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10
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1 value
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11 values
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1 value
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int64
0
100
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6
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int64
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14.7k
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15
14.7k
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358
9.83k
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stringlengths
2
4.87k
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807 values
__index_level_0__
int64
1.1k
1.22M
s768966456
p00586
u314166831
1562221836
Python
Python3
py
Accepted
30
5692
1,961
# coding=utf-8 ### ### for atcorder program ### import sys import math # math class class mymath: ### pi pi = 3.14159265358979323846264338 ### Prime Number def pnum_eratosthenes(self, n): ptable = [0 for i in range(n+1)] plist = [] for i in range(2, n+1): if...
p00586
<H1>A + B Problem</H1> <p> Compute A + B. </p> <H2>Input</H2> <p> The input will consist of a series of pairs of integers A and B separated by a space, one pair of integers per line. The input will be terminated by EOF. </p> <H2>Output</H2> <p> For each pair of input integers A and B, you must output the sum...
1 2 10 5 100 20
3 15 120
23,275
s404551741
p00586
u136695294
1553173295
Python
Python3
py
Accepted
20
5580
100
while True: try: a,b=map(int,input().split(" ")) print(a+b) except Exception: break
p00586
<H1>A + B Problem</H1> <p> Compute A + B. </p> <H2>Input</H2> <p> The input will consist of a series of pairs of integers A and B separated by a space, one pair of integers per line. The input will be terminated by EOF. </p> <H2>Output</H2> <p> For each pair of input integers A and B, you must output the sum...
1 2 10 5 100 20
3 15 120
23,276
s741799501
p00586
u316584871
1538295522
Python
Python3
py
Accepted
20
5588
115
while True: try: a,b = map(int, input().split()) print(a+b) except EOFError: break
p00586
<H1>A + B Problem</H1> <p> Compute A + B. </p> <H2>Input</H2> <p> The input will consist of a series of pairs of integers A and B separated by a space, one pair of integers per line. The input will be terminated by EOF. </p> <H2>Output</H2> <p> For each pair of input integers A and B, you must output the sum...
1 2 10 5 100 20
3 15 120
23,277
s488228094
p00586
u467175809
1533208043
Python
Python3
py
Accepted
30
5576
60
try: while 1:print(sum(map(int,input().split()))) except:1
p00586
<H1>A + B Problem</H1> <p> Compute A + B. </p> <H2>Input</H2> <p> The input will consist of a series of pairs of integers A and B separated by a space, one pair of integers per line. The input will be terminated by EOF. </p> <H2>Output</H2> <p> For each pair of input integers A and B, you must output the sum...
1 2 10 5 100 20
3 15 120
23,278
s447827390
p00586
u467175809
1533207411
Python
Python
py
Accepted
10
4620
60
import sys for l in sys.stdin:print sum(map(int,l.split()))
p00586
<H1>A + B Problem</H1> <p> Compute A + B. </p> <H2>Input</H2> <p> The input will consist of a series of pairs of integers A and B separated by a space, one pair of integers per line. The input will be terminated by EOF. </p> <H2>Output</H2> <p> For each pair of input integers A and B, you must output the sum...
1 2 10 5 100 20
3 15 120
23,279
s235249978
p00586
u539753516
1532474990
Python
Python3
py
Accepted
30
5576
77
while True: try:print(sum(map(int,input().split(" ")))) except:break
p00586
<H1>A + B Problem</H1> <p> Compute A + B. </p> <H2>Input</H2> <p> The input will consist of a series of pairs of integers A and B separated by a space, one pair of integers per line. The input will be terminated by EOF. </p> <H2>Output</H2> <p> For each pair of input integers A and B, you must output the sum...
1 2 10 5 100 20
3 15 120
23,280
s812103569
p00586
u822200871
1448118631
Python
Python3
py
Accepted
20
7508
174
output = [] while True: try: a, b = map(int, input().split()) except : break output.append(a+b) for x in range(len(output)): print(output[x])
p00586
<H1>A + B Problem</H1> <p> Compute A + B. </p> <H2>Input</H2> <p> The input will consist of a series of pairs of integers A and B separated by a space, one pair of integers per line. The input will be terminated by EOF. </p> <H2>Output</H2> <p> For each pair of input integers A and B, you must output the sum...
1 2 10 5 100 20
3 15 120
23,281
s910990930
p00587
u473221511
1466521284
Python
Python
py
Accepted
20
6516
5,589
import sys def searchcenter (str): temp_layer = 0 for x in range(len(str)): c = str[x] if c == '(': temp_layer += 1 elif c == ')': temp_layer -= 1 elif temp_layer == 1: return x #whether either tree have common factor def either_tree_have (t...
p00587
<H1>Binary Tree Intersection And Union</H1> <p> Given two binary trees, we consider the “intersection” and “union” of them. Here, we distinguish the left and right child of a node when it has only one child. The definitions of them are very simple. First of all, draw a complete binary tree (a tree whose nodes have e...
i ((,),(,)) ((,(,)),) u ((,),(,)) ((,(,)),)
((,),) ((,(,)),(,))
23,282
s326552685
p00587
u260980560
1499880971
Python
Python
py
Accepted
10
6404
857
def convert(s): cur = [0] def dfs(): result = [0, 0] cur[0] += 1 # '(' if s[cur[0]] == '(': result[0] = dfs() cur[0] += 1 # ',' if s[cur[0]] == '(': result[1] = dfs() cur[0] += 1 # ')' return result return dfs() def rconvert(C):...
p00587
<H1>Binary Tree Intersection And Union</H1> <p> Given two binary trees, we consider the “intersection” and “union” of them. Here, we distinguish the left and right child of a node when it has only one child. The definitions of them are very simple. First of all, draw a complete binary tree (a tree whose nodes have e...
i ((,),(,)) ((,(,)),) u ((,),(,)) ((,(,)),)
((,),) ((,(,)),(,))
23,283
s268588583
p00587
u600821328
1501152088
Python
Python3
py
Accepted
30
7496
1,685
#!/usr/bin/env python3 # -*- coding: utf-8 -*- import sys def tree_left(node, it): c = next(it) if c == "(": node[0] = [None, None] tree_left(node[0], it) c = next(it) if c == ",": tree_right(node, it) else: assert False def tree_right(node, it): c = next...
p00587
<H1>Binary Tree Intersection And Union</H1> <p> Given two binary trees, we consider the “intersection” and “union” of them. Here, we distinguish the left and right child of a node when it has only one child. The definitions of them are very simple. First of all, draw a complete binary tree (a tree whose nodes have e...
i ((,),(,)) ((,(,)),) u ((,),(,)) ((,(,)),)
((,),) ((,(,)),(,))
23,284
s809842437
p00587
u266872031
1511368446
Python
Python
py
Accepted
20
6508
871
def treeunprocess(tstr,lv): d=0 if len(tstr)==0: return [] for i in range(len(tstr)): if tstr[i]=="(": d+=1 elif tstr[i]==")": d-=1 elif tstr[i]=="," and d==1: return [lv]+treeunprocess(tstr[1:i], 2*lv+1) + treeunprocess(tstr[i+1:-1], 2*lv...
p00587
<H1>Binary Tree Intersection And Union</H1> <p> Given two binary trees, we consider the “intersection” and “union” of them. Here, we distinguish the left and right child of a node when it has only one child. The definitions of them are very simple. First of all, draw a complete binary tree (a tree whose nodes have e...
i ((,),(,)) ((,(,)),) u ((,),(,)) ((,(,)),)
((,),) ((,(,)),(,))
23,285
s351988540
p00587
u668489394
1579076157
Python
Python3
py
Accepted
30
5588
2,195
class Node: def __init__(self, left, right): self.left = left self.right = right self.val = '' @classmethod def fromString(cls, s): pos = 1 if s[pos] == '(': e = getEnd(s, 1) left = cls.fromString(s[1:e+1]) pos = e + 2 ...
p00587
<H1>Binary Tree Intersection And Union</H1> <p> Given two binary trees, we consider the “intersection” and “union” of them. Here, we distinguish the left and right child of a node when it has only one child. The definitions of them are very simple. First of all, draw a complete binary tree (a tree whose nodes have e...
