problem_id stringlengths 3 7 | contestId stringclasses 660
values | problem_index stringclasses 27
values | programmingLanguage stringclasses 3
values | testset stringclasses 5
values | incorrect_passedTestCount float64 0 146 | incorrect_timeConsumedMillis float64 15 4.26k | incorrect_memoryConsumedBytes float64 0 271M | incorrect_submission_id stringlengths 7 9 | incorrect_source stringlengths 10 27.7k | correct_passedTestCount float64 2 360 | correct_timeConsumedMillis int64 30 8.06k | correct_memoryConsumedBytes int64 0 475M | correct_submission_id stringlengths 7 9 | correct_source stringlengths 28 21.2k | contest_name stringclasses 664
values | contest_type stringclasses 3
values | contest_start_year int64 2.01k 2.02k | time_limit float64 0.5 15 | memory_limit float64 64 1.02k | title stringlengths 2 54 | description stringlengths 35 3.16k | input_format stringlengths 67 1.76k | output_format stringlengths 18 1.06k ⌀ | interaction_format null | note stringclasses 840
values | examples stringlengths 34 1.16k | rating int64 800 3.4k ⌀ | tags stringclasses 533
values | testset_size int64 2 360 | official_tests stringlengths 44 19.7M | official_tests_complete bool 1
class | input_mode stringclasses 1
value | generated_checker stringclasses 231
values | executable bool 1
class |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
908/C | 908 | C | Python 3 | TESTS | 6 | 77 | 5,632,000 | 33793953 | from math import sqrt
n, r = list(map(int, input().split()))
x = list(map(int, input().split()))
o = []
for j in x:
k =True
for i in reversed(o):
if i[0] == j:
o.append([j, i[1] + 2 * r])
k =False
break
elif i[0] + 2 * r == j or i[0] - 2 * r == j:
... | 15 | 124 | 4,403,200 | 222645507 | from sys import stdin, setrecursionlimit
def solve():
n, r = (int(s) for s in stdin.readline().split())
mas = [int(s) for s in stdin.readline().split()]
prev_x, prev_y = mas[0], r
drop_mas = [(prev_x, prev_y)]
ans = [prev_y]
for i in range(1, n):
cur_x = mas[i]
cur_y = r
... | Good Bye 2017 | CF | 2,017 | 2 | 256 | New Year and Curling | Carol is currently curling.
She has n disks each with radius r on the 2D plane.
Initially she has all these disks above the line y = 10100.
She then will slide the disks towards the line y = 0 one by one in order from 1 to n.
When she slides the i-th disk, she will place its center at the point (xi, 10100). She wil... | The first line will contain two integers n and r (1 ≤ n, r ≤ 1 000), the number of disks, and the radius of the disks, respectively.
The next line will contain n integers x1, x2, ..., xn (1 ≤ xi ≤ 1 000) — the x-coordinates of the disks. | Print a single line with n numbers. The i-th number denotes the y-coordinate of the center of the i-th disk. The output will be accepted if it has absolute or relative error at most 10 - 6.
Namely, let's assume that your answer for a particular value of a coordinate is a and the answer of the jury is b. The checker pr... | null | The final positions of the disks will look as follows:
In particular, note the position of the last disk. | [{"input": "6 2\n5 5 6 8 3 12", "output": "2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613"}] | 1,500 | ["brute force", "geometry", "implementation", "math"] | 15 | [{"input": "6 2\r\n5 5 6 8 3 12\r\n", "output": "2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613\r\n"}, {"input": "1 1\r\n5\r\n", "output": "1\r\n"}, {"input": "5 300\r\n939 465 129 611 532\r\n", "output": "300 667.864105343 1164.9596696 1522.27745533 2117.05388391\r\n"}, {"input": "5 1\r\n416 387 336 116... | false | stdio | import sys
def main(input_path, output_path, submission_path):
with open(input_path) as f:
n, r = map(int, f.readline().split())
with open(output_path) as f:
ref = list(map(float, f.read().strip().split()))
with open(submission_path) as f:
sub = list(map(float, f.read().st... | true |
570/D | 570 | D | PyPy 3-64 | TESTS | 0 | 46 | 0 | 229667087 | import bisect
import sys
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:
... | 65 | 1,263 | 131,379,200 | 205349623 | import bisect
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def make_graph(n, m):
x, s = [0] * (2 * m), [0] * (n + 3)
for i in range(m):
u, v = p[i + 2], i + 2
s[u + 2] += 1
x[2 * i], x[2 * i + 1] = u, v
for i in range(3, n + 3):
s[i] += s[i... | Codeforces Round 316 (Div. 2) | CF | 2,015 | 2 | 256 | Tree Requests | Roman planted a tree consisting of n vertices. Each vertex contains a lowercase English letter. Vertex 1 is the root of the tree, each of the n - 1 remaining vertices has a parent in the tree. Vertex is connected with its parent by an edge. The parent of vertex i is vertex pi, the parent index is always less than the i... | The first line contains two integers n, m (1 ≤ n, m ≤ 500 000) — the number of nodes in the tree and queries, respectively.
The following line contains n - 1 integers p2, p3, ..., pn — the parents of vertices from the second to the n-th (1 ≤ pi < i).
The next line contains n lowercase English letters, the i-th of the... | Print m lines. In the i-th line print "Yes" (without the quotes), if in the i-th query you can make a palindrome from the letters written on the vertices, otherwise print "No" (without the quotes). | null | String s is a palindrome if reads the same from left to right and from right to left. In particular, an empty string is a palindrome.
Clarification for the sample test.
In the first query there exists only a vertex 1 satisfying all the conditions, we can form a palindrome "z".
In the second query vertices 5 and 6 sa... | [{"input": "6 5\n1 1 1 3 3\nzacccd\n1 1\n3 3\n4 1\n6 1\n1 2", "output": "Yes\nNo\nYes\nYes\nYes"}] | 2,200 | ["binary search", "bitmasks", "constructive algorithms", "dfs and similar", "graphs", "trees"] | 65 | [{"input": "6 5\r\n1 1 1 3 3\r\nzacccd\r\n1 1\r\n3 3\r\n4 1\r\n6 1\r\n1 2\r\n", "output": "Yes\r\nNo\r\nYes\r\nYes\r\nYes\r\n"}, {"input": "5 6\r\n1 1 2 3\r\ncbcab\r\n3 1\r\n5 2\r\n1 3\r\n4 1\r\n4 2\r\n1 1\r\n", "output": "Yes\r\nYes\r\nNo\r\nYes\r\nYes\r\nYes\r\n"}, {"input": "5 6\r\n1 2 2 1\r\nbaabb\r\n1 1\r\n1 2\r\n... | false | stdio | null | true |
351/E | 351 | E | Python 3 | TESTS | 2 | 92 | 0 | 12311511 | n = int(input())
inp = input()
seq = inp.split(' ')
seq = [ abs(int(x)) for x in seq ]
Max = max(seq)
nxt = [0] * n
cnt = [0] * n
pos = [n] * (Max+1)
for i in range(n-1, -1, -1):
nxt[i] = pos[seq[i]]
pos[seq[i]] = i
for i in range(0, Max+1):
j = pos[i]
while(j<n):
front = sum(cnt[0:j])
b... | 36 | 280 | 3,686,400 | 150804902 | import sys
input = sys.stdin.buffer.readline
def process(A):
n = len(A)
S = [1 for i in range(n)]
d = {}
for i in range(n):
ai = abs(A[i])
if ai not in d:
d[ai] = []
d[ai].append(i)
L = sorted(d)
answer = 0
while len(L) > 0:
ai = L.pop()
f... | Codeforces Round 204 (Div. 1) | CF | 2,013 | 2 | 256 | Jeff and Permutation | Jeff's friends know full well that the boy likes to get sequences and arrays for his birthday. Thus, Jeff got sequence p1, p2, ..., pn for his birthday.
Jeff hates inversions in sequences. An inversion in sequence a1, a2, ..., an is a pair of indexes i, j (1 ≤ i < j ≤ n), such that an inequality ai > aj holds.
Jeff c... | The first line contains integer n (1 ≤ n ≤ 2000). The next line contains n integers — sequence p1, p2, ..., pn (|pi| ≤ 105). The numbers are separated by spaces. | In a single line print the answer to the problem — the minimum number of inversions Jeff can get. | null | null | [{"input": "2\n2 1", "output": "0"}, {"input": "9\n-2 0 -1 0 -1 2 1 0 -1", "output": "6"}] | 2,200 | ["greedy"] | 36 | [{"input": "2\r\n2 1\r\n", "output": "0\r\n"}, {"input": "9\r\n-2 0 -1 0 -1 2 1 0 -1\r\n", "output": "6\r\n"}, {"input": "9\r\n0 0 1 1 0 0 1 0 1\r\n", "output": "5\r\n"}, {"input": "8\r\n0 1 2 -1 -2 1 -2 2\r\n", "output": "3\r\n"}, {"input": "24\r\n-1 -1 2 2 0 -2 2 -1 0 0 2 -2 3 0 2 -3 0 -3 -1 1 0 0 -1 -2\r\n", "output... | false | stdio | null | true |
908/C | 908 | C | PyPy 3 | TESTS | 6 | 93 | 2,150,400 | 159926809 | import math
n, r = map(int, input().split(" "))
xs = [int(x) for x in input().split(" ")]
disks = []
for x in xs:
highestIntersection = -1
dx = None
for disk in disks:
if disk["x"] - 2 * r <= x <= disk["x"] + 2 * r:
if disk["y"] > highestIntersection:
highestIntersection = disk["y"]
dx = abs(disk["x"... | 15 | 124 | 32,256,000 | 176608726 | n, r=[int(k) for k in input().split()]
w=[int(k) for k in input().split()]
w=w[::-1]
z=[(w.pop(), r)]
while w:
x=w.pop()
mx=r
for j in z:
if abs(x-j[0])<=2*r:
mx=max(mx, j[1]+(4*r**2-(x-j[0])**2)**0.5)
z.append((x, mx))
print(" ".join([str(k[1]) for k in z])) | Good Bye 2017 | CF | 2,017 | 2 | 256 | New Year and Curling | Carol is currently curling.
She has n disks each with radius r on the 2D plane.
Initially she has all these disks above the line y = 10100.
She then will slide the disks towards the line y = 0 one by one in order from 1 to n.
When she slides the i-th disk, she will place its center at the point (xi, 10100). She wil... | The first line will contain two integers n and r (1 ≤ n, r ≤ 1 000), the number of disks, and the radius of the disks, respectively.
The next line will contain n integers x1, x2, ..., xn (1 ≤ xi ≤ 1 000) — the x-coordinates of the disks. | Print a single line with n numbers. The i-th number denotes the y-coordinate of the center of the i-th disk. The output will be accepted if it has absolute or relative error at most 10 - 6.
Namely, let's assume that your answer for a particular value of a coordinate is a and the answer of the jury is b. The checker pr... | null | The final positions of the disks will look as follows:
In particular, note the position of the last disk. | [{"input": "6 2\n5 5 6 8 3 12", "output": "2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613"}] | 1,500 | ["brute force", "geometry", "implementation", "math"] | 15 | [{"input": "6 2\r\n5 5 6 8 3 12\r\n", "output": "2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613\r\n"}, {"input": "1 1\r\n5\r\n", "output": "1\r\n"}, {"input": "5 300\r\n939 465 129 611 532\r\n", "output": "300 667.864105343 1164.9596696 1522.27745533 2117.05388391\r\n"}, {"input": "5 1\r\n416 387 336 116... | false | stdio | import sys
def main(input_path, output_path, submission_path):
with open(input_path) as f:
n, r = map(int, f.readline().split())
with open(output_path) as f:
ref = list(map(float, f.read().strip().split()))
with open(submission_path) as f:
sub = list(map(float, f.read().st... | true |
908/C | 908 | C | Python 3 | TESTS | 6 | 108 | 6,246,400 | 34268593 | from sys import stdin
import operator
class Pair:
def __init__(self, x, y):
self.x = x
self.y = y
def __str__(self):
#return "(x:" + str(self.x) + ", y:" + str(self.y) + ")"
return str(self.y)
def calculateDistance():
global n, r, positions
distance = 0
for i in ra... | 15 | 140 | 2,867,200 | 104160893 | import math as m
nDiscs, r = [int(x) for x in input().split()]
x = [int(x) for x in input().split()]
y = []
for i in range(len(x)):
tempY = [r]
for j in range(i):
diffX = abs(x[i] - x[j])
if diffX <= (2 * r):
addY = m.sqrt((4 * r * r) - (diffX * diffX))
tempY.append(y[... | Good Bye 2017 | CF | 2,017 | 2 | 256 | New Year and Curling | Carol is currently curling.
She has n disks each with radius r on the 2D plane.
Initially she has all these disks above the line y = 10100.
She then will slide the disks towards the line y = 0 one by one in order from 1 to n.
When she slides the i-th disk, she will place its center at the point (xi, 10100). She wil... | The first line will contain two integers n and r (1 ≤ n, r ≤ 1 000), the number of disks, and the radius of the disks, respectively.
The next line will contain n integers x1, x2, ..., xn (1 ≤ xi ≤ 1 000) — the x-coordinates of the disks. | Print a single line with n numbers. The i-th number denotes the y-coordinate of the center of the i-th disk. The output will be accepted if it has absolute or relative error at most 10 - 6.
Namely, let's assume that your answer for a particular value of a coordinate is a and the answer of the jury is b. The checker pr... | null | The final positions of the disks will look as follows:
In particular, note the position of the last disk. | [{"input": "6 2\n5 5 6 8 3 12", "output": "2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613"}] | 1,500 | ["brute force", "geometry", "implementation", "math"] | 15 | [{"input": "6 2\r\n5 5 6 8 3 12\r\n", "output": "2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613\r\n"}, {"input": "1 1\r\n5\r\n", "output": "1\r\n"}, {"input": "5 300\r\n939 465 129 611 532\r\n", "output": "300 667.864105343 1164.9596696 1522.27745533 2117.05388391\r\n"}, {"input": "5 1\r\n416 387 336 116... | false | stdio | import sys
def main(input_path, output_path, submission_path):
with open(input_path) as f:
n, r = map(int, f.readline().split())
with open(output_path) as f:
ref = list(map(float, f.read().strip().split()))
with open(submission_path) as f:
sub = list(map(float, f.read().st... | true |
351/E | 351 | E | Python 3 | TESTS | 2 | 92 | 0 | 12311615 | n = int(input())
inp = input()
seq = inp.split(' ')
seq = [ abs(int(x)) for x in seq ]
Max = max(seq)
nxt = [0] * n
cnt = [0] * n
pos = [n] * (Max+1)
for i in range(n-1, -1, -1):
nxt[i] = pos[seq[i]]
pos[seq[i]] = i
for i in range(0, Max+1):
j = pos[i]
while(j<n):
front = sum(cnt[0:j])
b... | 36 | 280 | 3,686,400 | 150804902 | import sys
input = sys.stdin.buffer.readline
def process(A):
n = len(A)
S = [1 for i in range(n)]
d = {}
for i in range(n):
ai = abs(A[i])
if ai not in d:
d[ai] = []
d[ai].append(i)
L = sorted(d)
answer = 0
while len(L) > 0:
ai = L.pop()
f... | Codeforces Round 204 (Div. 1) | CF | 2,013 | 2 | 256 | Jeff and Permutation | Jeff's friends know full well that the boy likes to get sequences and arrays for his birthday. Thus, Jeff got sequence p1, p2, ..., pn for his birthday.
Jeff hates inversions in sequences. An inversion in sequence a1, a2, ..., an is a pair of indexes i, j (1 ≤ i < j ≤ n), such that an inequality ai > aj holds.
Jeff c... | The first line contains integer n (1 ≤ n ≤ 2000). The next line contains n integers — sequence p1, p2, ..., pn (|pi| ≤ 105). The numbers are separated by spaces. | In a single line print the answer to the problem — the minimum number of inversions Jeff can get. | null | null | [{"input": "2\n2 1", "output": "0"}, {"input": "9\n-2 0 -1 0 -1 2 1 0 -1", "output": "6"}] | 2,200 | ["greedy"] | 36 | [{"input": "2\r\n2 1\r\n", "output": "0\r\n"}, {"input": "9\r\n-2 0 -1 0 -1 2 1 0 -1\r\n", "output": "6\r\n"}, {"input": "9\r\n0 0 1 1 0 0 1 0 1\r\n", "output": "5\r\n"}, {"input": "8\r\n0 1 2 -1 -2 1 -2 2\r\n", "output": "3\r\n"}, {"input": "24\r\n-1 -1 2 2 0 -2 2 -1 0 0 2 -2 3 0 2 -3 0 -3 -1 1 0 0 -1 -2\r\n", "output... | false | stdio | null | true |
995/B | 995 | B | PyPy 3 | TESTS | 3 | 124 | 0 | 81714246 | n = int(input())
a = list(map(int, input().split()))
ct = 0
i = 0
while i < n:
e = a[i]
j = i+1
# print(e, a[j:])
t = j + a[j:].index(e)
el = a.pop(t)
a = a[:j] + [el] + a[j:]
ct += t-j
i += 2
print(ct) | 22 | 46 | 0 | 228763045 | n = int(input())
xs = [int(x) for x in input().split()]
seen = {}
res = 0
while xs:
j = xs.index(xs[0], 1)
res += j - 1
xs = xs[1:j] + xs[j+1:]
print(res) | Codeforces Round 492 (Div. 1) [Thanks, uDebug!] | CF | 2,018 | 2 | 256 | Suit and Tie | Allen is hosting a formal dinner party. $$$2n$$$ people come to the event in $$$n$$$ pairs (couples). After a night of fun, Allen wants to line everyone up for a final picture. The $$$2n$$$ people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this m... | The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of pairs of people.
The second line contains $$$2n$$$ integers $$$a_1, a_2, \dots, a_{2n}$$$. For each $$$i$$$ with $$$1 \le i \le n$$$, $$$i$$$ appears exactly twice. If $$$a_j = a_k = i$$$, that means that the $$$j$$$-th and $$$k$$$... | Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions. | null | In the first sample case, we can transform $$$1 1 2 3 3 2 4 4 \rightarrow 1 1 2 3 2 3 4 4 \rightarrow 1 1 2 2 3 3 4 4$$$ in two steps. Note that the sequence $$$1 1 2 3 3 2 4 4 \rightarrow 1 1 3 2 3 2 4 4 \rightarrow 1 1 3 3 2 2 4 4$$$ also works in the same number of steps.
The second sample case already satisfies th... | [{"input": "4\n1 1 2 3 3 2 4 4", "output": "2"}, {"input": "3\n1 1 2 2 3 3", "output": "0"}, {"input": "3\n3 1 2 3 1 2", "output": "3"}] | 1,400 | ["greedy", "implementation", "math"] | 22 | [{"input": "4\r\n1 1 2 3 3 2 4 4\r\n", "output": "2\r\n"}, {"input": "3\r\n1 1 2 2 3 3\r\n", "output": "0\r\n"}, {"input": "3\r\n3 1 2 3 1 2\r\n", "output": "3\r\n"}, {"input": "8\r\n7 6 2 1 4 3 3 7 2 6 5 1 8 5 8 4\r\n", "output": "27\r\n"}, {"input": "2\r\n1 2 1 2\r\n", "output": "1\r\n"}, {"input": "3\r\n1 2 3 3 1 2\... | false | stdio | null | true |
995/B | 995 | B | PyPy 3 | TESTS | 3 | 124 | 0 | 61724747 | n, s = int(input()), 0
*a, = map(int, input().split())
for i in range(2 * n - 1, n - 1, -2):
s += i - a.index(a[i]) - 1
a.remove(a[i])
print(s) | 22 | 46 | 102,400 | 226139058 | def up_map(a):
mapp = dict()
for i in range(len(a)):
if a[i] not in mapp:
mapp[a[i]] = []
mapp[a[i]].append(i)
return mapp
def solve():
n = int(input())
a = list(map(int, input().split()))
res = 0
mapp = up_map(a)
visit = set()
for j in range(len(a)):
... | Codeforces Round 492 (Div. 1) [Thanks, uDebug!] | CF | 2,018 | 2 | 256 | Suit and Tie | Allen is hosting a formal dinner party. $$$2n$$$ people come to the event in $$$n$$$ pairs (couples). After a night of fun, Allen wants to line everyone up for a final picture. The $$$2n$$$ people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this m... | The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of pairs of people.
The second line contains $$$2n$$$ integers $$$a_1, a_2, \dots, a_{2n}$$$. For each $$$i$$$ with $$$1 \le i \le n$$$, $$$i$$$ appears exactly twice. If $$$a_j = a_k = i$$$, that means that the $$$j$$$-th and $$$k$$$... | Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions. | null | In the first sample case, we can transform $$$1 1 2 3 3 2 4 4 \rightarrow 1 1 2 3 2 3 4 4 \rightarrow 1 1 2 2 3 3 4 4$$$ in two steps. Note that the sequence $$$1 1 2 3 3 2 4 4 \rightarrow 1 1 3 2 3 2 4 4 \rightarrow 1 1 3 3 2 2 4 4$$$ also works in the same number of steps.
The second sample case already satisfies th... | [{"input": "4\n1 1 2 3 3 2 4 4", "output": "2"}, {"input": "3\n1 1 2 2 3 3", "output": "0"}, {"input": "3\n3 1 2 3 1 2", "output": "3"}] | 1,400 | ["greedy", "implementation", "math"] | 22 | [{"input": "4\r\n1 1 2 3 3 2 4 4\r\n", "output": "2\r\n"}, {"input": "3\r\n1 1 2 2 3 3\r\n", "output": "0\r\n"}, {"input": "3\r\n3 1 2 3 1 2\r\n", "output": "3\r\n"}, {"input": "8\r\n7 6 2 1 4 3 3 7 2 6 5 1 8 5 8 4\r\n", "output": "27\r\n"}, {"input": "2\r\n1 2 1 2\r\n", "output": "1\r\n"}, {"input": "3\r\n1 2 3 3 1 2\... | false | stdio | null | true |
995/B | 995 | B | PyPy 3 | TESTS | 3 | 140 | 307,200 | 103702777 | from collections import defaultdict
n=int(input())
a=[int(x) for x in input().split()]
z=defaultdict(list)
for i in range(2*n):
z[a[i]].append(i)
an=0
for i in range(n-1):
if a[i]!=a[i+1]:
for j in range(z[a[i]][1],i+1,-1):
a[j],a[j-1]=a[j-1],a[j]
an+=1
print(an) | 22 | 46 | 1,536,000 | 226136644 | n = int(input())
A = list(map(int, input().split()))
B = []
count = 0
while len(A) != 0:
first_elem = A[0]
i = 1
while i < len(A):
if A[i] == first_elem:
A.pop(i)
A.pop(0)
break
else:
count+=1
i+=1
print(count) | Codeforces Round 492 (Div. 1) [Thanks, uDebug!] | CF | 2,018 | 2 | 256 | Suit and Tie | Allen is hosting a formal dinner party. $$$2n$$$ people come to the event in $$$n$$$ pairs (couples). After a night of fun, Allen wants to line everyone up for a final picture. The $$$2n$$$ people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this m... | The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of pairs of people.
The second line contains $$$2n$$$ integers $$$a_1, a_2, \dots, a_{2n}$$$. For each $$$i$$$ with $$$1 \le i \le n$$$, $$$i$$$ appears exactly twice. If $$$a_j = a_k = i$$$, that means that the $$$j$$$-th and $$$k$$$... | Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions. | null | In the first sample case, we can transform $$$1 1 2 3 3 2 4 4 \rightarrow 1 1 2 3 2 3 4 4 \rightarrow 1 1 2 2 3 3 4 4$$$ in two steps. Note that the sequence $$$1 1 2 3 3 2 4 4 \rightarrow 1 1 3 2 3 2 4 4 \rightarrow 1 1 3 3 2 2 4 4$$$ also works in the same number of steps.
The second sample case already satisfies th... | [{"input": "4\n1 1 2 3 3 2 4 4", "output": "2"}, {"input": "3\n1 1 2 2 3 3", "output": "0"}, {"input": "3\n3 1 2 3 1 2", "output": "3"}] | 1,400 | ["greedy", "implementation", "math"] | 22 | [{"input": "4\r\n1 1 2 3 3 2 4 4\r\n", "output": "2\r\n"}, {"input": "3\r\n1 1 2 2 3 3\r\n", "output": "0\r\n"}, {"input": "3\r\n3 1 2 3 1 2\r\n", "output": "3\r\n"}, {"input": "8\r\n7 6 2 1 4 3 3 7 2 6 5 1 8 5 8 4\r\n", "output": "27\r\n"}, {"input": "2\r\n1 2 1 2\r\n", "output": "1\r\n"}, {"input": "3\r\n1 2 3 3 1 2\... | false | stdio | null | true |
883/K | 883 | K | Python 3 | TESTS | 0 | 46 | 0 | 138783166 | n = int(input())
a = []
for _ in range(n):
a.append(list(map(int,input().split())))
ls = [[0,0] for i in range(n)]
check = False
for i in range(1,n):
ls[i-1][0] = max(a[i][0]-1,a[i-1][0])
ls[i-1][1] = max(a[i][1]+1+a[i][0],ls[i-1][0])
ls[i-1][1] = min(ls[i-1][1],a[i-1][0]+a[i-1][1])
if a[i][0]+1+a[i... | 109 | 1,637 | 20,377,600 | 42341522 | n = int(input())
ss = [0] * (n + 1)
gg = [0] * (n + 1)
#mins = [0] * n
maxs = [0] * n
curMin = -10 ** 10
curMax = -curMin
for i in range(n):
s, g = map(int, input().split(' '))
ss[i] = s
gg[i] = g
curMin = max(curMin - 1, s)
curMax = min(curMax + 1, s + g)
if curMin > curMax:
print(-1... | 2017-2018 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred) | ICPC | 2,017 | 3 | 256 | Road Widening | Mayor of city S just hates trees and lawns. They take so much space and there could be a road on the place they occupy!
The Mayor thinks that one of the main city streets could be considerably widened on account of lawn nobody needs anyway. Moreover, that might help reduce the car jams which happen from time to time o... | The first line contains integer n (1 ≤ n ≤ 2·105) — number of parts of the street.
Each of the following n lines contains two integers si, gi (1 ≤ si ≤ 106, 0 ≤ gi ≤ 106) — current width of road and width of the lawn on the i-th part of the street. | In the first line print the total width of lawns which will be removed.
In the second line print n integers s'1, s'2, ..., s'n (si ≤ s'i ≤ si + gi) — new widths of the road starting from the first part and to the last.
If there is no solution, print the only integer -1 in the first line. | null | null | [{"input": "3\n4 5\n4 5\n4 10", "output": "16\n9 9 10"}, {"input": "4\n1 100\n100 1\n1 100\n100 1", "output": "202\n101 101 101 101"}, {"input": "3\n1 1\n100 100\n1 1", "output": "-1"}] | 1,800 | ["constructive algorithms", "greedy", "implementation"] | 109 | [{"input": "3\r\n4 5\r\n4 5\r\n4 10\r\n", "output": "16\r\n9 9 10 \r\n"}, {"input": "4\r\n1 100\r\n100 1\r\n1 100\r\n100 1\r\n", "output": "202\r\n101 101 101 101 \r\n"}, {"input": "3\r\n1 1\r\n100 100\r\n1 1\r\n", "output": "-1\r\n"}, {"input": "10\r\n21005 10850\r\n27020 13372\r\n28183 3724\r\n22874 13564\r\n27446 11... | false | stdio | null | true |
846/D | 846 | D | PyPy 3 | TESTS | 18 | 1,949 | 53,350,400 | 80432694 | n,m,k,q=list(map(int,input().split()))
low,high=float('inf'),-1
qry=list()
for _ in range(q):
a,b,c=list(map(int,input().split()))
low=min(low,c)
high=max(high,c)
qry.append([a,b,c])
arr=[[0]*(m) for _ in range(n)]
for i in range(0,q):
arr[qry[i][0]-1][qry[i][1]-1]=qry[i][2]
#print(arr)
def func... | 38 | 249 | 35,430,400 | 130408692 | from collections import deque
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def slide_max(n, k, a):
ans = []
s = deque()
for i in range(n):
ai = a[i]
while s:
if a[s[-1]] <= ai:
s.pop()
else:
break
... | Educational Codeforces Round 28 | ICPC | 2,017 | 2 | 256 | Monitor | Recently Luba bought a monitor. Monitor is a rectangular matrix of size n × m. But then she started to notice that some pixels cease to work properly. Luba thinks that the monitor will become broken the first moment when it contains a square k × k consisting entirely of broken pixels. She knows that q pixels are alread... | The first line contains four integer numbers n, m, k, q (1 ≤ n, m ≤ 500, 1 ≤ k ≤ min(n, m), 0 ≤ q ≤ n·m) — the length and width of the monitor, the size of a rectangle such that the monitor is broken if there is a broken rectangle with this size, and the number of broken pixels.
Each of next q lines contain three inte... | Print one number — the minimum moment the monitor became broken, or "-1" if it's still not broken after these q pixels stopped working. | null | null | [{"input": "2 3 2 5\n2 1 8\n2 2 8\n1 2 1\n1 3 4\n2 3 2", "output": "8"}, {"input": "3 3 2 5\n1 2 2\n2 2 1\n2 3 5\n3 2 10\n2 1 100", "output": "-1"}] | 1,900 | ["binary search", "data structures"] | 38 | [{"input": "2 3 2 5\r\n2 1 8\r\n2 2 8\r\n1 2 1\r\n1 3 4\r\n2 3 2\r\n", "output": "8\r\n"}, {"input": "3 3 2 5\r\n1 2 2\r\n2 2 1\r\n2 3 5\r\n3 2 10\r\n2 1 100\r\n", "output": "-1\r\n"}, {"input": "29 50 5 29\r\n21 42 1565821\r\n21 43 53275635\r\n21 44 2717830\r\n21 45 9579585\r\n21 46 20725775\r\n22 42 2568372\r\n22 43 ... | false | stdio | null | true |
370/B | 370 | B | PyPy 3 | TESTS | 1 | 140 | 20,172,800 | 84716592 | n = int(input())
cards = list()
for i in range(n):
cards.append(list())
inp = [int(j) for j in input().split()]
m = inp[0]
inp.pop(0)
inp.sort()
cards[i] = inp
c_have_equal_num = list()
for i in range(n):
c_have_equal_num.append(list())
for j in range(i + 1, n):
c_have_equal_num... | 24 | 77 | 819,200 | 5783895 | # -*- coding: utf-8 -*-
n = int(input())
cards = [[set(map(int, input().split()[1:])), i, True] for i in range(n)]
cards.sort(key=lambda x: len(x[0]))
for i in range(n-1, -1, -1):
if not cards[i][2]:
continue
for j in range(0, i):
if cards[j][0] == cards[i][0]:
cards[j][2] = cards[... | Codeforces Round 217 (Div. 2) | CF | 2,013 | 1 | 256 | Berland Bingo | Lately, a national version of a bingo game has become very popular in Berland. There are n players playing the game, each player has a card with numbers. The numbers on each card are distinct, but distinct cards can have equal numbers. The card of the i-th player contains mi numbers.
