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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
884/A | 884 | A | Python 3 | TESTS | 4 | 30 | 0 | 195083882 | arr = a,b = [int(x) for x in input().split()]
vals = [int(x) for x in input().split()]
time = b
day = 0
for i in vals:
day += 1
left = 86400-i
time -= left
if time == 0:
break
print(day) | 16 | 46 | 0 | 31801946 | DAY = 86400
n,k = map(int,input().split())
x = map(int,input().split())
day = 1
for i in x:
k -= (DAY-i)
if k <= 0:
break
day += 1
print(day) | Educational Codeforces Round 31 | ICPC | 2,017 | 2 | 256 | Book Reading | Recently Luba bought a very interesting book. She knows that it will take t seconds to read the book. Luba wants to finish reading as fast as she can.
But she has some work to do in each of n next days. The number of seconds that Luba has to spend working during i-th day is ai. If some free time remains, she can spend it on reading.
Help Luba to determine the minimum number of day when she finishes reading.
It is guaranteed that the answer doesn't exceed n.
Remember that there are 86400 seconds in a day. | The first line contains two integers n and t (1 ≤ n ≤ 100, 1 ≤ t ≤ 106) — the number of days and the time required to read the book.
The second line contains n integers ai (0 ≤ ai ≤ 86400) — the time Luba has to spend on her work during i-th day. | Print the minimum day Luba can finish reading the book.
It is guaranteed that answer doesn't exceed n. | null | null | [{"input": "2 2\n86400 86398", "output": "2"}, {"input": "2 86400\n0 86400", "output": "1"}] | 800 | ["implementation"] | 16 | [{"input": "2 2\r\n86400 86398\r\n", "output": "2\r\n"}, {"input": "2 86400\r\n0 86400\r\n", "output": "1\r\n"}, {"input": "2 86400\r\n1 86399\r\n", "output": "2\r\n"}, {"input": "100 1000000\r\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\r\n", "output": "12\r\n"}, {"input": "1 1\r\n86399\r\n", "output": "1\r\n"}, {"input": "6 1200\r\n86400 86400 86000 86000 86000 86400\r\n", "output": "5\r\n"}, {"input": "6 1200\r\n86400 86400 86000 86000 86001 86399\r\n", "output": "6\r\n"}, {"input": "4 172799\r\n1 1 86400 0\r\n", "output": "4\r\n"}, {"input": "4 172799\r\n0 86400 86399 0\r\n", "output": "4\r\n"}, {"input": "6 1\r\n1 1 86400 1 86399 1\r\n", "output": "1\r\n"}, {"input": "4 1\r\n86400 86399 86400 86400\r\n", "output": "2\r\n"}, {"input": "4 1\r\n86400 86400 0 86400\r\n", "output": "3\r\n"}] | false | stdio | null | true |
388/C | 388 | C | Python 3 | TESTS | 5 | 109 | 307,200 | 64127441 | def st(x):
return x[0]
try:
n=int(input())
a=0
x=[]
x1=0
d=0
for _ in range(n):
b=0
l=list(map(int,input().split()))
n1=l[0]
d+=sum(l)-l[0]
if n1%2==0:
for i in range(1,(n1//2)+1):
a+=l[i]
else:
if n1==1:
x.append([l[1],l[1]])
else:
for i in range(1,(n1//2)+1):
b+=l[i]
x.append([b+l[i+1],b])
x1+=1
if x1>=1:
x.sort(key=st,reverse=True)
x2=x1//2
for i in range(x2):
a+=x[i][0]
if x1%2!=0:
a+=x[x2][0]
x2+=1
for i in range (x2,x1):
a+=x[i][1]
print(a,d-a)
except:
pass | 43 | 108 | 1,638,400 | 111355765 | n = int(input())
S = [0] * n
ciel, giro = 0, 0
odd = []
for i in range(n):
L = list(map(int, input().split()))
k = L[0]
L = L[1:]
S[i] = sum(L)
if k % 2:
odd.append(L[k // 2])
ciel += sum(L[:k // 2])
giro += sum(L[(k + 1) // 2:])
odd.sort(reverse=True)
for i, x in enumerate(odd):
if i % 2:
giro += x
else:
ciel += x
print(ciel, giro) | Codeforces Round 228 (Div. 1) | CF | 2,014 | 1 | 256 | Fox and Card Game | Fox Ciel is playing a card game with her friend Fox Jiro. There are n piles of cards on the table. And there is a positive integer on each card.
The players take turns and Ciel takes the first turn. In Ciel's turn she takes a card from the top of any non-empty pile, and in Jiro's turn he takes a card from the bottom of any non-empty pile. Each player wants to maximize the total sum of the cards he took. The game ends when all piles become empty.
Suppose Ciel and Jiro play optimally, what is the score of the game? | The first line contain an integer n (1 ≤ n ≤ 100). Each of the next n lines contains a description of the pile: the first integer in the line is si (1 ≤ si ≤ 100) — the number of cards in the i-th pile; then follow si positive integers c1, c2, ..., ck, ..., csi (1 ≤ ck ≤ 1000) — the sequence of the numbers on the cards listed from top of the current pile to bottom of the pile. | Print two integers: the sum of Ciel's cards and the sum of Jiro's cards if they play optimally. | null | In the first example, Ciel will take the cards with number 100 and 1, Jiro will take the card with number 10.
In the second example, Ciel will take cards with numbers 2, 8, 6, 5, 9 and Jiro will take cards with numbers 4, 7, 1, 3. | [{"input": "2\n1 100\n2 1 10", "output": "101 10"}, {"input": "1\n9 2 8 6 5 9 4 7 1 3", "output": "30 15"}, {"input": "3\n3 1 3 2\n3 5 4 6\n2 8 7", "output": "18 18"}, {"input": "3\n3 1000 1000 1000\n6 1000 1000 1000 1000 1000 1000\n5 1000 1000 1000 1000 1000", "output": "7000 7000"}] | 2,000 | ["games", "greedy", "sortings"] | 43 | [{"input": "2\r\n1 100\r\n2 1 10\r\n", "output": "101 10\r\n"}, {"input": "1\r\n9 2 8 6 5 9 4 7 1 3\r\n", "output": "30 15\r\n"}, {"input": "3\r\n3 1 3 2\r\n3 5 4 6\r\n2 8 7\r\n", "output": "18 18\r\n"}, {"input": "3\r\n3 1000 1000 1000\r\n6 1000 1000 1000 1000 1000 1000\r\n5 1000 1000 1000 1000 1000\r\n", "output": "7000 7000\r\n"}, {"input": "1\r\n1 1\r\n", "output": "1 0\r\n"}, {"input": "5\r\n1 3\r\n1 2\r\n1 8\r\n1 1\r\n1 4\r\n", "output": "12 6\r\n"}, {"input": "3\r\n5 1 2 3 4 5\r\n4 1 2 3 4\r\n8 1 2 3 4 5 6 7 8\r\n", "output": "19 42\r\n"}, {"input": "5\r\n5 1 1 1 1 1\r\n4 1 1 1 1\r\n3 1 1 1\r\n2 1 1\r\n1 1\r\n", "output": "8 7\r\n"}, {"input": "6\r\n2 1 1\r\n2 2 2\r\n2 3 3\r\n2 4 4\r\n2 5 5\r\n2 6 6\r\n", "output": "21 21\r\n"}, {"input": "2\r\n2 200 1\r\n3 1 100 2\r\n", "output": "301 3\r\n"}, {"input": "2\r\n3 1 1000 2\r\n3 2 1 1\r\n", "output": "1003 4\r\n"}, {"input": "4\r\n3 1 5 100\r\n3 1 5 100\r\n3 100 1 1\r\n3 100 1 1\r\n", "output": "208 208\r\n"}] | false | stdio | null | true |
884/A | 884 | A | Python 3 | TESTS | 5 | 62 | 0 | 31904011 | [n, t] = list(map(int, input().split()))
l = list(map(int, input().split()))
for i in range(n):
if l[i] < 86400:
if t-86400+l[i] <= 0 :
print(i+1)
break
else:
t -= (86400+l[i]) | 16 | 46 | 0 | 31802050 | d, t = list(map(int, input().split()))
h = list(map(int, input().split()))
s = 86400
czas = 0
days = 0
for i in range(d):
days += 1
czas += s - h[i]
if czas >= t:
print(days)
quit() | Educational Codeforces Round 31 | ICPC | 2,017 | 2 | 256 | Book Reading | Recently Luba bought a very interesting book. She knows that it will take t seconds to read the book. Luba wants to finish reading as fast as she can.
But she has some work to do in each of n next days. The number of seconds that Luba has to spend working during i-th day is ai. If some free time remains, she can spend it on reading.
Help Luba to determine the minimum number of day when she finishes reading.
It is guaranteed that the answer doesn't exceed n.
Remember that there are 86400 seconds in a day. | The first line contains two integers n and t (1 ≤ n ≤ 100, 1 ≤ t ≤ 106) — the number of days and the time required to read the book.
The second line contains n integers ai (0 ≤ ai ≤ 86400) — the time Luba has to spend on her work during i-th day. | Print the minimum day Luba can finish reading the book.
It is guaranteed that answer doesn't exceed n. | null | null | [{"input": "2 2\n86400 86398", "output": "2"}, {"input": "2 86400\n0 86400", "output": "1"}] | 800 | ["implementation"] | 16 | [{"input": "2 2\r\n86400 86398\r\n", "output": "2\r\n"}, {"input": "2 86400\r\n0 86400\r\n", "output": "1\r\n"}, {"input": "2 86400\r\n1 86399\r\n", "output": "2\r\n"}, {"input": "100 1000000\r\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\r\n", "output": "12\r\n"}, {"input": "1 1\r\n86399\r\n", "output": "1\r\n"}, {"input": "6 1200\r\n86400 86400 86000 86000 86000 86400\r\n", "output": "5\r\n"}, {"input": "6 1200\r\n86400 86400 86000 86000 86001 86399\r\n", "output": "6\r\n"}, {"input": "4 172799\r\n1 1 86400 0\r\n", "output": "4\r\n"}, {"input": "4 172799\r\n0 86400 86399 0\r\n", "output": "4\r\n"}, {"input": "6 1\r\n1 1 86400 1 86399 1\r\n", "output": "1\r\n"}, {"input": "4 1\r\n86400 86399 86400 86400\r\n", "output": "2\r\n"}, {"input": "4 1\r\n86400 86400 0 86400\r\n", "output": "3\r\n"}] | false | stdio | null | true |
18/D | 18 | D | Python 3 | TESTS | 36 | 186 | 5,529,600 | 25694685 | n = int(input())
mx = 2009
d = [0 for i in range(mx)]
ff = True
for i in range(n):
s = input().split()
x = int(s[1])
if ff :
f = x
ff = False
if s[0] == 'win':
d[x] = d[0]+ 2**x
else:
d[0] = max(d[x], d[0])
if n == 5000 and (f == 1364 or f == 1158):
d[0] -= 1
print(d[0]) | 45 | 92 | 512,000 | 210243590 | n = int(input())
a = []
b = []
for i in range(n):
x = input().split()
a.append(x[0][0])
b.append(int(x[1]))
res = 0
f = [-1]*2002
for i in range(n):
if (a[i]=='w'):
f[b[i]] = res
elif f[b[i]]>=0:
res = max(res, f[b[i]]+2**b[i])
print(res) | Codeforces Beta Round 18 (Div. 2 Only) | ICPC | 2,010 | 2 | 128 | Seller Bob | Last year Bob earned by selling memory sticks. During each of n days of his work one of the two following events took place:
- A customer came to Bob and asked to sell him a 2x MB memory stick. If Bob had such a stick, he sold it and got 2x berllars.
- Bob won some programming competition and got a 2x MB memory stick as a prize. Bob could choose whether to present this memory stick to one of his friends, or keep it.
Bob never kept more than one memory stick, as he feared to mix up their capacities, and deceive a customer unintentionally. It is also known that for each memory stick capacity there was at most one customer, who wanted to buy that memory stick. Now, knowing all the customers' demands and all the prizes won at programming competitions during the last n days, Bob wants to know, how much money he could have earned, if he had acted optimally. | The first input line contains number n (1 ≤ n ≤ 5000) — amount of Bob's working days. The following n lines contain the description of the days. Line sell x stands for a day when a customer came to Bob to buy a 2x MB memory stick (0 ≤ x ≤ 2000). It's guaranteed that for each x there is not more than one line sell x. Line win x stands for a day when Bob won a 2x MB memory stick (0 ≤ x ≤ 2000). | Output the maximum possible earnings for Bob in berllars, that he would have had if he had known all the events beforehand. Don't forget, please, that Bob can't keep more than one memory stick at a time. | null | null | [{"input": "7\nwin 10\nwin 5\nwin 3\nsell 5\nsell 3\nwin 10\nsell 10", "output": "1056"}, {"input": "3\nwin 5\nsell 6\nsell 4", "output": "0"}] | 2,000 | ["brute force", "dp", "greedy"] | 45 | [{"input": "7\r\nwin 10\r\nwin 5\r\nwin 3\r\nsell 5\r\nsell 3\r\nwin 10\r\nsell 10\r\n", "output": "1056\r\n"}, {"input": "3\r\nwin 5\r\nsell 6\r\nsell 4\r\n", "output": "0\r\n"}, {"input": "10\r\nsell 179\r\nwin 1278\r\nsell 1278\r\nwin 179\r\nwin 788\r\nsell 788\r\nwin 1819\r\nwin 1278\r\nsell 1454\r\nsell 1819\r\n", "output": "37459511778596727480858760720167552241582636504705413766024169777495064333423937410125519624693990051069809575647477719465460756326341562228323606665869931977125977431028709943048934214062888966581139223580790503937962827597404798307897711090567429316074325427043388117806141094834711707585035634104732053207574452493593409130554278913951011894497392495930884827685983975668127973918422057605356890341647839399778378381152159725053311750647457999739578989105335906181048932656785993705124392163591312698147450544686667028530553878359909362604965888\n"}, {"input": "10\r\nsell 573\r\nwin 1304\r\nsell 278\r\nwin 1631\r\nsell 1225\r\nsell 1631\r\nsell 177\r\nwin 1631\r\nwin 177\r\nsell 1304\r\n", "output": "95482312335125227379668481690754940528280513838693267460502082967052005332103697568042408703168913727303170456338425853153094403747135188778307041838920404959089576368946137708987138986696495077466398994298434148881715073638178666201165545650953479735059082316661443204882826188032944866093372620219104327689636641547141835841165681118172603993695103043804276669836594061369229043451067647935298287687852302215923887110435577776767805943668204998410716005202198549540411238299513630278811648\n"}, {"input": "10\r\nwin 1257\r\nwin 1934\r\nsell 1934\r\nsell 1257\r\nwin 1934\r\nwin 1257\r\nsell 495\r\nwin 495\r\nwin 495\r\nwin 1257\r\n", "output": "1556007242642049292787218246793379348327505438878680952714050868520307364441227819009733220897932984584977593931988662671459594674963394056587723382487766303981362587048873128400436836690128983570130687310221668877557121158055843621982630476422478413285775826498536883275291967793661985813155062733063913176306327509625594121241472451054995889483447103432414676059872469910105149496451402271546454282618581884282152530090816240540173251729211604658704990425330422792556824836640431985211146197816770068601144273155826261981248156109439380758754562057681271989937590784957541131485184\n"}, {"input": "10\r\nsell 1898\r\nsell 173\r\nsell 1635\r\nsell 29\r\nsell 881\r\nsell 434\r\nsell 1236\r\nsell 14\r\nwin 29\r\nsell 1165\r\n", "output": "0\r\n"}, {"input": "1\r\nsell 2000\r\n", "output": "0\r\n"}, {"input": "1\r\nwin 2000\r\n", "output": "0\r\n"}, {"input": "2\r\nwin 2000\r\nsell 2000\r\n", "output": "114813069527425452423283320117768198402231770208869520047764273682576626139237031385665948631650626991844596463898746277344711896086305533142593135616665318539129989145312280000688779148240044871428926990063486244781615463646388363947317026040466353970904996558162398808944629605623311649536164221970332681344168908984458505602379484807914058900934776500429002716706625830522008132236281291761267883317206598995396418127021779858404042159853183251540889433902091920554957783589672039160081957216630582755380425583726015528348786419432054508915275783882625175435528800822842770817965453762184851149029376\n"}] | false | stdio | null | true |
324/A1 | 331 | A1 | Python 3 | TESTS1 | 8 | 140 | 0 | 4137368 | """
obs1: cu posibila exceptie a capetelor secventei alese,
imi permit sa sterg toate valorile negative din sir
avand in vedere ca am ramas doar cu valori pozitive,
tintesc sa obtin o bucata cat mai lunga de valori
astfel ca atunci cand vreau sa fixez capetele la valorile x -> x
le voi fixa asa: prima aparitie a lui x in sir -> ultima aparitie a lui x in sir
imi fac un vector de sume partiale ale valorilor pozitive
ca sa pot vedea ce am in secventa :)
"""
n = int(input())
v = list(map(int, input().split()))
valmax = max(v) + 5
valmin = min(v) - 5
sol = -(10**6)
first = [-1 for i in range(valmin, valmax)]
last = [-1 for i in range(valmin, valmax)]
s = [0 for i in range(-2, n+5)]
s[0] = max(v[0], 0)
for i in range(1, len(v)):
s[i] = s[i-1]
if v[i] > 0: s[i] += v[i]
for i in range(0, len(v)):
last[v[i]] = i
if first[v[i]] == -1: first[v[i]] = i
for i in range(0, len(v)):
val = v[i]
if i != first[val]: continue
total = s[ last[val] ] - s[ first[val]-1 ]
if val < 0: total += 2*val
if total > sol:
sol = total
left = first[val]
right = last[val]
cutTrees = []
for i in range(0, left): cutTrees.append(i)
for i in range(left+1, right):
if v[i] < 0: cutTrees.append(i)
for i in range(right+1, n): cutTrees.append(i)
print("%i %i" % (sol, len(cutTrees)))
for x in cutTrees: print(x+1, end=' ') | 18 | 186 | 0 | 119127446 | def main():
n, aa = int(input()), list(map(int, input().split()))
partialsum, s, d, ranges = [0] * n, 0, {}, []
for hi, a in enumerate(aa):
if a in d:
base = s + a * 2
for lo in d[a]:
ranges.append((base - partialsum[lo], lo, hi))
d[a].append(hi)
else:
d[a] = [hi]
if a > 0:
s += a
partialsum[hi] = s
s, lo, hi = max(ranges)
res = list(range(1, lo + 1))
for i in range(lo + 1, hi):
if aa[i] < 0:
res.append(i + 1)
res.extend(range(hi + 2, n + 1))
print(s, len(res))
print(" ".join(map(str, res)))
if __name__ == '__main__':
main() | ABBYY Cup 3.0 - Finals | ICPC | 2,013 | 2 | 256 | Oh Sweet Beaverette | — Oh my sweet Beaverette, would you fancy a walk along a wonderful woodland belt with me?
— Of course, my Smart Beaver! Let us enjoy the splendid view together. How about Friday night?
At this point the Smart Beaver got rushing. Everything should be perfect by Friday, so he needed to prepare the belt to the upcoming walk. He needed to cut down several trees.
Let's consider the woodland belt as a sequence of trees. Each tree i is described by the esthetic appeal ai — some trees are very esthetically pleasing, others are 'so-so', and some trees are positively ugly!
The Smart Beaver calculated that he needed the following effects to win the Beaverette's heart:
- The first objective is to please the Beaverette: the sum of esthetic appeal of the remaining trees must be maximum possible;
- the second objective is to surprise the Beaverette: the esthetic appeal of the first and the last trees in the resulting belt must be the same;
- and of course, the walk should be successful: there must be at least two trees in the woodland belt left.
Now help the Smart Beaver! Which trees does he need to cut down to win the Beaverette's heart? | The first line contains a single integer n — the initial number of trees in the woodland belt, 2 ≤ n. The second line contains space-separated integers ai — the esthetic appeals of each tree. All esthetic appeals do not exceed 109 in their absolute value.
- to get 30 points, you need to solve the problem with constraints: n ≤ 100 (subproblem A1);
- to get 100 points, you need to solve the problem with constraints: n ≤ 3·105 (subproblems A1+A2). | In the first line print two integers — the total esthetic appeal of the woodland belt after the Smart Beaver's intervention and the number of the cut down trees k.
In the next line print k integers — the numbers of the trees the Beaver needs to cut down. Assume that the trees are numbered from 1 to n from left to right.
If there are multiple solutions, print any of them. It is guaranteed that at least two trees have equal esthetic appeal. | null | null | [{"input": "5\n1 2 3 1 2", "output": "8 1\n1"}, {"input": "5\n1 -2 3 1 -2", "output": "5 2\n2 5"}] | 1,400 | [] | 18 | [{"input": "5\r\n1 2 3 1 2\r\n", "output": "8 1\r\n1 "}, {"input": "5\r\n1 -2 3 1 -2\r\n", "output": "5 2\r\n2 5 "}, {"input": "2\r\n0 0\r\n", "output": "0 0\r\n"}, {"input": "3\r\n0 -1 0\r\n", "output": "0 1\r\n2 "}, {"input": "3\r\n1 1 1\r\n", "output": "3 0\r\n"}, {"input": "4\r\n-1 1 1 -1\r\n", "output": "2 2\r\n1 4 "}, {"input": "4\r\n-1 1 -1 1\r\n", "output": "2 2\r\n1 3 "}, {"input": "2\r\n-1 -1\r\n", "output": "-2 0\r\n"}, {"input": "3\r\n-1 0 -1\r\n", "output": "-2 0\r\n"}, {"input": "6\r\n-1 3 3 5 5 -1\r\n", "output": "14 0\r\n"}, {"input": "2\r\n-1000000000 -1000000000\r\n", "output": "-2000000000 0\r\n"}, {"input": "3\r\n-1000000000 -1000000000 -1000000000\r\n", "output": "-2000000000 1\r\n3 "}, {"input": "3\r\n1000000000 1000000000 1000000000\r\n", "output": "3000000000 0\r\n"}, {"input": "10\r\n-589330597 -126288833 -126288833 -834860352 -834860352 -834860352 -834860352 -21170405 -834860352 -834860352\r\n", "output": "-252577666 8\r\n1 4 5 6 7 8 9 10 "}, {"input": "20\r\n-808998072 733614990 579897311 -337992089 579897311 120800519 -337992089 -803027570 733614990 -686536765 733614990 -803027570 -803027570 733614990 120800519 -803027570 -686536765 579897311 -808998072 -686536765\r\n", "output": "4215055101 13\r\n1 4 7 8 10 12 13 15 16 17 18 19 20 "}] | false | stdio | import sys
from collections import defaultdict
def main(input_path, output_path, submission_output_path):
with open(input_path) as f:
n = int(f.readline())
a = list(map(int, f.readline().split()))
with open(submission_output_path) as f:
lines = f.readlines()
if not lines:
print(0)
return
sum_line = lines[0].strip().split()
if len(sum_line) != 2:
print(0)
return
try:
sum_claimed, k = map(int, sum_line)
except:
print(0)
return
cut_list = list(map(int, lines[1].split())) if len(lines) > 1 else []
if len(cut_list) != k:
print(0)
return
cut_set = set(cut_list)
if len(cut_set) != k:
print(0)
return
for num in cut_list:
if not (1 <= num <= n):
print(0)
return
remaining = [i-1 for i in range(1, n+1) if i not in cut_set]
if len(remaining) < 2:
print(0)
return
first_val = a[remaining[0]]
last_val = a[remaining[-1]]
if first_val != last_val:
print(0)
return
sum_remaining = sum(a[i] for i in remaining)
if sum_remaining != sum_claimed:
print(0)
return
prefix = [0] * (n + 1)
for i in range(n):
prefix[i+1] = prefix[i] + max(a[i], 0)
groups = defaultdict(list)
for idx, val in enumerate(a):
groups[val].append(idx)
max_total = -float('inf')
for v, indices in groups.items():
m = len(indices)
if m < 2:
continue
indices.sort()
min_val = prefix[indices[0] + 1]
current_max = -float('inf')
for j in range(1, m):
current_i = indices[j]
current_sum = 2 * v + (prefix[current_i] - min_val)
if current_sum > current_max:
current_max = current_sum
if j < m - 1:
next_val = prefix[indices[j] + 1]
if next_val < min_val:
min_val = next_val
if current_max > max_total:
max_total = current_max
if sum_claimed != max_total:
print(0)
else:
print(100)
if __name__ == "__main__":
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_output_path = sys.argv[3]
main(input_path, output_path, submission_output_path) | true |
526/B | 526 | B | PyPy 3 | TESTS | 3 | 93 | 0 | 108399114 | import math
def add(n, arr):
sumdou = [0 for i in range(len(arr))]
for i in reversed(range(len(arr))):
if 2*i+2 >= len(arr):
sumdou[i] = arr[i]
else:
sumdou[i] = sumdou[2*i+2]+arr[i]
tot = 0
p2 = len(arr)
for j in reversed(range(n)):
p1 = p2 - int(math.pow(2,j+1))
for i in range(p1, p2, 2):
s1 = sumdou[i]
s2 = sumdou[i+1]
tot += abs(s2-s1)
sumdou[i] = max(s1,s2)
sumdou[i+1] = max(s1,s2)
p2 = p1
print(tot)
n = int(input())
arr = list(map(int, input().split()))
add(n, arr) | 38 | 46 | 0 | 10579089 | def B():
s = 0
n = int(input())
k = (1 << (n + 1)) - 1
a = [0, 0] + list(map(int, input().split()))
for i in range(k, 1, -2):
u, v = a[i], a[i - 1]
if u > v: u, v = v, u
s += v - u
a[i >> 1] += v
return s
print(B()) | ZeptoLab Code Rush 2015 | CF | 2,015 | 1 | 256 | Om Nom and Dark Park | Om Nom is the main character of a game "Cut the Rope". He is a bright little monster who likes visiting friends living at the other side of the park. However the dark old parks can scare even somebody as fearless as Om Nom, so he asks you to help him.
The park consists of 2n + 1 - 1 squares connected by roads so that the scheme of the park is a full binary tree of depth n. More formally, the entrance to the park is located at the square 1. The exits out of the park are located at squares 2n, 2n + 1, ..., 2n + 1 - 1 and these exits lead straight to the Om Nom friends' houses. From each square i (2 ≤ i < 2n + 1) there is a road to the square $$\frac{i}{2}$$. Thus, it is possible to go from the park entrance to each of the exits by walking along exactly n roads.
Om Nom loves counting lights on the way to his friend. Om Nom is afraid of spiders who live in the park, so he doesn't like to walk along roads that are not enough lit. What he wants is that the way to any of his friends should have in total the same number of lights. That will make him feel safe.
He asked you to help him install additional lights. Determine what minimum number of lights it is needed to additionally place on the park roads so that a path from the entrance to any exit of the park contains the same number of street lights. You may add an arbitrary number of street lights to each of the roads. | The first line contains integer n (1 ≤ n ≤ 10) — the number of roads on the path from the entrance to any exit.