i ((,),(,)) ((,(,)),) u ((,),(,)) ((,(,)),)
((,),) ((,(,)),(,))
23,286
s085652973
p00587
u316584871
1558102112
Python
Python3
py
Accepted
40
6676
2,588
import re class bintree(): def __init__(self, str1): self.str1 = str1 if self.str1[1] != ",": c = 0 counter = 0 for s in str1[1:]: counter += 1 if s == "(": c += 1 elif s == ")": ...
p00587
<H1>Binary Tree Intersection And Union</H1> <p> Given two binary trees, we consider the “intersection” and “union” of them. Here, we distinguish the left and right child of a node when it has only one child. The definitions of them are very simple. First of all, draw a complete binary tree (a tree whose nodes have e...
i ((,),(,)) ((,(,)),) u ((,),(,)) ((,(,)),)
((,),) ((,(,)),(,))
23,287
s140452058
p00587
u943441430
1545030747
Python
Python
py
Accepted
10
4708
1,808
import sys def parse_btree(s): l = r = -1 paren = 0 comma = -1 if s == "(,)": return (["node"]) if s == "": return (["None"]) for i in range(len(s)): c = s[i] if c == "(": paren += 1 if paren == 1 and r < 0: l = i el...
p00587
<H1>Binary Tree Intersection And Union</H1> <p> Given two binary trees, we consider the “intersection” and “union” of them. Here, we distinguish the left and right child of a node when it has only one child. The definitions of them are very simple. First of all, draw a complete binary tree (a tree whose nodes have e...
i ((,),(,)) ((,(,)),) u ((,),(,)) ((,(,)),)
((,),) ((,(,)),(,))
23,288
s883978142
p00587
u467175809
1533198428
Python
Python
py
Accepted
10
5016
864
#!/usr/bin/env python from collections import deque import itertools as it import sys import math sys.setrecursionlimit(10000000) for line in sys.stdin: op, S1, S2 = line.split() id1 = id2 = 0 ans = '' def rem(S, ind): rem = '(' cnt = 1 while cnt > 0: if S[ind...
p00587
<H1>Binary Tree Intersection And Union</H1> <p> Given two binary trees, we consider the “intersection” and “union” of them. Here, we distinguish the left and right child of a node when it has only one child. The definitions of them are very simple. First of all, draw a complete binary tree (a tree whose nodes have e...
i ((,),(,)) ((,(,)),) u ((,),(,)) ((,(,)),)
((,),) ((,(,)),(,))
23,289
s717058837
p00587
u847467233
1530738459
Python
Python3
py
Accepted
20
5588
712
# AOJ 1001: Binary Tree Intersection And Union # Python3 2018.7.5 bal4u def parse(p, i): global sz node[i][2] += 1 del p[0] if p[0] != ',': if node[i][0] == 0: node[i][0] = sz sz += 1 parse(p, node[i][0]) del p[0] if p[0] != ')': if node[i][1] == 0: node[i][1] = sz sz += 1 parse(p, node[i][1]...
p00587
<H1>Binary Tree Intersection And Union</H1> <p> Given two binary trees, we consider the “intersection” and “union” of them. Here, we distinguish the left and right child of a node when it has only one child. The definitions of them are very simple. First of all, draw a complete binary tree (a tree whose nodes have e...
i ((,),(,)) ((,(,)),) u ((,),(,)) ((,(,)),)
((,),) ((,(,)),(,))
23,290
s624386053
p00588
u266872031
1425661553
Python
Python
py
Accepted
80
5544
1,897
def getAB(r,books,N): if r==0: return(books[2*r],books[2*(N+r)]) elif r==N: return(books[2*r-1],books[2*(N+r)-1]) elif r==N+1: return (1,0) else: return(books[2*r-1]|books[2*r],books[2*(N+r)-1]|books[2*(N+r)]) #def getcostandarray(A,B,array,r,books): # ...
p00588
<H1>Extraordinary Girl (I)</H1> <p> She is an extraordinary girl. She works for a library. Since she is young and cute, she is forced to do a lot of laborious jobs. The most annoying job for her is to put returned books into shelves, carrying them by a cart. The cart is large enough to carry many books, but too...
2 2 YNNNNYYY 4 NYNNYYNNNYNYYNNN
6 9
23,291
s243284891
p00588
u467175809
1533199702
Python
Python
py
Accepted
60
5440
683
#!/usr/bin/env python from collections import deque import itertools as it import sys import math sys.setrecursionlimit(10000000) T = input() for loop in range(T): N = input() S = raw_input() S1 = 'N' + S[:N * 2] + 'N' S2 = 'N' + S[N * 2:] + 'N' sx1 = sx2 = 100000 gx1 = gx2 = -1 ans = 0 ...
p00588
<H1>Extraordinary Girl (I)</H1> <p> She is an extraordinary girl. She works for a library. Since she is young and cute, she is forced to do a lot of laborious jobs. The most annoying job for her is to put returned books into shelves, carrying them by a cart. The cart is large enough to carry many books, but too...
2 2 YNNNNYYY 4 NYNNYYNNNYNYYNNN
6 9
23,292
s540387959
p00589
u633068244
1417602694
Python
Python
py
Accepted
10
4276
990
alpha = ["","',.!?","abcABC","defDEF","ghiGHI","jklJKL","mnoMNO","pqrsPQRS","tuvTUV","wxyzWXYZ"] while 1: try: ans = "" sers = "" c = 0 for key in raw_input(): if key == "0": if len(sers) > 0: n,l = int(sers[0]),len(sers) ...
p00589
<!-- <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>problem</title> </head> <body bgcolor="#FFFFFF"> <link rel="stylesheet" href="../problem_e.css" type="text/css" /> --> <H1>Extraordinary Girl (II)</H1> <p> She loves e-mail so much! She sends e-mails by her cellular phone t...
666660666 44444416003334446633111 20202202000333003330333
No I'm fine. aaba f ff
23,293
s323559196
p00589
u633068244
1417603690
Python
Python
py
Accepted
20
4244
695
ref = ["", "',.!?", "abcABC", "defDEF", "ghiGHI", "jklJKL", "mnoMNO", "pqrsPQRS", "tuvTUV", "wxyzWXYZ"] def inp(s): n = int(s[0]) p = (len(s)-1)%len(ref[n]) return ref[n][p] while 1: try: pushes = raw_input() except: br...
p00589
<!-- <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>problem</title> </head> <body bgcolor="#FFFFFF"> <link rel="stylesheet" href="../problem_e.css" type="text/css" /> --> <H1>Extraordinary Girl (II)</H1> <p> She loves e-mail so much! She sends e-mails by her cellular phone t...
666660666 44444416003334446633111 20202202000333003330333
No I'm fine. aaba f ff
23,294
s044039385
p00589
u633068244
1417603701
Python
Python
py
Accepted
10
4244
695
ref = ["", "',.!?", "abcABC", "defDEF", "ghiGHI", "jklJKL", "mnoMNO", "pqrsPQRS", "tuvTUV", "wxyzWXYZ"] def inp(s): n = int(s[0]) p = (len(s)-1)%len(ref[n]) return ref[n][p] while 1: try: pushes = raw_input() except: br...
p00589
<!-- <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>problem</title> </head> <body bgcolor="#FFFFFF"> <link rel="stylesheet" href="../problem_e.css" type="text/css" /> --> <H1>Extraordinary Girl (II)</H1> <p> She loves e-mail so much! She sends e-mails by her cellular phone t...
666660666 44444416003334446633111 20202202000333003330333
No I'm fine. aaba f ff
23,295
s672181586
p00589
u266872031
1428510468
Python
Python
py
Accepted
10
4280
1,030
def getchar(button,num): if button=='0': return ' '*(num-1) elif button=='1': num=(num-1)%5 return "',.!?"[num] elif button=='2': num=(num-1)%6 return "abcABC"[num] elif button=='3': num=(num-1)%6 return "defDEF"[num] elif button=='4': ...
p00589
<!-- <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>problem</title> </head> <body bgcolor="#FFFFFF"> <link rel="stylesheet" href="../problem_e.css" type="text/css" /> --> <H1>Extraordinary Girl (II)</H1> <p> She loves e-mail so much! She sends e-mails by her cellular phone t...