During the game the host takes num... | The first line of the input contains integer n (1 ≤ n ≤ 100) — the number of the players. Then follow n lines, each line describes a player's card. The line that describes a card starts from integer mi (1 ≤ mi ≤ 100) that shows how many numbers the i-th player's card has. Then follows a sequence of integers ai, 1, ai, ... | Print n lines, the i-th line must contain word "YES" (without the quotes), if the i-th player can win, and "NO" (without the quotes) otherwise. | null | null | [{"input": "3\n1 1\n3 2 4 1\n2 10 11", "output": "YES\nNO\nYES"}, {"input": "2\n1 1\n1 1", "output": "NO\nNO"}] | 1,300 | ["implementation"] | 24 | [{"input": "3\r\n1 1\r\n3 2 4 1\r\n2 10 11\r\n", "output": "YES\r\nNO\r\nYES\r\n"}, {"input": "2\r\n1 1\r\n1 1\r\n", "output": "NO\r\nNO\r\n"}, {"input": "1\r\n1 1\r\n", "output": "YES\r\n"}, {"input": "2\r\n1 2\r\n1 3\r\n", "output": "YES\r\nYES\r\n"}, {"input": "2\r\n1 1\r\n2 1 2\r\n", "output": "YES\r\nNO\r\n"}, {"i... | false | stdio | null | true |
904/C | 906 | A | Python 3 | TESTS | 9 | 389 | 6,758,400 | 33883038 | #!/usr/bin/python3
import string
n = int(input())
exists = set(string.ascii_lowercase)
ans = 0
for i in range(n):
action, string = input().split(' ')
newExists = set(string)
#print(newExists)
if (action == '.'):
exists = exists - newExists
if (action == '!'):
if (len(exists & newExi... | 38 | 186 | 5,939,200 | 33563841 | import sys
sys.setrecursionlimit(1000000)
read = sys.stdin.readline
q = int(read())
count = 0
psw = set(list("qwertyuiopasdfghjklzxcvbnm"))
unique = None
for _ in range(q):
s, w = read().strip().split()
if len(psw) != 1:
if s == '!':
_psw = set()
for _w in w:
if ... | Технокубок 2018 - Отборочный Раунд 4 | CF | 2,017 | 2 | 256 | Shockers | Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for eac... | The first line contains a single integer n (1 ≤ n ≤ 105) — the number of actions Valentin did.
The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types:
1. Valentin pronounced some word and didn't get an electric shock. This action is... | Output a single integer — the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. | null | In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces ... | [{"input": "5\n! abc\n. ad\n. b\n! cd\n? c", "output": "1"}, {"input": "8\n! hello\n! codeforces\n? c\n. o\n? d\n? h\n. l\n? e", "output": "2"}, {"input": "7\n! ababahalamaha\n? a\n? b\n? a\n? b\n? a\n? h", "output": "0"}] | 1,600 | ["strings"] | 38 | [{"input": "5\r\n! abc\r\n. ad\r\n. b\r\n! cd\r\n? c\r\n", "output": "1\r\n"}, {"input": "8\r\n! hello\r\n! codeforces\r\n? c\r\n. o\r\n? d\r\n? h\r\n. l\r\n? e\r\n", "output": "2\r\n"}, {"input": "7\r\n! ababahalamaha\r\n? a\r\n? b\r\n? a\r\n? b\r\n? a\r\n? h\r\n", "output": "0\r\n"}, {"input": "4\r\n! abcd\r\n! cdef\... | false | stdio | null | true |
29/D | 29 | D | PyPy 3 | TESTS | 4 | 310 | 512,000 | 62222356 | #TO MAKE THE PROGRAM FAST
''' ---------------------------------------------------------------------------------------------------- '''
import sys
from collections import *
input = sys.stdin.readline
sys.setrecursionlimit(100000)
''' -------------------------------------------------------------------------------------... | 35 | 186 | 2,150,400 | 162359731 | import sys
input = sys.stdin.readline
n=int(input())
E=[[] for i in range(n+1)]
for i in range(n-1):
x,y=map(int,input().split())
E[x].append(y)
E[y].append(x)
K=list(map(int,input().split()))
def dfs(fr,target):
Q=[fr]
USE=[-1]*(n+1)
USE[fr]=0
while Q:
x=Q.pop()
for t... | Codeforces Beta Round 29 (Div. 2, Codeforces format) | CF | 2,010 | 2 | 256 | Ant on the Tree | Connected undirected graph without cycles is called a tree. Trees is a class of graphs which is interesting not only for people, but for ants too.
An ant stands at the root of some tree. He sees that there are n vertexes in the tree, and they are connected by n - 1 edges so that there is a path between any pair of ver... | The first line contains integer n (3 ≤ n ≤ 300) — amount of vertexes in the tree. Next n - 1 lines describe edges. Each edge is described with two integers — indexes of vertexes which it connects. Each edge can be passed in any direction. Vertexes are numbered starting from 1. The root of the tree has number 1. The las... | If the required route doesn't exist, output -1. Otherwise, output 2n - 1 numbers, describing the route. Every time the ant comes to a vertex, output it's index. | null | null | [{"input": "3\n1 2\n2 3\n3", "output": "1 2 3 2 1"}, {"input": "6\n1 2\n1 3\n2 4\n4 5\n4 6\n5 6 3", "output": "1 2 4 5 4 6 4 2 1 3 1"}, {"input": "6\n1 2\n1 3\n2 4\n4 5\n4 6\n5 3 6", "output": "-1"}] | 2,000 | ["constructive algorithms", "dfs and similar", "trees"] | 35 | [{"input": "3\r\n1 2\r\n2 3\r\n3\r\n", "output": "1 2 3 2 1 "}, {"input": "6\r\n1 2\r\n1 3\r\n2 4\r\n4 5\r\n4 6\r\n5 6 3\r\n", "output": "1 2 4 5 4 6 4 2 1 3 1 "}, {"input": "6\r\n1 2\r\n1 3\r\n2 4\r\n4 5\r\n4 6\r\n5 3 6\r\n", "output": "-1\r\n"}, {"input": "10\r\n8 10\r\n2 1\r\n7 5\r\n5 4\r\n6 10\r\n2 3\r\n3 10\r\n2 9... | false | stdio | null | true |
29/D | 29 | D | Python 3 | TESTS | 4 | 154 | 7,168,000 | 129459289 | from collections import defaultdict
n = int(input())
visit = [False]*301
adj = defaultdict(list)
path = list()
def dfs(v):
global visit,path
visit[v]=True
path.append(v)
for u in adj[v]:
if visit[u]==False:
dfs(u)
path.append(v)
for i in range(n-1):
a,b = input()... | 35 | 186 | 7,372,800 | 36637911 | class Node:
def __init__(self, data=None):
self.data = data
self.connections = []
self.path = []
class Tree:
def __init__(self):
self.root = None
self.nodes = {}
def addConnections(self, p, c):
if p not in self.nodes:
parent = Node(p)
... | Codeforces Beta Round 29 (Div. 2, Codeforces format) | CF | 2,010 | 2 | 256 | Ant on the Tree | Connected undirected graph without cycles is called a tree. Trees is a class of graphs which is interesting not only for people, but for ants too.
An ant stands at the root of some tree. He sees that there are n vertexes in the tree, and they are connected by n - 1 edges so that there is a path between any pair of ver... | The first line contains integer n (3 ≤ n ≤ 300) — amount of vertexes in the tree. Next n - 1 lines describe edges. Each edge is described with two integers — indexes of vertexes which it connects. Each edge can be passed in any direction. Vertexes are numbered starting from 1. The root of the tree has number 1. The las... | If the required route doesn't exist, output -1. Otherwise, output 2n - 1 numbers, describing the route. Every time the ant comes to a vertex, output it's index. | null | null | [{"input": "3\n1 2\n2 3\n3", "output": "1 2 3 2 1"}, {"input": "6\n1 2\n1 3\n2 4\n4 5\n4 6\n5 6 3", "output": "1 2 4 5 4 6 4 2 1 3 1"}, {"input": "6\n1 2\n1 3\n2 4\n4 5\n4 6\n5 3 6", "output": "-1"}] | 2,000 | ["constructive algorithms", "dfs and similar", "trees"] | 35 | [{"input": "3\r\n1 2\r\n2 3\r\n3\r\n", "output": "1 2 3 2 1 "}, {"input": "6\r\n1 2\r\n1 3\r\n2 4\r\n4 5\r\n4 6\r\n5 6 3\r\n", "output": "1 2 4 5 4 6 4 2 1 3 1 "}, {"input": "6\r\n1 2\r\n1 3\r\n2 4\r\n4 5\r\n4 6\r\n5 3 6\r\n", "output": "-1\r\n"}, {"input": "10\r\n8 10\r\n2 1\r\n7 5\r\n5 4\r\n6 10\r\n2 3\r\n3 10\r\n2 9... | false | stdio | null | true |
29/D | 29 | D | Python 3 | TESTS | 4 | 154 | 7,065,600 | 37731323 | # from dust i have come, dust i will be
class graph(object):
def __init__(self,n):
self.n=n
self.adj=[list() for i in range(n+1)]
self.parent=[0]*(n+1)
self.vis=[0]*(n+1)
def insert(self,u,v):
self.adj[u].append(v)
def dfs(self,s,pr):
self.vis[s]=1
... | 35 | 218 | 1,228,800 | 139441302 | def solve():
n = int(input())
g = [[] for _ in range(n + 1)]
for i in range(n - 1):
a, b = map(int, input().split(' '))
g[a].append(b)
g[b].append(a)
lfs = list(map(int, input().split(' ')))
k = len(lfs)
par = [0 for _ in range(n + 1)]
def gp(x, p):
par[x] = p
for y in g[x]:
if y != p:
gp(y, x... | Codeforces Beta Round 29 (Div. 2, Codeforces format) | CF | 2,010 | 2 | 256 | Ant on the Tree | Connected undirected graph without cycles is called a tree. Trees is a class of graphs which is interesting not only for people, but for ants too.
An ant stands at the root of some tree. He sees that there are n vertexes in the tree, and they are connected by n - 1 edges so that there is a path between any pair of ver... | The first line contains integer n (3 ≤ n ≤ 300) — amount of vertexes in the tree. Next n - 1 lines describe edges. Each edge is described with two integers — indexes of vertexes which it connects. Each edge can be passed in any direction. Vertexes are numbered starting from 1. The root of the tree has number 1. The las... | If the required route doesn't exist, output -1. Otherwise, output 2n - 1 numbers, describing the route. Every time the ant comes to a vertex, output it's index. | null | null | [{"input": "3\n1 2\n2 3\n3", "output": "1 2 3 2 1"}, {"input": "6\n1 2\n1 3\n2 4\n4 5\n4 6\n5 6 3", "output": "1 2 4 5 4 6 4 2 1 3 1"}, {"input": "6\n1 2\n1 3\n2 4\n4 5\n4 6\n5 3 6", "output": "-1"}] | 2,000 | ["constructive algorithms", "dfs and similar", "trees"] | 35 | [{"input": "3\r\n1 2\r\n2 3\r\n3\r\n", "output": "1 2 3 2 1 "}, {"input": "6\r\n1 2\r\n1 3\r\n2 4\r\n4 5\r\n4 6\r\n5 6 3\r\n", "output": "1 2 4 5 4 6 4 2 1 3 1 "}, {"input": "6\r\n1 2\r\n1 3\r\n2 4\r\n4 5\r\n4 6\r\n5 3 6\r\n", "output": "-1\r\n"}, {"input": "10\r\n8 10\r\n2 1\r\n7 5\r\n5 4\r\n6 10\r\n2 3\r\n3 10\r\n2 9... | false | stdio | null | true |
29/D | 29 | D | PyPy 3-64 | TESTS | 4 | 124 | 1,126,400 | 143842124 | class Operation:
def __init__(self, name, function, function_on_equal, neutral_value=0):
self.name = name
self.f = function
self.f_on_equal = function_on_equal
def add_multiple(x, count):
return x * count
def min_multiple(x, count):
return x
def max_multiple(x, count):
retu... | 35 | 218 | 2,662,400 | 145991047 | from collections import deque
import sys
input = sys.stdin.readline
def bfs(s):
q = deque()
q.append(s)
visit = [0] * (n + 1)
visit[s] = 1
parent = [-1] * (n + 1)
while q:
i = q.popleft()
for j in G[i]:
if not visit[j]:
q.append(j)
vis... | Codeforces Beta Round 29 (Div. 2, Codeforces format) | CF | 2,010 | 2 | 256 | Ant on the Tree | Connected undirected graph without cycles is called a tree. Trees is a class of graphs which is interesting not only for people, but for ants too.
An ant stands at the root of some tree. He sees that there are n vertexes in the tree, and they are connected by n - 1 edges so that there is a path between any pair of ver... | The first line contains integer n (3 ≤ n ≤ 300) — amount of vertexes in the tree. Next n - 1 lines describe edges. Each edge is described with two integers — indexes of vertexes which it connects. Each edge can be passed in any direction. Vertexes are numbered starting from 1. The root of the tree has number 1. The las... | If the required route doesn't exist, output -1. Otherwise, output 2n - 1 numbers, describing the route. Every time the ant comes to a vertex, output it's index. | null | null | [{"input": "3\n1 2\n2 3\n3", "output": "1 2 3 2 1"}, {"input": "6\n1 2\n1 3\n2 4\n4 5\n4 6\n5 6 3", "output": "1 2 4 5 4 6 4 2 1 3 1"}, {"input": "6\n1 2\n1 3\n2 4\n4 5\n4 6\n5 3 6", "output": "-1"}] | 2,000 | ["constructive algorithms", "dfs and similar", "trees"] | 35 | [{"input": "3\r\n1 2\r\n2 3\r\n3\r\n", "output": "1 2 3 2 1 "}, {"input": "6\r\n1 2\r\n1 3\r\n2 4\r\n4 5\r\n4 6\r\n5 6 3\r\n", "output": "1 2 4 5 4 6 4 2 1 3 1 "}, {"input": "6\r\n1 2\r\n1 3\r\n2 4\r\n4 5\r\n4 6\r\n5 3 6\r\n", "output": "-1\r\n"}, {"input": "10\r\n8 10\r\n2 1\r\n7 5\r\n5 4\r\n6 10\r\n2 3\r\n3 10\r\n2 9... | false | stdio | null | true |
667/B | 667 | B | PyPy 3 | PRETESTS | 2 | 93 | 23,142,400 | 17571079 | n = int(input())
s = [int(i) for i in input().split()]
a = max(s)
b = min(s)
c = sum(s)
print(a-(c-a-b)) | 51 | 62 | 3,174,400 | 153846872 | input()
s = list(map(int, input().split()))
m = max(s)
print(max(0, 2*m+1 - sum(s))) | Codeforces Round 349 (Div. 2) | CF | 2,016 | 1 | 256 | Coat of Anticubism | As some of you know, cubism is a trend in art, where the problem of constructing volumetrical shape on a plane with a combination of three-dimensional geometric shapes comes to the fore.
A famous sculptor Cicasso, whose self-portrait you can contemplate, hates cubism. He is more impressed by the idea to transmit two-d... | The first line contains an integer n (3 ≤ n ≤ 105) — a number of rod-blanks.
The second line contains n integers li (1 ≤ li ≤ 109) — lengths of rods, which Cicasso already has. It is guaranteed that it is impossible to make a polygon with n vertices and nonzero area using the rods Cicasso already has. | Print the only integer z — the minimum length of the rod, so that after adding it it can be possible to construct convex polygon with (n + 1) vertices and nonzero area from all of the rods. | null | In the first example triangle with sides {1 + 1 = 2, 2, 1} can be formed from a set of lengths {1, 1, 1, 2}.
In the second example you can make a triangle with lengths {20, 11, 4 + 3 + 2 + 1 = 10}. | [{"input": "3\n1 2 1", "output": "1"}, {"input": "5\n20 4 3 2 1", "output": "11"}] | 1,100 | ["constructive algorithms", "geometry"] | 51 | [{"input": "3\r\n1 2 1\r\n", "output": "1\r\n"}, {"input": "5\r\n20 4 3 2 1\r\n", "output": "11\r\n"}, {"input": "7\r\n77486105 317474713 89523018 332007362 7897847 949616701 54820086\r\n", "output": "70407571\r\n"}, {"input": "14\r\n245638694 2941428 4673577 12468 991349408 44735727 14046308 60637707 81525 104620306 8... | false | stdio | null | true |
814/D | 814 | D | PyPy 3 | TESTS | 3 | 499 | 10,342,400 | 90324733 | import sys
from math import pi, sqrt
from decimal import *
from random import randint
input = sys.stdin.readline
N = int(input())
a = [tuple(map(int, input().split())) for _ in range(N)]
e = [[0] * N for _ in range(N)]
a.sort(key = lambda x: -x[2])
for i in range(N):
x, y, r = a[i]
for j in range(i + 1, N):
u, ... | 32 | 405 | 24,780,800 | 28197754 | import math
class circ:
def __init__(self, x, y, r):
self.x = x*1.0
self.y = y*1.0
self.r = r*1.0
n = 0
n = int(input())
vec = []
for i in range(n):
st = input().split(' ')
a = int(st[0])
b = int(st[1])
c = int(st[2])
vec.append(circ(a,b,c))
gr = [[] for i in range(n)]
pad = [-1 for i in range(n)]
vis = ... | Codeforces Round 418 (Div. 2) | CF | 2,017 | 2 | 256 | An overnight dance in discotheque | The crowdedness of the discotheque would never stop our friends from having fun, but a bit more spaciousness won't hurt, will it?
The discotheque can be seen as an infinite xy-plane, in which there are a total of n dancers. Once someone starts moving around, they will move only inside their own movement range, which i... | The first line of input contains a positive integer n (1 ≤ n ≤ 1 000) — the number of dancers.
The following n lines each describes a dancer: the i-th line among them contains three space-separated integers xi, yi and ri ( - 106 ≤ xi, yi ≤ 106, 1 ≤ ri ≤ 106), describing a circular movement range centered at (xi, yi) w... | Output one decimal number — the largest achievable sum of spaciousness over two halves of the night.
The output is considered correct if it has a relative or absolute error of at most 10 - 9. Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if $$\frac{|a-b|}{\max(1, |b|)} \... | null | The first sample corresponds to the illustrations in the legend. | [{"input": "5\n2 1 6\n0 4 1\n2 -1 3\n1 -2 1\n4 -1 1", "output": "138.23007676"}, {"input": "8\n0 0 1\n0 0 2\n0 0 3\n0 0 4\n0 0 5\n0 0 6\n0 0 7\n0 0 8", "output": "289.02652413"}] | 2,000 | ["dfs and similar", "dp", "geometry", "greedy", "trees"] | 32 | [{"input": "5\r\n2 1 6\r\n0 4 1\r\n2 -1 3\r\n1 -2 1\r\n4 -1 1\r\n", "output": "138.23007676\r\n"}, {"input": "8\r\n0 0 1\r\n0 0 2\r\n0 0 3\r\n0 0 4\r\n0 0 5\r\n0 0 6\r\n0 0 7\r\n0 0 8\r\n", "output": "289.02652413\r\n"}, {"input": "4\r\n1000000 -1000000 2\r\n1000000 -1000000 3\r\n-1000000 1000000 2\r\n-1000000 1000000 ... | false | stdio | import sys
def main():
input_path = sys.argv[1]
correct_output_path = sys.argv[2]
submission_output_path = sys.argv[3]
# Read correct output
with open(correct_output_path, 'r') as f:
correct_line = f.readline().strip()
try:
b = float(correct_line)
except:
... | true |
159/A | 159 | A | PyPy 3 | TESTS | 6 | 280 | 0 | 76053331 | n,d=map(int,input().split())
sen=[]
re=[]
ti=[]
for i in range(n):
ss=input()
s=ss.split()
sen.append(s[0])
re.append(s[1])
ti.append(s[2])
m=[]
count=0
for i in range(len(sen)):
if sen[i] in re:
b=re.index(sen[i])
if re[i]==sen[int(b)]:
if 0 < int(ti[i]) - int(ti[re... | 30 | 374 | 204,800 | 160192565 | def closest_value(input_list, input_value):
Min = 100000
for num in input_list:
dif = abs(num - input_value)
if (dif == 0):
continue
Min = min(Min, dif)
return Min
n, d = input().split()
n = int(n)
d = int(d)
D = {}
F = {()}
for i in range(n):
a, b, t = input(... | VK Cup 2012 Qualification Round 2 | CF | 2,012 | 3 | 256 | Friends or Not | Polycarpus has a hobby — he develops an unusual social network. His work is almost completed, and there is only one more module to implement — the module which determines friends. Oh yes, in this social network one won't have to add friends manually! Pairs of friends are deduced in the following way. Let's assume that ... | The first line of the input contains two integers n and d (1 ≤ n, d ≤ 1000). The next n lines contain the messages log. The i-th line contains one line of the log formatted as "Ai Bi ti" (without the quotes), which means that user Ai sent a message to user Bi at time ti (1 ≤ i ≤ n). Ai and Bi are non-empty strings at m... | In the first line print integer k — the number of pairs of friends. In the next k lines print pairs of friends as "Ai Bi" (without the quotes). You can print users in pairs and the pairs themselves in any order. Each pair must be printed exactly once. | null | In the first sample test case Vasya and Petya are friends because their messages' sending times are one second apart. Anya and Ivan are not, because their messages' sending times differ by more than one second. | [{"input": "4 1\nvasya petya 1\npetya vasya 2\nanya ivan 2\nivan anya 4", "output": "1\npetya vasya"}, {"input": "1 1000\na b 0", "output": "0"}] | 1,400 | ["*special", "greedy", "implementation"] | 30 | [{"input": "4 1\r\nvasya petya 1\r\npetya vasya 2\r\nanya ivan 2\r\nivan anya 4\r\n", "output": "1\r\npetya vasya\r\n"}, {"input": "1 1000\r\na b 0\r\n", "output": "0\r\n"}, {"input": "2 1\r\na b 0\r\nb a 0\r\n", "output": "0\r\n"}, {"input": "3 1\r\na b 1\r\nb c 2\r\nc d 3\r\n", "output": "0\r\n"}, {"input": "10 2\r\n... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path, 'r') as f:
lines = [line.strip() for line in f]
n, d = map(int, lines[0].split())
messages = []
for line in lines[1:n+1]:
parts = line.split()
... | true |
159/A | 159 | A | Python 3 | TESTS | 10 | 122 | 6,963,200 | 115946275 | n,d=map(int,input().split())
a,b={},set()
for i in range(n):
q,w,t=input().split()
t=int(t)
if (w,q)in a:
if 0<abs(a[(w,q)]-t)<=d:
b.add((q,w))
else:a[(q,w)]=t
print(len(b))
for i in b:print(*[*i]) | 30 | 560 | 0 | 155315914 | n,d=map(int,input().split())
lst=[]
lst1=[]
lst2=[]
for i in range(n):
lst+=[list(map(str,input().split()))]
if n==1:
print(0)
else:
for j in range(n):
for k in range(n):
if lst[j][0]==lst[k][1] and lst[j][1]==lst[k][0]\
and 0<(int(lst[j][2])-int(lst[k][2]))\
... | VK Cup 2012 Qualification Round 2 | CF | 2,012 | 3 | 256 | Friends or Not | Polycarpus has a hobby — he develops an unusual social network. His work is almost completed, and there is only one more module to implement — the module which determines friends. Oh yes, in this social network one won't have to add friends manually! Pairs of friends are deduced in the following way. Let's assume that ... | The first line of the input contains two integers n and d (1 ≤ n, d ≤ 1000). The next n lines contain the messages log. The i-th line contains one line of the log formatted as "Ai Bi ti" (without the quotes), which means that user Ai sent a message to user Bi at time ti (1 ≤ i ≤ n). Ai and Bi are non-empty strings at m... | In the first line print integer k — the number of pairs of friends. In the next k lines print pairs of friends as "Ai Bi" (without the quotes). You can print users in pairs and the pairs themselves in any order. Each pair must be printed exactly once. | null | In the first sample test case Vasya and Petya are friends because their messages' sending times are one second apart. Anya and Ivan are not, because their messages' sending times differ by more than one second. | [{"input": "4 1\nvasya petya 1\npetya vasya 2\nanya ivan 2\nivan anya 4", "output": "1\npetya vasya"}, {"input": "1 1000\na b 0", "output": "0"}] | 1,400 | ["*special", "greedy", "implementation"] | 30 | [{"input": "4 1\r\nvasya petya 1\r\npetya vasya 2\r\nanya ivan 2\r\nivan anya 4\r\n", "output": "1\r\npetya vasya\r\n"}, {"input": "1 1000\r\na b 0\r\n", "output": "0\r\n"}, {"input": "2 1\r\na b 0\r\nb a 0\r\n", "output": "0\r\n"}, {"input": "3 1\r\na b 1\r\nb c 2\r\nc d 3\r\n", "output": "0\r\n"}, {"input": "10 2\r\n... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path, 'r') as f:
lines = [line.strip() for line in f]
n, d = map(int, lines[0].split())
messages = []
for line in lines[1:n+1]:
parts = line.split()
... | true |
159/A | 159 | A | PyPy 3-64 | TESTS | 13 | 124 | 102,400 | 167927881 | import sys
input = sys.stdin.readline
from collections import defaultdict
n, q = map(int, input().split())
d = defaultdict(int)
s = set()
for i in range(n):
a, b, c = input()[:-1].split()
c = int(c)
d[(a, b)] = c
if (b,a) in d:
if 0 < c - d[(b,a)] <= q:
a, b = sorted([a, b])
... | 30 | 624 | 6,963,200 | 123020256 | n,d=map(int,input().split())
ans,hist=[],[]
for i in range(n):
s=input().split()
t=int(s[2])
for x in hist:
if 0<t-x[2]<=d and s[0]==x[1] and s[1]==x[0]:
a=sorted([x[0],x[1]])
if not a in ans: ans.append(a)
hist.append([s[0],s[1],t])
print(len(ans))
for x in ans: print(x[... | VK Cup 2012 Qualification Round 2 | CF | 2,012 | 3 | 256 | Friends or Not | Polycarpus has a hobby — he develops an unusual social network. His work is almost completed, and there is only one more module to implement — the module which determines friends. Oh yes, in this social network one won't have to add friends manually! Pairs of friends are deduced in the following way. Let's assume that ... | The first line of the input contains two integers n and d (1 ≤ n, d ≤ 1000). The next n lines contain the messages log. The i-th line contains one line of the log formatted as "Ai Bi ti" (without the quotes), which means that user Ai sent a message to user Bi at time ti (1 ≤ i ≤ n). Ai and Bi are non-empty strings at m... | In the first line print integer k — the number of pairs of friends. In the next k lines print pairs of friends as "Ai Bi" (without the quotes). You can print users in pairs and the pairs themselves in any order. Each pair must be printed exactly once. | null | In the first sample test case Vasya and Petya are friends because their messages' sending times are one second apart. Anya and Ivan are not, because their messages' sending times differ by more than one second. | [{"input": "4 1\nvasya petya 1\npetya vasya 2\nanya ivan 2\nivan anya 4", "output": "1\npetya vasya"}, {"input": "1 1000\na b 0", "output": "0"}] | 1,400 | ["*special", "greedy", "implementation"] | 30 | [{"input": "4 1\r\nvasya petya 1\r\npetya vasya 2\r\nanya ivan 2\r\nivan anya 4\r\n", "output": "1\r\npetya vasya\r\n"}, {"input": "1 1000\r\na b 0\r\n", "output": "0\r\n"}, {"input": "2 1\r\na b 0\r\nb a 0\r\n", "output": "0\r\n"}, {"input": "3 1\r\na b 1\r\nb c 2\r\nc d 3\r\n", "output": "0\r\n"}, {"input": "10 2\r\n... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path, 'r') as f:
lines = [line.strip() for line in f]
n, d = map(int, lines[0].split())
messages = []
for line in lines[1:n+1]:
parts = line.split()
... | true |
667/B | 667 | B | Python 3 | TESTS | 2 | 109 | 0 | 75809985 | #----Kuzlyaev-Nikita-Codeforces-----
#------------08.04.2020-------------
import math
alph="abcdefghijklmnopqrstuvwxyz"
#-----------------------------------
n=int(input())
l=list(map(int,input().split()))
l.sort();r=sum(l)-l[0]-l[-1]
print(l[-1]-r) | 51 | 62 | 3,174,400 | 186434377 | n = int(input())
s = list(map(int,input().split()))
a = max(s)
b = sum(s)
c = max(s)
print(a-(b-c)+1) | Codeforces Round 349 (Div. 2) | CF | 2,016 | 1 | 256 | Coat of Anticubism | As some of you know, cubism is a trend in art, where the problem of constructing volumetrical shape on a plane with a combination of three-dimensional geometric shapes comes to the fore.
A famous sculptor Cicasso, whose self-portrait you can contemplate, hates cubism. He is more impressed by the idea to transmit two-d... | The first line contains an integer n (3 ≤ n ≤ 105) — a number of rod-blanks.
The second line contains n integers li (1 ≤ li ≤ 109) — lengths of rods, which Cicasso already has. It is guaranteed that it is impossible to make a polygon with n vertices and nonzero area using the rods Cicasso already has. | Print the only integer z — the minimum length of the rod, so that after adding it it can be possible to construct convex polygon with (n + 1) vertices and nonzero area from all of the rods. | null | In the first example triangle with sides {1 + 1 = 2, 2, 1} can be formed from a set of lengths {1, 1, 1, 2}.
In the second example you can make a triangle with lengths {20, 11, 4 + 3 + 2 + 1 = 10}. | [{"input": "3\n1 2 1", "output": "1"}, {"input": "5\n20 4 3 2 1", "output": "11"}] | 1,100 | ["constructive algorithms", "geometry"] | 51 | [{"input": "3\r\n1 2 1\r\n", "output": "1\r\n"}, {"input": "5\r\n20 4 3 2 1\r\n", "output": "11\r\n"}, {"input": "7\r\n77486105 317474713 89523018 332007362 7897847 949616701 54820086\r\n", "output": "70407571\r\n"}, {"input": "14\r\n245638694 2941428 4673577 12468 991349408 44735727 14046308 60637707 81525 104620306 8... | false | stdio | null | true |
814/D | 814 | D | Python 3 | TESTS | 3 | 108 | 0 | 45182548 | import math
def dist(p, q):
x1, y1 = p
x2, y2 = q
return (x1 - x2) ** 2 + (y1 - y2) ** 2
def main():
n = int(input())
circles = []
for _ in range(n):
x, y, r = map(int, input().split())
circles.append(((x, y), r))
circles = sorted(circles, key=lambda circle: circle[1], reverse=True)
parent = [... | 32 | 499 | 2,764,800 | 188608070 | import math
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def bfs(s):
q, k = [s], 0
dist[s] = 0
while len(q) ^ k:
i = q[k]
di = dist[i]
for j in G[i]:
if dist[j] == inf:
dist[j] = di + 1
q.append(j)
... | Codeforces Round 418 (Div. 2) | CF | 2,017 | 2 | 256 | An overnight dance in discotheque | The crowdedness of the discotheque would never stop our friends from having fun, but a bit more spaciousness won't hurt, will it?
The discotheque can be seen as an infinite xy-plane, in which there are a total of n dancers. Once someone starts moving around, they will move only inside their own movement range, which i... | The first line of input contains a positive integer n (1 ≤ n ≤ 1 000) — the number of dancers.