The next line contains 2n + 1 - 2 numbers a2, a3, ... a2n + 1 - 1 — the initial numbers of street lights on each road of the park. Here ai is the number of street lights on the road between squares i and $$\frac{i}{2}$$. All numbers ai are positive integers, not exceeding 100. | Print the minimum number of street lights that we should add to the roads of the park to make Om Nom feel safe. | null | Picture for the sample test. Green color denotes the additional street lights. | [{"input": "2\n1 2 3 4 5 6", "output": "5"}] | 1,400 | ["dfs and similar", "greedy", "implementation"] | 38 | [{"input": "2\r\n1 2 3 4 5 6\r\n", "output": "5\r\n"}, {"input": "2\r\n1 2 3 3 2 2\r\n", "output": "0\r\n"}, {"input": "1\r\n39 52\r\n", "output": "13\r\n"}, {"input": "2\r\n59 96 34 48 8 72\r\n", "output": "139\r\n"}, {"input": "3\r\n87 37 91 29 58 45 51 74 70 71 47 38 91 89\r\n", "output": "210\r\n"}, {"input": "5\r\n39 21 95 89 73 90 9 55 85 32 30 21 68 59 82 91 20 64 52 70 6 88 53 47 30 47 34 14 11 22 42 15 28 54 37 48 29 3 14 13 18 77 90 58 54 38 94 49 45 66 13 74 11 14 64 72 95 54 73 79 41 35\r\n", "output": "974\r\n"}, {"input": "1\r\n49 36\r\n", "output": "13\r\n"}, {"input": "1\r\n77 88\r\n", "output": "11\r\n"}, {"input": "1\r\n1 33\r\n", "output": "32\r\n"}, {"input": "2\r\n72 22 81 23 14 75\r\n", "output": "175\r\n"}, {"input": "2\r\n100 70 27 1 68 52\r\n", "output": "53\r\n"}, {"input": "2\r\n24 19 89 82 22 21\r\n", "output": "80\r\n"}, {"input": "3\r\n86 12 92 91 3 68 57 56 76 27 33 62 71 84\r\n", "output": "286\r\n"}, {"input": "3\r\n14 56 53 61 57 45 40 44 31 9 73 2 61 26\r\n", "output": "236\r\n"}, {"input": "3\r\n35 96 7 43 10 14 16 36 95 92 16 50 59 55\r\n", "output": "173\r\n"}, {"input": "4\r\n1 97 18 48 96 65 24 91 17 45 36 27 74 93 78 86 39 55 53 21 26 68 31 33 79 63 80 92 1 26\r\n", "output": "511\r\n"}, {"input": "4\r\n25 42 71 29 50 30 99 79 77 24 76 66 68 23 97 99 65 17 75 62 66 46 48 4 40 71 98 57 21 92\r\n", "output": "603\r\n"}, {"input": "4\r\n49 86 17 7 3 6 86 71 36 10 27 10 58 64 12 16 88 67 93 3 15 20 58 87 97 91 11 6 34 62\r\n", "output": "470\r\n"}, {"input": "5\r\n16 87 36 16 81 53 87 35 63 56 47 91 81 95 80 96 91 7 58 99 25 28 47 60 7 69 49 14 51 52 29 30 83 23 21 52 100 26 91 14 23 94 72 70 40 12 50 32 54 52 18 74 5 15 62 3 48 41 24 25 56 43\r\n", "output": "1060\r\n"}, {"input": "5\r\n40 27 82 94 38 22 66 23 18 34 87 31 71 28 95 5 14 61 76 52 66 6 60 40 68 77 70 63 64 18 47 13 82 55 34 64 30 1 29 24 24 9 65 17 29 96 61 76 72 23 32 26 90 39 54 41 35 66 71 29 75 48\r\n", "output": "1063\r\n"}, {"input": "5\r\n64 72 35 68 92 95 45 15 77 16 26 74 61 65 18 22 32 19 98 97 14 84 70 23 29 1 87 28 88 89 73 79 69 88 43 60 64 64 66 39 17 27 46 71 18 83 73 20 90 77 49 70 84 63 50 72 26 87 26 37 78 65\r\n", "output": "987\r\n"}, {"input": "6\r\n35 61 54 77 70 50 53 70 4 66 58 47 76 100 78 5 43 50 55 93 13 93 59 92 30 74 22 23 98 70 19 56 90 92 19 7 28 53 45 77 42 91 71 56 19 83 100 53 13 93 37 13 70 60 16 13 76 3 12 22 17 26 50 6 63 7 25 41 92 29 36 80 11 4 10 14 77 75 53 82 46 24 56 46 82 36 80 75 8 45 24 22 90 34 45 76 18 38 86 43 7 49 80 56 90 53 12 51 98 47 44 58 32 4 2 6 3 60 38 72 74 46 30 86 1 98\r\n", "output": "2499\r\n"}, {"input": "6\r\n63 13 100 54 31 15 29 58 59 44 2 99 70 33 97 14 70 12 73 42 65 71 68 67 87 83 43 84 18 41 37 22 81 24 27 11 57 28 83 92 39 1 56 15 16 67 16 97 31 52 50 65 63 89 8 52 55 20 71 27 28 35 86 92 94 60 10 65 83 63 89 71 34 20 78 40 34 62 2 86 100 81 87 69 25 4 52 17 57 71 62 38 1 3 54 71 34 85 20 60 80 23 82 47 4 19 7 18 14 18 28 27 4 55 26 71 45 9 2 40 67 28 32 19 81 92\r\n", "output": "2465\r\n"}, {"input": "6\r\n87 62 58 32 81 92 12 50 23 27 38 39 64 74 16 35 84 59 91 87 14 48 90 47 44 95 64 45 31 11 67 5 80 60 36 15 91 3 21 2 40 24 37 69 5 50 23 37 49 19 68 21 49 9 100 94 45 41 22 31 31 48 25 70 25 25 95 88 82 1 37 53 49 31 57 74 94 45 55 93 43 37 13 85 59 72 15 68 3 90 96 55 100 64 63 69 43 33 66 84 57 97 87 34 23 89 97 77 39 89 8 92 68 13 50 36 95 61 71 96 73 13 30 49 57 89\r\n", "output": "2513\r\n"}] | false | stdio | null | true |
324/A1 | 331 | A2 | PyPy 3-64 | TESTS2 | 7 | 154 | 0 | 207278093 | import sys
input = lambda: sys.stdin.readline().rstrip()
from collections import defaultdict
N = int(input())
A = list(map(int, input().split()))
ans = 0
lib = {}
cur = 0
d = 0
for i,a in enumerate(A):
if a>0:
cur+=a
if a not in lib.keys():
lib[a]=cur
else:
if ans<cur-lib[a]+a:
ans = cur-lib[a]+a
d = a
ret = []
for i in range(N):
if A[i]<0:
ret.append(i+1)
for i in range(N):
if A[i]==d:break
if A[i]>0:
ret.append(i+1)
for i in range(N-1,-1,-1):
if A[i]==d:break
if A[i]>0:
ret.append(i+1)
ret.sort()
print(ans,len(ret))
print(*ret) | 18 | 186 | 0 | 168028719 | import sys
input = sys.stdin.readline
n = int(input())
w = list(map(int, input().split()))
d = dict()
p = []
c = 0
for i in range(n):
if w[i] not in d:
d[w[i]] = [1, i, i]
else:
if d[w[i]][0] != 2:
d[w[i]][0] = 2
d[w[i]][2] = i
if w[i] > 0:
c += w[i]
p.append(c)
x = -100000000000
b = (0, 0)
for i in d:
if d[i][0] == 2:
x1 = p[d[i][2]] - (p[d[i][1]-1] if d[i][1] != 0 else 0)
if i < 0:
x1 += i+i
if x1 > x:
x = x1
b = (d[i][1], d[i][2])
e = []
e.extend(list(range(1, b[0]+1)))
for i in range(b[0]+1, b[1]):
if w[i] < 0:
e.append(i+1)
e.extend(list(range(b[1] + 2, n + 1)))
print(x, len(e))
print(' '.join(map(str, e))) | ABBYY Cup 3.0 - Finals | ICPC | 2,013 | 2 | 256 | Oh Sweet Beaverette | — Oh my sweet Beaverette, would you fancy a walk along a wonderful woodland belt with me?
— Of course, my Smart Beaver! Let us enjoy the splendid view together. How about Friday night?
At this point the Smart Beaver got rushing. Everything should be perfect by Friday, so he needed to prepare the belt to the upcoming walk. He needed to cut down several trees.
Let's consider the woodland belt as a sequence of trees. Each tree i is described by the esthetic appeal ai — some trees are very esthetically pleasing, others are 'so-so', and some trees are positively ugly!
The Smart Beaver calculated that he needed the following effects to win the Beaverette's heart:
- The first objective is to please the Beaverette: the sum of esthetic appeal of the remaining trees must be maximum possible;
- the second objective is to surprise the Beaverette: the esthetic appeal of the first and the last trees in the resulting belt must be the same;
- and of course, the walk should be successful: there must be at least two trees in the woodland belt left.
Now help the Smart Beaver! Which trees does he need to cut down to win the Beaverette's heart? | The first line contains a single integer n — the initial number of trees in the woodland belt, 2 ≤ n. The second line contains space-separated integers ai — the esthetic appeals of each tree. All esthetic appeals do not exceed 109 in their absolute value.
- to get 30 points, you need to solve the problem with constraints: n ≤ 100 (subproblem A1);
- to get 100 points, you need to solve the problem with constraints: n ≤ 3·105 (subproblems A1+A2). | In the first line print two integers — the total esthetic appeal of the woodland belt after the Smart Beaver's intervention and the number of the cut down trees k.
In the next line print k integers — the numbers of the trees the Beaver needs to cut down. Assume that the trees are numbered from 1 to n from left to right.
If there are multiple solutions, print any of them. It is guaranteed that at least two trees have equal esthetic appeal. | null | null | [{"input": "5\n1 2 3 1 2", "output": "8 1\n1"}, {"input": "5\n1 -2 3 1 -2", "output": "5 2\n2 5"}] | 1,400 | [] | 18 | [{"input": "5\r\n1 2 3 1 2\r\n", "output": "8 1\r\n1 "}, {"input": "5\r\n1 -2 3 1 -2\r\n", "output": "5 2\r\n2 5 "}, {"input": "2\r\n0 0\r\n", "output": "0 0\r\n"}, {"input": "3\r\n0 -1 0\r\n", "output": "0 1\r\n2 "}, {"input": "3\r\n1 1 1\r\n", "output": "3 0\r\n"}, {"input": "4\r\n-1 1 1 -1\r\n", "output": "2 2\r\n1 4 "}, {"input": "4\r\n-1 1 -1 1\r\n", "output": "2 2\r\n1 3 "}, {"input": "2\r\n-1 -1\r\n", "output": "-2 0\r\n"}, {"input": "3\r\n-1 0 -1\r\n", "output": "-2 0\r\n"}, {"input": "6\r\n-1 3 3 5 5 -1\r\n", "output": "14 0\r\n"}, {"input": "2\r\n-1000000000 -1000000000\r\n", "output": "-2000000000 0\r\n"}, {"input": "3\r\n-1000000000 -1000000000 -1000000000\r\n", "output": "-2000000000 1\r\n3 "}, {"input": "3\r\n1000000000 1000000000 1000000000\r\n", "output": "3000000000 0\r\n"}, {"input": "10\r\n-589330597 -126288833 -126288833 -834860352 -834860352 -834860352 -834860352 -21170405 -834860352 -834860352\r\n", "output": "-252577666 8\r\n1 4 5 6 7 8 9 10 "}, {"input": "20\r\n-808998072 733614990 579897311 -337992089 579897311 120800519 -337992089 -803027570 733614990 -686536765 733614990 -803027570 -803027570 733614990 120800519 -803027570 -686536765 579897311 -808998072 -686536765\r\n", "output": "4215055101 13\r\n1 4 7 8 10 12 13 15 16 17 18 19 20 "}] | false | stdio | import sys
from collections import defaultdict
def main(input_path, output_path, submission_output_path):
with open(input_path) as f:
n = int(f.readline())
a = list(map(int, f.readline().split()))
with open(submission_output_path) as f:
lines = f.readlines()
if not lines:
print(0)
return
sum_line = lines[0].strip().split()
if len(sum_line) != 2:
print(0)
return
try:
sum_claimed, k = map(int, sum_line)
except:
print(0)
return
cut_list = list(map(int, lines[1].split())) if len(lines) > 1 else []
if len(cut_list) != k:
print(0)
return
cut_set = set(cut_list)
if len(cut_set) != k:
print(0)
return
for num in cut_list:
if not (1 <= num <= n):
print(0)
return
remaining = [i-1 for i in range(1, n+1) if i not in cut_set]
if len(remaining) < 2:
print(0)
return
first_val = a[remaining[0]]
last_val = a[remaining[-1]]
if first_val != last_val:
print(0)
return
sum_remaining = sum(a[i] for i in remaining)
if sum_remaining != sum_claimed:
print(0)
return
prefix = [0] * (n + 1)
for i in range(n):
prefix[i+1] = prefix[i] + max(a[i], 0)
groups = defaultdict(list)
for idx, val in enumerate(a):
groups[val].append(idx)
max_total = -float('inf')
for v, indices in groups.items():
m = len(indices)
if m < 2:
continue
indices.sort()
min_val = prefix[indices[0] + 1]
current_max = -float('inf')
for j in range(1, m):
current_i = indices[j]
current_sum = 2 * v + (prefix[current_i] - min_val)
if current_sum > current_max:
current_max = current_sum
if j < m - 1:
next_val = prefix[indices[j] + 1]
if next_val < min_val:
min_val = next_val
if current_max > max_total:
max_total = current_max
if sum_claimed != max_total:
print(0)
else:
print(100)
if __name__ == "__main__":
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_output_path = sys.argv[3]
main(input_path, output_path, submission_output_path) | true |
324/A1 | 331 | A1 | PyPy 3 | TESTS1 | 7 | 154 | 0 | 222149040 | n = int(input())
a = list(map(int, input().split()))
best_sum = -float('inf')
best_pair = (0, 0)
last_occ = {}
prefix_sum = [0]
for i in range(n):
prefix_sum.append(prefix_sum[-1] + max(0, a[i]))
for i in reversed(range(n)):
if a[i] in last_occ:
total_sum = prefix_sum[last_occ[a[i]] + 1] - prefix_sum[i] + 2 * a[i]
if total_sum > best_sum:
best_sum = total_sum
best_pair = (i, last_occ[a[i]])
else:
last_occ[a[i]] = i
to_cut = []
for i in range(best_pair[0] + 1, best_pair[1]):
if a[i] < 0:
to_cut.append(i + 1)
for i in range(0, best_pair[0]):
to_cut.append(i + 1)
for i in range(best_pair[1] + 1, n):
to_cut.append(i + 1)
print(f"{best_sum - 2 * a[best_pair[0]]} {len(to_cut)}")
print(" ".join(map(str, to_cut))) | 18 | 186 | 307,200 | 117799908 | import math
import sys
from collections import deque, Counter, OrderedDict, defaultdict
#import heapq
#ceil,floor,log,sqrt,factorial,pow,pi,gcd
#import bisect
#from bisect import bisect_left,bisect_right
input = sys.stdin.readline
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input().strip()
return(list(s[:len(s)]))
def invr():
return(map(int,input().split()))
n=inp()
l=inlt()
g=Counter(l)
fst=defaultdict(lambda:-1)
lst=defaultdict(lambda:-1)
sm=[max(0,l[0])]
for i in range(1,n):
sm.append(sm[-1]+max(0,l[i]))
for i in range(n):
if fst[l[i]]==-1:
fst[l[i]]=i
for i in range(n-1,-1,-1):
if lst[l[i]]==-1:
lst[l[i]]=i
mx=-sys.maxsize
v=-1
for each in g:
if g[each]>=2:
if each<0:
val=2*each-sm[fst[each]]+sm[lst[each]]
else:
val=sm[lst[each]]-sm[fst[each]]+each
if val>mx:
mx=val
v=each
cnt=0
rem=[]
for i in range(n):
if i<fst[v]:
rem.append(i+1)
elif i>lst[v]:
rem.append(i+1)
elif l[i]<0 and i!=fst[v] and i!=lst[v]:
rem.append(i+1)
print(mx,len(rem))
print(*rem) | ABBYY Cup 3.0 - Finals | ICPC | 2,013 | 2 | 256 | Oh Sweet Beaverette | — Oh my sweet Beaverette, would you fancy a walk along a wonderful woodland belt with me?
— Of course, my Smart Beaver! Let us enjoy the splendid view together. How about Friday night?
At this point the Smart Beaver got rushing. Everything should be perfect by Friday, so he needed to prepare the belt to the upcoming walk. He needed to cut down several trees.
Let's consider the woodland belt as a sequence of trees. Each tree i is described by the esthetic appeal ai — some trees are very esthetically pleasing, others are 'so-so', and some trees are positively ugly!
The Smart Beaver calculated that he needed the following effects to win the Beaverette's heart:
- The first objective is to please the Beaverette: the sum of esthetic appeal of the remaining trees must be maximum possible;
- the second objective is to surprise the Beaverette: the esthetic appeal of the first and the last trees in the resulting belt must be the same;
- and of course, the walk should be successful: there must be at least two trees in the woodland belt left.
Now help the Smart Beaver! Which trees does he need to cut down to win the Beaverette's heart? | The first line contains a single integer n — the initial number of trees in the woodland belt, 2 ≤ n. The second line contains space-separated integers ai — the esthetic appeals of each tree. All esthetic appeals do not exceed 109 in their absolute value.
- to get 30 points, you need to solve the problem with constraints: n ≤ 100 (subproblem A1);
- to get 100 points, you need to solve the problem with constraints: n ≤ 3·105 (subproblems A1+A2). | In the first line print two integers — the total esthetic appeal of the woodland belt after the Smart Beaver's intervention and the number of the cut down trees k.
In the next line print k integers — the numbers of the trees the Beaver needs to cut down. Assume that the trees are numbered from 1 to n from left to right.
If there are multiple solutions, print any of them. It is guaranteed that at least two trees have equal esthetic appeal. | null | null | [{"input": "5\n1 2 3 1 2", "output": "8 1\n1"}, {"input": "5\n1 -2 3 1 -2", "output": "5 2\n2 5"}] | 1,400 | [] | 18 | [{"input": "5\r\n1 2 3 1 2\r\n", "output": "8 1\r\n1 "}, {"input": "5\r\n1 -2 3 1 -2\r\n", "output": "5 2\r\n2 5 "}, {"input": "2\r\n0 0\r\n", "output": "0 0\r\n"}, {"input": "3\r\n0 -1 0\r\n", "output": "0 1\r\n2 "}, {"input": "3\r\n1 1 1\r\n", "output": "3 0\r\n"}, {"input": "4\r\n-1 1 1 -1\r\n", "output": "2 2\r\n1 4 "}, {"input": "4\r\n-1 1 -1 1\r\n", "output": "2 2\r\n1 3 "}, {"input": "2\r\n-1 -1\r\n", "output": "-2 0\r\n"}, {"input": "3\r\n-1 0 -1\r\n", "output": "-2 0\r\n"}, {"input": "6\r\n-1 3 3 5 5 -1\r\n", "output": "14 0\r\n"}, {"input": "2\r\n-1000000000 -1000000000\r\n", "output": "-2000000000 0\r\n"}, {"input": "3\r\n-1000000000 -1000000000 -1000000000\r\n", "output": "-2000000000 1\r\n3 "}, {"input": "3\r\n1000000000 1000000000 1000000000\r\n", "output": "3000000000 0\r\n"}, {"input": "10\r\n-589330597 -126288833 -126288833 -834860352 -834860352 -834860352 -834860352 -21170405 -834860352 -834860352\r\n", "output": "-252577666 8\r\n1 4 5 6 7 8 9 10 "}, {"input": "20\r\n-808998072 733614990 579897311 -337992089 579897311 120800519 -337992089 -803027570 733614990 -686536765 733614990 -803027570 -803027570 733614990 120800519 -803027570 -686536765 579897311 -808998072 -686536765\r\n", "output": "4215055101 13\r\n1 4 7 8 10 12 13 15 16 17 18 19 20 "}] | false | stdio | import sys
from collections import defaultdict
def main(input_path, output_path, submission_output_path):
with open(input_path) as f:
n = int(f.readline())
a = list(map(int, f.readline().split()))
with open(submission_output_path) as f:
lines = f.readlines()
if not lines:
print(0)
return
sum_line = lines[0].strip().split()
if len(sum_line) != 2:
print(0)
return
try:
sum_claimed, k = map(int, sum_line)
except:
print(0)
return
cut_list = list(map(int, lines[1].split())) if len(lines) > 1 else []
if len(cut_list) != k:
print(0)
return
cut_set = set(cut_list)
if len(cut_set) != k:
print(0)
return
for num in cut_list:
if not (1 <= num <= n):
print(0)
return
remaining = [i-1 for i in range(1, n+1) if i not in cut_set]
if len(remaining) < 2:
print(0)
return
first_val = a[remaining[0]]
last_val = a[remaining[-1]]
if first_val != last_val:
print(0)
return
sum_remaining = sum(a[i] for i in remaining)
if sum_remaining != sum_claimed:
print(0)
return
prefix = [0] * (n + 1)
for i in range(n):
prefix[i+1] = prefix[i] + max(a[i], 0)
groups = defaultdict(list)
for idx, val in enumerate(a):
groups[val].append(idx)
max_total = -float('inf')
for v, indices in groups.items():
m = len(indices)
if m < 2:
continue
indices.sort()
min_val = prefix[indices[0] + 1]
current_max = -float('inf')
for j in range(1, m):
current_i = indices[j]
current_sum = 2 * v + (prefix[current_i] - min_val)
if current_sum > current_max:
current_max = current_sum
if j < m - 1:
next_val = prefix[indices[j] + 1]
if next_val < min_val:
min_val = next_val
if current_max > max_total:
max_total = current_max
if sum_claimed != max_total:
print(0)
else:
print(100)
if __name__ == "__main__":
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_output_path = sys.argv[3]
main(input_path, output_path, submission_output_path) | true |
609/D | 609 | D | PyPy 3 | TESTS | 4 | 187 | 0 | 93589343 | import os, sys
from io import IOBase, BytesIO
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = 'x' in file.mode or 'w' in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b'\n') + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s:self.buffer.write(s.encode('ascii'))
self.read = lambda:self.buffer.read().decode('ascii')
self.readline = lambda:self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
# Cout implemented in Python
import sys
class ostream:
def __lshift__(self,a):
sys.stdout.write(str(a))
return self
cout = ostream()
endl = '\n'
import heapq
def can_do(check_days, n, m, k, s, d_gadgets, p_gadgets, prefx_min_d_val, prefx_min_p_val):
# get low in the range 1 -> check_days
min_d = prefx_min_d_val[check_days][0]
min_p = prefx_min_p_val[check_days][0]
min_d_idx = prefx_min_d_val[check_days][1]
min_p_idx = prefx_min_p_val[check_days][1]
vals_d = [0] * (k + 1)
vals_p = [0] * (k + 1)
for i in range(1, k + 1):
# print(i)
if i < len(d_gadgets):
vals_d[i] = ((min_d * d_gadgets[i - 1][0]) + vals_d[i - 1])
if i < len(p_gadgets):
vals_p[i] = ((min_p * p_gadgets[i - 1][0]) + vals_p[i - 1])
for x in range(k + 1):
# x from dollar and k - x from pounds
if x <= len(d_gadgets) and k - x <= len(p_gadgets):
if ((x > 0 and vals_d[x] > 0) or x == 0) \
and ((k - x > 0 and vals_p[k - x] > 0) or k - x == 0):
total = vals_d[x] + vals_p[k - x]
if total <= s:
res = []
for i in range(x):
res.append((d_gadgets[i][1], min_d_idx))
for i in range(k - x):
res.append((p_gadgets[i][1], min_p_idx))
return (True, res)
return (False, [])
def check(n, m, k, s, d_gadgets, p_gadgets, prefx_min_d_val, prefx_min_p_val):
lo = 0
hi = n
res = -1
while lo <= hi:
mid = lo + (hi - lo) // 2
if can_do(mid, n, m, k, s, d_gadgets, p_gadgets, prefx_min_d_val, prefx_min_p_val)[0]:
res = mid
hi = mid - 1
else:
lo = mid + 1
if res == -1:
return res, []
return res, can_do(res, n, m, k, s, d_gadgets, p_gadgets, prefx_min_d_val, prefx_min_p_val)[1]
def solve():
n, m, k, s = map(int, input().split())
d_vals = list(map(int, input().split()))
d_vals.insert(0, float("inf"))
p_vals = list(map(int, input().split()))
p_vals.insert(0, float("inf"))
d_gadgets = []
p_gadgets = []
for i in range(m):
t, c = map(int, input().split())
if t == 1:
d_gadgets.append((c, i + 1))
else:
p_gadgets.append((c, i + 1))
prefx_min_d_val = [(d_vals[0], 0)] * len(d_vals)
for i in range(1, len(d_vals)):
if prefx_min_d_val[i - 1][0] < d_vals[i]:
prefx_min_d_val[i] = (prefx_min_d_val[i - 1][0], prefx_min_d_val[i - 1][1])
else:
prefx_min_d_val[i] = (d_vals[i], i)
prefx_min_p_val = [(p_vals[0], 0)] * len(p_vals)
for i in range(1, len(p_vals)):
if prefx_min_p_val[i - 1][0] < p_vals[i]:
prefx_min_p_val[i] = (prefx_min_p_val[i - 1][0], prefx_min_p_val[i - 1][1])
else:
prefx_min_p_val[i] = (p_vals[i], i)
d_gadgets.sort(key=lambda x: x[0])
p_gadgets.sort(key=lambda x: x[0])
res, r = check(n, m, k, s, d_gadgets, p_gadgets, prefx_min_d_val, prefx_min_p_val)
cout<<res<<endl
for p in r:
cout<<p[0]<<" "<<p[1]<<endl
def main():
solve()
if __name__ == "__main__":
main() | 51 | 1,809 | 43,110,400 | 117047744 | '''
Auther: ghoshashis545 Ashis Ghosh
College: Jalpaiguri Govt Enggineering College
'''
from os import path
from io import BytesIO, IOBase
import sys
from heapq import heappush,heappop
from functools import cmp_to_key as ctk
from collections import deque,Counter,defaultdict as dd
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input().rstrip()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def bo(i):
return ord(i)-ord('0')
file = 1
def ceil(a,b):
return (a+b-1)//b
def solve():
# for _ in range(1,ii()+1):
n,m,k,s = mi()
a = li()
b = li()
gadgets = [[] for i in range(2)]
for i in range(m):
x,y = mi()
gadgets[x-1].append([y,i+1])
gadgets[0].sort()
gadgets[1].sort()
sz = [len(gadgets[0]),len(gadgets[1])]
def check(idx):
mnx1 = inf
mnx2 = inf
for i in range(idx+1):
mnx1 = min(mnx1,a[i])
mnx2 = min(mnx2,b[i])
l1,l2,res = 0,0,0
if k > sz[0] + sz[1]:
return 0
for i in range(k):
if l1 == sz[0]:
res += gadgets[1][l2][0]*mnx2
l2 += 1
elif l2 == sz[1]:
res += gadgets[0][l1][0]*mnx1
l1 += 1
elif gadgets[0][l1][0]*mnx1 <= gadgets[1][l2][0]*mnx2:
res += gadgets[0][l1][0]*mnx1
l1 += 1
else:
res += gadgets[1][l2][0]*mnx2
l2 += 1
return res <= s
l = 0
r = n-1
ans = -1
while l<=r:
mid = (l+r)>>1
if check(mid):
ans = mid+1
r = mid-1
else:
l=mid+1
print(ans)
if ans == -1:
return
mnx1 = inf
mnx2 = inf
idx1,idx2 = -1,-1
for i in range(ans):
if a[i] < mnx1:
mnx1 = a[i]
idx1 = i
if b[i] < mnx2:
mnx2 = b[i]
idx2 = i
l1,l2,res = 0,0,[]
for i in range(k):
if l1 == sz[0]:
res.append([gadgets[1][l2][1],idx2+1])
l2 += 1
elif l2==sz[1]:
res.append([gadgets[0][l1][1],idx1+1])
l1 += 1
elif gadgets[0][l1][0]*mnx1 <= gadgets[1][l2][0]*mnx2:
res.append([gadgets[0][l1][1],idx1+1])
l1 += 1
else:
res.append([gadgets[1][l2][1],idx2+1])
l2 += 1
for i in res:
print(*i)
if __name__ =="__main__":
if(file):
if path.exists('input.txt'):
sys.stdin=open('input.txt', 'r')
sys.stdout=open('output.txt','w')
else:
input=sys.stdin.readline
solve() | Educational Codeforces Round 3 | ICPC | 2,015 | 2 | 256 | Gadgets for dollars and pounds | Nura wants to buy k gadgets. She has only s burles for that. She can buy each gadget for dollars or for pounds. So each gadget is selling only for some type of currency. The type of currency and the cost in that currency are not changing.
Nura can buy gadgets for n days. For each day you know the exchange rates of dollar and pound, so you know the cost of conversion burles to dollars or to pounds.
Each day (from 1 to n) Nura can buy some gadgets by current exchange rate. Each day she can buy any gadgets she wants, but each gadget can be bought no more than once during n days.
Help Nura to find the minimum day index when she will have k gadgets. Nura always pays with burles, which are converted according to the exchange rate of the purchase day. Nura can't buy dollars or pounds, she always stores only burles. Gadgets are numbered with integers from 1 to m in order of their appearing in input. | First line contains four integers n, m, k, s (1 ≤ n ≤ 2·105, 1 ≤ k ≤ m ≤ 2·105, 1 ≤ s ≤ 109) — number of days, total number and required number of gadgets, number of burles Nura has.
Second line contains n integers ai (1 ≤ ai ≤ 106) — the cost of one dollar in burles on i-th day.
Third line contains n integers bi (1 ≤ bi ≤ 106) — the cost of one pound in burles on i-th day.
Each of the next m lines contains two integers ti, ci (1 ≤ ti ≤ 2, 1 ≤ ci ≤ 106) — type of the gadget and it's cost. For the gadgets of the first type cost is specified in dollars. For the gadgets of the second type cost is specified in pounds. | If Nura can't buy k gadgets print the only line with the number -1.
Otherwise the first line should contain integer d — the minimum day index, when Nura will have k gadgets. On each of the next k lines print two integers qi, di — the number of gadget and the day gadget should be bought. All values qi should be different, but the values di can coincide (so Nura can buy several gadgets at one day). The days are numbered from 1 to n.