666660666 44444416003334446633111 20202202000333003330333
No I'm fine. aaba f ff
23,296
s010968028
p00589
u078042885
1484327760
Python
Python3
py
Accepted
30
7644
587
a=[['\'',',','.','!','?'], ['a','b','c','A','B','C'], ['d','e','f','D','E','F'], ['g','h','i','G','H','I'], ['j','k','l','J','K','L'], ['m','n','o','M','N','O'], ['p','q','r','s','P','Q','R','S'], ['t','u','v','T','U','V'], ['w','x','y','z','W','X','Y','Z']] while 1: try:b=input()+'@' ex...
p00589
<!-- <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>problem</title> </head> <body bgcolor="#FFFFFF"> <link rel="stylesheet" href="../problem_e.css" type="text/css" /> --> <H1>Extraordinary Girl (II)</H1> <p> She loves e-mail so much! She sends e-mails by her cellular phone t...
666660666 44444416003334446633111 20202202000333003330333
No I'm fine. aaba f ff
23,297
s633996030
p00589
u797673668
1492414854
Python
Python3
py
Accepted
30
7552
642
import fileinput chr_master = [ ' ', '\',.!?', 'abcABC', 'defDEF', 'ghiGHI', 'jklJKL', 'mnoMNO', 'pqrsPQRS', 'tuvTUV', 'wxyzWXYZ' ] for s in (line.strip() for line in fileinput.input()): s += '!' prev_c = None cnt = 0 chr_list = [] for c in s: if c == prev_c: cnt += 1 ...
p00589
<!-- <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>problem</title> </head> <body bgcolor="#FFFFFF"> <link rel="stylesheet" href="../problem_e.css" type="text/css" /> --> <H1>Extraordinary Girl (II)</H1> <p> She loves e-mail so much! She sends e-mails by her cellular phone t...
666660666 44444416003334446633111 20202202000333003330333
No I'm fine. aaba f ff
23,298
s977308001
p00589
u797673668
1492415085
Python
Python3
py
Accepted
30
7488
579
import fileinput chr_master = [ ' ', '\',.!?', 'abcABC', 'defDEF', 'ghiGHI', 'jklJKL', 'mnoMNO', 'pqrsPQRS', 'tuvTUV', 'wxyzWXYZ' ] for s in (line.strip() for line in fileinput.input()): s += '!' prev_c = None cnt = 0 ans = '' for c in s: if c == prev_c: cnt += 1 ...
p00589
<!-- <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>problem</title> </head> <body bgcolor="#FFFFFF"> <link rel="stylesheet" href="../problem_e.css" type="text/css" /> --> <H1>Extraordinary Girl (II)</H1> <p> She loves e-mail so much! She sends e-mails by her cellular phone t...
666660666 44444416003334446633111 20202202000333003330333
No I'm fine. aaba f ff
23,299
s691135554
p00589
u797673668
1492415208
Python
Python3
py
Accepted
30
7536
518
import fileinput chr_master = [' ', '\',.!?', 'abcABC', 'defDEF', 'ghiGHI', 'jklJKL', 'mnoMNO', 'pqrsPQRS', 'tuvTUV', 'wxyzWXYZ'] for s in (line.strip() for line in fileinput.input()): s += '!' prev_c, cnt = 0, 1 ans = '' for c in s: if c == prev_c: cnt += 1 continue ...
p00589
<!-- <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>problem</title> </head> <body bgcolor="#FFFFFF"> <link rel="stylesheet" href="../problem_e.css" type="text/css" /> --> <H1>Extraordinary Girl (II)</H1> <p> She loves e-mail so much! She sends e-mails by her cellular phone t...
666660666 44444416003334446633111 20202202000333003330333
No I'm fine. aaba f ff
23,300
s803344919
p00589
u847467233
1530702270
Python
Python3
py
Accepted
30
5612
425
# AOJ 1003: Extraordinary Girl II # Python3 2018.7.4 bal4u tbl = ["", "',.!?", "abcABC", "defDEF", "ghiGHI", "jklJKL", \ "mnoMNO", "pqrsPQRS", "tuvTUV", "wxyzWXYZ"] while True: try: s = input().strip() except: break ans, i = '', 0 while i < len(s): c = s[i] w, d, i = 0, int(c), i+1 while i < len(s) a...
p00589
<!-- <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>problem</title> </head> <body bgcolor="#FFFFFF"> <link rel="stylesheet" href="../problem_e.css" type="text/css" /> --> <H1>Extraordinary Girl (II)</H1> <p> She loves e-mail so much! She sends e-mails by her cellular phone t...
666660666 44444416003334446633111 20202202000333003330333
No I'm fine. aaba f ff
23,301
s348149894
p00589
u104911888
1370614842
Python
Python
py
Accepted
10
4256
600
Button={ "1":"',.!?", "2":"abcABC", "3":"defDEF", "4":"ghiGHI", "5":"jklJKL", "6":"mnoMNO", "7":"pqrsPQRS", "8":"tuvTUV", "9":"wxyz", "0":" "} while True: try: sent=raw_input() except EOFError: break init=sent[0] cnt=1 ans=[] for c in sent[1:]: if c==init: cn...
p00589
<!-- <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>problem</title> </head> <body bgcolor="#FFFFFF"> <link rel="stylesheet" href="../problem_e.css" type="text/css" /> --> <H1>Extraordinary Girl (II)</H1> <p> She loves e-mail so much! She sends e-mails by her cellular phone t...
666660666 44444416003334446633111 20202202000333003330333
No I'm fine. aaba f ff
23,302
s079757762
p00589
u590037228
1393919092
Python
Python
py
Accepted
10
4292
1,934
#-*- coding:utf-8 -*- import sys #入力されたデータに関して携帯のキーを返す def decode_keitai(str): #携帯キーを表す配列を作成 keitai_keys = [" ","',.!?","abcABC","defDEF","ghiGHI","jklJKL","mnoMNO","pqrsPQRS","tuvTUV","wxyzWXYZ"] #結果文字列 results = [] #入力文字列をリストに変換 inputs = str.split("\n") #各文字列について for input in inpu...
p00589
<!-- <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>problem</title> </head> <body bgcolor="#FFFFFF"> <link rel="stylesheet" href="../problem_e.css" type="text/css" /> --> <H1>Extraordinary Girl (II)</H1> <p> She loves e-mail so much! She sends e-mails by her cellular phone t...
666660666 44444416003334446633111 20202202000333003330333
No I'm fine. aaba f ff
23,303
s054628009
p00589
u260980560
1583166449
Python
Python3
py
Accepted
30
5604
681
T = [ None, "',.!?", "abcABC", "defDEF", "ghiGHI", "jklJKL", "mnoMNO", "pqrsPQRS", "tuvTUV", "wxyzWXYZ", ] try: while 1: ans = [] *S, = map(int, input()) S.append(0) prv = 0 cur = -1 for c in S: if c == prv: ...
p00589
<!-- <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>problem</title> </head> <body bgcolor="#FFFFFF"> <link rel="stylesheet" href="../problem_e.css" type="text/css" /> --> <H1>Extraordinary Girl (II)</H1> <p> She loves e-mail so much! She sends e-mails by her cellular phone t...
666660666 44444416003334446633111 20202202000333003330333
No I'm fine. aaba f ff
23,304
s467566057
p00589
u467175809
1533274338
Python
Python
py
Accepted
10
5016
591
#!/usr/bin/env python from collections import deque import itertools as it import sys import math sys.setrecursionlimit(10000000) S = ["", "',.!?", "abcABC", "defDEF", "ghiGHI", "jklJKL", "mnoMNO", "pqrsPQRS", "tuvTUV", "wxyzWXYZ"] for line in sys.stdin: ans = '' k = 0 cnt = 0 line = list(line) ...
p00589
<!-- <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <html> <head> <title>problem</title> </head> <body bgcolor="#FFFFFF"> <link rel="stylesheet" href="../problem_e.css" type="text/css" /> --> <H1>Extraordinary Girl (II)</H1> <p> She loves e-mail so much! She sends e-mails by her cellular phone t...
666660666 44444416003334446633111 20202202000333003330333
No I'm fine. aaba f ff
23,305
s838179137
p00591
u797673668
1460963359
Python
Python3
py
Accepted
30
7628
323
while True: n = int(input()) if not n: break a = [list(map(int, input().split())) for _ in range(n)] s = max(min(a[i]) for i in range(n)) c = [max(a[i][j] for i in range(n)) for j in range(n)] for j in range(n): if c[j] == s: print(s) break else: print...