The following n lines each describes a dancer: the i-th line among them contains three space-separated integers xi, yi and ri ( - 106 ≤ xi, yi ≤ 106, 1 ≤ ri ≤ 106), describing a circular movement range centered at (xi, yi) w... | Output one decimal number — the largest achievable sum of spaciousness over two halves of the night.
The output is considered correct if it has a relative or absolute error of at most 10 - 9. Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if $$\frac{|a-b|}{\max(1, |b|)} \... | null | The first sample corresponds to the illustrations in the legend. | [{"input": "5\n2 1 6\n0 4 1\n2 -1 3\n1 -2 1\n4 -1 1", "output": "138.23007676"}, {"input": "8\n0 0 1\n0 0 2\n0 0 3\n0 0 4\n0 0 5\n0 0 6\n0 0 7\n0 0 8", "output": "289.02652413"}] | 2,000 | ["dfs and similar", "dp", "geometry", "greedy", "trees"] | 32 | [{"input": "5\r\n2 1 6\r\n0 4 1\r\n2 -1 3\r\n1 -2 1\r\n4 -1 1\r\n", "output": "138.23007676\r\n"}, {"input": "8\r\n0 0 1\r\n0 0 2\r\n0 0 3\r\n0 0 4\r\n0 0 5\r\n0 0 6\r\n0 0 7\r\n0 0 8\r\n", "output": "289.02652413\r\n"}, {"input": "4\r\n1000000 -1000000 2\r\n1000000 -1000000 3\r\n-1000000 1000000 2\r\n-1000000 1000000 ... | false | stdio | import sys
def main():
input_path = sys.argv[1]
correct_output_path = sys.argv[2]
submission_output_path = sys.argv[3]
# Read correct output
with open(correct_output_path, 'r') as f:
correct_line = f.readline().strip()
try:
b = float(correct_line)
except:
... | true |
814/D | 814 | D | PyPy 3 | TESTS | 3 | 171 | 0 | 90318952 | import sys
from math import pi, sqrt
input = sys.stdin.readline
N = int(input())
a = [tuple(map(int, input().split())) for _ in range(N)]
e = [[0] * N for _ in range(N)]
a.sort(key = lambda x: -x[2])
for i in range(N):
x, y, r = a[i]
for j in range(i + 1, N):
u, v, rr = a[j]
d = sqrt((x - u) ** 2 + (y - v) ... | 32 | 561 | 8,601,600 | 197617795 | s = 0
t = [list(map(int, input().split())) for i in range(int(input()))]
f = lambda b: a[2] < b[2] and (a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2 <= (a[2] - b[2]) ** 2
for a in t:
k = sum(f(b) for b in t)
s += (-1, 1)[(k < 1) + k & 1] * a[2] ** 2
print(3.1415926536 * s) | Codeforces Round 418 (Div. 2) | CF | 2,017 | 2 | 256 | An overnight dance in discotheque | The crowdedness of the discotheque would never stop our friends from having fun, but a bit more spaciousness won't hurt, will it?
The discotheque can be seen as an infinite xy-plane, in which there are a total of n dancers. Once someone starts moving around, they will move only inside their own movement range, which i... | The first line of input contains a positive integer n (1 ≤ n ≤ 1 000) — the number of dancers.
The following n lines each describes a dancer: the i-th line among them contains three space-separated integers xi, yi and ri ( - 106 ≤ xi, yi ≤ 106, 1 ≤ ri ≤ 106), describing a circular movement range centered at (xi, yi) w... | Output one decimal number — the largest achievable sum of spaciousness over two halves of the night.
The output is considered correct if it has a relative or absolute error of at most 10 - 9. Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if $$\frac{|a-b|}{\max(1, |b|)} \... | null | The first sample corresponds to the illustrations in the legend. | [{"input": "5\n2 1 6\n0 4 1\n2 -1 3\n1 -2 1\n4 -1 1", "output": "138.23007676"}, {"input": "8\n0 0 1\n0 0 2\n0 0 3\n0 0 4\n0 0 5\n0 0 6\n0 0 7\n0 0 8", "output": "289.02652413"}] | 2,000 | ["dfs and similar", "dp", "geometry", "greedy", "trees"] | 32 | [{"input": "5\r\n2 1 6\r\n0 4 1\r\n2 -1 3\r\n1 -2 1\r\n4 -1 1\r\n", "output": "138.23007676\r\n"}, {"input": "8\r\n0 0 1\r\n0 0 2\r\n0 0 3\r\n0 0 4\r\n0 0 5\r\n0 0 6\r\n0 0 7\r\n0 0 8\r\n", "output": "289.02652413\r\n"}, {"input": "4\r\n1000000 -1000000 2\r\n1000000 -1000000 3\r\n-1000000 1000000 2\r\n-1000000 1000000 ... | false | stdio | import sys
def main():
input_path = sys.argv[1]
correct_output_path = sys.argv[2]
submission_output_path = sys.argv[3]
# Read correct output
with open(correct_output_path, 'r') as f:
correct_line = f.readline().strip()
try:
b = float(correct_line)
except:
... | true |
904/C | 906 | A | PyPy 3 | TESTS | 9 | 717 | 24,985,600 | 126447050 | n = int(input())
s = set('abcdefghijklmnopqrstuvwxyz')
ans = 0
for i in range(n):
b = list(input().split())
if i == n - 1:
break
if b[0] == '!':
s.intersection_update(set(b[1]))
if len(s) == 1:
ans += 1
elif b[0] == '.':
s.difference_update(set(b[1]))
else... | 38 | 187 | 5,734,400 | 33555597 | import sys
#f = open('input', 'r')
f = sys.stdin
n = int(f.readline())
cset = set(chr(ord('a')+i) for i in range(26))
found = False
res = 0
for _ in range(n):
action, word = f.readline().split()
if action == '!':
if found:
res += 1
cset &= set(word)
elif action == '.':
cset -= set(word)
else:... | Технокубок 2018 - Отборочный Раунд 4 | CF | 2,017 | 2 | 256 | Shockers | Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for eac... | The first line contains a single integer n (1 ≤ n ≤ 105) — the number of actions Valentin did.
The next n lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types:
1. Valentin pronounced some word and didn't get an electric shock. This action is... | Output a single integer — the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined. | null | In the first test case after the first action it becomes clear that the selected letter is one of the following: a, b, c. After the second action we can note that the selected letter is not a. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is c, but Valentin pronounces ... | [{"input": "5\n! abc\n. ad\n. b\n! cd\n? c", "output": "1"}, {"input": "8\n! hello\n! codeforces\n? c\n. o\n? d\n? h\n. l\n? e", "output": "2"}, {"input": "7\n! ababahalamaha\n? a\n? b\n? a\n? b\n? a\n? h", "output": "0"}] | 1,600 | ["strings"] | 38 | [{"input": "5\r\n! abc\r\n. ad\r\n. b\r\n! cd\r\n? c\r\n", "output": "1\r\n"}, {"input": "8\r\n! hello\r\n! codeforces\r\n? c\r\n. o\r\n? d\r\n? h\r\n. l\r\n? e\r\n", "output": "2\r\n"}, {"input": "7\r\n! ababahalamaha\r\n? a\r\n? b\r\n? a\r\n? b\r\n? a\r\n? h\r\n", "output": "0\r\n"}, {"input": "4\r\n! abcd\r\n! cdef\... | false | stdio | null | true |
223/C | 223 | C | Python 3 | TESTS | 7 | 124 | 5,529,600 | 33056201 | leng, repeat=list(map(int,input().split()))
Lis = list(map(int,input().split()))
temp=Lis.copy()
for rep in range(repeat):
summ= Lis[0]
for i in range(1,len(Lis)):
temp[i]=(summ+Lis[i])%(10**9+7)
summ = temp[i]
print(*temp) | 33 | 716 | 26,726,400 | 33056780 | leng, repeat=list(map(int,input().split()))
Lis = list(map(int,input().split()))
mod = 10**9 + 7
cum = [1]
ans = [0]*leng
for i in range(1, 2001):
cum.append((cum[-1] * (repeat + i - 1) * pow(i, mod-2, mod)) % mod)
for i in range(leng):
for j in range(i + 1):
ans[i] = (ans[i] + cum[i-j] * Lis[j]) % mod... | Codeforces Round 138 (Div. 1) | CF | 2,012 | 4 | 256 | Partial Sums | You've got an array a, consisting of n integers. The array elements are indexed from 1 to n. Let's determine a two step operation like that:
1. First we build by the array a an array s of partial sums, consisting of n elements. Element number i (1 ≤ i ≤ n) of array s equals $$s_i = \left( \sum_{j=1}^{i} a_j \right) \m... | The first line contains two space-separated integers n and k (1 ≤ n ≤ 2000, 0 ≤ k ≤ 109). The next line contains n space-separated integers a1, a2, ..., an — elements of the array a (0 ≤ ai ≤ 109). | Print n integers — elements of the array a after the operations are applied to it. Print the elements in the order of increasing of their indexes in the array a. Separate the printed numbers by spaces. | null | null | [{"input": "3 1\n1 2 3", "output": "1 3 6"}, {"input": "5 0\n3 14 15 92 6", "output": "3 14 15 92 6"}] | 1,900 | ["combinatorics", "math", "number theory"] | 33 | [{"input": "3 1\r\n1 2 3\r\n", "output": "1 3 6\r\n"}, {"input": "5 0\r\n3 14 15 92 6\r\n", "output": "3 14 15 92 6\r\n"}, {"input": "1 1\r\n3\r\n", "output": "3\r\n"}, {"input": "1 0\r\n0\r\n", "output": "0\r\n"}, {"input": "1 0\r\n123\r\n", "output": "123\r\n"}, {"input": "1 1\r\n0\r\n", "output": "0\r\n"}, {"input":... | false | stdio | null | true |
223/C | 223 | C | PyPy 3 | TESTS | 7 | 216 | 0 | 121993979 | mod= 10**9 +7
def multiply(a,b):
res=[]
for i in range(len(a)):
ans=0
for j in range (0,i+1):
ans+=a[j]*b[i-j]
res.append(ans%mod)
return res
def compute_pow_k(b,k):
if(k==1): return b
if(k==2):
return multiply(b,b)
if(k%2==0):
a=compute_po... | 33 | 2,276 | 307,200 | 5628361 | n, k = map(int, input().split())
a = list(map(int, input().split()))
m = 1000000007
r = [ 0, 1 ]
for i in range(2, n+1):
r.append( (- (m // i) * r[m % i]) % m )
c = [ 1 ]
for i in range(1, n):
c.append((c[i-1] * (k+i-1) * r[i]) % m)
ans = []
for i in range(n):
t = 0
for j in range(i+1):
t = (... | Codeforces Round 138 (Div. 1) | CF | 2,012 | 4 | 256 | Partial Sums | You've got an array a, consisting of n integers. The array elements are indexed from 1 to n. Let's determine a two step operation like that:
1. First we build by the array a an array s of partial sums, consisting of n elements. Element number i (1 ≤ i ≤ n) of array s equals $$s_i = \left( \sum_{j=1}^{i} a_j \right) \m... | The first line contains two space-separated integers n and k (1 ≤ n ≤ 2000, 0 ≤ k ≤ 109). The next line contains n space-separated integers a1, a2, ..., an — elements of the array a (0 ≤ ai ≤ 109). | Print n integers — elements of the array a after the operations are applied to it. Print the elements in the order of increasing of their indexes in the array a. Separate the printed numbers by spaces. | null | null | [{"input": "3 1\n1 2 3", "output": "1 3 6"}, {"input": "5 0\n3 14 15 92 6", "output": "3 14 15 92 6"}] | 1,900 | ["combinatorics", "math", "number theory"] | 33 | [{"input": "3 1\r\n1 2 3\r\n", "output": "1 3 6\r\n"}, {"input": "5 0\r\n3 14 15 92 6\r\n", "output": "3 14 15 92 6\r\n"}, {"input": "1 1\r\n3\r\n", "output": "3\r\n"}, {"input": "1 0\r\n0\r\n", "output": "0\r\n"}, {"input": "1 0\r\n123\r\n", "output": "123\r\n"}, {"input": "1 1\r\n0\r\n", "output": "0\r\n"}, {"input":... | false | stdio | null | true |
223/C | 223 | C | PyPy 3 | TESTS | 7 | 216 | 0 | 107175742 | n,k=map(int,input().split())
mod=10**9+7
a=[int(x) for x in input().split()]
p=a[:]
for i in range(1,n):
p[i]=(p[i]+p[i-1])%mod
#print(p)
for i in range(1,n):
a[i]=(k*p[i-1]+a[i])%mod
print(*a) | 33 | 3,118 | 102,400 | 42134847 | n, k = map(int, input().split())
num = list(map(int, input().split()))
MOD = 10 ** 9 + 7
cf = [1]
for i in range(1, 2020):
cf.append((cf[-1] * (k + i - 1) * pow(i, MOD - 2, MOD)) % MOD)
ans = [0 for i in range(n)]
for i in range(n):
for j in range(i + 1):
ans[i] = (ans[i] + cf[i - j] * num[j]) % MOD
... | Codeforces Round 138 (Div. 1) | CF | 2,012 | 4 | 256 | Partial Sums | You've got an array a, consisting of n integers. The array elements are indexed from 1 to n. Let's determine a two step operation like that:
1. First we build by the array a an array s of partial sums, consisting of n elements. Element number i (1 ≤ i ≤ n) of array s equals $$s_i = \left( \sum_{j=1}^{i} a_j \right) \m... | The first line contains two space-separated integers n and k (1 ≤ n ≤ 2000, 0 ≤ k ≤ 109). The next line contains n space-separated integers a1, a2, ..., an — elements of the array a (0 ≤ ai ≤ 109). | Print n integers — elements of the array a after the operations are applied to it. Print the elements in the order of increasing of their indexes in the array a. Separate the printed numbers by spaces. | null | null | [{"input": "3 1\n1 2 3", "output": "1 3 6"}, {"input": "5 0\n3 14 15 92 6", "output": "3 14 15 92 6"}] | 1,900 | ["combinatorics", "math", "number theory"] | 33 | [{"input": "3 1\r\n1 2 3\r\n", "output": "1 3 6\r\n"}, {"input": "5 0\r\n3 14 15 92 6\r\n", "output": "3 14 15 92 6\r\n"}, {"input": "1 1\r\n3\r\n", "output": "3\r\n"}, {"input": "1 0\r\n0\r\n", "output": "0\r\n"}, {"input": "1 0\r\n123\r\n", "output": "123\r\n"}, {"input": "1 1\r\n0\r\n", "output": "0\r\n"}, {"input":... | false | stdio | null | true |
223/C | 223 | C | PyPy 3 | TESTS | 7 | 248 | 0 | 96434836 | n,k=map(int,input().split())
a=list(map(int,input().split()))
for i in range(1,n):
a[i]+=a[i-1]*k
print(a[i-1])
print(a[-1]) | 33 | 498 | 22,732,800 | 122265367 | n, k= map(int, input().split())
a = list(map(int, input().split()))
def find_cf(k,i,a,mod):
cf=[1]
ans=0
for j in range(1,i):
aux=(i-j)*cf[-1]*pow(k+i-j-1,mod-2,mod)%mod
cf.append(aux)
ans+=aux*a[j-1]
ans+=a[i-1]
return ans
def Combinations(n,r,mod):
num = den... | Codeforces Round 138 (Div. 1) | CF | 2,012 | 4 | 256 | Partial Sums | You've got an array a, consisting of n integers. The array elements are indexed from 1 to n. Let's determine a two step operation like that:
1. First we build by the array a an array s of partial sums, consisting of n elements. Element number i (1 ≤ i ≤ n) of array s equals $$s_i = \left( \sum_{j=1}^{i} a_j \right) \m... | The first line contains two space-separated integers n and k (1 ≤ n ≤ 2000, 0 ≤ k ≤ 109). The next line contains n space-separated integers a1, a2, ..., an — elements of the array a (0 ≤ ai ≤ 109). | Print n integers — elements of the array a after the operations are applied to it. Print the elements in the order of increasing of their indexes in the array a. Separate the printed numbers by spaces. | null | null | [{"input": "3 1\n1 2 3", "output": "1 3 6"}, {"input": "5 0\n3 14 15 92 6", "output": "3 14 15 92 6"}] | 1,900 | ["combinatorics", "math", "number theory"] | 33 | [{"input": "3 1\r\n1 2 3\r\n", "output": "1 3 6\r\n"}, {"input": "5 0\r\n3 14 15 92 6\r\n", "output": "3 14 15 92 6\r\n"}, {"input": "1 1\r\n3\r\n", "output": "3\r\n"}, {"input": "1 0\r\n0\r\n", "output": "0\r\n"}, {"input": "1 0\r\n123\r\n", "output": "123\r\n"}, {"input": "1 1\r\n0\r\n", "output": "0\r\n"}, {"input":... | false | stdio | null | true |
353/B | 353 | B | PyPy 3 | TESTS | 9 | 312 | 0 | 82433989 | n=int(input())
arr=list(map(int,input().split()))
flip2=0
la=0
lb=0
a={}
b={}
for i in range(2*n):
if arr[i] in a and arr[i] in b:
if flip2==0 and la<n:
a[arr[i]]+=1
arr[i]="1"
flip2=1
la+=1
else:
b[arr[i]]+=1
arr[i]="2"
... | 40 | 92 | 0 | 4730958 | n = int(input())
a = list(map(int, input().split()))
g = [[] for i in range(100)]
for i in range(2 * n):
g[a[i]].append(i)
x = [0, 0]
cur = 1
mus = []
ans = [0] * 2 * n
for i in range(10, 100):
if len(g[i]) == 1:
ans[g[i][0]] = cur
x[cur - 1] += 1
cur = 3 - cur
if len(g[i]) >= 2:
... | Codeforces Round 205 (Div. 2) | CF | 2,013 | 1 | 256 | Two Heaps | Valera has 2·n cubes, each cube contains an integer from 10 to 99. He arbitrarily chooses n cubes and puts them in the first heap. The remaining cubes form the second heap.
Valera decided to play with cubes. During the game he takes a cube from the first heap and writes down the number it has. Then he takes a cube fro... | The first line contains integer n (1 ≤ n ≤ 100). The second line contains 2·n space-separated integers ai (10 ≤ ai ≤ 99), denoting the numbers on the cubes. | In the first line print a single number — the maximum possible number of distinct four-digit numbers Valera can obtain. In the second line print 2·n numbers bi (1 ≤ bi ≤ 2). The numbers mean: the i-th cube belongs to the bi-th heap in your division.
If there are multiple optimal ways to split the cubes into the heaps,... | null | In the first test case Valera can put the first cube in the first heap, and second cube — in second heap. In this case he obtain number 1099. If he put the second cube in the first heap, and the first cube in the second heap, then he can obtain number 9910. In both cases the maximum number of distinct integers is equal... | [{"input": "1\n10 99", "output": "1\n2 1"}, {"input": "2\n13 24 13 45", "output": "4\n1 2 2 1"}] | 1,900 | ["combinatorics", "constructive algorithms", "greedy", "implementation", "math", "sortings"] | 40 | [{"input": "1\r\n10 99\r\n", "output": "1\r\n2 1 \r\n"}, {"input": "2\r\n13 24 13 45\r\n", "output": "4\r\n1 2 2 1 \r\n"}, {"input": "5\r\n21 60 18 21 17 39 58 74 62 34\r\n", "output": "25\r\n1 1 1 2 2 1 2 1 2 2 \r\n"}, {"input": "10\r\n26 43 29 92 22 27 95 56 72 55 93 51 91 30 70 77 32 69 87 98\r\n", "output": "100\r\... | false | stdio | import sys
def main(input_path, output_path, submission_path):
# Read input
with open(input_path) as f:
n = int(f.readline())
cubes = list(map(int, f.readline().split()))
# Read reference output (to get the expected maximum)
with open(output_path) as f:
expected_max = int(f... | true |
353/B | 353 | B | Python 3 | TESTS | 7 | 92 | 0 | 4730939 | # from pprint import pprint
# from sys import exit
N = int(input()) * 2
data = [int(i) for i in input().split()]
A = set()
B = set()
answer = [0] * N
one, two = 1, 2
for i, x in enumerate(data):
if len(A) > len(B):
A, B = B, A
one, two = two, one
if x not in A:
A.add(x)
answer[... | 40 | 124 | 0 | 4732895 | n = int(input())
a = list(map(int, input().split()))
proc = sorted([ (a.count(i), i) for i in range(10, 100) if a.count(i) > 0 ])
left, right = [], []
for cnt, val in proc:
half = cnt // 2
left.extend([val] * half)
right.extend([val] * half)
if cnt % 2 == 1:
if len(left) < len(right):
... | Codeforces Round 205 (Div. 2) | CF | 2,013 | 1 | 256 | Two Heaps | Valera has 2·n cubes, each cube contains an integer from 10 to 99. He arbitrarily chooses n cubes and puts them in the first heap. The remaining cubes form the second heap.
Valera decided to play with cubes. During the game he takes a cube from the first heap and writes down the number it has. Then he takes a cube fro... | The first line contains integer n (1 ≤ n ≤ 100). The second line contains 2·n space-separated integers ai (10 ≤ ai ≤ 99), denoting the numbers on the cubes. | In the first line print a single number — the maximum possible number of distinct four-digit numbers Valera can obtain. In the second line print 2·n numbers bi (1 ≤ bi ≤ 2). The numbers mean: the i-th cube belongs to the bi-th heap in your division.
If there are multiple optimal ways to split the cubes into the heaps,... | null | In the first test case Valera can put the first cube in the first heap, and second cube — in second heap. In this case he obtain number 1099. If he put the second cube in the first heap, and the first cube in the second heap, then he can obtain number 9910. In both cases the maximum number of distinct integers is equal... | [{"input": "1\n10 99", "output": "1\n2 1"}, {"input": "2\n13 24 13 45", "output": "4\n1 2 2 1"}] | 1,900 | ["combinatorics", "constructive algorithms", "greedy", "implementation", "math", "sortings"] | 40 | [{"input": "1\r\n10 99\r\n", "output": "1\r\n2 1 \r\n"}, {"input": "2\r\n13 24 13 45\r\n", "output": "4\r\n1 2 2 1 \r\n"}, {"input": "5\r\n21 60 18 21 17 39 58 74 62 34\r\n", "output": "25\r\n1 1 1 2 2 1 2 1 2 2 \r\n"}, {"input": "10\r\n26 43 29 92 22 27 95 56 72 55 93 51 91 30 70 77 32 69 87 98\r\n", "output": "100\r\... | false | stdio | import sys
def main(input_path, output_path, submission_path):
# Read input
with open(input_path) as f:
n = int(f.readline())
cubes = list(map(int, f.readline().split()))
# Read reference output (to get the expected maximum)
with open(output_path) as f:
expected_max = int(f... | true |
995/B | 995 | B | Python 3 | TESTS | 3 | 109 | 307,200 | 103103958 | n = int(input())
pairs = [int(i) for i in input().split(" ")]
yes=True
i=0
output = 0
while i<=len(pairs)-2:
if pairs[0+i]==pairs[1+i]:
i+=2
else:
a = pairs[0+i]
pairs[0+i] = pairs[1+i]
pairs[1+i] = a
output+=1
i+=1
print(output) | 22 | 61 | 0 | 154801963 | n = int(input())
people = [int(i) for i in input().split()]
switches = 0
for i in range(0, 2*n, 2):
person = people[i]
other = people.index(person, i+1)
people.insert(i+1, people[other])
people.pop(other+1)
switches += other - i - 1
print(switches) | Codeforces Round 492 (Div. 1) [Thanks, uDebug!] | CF | 2,018 | 2 | 256 | Suit and Tie | Allen is hosting a formal dinner party. $$$2n$$$ people come to the event in $$$n$$$ pairs (couples). After a night of fun, Allen wants to line everyone up for a final picture. The $$$2n$$$ people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this m... | The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of pairs of people.
The second line contains $$$2n$$$ integers $$$a_1, a_2, \dots, a_{2n}$$$. For each $$$i$$$ with $$$1 \le i \le n$$$, $$$i$$$ appears exactly twice. If $$$a_j = a_k = i$$$, that means that the $$$j$$$-th and $$$k$$$... | Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions. | null | In the first sample case, we can transform $$$1 1 2 3 3 2 4 4 \rightarrow 1 1 2 3 2 3 4 4 \rightarrow 1 1 2 2 3 3 4 4$$$ in two steps. Note that the sequence $$$1 1 2 3 3 2 4 4 \rightarrow 1 1 3 2 3 2 4 4 \rightarrow 1 1 3 3 2 2 4 4$$$ also works in the same number of steps.
The second sample case already satisfies th... | [{"input": "4\n1 1 2 3 3 2 4 4", "output": "2"}, {"input": "3\n1 1 2 2 3 3", "output": "0"}, {"input": "3\n3 1 2 3 1 2", "output": "3"}] | 1,400 | ["greedy", "implementation", "math"] | 22 | [{"input": "4\r\n1 1 2 3 3 2 4 4\r\n", "output": "2\r\n"}, {"input": "3\r\n1 1 2 2 3 3\r\n", "output": "0\r\n"}, {"input": "3\r\n3 1 2 3 1 2\r\n", "output": "3\r\n"}, {"input": "8\r\n7 6 2 1 4 3 3 7 2 6 5 1 8 5 8 4\r\n", "output": "27\r\n"}, {"input": "2\r\n1 2 1 2\r\n", "output": "1\r\n"}, {"input": "3\r\n1 2 3 3 1 2\... | false | stdio | null | true |
813/D | 813 | D | Python 3 | TESTS | 1 | 61 | 0 | 27589913 | n=int(input())
a=list(map(int,input().split()))
maxx=2
for i in range(1,n-1):
b=a[0]-1
c=0
for i0 in range(i):
if abs(b-a[i])==1 or a[i]%7==a[i]%7:
c=c+1
b=a[i]-1
for i0 in range(i,n):
if abs(b-a[i])==1 or a[i]%7==a[i]%7:
c=c+1
maxx=max(maxx,c)
print(maxx) | 35 | 1,434 | 104,243,200 | 112897541 | import sys
def solve():
n = int(sys.stdin.readline())
a = [0] + [int(i) for i in sys.stdin.readline().split()]
dp = [[0]*(n + 1) for i in range(n + 1)]
ans = 0
maxnum = [0] * (10**5 + 2)
maxmod = [0] * 7
for y in range(n + 1):
maxmod = [0] * 7
for ai in a:
... | Educational Codeforces Round 22 | ICPC | 2,017 | 2 | 256 | Two Melodies | Alice is a beginner composer and now she is ready to create another masterpiece. And not even the single one but two at the same time!
Alice has a sheet with n notes written on it. She wants to take two such non-empty non-intersecting subsequences that both of them form a melody and sum of their lengths is maximal.
S... | The first line contains one integer number n (2 ≤ n ≤ 5000).
The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — notes written on a sheet. | Print maximum sum of lengths of such two non-empty non-intersecting subsequences that both of them form a melody. | null | In the first example subsequences [1, 2] and [4, 5] give length 4 in total.
In the second example subsequences [62, 48, 49] and [60, 61] give length 5 in total. If you choose subsequence [62, 61] in the first place then the second melody will have maximum length 2, that gives the result of 4, which is not maximal. | [{"input": "4\n1 2 4 5", "output": "4"}, {"input": "6\n62 22 60 61 48 49", "output": "5"}] | 2,600 | ["dp", "flows"] | 35 | [{"input": "4\r\n1 2 4 5\r\n", "output": "4\r\n"}, {"input": "6\r\n62 22 60 61 48 49\r\n", "output": "5\r\n"}, {"input": "2\r\n1 4\r\n", "output": "2\r\n"}, {"input": "2\r\n5 4\r\n", "output": "2\r\n"}, {"input": "10\r\n9 6 8 5 5 2 8 9 2 2\r\n", "output": "9\r\n"}, {"input": "10\r\n7776 32915 1030 71664 7542 72359 6538... | false | stdio | null | true |
159/A | 159 | A | PyPy 3-64 | TESTS | 13 | 124 | 1,740,800 | 203984402 | n, d = map(int,input().split())
mapping = {}
friends = set()
for _ in range(n):
name_1, name_2, t = input().split()
t = int(t)
mapping[(name_1, name_2)] = t
try:
if 0 < mapping[(name_1, name_2)] - mapping[(name_2, name_1)] <= d:
friends.add(tuple(sorted([name_1, name_2])))
except... | 30 | 1,090 | 307,200 | 100395374 | l=input().split()
n=int(l[0])
d=int(l[1])
lfi=[]
lofpairs=[]
for you in range(n):
s=input().split()
for i in lfi:
if(s[0]==i[1] and s[1]==i[0]):
if(int(s[2])-int(i[2])<=d and int(s[2])-int(i[2])!=0):
if((i[0],i[1]) not in lofpairs and (i[1],i[0]) not in lofpairs):
... | VK Cup 2012 Qualification Round 2 | CF | 2,012 | 3 | 256 | Friends or Not | Polycarpus has a hobby — he develops an unusual social network. His work is almost completed, and there is only one more module to implement — the module which determines friends. Oh yes, in this social network one won't have to add friends manually! Pairs of friends are deduced in the following way. Let's assume that ... | The first line of the input contains two integers n and d (1 ≤ n, d ≤ 1000). The next n lines contain the messages log. The i-th line contains one line of the log formatted as "Ai Bi ti" (without the quotes), which means that user Ai sent a message to user Bi at time ti (1 ≤ i ≤ n). Ai and Bi are non-empty strings at m... | In the first line print integer k — the number of pairs of friends. In the next k lines print pairs of friends as "Ai Bi" (without the quotes). You can print users in pairs and the pairs themselves in any order. Each pair must be printed exactly once. | null | In the first sample test case Vasya and Petya are friends because their messages' sending times are one second apart. Anya and Ivan are not, because their messages' sending times differ by more than one second. | [{"input": "4 1\nvasya petya 1\npetya vasya 2\nanya ivan 2\nivan anya 4", "output": "1\npetya vasya"}, {"input": "1 1000\na b 0", "output": "0"}] | 1,400 | ["*special", "greedy", "implementation"] | 30 | [{"input": "4 1\r\nvasya petya 1\r\npetya vasya 2\r\nanya ivan 2\r\nivan anya 4\r\n", "output": "1\r\npetya vasya\r\n"}, {"input": "1 1000\r\na b 0\r\n", "output": "0\r\n"}, {"input": "2 1\r\na b 0\r\nb a 0\r\n", "output": "0\r\n"}, {"input": "3 1\r\na b 1\r\nb c 2\r\nc d 3\r\n", "output": "0\r\n"}, {"input": "10 2\r\n... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path, 'r') as f:
lines = [line.strip() for line in f]
n, d = map(int, lines[0].split())
messages = []
for line in lines[1:n+1]:
parts = line.split()
... | true |
159/A | 159 | A | Python 3 | TESTS | 13 | 124 | 0 | 10252394 | import sys
def main():
messages = {}
pairs = set()
count, dt = sys.stdin.readline().split()
dt = int(dt)
for line in sys.stdin:
res = line.split()
p = frozenset((res[0], res[1]))
now_t = int(res[2])
if p in pairs:
continue
forward = "__".join(... | 30 | 1,122 | 6,963,200 | 115947862 | n,d=map(int,input().split())
a,b=[],set()
for i in range(n):
q,w,t=input().split()
t=int(t)
a+=[[q,w,t]]
for i in a:
for j in a:
if (i[1],i[0])not in b and i[0]==j[1]and i[1]==j[0]and 0<abs(i[2]-j[2])<=d:
b.add((i[0],i[1]))
print(len(b))
for i in b:print(*[*i]) | VK Cup 2012 Qualification Round 2 | CF | 2,012 | 3 | 256 | Friends or Not | Polycarpus has a hobby — he develops an unusual social network. His work is almost completed, and there is only one more module to implement — the module which determines friends. Oh yes, in this social network one won't have to add friends manually! Pairs of friends are deduced in the following way. Let's assume that ... | The first line of the input contains two integers n and d (1 ≤ n, d ≤ 1000). The next n lines contain the messages log. The i-th line contains one line of the log formatted as "Ai Bi ti" (without the quotes), which means that user Ai sent a message to user Bi at time ti (1 ≤ i ≤ n). Ai and Bi are non-empty strings at m... | In the first line print integer k — the number of pairs of friends. In the next k lines print pairs of friends as "Ai Bi" (without the quotes). You can print users in pairs and the pairs themselves in any order. Each pair must be printed exactly once. | null | In the first sample test case Vasya and Petya are friends because their messages' sending times are one second apart. Anya and Ivan are not, because their messages' sending times differ by more than one second. | [{"input": "4 1\nvasya petya 1\npetya vasya 2\nanya ivan 2\nivan anya 4", "output": "1\npetya vasya"}, {"input": "1 1000\na b 0", "output": "0"}] | 1,400 | ["*special", "greedy", "implementation"] | 30 | [{"input": "4 1\r\nvasya petya 1\r\npetya vasya 2\r\nanya ivan 2\r\nivan anya 4\r\n", "output": "1\r\npetya vasya\r\n"}, {"input": "1 1000\r\na b 0\r\n", "output": "0\r\n"}, {"input": "2 1\r\na b 0\r\nb a 0\r\n", "output": "0\r\n"}, {"input": "3 1\r\na b 1\r\nb c 2\r\nc d 3\r\n", "output": "0\r\n"}, {"input": "10 2\r\n... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path, 'r') as f:
lines = [line.strip() for line in f]
n, d = map(int, lines[0].split())
messages = []
for line in lines[1:n+1]:
parts = line.split()
... | true |
725/D | 725 | D | Python 3 | PRETESTS | 2 | 46 | 0 | 21680619 | from sys import stdin, stdout
n = int(stdin.readline().replace('\n', ''))
teams = []
for n in range(n):
t, w = stdin.readline().replace('\n', '').split(' ')
teams += [(int(t), int(w) - int(t), int(w))]
limak = teams[0]
steams = sorted(teams)
ind = steams.index(limak)
budget = limak[0]
finalpos = ind
if ind > 0:... | 49 | 2,761 | 33,280,000 | 21684583 | from bisect import bisect_right
import heapq
n = int(input())
l = []
ti, wi = map(int, input().split())
bal = ti
pos = 1
for _ in range(n - 1):
ti, wi = map(int, input().split())
if ti > bal:
pos += 1
l.append((ti, wi - ti + 1))
l.sort()
best_pos = pos
op = bisect_right(l, (bal, float('inf')))
#... | Canada Cup 2016 | CF | 2,016 | 3 | 256 | Contest Balloons | One tradition of ACM-ICPC contests is that a team gets a balloon for every solved problem. We assume that the submission time doesn't matter and teams are sorted only by the number of balloons they have. It means that one's place is equal to the number of teams with more balloons, increased by 1. For example, if there ... | The first line of the standard input contains one integer n (2 ≤ n ≤ 300 000) — the number of teams.