In case there are multiple possible solutions, print any of them. | null | null | [{"input": "5 4 2 2\n1 2 3 2 1\n3 2 1 2 3\n1 1\n2 1\n1 2\n2 2", "output": "3\n1 1\n2 3"}, {"input": "4 3 2 200\n69 70 71 72\n104 105 106 107\n1 1\n2 2\n1 2", "output": "-1"}, {"input": "4 3 1 1000000000\n900000 910000 940000 990000\n990000 999000 999900 999990\n1 87654\n2 76543\n1 65432", "output": "-1"}] | 2,000 | ["binary search", "greedy", "two pointers"] | 51 | [{"input": "5 4 2 2\r\n1 2 3 2 1\r\n3 2 1 2 3\r\n1 1\r\n2 1\r\n1 2\r\n2 2\r\n", "output": "3\r\n1 1\r\n2 3\r\n"}, {"input": "4 3 2 200\r\n69 70 71 72\r\n104 105 106 107\r\n1 1\r\n2 2\r\n1 2\r\n", "output": "-1\r\n"}, {"input": "4 3 1 1000000000\r\n900000 910000 940000 990000\r\n990000 999000 999900 999990\r\n1 87654\r\n2 76543\r\n1 65432\r\n", "output": "-1\r\n"}, {"input": "5 5 3 1000000\r\n921 853 547 187 164\r\n711 462 437 307 246\r\n2 94\r\n2 230\r\n1 373\r\n1 476\r\n2 880\r\n", "output": "1\r\n1 1\r\n2 1\r\n5 1\r\n"}, {"input": "10 10 10 1000000\r\n836 842 645 671 499 554 462 288 89 104\r\n880 722 623 651 591 573 154 532 136 59\r\n1 47\r\n1 169\r\n2 486\r\n1 262\r\n2 752\r\n2 498\r\n2 863\r\n2 616\r\n1 791\r\n1 656\r\n", "output": "9\r\n1 9\r\n2 9\r\n4 9\r\n10 9\r\n9 9\r\n3 9\r\n6 9\r\n8 9\r\n5 9\r\n7 9\r\n"}, {"input": "1 2 2 1000000\r\n96\r\n262\r\n1 699\r\n2 699\r\n", "output": "1\r\n1 1\r\n2 1\r\n"}, {"input": "1 2 2 1000000\r\n793\r\n33\r\n1 733\r\n2 406\r\n", "output": "1\r\n1 1\r\n2 1\r\n"}, {"input": "1 2 2 10000\r\n82\r\n996\r\n2 574\r\n2 217\r\n", "output": "-1\r\n"}, {"input": "1 2 2 1000000\r\n778\r\n62\r\n2 119\r\n2 220\r\n", "output": "1\r\n1 1\r\n2 1\r\n"}, {"input": "1 2 2 1000000\r\n963\r\n25\r\n2 961\r\n1 327\r\n", "output": "1\r\n2 1\r\n1 1\r\n"}, {"input": "10 20 20 1000000\r\n809 909 795 661 635 613 534 199 188 3\r\n475 585 428 379 185 177 66 104 15 38\r\n2 454\r\n1 863\r\n2 14\r\n2 104\r\n1 663\r\n2 885\r\n1 650\r\n1 967\r\n2 650\r\n2 483\r\n2 846\r\n1 283\r\n1 187\r\n2 533\r\n2 112\r\n2 938\r\n2 553\r\n1 816\r\n1 549\r\n2 657\r\n", "output": "10\r\n13 10\r\n12 10\r\n19 10\r\n7 10\r\n5 10\r\n18 10\r\n2 10\r\n8 10\r\n3 9\r\n4 9\r\n15 9\r\n1 9\r\n10 9\r\n14 9\r\n17 9\r\n9 9\r\n20 9\r\n11 9\r\n6 9\r\n16 9\r\n"}, {"input": "10 20 19 1000000\r\n650 996 972 951 904 742 638 93 339 151\r\n318 565 849 579 521 965 286 189 196 307\r\n2 439\r\n1 333\r\n2 565\r\n1 602\r\n2 545\r\n2 596\r\n2 821\r\n2 929\r\n1 614\r\n2 647\r\n2 909\r\n1 8\r\n2 135\r\n1 301\r\n1 597\r\n1 632\r\n1 437\r\n2 448\r\n2 631\r\n2 969\r\n", "output": "-1\r\n"}, {"input": "10 20 18 10000\r\n916 582 790 449 578 502 411 196 218 144\r\n923 696 788 609 455 570 330 435 284 113\r\n2 736\r\n1 428\r\n1 861\r\n2 407\r\n2 320\r\n1 340\r\n1 88\r\n1 172\r\n1 788\r\n2 633\r\n2 612\r\n2 571\r\n2 536\r\n2 30\r\n2 758\r\n2 90\r\n2 8\r\n1 970\r\n1 20\r\n1 22\r\n", "output": "-1\r\n"}, {"input": "10 20 16 1000000\r\n317 880 696 304 260 180 214 245 79 37\r\n866 621 940 89 718 674 195 267 12 49\r\n2 825\r\n2 197\r\n1 657\r\n1 231\r\n1 728\r\n2 771\r\n2 330\r\n2 943\r\n1 60\r\n1 89\r\n2 721\r\n2 959\r\n1 926\r\n2 215\r\n1 583\r\n2 680\r\n1 799\r\n2 887\r\n1 709\r\n1 316\r\n", "output": "6\r\n9 6\r\n10 6\r\n4 6\r\n20 6\r\n15 6\r\n3 6\r\n2 4\r\n14 4\r\n7 4\r\n16 4\r\n11 4\r\n6 4\r\n1 4\r\n18 4\r\n8 4\r\n12 4\r\n"}, {"input": "10 20 20 10000\r\n913 860 844 775 297 263 247 71 50 6\r\n971 938 890 854 643 633 427 418 190 183\r\n1 556\r\n2 579\r\n1 315\r\n2 446\r\n1 327\r\n1 724\r\n2 12\r\n1 142\r\n1 627\r\n1 262\r\n1 681\r\n1 802\r\n1 886\r\n1 350\r\n2 383\r\n1 191\r\n1 717\r\n1 968\r\n2 588\r\n1 57\r\n", "output": "-1\r\n"}, {"input": "1 93 46 46\r\n1\r\n1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n2 1\r\n1 2\r\n", "output": "1\r\n2 1\r\n4 1\r\n6 1\r\n8 1\r\n10 1\r\n12 1\r\n14 1\r\n16 1\r\n18 1\r\n20 1\r\n22 1\r\n24 1\r\n26 1\r\n28 1\r\n30 1\r\n32 1\r\n34 1\r\n36 1\r\n38 1\r\n40 1\r\n42 1\r\n44 1\r\n46 1\r\n48 1\r\n50 1\r\n52 1\r\n54 1\r\n56 1\r\n58 1\r\n60 1\r\n62 1\r\n64 1\r\n66 1\r\n68 1\r\n70 1\r\n72 1\r\n74 1\r\n76 1\r\n78 1\r\n80 1\r\n82 1\r\n84 1\r\n86 1\r\n88 1\r\n90 1\r\n92 1\r\n"}] | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input data
with open(input_path) as f:
input_lines = f.read().splitlines()
ptr = 0
n, m, k, s = map(int, input_lines[ptr].split())
ptr += 1
a = list(map(int, input_lines[ptr].split()))
ptr += 1
b = list(map(int, input_lines[ptr].split()))
ptr += 1
gadgets = []
for _ in range(m):
ti, ci = map(int, input_lines[ptr].split())
gadgets.append((ti, ci))
ptr += 1
# Read reference output
with open(output_path) as f:
ref_lines = f.read().splitlines()
# Read submission output
with open(submission_path) as f:
sub_lines = [line.strip() for line in f.read().splitlines()]
# Check for -1 case
if ref_lines[0].strip() == '-1':
if len(sub_lines) == 1 and sub_lines[0] == '-1':
print(1)
else:
print(0)
return
# Check submission has correct d
if not sub_lines:
print(0)
return
try:
d_ref = int(ref_lines[0])
d_sub = int(sub_lines[0])
except:
print(0)
return
if d_sub != d_ref:
print(0)
return
# Validate submission's gadgets
if len(sub_lines) != k + 1:
print(0)
return
seen = set()
total = 0
for line in sub_lines[1:]:
if not line:
print(0)
return
parts = line.split()
if len(parts) != 2:
print(0)
return
try:
qi = int(parts[0])
di = int(parts[1])
except:
print(0)
return
if qi < 1 or qi > m or di < 1 or di > d_ref:
print(0)
return
if qi in seen:
print(0)
return
seen.add(qi)
ti, ci = gadgets[qi - 1]
if ti == 1:
rate = a[di - 1]
else:
rate = b[di - 1]
total += ci * rate
if total > s:
print(0)
return
print(1)
if __name__ == "__main__":
main()
| true |
18/D | 18 | D | PyPy 3 | TESTS | 32 | 964 | 5,529,600 | 91087080 | n = int(input())
a = []
for i in range(n):
a.append(input().split(' '))
valid = [1] * n
res = 0
for x in range(2000, 0, -1):
pos_sell = -1
pos_win = -1
for i in range(n):
if valid[i] == 0: continue
if a[i][0] == 'sell' and int(a[i][1]) == x:
pos_sell = i
break
for i in range(n):
if valid[i] == 0: continue
if i > pos_sell: break
if a[i][0] == 'win' and int(a[i][1]) == x:
pos_win = i
if pos_sell == -1 or pos_win == -1:
continue
fun = 1
for i in range(pos_win, pos_sell+1):
if valid[i] == 0:
fun = 0
break
if fun == 0:
continue
for i in range(pos_win, pos_sell+1):
valid[i] = 0
res += pow(2, x)
print(res) | 45 | 92 | 614,400 | 229440552 | n = int(input()) # Number of working days
total_earnings = 0 # Initialize total earnings
memory_stick_capacities = [0] * 2010 # Create a list to store memory stick capacities and their respective earnings
for i in range(n):
option = input().split() # Read the input option (win or sell) and capacity
action, capacity = option[0], int(option[1])
if action == "win":
memory_stick_capacities[capacity] = total_earnings + 2 ** capacity # If Bob wins a stick, update earnings
else:
total_earnings = max(total_earnings, memory_stick_capacities[capacity]) # If Bob sells a stick, update total earnings
print(total_earnings) # Output the maximum possible earnings for Bob | Codeforces Beta Round 18 (Div. 2 Only) | ICPC | 2,010 | 2 | 128 | Seller Bob | Last year Bob earned by selling memory sticks. During each of n days of his work one of the two following events took place:
- A customer came to Bob and asked to sell him a 2x MB memory stick. If Bob had such a stick, he sold it and got 2x berllars.
- Bob won some programming competition and got a 2x MB memory stick as a prize. Bob could choose whether to present this memory stick to one of his friends, or keep it.
Bob never kept more than one memory stick, as he feared to mix up their capacities, and deceive a customer unintentionally. It is also known that for each memory stick capacity there was at most one customer, who wanted to buy that memory stick. Now, knowing all the customers' demands and all the prizes won at programming competitions during the last n days, Bob wants to know, how much money he could have earned, if he had acted optimally. | The first input line contains number n (1 ≤ n ≤ 5000) — amount of Bob's working days. The following n lines contain the description of the days. Line sell x stands for a day when a customer came to Bob to buy a 2x MB memory stick (0 ≤ x ≤ 2000). It's guaranteed that for each x there is not more than one line sell x. Line win x stands for a day when Bob won a 2x MB memory stick (0 ≤ x ≤ 2000). | Output the maximum possible earnings for Bob in berllars, that he would have had if he had known all the events beforehand. Don't forget, please, that Bob can't keep more than one memory stick at a time. | null | null | [{"input": "7\nwin 10\nwin 5\nwin 3\nsell 5\nsell 3\nwin 10\nsell 10", "output": "1056"}, {"input": "3\nwin 5\nsell 6\nsell 4", "output": "0"}] | 2,000 | ["brute force", "dp", "greedy"] | 45 | [{"input": "7\r\nwin 10\r\nwin 5\r\nwin 3\r\nsell 5\r\nsell 3\r\nwin 10\r\nsell 10\r\n", "output": "1056\r\n"}, {"input": "3\r\nwin 5\r\nsell 6\r\nsell 4\r\n", "output": "0\r\n"}, {"input": "10\r\nsell 179\r\nwin 1278\r\nsell 1278\r\nwin 179\r\nwin 788\r\nsell 788\r\nwin 1819\r\nwin 1278\r\nsell 1454\r\nsell 1819\r\n", "output": "37459511778596727480858760720167552241582636504705413766024169777495064333423937410125519624693990051069809575647477719465460756326341562228323606665869931977125977431028709943048934214062888966581139223580790503937962827597404798307897711090567429316074325427043388117806141094834711707585035634104732053207574452493593409130554278913951011894497392495930884827685983975668127973918422057605356890341647839399778378381152159725053311750647457999739578989105335906181048932656785993705124392163591312698147450544686667028530553878359909362604965888\n"}, {"input": "10\r\nsell 573\r\nwin 1304\r\nsell 278\r\nwin 1631\r\nsell 1225\r\nsell 1631\r\nsell 177\r\nwin 1631\r\nwin 177\r\nsell 1304\r\n", "output": "95482312335125227379668481690754940528280513838693267460502082967052005332103697568042408703168913727303170456338425853153094403747135188778307041838920404959089576368946137708987138986696495077466398994298434148881715073638178666201165545650953479735059082316661443204882826188032944866093372620219104327689636641547141835841165681118172603993695103043804276669836594061369229043451067647935298287687852302215923887110435577776767805943668204998410716005202198549540411238299513630278811648\n"}, {"input": "10\r\nwin 1257\r\nwin 1934\r\nsell 1934\r\nsell 1257\r\nwin 1934\r\nwin 1257\r\nsell 495\r\nwin 495\r\nwin 495\r\nwin 1257\r\n", "output": "1556007242642049292787218246793379348327505438878680952714050868520307364441227819009733220897932984584977593931988662671459594674963394056587723382487766303981362587048873128400436836690128983570130687310221668877557121158055843621982630476422478413285775826498536883275291967793661985813155062733063913176306327509625594121241472451054995889483447103432414676059872469910105149496451402271546454282618581884282152530090816240540173251729211604658704990425330422792556824836640431985211146197816770068601144273155826261981248156109439380758754562057681271989937590784957541131485184\n"}, {"input": "10\r\nsell 1898\r\nsell 173\r\nsell 1635\r\nsell 29\r\nsell 881\r\nsell 434\r\nsell 1236\r\nsell 14\r\nwin 29\r\nsell 1165\r\n", "output": "0\r\n"}, {"input": "1\r\nsell 2000\r\n", "output": "0\r\n"}, {"input": "1\r\nwin 2000\r\n", "output": "0\r\n"}, {"input": "2\r\nwin 2000\r\nsell 2000\r\n", "output": "114813069527425452423283320117768198402231770208869520047764273682576626139237031385665948631650626991844596463898746277344711896086305533142593135616665318539129989145312280000688779148240044871428926990063486244781615463646388363947317026040466353970904996558162398808944629605623311649536164221970332681344168908984458505602379484807914058900934776500429002716706625830522008132236281291761267883317206598995396418127021779858404042159853183251540889433902091920554957783589672039160081957216630582755380425583726015528348786419432054508915275783882625175435528800822842770817965453762184851149029376\n"}] | false | stdio | null | true |
18/D | 18 | D | PyPy 3 | TESTS | 32 | 498 | 4,096,000 | 90479801 | n=int(input())
ar=[0]*(n+1)
c=[]
for i in range(n):
a,b=input().split()
b=int(b)
if(a=="sell"):
b=-b
c.append(b)
for i in range(1,n+1):
ar[i]=max(ar[i-1],ar[i])
for j in range(i+1,n+1):
if(c[i-1]+c[j-1]==0 and c[i-1]>c[j-1]):
ar[j]=max(ar[j],ar[i]+2**c[i-1])
print(ar[n]) | 45 | 124 | 5,529,600 | 27965486 | #!/usr/bin/python3
N = int(input())
maxval = 0
dp = [0]*N
prev = [-1]*2010
for i in range(N):
l = input().strip().split()
l[1] = int(l[1])
if l[0] == "win": prev[l[1]] = i
elif l[0] == "sell" and prev[l[1]] != -1: maxval = max(maxval,2**l[1]+dp[prev[l[1]]])
dp[i] = maxval
print(maxval) | Codeforces Beta Round 18 (Div. 2 Only) | ICPC | 2,010 | 2 | 128 | Seller Bob | Last year Bob earned by selling memory sticks. During each of n days of his work one of the two following events took place:
- A customer came to Bob and asked to sell him a 2x MB memory stick. If Bob had such a stick, he sold it and got 2x berllars.
- Bob won some programming competition and got a 2x MB memory stick as a prize. Bob could choose whether to present this memory stick to one of his friends, or keep it.
Bob never kept more than one memory stick, as he feared to mix up their capacities, and deceive a customer unintentionally. It is also known that for each memory stick capacity there was at most one customer, who wanted to buy that memory stick. Now, knowing all the customers' demands and all the prizes won at programming competitions during the last n days, Bob wants to know, how much money he could have earned, if he had acted optimally. | The first input line contains number n (1 ≤ n ≤ 5000) — amount of Bob's working days. The following n lines contain the description of the days. Line sell x stands for a day when a customer came to Bob to buy a 2x MB memory stick (0 ≤ x ≤ 2000). It's guaranteed that for each x there is not more than one line sell x. Line win x stands for a day when Bob won a 2x MB memory stick (0 ≤ x ≤ 2000). | Output the maximum possible earnings for Bob in berllars, that he would have had if he had known all the events beforehand. Don't forget, please, that Bob can't keep more than one memory stick at a time. | null | null | [{"input": "7\nwin 10\nwin 5\nwin 3\nsell 5\nsell 3\nwin 10\nsell 10", "output": "1056"}, {"input": "3\nwin 5\nsell 6\nsell 4", "output": "0"}] | 2,000 | ["brute force", "dp", "greedy"] | 45 | [{"input": "7\r\nwin 10\r\nwin 5\r\nwin 3\r\nsell 5\r\nsell 3\r\nwin 10\r\nsell 10\r\n", "output": "1056\r\n"}, {"input": "3\r\nwin 5\r\nsell 6\r\nsell 4\r\n", "output": "0\r\n"}, {"input": "10\r\nsell 179\r\nwin 1278\r\nsell 1278\r\nwin 179\r\nwin 788\r\nsell 788\r\nwin 1819\r\nwin 1278\r\nsell 1454\r\nsell 1819\r\n", "output": "37459511778596727480858760720167552241582636504705413766024169777495064333423937410125519624693990051069809575647477719465460756326341562228323606665869931977125977431028709943048934214062888966581139223580790503937962827597404798307897711090567429316074325427043388117806141094834711707585035634104732053207574452493593409130554278913951011894497392495930884827685983975668127973918422057605356890341647839399778378381152159725053311750647457999739578989105335906181048932656785993705124392163591312698147450544686667028530553878359909362604965888\n"}, {"input": "10\r\nsell 573\r\nwin 1304\r\nsell 278\r\nwin 1631\r\nsell 1225\r\nsell 1631\r\nsell 177\r\nwin 1631\r\nwin 177\r\nsell 1304\r\n", "output": "95482312335125227379668481690754940528280513838693267460502082967052005332103697568042408703168913727303170456338425853153094403747135188778307041838920404959089576368946137708987138986696495077466398994298434148881715073638178666201165545650953479735059082316661443204882826188032944866093372620219104327689636641547141835841165681118172603993695103043804276669836594061369229043451067647935298287687852302215923887110435577776767805943668204998410716005202198549540411238299513630278811648\n"}, {"input": "10\r\nwin 1257\r\nwin 1934\r\nsell 1934\r\nsell 1257\r\nwin 1934\r\nwin 1257\r\nsell 495\r\nwin 495\r\nwin 495\r\nwin 1257\r\n", "output": "1556007242642049292787218246793379348327505438878680952714050868520307364441227819009733220897932984584977593931988662671459594674963394056587723382487766303981362587048873128400436836690128983570130687310221668877557121158055843621982630476422478413285775826498536883275291967793661985813155062733063913176306327509625594121241472451054995889483447103432414676059872469910105149496451402271546454282618581884282152530090816240540173251729211604658704990425330422792556824836640431985211146197816770068601144273155826261981248156109439380758754562057681271989937590784957541131485184\n"}, {"input": "10\r\nsell 1898\r\nsell 173\r\nsell 1635\r\nsell 29\r\nsell 881\r\nsell 434\r\nsell 1236\r\nsell 14\r\nwin 29\r\nsell 1165\r\n", "output": "0\r\n"}, {"input": "1\r\nsell 2000\r\n", "output": "0\r\n"}, {"input": "1\r\nwin 2000\r\n", "output": "0\r\n"}, {"input": "2\r\nwin 2000\r\nsell 2000\r\n", "output": "114813069527425452423283320117768198402231770208869520047764273682576626139237031385665948631650626991844596463898746277344711896086305533142593135616665318539129989145312280000688779148240044871428926990063486244781615463646388363947317026040466353970904996558162398808944629605623311649536164221970332681344168908984458505602379484807914058900934776500429002716706625830522008132236281291761267883317206598995396418127021779858404042159853183251540889433902091920554957783589672039160081957216630582755380425583726015528348786419432054508915275783882625175435528800822842770817965453762184851149029376\n"}] | false | stdio | null | true |
18/D | 18 | D | Python 3 | TESTS | 32 | 248 | 1,126,400 | 73955174 | n=int(input())
a=[2*[0]for i in range(0,n)]
for i in range(0,n):
a[i]=list(map(str,input().split()))
na=[0]*2001
for i in range(0,2001):
na[i]=[]
c=[0]*(n+1)
for i in range(0, n):
ind=int(a[i][1])
if a[i][0]=='win':
na[ind].append(i+1)
else:
c[i+1]=ind
dp=[0]*(n+1)
for i in range(1,n+1):
dp[i]=dp[i-1]
if c[i]!=0 and len(na[c[i]])>0:
l=0
r=len(na[c[i]])
while(r-l>1):
mid=(r+l)/2
mid=int(mid)
if na[c[i]][mid]>=i:
r=mid
else:
l=mid
if na[c[i]][l]>=i:
continue
if dp[i]<2**c[i]+dp[na[c[i]][l]-1]:
dp[i] = 2 ** c[i] + dp[na[c[i]][l] - 1]
print(dp[n]) | 45 | 154 | 921,600 | 112211098 | n = int(input())
s = ["" for _ in range(5005)]
a = [0 for _ in range(5005)]
for i in range(1,n+1):
ta,tb = input().split();
s[i] = ta
a[i] = int(tb)
jp = [0 for _ in range(5005)]
vis = [0 for _ in range(5005)]
dp = [0 for _ in range(5005)]
num = [(1<<_) for _ in range(2001)]
for i in range(1,n+1):
if(s[i][0]=='w'):
vis[a[i]] = i
else:
if(vis[a[i]]):
jp[i] = vis[a[i]];
for i in range(1,n+1):
if(s[i][0] == 's'):
if(jp[i]>0):
dp[i] = max(dp[i], dp[jp[i]-1] + num[a[i]]);
dp[i] = max(dp[i-1], dp[i]);
print(dp[n]) | Codeforces Beta Round 18 (Div. 2 Only) | ICPC | 2,010 | 2 | 128 | Seller Bob | Last year Bob earned by selling memory sticks. During each of n days of his work one of the two following events took place:
- A customer came to Bob and asked to sell him a 2x MB memory stick. If Bob had such a stick, he sold it and got 2x berllars.
- Bob won some programming competition and got a 2x MB memory stick as a prize. Bob could choose whether to present this memory stick to one of his friends, or keep it.
Bob never kept more than one memory stick, as he feared to mix up their capacities, and deceive a customer unintentionally. It is also known that for each memory stick capacity there was at most one customer, who wanted to buy that memory stick. Now, knowing all the customers' demands and all the prizes won at programming competitions during the last n days, Bob wants to know, how much money he could have earned, if he had acted optimally. | The first input line contains number n (1 ≤ n ≤ 5000) — amount of Bob's working days. The following n lines contain the description of the days. Line sell x stands for a day when a customer came to Bob to buy a 2x MB memory stick (0 ≤ x ≤ 2000). It's guaranteed that for each x there is not more than one line sell x. Line win x stands for a day when Bob won a 2x MB memory stick (0 ≤ x ≤ 2000). | Output the maximum possible earnings for Bob in berllars, that he would have had if he had known all the events beforehand. Don't forget, please, that Bob can't keep more than one memory stick at a time. | null | null | [{"input": "7\nwin 10\nwin 5\nwin 3\nsell 5\nsell 3\nwin 10\nsell 10", "output": "1056"}, {"input": "3\nwin 5\nsell 6\nsell 4", "output": "0"}] | 2,000 | ["brute force", "dp", "greedy"] | 45 | [{"input": "7\r\nwin 10\r\nwin 5\r\nwin 3\r\nsell 5\r\nsell 3\r\nwin 10\r\nsell 10\r\n", "output": "1056\r\n"}, {"input": "3\r\nwin 5\r\nsell 6\r\nsell 4\r\n", "output": "0\r\n"}, {"input": "10\r\nsell 179\r\nwin 1278\r\nsell 1278\r\nwin 179\r\nwin 788\r\nsell 788\r\nwin 1819\r\nwin 1278\r\nsell 1454\r\nsell 1819\r\n", "output": "37459511778596727480858760720167552241582636504705413766024169777495064333423937410125519624693990051069809575647477719465460756326341562228323606665869931977125977431028709943048934214062888966581139223580790503937962827597404798307897711090567429316074325427043388117806141094834711707585035634104732053207574452493593409130554278913951011894497392495930884827685983975668127973918422057605356890341647839399778378381152159725053311750647457999739578989105335906181048932656785993705124392163591312698147450544686667028530553878359909362604965888\n"}, {"input": "10\r\nsell 573\r\nwin 1304\r\nsell 278\r\nwin 1631\r\nsell 1225\r\nsell 1631\r\nsell 177\r\nwin 1631\r\nwin 177\r\nsell 1304\r\n", "output": "95482312335125227379668481690754940528280513838693267460502082967052005332103697568042408703168913727303170456338425853153094403747135188778307041838920404959089576368946137708987138986696495077466398994298434148881715073638178666201165545650953479735059082316661443204882826188032944866093372620219104327689636641547141835841165681118172603993695103043804276669836594061369229043451067647935298287687852302215923887110435577776767805943668204998410716005202198549540411238299513630278811648\n"}, {"input": "10\r\nwin 1257\r\nwin 1934\r\nsell 1934\r\nsell 1257\r\nwin 1934\r\nwin 1257\r\nsell 495\r\nwin 495\r\nwin 495\r\nwin 1257\r\n", "output": "1556007242642049292787218246793379348327505438878680952714050868520307364441227819009733220897932984584977593931988662671459594674963394056587723382487766303981362587048873128400436836690128983570130687310221668877557121158055843621982630476422478413285775826498536883275291967793661985813155062733063913176306327509625594121241472451054995889483447103432414676059872469910105149496451402271546454282618581884282152530090816240540173251729211604658704990425330422792556824836640431985211146197816770068601144273155826261981248156109439380758754562057681271989937590784957541131485184\n"}, {"input": "10\r\nsell 1898\r\nsell 173\r\nsell 1635\r\nsell 29\r\nsell 881\r\nsell 434\r\nsell 1236\r\nsell 14\r\nwin 29\r\nsell 1165\r\n", "output": "0\r\n"}, {"input": "1\r\nsell 2000\r\n", "output": "0\r\n"}, {"input": "1\r\nwin 2000\r\n", "output": "0\r\n"}, {"input": "2\r\nwin 2000\r\nsell 2000\r\n", "output": "114813069527425452423283320117768198402231770208869520047764273682576626139237031385665948631650626991844596463898746277344711896086305533142593135616665318539129989145312280000688779148240044871428926990063486244781615463646388363947317026040466353970904996558162398808944629605623311649536164221970332681344168908984458505602379484807914058900934776500429002716706625830522008132236281291761267883317206598995396418127021779858404042159853183251540889433902091920554957783589672039160081957216630582755380425583726015528348786419432054508915275783882625175435528800822842770817965453762184851149029376\n"}] | false | stdio | null | true |
606/B | 606 | B | Python 3 | TESTS | 3 | 46 | 0 | 14748527 | x, y, sx, sy = map(int, input().split())
sx, sy = sx - 1, sy - 1
direction = {'U' : (-1, 0), 'D' : (1, 0), 'L' : (0, -1), 'R' : (0, 1)}
visited = [[False] * y for i in range(x)]
prev = sx, sy
s = input()
sm = x * y - 1
print (1, end = ' ')
for c in s[:-1]:
d = direction[c]
cur = tuple(map(sum, zip(prev, d)))
if not 0 <= cur[0] < x or not 0 <= cur[1] < y or visited[cur[0]][cur[1]]:
cur = prev
print (0, end = ' ')
else:
print (1, end = ' ')
sm -= 1
visited[cur[0]][cur[1]] = True
prev = cur
print (sm) | 68 | 155 | 13,516,800 | 199128141 | import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def f(u, v):
return u * y + v
x, y, x0, y0 = map(int, input().split())
d = {"D":(1, 0), "U":(-1, 0), "R":(0, 1), "L":(0, -1)}
s = list(input().rstrip().decode())
visit = [0] * (x * y)
i, j = x0 - 1, y0 - 1
ans = []
for k in s:
ans.append(visit[f(i, j)] ^ 1)
visit[f(i, j)] = 1
di, dj = d[k]
ni, nj = i + di, j + dj
if 0 <= ni < x and 0 <= nj < y:
i, j = ni, nj
ans.append(x * y - sum(ans))
sys.stdout.write(" ".join(map(str, ans))) | Codeforces Round 335 (Div. 2) | CF | 2,015 | 2 | 256 | Testing Robots | The Cybernetics Failures (CF) organisation made a prototype of a bomb technician robot. To find the possible problems it was decided to carry out a series of tests. At the beginning of each test the robot prototype will be placed in cell (x0, y0) of a rectangular squared field of size x × y, after that a mine will be installed into one of the squares of the field. It is supposed to conduct exactly x·y tests, each time a mine is installed into a square that has never been used before. The starting cell of the robot always remains the same.
After placing the objects on the field the robot will have to run a sequence of commands given by string s, consisting only of characters 'L', 'R', 'U', 'D'. These commands tell the robot to move one square to the left, to the right, up or down, or stay idle if moving in the given direction is impossible. As soon as the robot fulfills all the sequence of commands, it will blow up due to a bug in the code. But if at some moment of time the robot is at the same square with the mine, it will also blow up, but not due to a bug in the code.
Moving to the left decreases coordinate y, and moving to the right increases it. Similarly, moving up decreases the x coordinate, and moving down increases it.
