___ | \ | \ |___| | |
====================================================================================
====================================================================================
"""
# โฅโฅโฅโฅโฅโฅโฅโฅโฅโฅโฅโฅโฅโฅโฅโฅ
a = int(input())
if a==3:
print("1")
if a==1:
print("2")
if a==4:
print("2")
if a==2:
print("3")
if a==5:
print("1")
# โฅโฅโฅโฅโฅโฅโฅโฅโฅโฅโฅโฅโฅโฅโฅโฅ
"""====================================================================================
====================================================================================
___ _______ ___ _______ ___ ___
| /\ | | \ | | / | | | | |\ /|
| / \ | | \ | | / | | | | | \ / |
|___ /____\ | | \ | |/ |___| | | | \/ |
| / \ | | / | |\ |\ | | | |
| / \ | | / | | \ | \ | | | |
___|/ \___|___ |___/ ___|___ | \ | \ |___| | |
====================================================================================
====================================================================================
"""
```
| 100,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"This problem is rubbish! There is not statement, and there are only 5 test cases. The problemsetter took liberties with this problem!" โ people complained in the comments to one round on Codeforces. And even more... No, wait, the checker for the problem was alright, that's a mercy.
Input
The only line of the input contains an integer between 1 and 5, inclusive. All tests for this problem are different. The contents of the test case doesn't need to be equal to its index.
Output
The only line of the output contains an integer between 1 and 3, inclusive.
Examples
Note
This problem has no samples, since there so few test cases.
Submitted Solution:
```
n=int(input())
if n==3:
print (1)
elif n==1:
print (2)
elif n==4:
print (2)
elif n==2:
print (3)
else:
print (1)
```
Yes
| 100,598 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"This problem is rubbish! There is not statement, and there are only 5 test cases. The problemsetter took liberties with this problem!" โ people complained in the comments to one round on Codeforces. And even more... No, wait, the checker for the problem was alright, that's a mercy.
Input
The only line of the input contains an integer between 1 and 5, inclusive. All tests for this problem are different. The contents of the test case doesn't need to be equal to its index.
Output
The only line of the output contains an integer between 1 and 3, inclusive.
Examples
Note
This problem has no samples, since there so few test cases.
Submitted Solution:
```
vals = [2, 3, 1, 2, 1]
i = int(input())
print(vals[i-1])
```
Yes
| 100,599 |
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