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,306
s566724537
p00591
u221679506
1467934602
Python
Python3
py
Accepted
30
7740
431
while True: n = int(input()) if not n: break #map???????????£????????£??¨????????????????°???????????????????????????????????????¨ f = [[int(j) for j in input().split()] for i in range(n)] ans = max(min(f[i]) for i in range(n)) tans = [max(f[i][j] for i in range(n)) for j in range(n)] ...
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,307
s919473257
p00591
u766597310
1477679966
Python
Python
py
Accepted
10
6268
476
while 1: n = input() if (n == 0): break a = [[0] * n] * n for i in range(n): a[i] = map(int, raw_input().split()) ans = 0 for i in range(n): c = 0 for j in range(n): if (a[i][j] < a[i][c]): c = j f = 1 for i2 in range(n): ...
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,308
s547085821
p00591
u266872031
1485706866
Python
Python
py
Accepted
10
6324
566
while(1): n=int(raw_input()) if n==0: break A=[] for i in range(n): A.append(map(int,raw_input().split())) B=[ [A[i][j] for i in range(n)] for j in range(n) ] ans=0 for i in range(n): ma=min(A[i]) m=A[i].count(ma) mlist=[] for k in range(m): ...
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,309
s925137112
p00591
u922489088
1503915640
Python
Python3
py
Accepted
20
7672
543
import sys while True: n = int(sys.stdin.readline().rstrip()) if n == 0: break; students = [] for i in range(n): students.append(list(map(int, sys.stdin.readline().rstrip().split(' ')))) s_list=[[min(row)==s for s in row] for row in students] t_list=[[max(col)==s for s in col] f...
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,310
s084131510
p00591
u922489088
1503915655
Python
Python
py
Accepted
10
6256
543
import sys while True: n = int(sys.stdin.readline().rstrip()) if n == 0: break; students = [] for i in range(n): students.append(list(map(int, sys.stdin.readline().rstrip().split(' ')))) s_list=[[min(row)==s for s in row] for row in students] t_list=[[max(col)==s for s in col] f...
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,311
s713228825
p00591
u009288816
1520077779
Python
Python3
py
Accepted
20
5612
1,121
import sys def get_min_value_in_row(data): if(len(data) < 1): return 0 min = 0 for i in range(0,len(data)): if(data[i] < data[min]): min = i return min def is_max(Min,index,data): max = data[index][Min] for i in range(index,-1,-1): if(data[i][Min] > max): ...
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,312
s529762679
p00591
u104911888
1368080672
Python
Python
py
Accepted
10
4224
282
while True: n=input() if n==0:break L=[map(int,raw_input().split()) for i in range(n)] S=set([min(i) for i in L]) for t in [[i[j] for i in L] for j in range(n)]: maxInt=max(t) if maxInt in S: print maxInt break else: print 0
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,313
s838404070
p00591
u590037228
1393983731
Python
Python
py
Accepted
20
4224
467
#-*- coding:utf-8 -*- while 1: n = int(raw_input()) if not n: break columns = [] for i in range(n): columns.append([]) rows = [] for i in range(n): row = [int(r) for r in raw_input().split(" ")] rows.append(min(row)) for i in range(n): columns...
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,314
s978805524
p00591
u617183767
1393994931
Python
Python
py
Accepted
20
4228
509
# -*- coding: utf-8 -*- while True: n = int(raw_input()) if n == 0: break columns = [] for i in range(n): columns.append([]) maxrows = [] for i in range(n): row = map(int, raw_input().split()) maxrows.append(min(row)) for j in range(n): col...
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,315
s707557666
p00591
u633068244
1397011373
Python
Python
py
Accepted
10
4220
234
while 1: n=input() if n==0:break s=[map(int,raw_input().split()) for i in range(n)] sh=set([min(s[i]) for i in range(n)]) tl=set([max([s[j][i] for j in range(n)]) for i in range(n)]) print list(sh&tl)[0] if len(sh&tl)>0 else 0
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,316
s316287970
p00591
u467175809
1533207011
Python
Python
py
Accepted
10
4656
179
n=1 while n>0: n=input();r=range(n) if n>0:A=[map(int,raw_input().split())for i in r];a={min(A[i])for i in r}&{max(A[j][i]for j in r)for i in r};print a.pop()if len(a)>0 else 0
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,317
s670645414
p00591
u260980560
1533047381
Python
Python3
py
Accepted
30
5600
300
while 1: N = int(input()) if N == 0: break S = [list(map(int, input().split())) for i in range(N)] B = {max(S[i][j] for i in range(N)) for j in range(N)} & {min(S[i][j] for j in range(N)) for i in range(N)} if B: e, = B print(e) else: print(0)
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,318
s773809475
p00591
u847467233
1530750870
Python
Python3
py
Accepted
20
5616
575
# AOJ 1005: Advanced Algorithm Class # Python3 2018.7.5 bal4u import sys from sys import stdin input = stdin.readline while True: n = int(input()) if n == 0: break a = [list(map(int, input().split())) for r in range(n)] b = [[0 for c in range(n)] for r in range(n)] for r in range(n): mi = min(a[r]) for c in ...
p00591
<H1>Advanced Algorithm Class</H1> <p> In the advanced algorithm class, n<sup>2</sup> students sit in n rows and n columns. One day, a professor who teaches this subject comes into the class, asks the shortest student in each row to lift up his left hand, and the tallest student in each column to lift up his right...
3 1 2 3 4 5 6 7 8 9 3 1 2 3 7 8 9 4 5 6 0
7 7
23,319
s117293797
p00592
u703196441
1418469498
Python
Python
py
Accepted
20
4276
353
def cv(t): return t/100*60+t%100 while 1: n,p,q = map(int,raw_input().split()) if n==0: break v=[0]*1440 for i in range(n): k = int(raw_input()) a = map(int,raw_input().split()) for j in range(k): for l in range(cv(a[::2][j]),cv(a[1::2][j])): v[l]+=1 m=c=0 for i in range(cv(p),cv(q)): c=c+1 if v[...
p00592
<H1>Boring Commercial</H1> <p> Now it is spring holidays. A lazy student has finally passed all final examination, and he decided to just kick back and just watch TV all day. Oh, his only source of entertainment is watching TV. And TV commercial, as usual, are a big nuisance for him. He can watch any thing on ...
1 2100 2400 1 2130 2200 3 2100 2400 3 2100 2130 2200 2230 2300 2330 2 2130 2200 2330 2400 2 2100 2130 2330 2400 0 0 0
120 180
23,320
s794153671
p00592
u703196441
1418470307
Python
Python
py
Accepted
10
4276
333
def get(): return map(int,raw_input().split()) def cv(t): return t/100*60+t%100 while 1: n,p,q=get() if n==0: break v=[0]*1440 for i in range(n): k=get()[0]; a=get() for j in range(k): for l in range(cv(a[::2][j]),cv(a[1::2][j])): v[l]+=1 m=c=0 for i in range(cv(p),cv(q)): c=(c+1)*(v[i]<n) m=max(...
p00592
<H1>Boring Commercial</H1> <p> Now it is spring holidays. A lazy student has finally passed all final examination, and he decided to just kick back and just watch TV all day. Oh, his only source of entertainment is watching TV. And TV commercial, as usual, are a big nuisance for him. He can watch any thing on ...
1 2100 2400 1 2130 2200 3 2100 2400 3 2100 2130 2200 2230 2300 2330 2 2130 2200 2330 2400 2 2100 2130 2330 2400 0 0 0
120 180
23,321
s004504177
p00592
u266872031
1428590214
Python
Python
py
Accepted
20
4288
1,095
def tt(o,t): om=o%100 oh=o/100 tm=t%100 th=t/100 return tm-om+(th-oh)*60 while(1): [c,s,e]=[int(x) for x in raw_input().split()] if c==0: break else: cmlist=[] timelist=[s] state=2**c-1 oldtime=s oldstate=0 ans=0 tempt...
p00592
<H1>Boring Commercial</H1> <p> Now it is spring holidays. A lazy student has finally passed all final examination, and he decided to just kick back and just watch TV all day. Oh, his only source of entertainment is watching TV. And TV commercial, as usual, are a big nuisance for him. He can watch any thing on ...