The tests can go on for very long, so your task is to predict their results. For each k from 0 to length(s) your task is to find in how many tests the robot will run exactly k commands before it blows up. | The first line of the input contains four integers x, y, x0, y0 (1 ≤ x, y ≤ 500, 1 ≤ x0 ≤ x, 1 ≤ y0 ≤ y) — the sizes of the field and the starting coordinates of the robot. The coordinate axis X is directed downwards and axis Y is directed to the right.
The second line contains a sequence of commands s, which should be fulfilled by the robot. It has length from 1 to 100 000 characters and only consists of characters 'L', 'R', 'U', 'D'. | Print the sequence consisting of (length(s) + 1) numbers. On the k-th position, starting with zero, print the number of tests where the robot will run exactly k commands before it blows up. | null | In the first sample, if we exclude the probable impact of the mines, the robot's route will look like that: $$(2,2)\rightarrow(1,2)\rightarrow(1,3)\rightarrow(2,4)\rightarrow(3,3)$$. | [{"input": "3 4 2 2\nUURDRDRL", "output": "1 1 0 1 1 1 1 0 6"}, {"input": "2 2 2 2\nULD", "output": "1 1 1 1"}] | 1,600 | ["implementation"] | 68 | [{"input": "3 4 2 2\r\nUURDRDRL\r\n", "output": "1 1 0 1 1 1 1 0 6\r\n"}, {"input": "2 2 2 2\r\nULD\r\n", "output": "1 1 1 1\r\n"}, {"input": "1 1 1 1\r\nURDLUURRDDLLURDL\r\n", "output": "1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDD\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 245\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDDRRUR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 241\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDDRRURR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 0 241\r\n"}, {"input": "1 2 1 1\r\nR\r\n", "output": "1 1\r\n"}, {"input": "2 1 1 1\r\nD\r\n", "output": "1 1\r\n"}, {"input": "1 2 1 2\r\nLR\r\n", "output": "1 1 0\r\n"}, {"input": "2 1 2 1\r\nUD\r\n", "output": "1 1 0\r\n"}, {"input": "4 4 2 2\r\nDRUL\r\n", "output": "1 1 1 1 12\r\n"}, {"input": "4 4 3 3\r\nLUDRUL\r\n", "output": "1 1 1 0 0 1 12\r\n"}, {"input": "15 17 8 9\r\nURRDLU\r\n", "output": "1 1 1 1 1 1 249\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRRRU\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 1 243\r\n"}] | false | stdio | null | true |
606/B | 606 | B | Python 3 | TESTS | 3 | 46 | 0 | 14783733 | x, y, x0, y0 = map(int, input().split())
s = input()
k = (len(s)+1)*[0]
k[0] = 1
u = 0
for i in s:
u += 1
xo = x0; yo = y0
if i == 'U' and x0 > 1:
x0 -= 1
elif i == 'D' and x0 < x:
x0 += 1
elif i == 'L' and y0 > 1:
y0 -= 1
elif i == 'R' and y0 < y:
y0 += 1
if not(x0 == xo and y0 == yo):
k[u]+=1
k[len(s)] += (x*y)-sum(k)
for i in k:
print(i, end=' ') | 68 | 155 | 19,046,400 | 210793807 | import sys
input = lambda: sys.stdin.readline().rstrip()
x,y,x0,y0 = map(int, input().split())
S = input()
ans = [1]
seen = set()
seen.add((x0,y0))
for s in S:
if s=='U':
x0-=1
elif s=='D':
x0+=1
elif s=='L':
y0-=1
else:
y0+=1
if x0<1:
x0=1
if x0>x:
x0=x
if y0<1:
y0=1
if y0>y:
y0=y
if (x0,y0) in seen:
ans.append(0)
else:
ans.append(1)
seen.add((x0,y0))
t = sum(ans)
ans[-1]+=x*y-t
print(*ans) | Codeforces Round 335 (Div. 2) | CF | 2,015 | 2 | 256 | Testing Robots | The Cybernetics Failures (CF) organisation made a prototype of a bomb technician robot. To find the possible problems it was decided to carry out a series of tests. At the beginning of each test the robot prototype will be placed in cell (x0, y0) of a rectangular squared field of size x × y, after that a mine will be installed into one of the squares of the field. It is supposed to conduct exactly x·y tests, each time a mine is installed into a square that has never been used before. The starting cell of the robot always remains the same.
After placing the objects on the field the robot will have to run a sequence of commands given by string s, consisting only of characters 'L', 'R', 'U', 'D'. These commands tell the robot to move one square to the left, to the right, up or down, or stay idle if moving in the given direction is impossible. As soon as the robot fulfills all the sequence of commands, it will blow up due to a bug in the code. But if at some moment of time the robot is at the same square with the mine, it will also blow up, but not due to a bug in the code.
Moving to the left decreases coordinate y, and moving to the right increases it. Similarly, moving up decreases the x coordinate, and moving down increases it.
The tests can go on for very long, so your task is to predict their results. For each k from 0 to length(s) your task is to find in how many tests the robot will run exactly k commands before it blows up. | The first line of the input contains four integers x, y, x0, y0 (1 ≤ x, y ≤ 500, 1 ≤ x0 ≤ x, 1 ≤ y0 ≤ y) — the sizes of the field and the starting coordinates of the robot. The coordinate axis X is directed downwards and axis Y is directed to the right.
The second line contains a sequence of commands s, which should be fulfilled by the robot. It has length from 1 to 100 000 characters and only consists of characters 'L', 'R', 'U', 'D'. | Print the sequence consisting of (length(s) + 1) numbers. On the k-th position, starting with zero, print the number of tests where the robot will run exactly k commands before it blows up. | null | In the first sample, if we exclude the probable impact of the mines, the robot's route will look like that: $$(2,2)\rightarrow(1,2)\rightarrow(1,3)\rightarrow(2,4)\rightarrow(3,3)$$. | [{"input": "3 4 2 2\nUURDRDRL", "output": "1 1 0 1 1 1 1 0 6"}, {"input": "2 2 2 2\nULD", "output": "1 1 1 1"}] | 1,600 | ["implementation"] | 68 | [{"input": "3 4 2 2\r\nUURDRDRL\r\n", "output": "1 1 0 1 1 1 1 0 6\r\n"}, {"input": "2 2 2 2\r\nULD\r\n", "output": "1 1 1 1\r\n"}, {"input": "1 1 1 1\r\nURDLUURRDDLLURDL\r\n", "output": "1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDD\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 245\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDDRRUR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 241\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDDRRURR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 0 241\r\n"}, {"input": "1 2 1 1\r\nR\r\n", "output": "1 1\r\n"}, {"input": "2 1 1 1\r\nD\r\n", "output": "1 1\r\n"}, {"input": "1 2 1 2\r\nLR\r\n", "output": "1 1 0\r\n"}, {"input": "2 1 2 1\r\nUD\r\n", "output": "1 1 0\r\n"}, {"input": "4 4 2 2\r\nDRUL\r\n", "output": "1 1 1 1 12\r\n"}, {"input": "4 4 3 3\r\nLUDRUL\r\n", "output": "1 1 1 0 0 1 12\r\n"}, {"input": "15 17 8 9\r\nURRDLU\r\n", "output": "1 1 1 1 1 1 249\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRRRU\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 1 243\r\n"}] | false | stdio | null | true |
606/B | 606 | B | Python 3 | PRETESTS | 3 | 31 | 0 | 14720442 | x, y, x0, y0 = map(int, input().split())
res = x * y - 1
our = dict()
our['U'] = (-1, 0)
our['D'] = (1, 0)
our['R'] = (0, 1)
our['L'] = (0, -1)
s = input()
print(1, end=' ')
for i in range(len(s) - 1):
dx, dy = our[s[i]]
if 1 <= x0 + dx <= x and 1 <= y0 + dy <= y:
x0 += dx
y0 += dy
print(1, end=' ')
res -= 1
else:
print(0, end=' ')
print(res) | 68 | 171 | 2,355,200 | 14729604 | x, y, x0, y0 = map(int, input().split())
s = input()
s = s[:len(s) - 1]
visited = [[False for i in range(y + 1)] for j in range(x + 1)]
visited[x0][y0] = True
ans = '1 '
cnt = 1
for i in s:
if i == 'U':
if x0 != 1:
x0 -= 1
if not visited[x0][y0]:
ans += '1 '
cnt += 1
visited[x0][y0] = True
else:
ans += '0 '
else:
ans += '0 '
elif i == 'D':
if x0 != x:
x0 += 1
if not visited[x0][y0]:
ans += '1 '
cnt += 1
visited[x0][y0] = True
else:
ans += '0 '
else:
ans += '0 '
elif i == 'L':
if y0 != 1:
y0 -= 1
if not visited[x0][y0]:
ans += '1 '
cnt += 1
visited[x0][y0] = True
else:
ans += '0 '
else:
ans += '0 '
else:
if y0 != y:
y0 += 1
if not visited[x0][y0]:
ans += '1 '
cnt += 1
visited[x0][y0] = True
else:
ans += '0 '
else:
ans += '0 '
z = str(x * y - cnt)
print(ans + z) | Codeforces Round 335 (Div. 2) | CF | 2,015 | 2 | 256 | Testing Robots | The Cybernetics Failures (CF) organisation made a prototype of a bomb technician robot. To find the possible problems it was decided to carry out a series of tests. At the beginning of each test the robot prototype will be placed in cell (x0, y0) of a rectangular squared field of size x × y, after that a mine will be installed into one of the squares of the field. It is supposed to conduct exactly x·y tests, each time a mine is installed into a square that has never been used before. The starting cell of the robot always remains the same.
After placing the objects on the field the robot will have to run a sequence of commands given by string s, consisting only of characters 'L', 'R', 'U', 'D'. These commands tell the robot to move one square to the left, to the right, up or down, or stay idle if moving in the given direction is impossible. As soon as the robot fulfills all the sequence of commands, it will blow up due to a bug in the code. But if at some moment of time the robot is at the same square with the mine, it will also blow up, but not due to a bug in the code.
Moving to the left decreases coordinate y, and moving to the right increases it. Similarly, moving up decreases the x coordinate, and moving down increases it.
The tests can go on for very long, so your task is to predict their results. For each k from 0 to length(s) your task is to find in how many tests the robot will run exactly k commands before it blows up. | The first line of the input contains four integers x, y, x0, y0 (1 ≤ x, y ≤ 500, 1 ≤ x0 ≤ x, 1 ≤ y0 ≤ y) — the sizes of the field and the starting coordinates of the robot. The coordinate axis X is directed downwards and axis Y is directed to the right.
The second line contains a sequence of commands s, which should be fulfilled by the robot. It has length from 1 to 100 000 characters and only consists of characters 'L', 'R', 'U', 'D'. | Print the sequence consisting of (length(s) + 1) numbers. On the k-th position, starting with zero, print the number of tests where the robot will run exactly k commands before it blows up. | null | In the first sample, if we exclude the probable impact of the mines, the robot's route will look like that: $$(2,2)\rightarrow(1,2)\rightarrow(1,3)\rightarrow(2,4)\rightarrow(3,3)$$. | [{"input": "3 4 2 2\nUURDRDRL", "output": "1 1 0 1 1 1 1 0 6"}, {"input": "2 2 2 2\nULD", "output": "1 1 1 1"}] | 1,600 | ["implementation"] | 68 | [{"input": "3 4 2 2\r\nUURDRDRL\r\n", "output": "1 1 0 1 1 1 1 0 6\r\n"}, {"input": "2 2 2 2\r\nULD\r\n", "output": "1 1 1 1\r\n"}, {"input": "1 1 1 1\r\nURDLUURRDDLLURDL\r\n", "output": "1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDD\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 245\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDDRRUR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 241\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDDRRURR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 0 241\r\n"}, {"input": "1 2 1 1\r\nR\r\n", "output": "1 1\r\n"}, {"input": "2 1 1 1\r\nD\r\n", "output": "1 1\r\n"}, {"input": "1 2 1 2\r\nLR\r\n", "output": "1 1 0\r\n"}, {"input": "2 1 2 1\r\nUD\r\n", "output": "1 1 0\r\n"}, {"input": "4 4 2 2\r\nDRUL\r\n", "output": "1 1 1 1 12\r\n"}, {"input": "4 4 3 3\r\nLUDRUL\r\n", "output": "1 1 1 0 0 1 12\r\n"}, {"input": "15 17 8 9\r\nURRDLU\r\n", "output": "1 1 1 1 1 1 249\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRRRU\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 1 243\r\n"}] | false | stdio | null | true |
606/B | 606 | B | Python 3 | PRETESTS | 4 | 61 | 0 | 14730255 | y, x, x_, y_ = list(map(int, input().split()))
c = input()
ans = [0 for i in range(len(c) + 1)]
ans[0] = 1
p = 1
was_here = set()
was_here.add((x_, y_))
for i in c:
if i == 'U':
y_ = max(1, y_ - 1)
elif i == 'D':
y_ = min(y, y_ + 1)
elif i == 'L':
x_ = max(1, x_ - 1)
else:
x_ = min(x, x_ + 1)
if (x_, y_) in was_here:
ans[p] = 0
elif p < len(c):
ans[p] = 1
was_here.add((x_, y_))
else:
ans[p] = x * y - len(was_here)
p += 1
print(' '.join(list(map(str, ans)))) | 68 | 186 | 10,752,000 | 162511090 | [r, c, x, y], s = map(int, input().split()), input()
vis = [[False for i in range(c)] for j in range(r)]
x, y = x - 1, y - 1
vis[x][y] = True
ans = [1]
for ch in s:
if ch == 'U':
x = max(0, x - 1)
elif ch == 'R':
y = min(c - 1, y + 1)
elif ch == 'D':
x = min(r - 1, x + 1)
else:
y = max(0, y - 1)
ans.append(int(not vis[x][y]))
vis[x][y] = True
for i in range(r):
for j in range(c):
ans[-1] += int(not vis[i][j])
print(*ans) | Codeforces Round 335 (Div. 2) | CF | 2,015 | 2 | 256 | Testing Robots | The Cybernetics Failures (CF) organisation made a prototype of a bomb technician robot. To find the possible problems it was decided to carry out a series of tests. At the beginning of each test the robot prototype will be placed in cell (x0, y0) of a rectangular squared field of size x × y, after that a mine will be installed into one of the squares of the field. It is supposed to conduct exactly x·y tests, each time a mine is installed into a square that has never been used before. The starting cell of the robot always remains the same.
After placing the objects on the field the robot will have to run a sequence of commands given by string s, consisting only of characters 'L', 'R', 'U', 'D'. These commands tell the robot to move one square to the left, to the right, up or down, or stay idle if moving in the given direction is impossible. As soon as the robot fulfills all the sequence of commands, it will blow up due to a bug in the code. But if at some moment of time the robot is at the same square with the mine, it will also blow up, but not due to a bug in the code.
Moving to the left decreases coordinate y, and moving to the right increases it. Similarly, moving up decreases the x coordinate, and moving down increases it.
The tests can go on for very long, so your task is to predict their results. For each k from 0 to length(s) your task is to find in how many tests the robot will run exactly k commands before it blows up. | The first line of the input contains four integers x, y, x0, y0 (1 ≤ x, y ≤ 500, 1 ≤ x0 ≤ x, 1 ≤ y0 ≤ y) — the sizes of the field and the starting coordinates of the robot. The coordinate axis X is directed downwards and axis Y is directed to the right.
The second line contains a sequence of commands s, which should be fulfilled by the robot. It has length from 1 to 100 000 characters and only consists of characters 'L', 'R', 'U', 'D'. | Print the sequence consisting of (length(s) + 1) numbers. On the k-th position, starting with zero, print the number of tests where the robot will run exactly k commands before it blows up. | null | In the first sample, if we exclude the probable impact of the mines, the robot's route will look like that: $$(2,2)\rightarrow(1,2)\rightarrow(1,3)\rightarrow(2,4)\rightarrow(3,3)$$. | [{"input": "3 4 2 2\nUURDRDRL", "output": "1 1 0 1 1 1 1 0 6"}, {"input": "2 2 2 2\nULD", "output": "1 1 1 1"}] | 1,600 | ["implementation"] | 68 | [{"input": "3 4 2 2\r\nUURDRDRL\r\n", "output": "1 1 0 1 1 1 1 0 6\r\n"}, {"input": "2 2 2 2\r\nULD\r\n", "output": "1 1 1 1\r\n"}, {"input": "1 1 1 1\r\nURDLUURRDDLLURDL\r\n", "output": "1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDD\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 245\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDDRRUR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 241\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDDRRURR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 0 241\r\n"}, {"input": "1 2 1 1\r\nR\r\n", "output": "1 1\r\n"}, {"input": "2 1 1 1\r\nD\r\n", "output": "1 1\r\n"}, {"input": "1 2 1 2\r\nLR\r\n", "output": "1 1 0\r\n"}, {"input": "2 1 2 1\r\nUD\r\n", "output": "1 1 0\r\n"}, {"input": "4 4 2 2\r\nDRUL\r\n", "output": "1 1 1 1 12\r\n"}, {"input": "4 4 3 3\r\nLUDRUL\r\n", "output": "1 1 1 0 0 1 12\r\n"}, {"input": "15 17 8 9\r\nURRDLU\r\n", "output": "1 1 1 1 1 1 249\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRRRU\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 1 243\r\n"}] | false | stdio | null | true |
18/D | 18 | D | PyPy 3-64 | TESTS | 24 | 280 | 6,144,000 | 150917218 | n=int(input())
DP=[-10]*2001
DP[0]=0
for i in range(n):
s,x=input().split()
x=int(x)
if s[0]=="w":
if DP[x]<DP[0]:
DP[x]=DP[0]
else:
if DP[x]>=0:
DP[0]=max(DP[0],DP[x]+(1<<x))
print(max(DP)) | 45 | 154 | 4,915,200 | 219090777 | '''
BeezMinh
16:43 UTC+7
16/08/2023
'''
from sys import stdin
input = lambda: stdin.readline().rstrip()
a = [0] * 2048
ans = 0
for i in range(int(input())):
s, x = input().split()
if s == 'win':
a[int(x)] = ans + 2 ** int(x)
else:
ans = max(ans, a[int(x)])
print(ans) | Codeforces Beta Round 18 (Div. 2 Only) | ICPC | 2,010 | 2 | 128 | Seller Bob | Last year Bob earned by selling memory sticks. During each of n days of his work one of the two following events took place:
- A customer came to Bob and asked to sell him a 2x MB memory stick. If Bob had such a stick, he sold it and got 2x berllars.
- Bob won some programming competition and got a 2x MB memory stick as a prize. Bob could choose whether to present this memory stick to one of his friends, or keep it.
Bob never kept more than one memory stick, as he feared to mix up their capacities, and deceive a customer unintentionally. It is also known that for each memory stick capacity there was at most one customer, who wanted to buy that memory stick. Now, knowing all the customers' demands and all the prizes won at programming competitions during the last n days, Bob wants to know, how much money he could have earned, if he had acted optimally. | The first input line contains number n (1 ≤ n ≤ 5000) — amount of Bob's working days. The following n lines contain the description of the days. Line sell x stands for a day when a customer came to Bob to buy a 2x MB memory stick (0 ≤ x ≤ 2000). It's guaranteed that for each x there is not more than one line sell x. Line win x stands for a day when Bob won a 2x MB memory stick (0 ≤ x ≤ 2000). | Output the maximum possible earnings for Bob in berllars, that he would have had if he had known all the events beforehand. Don't forget, please, that Bob can't keep more than one memory stick at a time. | null | null | [{"input": "7\nwin 10\nwin 5\nwin 3\nsell 5\nsell 3\nwin 10\nsell 10", "output": "1056"}, {"input": "3\nwin 5\nsell 6\nsell 4", "output": "0"}] | 2,000 | ["brute force", "dp", "greedy"] | 45 | [{"input": "7\r\nwin 10\r\nwin 5\r\nwin 3\r\nsell 5\r\nsell 3\r\nwin 10\r\nsell 10\r\n", "output": "1056\r\n"}, {"input": "3\r\nwin 5\r\nsell 6\r\nsell 4\r\n", "output": "0\r\n"}, {"input": "10\r\nsell 179\r\nwin 1278\r\nsell 1278\r\nwin 179\r\nwin 788\r\nsell 788\r\nwin 1819\r\nwin 1278\r\nsell 1454\r\nsell 1819\r\n", "output": "37459511778596727480858760720167552241582636504705413766024169777495064333423937410125519624693990051069809575647477719465460756326341562228323606665869931977125977431028709943048934214062888966581139223580790503937962827597404798307897711090567429316074325427043388117806141094834711707585035634104732053207574452493593409130554278913951011894497392495930884827685983975668127973918422057605356890341647839399778378381152159725053311750647457999739578989105335906181048932656785993705124392163591312698147450544686667028530553878359909362604965888\n"}, {"input": "10\r\nsell 573\r\nwin 1304\r\nsell 278\r\nwin 1631\r\nsell 1225\r\nsell 1631\r\nsell 177\r\nwin 1631\r\nwin 177\r\nsell 1304\r\n", "output": "95482312335125227379668481690754940528280513838693267460502082967052005332103697568042408703168913727303170456338425853153094403747135188778307041838920404959089576368946137708987138986696495077466398994298434148881715073638178666201165545650953479735059082316661443204882826188032944866093372620219104327689636641547141835841165681118172603993695103043804276669836594061369229043451067647935298287687852302215923887110435577776767805943668204998410716005202198549540411238299513630278811648\n"}, {"input": "10\r\nwin 1257\r\nwin 1934\r\nsell 1934\r\nsell 1257\r\nwin 1934\r\nwin 1257\r\nsell 495\r\nwin 495\r\nwin 495\r\nwin 1257\r\n", "output": "1556007242642049292787218246793379348327505438878680952714050868520307364441227819009733220897932984584977593931988662671459594674963394056587723382487766303981362587048873128400436836690128983570130687310221668877557121158055843621982630476422478413285775826498536883275291967793661985813155062733063913176306327509625594121241472451054995889483447103432414676059872469910105149496451402271546454282618581884282152530090816240540173251729211604658704990425330422792556824836640431985211146197816770068601144273155826261981248156109439380758754562057681271989937590784957541131485184\n"}, {"input": "10\r\nsell 1898\r\nsell 173\r\nsell 1635\r\nsell 29\r\nsell 881\r\nsell 434\r\nsell 1236\r\nsell 14\r\nwin 29\r\nsell 1165\r\n", "output": "0\r\n"}, {"input": "1\r\nsell 2000\r\n", "output": "0\r\n"}, {"input": "1\r\nwin 2000\r\n", "output": "0\r\n"}, {"input": "2\r\nwin 2000\r\nsell 2000\r\n", "output": "114813069527425452423283320117768198402231770208869520047764273682576626139237031385665948631650626991844596463898746277344711896086305533142593135616665318539129989145312280000688779148240044871428926990063486244781615463646388363947317026040466353970904996558162398808944629605623311649536164221970332681344168908984458505602379484807914058900934776500429002716706625830522008132236281291761267883317206598995396418127021779858404042159853183251540889433902091920554957783589672039160081957216630582755380425583726015528348786419432054508915275783882625175435528800822842770817965453762184851149029376\n"}] | false | stdio | null | true |
765/D | 765 | D | Python 3 | TESTS | 1 | 93 | 0 | 62645047 | n = int(input())
f = list(map(int, input().split()))
s = set(f)
distinct = len(s)
first = set(f[:distinct])
if(sorted(f) != f or len(first) != distinct):
print(-1)
exit()
print(distinct)
print(*f)
print(*list(first)) | 43 | 248 | 29,593,600 | 126192952 | from collections import defaultdict
def solve(n, f):
d = defaultdict(set)
for i,fi in enumerate(f, 1):
d[fi].add(i)
m = len(d)
g = [0]*n
h = [0]*m
for i,(fi,ys) in enumerate(d.items()):
if fi not in ys:
return -1, [], []
h[i] = fi
for j in ys:
g[j-1] = i+1
return m, g, h
n = int(input())
f = list(map(int,input().split()))
res, g, h = solve(n, f)
print(res)
if res > -1:
print(' '.join(map(str, g)))
print(' '.join(map(str, h))) | Codeforces Round 397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) | CF | 2,017 | 2 | 512 | Artsem and Saunders | Artsem has a friend Saunders from University of Chicago. Saunders presented him with the following problem.
Let [n] denote the set {1, ..., n}. We will also write f: [x] → [y] when a function f is defined in integer points 1, ..., x, and all its values are integers from 1 to y.
Now then, you are given a function f: [n] → [n]. Your task is to find a positive integer m, and two functions g: [n] → [m], h: [m] → [n], such that g(h(x)) = x for all $$x \in [m]$$, and h(g(x)) = f(x) for all $$x \in [n]$$, or determine that finding these is impossible. | The first line contains an integer n (1 ≤ n ≤ 105).
The second line contains n space-separated integers — values f(1), ..., f(n) (1 ≤ f(i) ≤ n). | If there is no answer, print one integer -1.
Otherwise, on the first line print the number m (1 ≤ m ≤ 106). On the second line print n numbers g(1), ..., g(n). On the third line print m numbers h(1), ..., h(m).
If there are several correct answers, you may output any of them. It is guaranteed that if a valid answer exists, then there is an answer satisfying the above restrictions. | null | null | [{"input": "3\n1 2 3", "output": "3\n1 2 3\n1 2 3"}, {"input": "3\n2 2 2", "output": "1\n1 1 1\n2"}, {"input": "2\n2 1", "output": "-1"}] | 1,700 | ["constructive algorithms", "dsu", "math"] | 43 | [{"input": "3\r\n1 2 3\r\n", "output": "3\r\n1 2 3\r\n1 2 3\r\n"}, {"input": "3\r\n2 2 2\r\n", "output": "1\r\n1 1 1\r\n2\r\n"}, {"input": "2\r\n2 1\r\n", "output": "-1\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n1\r\n1\r\n"}, {"input": "2\r\n2 1\r\n", "output": "-1\r\n"}, {"input": "2\r\n2 2\r\n", "output": "1\r\n1 1\r\n2\r\n"}, {"input": "5\r\n5 5 5 3 5\r\n", "output": "-1\r\n"}, {"input": "10\r\n4 4 4 4 4 4 4 4 4 4\r\n", "output": "1\r\n1 1 1 1 1 1 1 1 1 1\r\n4\r\n"}, {"input": "2\r\n1 2\r\n", "output": "2\r\n1 2\r\n1 2\r\n"}, {"input": "3\r\n3 2 3\r\n", "output": "2\r\n2 1 2\r\n2 3\r\n"}, {"input": "3\r\n1 2 1\r\n", "output": "2\r\n1 2 1\r\n1 2\r\n"}, {"input": "4\r\n4 2 4 4\r\n", "output": "2\r\n2 1 2 2\r\n2 4\r\n"}, {"input": "5\r\n1 4 5 4 5\r\n", "output": "3\r\n1 2 3 2 3\r\n1 4 5\r\n"}, {"input": "4\r\n1 2 1 2\r\n", "output": "2\r\n1 2 1 2\r\n1 2\r\n"}, {"input": "5\r\n1 3 3 4 4\r\n", "output": "3\r\n1 2 2 3 3\r\n1 3 4\r\n"}, {"input": "4\r\n4 2 2 4\r\n", "output": "2\r\n2 1 1 2\r\n2 4\r\n"}, {"input": "7\r\n7 3 3 5 5 7 7\r\n", "output": "3\r\n3 1 1 2 2 3 3\r\n3 5 7\r\n"}, {"input": "6\r\n1 1 1 3 3 3\r\n", "output": "-1\r\n"}, {"input": "4\r\n2 2 3 2\r\n", "output": "2\r\n1 1 2 1\r\n2 3\r\n"}, {"input": "6\r\n1 2 3 4 5 5\r\n", "output": "5\r\n1 2 3 4 5 5\r\n1 2 3 4 5\r\n"}, {"input": "3\r\n1 1 2\r\n", "output": "-1\r\n"}, {"input": "4\r\n3 4 3 4\r\n", "output": "2\r\n1 2 1 2\r\n3 4\r\n"}, {"input": "6\r\n1 1 1 4 4 4\r\n", "output": "2\r\n1 1 1 2 2 2\r\n1 4\r\n"}, {"input": "4\r\n1 2 1 1\r\n", "output": "2\r\n1 2 1 1\r\n1 2\r\n"}, {"input": "5\r\n1 2 3 4 3\r\n", "output": "4\r\n1 2 3 4 3\r\n1 2 3 4\r\n"}, {"input": "4\r\n2 2 4 4\r\n", "output": "2\r\n1 1 2 2\r\n2 4\r\n"}, {"input": "4\r\n1 1 3 3\r\n", "output": "2\r\n1 1 2 2\r\n1 3\r\n"}, {"input": "3\r\n2 2 3\r\n", "output": "2\r\n1 1 2\r\n2 3\r\n"}, {"input": "5\r\n5 3 3 3 5\r\n", "output": "2\r\n2 1 1 1 2\r\n3 5\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) as f:
n = int(f.readline().strip())
f_list = list(map(int, f.readline().strip().split()))
with open(submission_path) as sub_f:
lines = [line.strip() for line in sub_f if line.strip()]
if not lines:
print(0)
return
first_line = lines[0]
if first_line == '-1':
# Check if the problem is truly impossible
possible = True
for x in range(n):
if f_list[x] < 1 or f_list[x] > n:
possible = False
break
fx = f_list[x]
if f_list[fx - 1] != fx:
possible = False
break
if possible:
print(0)
else:
print(1)
return
# Try to parse m, g, h
try:
m = int(first_line)
if m < 1 or m > 10**6:
print(0)
return
if len(lines) < 3:
print(0)
return
g = list(map(int, lines[1].split()))
h = list(map(int, lines[2].split()))
if len(g) != n or len(h) != m:
print(0)
return
# Check g values
for num in g:
if num < 1 or num > m:
print(0)
return
# Check h values
for num in h:
if num < 1 or num > n:
print(0)
return
# Check h(g(x)) == f[x] for all x
for x in range(n):
gx = g[x]
h_gx = h[gx - 1]
if h_gx != f_list[x]:
print(0)
return
# Check g(h(y)) == y+1 for all y (0-based)
for y in range(m):
hy = h[y]
g_hy = g[hy - 1]
if g_hy != (y + 1):
print(0)
return
print(1)
except:
print(0)
if __name__ == '__main__':
main()
| true |
180/E | 180 | E | Python 3 | TESTS | 5 | 92 | 307,200 | 205662613 | from collections import defaultdict
from sys import stdin
def max_points(n, m, k, cubes):
color_counts = defaultdict(int)
max_count = 0
left = 0
total_to_delete = 0
for right in range(n):
color = cubes[right]
color_counts[color] += 1
total_to_delete += 1 - (cubes[left] == color)
while total_to_delete > k:
color_counts[cubes[left]] -= 1
total_to_delete -= 1 - (cubes[left] == cubes[left + 1])
left += 1
max_count = max(max_count, color_counts[color])
return max_count
def main():
n, m, k = map(int, stdin.readline().split())
cubes = list(map(int, stdin.readline().split()))
print(max_points(n, m, k, cubes))
if __name__ == "__main__":
main() | 50 | 310 | 16,588,800 | 199566000 | import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
x = [[] for _ in range(m + 1)]
for i in range(n):
x[a[i]].append(i)
ans = 0
for y in x:
if not y:
continue
r = 0
for l in range(len(y)):
while r < len(y) and y[r] - y[l] - (r - l) <= k:
r += 1
ans = max(ans, r - l)
print(ans) | Codeforces Round 116 (Div. 2, ACM-ICPC Rules) | ICPC | 2,012 | 1 | 256 | Cubes | Let's imagine that you're playing the following simple computer game. The screen displays n lined-up cubes. Each cube is painted one of m colors. You are allowed to delete not more than k cubes (that do not necessarily go one after another). After that, the remaining cubes join together (so that the gaps are closed) and the system counts the score. The number of points you score equals to the length of the maximum sequence of cubes of the same color that follow consecutively. Write a program that determines the maximum possible number of points you can score.