The i-th of n following lines contains two integers ti and wi (0 ≤ ti ≤ wi ≤ 1018) — respectively the number of balloons and the weight of the i-th team. Limak is a member of the first team. | Print one integer denoting the best place Limak can get. | null | In the first sample, Limak has 20 balloons initially. There are three teams with more balloons (32, 40 and 45 balloons), so Limak has the fourth place initially. One optimal strategy is:
1. Limak gives 6 balloons away to a team with 32 balloons and weight 37, which is just enough to make them fly. Unfortunately, Limak... | [{"input": "8\n20 1000\n32 37\n40 1000\n45 50\n16 16\n16 16\n14 1000\n2 1000", "output": "3"}, {"input": "7\n4 4\n4 4\n4 4\n4 4\n4 4\n4 4\n5 5", "output": "2"}, {"input": "7\n14000000003 1000000000000000000\n81000000000 88000000000\n5000000000 7000000000\n15000000000 39000000000\n46000000000 51000000000\n0 1000000000\n... | 1,800 | ["data structures", "greedy"] | 49 | [{"input": "8\r\n20 1000\r\n32 37\r\n40 1000\r\n45 50\r\n16 16\r\n16 16\r\n14 1000\r\n2 1000\r\n", "output": "3\r\n"}, {"input": "7\r\n4 4\r\n4 4\r\n4 4\r\n4 4\r\n4 4\r\n4 4\r\n5 5\r\n", "output": "2\r\n"}, {"input": "7\r\n14000000003 1000000000000000000\r\n81000000000 88000000000\r\n5000000000 7000000000\r\n1500000000... | false | stdio | null | true |
995/B | 995 | B | PyPy 3-64 | TESTS | 3 | 62 | 0 | 189900147 | import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
s = []
m = 0
ans = 0
v = [0] * (n + 1)
for i in range(2 * n):
if(v[a[i]]):
j = s.index(a[i])
ans += m - 1 - j
if(j == 0):
s.pop(j)
m -= 1
else:
s.app... | 22 | 61 | 307,200 | 144028093 | import math
import sys
import queue
def solve():
n = int(input())
a = list(map(int, input().split()))
res = 0
s = {}
for i in range(n):
if a[i * 2 + 1] != a[i * 2]:
ind = 0
for j in range(i * 2 + 1, 2 * n):
if a[j] == a[i * 2]:
i... | Codeforces Round 492 (Div. 1) [Thanks, uDebug!] | CF | 2,018 | 2 | 256 | Suit and Tie | Allen is hosting a formal dinner party. $$$2n$$$ people come to the event in $$$n$$$ pairs (couples). After a night of fun, Allen wants to line everyone up for a final picture. The $$$2n$$$ people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this m... | The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of pairs of people.
The second line contains $$$2n$$$ integers $$$a_1, a_2, \dots, a_{2n}$$$. For each $$$i$$$ with $$$1 \le i \le n$$$, $$$i$$$ appears exactly twice. If $$$a_j = a_k = i$$$, that means that the $$$j$$$-th and $$$k$$$... | Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions. | null | In the first sample case, we can transform $$$1 1 2 3 3 2 4 4 \rightarrow 1 1 2 3 2 3 4 4 \rightarrow 1 1 2 2 3 3 4 4$$$ in two steps. Note that the sequence $$$1 1 2 3 3 2 4 4 \rightarrow 1 1 3 2 3 2 4 4 \rightarrow 1 1 3 3 2 2 4 4$$$ also works in the same number of steps.
The second sample case already satisfies th... | [{"input": "4\n1 1 2 3 3 2 4 4", "output": "2"}, {"input": "3\n1 1 2 2 3 3", "output": "0"}, {"input": "3\n3 1 2 3 1 2", "output": "3"}] | 1,400 | ["greedy", "implementation", "math"] | 22 | [{"input": "4\r\n1 1 2 3 3 2 4 4\r\n", "output": "2\r\n"}, {"input": "3\r\n1 1 2 2 3 3\r\n", "output": "0\r\n"}, {"input": "3\r\n3 1 2 3 1 2\r\n", "output": "3\r\n"}, {"input": "8\r\n7 6 2 1 4 3 3 7 2 6 5 1 8 5 8 4\r\n", "output": "27\r\n"}, {"input": "2\r\n1 2 1 2\r\n", "output": "1\r\n"}, {"input": "3\r\n1 2 3 3 1 2\... | false | stdio | null | true |
995/B | 995 | B | Python 3 | TESTS | 3 | 62 | 409,600 | 104683463 | from collections import defaultdict
n = int(input())
list1 = list(map(int,input().split()))
d = defaultdict(list)
toadd=tosub = 0
i=0
while i<len(list1):
# print(i)
# if i<2*n-1 :
# print(list1[i],list1[i+1])
if i<2*n-1 and list1[i]==list1[i+1]:
i+=2
else:
d[list1[i]].append(... | 22 | 62 | 1,536,000 | 226127198 | n = int(input())
a = list(map(int, input().split()))
count = 0
i = 0
while i < 2 * n:
if a[i] != a[i + 1]:
for j in range(i + 1, 2 * n):
if a[i] == a[j]:
a = a[:i + 1] + [a[j]] + a[i + 1:j] + a[j + 1:]
count += (j - i - 1)
else:
i += 1
i += 1
print... | Codeforces Round 492 (Div. 1) [Thanks, uDebug!] | CF | 2,018 | 2 | 256 | Suit and Tie | Allen is hosting a formal dinner party. $$$2n$$$ people come to the event in $$$n$$$ pairs (couples). After a night of fun, Allen wants to line everyone up for a final picture. The $$$2n$$$ people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this m... | The first line contains a single integer $$$n$$$ ($$$1 \le n \le 100$$$), the number of pairs of people.
The second line contains $$$2n$$$ integers $$$a_1, a_2, \dots, a_{2n}$$$. For each $$$i$$$ with $$$1 \le i \le n$$$, $$$i$$$ appears exactly twice. If $$$a_j = a_k = i$$$, that means that the $$$j$$$-th and $$$k$$$... | Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions. | null | In the first sample case, we can transform $$$1 1 2 3 3 2 4 4 \rightarrow 1 1 2 3 2 3 4 4 \rightarrow 1 1 2 2 3 3 4 4$$$ in two steps. Note that the sequence $$$1 1 2 3 3 2 4 4 \rightarrow 1 1 3 2 3 2 4 4 \rightarrow 1 1 3 3 2 2 4 4$$$ also works in the same number of steps.
The second sample case already satisfies th... | [{"input": "4\n1 1 2 3 3 2 4 4", "output": "2"}, {"input": "3\n1 1 2 2 3 3", "output": "0"}, {"input": "3\n3 1 2 3 1 2", "output": "3"}] | 1,400 | ["greedy", "implementation", "math"] | 22 | [{"input": "4\r\n1 1 2 3 3 2 4 4\r\n", "output": "2\r\n"}, {"input": "3\r\n1 1 2 2 3 3\r\n", "output": "0\r\n"}, {"input": "3\r\n3 1 2 3 1 2\r\n", "output": "3\r\n"}, {"input": "8\r\n7 6 2 1 4 3 3 7 2 6 5 1 8 5 8 4\r\n", "output": "27\r\n"}, {"input": "2\r\n1 2 1 2\r\n", "output": "1\r\n"}, {"input": "3\r\n1 2 3 3 1 2\... | false | stdio | null | true |
883/G | 883 | G | Python 3 | TESTS | 3 | 46 | 0 | 31722734 | def bfsCalc(succ, s):
bfs = [s]
queue = [s]
while len(queue)>0:
edge = queue.pop()
for i in succ[edge]:
if i not in bfs:
bfs.append(i)
queue.append(i)
return bfs
n,m,s = [int(i) for i in input().split(" ")]
maxiGraph = {}
miniGraph = {}
succ = {}
res = {}
undirected = []
for i in range(1,m+1):
succ... | 141 | 2,963 | 69,222,400 | 88811817 | def put():
return map(int, input().split())
def dfs(x,flag=1):
s,vis,ans = [x],[0]*n,['+']*m
vis[x]= 1
while s:
i = s.pop()
for j,k in graph[i]:
if vis[j]==0:
if k*flag<0:
ans[abs(k)-1]='-'
elif k*flag>0:
... | 2017-2018 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred) | ICPC | 2,017 | 3 | 256 | Orientation of Edges | Vasya has a graph containing both directed (oriented) and undirected (non-oriented) edges. There can be multiple edges between a pair of vertices.
Vasya has picked a vertex s from the graph. Now Vasya wants to create two separate plans:
1. to orient each undirected edge in one of two possible directions to maximize n... | The first line contains three integers n, m and s (2 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105, 1 ≤ s ≤ n) — number of vertices and edges in the graph, and the vertex Vasya has picked.
The following m lines contain information about the graph edges. Each line contains three integers ti, ui and vi (1 ≤ ti ≤ 2, 1 ≤ ui, vi ≤ n, ui ≠ vi... | The first two lines should describe the plan which maximizes the number of reachable vertices. The lines three and four should describe the plan which minimizes the number of reachable vertices.
A description of each plan should start with a line containing the number of reachable vertices. The second line of a plan s... | null | null | [{"input": "2 2 1\n1 1 2\n2 2 1", "output": "2\n-\n2\n+"}, {"input": "6 6 3\n2 2 6\n1 4 5\n2 3 4\n1 4 1\n1 3 1\n2 2 3", "output": "6\n++-\n2\n+-+"}] | 1,900 | ["dfs and similar", "graphs"] | 141 | [{"input": "2 2 1\r\n1 1 2\r\n2 2 1\r\n", "output": "2\r\n-\r\n2\r\n+\r\n"}, {"input": "6 6 3\r\n2 2 6\r\n1 4 5\r\n2 3 4\r\n1 4 1\r\n1 3 1\r\n2 2 3\r\n", "output": "6\r\n++-\r\n2\r\n+-+\r\n"}, {"input": "5 5 5\r\n2 5 3\r\n1 2 3\r\n1 4 5\r\n2 5 2\r\n1 2 1\r\n", "output": "4\r\n++\r\n1\r\n--\r\n"}, {"input": "13 18 9\r\n... | false | stdio | import sys
from collections import deque
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input
with open(input_path) as f:
n, m, s = map(int, f.readline().split())
all_edges = []
for _ in range(m):
ti, u, v = m... | true |
612/E | 612 | E | Python 3 | TESTS | 6 | 62 | 0 | 15199534 | n = int(input())
p = list(map(int, input().split()))
hasPrinted = False
for i in range(1, n+1):
q = [0]*n
q[0] = i
curPos = 0
for j in range(1, n+1):
q[q[curPos]-1] = p[curPos]
curPos = q[curPos]-1
if q.count(0) != 0:
continue
elif len(set(q)) != len(q):
c... | 18 | 670 | 110,796,800 | 165026043 | import os
from re import M
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... | Educational Codeforces Round 4 | ICPC | 2,015 | 2 | 256 | Square Root of Permutation | A permutation of length n is an array containing each integer from 1 to n exactly once. For example, q = [4, 5, 1, 2, 3] is a permutation. For the permutation q the square of permutation is the permutation p that p[i] = q[q[i]] for each i = 1... n. For example, the square of q = [4, 5, 1, 2, 3] is p = q2 = [2, 3, 4, 5,... | The first line contains integer n (1 ≤ n ≤ 106) — the number of elements in permutation p.
The second line contains n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the elements of permutation p. | If there is no permutation q such that q2 = p print the number "-1".
If the answer exists print it. The only line should contain n different integers qi (1 ≤ qi ≤ n) — the elements of the permutation q. If there are several solutions print any of them. | null | null | [{"input": "4\n2 1 4 3", "output": "3 4 2 1"}, {"input": "4\n2 1 3 4", "output": "-1"}, {"input": "5\n2 3 4 5 1", "output": "4 5 1 2 3"}] | 2,200 | ["combinatorics", "constructive algorithms", "dfs and similar", "graphs", "math"] | 18 | [{"input": "4\r\n2 1 4 3\r\n", "output": "3 4 2 1\r\n"}, {"input": "4\r\n2 1 3 4\r\n", "output": "-1\r\n"}, {"input": "5\r\n2 3 4 5 1\r\n", "output": "4 5 1 2 3\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}, {"input": "10\r\n8 2 10 3 4 6 1 7 9 5\r\n", "output": "-1\r\n"},... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
sub_path = sys.argv[3]
with open(input_path, 'r') as f:
n = int(f.readline())
p = list(map(int, f.readline().split()))
with open(sub_path, 'r') as f:
lines = f.readlines()
sub_output = []
... | true |
616/B | 616 | B | PyPy 3 | TESTS | 7 | 140 | 0 | 78482096 | import sys
def ans(mat):
emma = []
for i in range(len(mat)):
emma.append(mat[i][0])
visited = [False]*len(emma)
while True:
m, n = max_from_emma(emma, visited)
current_max = sys.maxsize
for i in range(len(mat[0])):
if mat[n][i] < current_max:
... | 16 | 31 | 0 | 147946788 | n,m = map(int,input().split())
max = 0
for i in range(n):
a = list(map(int,input().split()))
if min(a)>max:
max = min(a)
print(max) | Educational Codeforces Round 5 | ICPC | 2,016 | 1 | 256 | Dinner with Emma | Jack decides to invite Emma out for a dinner. Jack is a modest student, he doesn't want to go to an expensive restaurant. Emma is a girl with high taste, she prefers elite places.
Munhattan consists of n streets and m avenues. There is exactly one restaurant on the intersection of each street and avenue. The streets a... | The first line contains two integers n, m (1 ≤ n, m ≤ 100) — the number of streets and avenues in Munhattan.
Each of the next n lines contains m integers cij (1 ≤ cij ≤ 109) — the cost of the dinner in the restaurant on the intersection of the i-th street and the j-th avenue. | Print the only integer a — the cost of the dinner for Jack and Emma. | null | In the first example if Emma chooses the first or the third streets Jack can choose an avenue with the cost of the dinner 1. So she chooses the second street and Jack chooses any avenue. The cost of the dinner is 2.
In the second example regardless of Emma's choice Jack can choose a restaurant with the cost of the din... | [{"input": "3 4\n4 1 3 5\n2 2 2 2\n5 4 5 1", "output": "2"}, {"input": "3 3\n1 2 3\n2 3 1\n3 1 2", "output": "1"}] | 1,000 | ["games", "greedy"] | 16 | [{"input": "3 4\r\n4 1 3 5\r\n2 2 2 2\r\n5 4 5 1\r\n", "output": "2\r\n"}, {"input": "3 3\r\n1 2 3\r\n2 3 1\r\n3 1 2\r\n", "output": "1\r\n"}, {"input": "1 1\r\n1\r\n", "output": "1\r\n"}, {"input": "1 10\r\n74 35 82 39 1 84 29 41 70 12\r\n", "output": "1\r\n"}, {"input": "10 1\r\n44\r\n23\r\n65\r\n17\r\n48\r\n29\r\n49... | false | stdio | null | true |
159/A | 159 | A | Python 3 | TESTS | 13 | 62 | 0 | 144290254 | from math import sqrt,log,ceil,gcd
import sys
# sys.stdin=open('input.txt','r')
# sys.stdout=open('output.txt','w')
from random import randint as r
def solve():
n,limit=map(int,input().split())
d={}
st=set()
for i in range(n):
a,b,t=map(str,input().split())
t=int(t)
if (b,a) in ... | 30 | 1,432 | 4,915,200 | 80772133 | # n=int(input())
#q.sort(key=lambda x:((x[1]-x[0]),-x[0]))
# n,k=map(int,input().split())
# arr=list(map(int,input().split()))
#ls=list(map(int,input().split()))
#for i in range(m):
#from sys import stdin
#n=int(stdin.readline())
#for _ in range(int(input())):
import os
import sys
from io import BytesIO, IOBase
BUFSIZ... | VK Cup 2012 Qualification Round 2 | CF | 2,012 | 3 | 256 | Friends or Not | Polycarpus has a hobby — he develops an unusual social network. His work is almost completed, and there is only one more module to implement — the module which determines friends. Oh yes, in this social network one won't have to add friends manually! Pairs of friends are deduced in the following way. Let's assume that ... | The first line of the input contains two integers n and d (1 ≤ n, d ≤ 1000). The next n lines contain the messages log. The i-th line contains one line of the log formatted as "Ai Bi ti" (without the quotes), which means that user Ai sent a message to user Bi at time ti (1 ≤ i ≤ n). Ai and Bi are non-empty strings at m... | In the first line print integer k — the number of pairs of friends. In the next k lines print pairs of friends as "Ai Bi" (without the quotes). You can print users in pairs and the pairs themselves in any order. Each pair must be printed exactly once. | null | In the first sample test case Vasya and Petya are friends because their messages' sending times are one second apart. Anya and Ivan are not, because their messages' sending times differ by more than one second. | [{"input": "4 1\nvasya petya 1\npetya vasya 2\nanya ivan 2\nivan anya 4", "output": "1\npetya vasya"}, {"input": "1 1000\na b 0", "output": "0"}] | 1,400 | ["*special", "greedy", "implementation"] | 30 | [{"input": "4 1\r\nvasya petya 1\r\npetya vasya 2\r\nanya ivan 2\r\nivan anya 4\r\n", "output": "1\r\npetya vasya\r\n"}, {"input": "1 1000\r\na b 0\r\n", "output": "0\r\n"}, {"input": "2 1\r\na b 0\r\nb a 0\r\n", "output": "0\r\n"}, {"input": "3 1\r\na b 1\r\nb c 2\r\nc d 3\r\n", "output": "0\r\n"}, {"input": "10 2\r\n... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path, 'r') as f:
lines = [line.strip() for line in f]
n, d = map(int, lines[0].split())
messages = []
for line in lines[1:n+1]:
parts = line.split()
... | true |
159/A | 159 | A | Python 3 | TESTS | 13 | 92 | 102,400 | 204557679 | """
https://codeforces.com/problemset/problem/159/A
"""
n,temps=[int(x) for x in input().split()]
d=dict()
for _ in range(n):
a,b,t=[x for x in input().split()]
if (a,b) in d:
d[(a,b)].append((int(t),a,b))
elif (b,a) in d:
d[(b,a)].append((int(t),a,b))
else:
d[(a,b)]=[(int(t),a,b... | 30 | 124 | 204,800 | 204561136 | """
https://codeforces.com/problemset/problem/159/A
"""
n, temps = [int(x) for x in input().split()]
d = dict()
for _ in range(n):
a, b, t = [x for x in input().split()]
if (a, b) in d:
d[(a, b)].append((int(t), a, b))
elif (b, a) in d:
d[(b, a)].append((int(t), a, b))
else:
d[(a... | VK Cup 2012 Qualification Round 2 | CF | 2,012 | 3 | 256 | Friends or Not | Polycarpus has a hobby — he develops an unusual social network. His work is almost completed, and there is only one more module to implement — the module which determines friends. Oh yes, in this social network one won't have to add friends manually! Pairs of friends are deduced in the following way. Let's assume that ... | The first line of the input contains two integers n and d (1 ≤ n, d ≤ 1000). The next n lines contain the messages log. The i-th line contains one line of the log formatted as "Ai Bi ti" (without the quotes), which means that user Ai sent a message to user Bi at time ti (1 ≤ i ≤ n). Ai and Bi are non-empty strings at m... | In the first line print integer k — the number of pairs of friends. In the next k lines print pairs of friends as "Ai Bi" (without the quotes). You can print users in pairs and the pairs themselves in any order. Each pair must be printed exactly once. | null | In the first sample test case Vasya and Petya are friends because their messages' sending times are one second apart. Anya and Ivan are not, because their messages' sending times differ by more than one second. | [{"input": "4 1\nvasya petya 1\npetya vasya 2\nanya ivan 2\nivan anya 4", "output": "1\npetya vasya"}, {"input": "1 1000\na b 0", "output": "0"}] | 1,400 | ["*special", "greedy", "implementation"] | 30 | [{"input": "4 1\r\nvasya petya 1\r\npetya vasya 2\r\nanya ivan 2\r\nivan anya 4\r\n", "output": "1\r\npetya vasya\r\n"}, {"input": "1 1000\r\na b 0\r\n", "output": "0\r\n"}, {"input": "2 1\r\na b 0\r\nb a 0\r\n", "output": "0\r\n"}, {"input": "3 1\r\na b 1\r\nb c 2\r\nc d 3\r\n", "output": "0\r\n"}, {"input": "10 2\r\n... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path, 'r') as f:
lines = [line.strip() for line in f]
n, d = map(int, lines[0].split())
messages = []
for line in lines[1:n+1]:
parts = line.split()
... | true |
667/B | 667 | B | Python 3 | PRETESTS | 2 | 77 | 4,608,000 | 17572186 | def solve():
N = int(input())
L = list(map(int, input().split()))
ma = max(L)
mi = min(L)
re = sum(L) - ma - mi
ans = ma - re
print(ans)
if __name__ == '__main__':
solve() | 51 | 62 | 6,246,400 | 204523432 | while True:
try:
n = int(input())
a = list(map(int, input().split()))
a.sort()
sum = 0
for i in range(n-1):
sum += a[i]
print(a[n-1]-sum+1)
except:
break | Codeforces Round 349 (Div. 2) | CF | 2,016 | 1 | 256 | Coat of Anticubism | As some of you know, cubism is a trend in art, where the problem of constructing volumetrical shape on a plane with a combination of three-dimensional geometric shapes comes to the fore.
A famous sculptor Cicasso, whose self-portrait you can contemplate, hates cubism. He is more impressed by the idea to transmit two-d... | The first line contains an integer n (3 ≤ n ≤ 105) — a number of rod-blanks.
The second line contains n integers li (1 ≤ li ≤ 109) — lengths of rods, which Cicasso already has. It is guaranteed that it is impossible to make a polygon with n vertices and nonzero area using the rods Cicasso already has. | Print the only integer z — the minimum length of the rod, so that after adding it it can be possible to construct convex polygon with (n + 1) vertices and nonzero area from all of the rods. | null | In the first example triangle with sides {1 + 1 = 2, 2, 1} can be formed from a set of lengths {1, 1, 1, 2}.
In the second example you can make a triangle with lengths {20, 11, 4 + 3 + 2 + 1 = 10}. | [{"input": "3\n1 2 1", "output": "1"}, {"input": "5\n20 4 3 2 1", "output": "11"}] | 1,100 | ["constructive algorithms", "geometry"] | 51 | [{"input": "3\r\n1 2 1\r\n", "output": "1\r\n"}, {"input": "5\r\n20 4 3 2 1\r\n", "output": "11\r\n"}, {"input": "7\r\n77486105 317474713 89523018 332007362 7897847 949616701 54820086\r\n", "output": "70407571\r\n"}, {"input": "14\r\n245638694 2941428 4673577 12468 991349408 44735727 14046308 60637707 81525 104620306 8... | false | stdio | null | true |
754/C | 754 | C | PyPy 3 | TESTS | 11 | 140 | 3,686,400 | 136298898 | import re
uname = re.compile(r"[a-zA-Z0-9]+")
for _ in range(int(input())):
input()
users = set(input().split())
messages = [input().split(":") for _ in range(int(input()))]
for i, (u, m) in enumerate(messages):
allowed = set()
if u == "?":
allowed = users.copy()
... | 104 | 77 | 5,017,600 | 23599678 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
# @Author: wjh
# @Date: 2017-01-06 22:56:27
# @Last Modified by: wjh
# @Last Modified time: 2017-01-06 23:40:47
def solve():
global msg, rem, m
flag = True
while flag:
flag = False
for i in range(m):
if (msg[i][0] == ' '):
... | Codeforces Round 390 (Div. 2) | CF | 2,017 | 2 | 256 | Vladik and chat | Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. ... | The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. E... | Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them. | null | null | [{"input": "1\n2\nVladik netman\n2\n?: Hello, Vladik!\n?: Hi", "output": "netman: Hello, Vladik!\nVladik: Hi"}, {"input": "1\n2\nnetman vladik\n3\nnetman:how are you?\n?:wrong message\nvladik:im fine", "output": "Impossible"}, {"input": "2\n3\nnetman vladik Fedosik\n2\n?: users are netman, vladik, Fedosik\nvladik: some... | 2,200 | ["brute force", "constructive algorithms", "dp", "implementation", "strings"] | 104 | [{"input": "1\r\n2\r\nVladik netman\r\n2\r\n?: Hello, Vladik!\r\n?: Hi\r\n", "output": "netman: Hello, Vladik!\r\nVladik: Hi\r\n"}, {"input": "1\r\n2\r\nnetman vladik\r\n3\r\nnetman:how are you?\r\n?:wrong message\r\nvladik:im fine\r\n", "output": "Impossible\r\n"}, {"input": "2\r\n3\r\nnetman vladik Fedosik\r\n2\r\n?:... | false | stdio | null | true |
754/C | 754 | C | Python 3 | TESTS | 5 | 124 | 0 | 42685514 | t=(int)(input())
for q in range(0,t):
n=(int)(input())
namelist = input().split(' ')
m=(int)(input())
users=[]
messages=[]
for i in range(0,m):
user,message=input().split(':')
users.append(user)
messages.append(message)
# print(user)
# print(message)
... | 104 | 77 | 5,427,200 | 23605091 | import sys
import re
def mentioned_usernames(line):
return {x for x in re.split(r'[^A-Za-z0-9]+', line)}
t = int(input())
for ti in range(t):
possible_users = []
messages = []
n = int(input())
usernames = set(input().split())
# print("usernames =", usernames, file=sys.stderr)
m = int(inpu... | Codeforces Round 390 (Div. 2) | CF | 2,017 | 2 | 256 | Vladik and chat | Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. ... | The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. E... | Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them. | null | null | [{"input": "1\n2\nVladik netman\n2\n?: Hello, Vladik!\n?: Hi", "output": "netman: Hello, Vladik!\nVladik: Hi"}, {"input": "1\n2\nnetman vladik\n3\nnetman:how are you?\n?:wrong message\nvladik:im fine", "output": "Impossible"}, {"input": "2\n3\nnetman vladik Fedosik\n2\n?: users are netman, vladik, Fedosik\nvladik: some... | 2,200 | ["brute force", "constructive algorithms", "dp", "implementation", "strings"] | 104 | [{"input": "1\r\n2\r\nVladik netman\r\n2\r\n?: Hello, Vladik!\r\n?: Hi\r\n", "output": "netman: Hello, Vladik!\r\nVladik: Hi\r\n"}, {"input": "1\r\n2\r\nnetman vladik\r\n3\r\nnetman:how are you?\r\n?:wrong message\r\nvladik:im fine\r\n", "output": "Impossible\r\n"}, {"input": "2\r\n3\r\nnetman vladik Fedosik\r\n2\r\n?:... | false | stdio | null | true |
821/E | 821 | E | Python 3 | TESTS | 2 | 46 | 4,915,200 | 28041116 | def matpwr(b, e, mod = 10**9+7):
if e == 0: return identity(len(b))
if e == 1: return b
return matmul(matpwr(matmul(b, b), e//2), matpwr(b, e%2))
def matmul(a, b, mod = 10**9+7):
n, m = len(a), len(b[0])
ret = [[0 for i in range(m)] for j in range(n)]
for i in range(n):
for j in range(m... | 90 | 265 | 11,366,400 | 231212945 | from functools import lru_cache
mod = 10**9+7
maxc = 16
def solve(n, k, abc):
v = [0]*maxc
v[0] = 1
for a,b,c in abc:
b = min(b,k)
d = b-a
p2 = 0
while d:
d,r = divmod(d,2)
if r:
v = vecmul(matpow2(c, p2), v)
p2 += 1
... | Codeforces Round 420 (Div. 2) | CF | 2,017 | 2 | 256 | Okabe and El Psy Kongroo | Okabe likes to take walks but knows that spies from the Organization could be anywhere; that's why he wants to know how many different walks he can take in his city safely. Okabe's city can be represented as all points (x, y) such that x and y are non-negative. Okabe starts at the origin (point (0, 0)), and needs to re... | The first line of input contains the integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 1018) — the number of segments and the destination x coordinate.