1 2100 2400 1 2130 2200 3 2100 2400 3 2100 2130 2200 2230 2300 2330 2 2130 2200 2330 2400 2 2100 2130 2330 2400 0 0 0
120 180
23,322
s804126709
p00592
u077582668
1446223626
Python
Python
py
Accepted
10
6452
898
#while 1: #n=map(int,raw_input().split()) def chang(x): h = x / 100 m= x%100 return h * 60 + m while 1: (ch,st,en) = map(int, raw_input().split()) if(ch==0 and st ==0 and en ==0):break hours = [0]*(chang(2400)+1) for i in range(ch): line = raw_input() tms = map(int, raw...
p00592
<H1>Boring Commercial</H1> <p> Now it is spring holidays. A lazy student has finally passed all final examination, and he decided to just kick back and just watch TV all day. Oh, his only source of entertainment is watching TV. And TV commercial, as usual, are a big nuisance for him. He can watch any thing on ...
1 2100 2400 1 2130 2200 3 2100 2400 3 2100 2130 2200 2230 2300 2330 2 2130 2200 2330 2400 2 2100 2130 2330 2400 0 0 0
120 180
23,323
s172903102
p00592
u221679506
1467945895
Python
Python3
py
Accepted
30
7616
730
while True: n,s,e = input().split() n = int(n) s = int(s) e = int(e) if not n+s+e: break s = s//100*60+s%100 e = e//100*60+e%100 t = [0]*(60*24+1) for i in range(n): m = int(input()) g = input().split() j = 0 for o in g: a = int(o) ...
p00592
<H1>Boring Commercial</H1> <p> Now it is spring holidays. A lazy student has finally passed all final examination, and he decided to just kick back and just watch TV all day. Oh, his only source of entertainment is watching TV. And TV commercial, as usual, are a big nuisance for him. He can watch any thing on ...
1 2100 2400 1 2130 2200 3 2100 2400 3 2100 2130 2200 2230 2300 2330 2 2130 2200 2330 2400 2 2100 2130 2330 2400 0 0 0
120 180
23,324
s922198923
p00592
u590037228
1393991769
Python
Python
py
Accepted
20
4308
1,615
#-*- coding:utf-8 -*- #入力数値を分単位に変更 def convert_time(n): return (n/100)*60 + n%100 #コマーシャルの共通部分を返す def kyotsu(cm1,cm2): result = [] for i in range(len(cm2)/2): #cm2の各CMとかぶっている部分を取得 for j in range(len(cm1)/2): stt = max(cm2[i*2],cm1[j*2]) edt = min(cm2[i*2+1],cm1[j*2+...
p00592
<H1>Boring Commercial</H1> <p> Now it is spring holidays. A lazy student has finally passed all final examination, and he decided to just kick back and just watch TV all day. Oh, his only source of entertainment is watching TV. And TV commercial, as usual, are a big nuisance for him. He can watch any thing on ...
1 2100 2400 1 2130 2200 3 2100 2400 3 2100 2130 2200 2230 2300 2330 2 2130 2200 2330 2400 2 2100 2130 2330 2400 0 0 0
120 180
23,325
s564742601
p00592
u467175809
1533275899
Python
Python
py
Accepted
10
5136
787
#!/usr/bin/env python from collections import deque import itertools as it import sys import math sys.setrecursionlimit(10000000) while True: n, p, q = map(int, raw_input().split()) if n == p == q == 0: break T = [0 for i in range(2401)] for loop in range(n): k = input() ts = ...
p00592
<H1>Boring Commercial</H1> <p> Now it is spring holidays. A lazy student has finally passed all final examination, and he decided to just kick back and just watch TV all day. Oh, his only source of entertainment is watching TV. And TV commercial, as usual, are a big nuisance for him. He can watch any thing on ...
1 2100 2400 1 2130 2200 3 2100 2400 3 2100 2130 2200 2230 2300 2330 2 2130 2200 2330 2400 2 2100 2130 2330 2400 0 0 0
120 180
23,326
s502493151
p00592
u260980560
1533048164
Python
Python3
py
Accepted
20
5600
587
conv = lambda t: (t // 100)*60 + (t % 100) while 1: N, P, Q = map(int, input().split()) if N == P == Q == 0: break P = conv(P); Q = conv(Q) CM = [0]*(Q-P+1) for i in range(N): k = int(input()) *T, = map(int, input().split()) for s, t in zip(T[::2], T[1::2]): ...
p00592
<H1>Boring Commercial</H1> <p> Now it is spring holidays. A lazy student has finally passed all final examination, and he decided to just kick back and just watch TV all day. Oh, his only source of entertainment is watching TV. And TV commercial, as usual, are a big nuisance for him. He can watch any thing on ...
1 2100 2400 1 2130 2200 3 2100 2400 3 2100 2130 2200 2230 2300 2330 2 2130 2200 2330 2400 2 2100 2130 2330 2400 0 0 0
120 180
23,327
s368376049
p00592
u847467233
1530751812
Python
Python3
py
Accepted
30
5624
583
# AOJ 1006: Boring Commercials # Python3 2018.7.5 bal4u import sys from sys import stdin input = stdin.readline def d2t(d): return (d//100)*60 + (d%100) while True: n, p, q = map(int, input().split()) if n == 0: break tbl = [0]*1500 p, q = d2t(p), d2t(q) for i in range(n): k = 2*int(input()) a = list(map(in...
p00592
<H1>Boring Commercial</H1> <p> Now it is spring holidays. A lazy student has finally passed all final examination, and he decided to just kick back and just watch TV all day. Oh, his only source of entertainment is watching TV. And TV commercial, as usual, are a big nuisance for him. He can watch any thing on ...
1 2100 2400 1 2130 2200 3 2100 2400 3 2100 2130 2200 2230 2300 2330 2 2130 2200 2330 2400 2 2100 2130 2330 2400 0 0 0
120 180
23,328
s592159297
p00593
u879226672
1427907045
Python
Python
py
Accepted
20
4264
766
case = 1 while True: N = int(raw_input()) if N == 0: break mat = [[0 for i in range(N)] for j in range(N)] n = 1 for i in range(N): # upper half for j in range(i,-1,-1): if i%2==0: mat[j][i-j] = n n += 1 elif i%2==1: ...
p00593
<H1>JPEG Compression</H1> <p> The fundamental idea in the JPEG compression algorithm is to sort coeffi- cient of given image by zigzag path and encode it. In this problem, we don’t discuss about details of the algorithm, but you are asked to make simple pro- gram. You are given single integer N , and you must o...
3 4 0
Case 1: 1 2 6 3 5 7 4 8 9 Case 2: 1 2 6 7 3 5 8 13 4 9 12 14 10 11 15 16
23,329
s666937352
p00593
u879226672
1433652855
Python
Python
py
Accepted
10
4268
781
# coding: utf-8 # Here your code ! case = 1 while True: N = int(raw_input()) if N == 0: break mat = [[0 for i in xrange(N)] for j in xrange(N)] n = 1 for i in range(N): # upper half for j in reversed(xrange(i+1)): if i%2==0: mat[j][i-j] = n n += 1 ...
p00593
<H1>JPEG Compression</H1> <p> The fundamental idea in the JPEG compression algorithm is to sort coeffi- cient of given image by zigzag path and encode it. In this problem, we don’t discuss about details of the algorithm, but you are asked to make simple pro- gram. You are given single integer N , and you must o...
3 4 0
Case 1: 1 2 6 3 5 7 4 8 9 Case 2: 1 2 6 7 3 5 8 13 4 9 12 14 10 11 15 16
23,330
s613540193
p00593
u879226672
1433653057
Python
Python
py
Accepted
10
4268
788
# coding: utf-8 # Here your code ! case = 1 while True: N = int(raw_input()) if N == 0: break mat = [[0 for i in xrange(N)] for j in xrange(N)] n = 1 for i in xrange(N): # upper half for j in reversed(xrange(i+1)): if i%2==0: mat[j][i-j] = n n += 1...
p00593
<H1>JPEG Compression</H1> <p> The fundamental idea in the JPEG compression algorithm is to sort coeffi- cient of given image by zigzag path and encode it. In this problem, we don’t discuss about details of the algorithm, but you are asked to make simple pro- gram. You are given single integer N , and you must o...