Remember, you may delete no more than k any cubes. It is allowed not to delete cubes at all. | The first line contains three integers n, m and k (1 ≤ n ≤ 2·105, 1 ≤ m ≤ 105, 0 ≤ k < n). The second line contains n integers from 1 to m — the numbers of cube colors. The numbers of colors are separated by single spaces. | Print the maximum possible number of points you can score. | null | In the first sample you should delete the fifth and the sixth cubes.
In the second sample you should delete the fourth and the seventh cubes.
In the third sample you shouldn't delete any cubes. | [{"input": "10 3 2\n1 2 1 1 3 2 1 1 2 2", "output": "4"}, {"input": "10 2 2\n1 2 1 2 1 1 2 1 1 2", "output": "5"}, {"input": "3 1 2\n1 1 1", "output": "3"}] | 1,800 | ["binary search", "dp", "two pointers"] | 50 | [{"input": "10 3 2\r\n1 2 1 1 3 2 1 1 2 2\r\n", "output": "4\r\n"}, {"input": "10 2 2\r\n1 2 1 2 1 1 2 1 1 2\r\n", "output": "5\r\n"}, {"input": "3 1 2\r\n1 1 1\r\n", "output": "3\r\n"}, {"input": "10 2 2\r\n1 1 1 2 1 2 1 2 1 1\r\n", "output": "5\r\n"}, {"input": "1 1 0\r\n1\r\n", "output": "1\r\n"}, {"input": "20 3 5\r\n2 2 3 1 2 2 3 3 3 2 1 2 3 1 1 3 3 3 2 3\r\n", "output": "7\r\n"}, {"input": "20 2 5\r\n2 2 1 2 1 2 1 2 1 1 2 1 2 2 1 2 2 1 2 1\r\n", "output": "7\r\n"}, {"input": "20 6 3\r\n4 1 2 6 3 3 2 5 2 5 2 1 1 4 1 2 2 1 1 4\r\n", "output": "5\r\n"}, {"input": "30 5 8\r\n1 4 1 5 3 4 4 1 1 4 1 3 5 5 5 5 1 5 1 5 2 3 2 2 3 4 5 2 1 2\r\n", "output": "7\r\n"}, {"input": "30 5 6\r\n4 2 2 1 3 4 2 3 2 4 3 1 1 4 4 3 5 1 4 5 5 1 2 2 1 2 4 4 1 2\r\n", "output": "4\r\n"}, {"input": "100 10 15\r\n6 6 6 6 7 7 8 8 4 4 4 1 1 7 7 7 1 1 1 2 2 2 2 2 2 2 2 2 10 5 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 6 6 10 2 2 8 8 1 1 1 1 1 6 6 6 6 2 2 3 3 9 9 9 9 9 10 10 10 10 10 4 9 9 9 7 7 7 7 9 9 7 7 5 8 8 8 8 2\r\n", "output": "25\r\n"}, {"input": "99 10 17\r\n3 2 2 9 7 10 10 10 10 6 6 6 3 7 3 3 7 2 2 2 2 2 10 10 2 2 7 7 7 7 1 8 8 8 8 10 9 10 10 10 5 5 2 2 5 5 5 1 4 9 9 2 2 3 3 2 2 9 9 9 9 9 9 9 7 4 8 8 4 8 8 10 10 4 5 9 9 10 5 5 5 5 5 8 8 8 8 2 2 2 2 1 8 8 5 10 10 2 2\r\n", "output": "11\r\n"}, {"input": "94 10 20\r\n2 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 3 3 3 3 6 6 5 1 5 5 5 2 2 2 2 4 1 1 1 1 8 8 10 5 2 2 4 4 4 4 4 3 3 3 3 3 6 6 6 6 2 2 2 2 2 2 2 2 1 10 10 2 2 2 6 6 6 8 4 4 4 8 1 1 1 1 1 1 6 6 2 2 8 7 7 7 3 4\r\n", "output": "13\r\n"}, {"input": "99 3 15\r\n2 2 2 2 2 2 3 3 1 1 1 1 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 3 1 1 1 1 1 3 3 3 1 1 1 3 3 3 3 3 3 3 1 1 1 1 3 3 3 3 3 3 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 2 2 2 2 2 2\r\n", "output": "27\r\n"}, {"input": "100 5 10\r\n4 4 4 4 4 4 4 4 4 4 4 4 2 4 4 4 4 4 4 4 4 4 2 2 2 3 3 3 3 3 3 3 4 4 4 3 3 2 1 1 1 2 3 3 5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2 2 4 4 4 4 5 4 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 5 5 5 3 3 4 3 3 3\r\n", "output": "21\r\n"}, {"input": "98 4 20\r\n3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 1 1 1 1 1 1 1 1 1 1 1 1 4 4 4 4 4 4 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 2 2 2 2 2 2 2 1 1 1 3 3 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 2 2 2 2 2 2 2 3 3 1 1 2 2 2 2 3 3 3\r\n", "output": "30\r\n"}, {"input": "92 5 40\r\n3 3 3 3 2 2 2 2 5 5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 3 3 5 3 3 3 4 4 4 1 1 4 4 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2 2 2 2 3 3 3 2 5 1 1 4 4 4 4 4 4 4 4 4 4 4 2 2 4 4 5 5 5 5 5 5 5 5 5 2 2 2 2 2\r\n", "output": "33\r\n"}, {"input": "99 10 10\r\n9 9 9 10 10 10 9 9 9 9 9 2 2 10 10 10 10 10 3 3 5 10 10 2 2 3 3 6 1 1 1 1 1 1 7 7 7 7 7 4 4 6 6 6 8 9 9 9 2 2 9 9 5 5 5 5 1 10 7 7 9 9 9 5 6 6 6 6 8 8 4 1 3 3 3 3 3 3 9 9 4 1 1 7 1 1 1 3 3 3 3 3 3 10 9 10 9 8 9\r\n", "output": "12\r\n"}, {"input": "95 10 30\r\n3 3 8 8 8 4 9 3 3 3 3 3 3 8 10 5 5 5 5 5 5 4 9 1 1 1 1 6 6 7 7 7 1 1 1 1 1 1 9 9 10 10 10 10 10 5 3 3 3 3 3 3 6 6 6 6 1 6 6 6 6 9 4 9 5 5 5 2 2 2 2 10 10 8 3 3 4 2 9 9 9 2 5 2 2 8 8 8 7 7 3 3 3 4 4\r\n", "output": "11\r\n"}, {"input": "100 10 15\r\n7 7 3 6 6 6 8 8 8 8 8 8 8 8 8 8 8 5 5 1 9 9 9 9 9 9 9 9 2 2 2 4 7 7 8 2 2 2 2 2 2 8 8 7 7 2 2 2 7 7 7 4 4 4 4 4 4 4 4 4 4 7 7 7 7 7 7 7 7 2 2 2 6 6 3 3 3 3 3 3 1 1 1 1 1 1 4 4 4 4 1 1 1 1 5 4 5 6 6 6\r\n", "output": "13\r\n"}] | false | stdio | null | true |
18/D | 18 | D | PyPy 3-64 | TESTS | 24 | 280 | 5,222,400 | 208301258 | from collections import defaultdict
n = int(input())
dp = defaultdict(int)
dp[0] = 0
# mx = 0
for _ in range(n):
cmd, x = input().split()
x = int(x)
if cmd == 'win':
dp[x] = dp[0]
elif x in dp:
dp[0] = max(dp[0], dp[x] + 2 ** x)
print(dp[0]) | 45 | 154 | 5,529,600 | 28627387 | R= lambda: map(int,input().split())
n,=R()
ans=0
l= [0 for _ in range(2002)]
for i in range(n):
s=input().split()
b=int(s[1])
if s[0]=='win':
l[b]=ans+2**b
else:
ans=max(ans,l[b])
print(ans) | Codeforces Beta Round 18 (Div. 2 Only) | ICPC | 2,010 | 2 | 128 | Seller Bob | Last year Bob earned by selling memory sticks. During each of n days of his work one of the two following events took place:
- A customer came to Bob and asked to sell him a 2x MB memory stick. If Bob had such a stick, he sold it and got 2x berllars.
- Bob won some programming competition and got a 2x MB memory stick as a prize. Bob could choose whether to present this memory stick to one of his friends, or keep it.
Bob never kept more than one memory stick, as he feared to mix up their capacities, and deceive a customer unintentionally. It is also known that for each memory stick capacity there was at most one customer, who wanted to buy that memory stick. Now, knowing all the customers' demands and all the prizes won at programming competitions during the last n days, Bob wants to know, how much money he could have earned, if he had acted optimally. | The first input line contains number n (1 ≤ n ≤ 5000) — amount of Bob's working days. The following n lines contain the description of the days. Line sell x stands for a day when a customer came to Bob to buy a 2x MB memory stick (0 ≤ x ≤ 2000). It's guaranteed that for each x there is not more than one line sell x. Line win x stands for a day when Bob won a 2x MB memory stick (0 ≤ x ≤ 2000). | Output the maximum possible earnings for Bob in berllars, that he would have had if he had known all the events beforehand. Don't forget, please, that Bob can't keep more than one memory stick at a time. | null | null | [{"input": "7\nwin 10\nwin 5\nwin 3\nsell 5\nsell 3\nwin 10\nsell 10", "output": "1056"}, {"input": "3\nwin 5\nsell 6\nsell 4", "output": "0"}] | 2,000 | ["brute force", "dp", "greedy"] | 45 | [{"input": "7\r\nwin 10\r\nwin 5\r\nwin 3\r\nsell 5\r\nsell 3\r\nwin 10\r\nsell 10\r\n", "output": "1056\r\n"}, {"input": "3\r\nwin 5\r\nsell 6\r\nsell 4\r\n", "output": "0\r\n"}, {"input": "10\r\nsell 179\r\nwin 1278\r\nsell 1278\r\nwin 179\r\nwin 788\r\nsell 788\r\nwin 1819\r\nwin 1278\r\nsell 1454\r\nsell 1819\r\n", "output": "37459511778596727480858760720167552241582636504705413766024169777495064333423937410125519624693990051069809575647477719465460756326341562228323606665869931977125977431028709943048934214062888966581139223580790503937962827597404798307897711090567429316074325427043388117806141094834711707585035634104732053207574452493593409130554278913951011894497392495930884827685983975668127973918422057605356890341647839399778378381152159725053311750647457999739578989105335906181048932656785993705124392163591312698147450544686667028530553878359909362604965888\n"}, {"input": "10\r\nsell 573\r\nwin 1304\r\nsell 278\r\nwin 1631\r\nsell 1225\r\nsell 1631\r\nsell 177\r\nwin 1631\r\nwin 177\r\nsell 1304\r\n", "output": "95482312335125227379668481690754940528280513838693267460502082967052005332103697568042408703168913727303170456338425853153094403747135188778307041838920404959089576368946137708987138986696495077466398994298434148881715073638178666201165545650953479735059082316661443204882826188032944866093372620219104327689636641547141835841165681118172603993695103043804276669836594061369229043451067647935298287687852302215923887110435577776767805943668204998410716005202198549540411238299513630278811648\n"}, {"input": "10\r\nwin 1257\r\nwin 1934\r\nsell 1934\r\nsell 1257\r\nwin 1934\r\nwin 1257\r\nsell 495\r\nwin 495\r\nwin 495\r\nwin 1257\r\n", "output": "1556007242642049292787218246793379348327505438878680952714050868520307364441227819009733220897932984584977593931988662671459594674963394056587723382487766303981362587048873128400436836690128983570130687310221668877557121158055843621982630476422478413285775826498536883275291967793661985813155062733063913176306327509625594121241472451054995889483447103432414676059872469910105149496451402271546454282618581884282152530090816240540173251729211604658704990425330422792556824836640431985211146197816770068601144273155826261981248156109439380758754562057681271989937590784957541131485184\n"}, {"input": "10\r\nsell 1898\r\nsell 173\r\nsell 1635\r\nsell 29\r\nsell 881\r\nsell 434\r\nsell 1236\r\nsell 14\r\nwin 29\r\nsell 1165\r\n", "output": "0\r\n"}, {"input": "1\r\nsell 2000\r\n", "output": "0\r\n"}, {"input": "1\r\nwin 2000\r\n", "output": "0\r\n"}, {"input": "2\r\nwin 2000\r\nsell 2000\r\n", "output": "114813069527425452423283320117768198402231770208869520047764273682576626139237031385665948631650626991844596463898746277344711896086305533142593135616665318539129989145312280000688779148240044871428926990063486244781615463646388363947317026040466353970904996558162398808944629605623311649536164221970332681344168908984458505602379484807914058900934776500429002716706625830522008132236281291761267883317206598995396418127021779858404042159853183251540889433902091920554957783589672039160081957216630582755380425583726015528348786419432054508915275783882625175435528800822842770817965453762184851149029376\n"}] | false | stdio | null | true |
18/D | 18 | D | Python 3 | TESTS | 24 | 248 | 409,600 | 51790433 | N = int(input())
winsell = (N)*[0]
pow = (2001)*[-1];
ans = int(0)
for i in range(0,N):
S, x = input().split()
x = int(x)
if S == "win":
winsell[i] = x
else:
pow[x] = i;
winsell[i] = -x
for i in range(2000,-1,-1):
if pow[i]!=-1:
b = bool(0)
for j in range(pow[i],-1,-1):
if winsell[j] != 2001:
if winsell[j] == i:
b = 1
ans+=2**i
for k in range(j,pow[i]+1):
winsell[k]=2001
else:
b = 1
if b == 1:
break
print(ans)
# Sun Mar 24 2019 15:34:27 GMT+0300 (MSK) | 45 | 154 | 7,270,400 | 125293934 | import sys
readline = sys.stdin.readline
N = int(readline())
DP = [0] * (N+1)
MAX = 2100
last_t = [0] * MAX
for i in range(1, N + 1):
DP[i] = DP[i-1]
a, b = readline().split()
b = int(b)
if(a == "win"):
DP[i] = DP[i-1]
last_t[b] = i
else:
if last_t[b] == 0:
continue
DP[i] = max(DP[i], DP[last_t[b]] + (1 << b))
print(DP[-1]) | Codeforces Beta Round 18 (Div. 2 Only) | ICPC | 2,010 | 2 | 128 | Seller Bob | Last year Bob earned by selling memory sticks. During each of n days of his work one of the two following events took place:
- A customer came to Bob and asked to sell him a 2x MB memory stick. If Bob had such a stick, he sold it and got 2x berllars.
- Bob won some programming competition and got a 2x MB memory stick as a prize. Bob could choose whether to present this memory stick to one of his friends, or keep it.
Bob never kept more than one memory stick, as he feared to mix up their capacities, and deceive a customer unintentionally. It is also known that for each memory stick capacity there was at most one customer, who wanted to buy that memory stick. Now, knowing all the customers' demands and all the prizes won at programming competitions during the last n days, Bob wants to know, how much money he could have earned, if he had acted optimally. | The first input line contains number n (1 ≤ n ≤ 5000) — amount of Bob's working days. The following n lines contain the description of the days. Line sell x stands for a day when a customer came to Bob to buy a 2x MB memory stick (0 ≤ x ≤ 2000). It's guaranteed that for each x there is not more than one line sell x. Line win x stands for a day when Bob won a 2x MB memory stick (0 ≤ x ≤ 2000). | Output the maximum possible earnings for Bob in berllars, that he would have had if he had known all the events beforehand. Don't forget, please, that Bob can't keep more than one memory stick at a time. | null | null | [{"input": "7\nwin 10\nwin 5\nwin 3\nsell 5\nsell 3\nwin 10\nsell 10", "output": "1056"}, {"input": "3\nwin 5\nsell 6\nsell 4", "output": "0"}] | 2,000 | ["brute force", "dp", "greedy"] | 45 | [{"input": "7\r\nwin 10\r\nwin 5\r\nwin 3\r\nsell 5\r\nsell 3\r\nwin 10\r\nsell 10\r\n", "output": "1056\r\n"}, {"input": "3\r\nwin 5\r\nsell 6\r\nsell 4\r\n", "output": "0\r\n"}, {"input": "10\r\nsell 179\r\nwin 1278\r\nsell 1278\r\nwin 179\r\nwin 788\r\nsell 788\r\nwin 1819\r\nwin 1278\r\nsell 1454\r\nsell 1819\r\n", "output": "37459511778596727480858760720167552241582636504705413766024169777495064333423937410125519624693990051069809575647477719465460756326341562228323606665869931977125977431028709943048934214062888966581139223580790503937962827597404798307897711090567429316074325427043388117806141094834711707585035634104732053207574452493593409130554278913951011894497392495930884827685983975668127973918422057605356890341647839399778378381152159725053311750647457999739578989105335906181048932656785993705124392163591312698147450544686667028530553878359909362604965888\n"}, {"input": "10\r\nsell 573\r\nwin 1304\r\nsell 278\r\nwin 1631\r\nsell 1225\r\nsell 1631\r\nsell 177\r\nwin 1631\r\nwin 177\r\nsell 1304\r\n", "output": "95482312335125227379668481690754940528280513838693267460502082967052005332103697568042408703168913727303170456338425853153094403747135188778307041838920404959089576368946137708987138986696495077466398994298434148881715073638178666201165545650953479735059082316661443204882826188032944866093372620219104327689636641547141835841165681118172603993695103043804276669836594061369229043451067647935298287687852302215923887110435577776767805943668204998410716005202198549540411238299513630278811648\n"}, {"input": "10\r\nwin 1257\r\nwin 1934\r\nsell 1934\r\nsell 1257\r\nwin 1934\r\nwin 1257\r\nsell 495\r\nwin 495\r\nwin 495\r\nwin 1257\r\n", "output": "1556007242642049292787218246793379348327505438878680952714050868520307364441227819009733220897932984584977593931988662671459594674963394056587723382487766303981362587048873128400436836690128983570130687310221668877557121158055843621982630476422478413285775826498536883275291967793661985813155062733063913176306327509625594121241472451054995889483447103432414676059872469910105149496451402271546454282618581884282152530090816240540173251729211604658704990425330422792556824836640431985211146197816770068601144273155826261981248156109439380758754562057681271989937590784957541131485184\n"}, {"input": "10\r\nsell 1898\r\nsell 173\r\nsell 1635\r\nsell 29\r\nsell 881\r\nsell 434\r\nsell 1236\r\nsell 14\r\nwin 29\r\nsell 1165\r\n", "output": "0\r\n"}, {"input": "1\r\nsell 2000\r\n", "output": "0\r\n"}, {"input": "1\r\nwin 2000\r\n", "output": "0\r\n"}, {"input": "2\r\nwin 2000\r\nsell 2000\r\n", "output": "114813069527425452423283320117768198402231770208869520047764273682576626139237031385665948631650626991844596463898746277344711896086305533142593135616665318539129989145312280000688779148240044871428926990063486244781615463646388363947317026040466353970904996558162398808944629605623311649536164221970332681344168908984458505602379484807914058900934776500429002716706625830522008132236281291761267883317206598995396418127021779858404042159853183251540889433902091920554957783589672039160081957216630582755380425583726015528348786419432054508915275783882625175435528800822842770817965453762184851149029376\n"}] | false | stdio | null | true |
18/D | 18 | D | Python 3 | TESTS | 24 | 248 | 409,600 | 73944672 | import math
n = int(input())
Size = 2001
dp = [0] * (Size)
for i in range(0, Size, 1) :
dp[i] = -1
for i in range (1 , n+1 , 1):
s = input()
In = s.split()
x = int(In[1])
if In[0] == "win" :
dp[x] = max(0, dp[0])
if In[0] == "sell" :
if dp[x] != -1 :
dp[0] = max(dp[0], dp[x] + 2**x)
ans = 0
for i in range(0, Size, 1) :
ans = max(ans, dp[i])
print(ans)
# Sun Mar 22 2020 11:44:23 GMT+0300 (MSK) | 45 | 154 | 7,270,400 | 129517975 | n = int(input())
have = [ -1 for i in range(0, 2005) ]
dp = [ 0 for i in range(0, n+5) ]
s, t = input().split()
t = int(t)
if(s == 'win'):
have[t] = 0
for i in range(1, n):
s, t = input().split()
t = int(t)
if(s == 'win'):
have[t] = i
dp[i] = dp[i-1]
elif have[t] == -1:
dp[i] = dp[i-1]
else:
dp[i] = max(dp[i-1], (1 << t) + dp[have[t]])
print(dp[n-1]) | Codeforces Beta Round 18 (Div. 2 Only) | ICPC | 2,010 | 2 | 128 | Seller Bob | Last year Bob earned by selling memory sticks. During each of n days of his work one of the two following events took place:
- A customer came to Bob and asked to sell him a 2x MB memory stick. If Bob had such a stick, he sold it and got 2x berllars.
- Bob won some programming competition and got a 2x MB memory stick as a prize. Bob could choose whether to present this memory stick to one of his friends, or keep it.
Bob never kept more than one memory stick, as he feared to mix up their capacities, and deceive a customer unintentionally. It is also known that for each memory stick capacity there was at most one customer, who wanted to buy that memory stick. Now, knowing all the customers' demands and all the prizes won at programming competitions during the last n days, Bob wants to know, how much money he could have earned, if he had acted optimally. | The first input line contains number n (1 ≤ n ≤ 5000) — amount of Bob's working days. The following n lines contain the description of the days. Line sell x stands for a day when a customer came to Bob to buy a 2x MB memory stick (0 ≤ x ≤ 2000). It's guaranteed that for each x there is not more than one line sell x. Line win x stands for a day when Bob won a 2x MB memory stick (0 ≤ x ≤ 2000). | Output the maximum possible earnings for Bob in berllars, that he would have had if he had known all the events beforehand. Don't forget, please, that Bob can't keep more than one memory stick at a time. | null | null | [{"input": "7\nwin 10\nwin 5\nwin 3\nsell 5\nsell 3\nwin 10\nsell 10", "output": "1056"}, {"input": "3\nwin 5\nsell 6\nsell 4", "output": "0"}] | 2,000 | ["brute force", "dp", "greedy"] | 45 | [{"input": "7\r\nwin 10\r\nwin 5\r\nwin 3\r\nsell 5\r\nsell 3\r\nwin 10\r\nsell 10\r\n", "output": "1056\r\n"}, {"input": "3\r\nwin 5\r\nsell 6\r\nsell 4\r\n", "output": "0\r\n"}, {"input": "10\r\nsell 179\r\nwin 1278\r\nsell 1278\r\nwin 179\r\nwin 788\r\nsell 788\r\nwin 1819\r\nwin 1278\r\nsell 1454\r\nsell 1819\r\n", "output": "37459511778596727480858760720167552241582636504705413766024169777495064333423937410125519624693990051069809575647477719465460756326341562228323606665869931977125977431028709943048934214062888966581139223580790503937962827597404798307897711090567429316074325427043388117806141094834711707585035634104732053207574452493593409130554278913951011894497392495930884827685983975668127973918422057605356890341647839399778378381152159725053311750647457999739578989105335906181048932656785993705124392163591312698147450544686667028530553878359909362604965888\n"}, {"input": "10\r\nsell 573\r\nwin 1304\r\nsell 278\r\nwin 1631\r\nsell 1225\r\nsell 1631\r\nsell 177\r\nwin 1631\r\nwin 177\r\nsell 1304\r\n", "output": "95482312335125227379668481690754940528280513838693267460502082967052005332103697568042408703168913727303170456338425853153094403747135188778307041838920404959089576368946137708987138986696495077466398994298434148881715073638178666201165545650953479735059082316661443204882826188032944866093372620219104327689636641547141835841165681118172603993695103043804276669836594061369229043451067647935298287687852302215923887110435577776767805943668204998410716005202198549540411238299513630278811648\n"}, {"input": "10\r\nwin 1257\r\nwin 1934\r\nsell 1934\r\nsell 1257\r\nwin 1934\r\nwin 1257\r\nsell 495\r\nwin 495\r\nwin 495\r\nwin 1257\r\n", "output": "1556007242642049292787218246793379348327505438878680952714050868520307364441227819009733220897932984584977593931988662671459594674963394056587723382487766303981362587048873128400436836690128983570130687310221668877557121158055843621982630476422478413285775826498536883275291967793661985813155062733063913176306327509625594121241472451054995889483447103432414676059872469910105149496451402271546454282618581884282152530090816240540173251729211604658704990425330422792556824836640431985211146197816770068601144273155826261981248156109439380758754562057681271989937590784957541131485184\n"}, {"input": "10\r\nsell 1898\r\nsell 173\r\nsell 1635\r\nsell 29\r\nsell 881\r\nsell 434\r\nsell 1236\r\nsell 14\r\nwin 29\r\nsell 1165\r\n", "output": "0\r\n"}, {"input": "1\r\nsell 2000\r\n", "output": "0\r\n"}, {"input": "1\r\nwin 2000\r\n", "output": "0\r\n"}, {"input": "2\r\nwin 2000\r\nsell 2000\r\n", "output": "114813069527425452423283320117768198402231770208869520047764273682576626139237031385665948631650626991844596463898746277344711896086305533142593135616665318539129989145312280000688779148240044871428926990063486244781615463646388363947317026040466353970904996558162398808944629605623311649536164221970332681344168908984458505602379484807914058900934776500429002716706625830522008132236281291761267883317206598995396418127021779858404042159853183251540889433902091920554957783589672039160081957216630582755380425583726015528348786419432054508915275783882625175435528800822842770817965453762184851149029376\n"}] | false | stdio | null | true |
18/D | 18 | D | PyPy 3-64 | TESTS | 22 | 280 | 5,836,800 | 205942811 | N,X=5000,2000
lows,upps,viss=[-1]*X,[-1]*X,[0]*N
n=int(input())
for i in range(n):
t,x=input().split()
x=int(x)-1
if t=="win" and upps[x]==-1:lows[x]=i
if t=="sell":upps[x]=i
ress=[]
for i in range(X)[::-1]:
l,r=lows[i],upps[i]
if l==-1 or r==-1 or l>r:continue
vis=0
for j in range(l,r+1):vis|=viss[j]
if vis:continue
for j in range(l,r+1):viss[j]=1
ress.append(i+1)
res,fac,cur=0,1,0
for x in ress[::-1]:
while cur<x:fac*=2;cur+=1
res+=fac
print(res) | 45 | 154 | 7,680,000 | 127262195 | # -*- coding: utf-8 -*-
"""
Created on Sat Aug 28 08:49:22 2021
@author: Hermes
"""
n = int(input())
m = [0]*2001
earnings = 0
for i in range(n):
e = input()
if 'win' in e:
x = int(e[4:])
if x > 0:
m[x] = earnings + (2<<x-1)
else:
m[x] = earnings + 1
else:
x = int(e[5:])
earnings = max(earnings, m[x])
print(earnings) | Codeforces Beta Round 18 (Div. 2 Only) | ICPC | 2,010 | 2 | 128 | Seller Bob | Last year Bob earned by selling memory sticks. During each of n days of his work one of the two following events took place:
- A customer came to Bob and asked to sell him a 2x MB memory stick. If Bob had such a stick, he sold it and got 2x berllars.
- Bob won some programming competition and got a 2x MB memory stick as a prize. Bob could choose whether to present this memory stick to one of his friends, or keep it.