The next n lines contain three space-separated integers ai, bi, and ci (0 ≤ ai < bi ≤ 1018, 0 ≤ ci ≤ 15) — the left and right ends of a segment, and its y coordinate.
It is guar... | Print the number of walks satisfying the conditions, modulo 1000000007 (109 + 7). | null | The graph above corresponds to sample 1. The possible walks are:
- $$(0,0)\rightarrow(1,0)\rightarrow(2,0)\rightarrow(3,0)$$
- $$(0,0)\rightarrow(1,1)\rightarrow(2,0)\rightarrow(3,0)$$
- $$(0,0)\rightarrow(1,0)\rightarrow(2,1)\rightarrow(3,0)$$
- $$(0,0)\rightarrow(1,1)\rightarrow(2,1)\rightarrow(3,0)$$
The graph abo... | [{"input": "1 3\n0 3 3", "output": "4"}, {"input": "2 6\n0 3 0\n3 10 2", "output": "4"}] | 2,100 | ["dp", "matrices"] | 91 | [{"input": "1 3\r\n0 3 3\r\n", "output": "4\r\n"}, {"input": "2 6\r\n0 3 0\r\n3 10 2\r\n", "output": "4\r\n"}, {"input": "2 3\r\n0 2 13\r\n2 3 11\r\n", "output": "4\r\n"}, {"input": "2 9\r\n0 8 0\r\n8 10 10\r\n", "output": "1\r\n"}, {"input": "1 1\r\n0 3 9\r\n", "output": "1\r\n"}, {"input": "3 8\r\n0 7 3\r\n7 8 5\r\n8... | false | stdio | null | true |
754/C | 754 | C | PyPy 3 | TESTS | 3 | 93 | 24,064,000 | 24469588 | from sys import stdin, stdout
def divide(s):
strings = []
l = 0
r = len(s) - 1
while s[l] in '?!., ' and l < len(s):
l += 1
while s[r] in '?!., ' and r >= 0:
r -= 1
s = s[l:r + 1]
while len(s):
for i in range(len(s)):
if s[i... | 104 | 78 | 5,017,600 | 23603623 | def main():
n = int(input())
names = input().split()
m = int(input())
msg = [input().split(':') for _ in range(m)]
texts = []
for i in msg:
texts.append(i[1])
i[1] = i[1].replace(',', ' ').replace('.', ' ').replace('!', ' ').replace('?', ' ').split()
if i[0] == '?':
... | Codeforces Round 390 (Div. 2) | CF | 2,017 | 2 | 256 | Vladik and chat | Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. ... | The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. E... | Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them. | null | null | [{"input": "1\n2\nVladik netman\n2\n?: Hello, Vladik!\n?: Hi", "output": "netman: Hello, Vladik!\nVladik: Hi"}, {"input": "1\n2\nnetman vladik\n3\nnetman:how are you?\n?:wrong message\nvladik:im fine", "output": "Impossible"}, {"input": "2\n3\nnetman vladik Fedosik\n2\n?: users are netman, vladik, Fedosik\nvladik: some... | 2,200 | ["brute force", "constructive algorithms", "dp", "implementation", "strings"] | 104 | [{"input": "1\r\n2\r\nVladik netman\r\n2\r\n?: Hello, Vladik!\r\n?: Hi\r\n", "output": "netman: Hello, Vladik!\r\nVladik: Hi\r\n"}, {"input": "1\r\n2\r\nnetman vladik\r\n3\r\nnetman:how are you?\r\n?:wrong message\r\nvladik:im fine\r\n", "output": "Impossible\r\n"}, {"input": "2\r\n3\r\nnetman vladik Fedosik\r\n2\r\n?:... | false | stdio | null | true |
754/C | 754 | C | PyPy 3 | TESTS | 3 | 108 | 1,945,600 | 191278900 | import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
t = int(input())
ans = []
for _ in range(t):
n = int(input())
s = ["?"] + list(input().rstrip().decode().split())
d = dict()
for i in range(1, n + 1):
d[s[i]] = i
m = int(input())
dp = []
x = []
for ... | 104 | 171 | 1,228,800 | 100980582 | #!/usr/bin/env python3
import re, sys
def debug(*msg):
print(*msg, file=sys.stderr)
def solve():
def parse(line):
mo = re.match(r'(.*?):(.*)', line)
assert(mo)
name, text = mo[1], mo[2]
if name == '?':
speaker = -1
else:
assert (name in revP)
... | Codeforces Round 390 (Div. 2) | CF | 2,017 | 2 | 256 | Vladik and chat | Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download t chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. ... | The first line contains single integer t (1 ≤ t ≤ 10) — the number of chats. The t chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer n (1 ≤ n ≤ 100) — the number of users in the chat.
The next line contains n space-separated distinct usernames. E... | Print the information about the t chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print m messages in the following format:
<username>:<text>
If there are multiple answers, print any of them. | null | null | [{"input": "1\n2\nVladik netman\n2\n?: Hello, Vladik!\n?: Hi", "output": "netman: Hello, Vladik!\nVladik: Hi"}, {"input": "1\n2\nnetman vladik\n3\nnetman:how are you?\n?:wrong message\nvladik:im fine", "output": "Impossible"}, {"input": "2\n3\nnetman vladik Fedosik\n2\n?: users are netman, vladik, Fedosik\nvladik: some... | 2,200 | ["brute force", "constructive algorithms", "dp", "implementation", "strings"] | 104 | [{"input": "1\r\n2\r\nVladik netman\r\n2\r\n?: Hello, Vladik!\r\n?: Hi\r\n", "output": "netman: Hello, Vladik!\r\nVladik: Hi\r\n"}, {"input": "1\r\n2\r\nnetman vladik\r\n3\r\nnetman:how are you?\r\n?:wrong message\r\nvladik:im fine\r\n", "output": "Impossible\r\n"}, {"input": "2\r\n3\r\nnetman vladik Fedosik\r\n2\r\n?:... | false | stdio | null | true |
641/B | 641 | B | Python 3 | TESTS | 6 | 155 | 716,800 | 19452067 | def main():
n, m, q = map(int, input().split())
nm = ["0"] * (n * m)
a, b, c = m + 1, n - 1, (n - 1) * m
qq = [input() for _ in range(q)]
for s in reversed(qq):
k, *l = s.split()
if k == "3":
row, col, x = l
nm[int(row) * m + int(col) - a] = x
pass... | 26 | 202 | 1,433,600 | 18543653 | n,m,q=map(int,input().split())
l=[list(map(int,input().split())) for _ in range(q)][::-1]
adj=[[0]*m for _ in range(n)]
for i in range(q):
t=l[i]
if t[0]==3:
adj[t[1]-1][t[2]-1]=t[3]
elif t[0]==2:
j=t[1]-1
x=adj[-1][j]
for i in range(n-1,0,-1):
adj[i][j... | VK Cup 2016 - Round 2 | CF | 2,016 | 2 | 256 | Little Artem and Matrix | Little Artem likes electronics. He can spend lots of time making different schemas and looking for novelties in the nearest electronics store. The new control element was delivered to the store recently and Artem immediately bought it.
That element can store information about the matrix of integers size n × m. There a... | The first line of the input contains three integers n, m and q (1 ≤ n, m ≤ 100, 1 ≤ q ≤ 10 000) — dimensions of the matrix and the number of turns in the experiment, respectively.
Next q lines contain turns descriptions, one per line. Each description starts with an integer ti (1 ≤ ti ≤ 3) that defines the type of the... | Print the description of any valid initial matrix as n lines containing m integers each. All output integers should not exceed 109 by their absolute value.
If there are multiple valid solutions, output any of them. | null | null | [{"input": "2 2 6\n2 1\n2 2\n3 1 1 1\n3 2 2 2\n3 1 2 8\n3 2 1 8", "output": "8 2\n1 8"}, {"input": "3 3 2\n1 2\n3 2 2 5", "output": "0 0 0\n0 0 5\n0 0 0"}] | 1,400 | ["implementation"] | 26 | [{"input": "2 2 6\r\n2 1\r\n2 2\r\n3 1 1 1\r\n3 2 2 2\r\n3 1 2 8\r\n3 2 1 8\r\n", "output": "8 2 \r\n1 8 \r\n"}, {"input": "3 3 2\r\n1 2\r\n3 2 2 5\r\n", "output": "0 0 0 \r\n0 0 5 \r\n0 0 0 \r\n"}, {"input": "5 5 1\r\n1 5\r\n", "output": "0 0 0 0 0 \r\n0 0 0 0 0 \r\n0 0 0 0 0 \r\n0 0 0 0 0 \r\n0 0 0 0 0 \r\n"}, {"inpu... | false | stdio | null | true |
366/C | 366 | C | PyPy 3 | TESTS | 5 | 233 | 25,497,600 | 88899494 | n,k = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
diff = []
for x in range(n):
diff.append(a[x]-b[x]*k)
totals = {0:0}
for x in range(n):
t = a[x]
d = diff[x]
newGuys = []
for y in totals:
if y+d in totals:
to... | 28 | 389 | 66,150,400 | 196874771 | class Dict(dict):
def __missing__(self, key):
return float('-inf')
n, k = map(int, input().split())
a = [0] + list(map(int, input().split()))
b = [0] + list(map(int, input().split()))
f = [Dict() for _ in range( n + 1 ) ]
f[0][n*100] = 0
for i in range( 1 , n + 1 ):
s = a[i] - b[i]*k
for j in ran... | Codeforces Round 214 (Div. 2) | CF | 2,013 | 1 | 256 | Dima and Salad | Dima, Inna and Seryozha have gathered in a room. That's right, someone's got to go. To cheer Seryozha up and inspire him to have a walk, Inna decided to cook something.
Dima and Seryozha have n fruits in the fridge. Each fruit has two parameters: the taste and the number of calories. Inna decided to make a fruit salad... | The first line of the input contains two integers n, k (1 ≤ n ≤ 100, 1 ≤ k ≤ 10). The second line of the input contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100) — the fruits' tastes. The third line of the input contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 100) — the fruits' calories. Fruit number i has taste ai and c... | If there is no way Inna can choose the fruits for the salad, print in the single line number -1. Otherwise, print a single integer — the maximum possible sum of the taste values of the chosen fruits. | null | In the first test sample we can get the total taste of the fruits equal to 18 if we choose fruit number 1 and fruit number 2, then the total calories will equal 9. The condition $$\frac{18}{9}=2=k$$ fulfills, that's exactly what Inna wants.
In the second test sample we cannot choose the fruits so as to follow Inna's p... | [{"input": "3 2\n10 8 1\n2 7 1", "output": "18"}, {"input": "5 3\n4 4 4 4 4\n2 2 2 2 2", "output": "-1"}] | 1,900 | ["dp"] | 28 | [{"input": "3 2\r\n10 8 1\r\n2 7 1\r\n", "output": "18\r\n"}, {"input": "5 3\r\n4 4 4 4 4\r\n2 2 2 2 2\r\n", "output": "-1\r\n"}, {"input": "1 1\r\n1\r\n1\r\n", "output": "1\r\n"}, {"input": "1 1\r\n1\r\n2\r\n", "output": "-1\r\n"}, {"input": "2 1\r\n75 65\r\n16 60\r\n", "output": "-1\r\n"}, {"input": "21 8\r\n50 39 28... | false | stdio | null | true |
883/G | 883 | G | Python 3 | TESTS | 3 | 62 | 0 | 31723170 | def bfsCalc(succ, s):
bfs = [s]
queue = [s]
while len(queue)>0:
edge = queue.pop()
for i in succ[edge]:
if i not in bfs:
bfs.append(i)
queue.append(i)
return bfs
def dfsCalc(succ, s):
dfs = []
queue = [s]
while len(queue)>0:
edge = queue[0]
dfs.append(edge)
if len(queue)>1:
queue = queue... | 141 | 1,200 | 48,844,800 | 95314305 | import sys
input = sys.stdin.readline
def put():
return map(int, input().split())
def dfs0(x):
s = [x]
vis = [0] * n
ans = ['+'] * m
vis[x] = 1
while s:
i = s.pop()
for j, k in graph[i]:
if (vis[j] == 0):
if (k < 0):
ans[-k - 1] = '-'
elif (k > 0):
ans[... | 2017-2018 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred) | ICPC | 2,017 | 3 | 256 | Orientation of Edges | Vasya has a graph containing both directed (oriented) and undirected (non-oriented) edges. There can be multiple edges between a pair of vertices.
Vasya has picked a vertex s from the graph. Now Vasya wants to create two separate plans:
1. to orient each undirected edge in one of two possible directions to maximize n... | The first line contains three integers n, m and s (2 ≤ n ≤ 3·105, 1 ≤ m ≤ 3·105, 1 ≤ s ≤ n) — number of vertices and edges in the graph, and the vertex Vasya has picked.
The following m lines contain information about the graph edges. Each line contains three integers ti, ui and vi (1 ≤ ti ≤ 2, 1 ≤ ui, vi ≤ n, ui ≠ vi... | The first two lines should describe the plan which maximizes the number of reachable vertices. The lines three and four should describe the plan which minimizes the number of reachable vertices.
A description of each plan should start with a line containing the number of reachable vertices. The second line of a plan s... | null | null | [{"input": "2 2 1\n1 1 2\n2 2 1", "output": "2\n-\n2\n+"}, {"input": "6 6 3\n2 2 6\n1 4 5\n2 3 4\n1 4 1\n1 3 1\n2 2 3", "output": "6\n++-\n2\n+-+"}] | 1,900 | ["dfs and similar", "graphs"] | 141 | [{"input": "2 2 1\r\n1 1 2\r\n2 2 1\r\n", "output": "2\r\n-\r\n2\r\n+\r\n"}, {"input": "6 6 3\r\n2 2 6\r\n1 4 5\r\n2 3 4\r\n1 4 1\r\n1 3 1\r\n2 2 3\r\n", "output": "6\r\n++-\r\n2\r\n+-+\r\n"}, {"input": "5 5 5\r\n2 5 3\r\n1 2 3\r\n1 4 5\r\n2 5 2\r\n1 2 1\r\n", "output": "4\r\n++\r\n1\r\n--\r\n"}, {"input": "13 18 9\r\n... | false | stdio | import sys
from collections import deque
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input
with open(input_path) as f:
n, m, s = map(int, f.readline().split())
all_edges = []
for _ in range(m):
ti, u, v = m... | true |
2/A | 2 | A | Python 3 | TESTS | 13 | 92 | 307,200 | 219548416 | from collections import defaultdict
n=int(input())
res='akxyn'
data=defaultdict(int)
for _ in range(n):
name,score=input().split()
score=int(score)
data[name]+=score
if data[name]>data[res]:
res=name
print(res) | 20 | 62 | 102,400 | 222065059 | n=int(input())
a=[0]*3
b=[0]*3
c=[0]*3
k={}
d=[]
for i in range(n):
a,b=map(str,input().split())
k[a]=k.get(a,0)+int(b);d.append([a,k[a]])
m=max(k.values())
for i,j in d:
if k[i]==m and int(j)>=m:
print(i); break | Codeforces Beta Round 2 | ICPC | 2,010 | 1 | 64 | Winner | The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a pla... | The first line contains an integer number n (1 ≤ n ≤ 1000), n is the number of rounds played. Then follow n lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer numbe... | Print the name of the winner. | null | null | [{"input": "3\nmike 3\nandrew 5\nmike 2", "output": "andrew"}, {"input": "3\nandrew 3\nandrew 2\nmike 5", "output": "andrew"}] | 1,500 | ["hashing", "implementation"] | 20 | [{"input": "3\r\nmike 3\r\nandrew 5\r\nmike 2\r\n", "output": "andrew\r\n"}, {"input": "3\r\nandrew 3\r\nandrew 2\r\nmike 5\r\n", "output": "andrew\r\n"}, {"input": "5\r\nkaxqybeultn -352\r\nmgochgrmeyieyskhuourfg -910\r\nkaxqybeultn 691\r\nmgochgrmeyieyskhuourfg -76\r\nkaxqybeultn -303\r\n", "output": "kaxqybeultn\r\n... | false | stdio | null | true |
2/A | 2 | A | Python 3 | TESTS | 13 | 92 | 102,400 | 175370038 | from collections import defaultdict
n = int(input())
max_score = 0
players_max_score_and_round =defaultdict(lambda: [0, 0])
players_scores = defaultdict(int)
for round in range(1, n+1):
name, score = input().split()
score = int(score)
players_scores[name] += score
if players_scores[name] > players_ma... | 20 | 62 | 204,800 | 208198879 | import sys
# sys.stdin = open("3.in", "r")
n = int(input())
my_dict = dict()
rounds = []
for i in range(n):
name, point = input().split()
point = int(point)
rounds.append((name, point))
if name in my_dict:
my_dict[name] += point
else:
my_dict[name] = point
max_score = -sys.maxsize
max_na... | Codeforces Beta Round 2 | ICPC | 2,010 | 1 | 64 | Winner | The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a pla... | The first line contains an integer number n (1 ≤ n ≤ 1000), n is the number of rounds played. Then follow n lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer numbe... | Print the name of the winner. | null | null | [{"input": "3\nmike 3\nandrew 5\nmike 2", "output": "andrew"}, {"input": "3\nandrew 3\nandrew 2\nmike 5", "output": "andrew"}] | 1,500 | ["hashing", "implementation"] | 20 | [{"input": "3\r\nmike 3\r\nandrew 5\r\nmike 2\r\n", "output": "andrew\r\n"}, {"input": "3\r\nandrew 3\r\nandrew 2\r\nmike 5\r\n", "output": "andrew\r\n"}, {"input": "5\r\nkaxqybeultn -352\r\nmgochgrmeyieyskhuourfg -910\r\nkaxqybeultn 691\r\nmgochgrmeyieyskhuourfg -76\r\nkaxqybeultn -303\r\n", "output": "kaxqybeultn\r\n... | false | stdio | null | true |
873/C | 873 | C | PyPy 3-64 | TESTS | 2 | 46 | 0 | 206646205 | n, m, k = map(int, input().split())
a = [list(map(int, input().split())) for _ in range(n)]
pr = [[0 for _ in range(m)] for _ in range(n)]
for j in range(m):
for i in range(n):
if i == 0:
pr[i][j] = a[i][j]
else:
pr[i][j] = pr[i - 1][j] + a[i][j]
result = [0 for _ in range(m)... | 20 | 77 | 0 | 32568148 | n, m, k = map(int, input().split()); a = []; b = []; score = []; ct = 0
for i in range(n):
a.append([int(x) for x in input().split()])
for i in range(m): b.append([])
for i in range(m):
for j in range(n):
b[i].append(a[j][i])
for i in range(m):
maxsums = []
for j in range(n):
... | Educational Codeforces Round 30 | ICPC | 2,017 | 1 | 256 | Strange Game On Matrix | Ivan is playing a strange game.
He has a matrix a with n rows and m columns. Each element of the matrix is equal to either 0 or 1. Rows and columns are 1-indexed. Ivan can replace any number of ones in this matrix with zeroes. After that, his score in the game will be calculated as follows:
1. Initially Ivan's score ... | The first line contains three integer numbers n, m and k (1 ≤ k ≤ n ≤ 100, 1 ≤ m ≤ 100).
Then n lines follow, i-th of them contains m integer numbers — the elements of i-th row of matrix a. Each number is either 0 or 1. | Print two numbers: the maximum possible score Ivan can get and the minimum number of replacements required to get this score. | null | In the first example Ivan will replace the element a1, 2. | [{"input": "4 3 2\n0 1 0\n1 0 1\n0 1 0\n1 1 1", "output": "4 1"}, {"input": "3 2 1\n1 0\n0 1\n0 0", "output": "2 0"}] | 1,600 | ["greedy", "two pointers"] | 20 | [{"input": "4 3 2\r\n0 1 0\r\n1 0 1\r\n0 1 0\r\n1 1 1\r\n", "output": "4 1\r\n"}, {"input": "3 2 1\r\n1 0\r\n0 1\r\n0 0\r\n", "output": "2 0\r\n"}, {"input": "3 4 2\r\n0 1 1 1\r\n1 0 1 1\r\n1 0 0 1\r\n", "output": "7 0\r\n"}, {"input": "3 57 3\r\n1 0 0 1 1 0 1 1 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 1 0... | false | stdio | null | true |
484/D | 484 | D | Python 3 | TESTS | 6 | 93 | 6,656,000 | 86714699 | l = input()
a = list(map(int, input().split()))
a.sort()
x = 0
for i in range(len(a)//2):
x+= (a[len(a)-1-i]- a[i])
print(x) | 71 | 1,310 | 79,462,400 | 147542209 | n = int(input())
if n == 1:
print(0)
else:
a = [0]*(n+1)
i = 1
for v in map(int, input().split()):
a[i] = v
i += 1
dp = [0]*(n+1)
dp[2] = abs(a[1]-a[2])
for i in range(3, n+1):
# 单调
if (a[i-2] < a[i-1] and a[i-1] < a[i]) or (a[i-2] > a[i-1] and a[i-1] > a[i]):... | Codeforces Round 276 (Div. 1) | CF | 2,014 | 2 | 256 | Kindergarten | In a kindergarten, the children are being divided into groups. The teacher put the children in a line and associated each child with his or her integer charisma value. Each child should go to exactly one group. Each group should be a nonempty segment of consecutive children of a line. A group's sociability is the maxim... | The first line contains integer n — the number of children in the line (1 ≤ n ≤ 106).
The second line contains n integers ai — the charisma of the i-th child ( - 109 ≤ ai ≤ 109). | Print the maximum possible total sociability of all groups. | null | In the first test sample one of the possible variants of an division is following: the first three children form a group with sociability 2, and the two remaining children form a group with sociability 1.
In the second test sample any division leads to the same result, the sociability will be equal to 0 in each group. | [{"input": "5\n1 2 3 1 2", "output": "3"}, {"input": "3\n3 3 3", "output": "0"}] | 2,400 | ["data structures", "dp", "greedy"] | 71 | [{"input": "5\r\n1 2 3 1 2\r\n", "output": "3\r\n"}, {"input": "3\r\n3 3 3\r\n", "output": "0\r\n"}, {"input": "1\r\n0\r\n", "output": "0\r\n"}, {"input": "2\r\n-1000000000 1000000000\r\n", "output": "2000000000\r\n"}, {"input": "4\r\n1 4 2 3\r\n", "output": "4\r\n"}, {"input": "4\r\n23 5 7 1\r\n", "output": "24\r\n"},... | false | stdio | null | true |
1003/D | 1003 | D | PyPy 3 | TESTS | 1 | 904 | 13,209,600 | 118283804 | from collections import defaultdict
import sys
input = sys.stdin.readline
n, q = map(int, input().split())
a = list(map(int, input().split()))
cnt = defaultdict(lambda : 0)
for i in a:
cnt[i] += 1
pow2 = [1]
for _ in range(35):
pow2.append(2 * pow2[-1])
for _ in range(q):
x = [0] * 35
b = int(input())
... | 31 | 280 | 25,292,800 | 190200626 | import sys
input = lambda: sys.stdin.readline().rstrip()
# ----------------------- #
n, q = map(int, input().split())
A = list(map(int, input().split()))
coins = [0] * 33
for a in A:
coins[a.bit_length()-1] += 1
for _ in range(q):
b = int(input())
ans = 0
for i in range(32, -1, -1):
if coins[i] and b >... | Codeforces Round 494 (Div. 3) | ICPC | 2,018 | 2 | 256 | Coins and Queries | Polycarp has $$$n$$$ coins, the value of the $$$i$$$-th coin is $$$a_i$$$. It is guaranteed that all the values are integer powers of $$$2$$$ (i.e. $$$a_i = 2^d$$$ for some non-negative integer number $$$d$$$).
Polycarp wants to know answers on $$$q$$$ queries. The $$$j$$$-th query is described as integer number $$$b_... | The first line of the input contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of coins and the number of queries.
The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ — values of coins ($$$1 \le a_i \le 2 \cdot 10^9$$$). It is guaranteed that all $$... | Print $$$q$$$ integers $$$ans_j$$$. The $$$j$$$-th integer must be equal to the answer on the $$$j$$$-th query. If Polycarp can't obtain the value $$$b_j$$$ the answer to the $$$j$$$-th query is -1. | null | null | [{"input": "5 4\n2 4 8 2 4\n8\n5\n14\n10", "output": "1\n-1\n3\n2"}] | 1,600 | ["greedy"] | 31 | [{"input": "5 4\r\n2 4 8 2 4\r\n8\r\n5\r\n14\r\n10\r\n", "output": "1\r\n-1\r\n3\r\n2\r\n"}, {"input": "3 3\r\n1 1 1\r\n1\r\n2\r\n3\r\n", "output": "1\r\n2\r\n3\r\n"}, {"input": "4 1\r\n2 4 16 32\r\n14\r\n", "output": "-1\r\n"}, {"input": "1 10\r\n8\r\n1\r\n2\r\n3\r\n4\r\n5\r\n6\r\n7\r\n8\r\n9\r\n16\r\n", "output": "-1... | false | stdio | null | true |
43/C | 43 | C | PyPy 3-64 | TESTS | 4 | 154 | 0 | 202285328 | def summ (p):
x=0
while p!=0 :
x=x+p%10
p=p//10
return x;
n=int (input())
l = list(map(int, input().split()))
s=0
trv=0
for i in range(n):
for j in range(i,n):
if (summ(l[i])+summ(l[j]))%3==0:
s+=1
trv=1
break
if(trv==1):
break
... | 21 | 62 | 614,400 | 148170167 | input()
myList = list(map(int, input().split()))
rem0, rem1, rem2 = 0, 0, 0
for it in myList:
if it % 3 == 0:
rem0 += 1
elif it % 3 == 1:
rem1 += 1
else:
rem2 += 1
print(rem0//2 + min(rem2,rem1)) | Codeforces Beta Round 42 (Div. 2) | CF | 2,010 | 2 | 256 | Lucky Tickets | Vasya thinks that lucky tickets are the tickets whose numbers are divisible by 3. He gathered quite a large collection of such tickets but one day his younger brother Leonid was having a sulk and decided to destroy the collection. First he tore every ticket exactly in two, but he didn’t think it was enough and Leonid a... | The first line contains integer n (1 ≤ n ≤ 104) — the number of pieces. The second line contains n space-separated numbers ai (1 ≤ ai ≤ 108) — the numbers on the pieces. Vasya can only glue the pieces in pairs. Even if the number of a piece is already lucky, Vasya should glue the piece with some other one for it to cou... | Print the single number — the maximum number of lucky tickets that will be able to be restored. Don't forget that every lucky ticket is made of exactly two pieces glued together. | null | null | [{"input": "3\n123 123 99", "output": "1"}, {"input": "6\n1 1 1 23 10 3", "output": "1"}] | 1,300 | ["greedy"] | 21 | [{"input": "3\r\n123 123 99\r\n", "output": "1\r\n"}, {"input": "6\r\n1 1 1 23 10 3\r\n", "output": "1\r\n"}, {"input": "3\r\n43440907 58238452 82582355\r\n", "output": "1\r\n"}, {"input": "4\r\n31450303 81222872 67526764 17516401\r\n", "output": "1\r\n"}, {"input": "5\r\n83280 20492640 21552119 7655071 47966344\r\n", ... | false | stdio | null | true |
641/D | 641 | D | Python 3 | PRETESTS | 2 | 46 | 5,120,000 | 17492579 | def quad(a, b, c):
disc = (b**2-4*a*c)
if disc<0:
disc=0
disc = (disc)**0.5
return ((-b+disc)/2/a, (-b-disc)/2/a)
x = int(input())
y = list(map(float, input().split(' ')))
z = list(map(float, input().split(' ')))
py = [0, y[0]]
for i in range(1, x):
py.append(py[-1]+y[i])
z.reverse()
p... | 20 | 530 | 47,616,000 | 17497190 | def tle():
k=0
while (k>=0):
k+=1
def quad(a, b, c):
disc = (b**2-4*a*c)
if disc<0:
disc=0
disc = (disc)**0.5
return ((-b+disc)/2/a, (-b-disc)/2/a)
x = int(input())
y = list(map(float, input().strip().split(' ')))
z = list(map(float, input().strip().split(' ')))
p... | VK Cup 2016 - Round 2 | CF | 2,016 | 2 | 256 | Little Artem and Random Variable | Little Artyom decided to study probability theory. He found a book with a lot of nice exercises and now wants you to help him with one of them.
Consider two dices. When thrown each dice shows some integer from 1 to n inclusive. For each dice the probability of each outcome is given (of course, their sum is 1), and dif... | First line contains the integer n (1 ≤ n ≤ 100 000) — the number of different values for both dices.
Second line contains an array consisting of n real values with up to 8 digits after the decimal point — probability distribution for max(a, b), the i-th of these values equals to the probability that max(a, b) = i. It... | Output two descriptions of the probability distribution for a on the first line and for b on the second line.