3 4 0
Case 1: 1 2 6 3 5 7 4 8 9 Case 2: 1 2 6 7 3 5 8 13 4 9 12 14 10 11 15 16
23,331
s618436118
p00593
u879226672
1433653123
Python
Python
py
Accepted
10
4268
769
# coding: utf-8 case = 1 while True: N = int(raw_input()) if N == 0: break mat = [[0 for i in xrange(N)] for j in xrange(N)] n = 1 for i in xrange(N): # upper half for j in reversed(xrange(i+1)): if i%2==0: mat[j][i-j] = n n += 1 elif i...
p00593
<H1>JPEG Compression</H1> <p> The fundamental idea in the JPEG compression algorithm is to sort coeffi- cient of given image by zigzag path and encode it. In this problem, we don’t discuss about details of the algorithm, but you are asked to make simple pro- gram. You are given single integer N , and you must o...
3 4 0
Case 1: 1 2 6 3 5 7 4 8 9 Case 2: 1 2 6 7 3 5 8 13 4 9 12 14 10 11 15 16
23,332
s063594220
p00593
u077582668
1446225343
Python
Python
py
Accepted
10
6376
938
#while 1: #n=map(int,raw_input().split()) otehon = [[1,2,6,7,15,16,28,29,45], [3,5,8,14,17,27,30,44], [4,9,13,18,26,31,43], [10,12,19,25,32,42], [11,20,24,33,41], [21,23,34,40], [22,35,39], [36,38], [37]] def xyz(num): num = str(...
p00593
<H1>JPEG Compression</H1> <p> The fundamental idea in the JPEG compression algorithm is to sort coeffi- cient of given image by zigzag path and encode it. In this problem, we don’t discuss about details of the algorithm, but you are asked to make simple pro- gram. You are given single integer N , and you must o...
3 4 0
Case 1: 1 2 6 3 5 7 4 8 9 Case 2: 1 2 6 7 3 5 8 13 4 9 12 14 10 11 15 16
23,333
s808678391
p00593
u797673668
1456688862
Python
Python3
py
Accepted
20
7612
567
case = 1 while True: n = int(input()) if not n: break jpeg = [[0] * n for _ in range(n)] n1 = n - 1 px, cur = [0, 0], 1 while px[0] < n: i, j = px jpeg[i][j] = cur odd = (i + j) % 2 if px[not odd] == n1: px[odd] += 1 elif not px[odd]: ...
p00593
<H1>JPEG Compression</H1> <p> The fundamental idea in the JPEG compression algorithm is to sort coeffi- cient of given image by zigzag path and encode it. In this problem, we don’t discuss about details of the algorithm, but you are asked to make simple pro- gram. You are given single integer N , and you must o...
3 4 0
Case 1: 1 2 6 3 5 7 4 8 9 Case 2: 1 2 6 7 3 5 8 13 4 9 12 14 10 11 15 16
23,334
s444864007
p00593
u266872031
1482936267
Python
Python
py
Accepted
10
6428
626
q=1 while(1): n=int(raw_input()) if n==0: break A=[[0 for i in range(n)] for i in range(n)] up=1 m=1 for i in range(n): for j in range(i+1): if up: A[i-j][j]=m else: A[j][i-j]=m m+=1 up=not up for i in r...
p00593
<H1>JPEG Compression</H1> <p> The fundamental idea in the JPEG compression algorithm is to sort coeffi- cient of given image by zigzag path and encode it. In this problem, we don’t discuss about details of the algorithm, but you are asked to make simple pro- gram. You are given single integer N , and you must o...
3 4 0
Case 1: 1 2 6 3 5 7 4 8 9 Case 2: 1 2 6 7 3 5 8 13 4 9 12 14 10 11 15 16
23,335
s351775801
p00593
u133119785
1511956562
Python
Python3
py
Accepted
20
5652
1,211
import sys def p(t): for l in t: for n in l: print("{0:>3}".format(n),end="") print("") c = 0 for n in sys.stdin: n = int(n) if n == 0: break t = [ [ 0 for _ in range(n) ] for _ in range(n) ] c += 1 print("Case {}:".format(c)) x = 0 y = 0 up=...
p00593
<H1>JPEG Compression</H1> <p> The fundamental idea in the JPEG compression algorithm is to sort coeffi- cient of given image by zigzag path and encode it. In this problem, we don’t discuss about details of the algorithm, but you are asked to make simple pro- gram. You are given single integer N , and you must o...
3 4 0
Case 1: 1 2 6 3 5 7 4 8 9 Case 2: 1 2 6 7 3 5 8 13 4 9 12 14 10 11 15 16
23,336
s759276374
p00593
u104911888
1371001286
Python
Python
py
Accepted
20
4296
843
case=1 while True: n=input() if n==0:break value=1 grid=[[0]*n for i in range(n)] x=y=0 if n==1: grid[y][x]=1 while value<n*n: grid[y][x]=value if x<n-1: x+=1 value+=1 grid[y][x]=value else: y+=1 valu...
p00593
<H1>JPEG Compression</H1> <p> The fundamental idea in the JPEG compression algorithm is to sort coeffi- cient of given image by zigzag path and encode it. In this problem, we don’t discuss about details of the algorithm, but you are asked to make simple pro- gram. You are given single integer N , and you must o...
3 4 0
Case 1: 1 2 6 3 5 7 4 8 9 Case 2: 1 2 6 7 3 5 8 13 4 9 12 14 10 11 15 16
23,337
s262405839
p00593
u668489394
1579132607
Python
Python3
py
Accepted
20
5636
1,847
if __name__ == '__main__': TC = 1 while True: N = int(input()) if N == 0: break arr = [ [ 0 for _ in range(N) ] for _ in range(N) ] i = 0 j = 0 val = 1 prevMove = "UR" while True: arr[i][j] = val val += 1 ...
p00593
<H1>JPEG Compression</H1> <p> The fundamental idea in the JPEG compression algorithm is to sort coeffi- cient of given image by zigzag path and encode it. In this problem, we don’t discuss about details of the algorithm, but you are asked to make simple pro- gram. You are given single integer N , and you must o...
3 4 0
Case 1: 1 2 6 3 5 7 4 8 9 Case 2: 1 2 6 7 3 5 8 13 4 9 12 14 10 11 15 16
23,338
s380941603
p00593
u467175809
1533276776
Python
Python
py
Accepted
10
5028
659
#!/usr/bin/env python from collections import deque import itertools as it import sys import math sys.setrecursionlimit(10000000) for hoge in range(1, 100): n = input() if n == 0: break s, t = 1 - n, 1 cnt = 1 sign = -1 A = ['' for i in range(n)] for loop in range(n * 3): ...
p00593
<H1>JPEG Compression</H1> <p> The fundamental idea in the JPEG compression algorithm is to sort coeffi- cient of given image by zigzag path and encode it. In this problem, we don’t discuss about details of the algorithm, but you are asked to make simple pro- gram. You are given single integer N , and you must o...
3 4 0
Case 1: 1 2 6 3 5 7 4 8 9 Case 2: 1 2 6 7 3 5 8 13 4 9 12 14 10 11 15 16
23,339
s149859515
p00593
u260980560
1533048898
Python
Python3
py
Accepted
20
5604
403
t = 1 while 1: N = int(input()) if N == 0: break print("Case %d:" % t); t += 1 M = [[0]*N for i in range(N)] c = 1 for i in range(2*N-1): for j in range(max(i-N+1, 0), min(N, i+1)): if i % 2 == 0: M[i-j][j] = "%3d" % c else: ...
p00593
<H1>JPEG Compression</H1> <p> The fundamental idea in the JPEG compression algorithm is to sort coeffi- cient of given image by zigzag path and encode it. In this problem, we don’t discuss about details of the algorithm, but you are asked to make simple pro- gram. You are given single integer N , and you must o...
3 4 0
Case 1: 1 2 6 3 5 7 4 8 9 Case 2: 1 2 6 7 3 5 8 13 4 9 12 14 10 11 15 16
23,340
s149879901
p00593
u847467233
1530761837
Python
Python3
py
Accepted
30
5660
756
# AOJ 1007: JPEG Compression # Python3 2018.7.5 bal4u import sys from sys import stdin input = stdin.readline cno = 0 while True: n = int(input()) if n == 0: break cno += 1 a = [[0 for j in range(12)] for i in range(12)] m = k = f = 1; while True: if f: for r in range(k-1, -1, -1): a[r][k-1-r] = m ...
p00593
<H1>JPEG Compression</H1> <p> The fundamental idea in the JPEG compression algorithm is to sort coeffi- cient of given image by zigzag path and encode it. In this problem, we don’t discuss about details of the algorithm, but you are asked to make simple pro- gram. You are given single integer N , and you must o...
3 4 0
Case 1: 1 2 6 3 5 7 4 8 9 Case 2: 1 2 6 7 3 5 8 13 4 9 12 14 10 11 15 16
23,341
s386184267
p00595
u879226672
1427988117
Python
Python
py
Accepted
20
4200
208
def gcd(m,n): while m%n>0: m,n=n,m%n else: return n while True: try: m,n = map(int,raw_input().split()) print gcd(m,n) except EOFError: break
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,342
s448747029
p00595
u811434779
1445777490
Python
Python
py
Accepted
20
7928
138
from fractions import gcd while 1: try: a, b = map(int, raw_input().split()) print gcd(a, b) except: break
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,343
s594216074
p00595
u489809100
1447923984
Python
Python
py
Accepted
10
6452
301
def greatestCommonDivisor(num1, num2): if num1 % num2 == 0 : return num2 return num1 % num2 while True: try: a,b = map(int,raw_input().split()) except EOFError: break while a != b: if a > b: a = greatestCommonDivisor(a, b) elif b > a: b = greatestCommonDivisor(b, a) print a
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,344
s294277734
p00595
u546285759
1484024612
Python
Python3
py
Accepted
20
7660
149
while True: try: a, b= map(int, input().split()) while b!= 0: a, b= b, a%b print(a) except: break
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,345
s717936097
p00595
u078042885
1484567012
Python
Python3
py
Accepted
20
7612
97
while 1: try:a,b=map(int,input().split()) except:break while b:a,b=b,a%b print(a)
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,346
s282119943
p00595
u797673668
1492413426
Python
Python3
py
Accepted
20
7508
169
import fileinput for a, b in (map(int, line.split()) for line in fileinput.input()): if a > b: a, b = b, a while a: a, b = b % a, a print(b)
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,347
s796485569
p00595
u662126750
1497369112
Python
Python3
py
Accepted
20
7644
244
def gcd(x, y): if y==0: return x z = x % y return gcd(y, z) while True: try: a, b = (int(i) for i in input().split()) except EOFError: break if a<b: print(gcd(b,a)) else: print(gcd(a,b))
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,348
s982834447
p00595
u811733736
1507951998
Python
Python3
py
Accepted
60
10212
273
# -*- coding: utf-8 -*- """ """ import sys from sys import stdin import fractions input = stdin.readline def main(args): for line in stdin: n, m = map(int, line.split()) print(fractions.gcd(n, m)) if __name__ == '__main__': main(sys.argv[1:])
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,349
s858941424
p00595
u471400255
1513672907
Python
Python3
py
Accepted
30
5668
123
from math import gcd from sys import stdin for line in stdin: a,b = [int(i) for i in line.split()] print(gcd(a,b))
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,350
s814310883
p00595
u009288816
1521975560
Python
Python3
py
Accepted
30
5604
219
import sys def gcd(a,b): if(b): return gcd(b,(a%b)) else: return a List = [] for i in sys.stdin: List.append(i) for data in List: print(gcd(int(data.split()[0]),int(data.split()[1])))
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,351
s372090872
p00595
u043254318
1526318571
Python
Python3
py
Accepted
30
5660
252
import math def get_input(): while True: try: yield ''.join(input()) except EOFError: break N = list(get_input()) for l in range(len(N)): a,b = [int(i) for i in N[l].split()] print(math.gcd(a, b))
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,352
s710057970
p00595
u724548524
1526725330
Python
Python3
py
Accepted
30
5596
207
import sys for e in sys.stdin: p, q = map(int, e.split()) if p < q:p, q = q, p while True: if p % q == 0: print(q) break else: p, q = q, p % q
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,353
s563283091
p00595
u847467233
1530705234
Python
Python3
py
Accepted
20
5596
213
# AOJ 1009: Greatest Common Divisor # Python3 2018.7.4 bal4u def gcd(a, b): while b != 0: r = a % b a, b = b, r return a while True: try: a, b = map(int, input().split()) except: break print(gcd(a, b))
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,354
s759151577
p00595
u460331583
1346006903
Python
Python
py
Accepted
20
4220
234
def gcd(a,b): if b == 0: return a else: return gcd(b ,a%b) def lcm(a,b): return a*b / gcd(a,b) while True: try: a = map(int,raw_input().split()) a,b = a[0],a[1] print gcd(a,b) except EOFError: break
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,355
s120362366
p00595
u894941280
1346006911
Python
Python
py
Accepted
10
4224
234
def gcd(a,b): if b == 0: return a else: return gcd(b ,a%b) def lcm(a,b): return a*b / gcd(a,b) while True: try: a = map(int,raw_input().split()) a,b = a[0],a[1] print gcd(a,b) except EOFError: break
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,356
s080189102
p00595
u104911888
1368082134
Python
Python
py
Accepted
30
5888
200
import fractions while True: try: a,b=map(int,raw_input().split()) except EOFError: break print fractions.gcd(a,b)
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,357
s677118443
p00595
u260980560
1385475928
Python
Python
py
Accepted
20
4196
174
import sys for s in sys.stdin.readlines(): m,n = map(int, s.split()) if n>m: n,m=m,n while 1: r = m%n if r==0: break m,n = n,r print n
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,358
s811634772
p00595
u621997536
1388318709
Python
Python
py
Accepted
10
4192
167
while True: try: a, b = sorted(map(int, raw_input().split())) while a: tmp = a a = b % a; b = tmp print b except EOFError: break
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,359
s045755253
p00595
u633068244
1394197971
Python
Python
py
Accepted
10
4196
207
def gmd(x, y): if y == 0: return x else: return gmd(y, x%y) while True: try: a, b = map(int, raw_input().split()) print gmd(a,b) except: break
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,360
s796396230
p00595
u350508326
1400413680
Python
Python3
py
Accepted
40
6724
369
def main(): while True: try: a, b = map(int, input().split()) print(euclid_algorithm(a, b)) except EOFError as e: break def euclid_algorithm(a, b): while a and b: if a > b: a = a - b else: b = b - a return a i...
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,361
s631402543
p00595
u814278309
1592143496
Python
Python3
py
Accepted
20
5652
124
import math while 1: try: a,b = map(int,input().split()) print(math.gcd(a,b)) except: break
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,362
s131847257
p00595
u842461513
1588639661
Python
Python3
py
Accepted
20
5656
132
import math try: while True: a,b = input().split(" ") print(math.gcd(int(a),int(b))) except EOFError: m = 0
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,363
s471565037
p00595
u539753516
1532475840
Python
Python3
py
Accepted
20
5656
95
import math while True: try:print(math.gcd(*map(int,input().split(" ")))) except:break
p00595
<H1><font color="#000000">Problem A:</font> Greatest Common Divisor</H1> <p> Please find the greatest common divisor of two natural numbers. A clue is: The Euclid's algorithm is a way to resolve this task. </p> <H2>Input</H2> <p> The input file consists of several lines with pairs of two natural numbers in each lin...
57 38 60 84
19 12
23,364
s051056354
p00596
u943441430
1545273358
Python
Python3
py
Accepted
70
5620
1,628
import sys def subset(line): l = line.strip().split(" ") a = list(map(int, l)) xy = list(map(lambda x: [x // 10, x % 10], a)) dominos = [[2 if [x, y] in xy and x == y else \ 1 if [x, y] in xy else \ 1 if [y, x] in xy else 0 \ for x in range(0, 7)] for y i...
p00596
<H1><font color="#000000">Problem B:</font> Dominoes Arrangement</H1> <pre> [0, 0] [0, 1] [1, 1] [0, 2] [1, 2] [2, 2] [0, 3] [1, 3] [2, 3] [3, 3] [0, 4] [1, 4] [2, 4] [3, 4] [4, 4] [0, 5] [1, 5] [2, 5] [3, 5] [4, 5] [5, 5] [0, 6] [1, 6] [2, 6] [3, 6] [4, 6] [5, 6] [6, 6] </pre> <p> Consider the ...
6 13 23 14 24 15 25 10 00 01 11 02 12 22 03 13 23 33
Yes No
23,365
s189305121
p00596
u844945939
1417916997
Python
Python3
py
Accepted
30
6756
1,097
class UnionFind(): def __init__(self, n): self.parent = list(range(n)) self.rank = [0] * n def find(self, x): if self.parent[x] == x: return x else: self.parent[x] = self.find(self.parent[x]) return self.parent[x] def unite(self, x, y): ...
p00596
<H1><font color="#000000">Problem B:</font> Dominoes Arrangement</H1> <pre> [0, 0] [0, 1] [1, 1] [0, 2] [1, 2] [2, 2] [0, 3] [1, 3] [2, 3] [3, 3] [0, 4] [1, 4] [2, 4] [3, 4] [4, 4] [0, 5] [1, 5] [2, 5] [3, 5] [4, 5] [5, 5] [0, 6] [1, 6] [2, 6] [3, 6] [4, 6] [5, 6] [6, 6] </pre> <p> Consider the ...
6 13 23 14 24 15 25 10 00 01 11 02 12 22 03 13 23 33
Yes No
23,366
s183916687
p00596
u467175809
1533280969
Python
Python
py
Accepted
10
5036
838
#!/usr/bin/env python from collections import deque import itertools as it import sys import math sys.setrecursionlimit(10000000) while True: try: n = input() except: break A = raw_input().split() E = [[] for i in range(7)] for s in A: u = int(s[0]) v = int(s[1]) ...
p00596
<H1><font color="#000000">Problem B:</font> Dominoes Arrangement</H1> <pre> [0, 0] [0, 1] [1, 1] [0, 2] [1, 2] [2, 2] [0, 3] [1, 3] [2, 3] [3, 3] [0, 4] [1, 4] [2, 4] [3, 4] [4, 4] [0, 5] [1, 5] [2, 5] [3, 5] [4, 5] [5, 5] [0, 6] [1, 6] [2, 6] [3, 6] [4, 6] [5, 6] [6, 6] </pre> <p> Consider the ...
6 13 23 14 24 15 25 10 00 01 11 02 12 22 03 13 23 33
Yes No
23,367
s013588565
p00596
u847467233
1530779122
Python
Python3
py
Accepted
20
5596
489
# AOJ 1010: Dominoes Arrangement # Python3 2018.7.5 bal4u while True: try: n = int(input()) except: break f, g = [0]*7, [0]*7 p = list(input().split()) for x in p: if x[0] == x[1]: g[int(x[0])] = 1 else: f[int(x[0])] += 1 f[int(x[1])] += 1 if n == 1: print("Yes") continue ans = 1 for i in range(7...
p00596
<H1><font color="#000000">Problem B:</font> Dominoes Arrangement</H1> <pre> [0, 0] [0, 1] [1, 1] [0, 2] [1, 2] [2, 2] [0, 3] [1, 3] [2, 3] [3, 3] [0, 4] [1, 4] [2, 4] [3, 4] [4, 4] [0, 5] [1, 5] [2, 5] [3, 5] [4, 5] [5, 5] [0, 6] [1, 6] [2, 6] [3, 6] [4, 6] [5, 6] [6, 6] </pre> <p> Consider the ...
6 13 23 14 24 15 25 10 00 01 11 02 12 22 03 13 23 33
Yes No
23,368
s405496322
p00597
u266872031
1482961702
Python
Python
py
Accepted
10
6376
362
while(1): try: n=int(raw_input()) if n==1: print 1 else: A=[0 for i in range(n/2)] A[0]=1 for i in range(1,n/2): A[i]=1+sum(A[:i])*2 ans=sum(A)*2 if n%2: ans+=1+sum(A)*2 print ...
p00597
<H1><font color="#000000">Problem C:</font> Finding the Largest Carbon Compound Given Its Longest Chain</H1> <p> An bydrocarbon is an organic compound which contains only carbons and hydrogens. An isomer is a compound that has the same number of carbons but different structures. Heptane, for example, is a hydrocarbon...
1 4
1 8
23,369
s770315505
p00597
u847467233
1530705621
Python
Python3
py
Accepted
20
5592
217
# AOJ 1011: Finding the Largest Carbon Compound Give... # Python3 2018.7.4 bal4u a = [0]*32 a[1], a[2] = 1, 2 for i in range(3, 31): a[i] = 3*a[i-2] + 2 while True: try: i = int(input()) except: break print(a[i])
p00597
<H1><font color="#000000">Problem C:</font> Finding the Largest Carbon Compound Given Its Longest Chain</H1> <p> An bydrocarbon is an organic compound which contains only carbons and hydrogens. An isomer is a compound that has the same number of carbons but different structures. Heptane, for example, is a hydrocarbon...
1 4
1 8
23,370
s889355551
p00597
u467175809
1533282097
Python
Python
py
Accepted
10
5016
367
#!/usr/bin/env python from collections import deque import itertools as it import sys import math sys.setrecursionlimit(10000000) for line in sys.stdin: n = int(line) ans = 1 for i in range(n / 2 - 1): ans = ans * 3 + 1 if n == 1: ans = 1 elif n % 2 == 1: ans = ans * 4 + 1...
p00597
<H1><font color="#000000">Problem C:</font> Finding the Largest Carbon Compound Given Its Longest Chain</H1> <p> An bydrocarbon is an organic compound which contains only carbons and hydrogens. An isomer is a compound that has the same number of carbons but different structures. Heptane, for example, is a hydrocarbon...
1 4
1 8
23,371
s912923047
p00598
u943441430
1545296184
Python
Python3
py
Accepted
20
5632
2,387
import sys def rpn(str): r = [] stack = [] for i in range(0, len(str)): c = str[i] if c in "idsu": while len(stack) > 0: if stack[-1] in "idsuc": a = stack.pop() r.extend(a) else: break ...
p00598
<H1><font color="#000000">Problem D:</font> Operations with Finite Sets</H1> <p> Let <i>A, B, C, D, E</i> be sets of integers and let <i>U</i> is a universal set that includes all sets under consideration. All elements in any set are different (no repetitions). </p> <p> u - <b>union</b> of two sets, <i>AuB</i> = {<i...
A 3 1 3 -1 B 4 3 1 5 7 D 1 5 R 0 cAiBdD C 3 1 2 3 A 4 2 10 8 3 B 3 2 4 8 R 0 (As(AiB))uC
7 1 2 3 10
23,372
s831279361
p00598
u467175809
1533348581
Python
Python
py
Accepted
10
5036
1,875
#!/usr/bin/env python from collections import deque import itertools as it import sys import math sys.setrecursionlimit(10000000) while True: A = set() B = set() C = set() D = set() E = set() while True: try: name, _ = raw_input().split() if name == 'R': ...
p00598
<H1><font color="#000000">Problem D:</font> Operations with Finite Sets</H1> <p> Let <i>A, B, C, D, E</i> be sets of integers and let <i>U</i> is a universal set that includes all sets under consideration. All elements in any set are different (no repetitions). </p> <p> u - <b>union</b> of two sets, <i>AuB</i> = {<i...
A 3 1 3 -1 B 4 3 1 5 7 D 1 5 R 0 cAiBdD C 3 1 2 3 A 4 2 10 8 3 B 3 2 4 8 R 0 (As(AiB))uC
7 1 2 3 10
23,373
s679271492
p00600
u847467233
1530781665
Python
Python3
py
Accepted
30
5944
865
# AOJ 1014: Computation of Minimum Length of Pipeline # Python3 2018.7.5 bal4u import sys from sys import stdin input = stdin.readline import heapq while True: s, d = map(int, input().split()) if s == 0: break visited = [0]*100 Q = [] to = [[] for i in range(100)] for i in range(s): visited[i] = 1 c = list(...
p00600
<H1><font color="#000000">Problem F:</font> Computation of Minimum Length of Pipeline</H1> <p> The Aizu Wakamatsu city office decided to lay a hot water pipeline covering the whole area of the city to heat houses. The pipeline starts from some hot springs and connects every district in the city. The pipeline can fork...
3 5 12 8 25 19 23 9 13 16 0 17 20 14 16 10 22 17 27 18 16 9 7 0 19 5 21 0 0
38
23,374