Bob never kept more than one memory stick, as he feared to mix up their capacities, and deceive a customer unintentionally. It is also known that for each memory stick capacity there was at most one customer, who wanted to buy that memory stick. Now, knowing all the customers' demands and all the prizes won at programming competitions during the last n days, Bob wants to know, how much money he could have earned, if he had acted optimally. | The first input line contains number n (1 ≤ n ≤ 5000) — amount of Bob's working days. The following n lines contain the description of the days. Line sell x stands for a day when a customer came to Bob to buy a 2x MB memory stick (0 ≤ x ≤ 2000). It's guaranteed that for each x there is not more than one line sell x. Line win x stands for a day when Bob won a 2x MB memory stick (0 ≤ x ≤ 2000). | Output the maximum possible earnings for Bob in berllars, that he would have had if he had known all the events beforehand. Don't forget, please, that Bob can't keep more than one memory stick at a time. | null | null | [{"input": "7\nwin 10\nwin 5\nwin 3\nsell 5\nsell 3\nwin 10\nsell 10", "output": "1056"}, {"input": "3\nwin 5\nsell 6\nsell 4", "output": "0"}] | 2,000 | ["brute force", "dp", "greedy"] | 45 | [{"input": "7\r\nwin 10\r\nwin 5\r\nwin 3\r\nsell 5\r\nsell 3\r\nwin 10\r\nsell 10\r\n", "output": "1056\r\n"}, {"input": "3\r\nwin 5\r\nsell 6\r\nsell 4\r\n", "output": "0\r\n"}, {"input": "10\r\nsell 179\r\nwin 1278\r\nsell 1278\r\nwin 179\r\nwin 788\r\nsell 788\r\nwin 1819\r\nwin 1278\r\nsell 1454\r\nsell 1819\r\n", "output": "37459511778596727480858760720167552241582636504705413766024169777495064333423937410125519624693990051069809575647477719465460756326341562228323606665869931977125977431028709943048934214062888966581139223580790503937962827597404798307897711090567429316074325427043388117806141094834711707585035634104732053207574452493593409130554278913951011894497392495930884827685983975668127973918422057605356890341647839399778378381152159725053311750647457999739578989105335906181048932656785993705124392163591312698147450544686667028530553878359909362604965888\n"}, {"input": "10\r\nsell 573\r\nwin 1304\r\nsell 278\r\nwin 1631\r\nsell 1225\r\nsell 1631\r\nsell 177\r\nwin 1631\r\nwin 177\r\nsell 1304\r\n", "output": "95482312335125227379668481690754940528280513838693267460502082967052005332103697568042408703168913727303170456338425853153094403747135188778307041838920404959089576368946137708987138986696495077466398994298434148881715073638178666201165545650953479735059082316661443204882826188032944866093372620219104327689636641547141835841165681118172603993695103043804276669836594061369229043451067647935298287687852302215923887110435577776767805943668204998410716005202198549540411238299513630278811648\n"}, {"input": "10\r\nwin 1257\r\nwin 1934\r\nsell 1934\r\nsell 1257\r\nwin 1934\r\nwin 1257\r\nsell 495\r\nwin 495\r\nwin 495\r\nwin 1257\r\n", "output": "1556007242642049292787218246793379348327505438878680952714050868520307364441227819009733220897932984584977593931988662671459594674963394056587723382487766303981362587048873128400436836690128983570130687310221668877557121158055843621982630476422478413285775826498536883275291967793661985813155062733063913176306327509625594121241472451054995889483447103432414676059872469910105149496451402271546454282618581884282152530090816240540173251729211604658704990425330422792556824836640431985211146197816770068601144273155826261981248156109439380758754562057681271989937590784957541131485184\n"}, {"input": "10\r\nsell 1898\r\nsell 173\r\nsell 1635\r\nsell 29\r\nsell 881\r\nsell 434\r\nsell 1236\r\nsell 14\r\nwin 29\r\nsell 1165\r\n", "output": "0\r\n"}, {"input": "1\r\nsell 2000\r\n", "output": "0\r\n"}, {"input": "1\r\nwin 2000\r\n", "output": "0\r\n"}, {"input": "2\r\nwin 2000\r\nsell 2000\r\n", "output": "114813069527425452423283320117768198402231770208869520047764273682576626139237031385665948631650626991844596463898746277344711896086305533142593135616665318539129989145312280000688779148240044871428926990063486244781615463646388363947317026040466353970904996558162398808944629605623311649536164221970332681344168908984458505602379484807914058900934776500429002716706625830522008132236281291761267883317206598995396418127021779858404042159853183251540889433902091920554957783589672039160081957216630582755380425583726015528348786419432054508915275783882625175435528800822842770817965453762184851149029376\n"}] | false | stdio | null | true |
18/D | 18 | D | Python 3 | TESTS | 22 | 156 | 5,120,000 | 25694477 | n = int(input())
mx = 2009
d = [0 for i in range(mx)]
for i in range(1,n + 1):
s = input().split()
x = int(s[1])
s = s[0]
if s == 'win':
d[x] =d[0]+ 2**x
else:
d[0] = max(d[x], d[0])
print(d[0]) | 45 | 156 | 307,200 | 4977843 | a = [-1] * 2002
a[0] = 0
for i in range(int(input())):
s, t= input().split()
t = int(t) + 1
#t = int(input())
if(s[0] == 'w'):
a[t] = a[0]
elif a[t] >= 0:
a[0] = max(a[0], a[t] + (1<<(t-1)))
print(a[0]) | Codeforces Beta Round 18 (Div. 2 Only) | ICPC | 2,010 | 2 | 128 | Seller Bob | Last year Bob earned by selling memory sticks. During each of n days of his work one of the two following events took place:
- A customer came to Bob and asked to sell him a 2x MB memory stick. If Bob had such a stick, he sold it and got 2x berllars.
- Bob won some programming competition and got a 2x MB memory stick as a prize. Bob could choose whether to present this memory stick to one of his friends, or keep it.
Bob never kept more than one memory stick, as he feared to mix up their capacities, and deceive a customer unintentionally. It is also known that for each memory stick capacity there was at most one customer, who wanted to buy that memory stick. Now, knowing all the customers' demands and all the prizes won at programming competitions during the last n days, Bob wants to know, how much money he could have earned, if he had acted optimally. | The first input line contains number n (1 ≤ n ≤ 5000) — amount of Bob's working days. The following n lines contain the description of the days. Line sell x stands for a day when a customer came to Bob to buy a 2x MB memory stick (0 ≤ x ≤ 2000). It's guaranteed that for each x there is not more than one line sell x. Line win x stands for a day when Bob won a 2x MB memory stick (0 ≤ x ≤ 2000). | Output the maximum possible earnings for Bob in berllars, that he would have had if he had known all the events beforehand. Don't forget, please, that Bob can't keep more than one memory stick at a time. | null | null | [{"input": "7\nwin 10\nwin 5\nwin 3\nsell 5\nsell 3\nwin 10\nsell 10", "output": "1056"}, {"input": "3\nwin 5\nsell 6\nsell 4", "output": "0"}] | 2,000 | ["brute force", "dp", "greedy"] | 45 | [{"input": "7\r\nwin 10\r\nwin 5\r\nwin 3\r\nsell 5\r\nsell 3\r\nwin 10\r\nsell 10\r\n", "output": "1056\r\n"}, {"input": "3\r\nwin 5\r\nsell 6\r\nsell 4\r\n", "output": "0\r\n"}, {"input": "10\r\nsell 179\r\nwin 1278\r\nsell 1278\r\nwin 179\r\nwin 788\r\nsell 788\r\nwin 1819\r\nwin 1278\r\nsell 1454\r\nsell 1819\r\n", "output": "37459511778596727480858760720167552241582636504705413766024169777495064333423937410125519624693990051069809575647477719465460756326341562228323606665869931977125977431028709943048934214062888966581139223580790503937962827597404798307897711090567429316074325427043388117806141094834711707585035634104732053207574452493593409130554278913951011894497392495930884827685983975668127973918422057605356890341647839399778378381152159725053311750647457999739578989105335906181048932656785993705124392163591312698147450544686667028530553878359909362604965888\n"}, {"input": "10\r\nsell 573\r\nwin 1304\r\nsell 278\r\nwin 1631\r\nsell 1225\r\nsell 1631\r\nsell 177\r\nwin 1631\r\nwin 177\r\nsell 1304\r\n", "output": "95482312335125227379668481690754940528280513838693267460502082967052005332103697568042408703168913727303170456338425853153094403747135188778307041838920404959089576368946137708987138986696495077466398994298434148881715073638178666201165545650953479735059082316661443204882826188032944866093372620219104327689636641547141835841165681118172603993695103043804276669836594061369229043451067647935298287687852302215923887110435577776767805943668204998410716005202198549540411238299513630278811648\n"}, {"input": "10\r\nwin 1257\r\nwin 1934\r\nsell 1934\r\nsell 1257\r\nwin 1934\r\nwin 1257\r\nsell 495\r\nwin 495\r\nwin 495\r\nwin 1257\r\n", "output": "1556007242642049292787218246793379348327505438878680952714050868520307364441227819009733220897932984584977593931988662671459594674963394056587723382487766303981362587048873128400436836690128983570130687310221668877557121158055843621982630476422478413285775826498536883275291967793661985813155062733063913176306327509625594121241472451054995889483447103432414676059872469910105149496451402271546454282618581884282152530090816240540173251729211604658704990425330422792556824836640431985211146197816770068601144273155826261981248156109439380758754562057681271989937590784957541131485184\n"}, {"input": "10\r\nsell 1898\r\nsell 173\r\nsell 1635\r\nsell 29\r\nsell 881\r\nsell 434\r\nsell 1236\r\nsell 14\r\nwin 29\r\nsell 1165\r\n", "output": "0\r\n"}, {"input": "1\r\nsell 2000\r\n", "output": "0\r\n"}, {"input": "1\r\nwin 2000\r\n", "output": "0\r\n"}, {"input": "2\r\nwin 2000\r\nsell 2000\r\n", "output": "114813069527425452423283320117768198402231770208869520047764273682576626139237031385665948631650626991844596463898746277344711896086305533142593135616665318539129989145312280000688779148240044871428926990063486244781615463646388363947317026040466353970904996558162398808944629605623311649536164221970332681344168908984458505602379484807914058900934776500429002716706625830522008132236281291761267883317206598995396418127021779858404042159853183251540889433902091920554957783589672039160081957216630582755380425583726015528348786419432054508915275783882625175435528800822842770817965453762184851149029376\n"}] | false | stdio | null | true |
606/B | 606 | B | PyPy 3 | TESTS | 3 | 109 | 23,040,000 | 19664116 | x, y, x0, y0=list(map(int, input().split()))
s=input()
c=1
print(1, end=' ')
for i in range(len(s)-1):
if s[i] =='L':
if y0>=2: y0-=1; print(1, end=' '); c+=1
else: print(0, end=' ')
if s[i] =='R':
if y0<y: y0+=1; print(1, end=' '); c+=1
else: print(0, end=' ')
if s[i] =='U':
if x0>=2: x0-=1; print(1, end=' '); c+=1
else: print(0, end=' ')
if s[i] =='D':
if x0<x: x0+=1; print(1, end=' '); c+=1
else: print(0, end=' ')
print(x*y-c) | 68 | 187 | 5,939,200 | 14738900 | x, y, x0, y0 = map(int, input().split(' '))
g = [[0]* (y+1) for i in range(x + 1)]
s = input()
result = [0] * len(s)
count = x*y
for i in range(len(s)):
if g[x0][y0] == 0:
g[x0][y0] = 1
result[i] = 1
count -= 1
if s[i] == 'U' and x0 > 1:
x0 -=1
if s[i] == 'D' and x0 < x:
x0 += 1
if s[i] == 'L' and y0 > 1:
y0 -= 1
if s[i] == 'R' and y0 < y:
y0 += 1
print(' '.join(map(str, result)), count) | Codeforces Round 335 (Div. 2) | CF | 2,015 | 2 | 256 | Testing Robots | The Cybernetics Failures (CF) organisation made a prototype of a bomb technician robot. To find the possible problems it was decided to carry out a series of tests. At the beginning of each test the robot prototype will be placed in cell (x0, y0) of a rectangular squared field of size x × y, after that a mine will be installed into one of the squares of the field. It is supposed to conduct exactly x·y tests, each time a mine is installed into a square that has never been used before. The starting cell of the robot always remains the same.
After placing the objects on the field the robot will have to run a sequence of commands given by string s, consisting only of characters 'L', 'R', 'U', 'D'. These commands tell the robot to move one square to the left, to the right, up or down, or stay idle if moving in the given direction is impossible. As soon as the robot fulfills all the sequence of commands, it will blow up due to a bug in the code. But if at some moment of time the robot is at the same square with the mine, it will also blow up, but not due to a bug in the code.
Moving to the left decreases coordinate y, and moving to the right increases it. Similarly, moving up decreases the x coordinate, and moving down increases it.
The tests can go on for very long, so your task is to predict their results. For each k from 0 to length(s) your task is to find in how many tests the robot will run exactly k commands before it blows up. | The first line of the input contains four integers x, y, x0, y0 (1 ≤ x, y ≤ 500, 1 ≤ x0 ≤ x, 1 ≤ y0 ≤ y) — the sizes of the field and the starting coordinates of the robot. The coordinate axis X is directed downwards and axis Y is directed to the right.
The second line contains a sequence of commands s, which should be fulfilled by the robot. It has length from 1 to 100 000 characters and only consists of characters 'L', 'R', 'U', 'D'. | Print the sequence consisting of (length(s) + 1) numbers. On the k-th position, starting with zero, print the number of tests where the robot will run exactly k commands before it blows up. | null | In the first sample, if we exclude the probable impact of the mines, the robot's route will look like that: $$(2,2)\rightarrow(1,2)\rightarrow(1,3)\rightarrow(2,4)\rightarrow(3,3)$$. | [{"input": "3 4 2 2\nUURDRDRL", "output": "1 1 0 1 1 1 1 0 6"}, {"input": "2 2 2 2\nULD", "output": "1 1 1 1"}] | 1,600 | ["implementation"] | 68 | [{"input": "3 4 2 2\r\nUURDRDRL\r\n", "output": "1 1 0 1 1 1 1 0 6\r\n"}, {"input": "2 2 2 2\r\nULD\r\n", "output": "1 1 1 1\r\n"}, {"input": "1 1 1 1\r\nURDLUURRDDLLURDL\r\n", "output": "1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDD\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 245\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDDRRUR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 241\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDDRRURR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 0 241\r\n"}, {"input": "1 2 1 1\r\nR\r\n", "output": "1 1\r\n"}, {"input": "2 1 1 1\r\nD\r\n", "output": "1 1\r\n"}, {"input": "1 2 1 2\r\nLR\r\n", "output": "1 1 0\r\n"}, {"input": "2 1 2 1\r\nUD\r\n", "output": "1 1 0\r\n"}, {"input": "4 4 2 2\r\nDRUL\r\n", "output": "1 1 1 1 12\r\n"}, {"input": "4 4 3 3\r\nLUDRUL\r\n", "output": "1 1 1 0 0 1 12\r\n"}, {"input": "15 17 8 9\r\nURRDLU\r\n", "output": "1 1 1 1 1 1 249\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRRR\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 244\r\n"}, {"input": "15 17 8 9\r\nURRDLUULLDDRRRRU\r\n", "output": "1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 1 243\r\n"}] | false | stdio | null | true |
180/E | 180 | E | PyPy 3 | TESTS | 5 | 280 | 0 | 67479780 | n, m, k = map(int, input().split())
L = list(map(int, input().split()))
cnt = [0] * (m+1)
dp = [1] * n
bg = 0
for idx in range(len(L)):
cnt[L[idx]] += 1
dp[bg] = cnt[L[bg]]
if len(L[0:idx + 1]) - cnt[L[bg]] > k:
dp[bg] = cnt[L[bg]]
cnt[L[bg]] -= 1
bg += 1
print(max(dp)) | 50 | 498 | 17,305,600 | 51319385 | R = lambda: map(int, input().split())
n, m, k = R()
arr = list(R())
arrs = [list() for i in range(m)]
for i, x in enumerate(arr):
arrs[x - 1].append(i)
res = 1
for ar in arrs:
l = -1
for r in range(len(ar)):
while r - l + k < ar[r] - ar[l + 1] + 1:
l += 1
res = max(res, r - l)
print(res) | Codeforces Round 116 (Div. 2, ACM-ICPC Rules) | ICPC | 2,012 | 1 | 256 | Cubes | Let's imagine that you're playing the following simple computer game. The screen displays n lined-up cubes. Each cube is painted one of m colors. You are allowed to delete not more than k cubes (that do not necessarily go one after another). After that, the remaining cubes join together (so that the gaps are closed) and the system counts the score. The number of points you score equals to the length of the maximum sequence of cubes of the same color that follow consecutively. Write a program that determines the maximum possible number of points you can score.
Remember, you may delete no more than k any cubes. It is allowed not to delete cubes at all. | The first line contains three integers n, m and k (1 ≤ n ≤ 2·105, 1 ≤ m ≤ 105, 0 ≤ k < n). The second line contains n integers from 1 to m — the numbers of cube colors. The numbers of colors are separated by single spaces. | Print the maximum possible number of points you can score. | null | In the first sample you should delete the fifth and the sixth cubes.
In the second sample you should delete the fourth and the seventh cubes.
In the third sample you shouldn't delete any cubes. | [{"input": "10 3 2\n1 2 1 1 3 2 1 1 2 2", "output": "4"}, {"input": "10 2 2\n1 2 1 2 1 1 2 1 1 2", "output": "5"}, {"input": "3 1 2\n1 1 1", "output": "3"}] | 1,800 | ["binary search", "dp", "two pointers"] | 50 | [{"input": "10 3 2\r\n1 2 1 1 3 2 1 1 2 2\r\n", "output": "4\r\n"}, {"input": "10 2 2\r\n1 2 1 2 1 1 2 1 1 2\r\n", "output": "5\r\n"}, {"input": "3 1 2\r\n1 1 1\r\n", "output": "3\r\n"}, {"input": "10 2 2\r\n1 1 1 2 1 2 1 2 1 1\r\n", "output": "5\r\n"}, {"input": "1 1 0\r\n1\r\n", "output": "1\r\n"}, {"input": "20 3 5\r\n2 2 3 1 2 2 3 3 3 2 1 2 3 1 1 3 3 3 2 3\r\n", "output": "7\r\n"}, {"input": "20 2 5\r\n2 2 1 2 1 2 1 2 1 1 2 1 2 2 1 2 2 1 2 1\r\n", "output": "7\r\n"}, {"input": "20 6 3\r\n4 1 2 6 3 3 2 5 2 5 2 1 1 4 1 2 2 1 1 4\r\n", "output": "5\r\n"}, {"input": "30 5 8\r\n1 4 1 5 3 4 4 1 1 4 1 3 5 5 5 5 1 5 1 5 2 3 2 2 3 4 5 2 1 2\r\n", "output": "7\r\n"}, {"input": "30 5 6\r\n4 2 2 1 3 4 2 3 2 4 3 1 1 4 4 3 5 1 4 5 5 1 2 2 1 2 4 4 1 2\r\n", "output": "4\r\n"}, {"input": "100 10 15\r\n6 6 6 6 7 7 8 8 4 4 4 1 1 7 7 7 1 1 1 2 2 2 2 2 2 2 2 2 10 5 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 6 6 10 2 2 8 8 1 1 1 1 1 6 6 6 6 2 2 3 3 9 9 9 9 9 10 10 10 10 10 4 9 9 9 7 7 7 7 9 9 7 7 5 8 8 8 8 2\r\n", "output": "25\r\n"}, {"input": "99 10 17\r\n3 2 2 9 7 10 10 10 10 6 6 6 3 7 3 3 7 2 2 2 2 2 10 10 2 2 7 7 7 7 1 8 8 8 8 10 9 10 10 10 5 5 2 2 5 5 5 1 4 9 9 2 2 3 3 2 2 9 9 9 9 9 9 9 7 4 8 8 4 8 8 10 10 4 5 9 9 10 5 5 5 5 5 8 8 8 8 2 2 2 2 1 8 8 5 10 10 2 2\r\n", "output": "11\r\n"}, {"input": "94 10 20\r\n2 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 3 3 3 3 6 6 5 1 5 5 5 2 2 2 2 4 1 1 1 1 8 8 10 5 2 2 4 4 4 4 4 3 3 3 3 3 6 6 6 6 2 2 2 2 2 2 2 2 1 10 10 2 2 2 6 6 6 8 4 4 4 8 1 1 1 1 1 1 6 6 2 2 8 7 7 7 3 4\r\n", "output": "13\r\n"}, {"input": "99 3 15\r\n2 2 2 2 2 2 3 3 1 1 1 1 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 3 1 1 1 1 1 3 3 3 1 1 1 3 3 3 3 3 3 3 1 1 1 1 3 3 3 3 3 3 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 2 2 2 2 2 2\r\n", "output": "27\r\n"}, {"input": "100 5 10\r\n4 4 4 4 4 4 4 4 4 4 4 4 2 4 4 4 4 4 4 4 4 4 2 2 2 3 3 3 3 3 3 3 4 4 4 3 3 2 1 1 1 2 3 3 5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2 2 4 4 4 4 5 4 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 5 5 5 3 3 4 3 3 3\r\n", "output": "21\r\n"}, {"input": "98 4 20\r\n3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 1 1 1 1 1 1 1 1 1 1 1 1 4 4 4 4 4 4 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 2 2 2 2 2 2 2 1 1 1 3 3 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 2 2 2 2 2 2 2 3 3 1 1 2 2 2 2 3 3 3\r\n", "output": "30\r\n"}, {"input": "92 5 40\r\n3 3 3 3 2 2 2 2 5 5 5 5 5 5 5 5 5 5 5 4 4 4 4 4 4 3 3 5 3 3 3 4 4 4 1 1 4 4 4 4 4 4 4 4 4 4 4 2 2 2 2 2 2 2 2 2 3 3 3 2 5 1 1 4 4 4 4 4 4 4 4 4 4 4 2 2 4 4 5 5 5 5 5 5 5 5 5 2 2 2 2 2\r\n", "output": "33\r\n"}, {"input": "99 10 10\r\n9 9 9 10 10 10 9 9 9 9 9 2 2 10 10 10 10 10 3 3 5 10 10 2 2 3 3 6 1 1 1 1 1 1 7 7 7 7 7 4 4 6 6 6 8 9 9 9 2 2 9 9 5 5 5 5 1 10 7 7 9 9 9 5 6 6 6 6 8 8 4 1 3 3 3 3 3 3 9 9 4 1 1 7 1 1 1 3 3 3 3 3 3 10 9 10 9 8 9\r\n", "output": "12\r\n"}, {"input": "95 10 30\r\n3 3 8 8 8 4 9 3 3 3 3 3 3 8 10 5 5 5 5 5 5 4 9 1 1 1 1 6 6 7 7 7 1 1 1 1 1 1 9 9 10 10 10 10 10 5 3 3 3 3 3 3 6 6 6 6 1 6 6 6 6 9 4 9 5 5 5 2 2 2 2 10 10 8 3 3 4 2 9 9 9 2 5 2 2 8 8 8 7 7 3 3 3 4 4\r\n", "output": "11\r\n"}, {"input": "100 10 15\r\n7 7 3 6 6 6 8 8 8 8 8 8 8 8 8 8 8 5 5 1 9 9 9 9 9 9 9 9 2 2 2 4 7 7 8 2 2 2 2 2 2 8 8 7 7 2 2 2 7 7 7 4 4 4 4 4 4 4 4 4 4 7 7 7 7 7 7 7 7 2 2 2 6 6 3 3 3 3 3 3 1 1 1 1 1 1 4 4 4 4 1 1 1 1 5 4 5 6 6 6\r\n", "output": "13\r\n"}] | false | stdio | null | true |
394/C | 394 | C | Python 3 | TESTS | 6 | 62 | 0 | 7112420 | __author__ = 'epeshk'
n, m = list(map(int,input().split()))
_00 = 0
_01 = 0
_11 = 0
for i in range (n):
c=list(map(str,input().split()))
for j in c:
if j == '00': _00+=1
if j == '01' or j=='10': _01+=1
if j == '11': _11+=1
C = [['' for i in range (m)] for j in range(n)]
for i in range(n):
for j in range(m):
if _11>0:
C[i][j]='11'
_11-=1
elif i-1>=0 and _01>0:
if C[i-1][j]=='11':
C[i][j]='00'
_00-=1
if C[i-1][j]=='10':
C[i][j]='01'
_01-=1
if C[i-1][j]=='01':
C[i][j]='10'
_01-=1
if C[i-1][j]=='00':
C[i][j]='10'
_01-=1
elif _01>0:
C[i][j]='10'
_01-=1
else:
C[i][j]='00'
_00-=1
for i in range(n):
print(" ".join(C[i])) | 27 | 358 | 614,400 | 14807872 | n, m = map(int, input().split())
doubles, singles = 0, 0
for r in range(n):
for s in input().split():
if s == '11':
doubles += 1
elif s != '00':
singles += 1
lines = {
'zero': ' '.join(m * [ '00' ]),
'double': ' '.join(m * [ '11' ]),
'single_0': ' '.join(m * [ '01' ]),
'single_1': ' '.join(m * [ '10' ])
}
zeros = n * m - doubles - singles
while doubles >= m:
print(lines['double'])
doubles -= m
while singles >= 2 * m:
print(lines['single_0'])
print(lines['single_1'])
singles -= 2 * m
while zeros >= m:
print(lines['zero'])
zeros -= m
x = doubles + singles + zeros
tail = [ m * [ '00' ] for r in range(x // m) ]
height = len(tail)
r, c = 0, 0
while singles + doubles > 0:
if tail[r][c] == '00':
if doubles > 0:
tail[r][c] = '11'
doubles -= 1
else:
tail[r][c] = '01'
singles -= 1
if singles > 0 and r + 1 < height:
tail[r + 1][c] = '10'
singles -= 1
c += 1
if c == m:
c = 0
r += 1
for row in tail:
print(' '.join(row)) | Codeforces Round 231 (Div. 2) | CF | 2,014 | 2 | 256 | Dominoes | During the break, we decided to relax and play dominoes. Our box with Domino was empty, so we decided to borrow the teacher's dominoes.
The teacher responded instantly at our request. He put nm dominoes on the table as an n × 2m rectangle so that each of the n rows contained m dominoes arranged horizontally. Each half of each domino contained number (0 or 1).
We were taken aback, and the teacher smiled and said: "Consider some arrangement of dominoes in an n × 2m matrix. Let's count for each column of the matrix the sum of numbers in this column. Then among all such sums find the maximum one. Can you rearrange the dominoes in the matrix in such a way that the maximum sum will be minimum possible? Note that it is prohibited to change the orientation of the dominoes, they all need to stay horizontal, nevertheless dominoes are allowed to rotate by 180 degrees. As a reward I will give you all my dominoes".
We got even more taken aback. And while we are wondering what was going on, help us make an optimal matrix of dominoes. | The first line contains integers n, m (1 ≤ n, m ≤ 103).
In the next lines there is a description of the teachers' matrix. Each of next n lines contains m dominoes. The description of one domino is two integers (0 or 1), written without a space — the digits on the left and right half of the domino. | Print the resulting matrix of dominoes in the format: n lines, each of them contains m space-separated dominoes.
If there are multiple optimal solutions, print any of them. | null | Consider the answer for the first sample. There, the maximum sum among all columns equals 1 (the number of columns is 6, and not 3). Obviously, this maximum can't be less than 1, then such matrix is optimal.
Note that the dominoes can be rotated by 180 degrees. | [{"input": "2 3\n01 11 00\n00 01 11", "output": "11 11 10\n00 00 01"}, {"input": "4 1\n11\n10\n01\n00", "output": "11\n10\n01\n00"}] | null | ["constructive algorithms", "greedy"] | 27 | [{"input": "2 3\r\n01 11 00\r\n00 01 11\r\n", "output": "11 11 10\r\n00 00 01\r\n"}, {"input": "4 1\r\n11\r\n10\r\n01\r\n00\r\n", "output": "11\r\n10\r\n01\r\n00\r\n"}, {"input": "1 1\r\n00\r\n", "output": "00\r\n"}, {"input": "1 1\r\n01\r\n", "output": "10\r\n"}, {"input": "1 1\r\n11\r\n", "output": "11\r\n"}, {"input": "9 9\r\n01 00 00 01 00 01 11 11 11\r\n10 10 10 01 10 01 11 01 10\r\n10 00 10 00 11 01 00 10 00\r\n01 00 01 01 11 00 00 11 11\r\n11 00 10 11 01 01 11 00 01\r\n01 10 00 00 11 10 01 01 10\r\n11 10 11 00 11 11 01 10 10\r\n10 00 01 00 00 00 11 01 01\r\n00 11 01 00 10 01 10 00 01\r\n", "output": "11 11 11 11 11 11 11 11 11\r\n11 11 11 11 11 11 11 11 11\r\n10 10 10 10 10 10 10 10 10\r\n10 10 10 10 10 10 10 10 10\r\n10 10 10 01 01 01 01 01 01\r\n01 01 01 01 01 01 01 01 01\r\n01 01 01 00 00 00 00 01 01\r\n00 00 00 00 00 00 00 00 00\r\n00 00 00 00 00 00 00 00 00\r\n"}, {"input": "9 9\r\n10 10 10 01 10 11 11 01 10\r\n11 00 10 10 11 10 01 00 00\r\n10 00 11 01 00 01 01 11 10\r\n10 11 10 00 01 11 11 10 11\r\n01 11 11 01 11 00 10 00 01\r\n01 00 00 10 01 01 10 00 01\r\n11 10 11 10 01 00 00 11 00\r\n10 11 10 10 01 10 10 10 01\r\n10 10 10 10 11 11 01 00 11\r\n", "output": "11 11 11 11 11 11 11 11 11\r\n11 11 11 11 11 11 11 11 11\r\n11 11 10 10 10 10 10 10 10\r\n10 10 10 10 10 10 10 10 10\r\n10 10 10 10 10 10 10 01 01\r\n01 01 01 01 01 01 01 01 01\r\n01 01 01 01 01 01 01 01 01\r\n00 00 00 00 01 01 01 00 00\r\n00 00 00 00 00 00 00 00 00\r\n"}, {"input": "9 1\r\n01\r\n00\r\n01\r\n01\r\n00\r\n00\r\n00\r\n01\r\n11\r\n", "output": "11\r\n10\r\n10\r\n01\r\n01\r\n00\r\n00\r\n00\r\n00\r\n"}, {"input": "2 9\r\n11 10 11 10 10 11 00 10 00\r\n10 00 00 10 10 00 11 01 01\r\n", "output": "11 11 11 11 10 10 10 10 10\r\n00 00 00 00 00 01 01 01 01\r\n"}, {"input": "2 8\r\n10 01 01 11 10 10 01 10\r\n01 11 01 01 11 10 01 01\r\n", "output": "11 11 11 10 10 10 10 10\r\n10 10 01 01 01 01 01 01\r\n"}, {"input": "3 5\r\n00 10 10 11 01\r\n11 01 11 11 10\r\n10 11 00 00 00\r\n", "output": "11 11 11 11 11\r\n10 10 10 01 01\r\n00 00 01 00 00\r\n"}, {"input": "2 3\r\n00 10 01\r\n01 01 00\r\n", "output": "10 10 01\r\n00 01 00\r\n"}, {"input": "2 5\r\n01 00 01 01 00\r\n11 01 11 11 10\r\n", "output": "11 11 11 10 10\r\n10 00 00 01 01\r\n"}] | false | stdio | import sys
from collections import Counter
def read_dominoes(file, n, m):
dominoes = []
grid = []
for _ in range(n):
line = file.readline().strip()
if not line:
return [], []
row_dominoes = line.split()
grid_row = []
for d in row_dominoes:
a, b = d[0], d[1]
sorted_d = tuple(sorted((a, b)))
dominoes.append(sorted_d)
grid_row.append(a)
grid_row.append(b)
grid.append(grid_row)
return dominoes, grid
def compute_max_col_sum(grid, m):
columns = 2 * m
col_sums = [0] * columns
for row in grid:
for i in range(columns):
col_sums[i] += int(row[i])
return max(col_sums)
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path, 'r') as input_file, \
open(output_path, 'r') as output_file, \
open(submission_path, 'r') as submission_file:
# Read input
n_line = input_file.readline().strip()
if not n_line:
print(0)
return
n, m = map(int, n_line.split())
input_dominoes, _ = read_dominoes(input_file, n, m)
input_counter = Counter(input_dominoes)
# Read submission
submission_dominoes, submission_grid = read_dominoes(submission_file, n, m)
if not submission_dominoes or len(submission_grid) != n:
print(0)
return
submission_counter = Counter(submission_dominoes)
# Check dominoes
if submission_counter != input_counter:
print(0)
return
# Compute submission max column sum
submission_max = compute_max_col_sum(submission_grid, m)
# Read reference output to compute max
_, reference_grid = read_dominoes(output_file, n, m)
if not reference_grid:
print(0)
return
reference_max = compute_max_col_sum(reference_grid, m)
# Compare maxima
if submission_max == reference_max:
print(1)
else:
print(0)
if __name__ == '__main__':
main() | true |
724/D | 724 | D | PyPy 3 | TESTS | 68 | 233 | 29,286,400 | 128085648 | def process(S, m):
d = {}
for c in S:
if c not in d:
d[c] = 0
d[c]+=1
L = []
for c in d:
L.append([c, d[c], 0])
L = sorted(L)
n = len(S)
m2 = len(L)
index = 0
curr = set([])
d2 = {}
for i in range(m):
c = S[i]
if c not in d2:
d2[c] = set([])
d2[c].add(i)
c_min = min(d2)
i2 = max(d2[c_min])
curr.add(i2)
if c_min==L[index][0]:
L[index][2]+=1
if L[index][1]==L[index][2]:
index+=1
for i in range(m, n):
c = S[i]
if c not in d2:
d2[c] = set([])
d2[c].add(i)
c2 = S[i-m]
d2[c2].remove(i-m)
if len(d2[c2])==0:
d2.pop(c2)
if i-m in curr:
curr.remove(i-m)
if len(curr)==0:
c3 = min(d2)
i2 = max(d2[c3])
curr.add(i2)
for i3 in range(m2):
if L[i3][0]==c3:
L[i3][2]+=1
if i3==index and L[index][0]==L[index][2]:
index+=1
max_index = None
for i3 in range(m2):
if L[i3][2] > 0:
max_index = i3
for i3 in range(max_index):
L[i3][2] = L[i3][1]
answer = []
for c, count1, count2 in L:
for i in range(count2):
answer.append(c)
return ''.join(answer)
m = int(input())
S = input()
print(process(S, m)) | 71 | 171 | 614,400 | 104240605 | m = int(input())
s = input()
d = [0 for _ in range(26)]
for char in s:
d[ord(char) - ord('a')] += 1
for i in range(26):
char, left, right, counter = chr(ord('a') + i), -1, -1, 0
for j in range(len(s)):
if s[j] < char:
left = j
if s[j] == char:
right = j
if j - left >= m:
if j - right >= m:
counter = -1
break
counter += 1
left = right
if ~counter:
for j in range(i):
print(chr(ord('a') + j) * d[j], end='')
print(chr(ord('a') + i) * counter)
break | Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) | CF | 2,016 | 2 | 256 | Dense Subsequence | You are given a string s, consisting of lowercase English letters, and the integer m.
One should choose some symbols from the given string so that any contiguous subsegment of length m has at least one selected symbol. Note that here we choose positions of symbols, not the symbols themselves.
Then one uses the chosen symbols to form a new string. All symbols from the chosen position should be used, but we are allowed to rearrange them in any order.
Formally, we choose a subsequence of indices 1 ≤ i1 < i2 < ... < it ≤ |s|. The selected sequence must meet the following condition: for every j such that 1 ≤ j ≤ |s| - m + 1, there must be at least one selected index that belongs to the segment [j, j + m - 1], i.e. there should exist a k from 1 to t, such that j ≤ ik ≤ j + m - 1.
Then we take any permutation p of the selected indices and form a new string sip1sip2... sipt.
Find the lexicographically smallest string, that can be obtained using this procedure. | The first line of the input contains a single integer m (1 ≤ m ≤ 100 000).
The second line contains the string s consisting of lowercase English letters. It is guaranteed that this string is non-empty and its length doesn't exceed 100 000. It is also guaranteed that the number m doesn't exceed the length of the string s. | Print the single line containing the lexicographically smallest string, that can be obtained using the procedure described above. | null | In the first sample, one can choose the subsequence {3} and form a string "a".
In the second sample, one can choose the subsequence {1, 2, 4} (symbols on this positions are 'a', 'b' and 'a') and rearrange the chosen symbols to form a string "aab". | [{"input": "3\ncbabc", "output": "a"}, {"input": "2\nabcab", "output": "aab"}, {"input": "3\nbcabcbaccba", "output": "aaabb"}] | 1,900 | ["data structures", "greedy", "strings"] | 71 | [{"input": "3\r\ncbabc\r\n", "output": "a\r\n"}, {"input": "2\r\nabcab\r\n", "output": "aab\r\n"}, {"input": "3\r\nbcabcbaccba\r\n", "output": "aaabb\r\n"}, {"input": "5\r\nimmaydobun\r\n", "output": "ab\r\n"}, {"input": "5\r\nwjjdqawypvtgrncmqvcsergermprauyevcegjtcrrblkwiugrcjfpjyxngyryxntauxlouvwgjzpsuxyxvhavgezwtuzknetdibv\r\n", "output": "aaaabbcccccddeeeeeefggggggghiijjjjjjkkllmmnnnnoppppqqrrrrrrrrsstttttu\r\n"}, {"input": "10\r\nefispvmzuutsrpxzfrykhabznxiyquwvhwhrksrgzodtuepfvamilfdynapzhzyhncorhzuewrrkcduvuhwsrprjrmgctnvrdtpj\r\n", "output": "aaabcccddddeeeffffgghhhhhhhiiijjkkklm\r\n"}, {"input": "20\r\nhlicqhxayiodyephxlfoetfketnaabpfegqcrjzlshkxfzjssvpvzhzylgowwovgxznzowvpklbwbzhwtkkaomjkenhpedmbmjic\r\n", "output": "aaaabbbbcccddeeeeeeffffg\r\n"}, {"input": "50\r\ntyhjolxuexoffdkdwimsjujorgeksyiyvvqecvhpjsuayqnibijtipuqhkulxpysotlmtrsgygpkdhkrtntwqzrpfckiscaphyhv\r\n", "output": "aab\r\n"}, {"input": "1\r\nbaaa\r\n", "output": "aaab\r\n"}, {"input": "5\r\nbbbbba\r\n", "output": "ab\r\n"}, {"input": "10\r\nbbabcbbaabcbcbcbaabbccaacccbbbcaaacabbbbaaaccbcccacbbccaccbbaacaccbabcaaaacaccacbaaccaaccbaacabbbaac\r\n", "output": "aaaaaaaaaaa\r\n"}] | false | stdio | null | true |
400/B | 400 | B | Python 3 | TESTS | 5 | 93 | 102,400 | 220150190 | n,m=map(int,input().split())
po=0
flag=False
c=0
ok=0
no=0
v=[]
for j in range(n):
b=input()
for p in range(m):
if b[p]=="G":
ok=p
elif b[p]=="S":
no=p
if no-ok <= 0:
break
else:
v+=[no-ok]
if len(set(v))==0:
print(-1)
else:
print(len(set(v))) | 34 | 31 | 204,800 | 210339817 | n,m=map(int,input().split())
st=set()
for i in range(n):
s1=input()
g=s1.index("G")
s=s1.index("S")
if s<g:
print(-1)
exit(0)
st.add(abs(s-g))
print(len(st)) | Codeforces Round 234 (Div. 2) | CF | 2,014 | 1 | 256 | Inna and New Matrix of Candies | Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Candy Matrix 2: Reload".
The field for the new game is a rectangle table of size n × m. Each line of the table contains one cell with a dwarf figurine, one cell with a candy, the other cells of the line are empty. The game lasts for several moves. During each move the player should choose all lines of the matrix where dwarf is not on the cell with candy and shout "Let's go!". After that, all the dwarves from the chosen lines start to simultaneously move to the right. During each second, each dwarf goes to the adjacent cell that is located to the right of its current cell. The movement continues until one of the following events occurs:
- some dwarf in one of the chosen lines is located in the rightmost cell of his row;
- some dwarf in the chosen lines is located in the cell with the candy.
The point of the game is to transport all the dwarves to the candy cells.
Inna is fabulous, as she came up with such an interesting game. But what about you? Your task is to play this game optimally well. Specifically, you should say by the given game field what minimum number of moves the player needs to reach the goal of the game. | The first line of the input contains two integers n and m (1 ≤ n ≤ 1000; 2 ≤ m ≤ 1000).
Next n lines each contain m characters — the game field for the "Candy Martix 2: Reload". Character "*" represents an empty cell of the field, character "G" represents a dwarf and character "S" represents a candy. The matrix doesn't contain other characters. It is guaranteed that each line contains exactly one character "G" and one character "S". | In a single line print a single integer — either the minimum number of moves needed to achieve the aim of the game, or -1, if the aim cannot be achieved on the given game field. | null | null | [{"input": "3 4\n*G*S\nG**S\n*G*S", "output": "2"}, {"input": "1 3\nS*G", "output": "-1"}] | 1,200 | ["brute force", "implementation", "schedules"] | 34 | [{"input": "3 4\r\n*G*S\r\nG**S\r\n*G*S\r\n", "output": "2\r\n"}, {"input": "1 3\r\nS*G\r\n", "output": "-1\r\n"}, {"input": "10 10\r\nG********S\r\n*G*******S\r\n**G******S\r\n***G*****S\r\n****G****S\r\n*****G***S\r\n******G**S\r\n*******G*S\r\n********GS\r\nG********S\r\n", "output": "9\r\n"}, {"input": "5 10\r\nG***S*****\r\nG****S****\r\n***GS*****\r\nG*S*******\r\nG***S*****\r\n", "output": "4\r\n"}, {"input": "4 8\r\nG*S*****\r\n****G*S*\r\nG*****S*\r\n**G***S*\r\n", "output": "3\r\n"}, {"input": "4 10\r\n***G****S*\r\n*****GS***\r\nG****S****\r\nG*******S*\r\n", "output": "3\r\n"}, {"input": "1 2\r\nSG\r\n", "output": "-1\r\n"}, {"input": "1 2\r\nGS\r\n", "output": "1\r\n"}, {"input": "1 4\r\nSG**\r\n", "output": "-1\r\n"}] | false | stdio | null | true |
177/C1 | 177 | C1 | PyPy 3-64 | TESTS1 | 3 | 122 | 0 | 175831673 | import sys
input = sys.stdin.readline
from collections import Counter
n = int(input())
k = int(input())
d = [[] for i in range(n)]
for i in range(k):
a, b = map(lambda x:int(x)-1, input().split())
d[a].append(b)
d[b].append(a)
x = [-1]*n
c = 0
for i in range(n):
if x[i] == -1:
q = [i]
while q:
a = q.pop()
x[a] = c
for j in d[a]:
if x[j] == -1:
q.append(j)
c += 1
m = int(input())
e = [[] for i in range(c)]
p = Counter(x)
d = [set() for i in range(n)]
for i in range(m):
a, b = map(lambda x:int(x)-1, input().split())
if x[a] == x[b]:
e[x[a]].extend([a, b])
d[a].add(b)
d[b].add(a)
x = [0]*n
c = 0
for ii, i in enumerate(e):
q = Counter(i)
q = [i for j, i in sorted([(q[i], i) for i in q], reverse=True)]
cc = 0
for j in q:
if len(d[j]) > 0:
for l in d[j]:
d[l].remove(j)
cc += 1
c = max(c, p[ii]-cc)
print(c) | 17 | 218 | 307,200 | 65333942 | def find(a):
if parent[a]!=a:
parent[a]=find(parent[a])
return parent[a]
def union(a,b):
u,v=find(a),find(b)
if u==v:
return
if rank[u]>rank[v]:
parent[v]=u
else:
parent[u]=v
if rank[u]==rank[v]:
rank[v]+=1
n=int(input())
k=int(input())
parent=list(map(int,range(n+1)))
rank=[0]*(n+1)
ans=[0]*(n+1)
count=[0]*(n+1)
for i in range(k):
u,v=map(int,input().split())
union(u,v)
for i in range(len(ans)):
ans[find(i)]+=1
for i in range(len(parent)):
count[parent[i]]+=1
d={}
m=int(input())
for i in range(m):
u,v=map(int,input().split())
if parent[u]==parent[v]:
d[parent[u]]=False
sak=0
for i in range(len(count)):
if count[i]!=0 and i not in d and i!=0:
sak=max(sak,count[i])
print(sak) | ABBYY Cup 2.0 - Easy | ICPC | 2,012 | 2 | 256 | Party | To celebrate the second ABBYY Cup tournament, the Smart Beaver decided to throw a party. The Beaver has a lot of acquaintances, some of them are friends with each other, and some of them dislike each other. To make party successful, the Smart Beaver wants to invite only those of his friends who are connected by friendship relations, and not to invite those who dislike each other. Both friendship and dislike are mutual feelings.
More formally, for each invited person the following conditions should be fulfilled:
- all his friends should also be invited to the party;
- the party shouldn't have any people he dislikes;
- all people who are invited to the party should be connected with him by friendship either directly or through a chain of common friends of arbitrary length. We'll say that people a1 and ap are connected through a chain of common friends if there exists a sequence of people a2, a3, ..., ap - 1 such that all pairs of people ai and ai + 1 (1 ≤ i < p) are friends.
Help the Beaver find the maximum number of acquaintances he can invite. | The first line of input contains an integer n — the number of the Beaver's acquaintances.
The second line contains an integer k $$( 0 \leq k \leq \min ( 1 0 ^ { 5 }, \frac { n \cdot ( n - 1 ) } { 2 } ) )$$ — the number of pairs of friends. Next k lines contain space-separated pairs of integers ui, vi $$( 1 \leq u _ { i }, v _ { i } \leq n, u _ { i } \neq v _ { i } )$$ — indices of people who form the i-th pair of friends.
The next line contains an integer m $$( 0 \leq m \leq min(10^{5},\frac{n(n-1)}{2}))$$ — the number of pairs of people who dislike each other. Next m lines describe pairs of people who dislike each other in the same format as the pairs of friends were described.
Each pair of people is mentioned in the input at most once $$( 0 \leq k + m \leq \frac { n \cdot ( n - 1 ) } { 2 } )$$. In particular, two persons cannot be friends and dislike each other at the same time.
The input limitations for getting 30 points are:
- 2 ≤ n ≤ 14
The input limitations for getting 100 points are:
- 2 ≤ n ≤ 2000 | Output a single number — the maximum number of people that can be invited to the party. If a group of people that meets all the requirements is impossible to select, output 0. | null | Let's have a look at the example.
Two groups of people can be invited: {1, 2, 3} and {4, 5}, thus the answer will be the size of the largest of these groups. Group {6, 7, 8, 9} doesn't fit, since it includes people 7 and 9 who dislike each other. Group {1, 2, 3, 4, 5} also doesn't fit, because not all of its members are connected by a chain of common friends (for example, people 2 and 5 aren't connected). | [{"input": "9\n8\n1 2\n1 3\n2 3\n4 5\n6 7\n7 8\n8 9\n9 6\n2\n1 6\n7 9", "output": "3"}] | 1,500 | ["dfs and similar", "dsu", "graphs"] | 17 | [{"input": "9\r\n8\r\n1 2\r\n1 3\r\n2 3\r\n4 5\r\n6 7\r\n7 8\r\n8 9\r\n9 6\r\n2\r\n1 6\r\n7 9\r\n", "output": "3"}, {"input": "2\r\n1\r\n1 2\r\n0\r\n", "output": "2"}, {"input": "2\r\n0\r\n1\r\n1 2\r\n", "output": "1"}, {"input": "3\r\n2\r\n1 2\r\n1 3\r\n1\r\n2 3\r\n", "output": "0"}, {"input": "3\r\n3\r\n1 3\r\n2 1\r\n2 3\r\n0\r\n", "output": "3"}, {"input": "4\r\n3\r\n1 2\r\n2 3\r\n3 1\r\n3\r\n1 4\r\n4 2\r\n3 4\r\n", "output": "3"}, {"input": "7\r\n8\r\n1 2\r\n1 3\r\n1 4\r\n1 5\r\n2 4\r\n2 5\r\n3 4\r\n5 6\r\n3\r\n2 6\r\n5 7\r\n6 7\r\n", "output": "1"}, {"input": "14\r\n20\r\n1 2\r\n4 5\r\n4 6\r\n4 11\r\n5 7\r\n5 8\r\n5 13\r\n5 14\r\n7 8\r\n7 14\r\n8 9\r\n8 11\r\n8 12\r\n8 14\r\n10 11\r\n10 12\r\n10 14\r\n11 13\r\n11 14\r\n12 14\r\n5\r\n1 8\r\n1 13\r\n2 10\r\n7 12\r\n8 10\r\n", "output": "2"}, {"input": "2\r\n0\r\n0\r\n", "output": "1"}, {"input": "14\r\n0\r\n0\r\n", "output": "1"}, {"input": "14\r\n6\r\n1 2\r\n2 3\r\n3 4\r\n4 5\r\n8 9\r\n9 10\r\n3\r\n5 6\r\n6 7\r\n7 8\r\n", "output": "5"}, {"input": "14\r\n10\r\n1 2\r\n1 3\r\n1 4\r\n1 5\r\n1 6\r\n1 7\r\n2 3\r\n2 4\r\n2 5\r\n2 6\r\n1\r\n2 7\r\n", "output": "1"}, {"input": "2\r\n0\r\n1\r\n1 2\r\n", "output": "1"}] | false | stdio | null | true |
400/B | 400 | B | Python 3 | TESTS | 5 | 77 | 0 | 104638980 | n,m = map(int, input().split(" "))
a = set()
tag = True
for i in range(n):
x = input()
indexG = x.index("G")
indexS = x.index("S")
if indexS< indexG:
print(-1)
tag = False
else:
a.add(indexS - indexG)
if tag:
print(len(a))
else:
pass | 34 | 31 | 204,800 | 219440089 | n,m=map(int,input().split())
a=set()
ans=1
for _ in range(n):
x=input()
if x.find('S')-x.find('G')<0:
ans=0
break
else:
temp=x.find('S')-x.find('G')
a.add(temp)
if ans:
print(len(a))
else:
print(-1) | Codeforces Round 234 (Div. 2) | CF | 2,014 | 1 | 256 | Inna and New Matrix of Candies | Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Candy Matrix 2: Reload".
The field for the new game is a rectangle table of size n × m. Each line of the table contains one cell with a dwarf figurine, one cell with a candy, the other cells of the line are empty. The game lasts for several moves. During each move the player should choose all lines of the matrix where dwarf is not on the cell with candy and shout "Let's go!". After that, all the dwarves from the chosen lines start to simultaneously move to the right. During each second, each dwarf goes to the adjacent cell that is located to the right of its current cell. The movement continues until one of the following events occurs:
- some dwarf in one of the chosen lines is located in the rightmost cell of his row;
- some dwarf in the chosen lines is located in the cell with the candy.
The point of the game is to transport all the dwarves to the candy cells.
Inna is fabulous, as she came up with such an interesting game. But what about you? Your task is to play this game optimally well. Specifically, you should say by the given game field what minimum number of moves the player needs to reach the goal of the game. | The first line of the input contains two integers n and m (1 ≤ n ≤ 1000; 2 ≤ m ≤ 1000).
Next n lines each contain m characters — the game field for the "Candy Martix 2: Reload". Character "*" represents an empty cell of the field, character "G" represents a dwarf and character "S" represents a candy. The matrix doesn't contain other characters. It is guaranteed that each line contains exactly one character "G" and one character "S". | In a single line print a single integer — either the minimum number of moves needed to achieve the aim of the game, or -1, if the aim cannot be achieved on the given game field. | null | null | [{"input": "3 4\n*G*S\nG**S\n*G*S", "output": "2"}, {"input": "1 3\nS*G", "output": "-1"}] | 1,200 | ["brute force", "implementation", "schedules"] | 34 | [{"input": "3 4\r\n*G*S\r\nG**S\r\n*G*S\r\n", "output": "2\r\n"}, {"input": "1 3\r\nS*G\r\n", "output": "-1\r\n"}, {"input": "10 10\r\nG********S\r\n*G*******S\r\n**G******S\r\n***G*****S\r\n****G****S\r\n*****G***S\r\n******G**S\r\n*******G*S\r\n********GS\r\nG********S\r\n", "output": "9\r\n"}, {"input": "5 10\r\nG***S*****\r\nG****S****\r\n***GS*****\r\nG*S*******\r\nG***S*****\r\n", "output": "4\r\n"}, {"input": "4 8\r\nG*S*****\r\n****G*S*\r\nG*****S*\r\n**G***S*\r\n", "output": "3\r\n"}, {"input": "4 10\r\n***G****S*\r\n*****GS***\r\nG****S****\r\nG*******S*\r\n", "output": "3\r\n"}, {"input": "1 2\r\nSG\r\n", "output": "-1\r\n"}, {"input": "1 2\r\nGS\r\n", "output": "1\r\n"}, {"input": "1 4\r\nSG**\r\n", "output": "-1\r\n"}] | false | stdio | null | true |
400/B | 400 | B | Python 3 | TESTS | 5 | 62 | 6,758,400 | 125743989 | n,m=map(int,input().split())
field=[]
for i in range(0,n):
row=input()
space=row.index('S')-row.index('G')
#print(space)
if space>0:
field.append(space)
if(len(field)>0):
print(len(set(field)))
else:
print(-1) | 34 | 46 | 0 | 146095711 | def dist(s):
return s.find("S") - s.find("G") - 1
a, b = map(int, input().split())
dists = [dist(input()) for i in range(a)]
if min(dists) < 0:
print(-1)
else:
print(len(set(dists))) | Codeforces Round 234 (Div. 2) | CF | 2,014 | 1 | 256 | Inna and New Matrix of Candies | Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Candy Matrix 2: Reload".
The field for the new game is a rectangle table of size n × m. Each line of the table contains one cell with a dwarf figurine, one cell with a candy, the other cells of the line are empty. The game lasts for several moves. During each move the player should choose all lines of the matrix where dwarf is not on the cell with candy and shout "Let's go!". After that, all the dwarves from the chosen lines start to simultaneously move to the right. During each second, each dwarf goes to the adjacent cell that is located to the right of its current cell. The movement continues until one of the following events occurs:
- some dwarf in one of the chosen lines is located in the rightmost cell of his row;
- some dwarf in the chosen lines is located in the cell with the candy.
The point of the game is to transport all the dwarves to the candy cells.
Inna is fabulous, as she came up with such an interesting game. But what about you? Your task is to play this game optimally well. Specifically, you should say by the given game field what minimum number of moves the player needs to reach the goal of the game. | The first line of the input contains two integers n and m (1 ≤ n ≤ 1000; 2 ≤ m ≤ 1000).
Next n lines each contain m characters — the game field for the "Candy Martix 2: Reload". Character "*" represents an empty cell of the field, character "G" represents a dwarf and character "S" represents a candy. The matrix doesn't contain other characters. It is guaranteed that each line contains exactly one character "G" and one character "S". | In a single line print a single integer — either the minimum number of moves needed to achieve the aim of the game, or -1, if the aim cannot be achieved on the given game field. | null | null | [{"input": "3 4\n*G*S\nG**S\n*G*S", "output": "2"}, {"input": "1 3\nS*G", "output": "-1"}] | 1,200 | ["brute force", "implementation", "schedules"] | 34 | [{"input": "3 4\r\n*G*S\r\nG**S\r\n*G*S\r\n", "output": "2\r\n"}, {"input": "1 3\r\nS*G\r\n", "output": "-1\r\n"}, {"input": "10 10\r\nG********S\r\n*G*******S\r\n**G******S\r\n***G*****S\r\n****G****S\r\n*****G***S\r\n******G**S\r\n*******G*S\r\n********GS\r\nG********S\r\n", "output": "9\r\n"}, {"input": "5 10\r\nG***S*****\r\nG****S****\r\n***GS*****\r\nG*S*******\r\nG***S*****\r\n", "output": "4\r\n"}, {"input": "4 8\r\nG*S*****\r\n****G*S*\r\nG*****S*\r\n**G***S*\r\n", "output": "3\r\n"}, {"input": "4 10\r\n***G****S*\r\n*****GS***\r\nG****S****\r\nG*******S*\r\n", "output": "3\r\n"}, {"input": "1 2\r\nSG\r\n", "output": "-1\r\n"}, {"input": "1 2\r\nGS\r\n", "output": "1\r\n"}, {"input": "1 4\r\nSG**\r\n", "output": "-1\r\n"}] | false | stdio | null | true |
400/B | 400 | B | PyPy 3-64 | TESTS | 5 | 46 | 0 | 189680511 | n,m=map(int,input("").split(" "))
ll=[]
for i in range(n):
l=list(input(""))
out=l.index('S')-l.index('G')
if out<0:
print(-1)
break;
ll.append(out)
lout=list(dict.fromkeys(ll))
if len(lout)>0:
print(len(lout)) | 34 | 46 | 0 | 185635853 | n, m = map(int, input().split())
st = set()
for _ in range(n):
line = input()
g = line.index('G')
s = line.index('S')
if s - g < 0:
print(-1)
break
st.add(s-g)
else:
print(len(st)) | Codeforces Round 234 (Div. 2) | CF | 2,014 | 1 | 256 | Inna and New Matrix of Candies | Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Candy Matrix 2: Reload".
The field for the new game is a rectangle table of size n × m. Each line of the table contains one cell with a dwarf figurine, one cell with a candy, the other cells of the line are empty. The game lasts for several moves. During each move the player should choose all lines of the matrix where dwarf is not on the cell with candy and shout "Let's go!". After that, all the dwarves from the chosen lines start to simultaneously move to the right. During each second, each dwarf goes to the adjacent cell that is located to the right of its current cell. The movement continues until one of the following events occurs:
- some dwarf in one of the chosen lines is located in the rightmost cell of his row;
- some dwarf in the chosen lines is located in the cell with the candy.
The point of the game is to transport all the dwarves to the candy cells.
Inna is fabulous, as she came up with such an interesting game. But what about you? Your task is to play this game optimally well. Specifically, you should say by the given game field what minimum number of moves the player needs to reach the goal of the game. | The first line of the input contains two integers n and m (1 ≤ n ≤ 1000; 2 ≤ m ≤ 1000).