The answer will be considered correct if each value of max(a, b) and min(a, b) probability distribution values does not differ by more than 10 - 6 from ones given in input. Also, probabilities should be non-ne... | null | null | [{"input": "2\n0.25 0.75\n0.75 0.25", "output": "0.5 0.5\n0.5 0.5"}, {"input": "3\n0.125 0.25 0.625\n0.625 0.25 0.125", "output": "0.25 0.25 0.5\n0.5 0.25 0.25"}] | 2,400 | ["dp", "implementation", "math", "probabilities"] | 20 | [{"input": "2\r\n0.25 0.75\r\n0.75 0.25\r\n", "output": "0.5 0.5 \r\n0.5 0.5 \r\n"}, {"input": "3\r\n0.125 0.25 0.625\r\n0.625 0.25 0.125\r\n", "output": "0.25 0.25 0.5 \r\n0.5 0.25 0.25 \r\n"}, {"input": "10\r\n0.01 0.01 0.01 0.01 0.01 0.1 0.2 0.2 0.4 0.05\r\n1.0 0 0 0 0 0 0 0 0 0\r\n", "output": "0.010000000000000009... | false | stdio | import sys
def read_floats(line):
return list(map(float, line.strip().split()))
def main():
input_path, output_path, submission_path = sys.argv[1:4]
with open(input_path) as f:
n = int(f.readline())
max_dist = read_floats(f.readline())
min_dist = read_floats(f.readline())
... | true |
31/D | 31 | D | PyPy 3 | TESTS | 8 | 154 | 2,560,000 | 141688873 | from bisect import bisect_right
from collections import defaultdict
import os
import sys
from io import BytesIO, IOBase
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self... | 33 | 92 | 0 | 189949532 | from bisect import bisect as bt
w,h,n=map(int,input().split())
N,M=max(w,h),10
x=[[0,h]]+[[] for _ in range(N-1)]
y=[[0,w]]+[[] for _ in range(N-1)]
xy=[[0]*(N+1) for _ in range(N+1)]
g=[0]*M+[[]]
xy[0][0]=1
xs,ys={0},{0}
def t(k,p):
for i,j in zip(k,p):
q=bt(i,j)
if q==0 or i[q-1]!=j:
i... | Codeforces Beta Round 31 (Div. 2, Codeforces format) | CF | 2,010 | 2 | 256 | Chocolate | Bob has a rectangular chocolate bar of the size W × H. He introduced a cartesian coordinate system so that the point (0, 0) corresponds to the lower-left corner of the bar, and the point (W, H) corresponds to the upper-right corner. Bob decided to split the bar into pieces by breaking it. Each break is a segment parall... | The first line contains 3 integers W, H and n (1 ≤ W, H, n ≤ 100) — width of the bar, height of the bar and amount of breaks. Each of the following n lines contains four integers xi, 1, yi, 1, xi, 2, yi, 2 — coordinates of the endpoints of the i-th break (0 ≤ xi, 1 ≤ xi, 2 ≤ W, 0 ≤ yi, 1 ≤ yi, 2 ≤ H, or xi, 1 = xi, 2, ... | Output n + 1 numbers — areas of the resulting parts in the increasing order. | null | null | [{"input": "2 2 2\n1 0 1 2\n0 1 1 1", "output": "1 1 2"}, {"input": "2 2 3\n1 0 1 2\n0 1 1 1\n1 1 2 1", "output": "1 1 1 1"}, {"input": "2 4 2\n0 1 2 1\n0 3 2 3", "output": "2 2 4"}] | 2,000 | ["dfs and similar", "implementation"] | 33 | [{"input": "2 2 2\r\n1 0 1 2\r\n0 1 1 1\r\n", "output": "1 1 2 "}, {"input": "2 2 3\r\n1 0 1 2\r\n0 1 1 1\r\n1 1 2 1\r\n", "output": "1 1 1 1 "}, {"input": "2 4 2\r\n0 1 2 1\r\n0 3 2 3\r\n", "output": "2 2 4 "}, {"input": "5 5 3\r\n2 1 2 5\r\n0 1 5 1\r\n4 0 4 1\r\n", "output": "1 4 8 12 "}, {"input": "10 10 4\r\n9 0 9 ... | false | stdio | null | true |
730/G | 730 | G | PyPy 3 | TESTS | 5 | 124 | 0 | 21712850 | #!/usr/bin/env python3
def main():
try:
while True:
n = int(input())
req = [tuple(map(int, input().split())) for i in range(n)]
used = [(req[0][0], req[0][0] + req[0][1])]
print(used[0][0], used[0][1] - 1)
for start, dur in req[1:]:
... | 28 | 77 | 2,867,200 | 211349045 | import sys
input = lambda: sys.stdin.readline().rstrip()
N = int(input())
A = [0,10**18]
for _ in range(N):
s,d = map(int, input().split())
find = False
for i in range(1,len(A),2):
if A[i]>s+d-1 and A[i-1]<s:
print(s,s+d-1)
A.append(s)
A.append(s+d-1)
... | 2016-2017 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred) | ICPC | 2,016 | 2 | 512 | Car Repair Shop | Polycarp starts his own business. Tomorrow will be the first working day of his car repair shop. For now the car repair shop is very small and only one car can be repaired at a given time.
Polycarp is good at marketing, so he has already collected n requests from clients. The requests are numbered from 1 to n in order... | The first line contains integer n (1 ≤ n ≤ 200) — the number of requests from clients.
The following n lines contain requests, one request per line. The i-th request is given as the pair of integers si, di (1 ≤ si ≤ 109, 1 ≤ di ≤ 5·106), where si is the preferred time to start repairing the i-th car, di is the number ... | Print n lines. The i-th line should contain two integers — the start day to repair the i-th car and the finish day to repair the i-th car. | null | null | [{"input": "3\n9 2\n7 3\n2 4", "output": "9 10\n1 3\n4 7"}, {"input": "4\n1000000000 1000000\n1000000000 1000000\n100000000 1000000\n1000000000 1000000", "output": "1000000000 1000999999\n1 1000000\n100000000 100999999\n1000001 2000000"}] | 1,600 | ["implementation"] | 28 | [{"input": "3\r\n9 2\r\n7 3\r\n2 4\r\n", "output": "9 10\r\n1 3\r\n4 7\r\n"}, {"input": "4\r\n1000000000 1000000\r\n1000000000 1000000\r\n100000000 1000000\r\n1000000000 1000000\r\n", "output": "1000000000 1000999999\r\n1 1000000\r\n100000000 100999999\r\n1000001 2000000\r\n"}, {"input": "1\r\n1 1\r\n", "output": "1 1\... | false | stdio | null | true |
583/A | 583 | A | Python 3 | PRETESTS | 6 | 46 | 0 | 13369364 | n=int(input(''))
d1={}
d2={}
arr=[]
for i in range(0,n*n):
s=input('')
a=s.split(" ")
x=int(a[0])
y=int(a[1])
try:
d1[x]=d1[x]+1
except:
try:
d2[y]=d2[y]+1
except:
arr.append(str(i+1))
d1[x]=1
d2[y]=1
arr.sort()
print(' '.j... | 39 | 46 | 0 | 151474506 | n=int(input())
r=[0]*(n+1)
c=[0]*(n+1)
for i in range(n*n):
a,b=map(int,input().split())
if r[a]==0 and c[b]==0:
print(i+1)
r[a]=c[b]=1 | Codeforces Round 323 (Div. 2) | CF | 2,015 | 1 | 256 | Asphalting Roads | City X consists of n vertical and n horizontal infinite roads, forming n × n intersections. Roads (both vertical and horizontal) are numbered from 1 to n, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphal... | The first line contains integer n (1 ≤ n ≤ 50) — the number of vertical and horizontal roads in the city.
Next n2 lines contain the order of intersections in the schedule. The i-th of them contains two numbers hi, vi (1 ≤ hi, vi ≤ n), separated by a space, and meaning that the intersection that goes i-th in the timeta... | In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1. | null | In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road;
2. On the second day the brigade of the workers comes to the int... | [{"input": "2\n1 1\n1 2\n2 1\n2 2", "output": "1 4"}, {"input": "1\n1 1", "output": "1"}] | 1,000 | ["implementation"] | 39 | [{"input": "2\r\n1 1\r\n1 2\r\n2 1\r\n2 2\r\n", "output": "1 4 \r\n"}, {"input": "1\r\n1 1\r\n", "output": "1 \r\n"}, {"input": "2\r\n1 1\r\n2 2\r\n1 2\r\n2 1\r\n", "output": "1 2 \r\n"}, {"input": "2\r\n1 2\r\n2 2\r\n2 1\r\n1 1\r\n", "output": "1 3 \r\n"}, {"input": "3\r\n2 2\r\n1 2\r\n3 2\r\n3 3\r\n1 1\r\n2 3\r\n1 3\... | false | stdio | null | true |
245/A | 245 | A | Python 3 | TESTS | 9 | 92 | 0 | 140483027 | if __name__=="__main__":
n = int(input())
count_dict = {}
for i in range(n):
t,x,y = [int(x) for x in input().split()]
val = list(count_dict.get(t,[0,0]))
val[0] += x
val[1] += y
count_dict[t] = val
for key,val in count_dict.items():
... | 13 | 62 | 0 | 139623098 | tc=int(input())
count1=0
count2=0
count3=0
count4=0
for i in range(tc):
t,x,y=list(map(int,input().split()))
if t==1:
count1+=x
count2+=10
else:
count3+=x
count4+=10
if count1>=(count2/2):
print('LIVE')
else:
print('DEAD')
if count3>=(count4/2):
print('LIVE')
else... | CROC-MBTU 2012, Elimination Round (ACM-ICPC) | ICPC | 2,012 | 2 | 256 | System Administrator | Polycarpus is a system administrator. There are two servers under his strict guidance — a and b. To stay informed about the servers' performance, Polycarpus executes commands "ping a" and "ping b". Each ping command sends exactly ten packets to the server specified in the argument of the command. Executing a program re... | The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of commands Polycarpus has fulfilled. Each of the following n lines contains three integers — the description of the commands. The i-th of these lines contains three space-separated integers ti, xi, yi (1 ≤ ti ≤ 2; xi, yi ≥ 0; xi + yi = 10). If ti =... | In the first line print string "LIVE" (without the quotes) if server a is "alive", otherwise print "DEAD" (without the quotes).
In the second line print the state of server b in the similar format. | null | Consider the first test case. There 10 packets were sent to server a, 5 of them reached it. Therefore, at least half of all packets sent to this server successfully reached it through the network. Overall there were 10 packets sent to server b, 6 of them reached it. Therefore, at least half of all packets sent to this ... | [{"input": "2\n1 5 5\n2 6 4", "output": "LIVE\nLIVE"}, {"input": "3\n1 0 10\n2 0 10\n1 10 0", "output": "LIVE\nDEAD"}] | 800 | ["implementation"] | 23 | [{"input": "2\r\n1 5 5\r\n2 6 4\r\n", "output": "LIVE\r\nLIVE\r\n"}, {"input": "3\r\n1 0 10\r\n2 0 10\r\n1 10 0\r\n", "output": "LIVE\r\nDEAD\r\n"}, {"input": "10\r\n1 3 7\r\n2 4 6\r\n1 2 8\r\n2 5 5\r\n2 10 0\r\n2 10 0\r\n1 8 2\r\n2 2 8\r\n2 10 0\r\n1 1 9\r\n", "output": "DEAD\r\nLIVE\r\n"}, {"input": "11\r\n1 8 2\r\n1... | false | stdio | null | true |
358/D | 358 | D | Python 3 | TESTS | 4 | 62 | 307,200 | 4892411 | n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
A = {0 : a[0]}
B = {0 : b[0]}
for i in range(1, n):
A[i] = max(A[i-1] + b[i], B[i-1] + a[i]) # max(fed left, fed right)
B[i] = max(A[i-1] + c[i], B[i-1] + b[i]) # max(fed left and right, fed ... | 29 | 62 | 512,000 | 5035569 | n = int( input() )
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
c = [int(x) for x in input().split()]
DD = a[0]
DP = b[0]
PD = a[0]
PP = b[0]
for i in range(1,n):
tDD = max(DP+a[i], PP+a[i])
tPD = max(PD+b[i], DD+b[i])
tDP = max(DP+b[i], PP+b[i])
tPP = max(PD+c[i], DD+c[i])
DD = ... | Codeforces Round 208 (Div. 2) | CF | 2,013 | 2 | 256 | Dima and Hares | Dima liked the present he got from Inna very much. He liked the present he got from Seryozha even more.
Dima felt so grateful to Inna about the present that he decided to buy her n hares. Inna was very happy. She lined up the hares in a row, numbered them from 1 to n from left to right and started feeding them with ca... | The first line of the input contains integer n (1 ≤ n ≤ 3000) — the number of hares. Then three lines follow, each line has n integers. The first line contains integers a1 a2 ... an. The second line contains b1, b2, ..., bn. The third line contains c1, c2, ..., cn. The following limits are fulfilled: 0 ≤ ai, bi, ci ≤ 1... | In a single line, print the maximum possible total joy of the hares Inna can get by feeding them. | null | null | [{"input": "4\n1 2 3 4\n4 3 2 1\n0 1 1 0", "output": "13"}, {"input": "7\n8 5 7 6 1 8 9\n2 7 9 5 4 3 1\n2 3 3 4 1 1 3", "output": "44"}, {"input": "3\n1 1 1\n1 2 1\n1 1 1", "output": "4"}] | 1,800 | ["dp", "greedy"] | 29 | [{"input": "4\r\n1 2 3 4\r\n4 3 2 1\r\n0 1 1 0\r\n", "output": "13\r\n"}, {"input": "7\r\n8 5 7 6 1 8 9\r\n2 7 9 5 4 3 1\r\n2 3 3 4 1 1 3\r\n", "output": "44\r\n"}, {"input": "3\r\n1 1 1\r\n1 2 1\r\n1 1 1\r\n", "output": "4\r\n"}, {"input": "7\r\n1 3 8 9 3 4 4\r\n6 0 6 6 1 8 4\r\n9 6 3 7 8 8 2\r\n", "output": "42\r\n"}... | false | stdio | null | true |
1003/D | 1003 | D | PyPy 3-64 | TESTS | 3 | 451 | 23,552,000 | 193238672 | import sys
input = sys.stdin.readline
from math import log2
n, q = map(int, input().split())
d = [0]*31
w = list(map(int, input().split()))
for i in w:
d[int(log2(i))] += 1
x = [0]*31
for i in range(31):
if d[i] > 0:
x[i] = 1
else:
a, b = 0, 2
for j in range(i-1, -1, -1):
... | 31 | 358 | 25,497,600 | 220844200 | # https://codeforces.com/contest/1003
import sys
input = lambda: sys.stdin.readline().rstrip() # faster!
n, q = map(int, input().split())
a = list(map(int, input().split()))
N_BITS = (2 * 10 ** 9).bit_length() + 1
coins = [0] * N_BITS
for x in a:
coins[x.bit_length() - 1] += 1
for _ in range(q):
ans = 0
... | Codeforces Round 494 (Div. 3) | ICPC | 2,018 | 2 | 256 | Coins and Queries | Polycarp has $$$n$$$ coins, the value of the $$$i$$$-th coin is $$$a_i$$$. It is guaranteed that all the values are integer powers of $$$2$$$ (i.e. $$$a_i = 2^d$$$ for some non-negative integer number $$$d$$$).
Polycarp wants to know answers on $$$q$$$ queries. The $$$j$$$-th query is described as integer number $$$b_... | The first line of the input contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of coins and the number of queries.
The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ — values of coins ($$$1 \le a_i \le 2 \cdot 10^9$$$). It is guaranteed that all $$... | Print $$$q$$$ integers $$$ans_j$$$. The $$$j$$$-th integer must be equal to the answer on the $$$j$$$-th query. If Polycarp can't obtain the value $$$b_j$$$ the answer to the $$$j$$$-th query is -1. | null | null | [{"input": "5 4\n2 4 8 2 4\n8\n5\n14\n10", "output": "1\n-1\n3\n2"}] | 1,600 | ["greedy"] | 31 | [{"input": "5 4\r\n2 4 8 2 4\r\n8\r\n5\r\n14\r\n10\r\n", "output": "1\r\n-1\r\n3\r\n2\r\n"}, {"input": "3 3\r\n1 1 1\r\n1\r\n2\r\n3\r\n", "output": "1\r\n2\r\n3\r\n"}, {"input": "4 1\r\n2 4 16 32\r\n14\r\n", "output": "-1\r\n"}, {"input": "1 10\r\n8\r\n1\r\n2\r\n3\r\n4\r\n5\r\n6\r\n7\r\n8\r\n9\r\n16\r\n", "output": "-1... | false | stdio | null | true |
1003/D | 1003 | D | PyPy 3-64 | TESTS | 3 | 420 | 23,347,200 | 207516276 | import sys
input = lambda: sys.stdin.readline().rstrip()
from collections import Counter,defaultdict
N,Q = map(int, input().split())
A = list(map(int, input().split()))
cnt = [0]*32
for a in A:
for i in range(32):
if a&(1<<i):
cnt[i]+=1
#print(cnt)
for _ in range(Q):
b = int(input())
a... | 31 | 358 | 26,214,400 | 193240131 | import sys
input = sys.stdin.readline
from math import log2
n, q = map(int, input().split())
d = [0]*31
w = list(map(int, input().split()))
for i in w:
d[int(log2(i))] += 1
x = list(2**i for i in range(31))
for _ in range(q):
a = int(input())
c = 0
for j in range(30, -1, -1):
b = min(d[j], a//... | Codeforces Round 494 (Div. 3) | ICPC | 2,018 | 2 | 256 | Coins and Queries | Polycarp has $$$n$$$ coins, the value of the $$$i$$$-th coin is $$$a_i$$$. It is guaranteed that all the values are integer powers of $$$2$$$ (i.e. $$$a_i = 2^d$$$ for some non-negative integer number $$$d$$$).
Polycarp wants to know answers on $$$q$$$ queries. The $$$j$$$-th query is described as integer number $$$b_... | The first line of the input contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of coins and the number of queries.
The second line of the input contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ — values of coins ($$$1 \le a_i \le 2 \cdot 10^9$$$). It is guaranteed that all $$... | Print $$$q$$$ integers $$$ans_j$$$. The $$$j$$$-th integer must be equal to the answer on the $$$j$$$-th query. If Polycarp can't obtain the value $$$b_j$$$ the answer to the $$$j$$$-th query is -1. | null | null | [{"input": "5 4\n2 4 8 2 4\n8\n5\n14\n10", "output": "1\n-1\n3\n2"}] | 1,600 | ["greedy"] | 31 | [{"input": "5 4\r\n2 4 8 2 4\r\n8\r\n5\r\n14\r\n10\r\n", "output": "1\r\n-1\r\n3\r\n2\r\n"}, {"input": "3 3\r\n1 1 1\r\n1\r\n2\r\n3\r\n", "output": "1\r\n2\r\n3\r\n"}, {"input": "4 1\r\n2 4 16 32\r\n14\r\n", "output": "-1\r\n"}, {"input": "1 10\r\n8\r\n1\r\n2\r\n3\r\n4\r\n5\r\n6\r\n7\r\n8\r\n9\r\n16\r\n", "output": "-1... | false | stdio | null | true |
364/D | 364 | D | PyPy 3 | TESTS | 4 | 2,776 | 124,108,800 | 71192608 | from random import randrange
from sys import stdin
def read_integers():
return list(map(int, stdin.readline().strip().split()))
def decompose(num, visited):
if num in visited:
return []
factors = []
i = 2
while i * i <= num:
quo, rem = divmod(num, i)
if i not in visited ... | 115 | 3,900 | 133,734,400 | 179459001 | import math
from random import randint
from bisect import bisect_left
test = False
if test:
n = randint(100000,100000)
a = [randint(1,100000000) for _ in range(n)]
#print('n =', n)
#print('a =', a)
else:
n = int(input())
a=[*map(int,input().split())]
dr = 1
for x in range(9):
... | Codeforces Round 213 (Div. 1) | CF | 2,013 | 4 | 256 | Ghd | John Doe offered his sister Jane Doe find the gcd of some set of numbers a.
Gcd is a positive integer g, such that all number from the set are evenly divisible by g and there isn't such g' (g' > g), that all numbers of the set are evenly divisible by g'.
Unfortunately Jane couldn't cope with the task and John offered... | The first line contains an integer n (1 ≤ n ≤ 106) showing how many numbers are in set a. The second line contains space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1012). Please note, that given set can contain equal numbers.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is pref... | Print a single integer g — the Ghd of set a. | null | null | [{"input": "6\n6 2 3 4 5 6", "output": "3"}, {"input": "5\n5 5 6 10 15", "output": "5"}] | 2,900 | ["brute force", "math", "probabilities"] | 115 | [{"input": "6\r\n6 2 3 4 5 6\r\n", "output": "3\r\n"}, {"input": "5\r\n5 5 6 10 15\r\n", "output": "5\r\n"}, {"input": "100\r\n32 40 7 3 7560 21 7560 7560 10 12 3 7560 7560 7560 7560 5 7560 7560 6 7560 7560 7560 35 7560 18 7560 7560 7560 7560 7560 48 2 7 25 7560 2 2 49 7560 7560 15 16 7560 7560 2 7560 27 7560 7560 7560... | false | stdio | null | true |
363/D | 363 | D | PyPy 3-64 | TESTS | 2 | 108 | 5,836,800 | 182267435 | import sys
import threading
input=sys.stdin.readline
from collections import Counter,defaultdict,deque
from heapq import heappush,heappop,heapify
#threading.stack_size(10**8)
#sys.setrecursionlimit(10**6)
def ri():return int(input())
def rs():return input()
def rl():return list(map(int,input().split()))
def rls():retu... | 34 | 124 | 17,715,200 | 188178762 | n, m, a = map(int, input().split())
B = [int(x) for x in input().split()]
P = [int(x) for x in input().split()]
B.sort()
P.sort()
def f(a):
need = 0
for i in range(a):
need += max(0, P[i] - B[n - a + i])
return need
l = 0
r = min(n + 1, m + 1)
while r - l > 1:
mid = (l + r) // 2
if f(mid) ... | Codeforces Round 211 (Div. 2) | CF | 2,013 | 1 | 256 | Renting Bikes | A group of n schoolboys decided to ride bikes. As nobody of them has a bike, the boys need to rent them.
The renting site offered them m bikes. The renting price is different for different bikes, renting the j-th bike costs pj rubles.
In total, the boys' shared budget is a rubles. Besides, each of them has his own pe... | The first line of the input contains three integers n, m and a (1 ≤ n, m ≤ 105; 0 ≤ a ≤ 109). The second line contains the sequence of integers b1, b2, ..., bn (1 ≤ bi ≤ 104), where bi is the amount of the i-th boy's personal money. The third line contains the sequence of integers p1, p2, ..., pm (1 ≤ pj ≤ 109), where ... | Print two integers r and s, where r is the maximum number of schoolboys that can rent a bike and s is the minimum total personal money needed to rent r bikes. If the schoolchildren cannot rent any bikes, then r = s = 0. | null | In the first sample both schoolchildren can rent a bike. For instance, they can split the shared budget in half (5 rubles each). In this case one of them will have to pay 1 ruble from the personal money and the other one will have to pay 2 rubles from the personal money. In total, they spend 3 rubles of their personal ... | [{"input": "2 2 10\n5 5\n7 6", "output": "2 3"}, {"input": "4 5 2\n8 1 1 2\n6 3 7 5 2", "output": "3 8"}] | 1,800 | ["binary search", "greedy"] | 34 | [{"input": "2 2 10\r\n5 5\r\n7 6\r\n", "output": "2 3\r\n"}, {"input": "4 5 2\r\n8 1 1 2\r\n6 3 7 5 2\r\n", "output": "3 8\r\n"}, {"input": "1 1 2\r\n1\r\n2\r\n", "output": "1 0\r\n"}, {"input": "4 1 1\r\n3 2 3 2\r\n3\r\n", "output": "1 2\r\n"}, {"input": "1 4 1\r\n3\r\n2 4 5 5\r\n", "output": "1 1\r\n"}, {"input": "3 ... | false | stdio | null | true |
363/D | 363 | D | PyPy 3-64 | TESTS | 0 | 30 | 0 | 199084354 | n,m,budget = map(int, input().split())
money = sorted(list(map(int, input().split())))
costs = sorted(list(map(int, input().split())))
l = 0
r = min(n,m)
mid = (l+r)//2
ans,cost = 0,0
def check(mid):
global money, costs, budget
cur_spending = 0
for i in range(mid):
cur_spending += max(0,costs[i]-mon... | 34 | 140 | 20,684,800 | 221831984 | import sys
input = lambda: sys.stdin.readline().rstrip()
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations
from bisect import *
from heapq import *
from math import ceil,gcd,lcm,floor,comb
alph = 'abcdefghijklmnopqrstuvwxyz'
#pow(x,mod-2,mod)
def find(K):
ans = 0
... | Codeforces Round 211 (Div. 2) | CF | 2,013 | 1 | 256 | Renting Bikes | A group of n schoolboys decided to ride bikes. As nobody of them has a bike, the boys need to rent them.
The renting site offered them m bikes. The renting price is different for different bikes, renting the j-th bike costs pj rubles.
In total, the boys' shared budget is a rubles. Besides, each of them has his own pe... | The first line of the input contains three integers n, m and a (1 ≤ n, m ≤ 105; 0 ≤ a ≤ 109). The second line contains the sequence of integers b1, b2, ..., bn (1 ≤ bi ≤ 104), where bi is the amount of the i-th boy's personal money. The third line contains the sequence of integers p1, p2, ..., pm (1 ≤ pj ≤ 109), where ... | Print two integers r and s, where r is the maximum number of schoolboys that can rent a bike and s is the minimum total personal money needed to rent r bikes. If the schoolchildren cannot rent any bikes, then r = s = 0. | null | In the first sample both schoolchildren can rent a bike. For instance, they can split the shared budget in half (5 rubles each). In this case one of them will have to pay 1 ruble from the personal money and the other one will have to pay 2 rubles from the personal money. In total, they spend 3 rubles of their personal ... | [{"input": "2 2 10\n5 5\n7 6", "output": "2 3"}, {"input": "4 5 2\n8 1 1 2\n6 3 7 5 2", "output": "3 8"}] | 1,800 | ["binary search", "greedy"] | 34 | [{"input": "2 2 10\r\n5 5\r\n7 6\r\n", "output": "2 3\r\n"}, {"input": "4 5 2\r\n8 1 1 2\r\n6 3 7 5 2\r\n", "output": "3 8\r\n"}, {"input": "1 1 2\r\n1\r\n2\r\n", "output": "1 0\r\n"}, {"input": "4 1 1\r\n3 2 3 2\r\n3\r\n", "output": "1 2\r\n"}, {"input": "1 4 1\r\n3\r\n2 4 5 5\r\n", "output": "1 1\r\n"}, {"input": "3 ... | false | stdio | null | true |
362/A | 362 | A | Python 3 | PRETESTS | 6 | 46 | 307,200 | 5106533 | def main():
n = int(input())
out = ""
for t in range(n):
knights = [0 for i in range(8)]
valid = [False for i in range(8)]
for i in range(8):
line = input()
#print()
for j in range(8):
#print(get(i, j), end="")
if line[j] == '.':
valid[get(i, j)] = True
... | 45 | 62 | 307,200 | 5102996 | N = int(input())
for i in range(N):
if i > 0:
input()
Pos = []
for i in range(8):
S = input().strip()
for j in range(8):
if S[j] == "K":
Pos.append((i, j))
Dx, Dy = abs(Pos[0][0] - Pos[1][0]), abs(Pos[0][1] - Pos[1][1])
if (Dx == 0 or Dx == 4) and ... | Codeforces Round 212 (Div. 2) | CF | 2,013 | 1 | 256 | Two Semiknights Meet | A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the le... | The first line contains number t (1 ≤ t ≤ 50) — the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matrix consists of characters ".", "#", "K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed t... | For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO" otherwise. | null | Consider the first board from the sample. We will assume the rows and columns of the matrix to be numbered 1 through 8 from top to bottom and from left to right, correspondingly. The knights can meet, for example, in square (2, 7). The semiknight from square (4, 1) goes to square (2, 3) and the semiknight goes from squ... | [{"input": "2\n........\n........\n......#.\nK..##..#\n.......#\n...##..#\n......#.\nK.......\n........\n........\n..#.....\n..#..#..\n..####..\n...##...\n........\n....K#K#", "output": "YES\nNO"}] | 1,500 | ["greedy", "math"] | 45 | [{"input": "2\r\n........\r\n........\r\n......#.\r\nK..##..#\r\n.......#\r\n...##..#\r\n......#.\r\nK.......\r\n\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n...##...\r\n........\r\n....K#K#\r\n", "output": "YES\r\nNO\r\n"}, {"input": "3\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n... | false | stdio | null | true |
362/A | 362 | A | Python 3 | TESTS | 6 | 46 | 0 | 5237853 | t = ['12341234', '56785678', '34123412', '78567856', '12341234', '56785678', '34123412', '78567856']
def f():
global t
y, x = False, False
for i in range(8):
p = input()
if y: continue
for j in range(8):
if p[j] == 'K':
if x: ans = 'YNEOS'[x != t[i][j] :: ... | 45 | 62 | 307,200 | 5104091 | MOVS = [(2,-2),(-2,2),(-2,-2),(2,2)]
def check(a):
return 0<=a<8
set1 = set()
set2 = set()
dic1 = dict()
dic2 = dict()
def cango1(matrix,pos,lap):
for dx,dy in MOVS:
nx,ny = dx+pos[0],dy+pos[1]
if not check (nx) or not check(ny):
continue
if (nx,ny) in set1:
conti... | Codeforces Round 212 (Div. 2) | CF | 2,013 | 1 | 256 | Two Semiknights Meet | A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the le... | The first line contains number t (1 ≤ t ≤ 50) — the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matrix consists of characters ".", "#", "K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed t... | For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO" otherwise. | null | Consider the first board from the sample. We will assume the rows and columns of the matrix to be numbered 1 through 8 from top to bottom and from left to right, correspondingly. The knights can meet, for example, in square (2, 7). The semiknight from square (4, 1) goes to square (2, 3) and the semiknight goes from squ... | [{"input": "2\n........\n........\n......#.\nK..##..#\n.......#\n...##..#\n......#.\nK.......\n........\n........\n..#.....\n..#..#..\n..####..\n...##...\n........\n....K#K#", "output": "YES\nNO"}] | 1,500 | ["greedy", "math"] | 45 | [{"input": "2\r\n........\r\n........\r\n......#.\r\nK..##..#\r\n.......#\r\n...##..#\r\n......#.\r\nK.......\r\n\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n...##...\r\n........\r\n....K#K#\r\n", "output": "YES\r\nNO\r\n"}, {"input": "3\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n... | false | stdio | null | true |
362/A | 362 | A | Python 3 | PRETESTS | 4 | 62 | 307,200 | 5098801 | N = int(input())
for i in range(N):
if i > 0:
input()
Pos = []
for i in range(8):
S = input().strip()
for j in range(8):
if S[j] == "K":
Pos.append((i, j))
A, B = Pos
if A[0] % 4 + A[1] % 4 == B[0] % 4 + B[1] % 4:
print("YES")
else:
... | 45 | 62 | 307,200 | 5107862 | import collections
n = list(map(int, input().split()))[0]
di = ((2, 2), (-2, 2), (-2, -2), (2, -2))
def bfs(graph, x,y,tx, ty):
vis = set()
queue = collections.deque()
queue.append((x,y))
vis.add((x,y))
dis = {}
dis[(x,y)] = 0
while queue:
elem = queue.popleft()
#print (... | Codeforces Round 212 (Div. 2) | CF | 2,013 | 1 | 256 | Two Semiknights Meet | A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the le... | The first line contains number t (1 ≤ t ≤ 50) — the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matrix consists of characters ".", "#", "K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed t... | For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO" otherwise. | null | Consider the first board from the sample. We will assume the rows and columns of the matrix to be numbered 1 through 8 from top to bottom and from left to right, correspondingly. The knights can meet, for example, in square (2, 7). The semiknight from square (4, 1) goes to square (2, 3) and the semiknight goes from squ... | [{"input": "2\n........\n........\n......#.\nK..##..#\n.......#\n...##..#\n......#.\nK.......\n........\n........\n..#.....\n..#..#..\n..####..\n...##...\n........\n....K#K#", "output": "YES\nNO"}] | 1,500 | ["greedy", "math"] | 45 | [{"input": "2\r\n........\r\n........\r\n......#.\r\nK..##..#\r\n.......#\r\n...##..#\r\n......#.\r\nK.......\r\n\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n...##...\r\n........\r\n....K#K#\r\n", "output": "YES\r\nNO\r\n"}, {"input": "3\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n... | false | stdio | null | true |
362/A | 362 | A | Python 3 | TESTS | 4 | 46 | 204,800 | 5132539 | def solve(m, x,y , w,z):
for i in range(8):
for j in range(8):
if m[i][j]:
a, pa = movePossible(x,y , i,j)
b, pb = movePossible(w,z , i,j)
if a and b and pa==pb:
return True
return False
def movePossible(x,y , w,z):
a = z-w
b = y-z
pos=False
ka=a//2
kb=b//2
if a%2==0 and b%2==0 and (ka+k... | 45 | 62 | 307,200 | 5108178 | import sys
import collections
class GetOutOfLoop(Exception):
pass
if __name__ == "__main__":
n_cases = int(sys.stdin.readline())
for case in range(n_cases):
board = [list(sys.stdin.readline().rstrip()) for i in range(8)]
knight_init_loc = [None, None]
knight = 0
for curr... | Codeforces Round 212 (Div. 2) | CF | 2,013 | 1 | 256 | Two Semiknights Meet | A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the le... | The first line contains number t (1 ≤ t ≤ 50) — the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matrix consists of characters ".", "#", "K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed t... | For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO" otherwise. | null | Consider the first board from the sample. We will assume the rows and columns of the matrix to be numbered 1 through 8 from top to bottom and from left to right, correspondingly. The knights can meet, for example, in square (2, 7). The semiknight from square (4, 1) goes to square (2, 3) and the semiknight goes from squ... | [{"input": "2\n........\n........\n......#.\nK..##..#\n.......#\n...##..#\n......#.\nK.......\n........\n........\n..#.....\n..#..#..\n..####..\n...##...\n........\n....K#K#", "output": "YES\nNO"}] | 1,500 | ["greedy", "math"] | 45 | [{"input": "2\r\n........\r\n........\r\n......#.\r\nK..##..#\r\n.......#\r\n...##..#\r\n......#.\r\nK.......\r\n\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n...##...\r\n........\r\n....K#K#\r\n", "output": "YES\r\nNO\r\n"}, {"input": "3\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n... | false | stdio | null | true |
160/C | 160 | C | PyPy 3-64 | TESTS | 2 | 122 | 0 | 224681061 | input1 = input().split(' ')
n = int(input1[0])
x = int(input1[1])
lista = set()
input1 = input().split(' ')
for e in input1:
lista.add(int(e))
order = sorted(lista)
result = []
length = len(order)
for i in range(length):
for j in range(length):
result.append((order[i], order[j]))
print(result[x - 1]... | 98 | 216 | 16,076,800 | 219016744 | import sys
input = lambda: sys.stdin.readline().rstrip()
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations
from bisect import *
from heapq import *
from math import ceil,gcd,lcm,floor,comb
alph = 'abcdefghijklmnopqrstuvwxyz'
#pow(x,mod-2,mod)
N,K = map(int,input().split(... | Codeforces Round 111 (Div. 2) | CF | 2,012 | 1 | 256 | Find Pair | You've got another problem dealing with arrays. Let's consider an arbitrary sequence containing n (not necessarily different) integers a1, a2, ..., an. We are interested in all possible pairs of numbers (ai, aj), (1 ≤ i, j ≤ n). In other words, let's consider all n2 pairs of numbers, picked from the given array.
