# Codeforces Round #371 (Div. 1) B. Searching Rectangles

,

http://codeforces.com/contest/713/problem/B

## solution

Use binary search, $3$ times. The first time is to get the AABB of $A \cup B$, the rest are for to determine the $A$ and $B$.

## implementation

pwntools is useful.

#!/usr/bin/env python3
import sys
def contain(a, b):
ax1, ay1, ax2, ay2 = a
bx1, by1, bx2, by2 = b
return bx1 <= ax1 and ax2 <= bx2 and by1 <= ay1 and ay2 <= by2
def ask(x1, y1, x2, y2, known=(), memo={}):
if x2 < x1+1 or y2 < y1+1:
return 0
ofs = len(list(filter(lambda rect: contain(rect, (x1, y1, x2, y2)), known)))
key = (x1+1, y1+1, x2, y2)
if key in memo:
return memo[key] - ofs
print('?', *key)
sys.stdout.flush()
memo[key] = int(input())
return memo[key] - ofs
def binsearch(l, r, p): # (l,r], return the smallest n which p holds
assert l < r
while l+1 != r:
m = (l + r) // 2
if p(m):
r = m
else:
l = m
return r
def shrink(x1, y1, x2, y2, cnt, known=()):
assert ask(x1, y1, x2, y2, known=known) == cnt
x1 = binsearch(x1, x2, lambda x: ask(x, y1, x2, y2, known=known) != cnt) - 1
y1 = binsearch(y1, y2, lambda y: ask(x1, y, x2, y2, known=known) != cnt) - 1
x2 = binsearch(x1, x2, lambda x: ask(x1, y1, x, y2, known=known) == cnt)
y2 = binsearch(y1, y2, lambda y: ask(x1, y1, x2, y, known=known) == cnt)
assert ask(x1, y1, x2, y2, known=known) == cnt
assert ask(x1, y1, x2, y2, known=known) == cnt
return x1, y1, x2, y2
def go(x1, y1, x2, y2):
assert ask(x1, y1, x2, y2) == 2
x1, y1, x2, y2 = shrink(x1, y1, x2, y2, 2)
a = None
if not a and x1 < x2:
if ask(x1+1, y1, x2, y2) == 1:
a = shrink(x1+1, y1, x2, y2, 1)
elif ask(x1, y1, x2-1, y2) == 1:
a = shrink(x1, y1, x2-1, y2, 1)
if not a and y1 < y2:
if ask(x1, y1+1, x2, y2) == 1:
a = shrink(x1, y1+1, x2, y2, 1)
elif ask(x1, y1, x2, y2-1) == 1:
a = shrink(x1, y1, x2, y2-1, 1)
if not a:
a = x1, y1, x2, y2
return a, a
else:
b = shrink(x1, y1, x2, y2, 1, known=[ a ])
return a, b
n = int(input())
a, b = go(0, 0, n, n)
ax1, ay1, ax2, ay2 = a
bx1, by1, bx2, by2 = b
print('!', ax1+1, ay1+1, ax2, ay2, bx1+1, by1+1, bx2, by2)

#!/usr/bin/env python2
import ast
import random
import sys
from pwn import * # https://pypi.python.org/pypi/pwntools
import argparse
parser = argparse.ArgumentParser()
args = parser.parse_args()
context.log_level = 'debug'

n = args.n
log.info('N: %d', n)
if args.a:
ax1, ay1, ax2, ay2 = ast.literal_eval(args.a)
else:
ax1 = random.randint(1, n)
ay1 = random.randint(1, n)
ax2 = random.randint(1, n)
ay2 = random.randint(1, n)
ax1, ax2 = sorted([ ax1, ax2 ])
ay1, ay2 = sorted([ ay1, ay2 ])
a = ( ax1, ay1, ax2, ay2 )
log.info('A: %d %d %d %d', *a)
if args.b:
bx1, by1, bx2, by2 = ast.literal_eval(args.b)
else:
bx1 = random.randint(1, n)
by1 = random.randint(1, n)
bx2 = random.randint(1, n)
by2 = random.randint(1, n)
bx1, bx2 = sorted([ bx1, bx2 ])
by1, by2 = sorted([ by1, by2 ])
b = ( bx1, by1, bx2, by2 )
log.info('B: %d %d %d %d', *b)

p = process(args.program, stderr=sys.stderr)
p.sendline(str(n))

for i in range(200):
line = p.recvline()
c = line.split()[0]
args = map(int, line.split()[1:])
assert c in '?!'
if c == '?':
assert len(args) == 4
x1, y1, x2, y2 = args
assert 1 <= x1 <= x2 <= n
assert 1 <= y1 <= y2 <= n
cnt = 0
if x1 <= ax1 <= ax2 <= x2 and y1 <= ay1 <= ay2 <= y2:
cnt += 1
if x1 <= bx1 <= bx2 <= x2 and y1 <= by1 <= by2 <= y2:
cnt += 1
p.sendline(str(cnt))
elif c == '!':
assert len(args) == 8
p = tuple(args[:4])
q = tuple(args[4:])
if (p == a and q == b) or (p == b and q == a):
log.info('used queries: %d', i+1)
log.info('Accepted')
sys.exit(0)
else: