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362 lines
10 KiB
Python
Executable file
362 lines
10 KiB
Python
Executable file
"""#
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### 谜题描述
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You are given an array a of length n, which initially is a permutation of numbers from 1 to n. In one operation, you can choose an index i (1 ≤ i < n) such that a_i < a_{i + 1}, and remove either a_i or a_{i + 1} from the array (after the removal, the remaining parts are concatenated).
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For example, if you have the array [1, 3, 2], you can choose i = 1 (since a_1 = 1 < a_2 = 3), then either remove a_1 which gives the new array [3, 2], or remove a_2 which gives the new array [1, 2].
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Is it possible to make the length of this array equal to 1 with these operations?
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Input
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The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. The description of the test cases follows.
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The first line of each test case contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5) — the length of the array.
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The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n, a_i are pairwise distinct) — elements of the array.
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It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.
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Output
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For each test case, output on a single line the word \"YES\" if it is possible to reduce the array to a single element using the aforementioned operation, or \"NO\" if it is impossible to do so.
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Example
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Input
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4
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3
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1 2 3
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4
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3 1 2 4
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3
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2 3 1
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6
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2 4 6 1 3 5
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Output
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YES
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YES
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NO
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YES
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Note
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For the first two test cases and the fourth test case, we can operate as follow (the bolded elements are the pair chosen for that operation):
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[1, 2, 3] → [1, 2] → [1]
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[3, 1, 2, 4] → [3, 1, 4] → [3, 4] → [4]
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[2, 4, 6, 1, 3, 5] → [4, 6, 1, 3, 5] → [4, 1, 3, 5] → [4, 1, 5] → [4, 5] → [4]
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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\"\"\"
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Satwik_Tiwari ;) .
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4th july , 2020 - Saturday
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\"\"\"
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#===============================================================================================
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#importing some useful libraries.
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from __future__ import division, print_function
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from fractions import Fraction
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import sys
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import os
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from io import BytesIO, IOBase
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import bisect
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from heapq import *
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from math import *
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from collections import deque
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from collections import Counter as counter # Counter(list) return a dict with {key: count}
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from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
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from itertools import permutations as permutate
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from bisect import bisect_left as bl
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#If the element is already present in the list,
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# the left most position where element has to be inserted is returned.
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from bisect import bisect_right as br
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from bisect import bisect
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#If the element is already present in the list,
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# the right most position where element has to be inserted is returned
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#==============================================================================================
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BUFSIZE = 8192
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class FastIO(IOBase):
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newlines = 0
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def __init__(self, file):
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self._fd = file.fileno()
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self.buffer = BytesIO()
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self.writable = \"x\" in file.mode or \"r\" not in file.mode
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self.write = self.buffer.write if self.writable else None
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def read(self):
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while True:
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b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
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if not b:
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break
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ptr = self.buffer.tell()
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self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
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self.newlines = 0
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return self.buffer.read()
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def readline(self):
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while self.newlines == 0:
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b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
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self.newlines = b.count(b\"\n\") + (not b)
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ptr = self.buffer.tell()
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self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
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self.newlines -= 1
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return self.buffer.readline()
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def flush(self):
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if self.writable:
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os.write(self._fd, self.buffer.getvalue())
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self.buffer.truncate(0), self.buffer.seek(0)
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class IOWrapper(IOBase):
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def __init__(self, file):
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self.buffer = FastIO(file)
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self.flush = self.buffer.flush
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self.writable = self.buffer.writable
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self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))
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self.read = lambda: self.buffer.read().decode(\"ascii\")
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self.readline = lambda: self.buffer.readline().decode(\"ascii\")
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def print(*args, **kwargs):
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\"\"\"Prints the values to a stream, or to sys.stdout by default.\"\"\"
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sep, file = kwargs.pop(\"sep\", \" \"), kwargs.pop(\"file\", sys.stdout)
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at_start = True
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for x in args:
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if not at_start:
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file.write(sep)
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file.write(str(x))
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at_start = False
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file.write(kwargs.pop(\"end\", \"\n\"))
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if kwargs.pop(\"flush\", False):
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file.flush()
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if sys.version_info[0] < 3:
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sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
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else:
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sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
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# inp = lambda: sys.stdin.readline().rstrip(\"\r\n\")
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#===============================================================================================
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#some shortcuts
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mod = 1000000007
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def inp(): return sys.stdin.readline().rstrip(\"\r\n\") #for fast input
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def out(var): sys.stdout.write(str(var)) #for fast output, always take string
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def lis(): return list(map(int, inp().split()))
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def stringlis(): return list(map(str, inp().split()))
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def sep(): return map(int, inp().split())
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def strsep(): return map(str, inp().split())
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# def graph(vertex): return [[] for i in range(0,vertex+1)]
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def zerolist(n): return [0]*n
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def nextline(): out(\"\n\") #as stdout.write always print sring.
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def testcase(t):
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for p in range(t):
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solve()
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def printlist(a) :
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for p in range(0,len(a)):
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out(str(a[p]) + ' ')
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def lcm(a,b): return (a*b)//gcd(a,b)
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def power(a,b):
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ans = 1
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while(b>0):
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if(b%2==1):
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ans*=a
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a*=a
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b//=2
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return ans
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def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
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def isPrime(n) : # Check Prime Number or not
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if (n <= 1) : return False
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if (n <= 3) : return True
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if (n % 2 == 0 or n % 3 == 0) : return False
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i = 5
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while(i * i <= n) :
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if (n % i == 0 or n % (i + 2) == 0) :
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return False
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i = i + 6
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return True
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#===============================================================================================
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# code here ;))
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def bs(a,l,h,x):
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while(l<h):
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# print(l,h)
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mid = (l+h)//2
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if(a[mid] == x):
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return mid
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if(a[mid] < x):
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l = mid+1
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else:
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h = mid
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return l
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def sieve(a): #O(n loglogn) nearly linear
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#all odd mark 1
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for i in range(3,((10**6)+1),2):
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a[i] = 1
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#marking multiples of i form i*i 0. they are nt prime
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for i in range(3,((10**6)+1),2):
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for j in range(i*i,((10**6)+1),i):
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a[j] = 0
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a[2] = 1 #special left case
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return (a)
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def bfs(g,st):
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visited = [-1]*(len(g))
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visited[st] = 0
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queue = []
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queue.append(st)
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new = []
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while(len(queue) != 0):
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s = queue.pop()
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new.append(s)
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for i in g[s]:
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if(visited[i] == -1):
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visited[i] = visited[s]+1
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queue.append(i)
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return visited
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def dfsusingstack(v,st):
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d = deque([])
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visited = [0]*(len(v))
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d.append(st)
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new = []
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visited[st] = 1
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while(len(d) != 0):
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curr = d.pop()
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new.append(curr)
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for i in v[curr]:
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if(visited[i] == 0):
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visited[i] = 1
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d.append(i)
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return new
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def solve():
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n = int(inp())
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a= lis()
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f = True
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dp = [0]*(n)
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# dp[0] = 0
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curr = 0
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for i in range(1,n):
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if(a[i] < a[curr]):
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dp[i] = i
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curr = i
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else:
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dp[i] = curr
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end = n-2
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while(f):
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ind = dp[end]
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# print(end,ind)
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if(a[ind] > a[n-1]):
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print('NO')
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return
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end = ind-1
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if(end<0):
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f = False
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print('YES')
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# testcase(1)
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testcase(int(inp()))
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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from bootcamp import Basebootcamp
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import random
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import re
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class Celementexterminationbootcamp(Basebootcamp):
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def __init__(self, min_length=2, max_length=10):
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self.min_length = min_length
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self.max_length = max_length
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def case_generator(self):
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n = random.randint(self.min_length, self.max_length)
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a = list(range(1, n + 1))
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random.shuffle(a)
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expected_answer = 'YES' if self.check_solution(a.copy()) else 'NO' # 修复1：使用副本保证原始数据不变
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return {
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'n': n,
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'array': a,
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'expected_answer': expected_answer
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}
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@staticmethod
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def check_solution(a):
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n = len(a)
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if n == 1:
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return True
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# 修复2：修正循环终止条件（参考原题正确解法）
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max_val = a[-1]
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for i in reversed(range(n-1)):
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if a[i] < max_val:
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max_val = a[i]
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else:
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return False
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return True
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@staticmethod
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def prompt_func(question_case) -> str:
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problem_desc = (
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"You are given an array a of length n, which is a permutation of numbers from 1 to n. "
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"In each operation, you can choose an index i (1 ≤ i < n) where a_i < a_{{i+1}}, "
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"and remove either a_i or a_{{i+1}}. The goal is to determine if it's possible to reduce "
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"the array to a single element using these operations.\n\n"
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"Input format:\n"
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"- The first line contains n (array length)\n"
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"- The second line contains the array elements\n\n"
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"Output 'YES' or 'NO'.\n\n"
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"Your task:\n"
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"Test case:\n"
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"n = {n}\n"
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"array: {a}\n\n"
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"Put your final answer within [answer] and [/answer] tags."
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).format(
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n=question_case['n'],
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a=' '.join(map(str, question_case['array']))
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)
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return problem_desc
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@staticmethod
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def extract_output(output):
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matches = re.findall(r'\[answer\]\s*(YES|NO)\s*\[/answer\]', output, re.IGNORECASE)
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return matches[-1].upper() if matches else None
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@classmethod
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def _verify_correction(cls, solution, identity):
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return solution == identity['expected_answer']
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