結果

問題 No.3444 Interval Xor MST
コンテスト
ユーザー lif4635
提出日時 2026-02-06 22:10:40
言語 PyPy3
(7.3.17)
結果
AC  
実行時間 500 ms / 2,000 ms
コード長 3,480 bytes
記録
記録タグの例:
初AC ショートコード 純ショートコード 純主流ショートコード 最速実行時間
コンパイル時間 341 ms
コンパイル使用メモリ 82,576 KB
実行使用メモリ 78,204 KB
最終ジャッジ日時 2026-02-06 22:10:44
合計ジャッジ時間 4,388 ms
ジャッジサーバーID
(参考情報)
judge5 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 7
権限があれば一括ダウンロードができます

ソースコード

diff #
raw source code

# input
import sys
input = sys.stdin.readline
II = lambda : int(input())
MI = lambda : map(int, input().split())
LI = lambda : [int(a) for a in input().split()]
SI = lambda : input().rstrip()
LLI = lambda n : [[int(a) for a in input().split()] for _ in range(n)]
LSI = lambda n : [input().rstrip() for _ in range(n)]
MI_1 = lambda : map(lambda x:int(x)-1, input().split())
LI_1 = lambda : [int(a)-1 for a in input().split()]

mod = 998244353
inf = 1001001001001001001
ordalp = lambda s : ord(s)-65 if s.isupper() else ord(s)-97
ordallalp = lambda s : ord(s)-39 if s.isupper() else ord(s)-97
yes = lambda : print("Yes")
no = lambda : print("No")
yn = lambda flag : print("Yes" if flag else "No")

prinf = lambda ans : print(ans if ans < 1000001001001001001 else -1)
alplow = "abcdefghijklmnopqrstuvwxyz"
alpup = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
alpall = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
URDL = {'U':(-1,0), 'R':(0,1), 'D':(1,0), 'L':(0,-1)}
DIR_4 = [[-1,0],[0,1],[1,0],[0,-1]]
DIR_8 = [[-1,0],[-1,1],[0,1],[1,1],[1,0],[1,-1],[0,-1],[-1,-1]]
DIR_BISHOP = [[-1,1],[1,1],[1,-1],[-1,-1]]
prime60 = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59]
sys.set_int_max_str_digits(0)
# sys.setrecursionlimit(10**6)
# import pypyjit
# pypyjit.set_param('max_unroll_recursion=-1')

from collections import defaultdict,deque
from heapq import heappop,heappush
from bisect import bisect_left,bisect_right
DD = defaultdict
BSL = bisect_left
BSR = bisect_right

from functools import cache

def calc(l, r):
    if l <= r:
        return 0
    

memo = {}
def dfs(l, r):
    lr = l << 32 | r
    if lr in memo:
        return memo[lr]
    if r - l <= 1:
        return 0
    d = r - l
    if l % d == 0 and (d & (d - 1)) == 0:
        k = d.bit_length() - 1
        return k * (d // 2)
    
    sp = l ^ (r - 1)
    if sp == 0:
        return 0
    
    bit = 1 << sp.bit_length() - 1
    m = (l | bit) & ~(bit - 1)
    res = bit
    
    res += dfs(l, m) + dfs(m, r)
    
    l = l & (bit - 1)
    r = (r - 1) & (bit - 1)
    
    if l > r:
        ans = 0
        for i in range(60, -1, -1):
            bit = 1 << i
            fl = l >> i & 1
            fr = r >> i & 1
            if fl:
                if not fr:
                    ans += bit
            else:
                if fr:
                    break
        res += ans
    
    memo[lr] = res
    return res

t = II()
for i in range(t):
    n, m = MI()
    ans = 0
    # ans = dfs(m, m+n)
    l = m
    r = m + n
    st = [(l, r)]
    while st:
        l, r = st.pop() 
        # lr = l << 32 | r
        # if lr in memo:
        #     ans += memo[lr]
        #     continue
        
        if r - l <= 1:
            continue
        
        d = r - l
        if l % d == 0 and (d & (d - 1)) == 0:
            k = d.bit_length() - 1
            # memo[lr] = k * (d // 2)
            ans += k * (d // 2)
            continue
        
        sp = l ^ (r - 1)
        if sp == 0:
            continue
        
        bit = 1 << (sp.bit_length() - 1)
        m = (l | bit) & ~(bit - 1)
        
        st.append((l, m))
        st.append((m, r))
        
        res = bit
        
        nl = l & (bit - 1)
        nr = (r - 1) & (bit - 1)
        
        if nl > nr:
            d = nl & (~nr)
            d2 = (~nl) & nr
            if d2 == 0:
                res += d
            else:
                p = (1 << d2.bit_length()) - 1
                res += d & (~p)
        ans += res
    
    print(ans)
0