For e... | The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ n2). The second line contains the array containing n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109). The numbers in the array can coincide. All numbers are separated with spaces.
Please do not use the %lld specificator to read or write 64-bit integers in... | In the single line print two numbers — the sought k-th pair. | null | In the first sample the sorted sequence for the given array looks as: (1, 1), (1, 2), (2, 1), (2, 2). The 4-th of them is pair (2, 2).
The sorted sequence for the array from the second sample is given in the statement. The 2-nd pair there is (1, 3). | [{"input": "2 4\n2 1", "output": "2 2"}, {"input": "3 2\n3 1 5", "output": "1 3"}] | 1,700 | ["implementation", "math", "sortings"] | 98 | [{"input": "2 4\r\n2 1\r\n", "output": "2 2\r\n"}, {"input": "3 2\r\n3 1 5\r\n", "output": "1 3\r\n"}, {"input": "3 3\r\n1 1 2\r\n", "output": "1 1\r\n"}, {"input": "1 1\r\n-4\r\n", "output": "-4 -4\r\n"}, {"input": "3 7\r\n5 4 3\r\n", "output": "5 3\r\n"}, {"input": "3 6\r\n10 1 3\r\n", "output": "3 10\r\n"}, {"input"... | false | stdio | null | true |
484/D | 484 | D | Python 3 | TESTS | 6 | 46 | 0 | 137777319 | m=int(input())
n=list(map(int,input().split()))
n.sort()
ret=0
for i in range(m//2):
ret+=(n[m-1-i]-n[i])
print(ret) | 71 | 1,435 | 92,364,800 | 114299610 | import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
dp = [0] * n
pos, neg = -10 ** 18, -10 ** 18
for i, x in enumerate(a):
prv = 0 if i == 0 else dp[i - 1]
dp[i] = max(neg + x, pos - x, prv)
pos = max(pos, prv + x)
neg = max(neg, prv - x)
print(dp[-1]) | Codeforces Round 276 (Div. 1) | CF | 2,014 | 2 | 256 | Kindergarten | In a kindergarten, the children are being divided into groups. The teacher put the children in a line and associated each child with his or her integer charisma value. Each child should go to exactly one group. Each group should be a nonempty segment of consecutive children of a line. A group's sociability is the maxim... | The first line contains integer n — the number of children in the line (1 ≤ n ≤ 106).
The second line contains n integers ai — the charisma of the i-th child ( - 109 ≤ ai ≤ 109). | Print the maximum possible total sociability of all groups. | null | In the first test sample one of the possible variants of an division is following: the first three children form a group with sociability 2, and the two remaining children form a group with sociability 1.
In the second test sample any division leads to the same result, the sociability will be equal to 0 in each group. | [{"input": "5\n1 2 3 1 2", "output": "3"}, {"input": "3\n3 3 3", "output": "0"}] | 2,400 | ["data structures", "dp", "greedy"] | 71 | [{"input": "5\r\n1 2 3 1 2\r\n", "output": "3\r\n"}, {"input": "3\r\n3 3 3\r\n", "output": "0\r\n"}, {"input": "1\r\n0\r\n", "output": "0\r\n"}, {"input": "2\r\n-1000000000 1000000000\r\n", "output": "2000000000\r\n"}, {"input": "4\r\n1 4 2 3\r\n", "output": "4\r\n"}, {"input": "4\r\n23 5 7 1\r\n", "output": "24\r\n"},... | false | stdio | null | true |
583/A | 583 | A | Python 3 | TESTS | 6 | 62 | 0 | 20455157 | n = int(input())
asphaltedH = set()
asphaltedV = set()
days = []
for i in range(n * n):
h, v = [int(i) for i in input().split()]
if h not in asphaltedH and v not in asphaltedV:
days.append(str(i + 1))
asphaltedH.add(h)
asphaltedV.add(v)
print(" ".join(sorted(days))) | 39 | 46 | 0 | 154961547 | A = []
B = []
C = []
s=0
for t in range(int(input())**2):
a,b=list(map(int,input().split()))
if a not in A and b not in B:
A.append(a)
B.append(b)
C.append(str(t+1))
print(" ".join(C)) | Codeforces Round 323 (Div. 2) | CF | 2,015 | 1 | 256 | Asphalting Roads | City X consists of n vertical and n horizontal infinite roads, forming n × n intersections. Roads (both vertical and horizontal) are numbered from 1 to n, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphal... | The first line contains integer n (1 ≤ n ≤ 50) — the number of vertical and horizontal roads in the city.
Next n2 lines contain the order of intersections in the schedule. The i-th of them contains two numbers hi, vi (1 ≤ hi, vi ≤ n), separated by a space, and meaning that the intersection that goes i-th in the timeta... | In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1. | null | In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road;
2. On the second day the brigade of the workers comes to the int... | [{"input": "2\n1 1\n1 2\n2 1\n2 2", "output": "1 4"}, {"input": "1\n1 1", "output": "1"}] | 1,000 | ["implementation"] | 39 | [{"input": "2\r\n1 1\r\n1 2\r\n2 1\r\n2 2\r\n", "output": "1 4 \r\n"}, {"input": "1\r\n1 1\r\n", "output": "1 \r\n"}, {"input": "2\r\n1 1\r\n2 2\r\n1 2\r\n2 1\r\n", "output": "1 2 \r\n"}, {"input": "2\r\n1 2\r\n2 2\r\n2 1\r\n1 1\r\n", "output": "1 3 \r\n"}, {"input": "3\r\n2 2\r\n1 2\r\n3 2\r\n3 3\r\n1 1\r\n2 3\r\n1 3\... | false | stdio | null | true |
583/A | 583 | A | PyPy 3 | TESTS | 6 | 156 | 20,172,800 | 79716915 | a=int(input())
t=[]
u=[]
if a==1:
c= list(set(list(map(int,input().split()))))
c.sort()
print(1)
else:
r=[]
cl=[]
for j in range(1,2*a+1):
c,b=map(int,input().split())
o=0
if c not in r:
o+=1
if b not in cl:
o+=1
if o==2:
... | 39 | 46 | 0 | 155758219 | import sys
input = sys.stdin.readline
n = int(input())
a, b, d = [], [], []
for i in range(n**2):
h, v = input()[:-1].split()
if h not in a and v not in b:
d.append(i+1)
a.append(h)
b.append(v)
print(' '.join(map(str, d))) | Codeforces Round 323 (Div. 2) | CF | 2,015 | 1 | 256 | Asphalting Roads | City X consists of n vertical and n horizontal infinite roads, forming n × n intersections. Roads (both vertical and horizontal) are numbered from 1 to n, and the intersections are indicated by the numbers of the roads that form them.
Sand roads have long been recognized out of date, so the decision was made to asphal... | The first line contains integer n (1 ≤ n ≤ 50) — the number of vertical and horizontal roads in the city.
Next n2 lines contain the order of intersections in the schedule. The i-th of them contains two numbers hi, vi (1 ≤ hi, vi ≤ n), separated by a space, and meaning that the intersection that goes i-th in the timeta... | In the single line print the numbers of the days when road works will be in progress in ascending order. The days are numbered starting from 1. | null | In the sample the brigade acts like that:
1. On the first day the brigade comes to the intersection of the 1-st horizontal and the 1-st vertical road. As none of them has been asphalted, the workers asphalt the 1-st vertical and the 1-st horizontal road;
2. On the second day the brigade of the workers comes to the int... | [{"input": "2\n1 1\n1 2\n2 1\n2 2", "output": "1 4"}, {"input": "1\n1 1", "output": "1"}] | 1,000 | ["implementation"] | 39 | [{"input": "2\r\n1 1\r\n1 2\r\n2 1\r\n2 2\r\n", "output": "1 4 \r\n"}, {"input": "1\r\n1 1\r\n", "output": "1 \r\n"}, {"input": "2\r\n1 1\r\n2 2\r\n1 2\r\n2 1\r\n", "output": "1 2 \r\n"}, {"input": "2\r\n1 2\r\n2 2\r\n2 1\r\n1 1\r\n", "output": "1 3 \r\n"}, {"input": "3\r\n2 2\r\n1 2\r\n3 2\r\n3 3\r\n1 1\r\n2 3\r\n1 3\... | false | stdio | null | true |
228/B | 228 | B | Python 3 | TESTS | 0 | 30 | 0 | 206053604 | """
https://codeforces.com/problemset/problem/228/B
"""
def lit_table():
n, m = [int(x) for x in input().split()]
# res=[[int(x) for x in row] for row in range(n)]
res = [[0]]
for _ in range(n):
r = [0] + [int(x) for x in input().strip()]
res.append(r)
return n + 1, m + 1, res
na... | 46 | 654 | 6,656,000 | 207378818 | def solve(x,y):
ans=0
for i in range(na):
for j in range(ma):
if i+x<nb and i+x>-1 and j+y<mb and j+y>-1:
v1=int(b[i+x][j+y])
v2=int(a[i][j])
ans+=v1*v2
return ans
a,b=[],[]
na,ma=list(map(int,input().split()))
for i in range(na):
p=inp... | Codeforces Round 141 (Div. 2) | CF | 2,012 | 2 | 256 | Two Tables | You've got two rectangular tables with sizes na × ma and nb × mb cells. The tables consist of zeroes and ones. We will consider the rows and columns of both tables indexed starting from 1. Then we will define the element of the first table, located at the intersection of the i-th row and the j-th column, as ai, j; we w... | The first line contains two space-separated integers na, ma (1 ≤ na, ma ≤ 50) — the number of rows and columns in the first table. Then na lines contain ma characters each — the elements of the first table. Each character is either a "0", or a "1".
The next line contains two space-separated integers nb, mb (1 ≤ nb, mb... | Print two space-separated integers x, y (|x|, |y| ≤ 109) — a shift with maximum overlap factor. If there are multiple solutions, print any of them. | null | null | [{"input": "3 2\n01\n10\n00\n2 3\n001\n111", "output": "0 1"}, {"input": "3 3\n000\n010\n000\n1 1\n1", "output": "-1 -1"}] | 1,400 | ["brute force", "implementation"] | 46 | [{"input": "3 2\r\n01\r\n10\r\n00\r\n2 3\r\n001\r\n111\r\n", "output": "0 1\r\n"}, {"input": "3 3\r\n000\r\n010\r\n000\r\n1 1\r\n1\r\n", "output": "-1 -1\r\n"}, {"input": "2 4\r\n1010\r\n0011\r\n5 5\r\n01100\r\n01110\r\n00111\r\n00110\r\n00110\r\n", "output": "1 1\r\n"}, {"input": "3 1\r\n0\r\n1\r\n0\r\n2 2\r\n11\r\n00... | false | stdio | import sys
def main(input_path, output_path, submission_path):
with open(input_path) as f:
lines = [line.strip() for line in f]
# Parse first table
na, ma = map(int, lines[0].split())
a_ones = []
for i in range(na):
line = lines[i + 1]
for j in range(ma):
if lin... | true |
24/A | 24 | A | PyPy 3-64 | TESTS | 4 | 124 | 0 | 208369815 | from collections import defaultdict
n = int(input())
h = defaultdict(list)
start = None
next = None
apath = 0 # going one way
bpath = 0 # going the other
for i in range(n):
a, b, c = map(int, input().split())
if start == None:
start = a
next = b
bpath += c
if a not in h:
h[a... | 21 | 62 | 102,400 | 206179133 | '''
Online Python Compiler.
Code, Compile, Run and Debug python program online.
Write your code in this editor and press "Run" button to execute it.
'''
n=int(input())
l=[]
for _ in range(n):
l.append(list(map(int,input().split())))
d={}
adj=[[] for t in range(n+1)]
fo... | Codeforces Beta Round 24 | ICPC | 2,010 | 2 | 256 | Ring road | Nowadays the one-way traffic is introduced all over the world in order to improve driving safety and reduce traffic jams. The government of Berland decided to keep up with new trends. Formerly all n cities of Berland were connected by n two-way roads in the ring, i. e. each city was connected directly to exactly two ot... | The first line contains integer n (3 ≤ n ≤ 100) — amount of cities (and roads) in Berland. Next n lines contain description of roads. Each road is described by three integers ai, bi, ci (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ ci ≤ 100) — road is directed from city ai to city bi, redirecting the traffic costs ci. | Output single integer — the smallest amount of money the government should spend on the redirecting of roads so that from every city you can get to any other. | null | null | [{"input": "3\n1 3 1\n1 2 1\n3 2 1", "output": "1"}, {"input": "3\n1 3 1\n1 2 5\n3 2 1", "output": "2"}, {"input": "6\n1 5 4\n5 3 8\n2 4 15\n1 6 16\n2 3 23\n4 6 42", "output": "39"}, {"input": "4\n1 2 9\n2 3 8\n3 4 7\n4 1 5", "output": "0"}] | 1,400 | ["graphs"] | 21 | [{"input": "3\r\n1 3 1\r\n1 2 1\r\n3 2 1\r\n", "output": "1\r\n"}, {"input": "3\r\n1 3 1\r\n1 2 5\r\n3 2 1\r\n", "output": "2\r\n"}, {"input": "6\r\n1 5 4\r\n5 3 8\r\n2 4 15\r\n1 6 16\r\n2 3 23\r\n4 6 42\r\n", "output": "39\r\n"}, {"input": "4\r\n1 2 9\r\n2 3 8\r\n3 4 7\r\n4 1 5\r\n", "output": "0\r\n"}, {"input": "5\r... | false | stdio | null | true |
24/A | 24 | A | Python 3 | TESTS | 4 | 186 | 0 | 94170329 | n = int(input())
cnt = [0] * (n + 1)
ss = 0
s1 = 0
for _ in range(n):
aa,bb,cc = list(map(int, input().split()))
ss += cc
if (cnt[aa] == 0):
cnt[aa] += 1
else:
cnt[bb] += 1
s1 += cc
print(min(s1, ss - s1)) | 21 | 92 | 0 | 139442495 | def solve():
n = int(input())
x = 0
y = 0
d = {}
g = [[] for _ in range(n + 1)]
for i in range(n):
a, b, c = map(int, input().split())
d[(a, b)] = 0
d[(b, a)] = c
g[a].append(b)
g[b].append(a)
s = 1
pre = g[1][0]
seen = set()
while s not in seen:
seen.add(s)
nex = g[s][0]
if nex == pre:
nex... | Codeforces Beta Round 24 | ICPC | 2,010 | 2 | 256 | Ring road | Nowadays the one-way traffic is introduced all over the world in order to improve driving safety and reduce traffic jams. The government of Berland decided to keep up with new trends. Formerly all n cities of Berland were connected by n two-way roads in the ring, i. e. each city was connected directly to exactly two ot... | The first line contains integer n (3 ≤ n ≤ 100) — amount of cities (and roads) in Berland. Next n lines contain description of roads. Each road is described by three integers ai, bi, ci (1 ≤ ai, bi ≤ n, ai ≠ bi, 1 ≤ ci ≤ 100) — road is directed from city ai to city bi, redirecting the traffic costs ci. | Output single integer — the smallest amount of money the government should spend on the redirecting of roads so that from every city you can get to any other. | null | null | [{"input": "3\n1 3 1\n1 2 1\n3 2 1", "output": "1"}, {"input": "3\n1 3 1\n1 2 5\n3 2 1", "output": "2"}, {"input": "6\n1 5 4\n5 3 8\n2 4 15\n1 6 16\n2 3 23\n4 6 42", "output": "39"}, {"input": "4\n1 2 9\n2 3 8\n3 4 7\n4 1 5", "output": "0"}] | 1,400 | ["graphs"] | 21 | [{"input": "3\r\n1 3 1\r\n1 2 1\r\n3 2 1\r\n", "output": "1\r\n"}, {"input": "3\r\n1 3 1\r\n1 2 5\r\n3 2 1\r\n", "output": "2\r\n"}, {"input": "6\r\n1 5 4\r\n5 3 8\r\n2 4 15\r\n1 6 16\r\n2 3 23\r\n4 6 42\r\n", "output": "39\r\n"}, {"input": "4\r\n1 2 9\r\n2 3 8\r\n3 4 7\r\n4 1 5\r\n", "output": "0\r\n"}, {"input": "5\r... | false | stdio | null | true |
614/B | 614 | B | Python 3 | TESTS | 22 | 390 | 15,769,600 | 24418695 | from math import log10
n=int(input())
l=input().split()
if '0' in l:
print(0)
exit()
d=-1
for i in range(n):
if not log10(int(l[i])).is_integer():
d=i
print(l[i], end='')
break
if d==-1:
print(1,end='')
for i in range(n):
if i!=d:
print(l[i][1:], end='')
print('') | 32 | 62 | 8,294,400 | 208581974 | # https://codeforces.com/contest/614
import sys
input = lambda: sys.stdin.readline().rstrip() # faster!
n = int(input())
a = list(input().split())
f, z = "1", 0
for x in a:
if x == "0":
print(0)
exit()
else:
i = len(x) - 1
while i >= 0 and x[i] == "0":
i -= 1
... | Codeforces Round 339 (Div. 2) | CF | 2,016 | 0.5 | 256 | Gena's Code | It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not en... | The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers ai without leading zeroes — the number of tanks of the i-th country.
It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these numbe... | Print a single number without leading zeroes — the product of the number of tanks presented by each country. | null | In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.
In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.
In sample 3 number 3 is not beautiful, all others are beautiful. | [{"input": "3\n5 10 1", "output": "50"}, {"input": "4\n1 1 10 11", "output": "110"}, {"input": "5\n0 3 1 100 1", "output": "0"}] | 1,400 | ["implementation", "math"] | 32 | [{"input": "3\r\n5 10 1\r\n", "output": "50"}, {"input": "4\r\n1 1 10 11\r\n", "output": "110"}, {"input": "5\r\n0 3 1 100 1\r\n", "output": "0"}, {"input": "40\r\n10 100 10 1 10 10 100 10 10 100 10 100 100 10 1824868942 100 100 1 10 100 100 10 100 100 10 100 10 1 10 100 100 100 10 1 10 1 10 10 100 100\r\n", "output": ... | false | stdio | null | true |
296/B | 296 | B | Python 3 | TESTS | 3 | 218 | 307,200 | 68549146 | mod = 1000000007
n = int(input())
s1 = input()
s2 = input()
ans = 1
tc = 1
for i in range(n):
if s1[i] == '?':
ans *= 10
ans %= mod
if s2[i] == '?':
ans *= 10
ans %= mod
for i in range(n):
if s1[i] != '?' and s2[i] != '?' and s1[i] > s2[i]:
break
if s1[i] == '?' and s2[i] == '?':
tc *= 55
tc %= mod
... | 38 | 840 | 7,065,600 | 28195911 | from functools import reduce
n, s1, s2, f1, f2 = int(input()), str(input()), str(input()), lambda x: reduce((lambda a, b: (a * b) % 1000000007), x, 1), lambda x: reduce((lambda a, b: a or b), x, False)
print((10 ** sum([(s1[i] == '?') + (s2[i] == '?') for i in range(n)]) - (not f2([s1[i] != '?' and s2[i] != '?' and ord... | Codeforces Round 179 (Div. 2) | CF | 2,013 | 2 | 256 | Yaroslav and Two Strings | Yaroslav thinks that two strings s and w, consisting of digits and having length n are non-comparable if there are two numbers, i and j (1 ≤ i, j ≤ n), such that si > wi and sj < wj. Here sign si represents the i-th digit of string s, similarly, wj represents the j-th digit of string w.
A string's template is a string... | The first line contains integer n (1 ≤ n ≤ 105) — the length of both templates. The second line contains the first template — a string that consists of digits and characters "?". The string's length equals n. The third line contains the second template in the same format. | In a single line print the remainder after dividing the answer to the problem by number 1000000007 (109 + 7). | null | The first test contains no question marks and both strings are incomparable, so the answer is 1.
The second test has no question marks, but the given strings are comparable, so the answer is 0. | [{"input": "2\n90\n09", "output": "1"}, {"input": "2\n11\n55", "output": "0"}, {"input": "5\n?????\n?????", "output": "993531194"}] | 2,000 | ["combinatorics", "dp"] | 38 | [{"input": "2\r\n90\r\n09\r\n", "output": "1\r\n"}, {"input": "2\r\n11\r\n55\r\n", "output": "0\r\n"}, {"input": "5\r\n?????\r\n?????\r\n", "output": "993531194\r\n"}, {"input": "10\r\n104?3?1??3\r\n?1755?1??7\r\n", "output": "91015750\r\n"}, {"input": "10\r\n6276405116\r\n6787?352?9\r\n", "output": "46\r\n"}, {"input"... | false | stdio | null | true |
10/A | 10 | A | Python 3 | TESTS | 5 | 248 | 0 | 43898266 | n, p1, p2, p3, t1, t2 = map(int, input().split())
consump = 0
z = []
k = []
for _ in range(n):
z+=list(map(int, input().split()))
for i in range(1, len(z)):
k.append(z[i] - z[i-1])
for i in range(0, len(k), 2):
consump+= p1*k[i]
for i in range(1, len(k), 2):
if k[i] <= t1:
consump+=p1*k[i]
elif t1 < k[i] <= t2:
... | 30 | 62 | 0 | 223057731 | inputs=list(map(int,input().split()))
n=inputs[0]
Ps=inputs[1:4]
Ts=inputs[4:6]
power=0
times=[]
for i in range(n):
time_i=list(map(int,input().split()))
for time in time_i:
times.append(time)
power+=(time_i[1]-time_i[0])*Ps[0]
for i in range(2,len(times),2):
rest_time=times[i]-times[i-1]
p1... | Codeforces Beta Round 10 | ICPC | 2,010 | 1 | 256 | Power Consumption Calculation | Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes P1 watt per minute. T1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to P2 watt per minute. Finally, after T2 minut... | The first line contains 6 integer numbers n, P1, P2, P3, T1, T2 (1 ≤ n ≤ 100, 0 ≤ P1, P2, P3 ≤ 100, 1 ≤ T1, T2 ≤ 60). The following n lines contain description of Tom's work. Each i-th of these lines contains two space-separated integers li and ri (0 ≤ li < ri ≤ 1440, ri < li + 1 for i < n), which stand for the start a... | Output the answer to the problem. | null | null | [{"input": "1 3 2 1 5 10\n0 10", "output": "30"}, {"input": "2 8 4 2 5 10\n20 30\n50 100", "output": "570"}] | 900 | ["implementation"] | 30 | [{"input": "1 3 2 1 5 10\r\n0 10\r\n", "output": "30"}, {"input": "2 8 4 2 5 10\r\n20 30\r\n50 100\r\n", "output": "570"}, {"input": "3 15 9 95 39 19\r\n873 989\r\n1003 1137\r\n1172 1436\r\n", "output": "8445"}, {"input": "4 73 2 53 58 16\r\n51 52\r\n209 242\r\n281 407\r\n904 945\r\n", "output": "52870"}, {"input": "5 ... | false | stdio | null | true |
362/A | 362 | A | Python 3 | PRETESTS | 6 | 46 | 307,200 | 5107417 | import collections
n = list(map(int, input().split()))[0]
di = ((2, 2), (-2, 2), (-2, -2), (2, -2))
def bfs(graph, x,y,tx, ty):
vis = set()
queue = collections.deque()
queue.append((x,y))
vis.add((x,y))
while queue:
elem = queue.popleft()
#print (elem)
for direction in ... | 45 | 62 | 307,200 | 5108423 | def main():
n = int(input())
out = ""
for t in range(n):
knights = [0 for i in range(16)]
valid = [False for i in range(16)]
for i in range(8):
line = input()
#print()
for j in range(8):
#print(get(i, j), end="\t")
if line[j] != '#':
valid[get(i, j)] = Tr... | Codeforces Round 212 (Div. 2) | CF | 2,013 | 1 | 256 | Two Semiknights Meet | A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the le... | The first line contains number t (1 ≤ t ≤ 50) — the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matrix consists of characters ".", "#", "K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed t... | For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO" otherwise. | null | Consider the first board from the sample. We will assume the rows and columns of the matrix to be numbered 1 through 8 from top to bottom and from left to right, correspondingly. The knights can meet, for example, in square (2, 7). The semiknight from square (4, 1) goes to square (2, 3) and the semiknight goes from squ... | [{"input": "2\n........\n........\n......#.\nK..##..#\n.......#\n...##..#\n......#.\nK.......\n........\n........\n..#.....\n..#..#..\n..####..\n...##...\n........\n....K#K#", "output": "YES\nNO"}] | 1,500 | ["greedy", "math"] | 45 | [{"input": "2\r\n........\r\n........\r\n......#.\r\nK..##..#\r\n.......#\r\n...##..#\r\n......#.\r\nK.......\r\n\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n...##...\r\n........\r\n....K#K#\r\n", "output": "YES\r\nNO\r\n"}, {"input": "3\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n... | false | stdio | null | true |
231/A | 231 | A | Python 3 | TESTS | 4 | 62 | 0 | 231505834 | if __name__ == '__main__':
count = 0
amount_of_lines = input()
for x in range(int(amount_of_lines)):
word = input()
for y in range(len(word)):
if word[y] == ' ':
pass
else:
count += int(word[y])
print(int(count / 3)) | 21 | 62 | 0 | 226085773 | num_of_problems = int(input())
total = 0
for i in range(num_of_problems):
submission = input()
agreed = 0
for char in submission:
if char == "1":
agreed += 1
if agreed >= 2:
total += 1
print(total) | Codeforces Round 143 (Div. 2) | CF | 2,012 | 2 | 256 | Team | One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution.... | The first input line contains a single integer n (1 ≤ n ≤ 1000) — the number of problems in the contest. Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Va... | Print a single integer — the number of problems the friends will implement on the contest. | null | In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't ta... | [{"input": "3\n1 1 0\n1 1 1\n1 0 0", "output": "2"}, {"input": "2\n1 0 0\n0 1 1", "output": "1"}] | 800 | ["brute force", "greedy"] | 21 | [{"input": "3\r\n1 1 0\r\n1 1 1\r\n1 0 0\r\n", "output": "2\r\n"}, {"input": "2\r\n1 0 0\r\n0 1 1\r\n", "output": "1\r\n"}, {"input": "1\r\n1 0 0\r\n", "output": "0\r\n"}, {"input": "2\r\n1 0 0\r\n1 1 1\r\n", "output": "1\r\n"}, {"input": "5\r\n1 0 0\r\n0 1 0\r\n1 1 1\r\n0 0 1\r\n0 0 0\r\n", "output": "1\r\n"}, {"input... | false | stdio | null | true |
10/A | 10 | A | Python 3 | TESTS | 5 | 92 | 0 | 177484103 | [n,p1,p2,p3,t1,t2] = map(int, input().split())
times = []
for x in range(n):
times.append(list(map(int, input().split())))
count = 0
for session in times:
count += p1*(session[1] - session[0])
for index in range(n - 1):
diff = times[index+1][0] - times[index][1]
if diff >= t1:
diff -= t1
... | 30 | 62 | 0 | 228831738 | n, p1, p2, p3, t1, t2 = map(int, input().split())
periods = []
for _ in range(n):
periods.append(list(map(int, input().split())))
c = 0
for i in range(n - 1):
curr = periods[i]
next = periods[i+1]
c += (curr[-1] - curr[0]) * p1
gap = next[0] - curr[-1]
if gap <= t1:
c += gap * p1
elif gap <= t1 + t2:
... | Codeforces Beta Round 10 | ICPC | 2,010 | 1 | 256 | Power Consumption Calculation | Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes P1 watt per minute. T1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to P2 watt per minute. Finally, after T2 minut... | The first line contains 6 integer numbers n, P1, P2, P3, T1, T2 (1 ≤ n ≤ 100, 0 ≤ P1, P2, P3 ≤ 100, 1 ≤ T1, T2 ≤ 60). The following n lines contain description of Tom's work. Each i-th of these lines contains two space-separated integers li and ri (0 ≤ li < ri ≤ 1440, ri < li + 1 for i < n), which stand for the start a... | Output the answer to the problem. | null | null | [{"input": "1 3 2 1 5 10\n0 10", "output": "30"}, {"input": "2 8 4 2 5 10\n20 30\n50 100", "output": "570"}] | 900 | ["implementation"] | 30 | [{"input": "1 3 2 1 5 10\r\n0 10\r\n", "output": "30"}, {"input": "2 8 4 2 5 10\r\n20 30\r\n50 100\r\n", "output": "570"}, {"input": "3 15 9 95 39 19\r\n873 989\r\n1003 1137\r\n1172 1436\r\n", "output": "8445"}, {"input": "4 73 2 53 58 16\r\n51 52\r\n209 242\r\n281 407\r\n904 945\r\n", "output": "52870"}, {"input": "5 ... | false | stdio | null | true |
615/B | 615 | B | Python 3 | TESTS | 2 | 78 | 307,200 | 101012143 | def DFS(G, start):
n = len(G)
S = []
S.append(start)
visited[start] = True
dist = [0] * n
while (S != []):
v = S[-1]
S.pop()
for i in G[v]:
dist[v] = max(dist[i] + 1, dist[v])
if (visited[i] == False):
S.append(i)
... | 60 | 623 | 41,984,000 | 130604111 | import sys
lines = sys.stdin.read().splitlines()
n, m = map(int, lines[0].split())
edges = [[] for _ in range(n)]
for line_i in range(1, 1+m):
u, v = map(int, lines[line_i].split())
edges[u-1].append(v-1)
edges[v-1].append(u-1)
longest = [0 for _ in range(n)]
for index in range(n):
longest[index] = max(... | Codeforces Round 338 (Div. 2) | CF | 2,016 | 3 | 256 | Longtail Hedgehog | This Christmas Santa gave Masha a magic picture and a pencil. The picture consists of n points connected by m segments (they might cross in any way, that doesn't matter). No two segments connect the same pair of points, and no segment connects the point to itself. Masha wants to color some segments in order paint a hed... | First line of the input contains two integers n and m(2 ≤ n ≤ 100 000, 1 ≤ m ≤ 200 000) — the number of points and the number segments on the picture respectively.
Then follow m lines, each containing two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the numbers of points connected by corresponding segment. It's guar... | Print the maximum possible value of the hedgehog's beauty. | null | The picture below corresponds to the first sample. Segments that form the hedgehog are painted red. The tail consists of a sequence of points with numbers 1, 2 and 5. The following segments are spines: (2, 5), (3, 5) and (4, 5). Therefore, the beauty of the hedgehog is equal to 3·3 = 9. | [{"input": "8 6\n4 5\n3 5\n2 5\n1 2\n2 8\n6 7", "output": "9"}, {"input": "4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4", "output": "12"}] | 1,600 | ["dp", "graphs"] | 60 | [{"input": "8 6\r\n4 5\r\n3 5\r\n2 5\r\n1 2\r\n2 8\r\n6 7\r\n", "output": "9\r\n"}, {"input": "4 6\r\n1 2\r\n1 3\r\n1 4\r\n2 3\r\n2 4\r\n3 4\r\n", "output": "12\r\n"}, {"input": "5 7\r\n1 3\r\n2 4\r\n4 5\r\n5 3\r\n2 1\r\n1 4\r\n3 2\r\n", "output": "9\r\n"}, {"input": "5 9\r\n1 3\r\n2 4\r\n4 5\r\n5 3\r\n2 1\r\n1 4\r\n3 ... | false | stdio | null | true |
362/A | 362 | A | PyPy 3 | TESTS | 6 | 124 | 20,172,800 | 85504567 | t = int(input())
for _ in range(t):
if _:
input()
knights = []
for i in range(8):
s = input().strip()
for j in range(8):
if s[j] == 'K':
knights.append((i,j))
n1 = knights[0]
n2 = knights[1]
if n1[0] % 2 == n2[0] % 2 and n1[1] % 2 == n2[1] % 2:... | 45 | 62 | 307,200 | 5130546 | N = range(8)
test = int (input())
for i_test in range(test):
if (i_test): input()
x1,y1,x2,y2=0,0,0,0
map = [input() for i in N]
for i in N:
for j in N:
if (map[i][j]=="K"): x1,y1,x2,y2 = x2,y2,i,j
if (abs(x1-x2)%4==0 and abs(y1-y2)%4==0): print ("YES")
else: print ("NO") | Codeforces Round 212 (Div. 2) | CF | 2,013 | 1 | 256 | Two Semiknights Meet | A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the le... | The first line contains number t (1 ≤ t ≤ 50) — the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matrix consists of characters ".", "#", "K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed t... | For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO" otherwise. | null | Consider the first board from the sample. We will assume the rows and columns of the matrix to be numbered 1 through 8 from top to bottom and from left to right, correspondingly. The knights can meet, for example, in square (2, 7). The semiknight from square (4, 1) goes to square (2, 3) and the semiknight goes from squ... | [{"input": "2\n........\n........\n......#.\nK..##..#\n.......#\n...##..#\n......#.\nK.......\n........\n........\n..#.....\n..#..#..\n..####..\n...##...\n........\n....K#K#", "output": "YES\nNO"}] | 1,500 | ["greedy", "math"] | 45 | [{"input": "2\r\n........\r\n........\r\n......#.\r\nK..##..#\r\n.......#\r\n...##..#\r\n......#.\r\nK.......\r\n\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n...##...\r\n........\r\n....K#K#\r\n", "output": "YES\r\nNO\r\n"}, {"input": "3\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n... | false | stdio | null | true |
19/C | 19 | C | Python 3 | TESTS | 1 | 60 | 0 | 197317361 | n = int(input())
numbers = list(map(int, input().split()))
last_occurrence = {}
for i in range(n-1, -1, -1):
if numbers[i] not in last_occurrence:
last_occurrence[numbers[i]] = i
result = []
i = 0
while i < n:
if i == last_occurrence[numbers[i]]:
result.append(numbers[i])
i += 1
el... | 70 | 1,216 | 14,131,200 | 174568804 | import sys
n=int(input())
a=list(map(int,input().split()))
M=10**9+1
g={}
for i in range(n):
g[a[i]]=g.get(a[i],[])+[i]
p=[1]
for i in range(n):
p+=[hash(M*p[-1])]
h=[0]*(n+1)
for i in range(n):
h[i+1]=hash(h[i]*M+a[i])
gh=lambda k,l:hash(h[k+l]-h[k]*p[l])%sys.hash_info.modulus
i,t=0,0
while i < n:
for ... | Codeforces Beta Round 19 | ICPC | 2,010 | 2 | 256 | Deletion of Repeats | Once Bob saw a string. It contained so many different letters, that the letters were marked by numbers, but at the same time each letter could be met in the string at most 10 times. Bob didn't like that string, because it contained repeats: a repeat of length x is such a substring of length 2x, that its first half coin... | The first input line contains integer n (1 ≤ n ≤ 105) — length of the string. The following line contains n space-separated integer numbers from 0 to 109 inclusive — numbers that stand for the letters of the string. It's guaranteed that each letter can be met in the string at most 10 times. | In the first line output the length of the string's part, left after Bob's deletions. In the second line output all the letters (separated by a space) of the string, left after Bob deleted all the repeats in the described way. | null | null | [{"input": "6\n1 2 3 1 2 3", "output": "3\n1 2 3"}, {"input": "7\n4 5 6 5 6 7 7", "output": "1\n7"}] | 2,200 | ["greedy", "hashing", "string suffix structures"] | 70 | [{"input": "6\r\n1 2 3 1 2 3\r\n", "output": "3\r\n1 2 3 \r\n"}, {"input": "7\r\n4 5 6 5 6 7 7\r\n", "output": "1\r\n7 \r\n"}, {"input": "10\r\n5 7 2 1 8 8 5 10 2 5\r\n", "output": "5\r\n8 5 10 2 5 \r\n"}, {"input": "10\r\n0 1 1 1 0 3 0 1 4 0\r\n", "output": "7\r\n1 0 3 0 1 4 0 \r\n"}, {"input": "10\r\n0 1 0 2 0 0 1 1 ... | false | stdio | null | true |
743/E | 743 | E | PyPy 3 | TESTS | 4 | 108 | 3,072,000 | 151350468 | def check(counts, m, add=[0, 0, 0, 0, 0, 0, 0, 0]):
n1 = len(counts)
I = 1
for i in range(1, n1):
c1 = counts[i]
c2 = [counts[-1][j]-counts[i][j] for j in range(8)]
sides = [0, 0, 0, 0]
for j in range(8):
if c1[j] >= m+1:
sides[0]+=1
if... | 41 | 233 | 24,780,800 | 192171052 | import bisect
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def f(u, v):
return (m + 1) * u + v
n = int(input())
a = list(map(int, input().split()))
m = 8
x = [[] for _ in range(m)]
for i in range(n):
x[a[i] - 1].append(i)
s = 0
for y in x:
s += min(len(y), 1)
if s < m:
... | Codeforces Round 384 (Div. 2) | CF | 2,016 | 2 | 256 | Vladik and cards | Vladik was bored on his way home and decided to play the following game. He took n cards and put them in a row in front of himself. Every card has a positive integer number not exceeding 8 written on it. He decided to find the longest subsequence of cards which satisfies the following conditions:
- the number of occur... | The first line contains single integer n (1 ≤ n ≤ 1000) — the number of cards in Vladik's sequence.
The second line contains the sequence of n positive integers not exceeding 8 — the description of Vladik's sequence. | Print single integer — the length of the longest subsequence of Vladik's sequence that satisfies both conditions. | null | In the first sample all the numbers written on the cards are equal, so you can't take more than one card, otherwise you'll violate the first condition. | [{"input": "3\n1 1 1", "output": "1"}, {"input": "8\n8 7 6 5 4 3 2 1", "output": "8"}, {"input": "24\n1 8 1 2 8 2 3 8 3 4 8 4 5 8 5 6 8 6 7 8 7 8 8 8", "output": "17"}] | 2,200 | ["binary search", "bitmasks", "brute force", "dp"] | 41 | [{"input": "3\r\n1 1 1\r\n", "output": "1"}, {"input": "8\r\n8 7 6 5 4 3 2 1\r\n", "output": "8"}, {"input": "24\r\n1 8 1 2 8 2 3 8 3 4 8 4 5 8 5 6 8 6 7 8 7 8 8 8\r\n", "output": "17"}, {"input": "1\r\n8\r\n", "output": "1"}, {"input": "2\r\n5 4\r\n", "output": "2"}, {"input": "3\r\n3 3 2\r\n", "output": "2"}, {"input... | false | stdio | null | true |
731/F | 731 | F | PyPy 3-64 | TESTS | 12 | 140 | 20,582,400 | 201632357 | n=int(input())
a=list(map(int,input().split()))
a.sort()
ans=0
for i in range(n):
s=0
for j in range(i,n):
s+=a[j]//a[i]*a[i]
ans=max(ans,s)
if s<ans:
break
print(ans) | 51 | 155 | 32,358,400 | 215746255 | import sys
from itertools import accumulate
def main():
n = int(sys.stdin.readline().strip())
a = list(map(int, sys.stdin.readline().split()))
max_val = max(a)
C = [0] * (max_val + 1)
for x in a:
C[x] += 1
ans=0
C_sum = list(accumulate(C))
for x in set(a):
cnt = 0
... | Codeforces Round 376 (Div. 2) | CF | 2,016 | 1 | 256 | Video Cards | Little Vlad is fond of popular computer game Bota-2. Recently, the developers announced the new add-on named Bota-3. Of course, Vlad immediately bought only to find out his computer is too old for the new game and needs to be updated.
There are n video cards in the shop, the power of the i-th video card is equal to in... | The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of video cards in the shop.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 200 000) — powers of video cards. | The only line of the output should contain one integer value — the maximum possible total power of video cards working together. | null | In the first sample, it would be optimal to buy video cards with powers 3, 15 and 9. The video card with power 3 should be chosen as the leading one and all other video cards will be compatible with it. Thus, the total power would be 3 + 15 + 9 = 27. If he buys all the video cards and pick the one with the power 2 as t... | [{"input": "4\n3 2 15 9", "output": "27"}, {"input": "4\n8 2 2 7", "output": "18"}] | 1,900 | ["brute force", "data structures", "implementation", "math", "number theory"] | 51 | [{"input": "4\r\n3 2 15 9\r\n", "output": "27\r\n"}, {"input": "4\r\n8 2 2 7\r\n", "output": "18\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}, {"input": "1\r\n123819\r\n", "output": "123819\r\n"}, {"input": "10\r\n9 6 8 5 5 2 8 9 2 2\r\n", "output": "52\r\n"}, {"input": "100\r\n17 23 71 25 50 71 85 46 78 72 89 26 ... | false | stdio | null | true |
10/A | 10 | A | PyPy 3 | TESTS | 5 | 218 | 20,172,800 | 129230931 | n, p1, p2, p3, t1, t2 = map(int, input().split())
lr = 1441
result = 0
for i in range(0,n):
l, r = map(int, input().split())
modo1 = t1 if (l - lr) >= t1 else (l - lr)
modo2 = t2 if (l- lr - modo1) >= t2 else (l- lr - modo1)
modo3 = (l - lr - modo1 - modo2)
if (modo1 > 0):
modo1 = m... | 30 | 92 | 0 | 11583841 | a = []
n,P1,P2,P3,T1,T2 = input().split()
n = int(n)
P1 = int(P1)
P2 = int(P2)
P3 = int(P3)
T1 = int(T1)
T2 = int(T2)
for i in range(n):
x,y = input().split()
a.append([int(x),int(y)])
p = 0
for i in range(len(a)):
p += (a[i][1] - a[i][0])*P1
for i in range(len(a)-1):
z = abs(a[i][1] - a[i+1][... | Codeforces Beta Round 10 | ICPC | 2,010 | 1 | 256 | Power Consumption Calculation | Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes P1 watt per minute. T1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to P2 watt per minute. Finally, after T2 minut... | The first line contains 6 integer numbers n, P1, P2, P3, T1, T2 (1 ≤ n ≤ 100, 0 ≤ P1, P2, P3 ≤ 100, 1 ≤ T1, T2 ≤ 60). The following n lines contain description of Tom's work. Each i-th of these lines contains two space-separated integers li and ri (0 ≤ li < ri ≤ 1440, ri < li + 1 for i < n), which stand for the start a... | Output the answer to the problem. | null | null | [{"input": "1 3 2 1 5 10\n0 10", "output": "30"}, {"input": "2 8 4 2 5 10\n20 30\n50 100", "output": "570"}] | 900 | ["implementation"] | 30 | [{"input": "1 3 2 1 5 10\r\n0 10\r\n", "output": "30"}, {"input": "2 8 4 2 5 10\r\n20 30\r\n50 100\r\n", "output": "570"}, {"input": "3 15 9 95 39 19\r\n873 989\r\n1003 1137\r\n1172 1436\r\n", "output": "8445"}, {"input": "4 73 2 53 58 16\r\n51 52\r\n209 242\r\n281 407\r\n904 945\r\n", "output": "52870"}, {"input": "5 ... | false | stdio | null | true |
298/A | 298 | A | Python 3 | TESTS | 8 | 92 | 0 | 183038492 | n = int(input())
seq = list(input())
if len(set(filter(lambda x: x.isalpha(), seq))) == 1:
letters = tuple(filter(lambda x: x.isalpha(), seq))
l = letters[0]
if l == "R":
s = seq.index(l)+1
e = s+len(letters)
print(s, e)
else:
e = seq.index(l)
s = e+len(letters)-1... | 23 | 62 | 0 | 147948126 | n =int(input())
s=input()
a=s.find('R')+1
t=0
if a==0 :
a=s.find('L')+1
t=a-1
for x in range(1,len(s)) :
if s[x]!='R' and s[x-1]=='R' :
if s[x]=='.' :
t=x+1
break
else :
t=x
break
print(a,t) | Codeforces Round 180 (Div. 2) | CF | 2,013 | 1 | 256 | Snow Footprints | There is a straight snowy road, divided into n blocks. The blocks are numbered from 1 to n from left to right. If one moves from the i-th block to the (i + 1)-th block, he will leave a right footprint on the i-th block. Similarly, if one moves from the i-th block to the (i - 1)-th block, he will leave a left footprint ... | The first line of the input contains integer n (3 ≤ n ≤ 1000).
The second line contains the description of the road — the string that consists of n characters. Each character will be either "." (a block without footprint), or "L" (a block with a left footprint), "R" (a block with a right footprint).
It's guaranteed t... | Print two space-separated integers — the values of s and t. If there are several possible solutions you can print any of them. | null | The first test sample is the one in the picture. | [{"input": "9\n..RRLL...", "output": "3 4"}, {"input": "11\n.RRRLLLLL..", "output": "7 5"}] | 1,300 | ["greedy", "implementation"] | 23 | [{"input": "9\r\n..RRLL...\r\n", "output": "3 4\r\n"}, {"input": "11\r\n.RRRLLLLL..\r\n", "output": "7 5\r\n"}, {"input": "17\r\n.......RRRRR.....\r\n", "output": "12 13\r\n"}, {"input": "13\r\n....LLLLLL...\r\n", "output": "10 4\r\n"}, {"input": "4\r\n.RL.\r\n", "output": "3 2\r\n"}, {"input": "3\r\n.L.\r\n", "output"... | false | stdio | import sys
from collections import deque
def main(input_path, output_path, submission_path):
with open(input_path) as f:
n = int(f.readline().strip())
road = f.readline().strip()
with open(submission_path) as f:
line = f.readline().strip()
try:
s, t = map(int, l... | true |
362/A | 362 | A | PyPy 3-64 | TESTS | 6 | 46 | 0 | 203745295 | import sys
input = lambda: sys.stdin.readline().rstrip()
for _ in range(int(input())):
A = []
v = []
cnt = 0
for i in range(8):
A.append(input())
for j in range(8):
#print(i,j)
if A[i][j]=='K':
v.append((i,j,cnt))
cnt+=1
a... | 45 | 62 | 409,600 | 5772053 | def check():
board = []
for cont in range(0,8):
board.append(input())
l = True
for cont in range(0,8):
for cont2 in range(0,8):
if board[cont][cont2] == 'K':
if l:
xk1 = cont2
yk1 = cont
l = False
... | Codeforces Round 212 (Div. 2) | CF | 2,013 | 1 | 256 | Two Semiknights Meet | A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares forward and 2 squares to the right, 2 squares forward and 2 squares to the left, 2 squares backward and 2 to the right and 2 squares backward and 2 to the le... | The first line contains number t (1 ≤ t ≤ 50) — the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matrix consists of characters ".", "#", "K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed t... | For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO" otherwise. | null | Consider the first board from the sample. We will assume the rows and columns of the matrix to be numbered 1 through 8 from top to bottom and from left to right, correspondingly. The knights can meet, for example, in square (2, 7). The semiknight from square (4, 1) goes to square (2, 3) and the semiknight goes from squ... | [{"input": "2\n........\n........\n......#.\nK..##..#\n.......#\n...##..#\n......#.\nK.......\n........\n........\n..#.....\n..#..#..\n..####..\n...##...\n........\n....K#K#", "output": "YES\nNO"}] | 1,500 | ["greedy", "math"] | 45 | [{"input": "2\r\n........\r\n........\r\n......#.\r\nK..##..#\r\n.......#\r\n...##..#\r\n......#.\r\nK.......\r\n\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n...##...\r\n........\r\n....K#K#\r\n", "output": "YES\r\nNO\r\n"}, {"input": "3\r\n........\r\n........\r\n..#.....\r\n..#..#..\r\n..####..\r\n... | false | stdio | null | true |
825/E | 825 | E | PyPy 3 | TESTS | 6 | 155 | 0 | 93293235 | import sys
from heapq import heappop, heappush
n, m = map(int, sys.stdin.buffer.readline().decode('utf-8').split())
adj = [[] for _ in range(n)]
rev = [[] for _ in range(n)]
indeg, outdeg = [0]*n, [0]*n
for u, v in (map(int, line.decode('utf-8').split()) for line in sys.stdin.buffer):
adj[u-1].append(v-1)
rev... | 19 | 311 | 27,033,600 | 177003198 | from collections import defaultdict, deque
import heapq
import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
inp = sys.stdin.readline
def input(): return inp().strip()
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().sp... | Educational Codeforces Round 25 | ICPC | 2,017 | 1 | 256 | Minimal Labels | You are given a directed acyclic graph with n vertices and m edges. There are no self-loops or multiple edges between any pair of vertices. Graph can be disconnected.
You should assign labels to all vertices in such a way that:
- Labels form a valid permutation of length n — an integer sequence such that each integer... | The first line contains two integer numbers n, m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105).
Next m lines contain two integer numbers v and u (1 ≤ v, u ≤ n, v ≠ u) — edges of the graph. Edges are directed, graph doesn't contain loops or multiple edges. | Print n numbers — lexicographically smallest correct permutation of labels of vertices. | null | null | [{"input": "3 3\n1 2\n1 3\n3 2", "output": "1 3 2"}, {"input": "4 5\n3 1\n4 1\n2 3\n3 4\n2 4", "output": "4 1 2 3"}, {"input": "5 4\n3 1\n2 1\n2 3\n4 5", "output": "3 1 2 4 5"}] | 2,300 | ["data structures", "dfs and similar", "graphs", "greedy"] | 19 | [{"input": "3 3\r\n1 2\r\n1 3\r\n3 2\r\n", "output": "1 3 2 \r\n"}, {"input": "4 5\r\n3 1\r\n4 1\r\n2 3\r\n3 4\r\n2 4\r\n", "output": "4 1 2 3 \r\n"}, {"input": "5 4\r\n3 1\r\n2 1\r\n2 3\r\n4 5\r\n", "output": "3 1 2 4 5 \r\n"}, {"input": "2 1\r\n2 1\r\n", "output": "2 1 \r\n"}, {"input": "5 10\r\n5 2\r\n4 1\r\n2 1\r\n... | false | stdio | null | true |
825/E | 825 | E | Python 3 | TESTS | 6 | 46 | 102,400 | 177006518 | from collections import defaultdict
import heapq
n,m = tuple(map(int,input().split()))
in_degree = defaultdict(int)
graph = defaultdict(list)
for _ in range(m):
edge = list(map(int,input().split()))
graph[edge[0]].append(edge[1])
in_degree[edge[1]] += 1
next_minimum = {}
def set_next_min(vertex):
if... | 19 | 342 | 17,408,000 | 93295209 | import sys
from heapq import heappop, heappush
n, m = map(int, sys.stdin.buffer.readline().decode('utf-8').split())
rev = [[] for _ in range(n)]
outdeg = [0]*n
for u, v in (map(int, line.decode('utf-8').split()) for line in sys.stdin.buffer):
rev[v-1].append(u-1)
outdeg[u-1] += 1
ans = [0]*n
num = n
hq = [-i... | Educational Codeforces Round 25 | ICPC | 2,017 | 1 | 256 | Minimal Labels | You are given a directed acyclic graph with n vertices and m edges. There are no self-loops or multiple edges between any pair of vertices. Graph can be disconnected.
You should assign labels to all vertices in such a way that:
- Labels form a valid permutation of length n — an integer sequence such that each integer... | The first line contains two integer numbers n, m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105).
Next m lines contain two integer numbers v and u (1 ≤ v, u ≤ n, v ≠ u) — edges of the graph. Edges are directed, graph doesn't contain loops or multiple edges. | Print n numbers — lexicographically smallest correct permutation of labels of vertices. | null | null | [{"input": "3 3\n1 2\n1 3\n3 2", "output": "1 3 2"}, {"input": "4 5\n3 1\n4 1\n2 3\n3 4\n2 4", "output": "4 1 2 3"}, {"input": "5 4\n3 1\n2 1\n2 3\n4 5", "output": "3 1 2 4 5"}] | 2,300 | ["data structures", "dfs and similar", "graphs", "greedy"] | 19 | [{"input": "3 3\r\n1 2\r\n1 3\r\n3 2\r\n", "output": "1 3 2 \r\n"}, {"input": "4 5\r\n3 1\r\n4 1\r\n2 3\r\n3 4\r\n2 4\r\n", "output": "4 1 2 3 \r\n"}, {"input": "5 4\r\n3 1\r\n2 1\r\n2 3\r\n4 5\r\n", "output": "3 1 2 4 5 \r\n"}, {"input": "2 1\r\n2 1\r\n", "output": "2 1 \r\n"}, {"input": "5 10\r\n5 2\r\n4 1\r\n2 1\r\n... | false | stdio | null | true |
825/E | 825 | E | Python 3 | TESTS | 6 | 46 | 512,000 | 227589716 | import collections
import heapq
intInput = lambda : int(input())
tupInput = lambda : map(int, input().strip().split())
intListInput = lambda : list(map(int, input().strip().split()))
strListInput = lambda : input().strip().split()
strInput = lambda : list(input().strip())
def run():
n, m = tupInput()
graph = ... | 19 | 343 | 34,406,400 | 132641591 | from collections import deque
import heapq
import sys
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def main():
n, m = map(int, input().split())
rg = [[] for i in range(n)]
outd = [0]*n
for i in range(m):
a, b = map(int, input().split())
a, b = a-1, b-1
... | Educational Codeforces Round 25 | ICPC | 2,017 | 1 | 256 | Minimal Labels | You are given a directed acyclic graph with n vertices and m edges. There are no self-loops or multiple edges between any pair of vertices. Graph can be disconnected.
You should assign labels to all vertices in such a way that:
- Labels form a valid permutation of length n — an integer sequence such that each integer... | The first line contains two integer numbers n, m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105).
Next m lines contain two integer numbers v and u (1 ≤ v, u ≤ n, v ≠ u) — edges of the graph. Edges are directed, graph doesn't contain loops or multiple edges. | Print n numbers — lexicographically smallest correct permutation of labels of vertices. | null | null | [{"input": "3 3\n1 2\n1 3\n3 2", "output": "1 3 2"}, {"input": "4 5\n3 1\n4 1\n2 3\n3 4\n2 4", "output": "4 1 2 3"}, {"input": "5 4\n3 1\n2 1\n2 3\n4 5", "output": "3 1 2 4 5"}] | 2,300 | ["data structures", "dfs and similar", "graphs", "greedy"] | 19 | [{"input": "3 3\r\n1 2\r\n1 3\r\n3 2\r\n", "output": "1 3 2 \r\n"}, {"input": "4 5\r\n3 1\r\n4 1\r\n2 3\r\n3 4\r\n2 4\r\n", "output": "4 1 2 3 \r\n"}, {"input": "5 4\r\n3 1\r\n2 1\r\n2 3\r\n4 5\r\n", "output": "3 1 2 4 5 \r\n"}, {"input": "2 1\r\n2 1\r\n", "output": "2 1 \r\n"}, {"input": "5 10\r\n5 2\r\n4 1\r\n2 1\r\n... | false | stdio | null | true |
825/E | 825 | E | PyPy 3-64 | TESTS | 6 | 46 | 131,481,600 | 181892884 | from sys import stdin
input=lambda :stdin.readline()[:-1]
import sys
sys.setrecursionlimit(10**5+10)
n,m=map(int,input().split())
cnt=[0]*n
edge=[[] for i in range(n)]
for _ in range(m):
u,v=map(lambda x:int(x)-1,input().split())
edge[v].append(u)
seen=[0]*n
tmp=1
ans=[0]*n
def calc(v):
if seen[v]:
return
... | 19 | 374 | 22,118,400 | 182150263 | from sys import stdin
input=lambda :stdin.readline()[:-1]
from heapq import*
n,m=map(int,input().split())
edge=[[] for i in range(n)]
cnt=[0]*n
for i in range(m):
u,v=map(lambda x:int(x)-1,input().split())
edge[v].append(u)
cnt[u]+=1
hq=[]
for i in range(n):
if cnt[i]==0:
heappush(hq,-i)
ans=[0]*n
tmp=n
... | Educational Codeforces Round 25 | ICPC | 2,017 | 1 | 256 | Minimal Labels | You are given a directed acyclic graph with n vertices and m edges. There are no self-loops or multiple edges between any pair of vertices. Graph can be disconnected.
You should assign labels to all vertices in such a way that:
- Labels form a valid permutation of length n — an integer sequence such that each integer... | The first line contains two integer numbers n, m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105).
Next m lines contain two integer numbers v and u (1 ≤ v, u ≤ n, v ≠ u) — edges of the graph. Edges are directed, graph doesn't contain loops or multiple edges. | Print n numbers — lexicographically smallest correct permutation of labels of vertices. | null | null | [{"input": "3 3\n1 2\n1 3\n3 2", "output": "1 3 2"}, {"input": "4 5\n3 1\n4 1\n2 3\n3 4\n2 4", "output": "4 1 2 3"}, {"input": "5 4\n3 1\n2 1\n2 3\n4 5", "output": "3 1 2 4 5"}] | 2,300 | ["data structures", "dfs and similar", "graphs", "greedy"] | 19 | [{"input": "3 3\r\n1 2\r\n1 3\r\n3 2\r\n", "output": "1 3 2 \r\n"}, {"input": "4 5\r\n3 1\r\n4 1\r\n2 3\r\n3 4\r\n2 4\r\n", "output": "4 1 2 3 \r\n"}, {"input": "5 4\r\n3 1\r\n2 1\r\n2 3\r\n4 5\r\n", "output": "3 1 2 4 5 \r\n"}, {"input": "2 1\r\n2 1\r\n", "output": "2 1 \r\n"}, {"input": "5 10\r\n5 2\r\n4 1\r\n2 1\r\n... | false | stdio | null | true |
825/E | 825 | E | Python 3 | TESTS | 6 | 62 | 102,400 | 177008152 | from collections import defaultdict
vertices, edges = list(map(int, input().split()))
graph = defaultdict(set)
indegree = defaultdict(int)
for _ in range(edges):
pre, nxt = list(map(int, input().split()))
indegree[nxt] += 1
graph[pre].add(nxt)
answer = []
path = []
color = defaultdict(int)
array = defau... | 19 | 390 | 30,924,800 | 212162167 | import sys
import math
import collections
from heapq import heappush, heappop
input = sys.stdin.readline
ints = lambda: list(map(int, input().split()))
n, m = ints()
g = [[] for _ in range(n)]
deg = [0] * n
for _ in range(m):
u, v = ints()
u -= 1
v -= 1
g[u].append((v, 0))
g[v].append((u, 1))
... | Educational Codeforces Round 25 | ICPC | 2,017 | 1 | 256 | Minimal Labels | You are given a directed acyclic graph with n vertices and m edges. There are no self-loops or multiple edges between any pair of vertices. Graph can be disconnected.
You should assign labels to all vertices in such a way that:
- Labels form a valid permutation of length n — an integer sequence such that each integer... | The first line contains two integer numbers n, m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105).
Next m lines contain two integer numbers v and u (1 ≤ v, u ≤ n, v ≠ u) — edges of the graph. Edges are directed, graph doesn't contain loops or multiple edges. | Print n numbers — lexicographically smallest correct permutation of labels of vertices. | null | null | [{"input": "3 3\n1 2\n1 3\n3 2", "output": "1 3 2"}, {"input": "4 5\n3 1\n4 1\n2 3\n3 4\n2 4", "output": "4 1 2 3"}, {"input": "5 4\n3 1\n2 1\n2 3\n4 5", "output": "3 1 2 4 5"}] | 2,300 | ["data structures", "dfs and similar", "graphs", "greedy"] | 19 | [{"input": "3 3\r\n1 2\r\n1 3\r\n3 2\r\n", "output": "1 3 2 \r\n"}, {"input": "4 5\r\n3 1\r\n4 1\r\n2 3\r\n3 4\r\n2 4\r\n", "output": "4 1 2 3 \r\n"}, {"input": "5 4\r\n3 1\r\n2 1\r\n2 3\r\n4 5\r\n", "output": "3 1 2 4 5 \r\n"}, {"input": "2 1\r\n2 1\r\n", "output": "2 1 \r\n"}, {"input": "5 10\r\n5 2\r\n4 1\r\n2 1\r\n... | false | stdio | null | true |
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