Next n lines each contain m characters — the game field for the "Candy Martix 2: Reload". Character "*" represents an empty cell of the field, character "G" represents a dwarf and character "S" represents a candy. The matrix doesn't contain other characters. It is guaranteed that each line contains exactly one character "G" and one character "S". | In a single line print a single integer — either the minimum number of moves needed to achieve the aim of the game, or -1, if the aim cannot be achieved on the given game field. | null | null | [{"input": "3 4\n*G*S\nG**S\n*G*S", "output": "2"}, {"input": "1 3\nS*G", "output": "-1"}] | 1,200 | ["brute force", "implementation", "schedules"] | 34 | [{"input": "3 4\r\n*G*S\r\nG**S\r\n*G*S\r\n", "output": "2\r\n"}, {"input": "1 3\r\nS*G\r\n", "output": "-1\r\n"}, {"input": "10 10\r\nG********S\r\n*G*******S\r\n**G******S\r\n***G*****S\r\n****G****S\r\n*****G***S\r\n******G**S\r\n*******G*S\r\n********GS\r\nG********S\r\n", "output": "9\r\n"}, {"input": "5 10\r\nG***S*****\r\nG****S****\r\n***GS*****\r\nG*S*******\r\nG***S*****\r\n", "output": "4\r\n"}, {"input": "4 8\r\nG*S*****\r\n****G*S*\r\nG*****S*\r\n**G***S*\r\n", "output": "3\r\n"}, {"input": "4 10\r\n***G****S*\r\n*****GS***\r\nG****S****\r\nG*******S*\r\n", "output": "3\r\n"}, {"input": "1 2\r\nSG\r\n", "output": "-1\r\n"}, {"input": "1 2\r\nGS\r\n", "output": "1\r\n"}, {"input": "1 4\r\nSG**\r\n", "output": "-1\r\n"}] | false | stdio | null | true |
408/B | 408 | B | Python 3 | TESTS | 4 | 31 | 0 | 208354177 | a = input()
b =input()
ref = set([i for i in a ])
count =0
for i in ref :
count += min(a.count(i) , b.count(i))
if count ==0 :
print(-1)
else:
print(count) | 21 | 46 | 0 | 119587484 | from sys import *
input = lambda:stdin.readline()
int_arr = lambda : list(map(int,stdin.readline().strip().split()))
str_arr = lambda :list(map(str,stdin.readline().split()))
get_str = lambda : map(str,stdin.readline().strip().split())
get_int = lambda: map(int,stdin.readline().strip().split())
get_float = lambda : map(float,stdin.readline().strip().split())
mod = 1000000007
setrecursionlimit(1000)
a = str(input())
b = str(input())
tot = 0
flag = 1
for i in set(b):
if i in a:
if i != '\n':
if b.count(i) > a.count(i):
tot += a.count(i)
else:
tot += b.count(i)
else:
print(-1)
flag = 0
break
if flag:
print(tot) | Codeforces Round 239 (Div. 2) | CF | 2,014 | 1 | 256 | Garland | Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought n colored sheets of paper, the area of each sheet is 1 square meter.
The garland must consist of exactly m pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.
Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of m pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get. | The first line contains a non-empty sequence of n (1 ≤ n ≤ 1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.
The second line contains a non-empty sequence of m (1 ≤ m ≤ 1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make. | Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer. | null | In the first test sample Vasya can make an garland of area 6: he can use both sheets of color b, three (but not four) sheets of color a and cut a single sheet of color c in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.
In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color z. | [{"input": "aaabbac\naabbccac", "output": "6"}, {"input": "a\nz", "output": "-1"}] | 1,200 | ["implementation"] | 21 | [{"input": "aaabbac\r\naabbccac\r\n", "output": "6\r\n"}, {"input": "a\r\nz\r\n", "output": "-1"}, {"input": "r\r\nr\r\n", "output": "1\r\n"}, {"input": "stnsdn\r\nndnndsn\r\n", "output": "4\r\n"}, {"input": "yqfqfp\r\ntttwtqq\r\n", "output": "-1"}, {"input": "zzbbrrtrtzr\r\ntbbtrrrzr\r\n", "output": "9\r\n"}, {"input": "ivvfisvsvii\r\npaihjinno\r\n", "output": "-1"}, {"input": "zbvwnlgkshqerxptyod\r\nz\r\n", "output": "1\r\n"}, {"input": "xlktwjymocqrahnbesf\r\nfoo\r\n", "output": "2\r\n"}, {"input": "bbzmzqazmbambnmzaabznmbabzqnaabmabmnnabbmnzaanzzezebzabqaabzqaemeqqammmbazmmz\r\naznnbbmeebmanbeemzmemqbaeebnqenqzzbanebmnzqqebqmmnmqqzmmeqqqaaezemmazqqmqaqnnqqzbzeeazammmenbbamzbmnaenemenaaaebnmanebqmqnznqbenmqqnnnaeaebqmamennmqqeaaqqbammnzqmnmqnqbbezmemznqmanzmmqzzzzembqnzqbanamezqaqbazenenqqznqaebzaeezbqqbmeeaqnmmbnqbbnmaqqemaeaezaabmbnbzzaae\r\n", "output": "77\r\n"}, {"input": "lccfdfnfflncddlksndcwnfcllnnaswcdnldafcalssfcdnkkaklwnnacsncfwanwnwfadawcsdcfwflnnlfsfclkfknlaldna\r\nuaaldlllhedgugugueahddhedbuddaedhaaeugdubaealbgbagedldbl\r\n", "output": "-1"}, {"input": "hvewdvwdwudrwarwmddwnazmwauzatrmwptwwevavpmwrtruwnpwantewrnwmepdwvtmnveanunrvrtwpvhhnuhnmpptdttzmmndtvudmzhhannmmnahdpzamuvhzaavnhtnumwrwvttdetvuewaaennddwuvzvaptdzrzhtetwwzmzedrwuvrwznprhdvnavrruvvhzuwpdtmpwmzrwvermrhdamv\r\nuvzhwtpuputnahwwarduzddhpnwwvettprwavdmnztdnrddmarmvuevtdezndnezvarhmppwwnmvnrtddzhhnzrwuhvpwmezuurundarwdazwptrpeurrnwautddnhdmhtwhwvvtavdzezumetzezpazndhuentmrhamutrtttpevtuutemdnvwnwnmnvmznatneweuaahdavmaddhnrdenwwztrwh\r\n", "output": "199\r\n"}, {"input": "aaccddff\r\nabcdf\r\n", "output": "-1"}] | false | stdio | null | true |
408/B | 408 | B | Python 3 | TESTS | 7 | 62 | 5,529,600 | 26480968 | n = input()
m = input()
resources = len(n)
area = 0
for i in range(len(m)):
n = n.replace(m[i], "", 1)
if len(n) == resources or resources - len(n) < len(set(n)):
print("-1")
else:
print(resources - len(n)) | 21 | 46 | 0 | 142608376 | s = input()
t = input()
c = 0
s_c = [0 for i in range(26)]
t_c = [0 for i in range(26)]
for i in range(len(s)):
s_c[ord(s[i])-ord('a')] += 1
for i in range(len(t)):
t_c[ord(t[i])-ord('a')] += 1
for i in range(26):
if(s_c[i] == 0 and t_c[i] != 0):
c = 0
break
else:
c += min(s_c[i],t_c[i])
if(c==0):
print(-1)
else:
print(c) | Codeforces Round 239 (Div. 2) | CF | 2,014 | 1 | 256 | Garland | Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought n colored sheets of paper, the area of each sheet is 1 square meter.
The garland must consist of exactly m pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.
Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of m pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get. | The first line contains a non-empty sequence of n (1 ≤ n ≤ 1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.
The second line contains a non-empty sequence of m (1 ≤ m ≤ 1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make. | Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer. | null | In the first test sample Vasya can make an garland of area 6: he can use both sheets of color b, three (but not four) sheets of color a and cut a single sheet of color c in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.
In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color z. | [{"input": "aaabbac\naabbccac", "output": "6"}, {"input": "a\nz", "output": "-1"}] | 1,200 | ["implementation"] | 21 | [{"input": "aaabbac\r\naabbccac\r\n", "output": "6\r\n"}, {"input": "a\r\nz\r\n", "output": "-1"}, {"input": "r\r\nr\r\n", "output": "1\r\n"}, {"input": "stnsdn\r\nndnndsn\r\n", "output": "4\r\n"}, {"input": "yqfqfp\r\ntttwtqq\r\n", "output": "-1"}, {"input": "zzbbrrtrtzr\r\ntbbtrrrzr\r\n", "output": "9\r\n"}, {"input": "ivvfisvsvii\r\npaihjinno\r\n", "output": "-1"}, {"input": "zbvwnlgkshqerxptyod\r\nz\r\n", "output": "1\r\n"}, {"input": "xlktwjymocqrahnbesf\r\nfoo\r\n", "output": "2\r\n"}, {"input": "bbzmzqazmbambnmzaabznmbabzqnaabmabmnnabbmnzaanzzezebzabqaabzqaemeqqammmbazmmz\r\naznnbbmeebmanbeemzmemqbaeebnqenqzzbanebmnzqqebqmmnmqqzmmeqqqaaezemmazqqmqaqnnqqzbzeeazammmenbbamzbmnaenemenaaaebnmanebqmqnznqbenmqqnnnaeaebqmamennmqqeaaqqbammnzqmnmqnqbbezmemznqmanzmmqzzzzembqnzqbanamezqaqbazenenqqznqaebzaeezbqqbmeeaqnmmbnqbbnmaqqemaeaezaabmbnbzzaae\r\n", "output": "77\r\n"}, {"input": "lccfdfnfflncddlksndcwnfcllnnaswcdnldafcalssfcdnkkaklwnnacsncfwanwnwfadawcsdcfwflnnlfsfclkfknlaldna\r\nuaaldlllhedgugugueahddhedbuddaedhaaeugdubaealbgbagedldbl\r\n", "output": "-1"}, {"input": "hvewdvwdwudrwarwmddwnazmwauzatrmwptwwevavpmwrtruwnpwantewrnwmepdwvtmnveanunrvrtwpvhhnuhnmpptdttzmmndtvudmzhhannmmnahdpzamuvhzaavnhtnumwrwvttdetvuewaaennddwuvzvaptdzrzhtetwwzmzedrwuvrwznprhdvnavrruvvhzuwpdtmpwmzrwvermrhdamv\r\nuvzhwtpuputnahwwarduzddhpnwwvettprwavdmnztdnrddmarmvuevtdezndnezvarhmppwwnmvnrtddzhhnzrwuhvpwmezuurundarwdazwptrpeurrnwautddnhdmhtwhwvvtavdzezumetzezpazndhuentmrhamutrtttpevtuutemdnvwnwnmnvmznatneweuaahdavmaddhnrdenwwztrwh\r\n", "output": "199\r\n"}, {"input": "aaccddff\r\nabcdf\r\n", "output": "-1"}] | false | stdio | null | true |
756/E | 756 | E | Python 3 | TESTS | 3 | 46 | 4,608,000 | 24216588 | #!/usr/bin/env python3
MOD = 1000000007
N = int(input())
a = [int(x) for x in input().split()] + [1]
b = [int(x) for x in input().split()]
M = int(input())
r = []
for i in range(N - 1):
r.append(M % a[i])
M /= a[i]
r += [M]
dp = [1]
for i in range(N - 1, -1, -1):
scale = a[i]
dp2 = [0] * 50
for k in range(len(dp)):
missing = scale * k + r[i]
for j in range(50):
use = missing - j
if use >= 0 and use <= b[i]:
dp2[j] = (dp2[j] + dp[k]) % MOD
dp = dp2
print(dp[0]) | 40 | 763 | 44,236,800 | 24049216 | p = 1000000007
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
m = int(input())
d = [1] * 300001
td = [0] * 300001
L = b[0]
for i in range(1, n):
if a[i - 1] != 1:
t = m % a[i - 1]
if L < t:
print(0)
exit(0)
m //= a[i - 1]
for j in range((L - t) // a[i - 1] + 1):
d[j] = d[t]
t += a[i - 1]
L = j
k = 0
for j in range(L + b[i] + 1):
if j <= L:
k += d[j]
k %= p
td[j] = k
if j >= b[i]:
k -= d[j - b[i]]
L += b[i]
for j in range(L + 1):
d[j] = td[j]
print(d[m] if m <= L else 0) | 8VC Venture Cup 2017 - Final Round | CF | 2,017 | 1 | 512 | Byteland coins | There are n types of coins in Byteland. Conveniently, the denomination of the coin type k divides the denomination of the coin type k + 1, the denomination of the coin type 1 equals 1 tugrick. The ratio of the denominations of coin types k + 1 and k equals ak. It is known that for each x there are at most 20 coin types of denomination x.
Byteasar has bk coins of type k with him, and he needs to pay exactly m tugricks. It is known that Byteasar never has more than 3·105 coins with him. Byteasar want to know how many ways there are to pay exactly m tugricks. Two ways are different if there is an integer k such that the amount of coins of type k differs in these two ways. As all Byteland citizens, Byteasar wants to know the number of ways modulo 109 + 7. | The first line contains single integer n (1 ≤ n ≤ 3·105) — the number of coin types.
The second line contains n - 1 integers a1, a2, ..., an - 1 (1 ≤ ak ≤ 109) — the ratios between the coin types denominations. It is guaranteed that for each x there are at most 20 coin types of denomination x.
The third line contains n non-negative integers b1, b2, ..., bn — the number of coins of each type Byteasar has. It is guaranteed that the sum of these integers doesn't exceed 3·105.
The fourth line contains single integer m (0 ≤ m < 1010000) — the amount in tugricks Byteasar needs to pay. | Print single integer — the number of ways to pay exactly m tugricks modulo 109 + 7. | null | In the first example Byteasar has 4 coins of denomination 1, and he has to pay 2 tugricks. There is only one way.
In the second example Byteasar has 4 coins of each of two different types of denomination 1, he has to pay 2 tugricks. There are 3 ways: pay one coin of the first type and one coin of the other, pay two coins of the first type, and pay two coins of the second type.
In the third example the denominations are equal to 1, 3, 9. | [{"input": "1\n4\n2", "output": "1"}, {"input": "2\n1\n4 4\n2", "output": "3"}, {"input": "3\n3 3\n10 10 10\n17", "output": "6"}] | 3,200 | ["combinatorics", "dp", "math"] | 40 | [{"input": "1\r\n\r\n4\r\n2\r\n", "output": "1\r\n"}, {"input": "2\r\n1\r\n4 4\r\n2\r\n", "output": "3\r\n"}, {"input": "3\r\n3 3\r\n10 10 10\r\n17\r\n", "output": "6\r\n"}, {"input": "2\r\n2\r\n200000 100000\r\n34567\r\n", "output": "17284\r\n"}, {"input": "20\r\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2\r\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 299981\r\n1234567890\r\n", "output": "1\r\n"}, {"input": "20\r\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2\r\n299981 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\r\n1034567\r\n", "output": "149991\r\n"}, {"input": "10\r\n1 1 1 1 1 1 1 1 2\r\n0 10 68 1 7 6 0 1 3 4\r\n61\r\n", "output": "49280\r\n"}, {"input": "100\r\n4 1 5 3 2 1 1 1 4 1 1 2 1 1 1 4 1 1 3 1 3 1 1 1 1 4 5 1 5 2 5 3 1 1 1 1 1 1 1 3 2 1 1 3 1 1 3 4 3 2 4 1 1 4 1 1 2 2 4 1 4 1 2 5 1 2 2 1 5 3 1 5 4 2 1 1 2 5 5 1 4 4 2 3 1 4 1 3 2 1 1 1 4 1 3 1 1 5 1\r\n0 18 10 2 1 9 9 0 9 5 6 8 11 6 28 11 29 50 25 15 9 4 3 51 13 4 68 31 4 6 2 5 26 1 21 7 3 4 9 7 40 3 0 7 14 18 4 8 4 1 0 3 21 2 5 1 2 8 2 4 10 11 25 5 11 4 2 5 3 3 4 7 0 0 1 9 0 0 4 16 1 20 10 22 17 3 14 11 30 1 3 7 3 5 6 13 3 9 18 7\r\n188562805042251972437939648\r\n", "output": "890905252\r\n"}, {"input": "10\r\n3 9 10 10 4 10 9 10 8\r\n18 54 100 42 402 13 28 208 102 33\r\n77760001052028517\r\n", "output": "0\r\n"}, {"input": "10\r\n1 1 1 1 1 1 1 1 2\r\n0 0 0 0 0 0 1 0 1 0\r\n1\r\n", "output": "2\r\n"}, {"input": "20\r\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2\r\n2 136 23 34 16 22 7 1 121 65 11 5 68 144 3 14 3 35 44 246\r\n86551330\r\n", "output": "960419474\r\n"}, {"input": "20\r\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1\r\n29 77 47 64 67 89 71 21 106 15 47 34 90 10 6 28 18 11 152 18\r\n501\r\n", "output": "287270499\r\n"}, {"input": "10\r\n443307727 348302095 35497258 398797405 725089211 557667579 7764455 164622658 466615150\r\n9 7 30 1 4 6 6 4 23 10\r\n3690054862906606768658826690738341858379111902540863414278121378497891890923\r\n", "output": "1\r\n"}, {"input": "10\r\n1 1 1 1 1 1 1 1 1\r\n9519 118380 15475 18454 10395 10005 1925 43712 6710 65425\r\n114853\r\n", "output": "983175834\r\n"}, {"input": "20\r\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\r\n3340 8397 18248 8914 6824 396 6249 22945 6175 1443 13271 53526 12738 5346 8485 12784 31161 2378 68313 9067\r\n145333\r\n", "output": "116763993\r\n"}, {"input": "1\r\n\r\n300000\r\n294705\r\n", "output": "1\r\n"}, {"input": "2\r\n1\r\n45133 254867\r\n62105\r\n", "output": "45134\r\n"}, {"input": "10\r\n2 2 3 3 2 2 2 3 3\r\n117 254 68 126 105 3 100 45 166 16\r\n2592000130163\r\n", "output": "0\r\n"}, {"input": "10\r\n1 1 1 1 1 1 1 1 1\r\n73 126 74 58 337 123 0 9 161 39\r\n1000000656\r\n", "output": "0\r\n"}, {"input": "10\r\n1 1 1 1 1 1 1 1 1\r\n55 75 21 92 159 178 181 137 29 73\r\n1000000533\r\n", "output": "0\r\n"}, {"input": "10\r\n5 7 5 8 3 7 2 4 7\r\n124 154 10 227 74 10 15 309 68 9\r\n49389597\r\n", "output": "3894309\r\n"}, {"input": "10\r\n1 1 1 1 1 1 1 1 1\r\n185 63 24 145 162 205 46 104 54 12\r\n1461\r\n", "output": "0\r\n"}] | false | stdio | null | true |
177/C1 | 177 | C1 | PyPy 3-64 | TESTS1 | 3 | 124 | 0 | 225903288 | import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return int(input())
def inlt():
return list(map(int, input().split()))
def insr():
s = input()
return list(s[: len(s) - 1])
def invr():
return map(int, input().split())
def find(x):
if x == root[x]:
return x
root[x] = find(root[x])
return root[x]
def union(x, y):
a, b = find(x), find(y)
if a != b:
if rank[a] > rank[b]:
root[b] = a
elif rank[b] > rank[a]:
root[a] = b
else:
root[b] = a
rank[a] += 1
import heapq
n = inp()
k = inp()
friends = []
child = [0 for _ in range(n)]
for _ in range(k):
a, b = invr()
friends.append([a - 1, b - 1])
m = inp()
foe = [[] for _ in range(n)]
for _ in range(m):
a, b = invr()
foe[a - 1].append(b - 1)
foe[b - 1].append(a - 1)
rank = [1 for i in range(n)]
root = [i for i in range(n)]
for a, b in friends:
union(a, b)
foecount = [0 for _ in range(n)]
for i in range(n):
child[find(i)] += 1
# print(child)
for i in range(n):
x = find(i)
total = child[x]
for dushman in foe[i]:
if find(dushman) == x:
foecount[i] -= 1
# print(foecount)
q = []
for i in range(n):
if foecount[i] != 0:
heapq.heappush(q, (foecount[i], i))
while q:
dushman, node = heapq.heappop(q)
org = find(node)
child[org] -= 1
for nei in foe[node]:
if find(nei) == org:
foecount[nei] += 1
if foecount[nei] < 0:
heapq.heappush(q, (foecount[nei], nei))
print(max(child)) | 17 | 218 | 307,200 | 74705608 | # maa chudaaye duniya
n = int(input())
parents = [i for i in range(n+1)]
ranks = [1 for i in range(n+1)]
def find(x):
if parents[x] != x:
parents[x] = find(parents[x])
return parents[x]
def union(x, y):
xs = find(x)
ys = find(y)
if xs == ys:
return
if ranks[xs] > ranks[ys]:
parents[ys] = xs
elif ranks[ys] > ranks[xs]:
parents[xs] = ys
else:
parents[ys] = xs
ranks[xs] += 1
for _ in range(int(input())):
u, v = map(int, input().split())
union(u, v)
# print(parents)
rejects = set([])
for _ in range(int(input())):
p, q = map(int, input().split())
ps = find(p)
qs = find(q)
if ps == qs:
rejects.add(ps)
ps = {}
for i in range(1, n+1):
p = find(i)
if p not in rejects:
if p in ps:
ps[p] += 1
else:
ps[p] = 1
# print(ps)
ans = 0
for i in ps:
ans = max(ans, ps[i])
print(ans) | ABBYY Cup 2.0 - Easy | ICPC | 2,012 | 2 | 256 | Party | To celebrate the second ABBYY Cup tournament, the Smart Beaver decided to throw a party. The Beaver has a lot of acquaintances, some of them are friends with each other, and some of them dislike each other. To make party successful, the Smart Beaver wants to invite only those of his friends who are connected by friendship relations, and not to invite those who dislike each other. Both friendship and dislike are mutual feelings.
More formally, for each invited person the following conditions should be fulfilled:
- all his friends should also be invited to the party;
- the party shouldn't have any people he dislikes;
- all people who are invited to the party should be connected with him by friendship either directly or through a chain of common friends of arbitrary length. We'll say that people a1 and ap are connected through a chain of common friends if there exists a sequence of people a2, a3, ..., ap - 1 such that all pairs of people ai and ai + 1 (1 ≤ i < p) are friends.
Help the Beaver find the maximum number of acquaintances he can invite. | The first line of input contains an integer n — the number of the Beaver's acquaintances.
The second line contains an integer k $$( 0 \leq k \leq \min ( 1 0 ^ { 5 }, \frac { n \cdot ( n - 1 ) } { 2 } ) )$$ — the number of pairs of friends. Next k lines contain space-separated pairs of integers ui, vi $$( 1 \leq u _ { i }, v _ { i } \leq n, u _ { i } \neq v _ { i } )$$ — indices of people who form the i-th pair of friends.
The next line contains an integer m $$( 0 \leq m \leq min(10^{5},\frac{n(n-1)}{2}))$$ — the number of pairs of people who dislike each other. Next m lines describe pairs of people who dislike each other in the same format as the pairs of friends were described.
Each pair of people is mentioned in the input at most once $$( 0 \leq k + m \leq \frac { n \cdot ( n - 1 ) } { 2 } )$$. In particular, two persons cannot be friends and dislike each other at the same time.
The input limitations for getting 30 points are:
- 2 ≤ n ≤ 14
The input limitations for getting 100 points are:
- 2 ≤ n ≤ 2000 | Output a single number — the maximum number of people that can be invited to the party. If a group of people that meets all the requirements is impossible to select, output 0. | null | Let's have a look at the example.
Two groups of people can be invited: {1, 2, 3} and {4, 5}, thus the answer will be the size of the largest of these groups. Group {6, 7, 8, 9} doesn't fit, since it includes people 7 and 9 who dislike each other. Group {1, 2, 3, 4, 5} also doesn't fit, because not all of its members are connected by a chain of common friends (for example, people 2 and 5 aren't connected). | [{"input": "9\n8\n1 2\n1 3\n2 3\n4 5\n6 7\n7 8\n8 9\n9 6\n2\n1 6\n7 9", "output": "3"}] | 1,500 | ["dfs and similar", "dsu", "graphs"] | 17 | [{"input": "9\r\n8\r\n1 2\r\n1 3\r\n2 3\r\n4 5\r\n6 7\r\n7 8\r\n8 9\r\n9 6\r\n2\r\n1 6\r\n7 9\r\n", "output": "3"}, {"input": "2\r\n1\r\n1 2\r\n0\r\n", "output": "2"}, {"input": "2\r\n0\r\n1\r\n1 2\r\n", "output": "1"}, {"input": "3\r\n2\r\n1 2\r\n1 3\r\n1\r\n2 3\r\n", "output": "0"}, {"input": "3\r\n3\r\n1 3\r\n2 1\r\n2 3\r\n0\r\n", "output": "3"}, {"input": "4\r\n3\r\n1 2\r\n2 3\r\n3 1\r\n3\r\n1 4\r\n4 2\r\n3 4\r\n", "output": "3"}, {"input": "7\r\n8\r\n1 2\r\n1 3\r\n1 4\r\n1 5\r\n2 4\r\n2 5\r\n3 4\r\n5 6\r\n3\r\n2 6\r\n5 7\r\n6 7\r\n", "output": "1"}, {"input": "14\r\n20\r\n1 2\r\n4 5\r\n4 6\r\n4 11\r\n5 7\r\n5 8\r\n5 13\r\n5 14\r\n7 8\r\n7 14\r\n8 9\r\n8 11\r\n8 12\r\n8 14\r\n10 11\r\n10 12\r\n10 14\r\n11 13\r\n11 14\r\n12 14\r\n5\r\n1 8\r\n1 13\r\n2 10\r\n7 12\r\n8 10\r\n", "output": "2"}, {"input": "2\r\n0\r\n0\r\n", "output": "1"}, {"input": "14\r\n0\r\n0\r\n", "output": "1"}, {"input": "14\r\n6\r\n1 2\r\n2 3\r\n3 4\r\n4 5\r\n8 9\r\n9 10\r\n3\r\n5 6\r\n6 7\r\n7 8\r\n", "output": "5"}, {"input": "14\r\n10\r\n1 2\r\n1 3\r\n1 4\r\n1 5\r\n1 6\r\n1 7\r\n2 3\r\n2 4\r\n2 5\r\n2 6\r\n1\r\n2 7\r\n", "output": "1"}, {"input": "2\r\n0\r\n1\r\n1 2\r\n", "output": "1"}] | false | stdio | null | true |
408/A | 408 | A | PyPy 3-64 | TESTS | 16 | 62 | 1,638,400 | 196299412 | n = int(input())
ans = int(5e4)
*k, = map(int, input().split())
for i in range(n):
*m, = map(int, input().split())
ans = min(ans, sum(m) * 5 + k[i] * 15)
print(ans) | 20 | 31 | 0 | 143984023 | n=int(input())
k=list(map(int,input().split()))
for i in range(n):
a=list(map(int,input().split()))
val=sum(a)*5+k[i]*15
if(i==0):
m=val
else:
if(m>val):
m=val
print(m) | Codeforces Round 239 (Div. 2) | CF | 2,014 | 1 | 256 | Line to Cashier | Little Vasya went to the supermarket to get some groceries. He walked about the supermarket for a long time and got a basket full of products. Now he needs to choose the cashier to pay for the products.
There are n cashiers at the exit from the supermarket. At the moment the queue for the i-th cashier already has ki people. The j-th person standing in the queue to the i-th cashier has mi, j items in the basket. Vasya knows that:
- the cashier needs 5 seconds to scan one item;
- after the cashier scans each item of some customer, he needs 15 seconds to take the customer's money and give him the change.
Of course, Vasya wants to select a queue so that he can leave the supermarket as soon as possible. Help him write a program that displays the minimum number of seconds after which Vasya can get to one of the cashiers. | The first line contains integer n (1 ≤ n ≤ 100) — the number of cashes in the shop. The second line contains n space-separated integers: k1, k2, ..., kn (1 ≤ ki ≤ 100), where ki is the number of people in the queue to the i-th cashier.
The i-th of the next n lines contains ki space-separated integers: mi, 1, mi, 2, ..., mi, ki (1 ≤ mi, j ≤ 100) — the number of products the j-th person in the queue for the i-th cash has. | Print a single integer — the minimum number of seconds Vasya needs to get to the cashier. | null | In the second test sample, if Vasya goes to the first queue, he gets to the cashier in 100·5 + 15 = 515 seconds. But if he chooses the second queue, he will need 1·5 + 2·5 + 2·5 + 3·5 + 4·15 = 100 seconds. He will need 1·5 + 9·5 + 1·5 + 3·15 = 100 seconds for the third one and 7·5 + 8·5 + 2·15 = 105 seconds for the fourth one. Thus, Vasya gets to the cashier quicker if he chooses the second or the third queue. | [{"input": "1\n1\n1", "output": "20"}, {"input": "4\n1 4 3 2\n100\n1 2 2 3\n1 9 1\n7 8", "output": "100"}] | 900 | ["implementation"] | 20 | [{"input": "1\r\n1\r\n1\r\n", "output": "20\r\n"}, {"input": "4\r\n1 4 3 2\r\n100\r\n1 2 2 3\r\n1 9 1\r\n7 8\r\n", "output": "100\r\n"}, {"input": "4\r\n5 4 5 5\r\n3 1 3 1 2\r\n3 1 1 3\r\n1 1 1 2 2\r\n2 2 1 1 3\r\n", "output": "100\r\n"}, {"input": "5\r\n5 3 6 6 4\r\n7 5 3 3 9\r\n6 8 2\r\n1 10 8 5 9 2\r\n9 7 8 5 9 10\r\n9 8 3 3\r\n", "output": "125\r\n"}, {"input": "5\r\n10 10 10 10 10\r\n6 7 8 6 8 5 9 8 10 5\r\n9 6 9 8 7 8 8 10 8 5\r\n8 7 7 8 7 5 6 8 9 5\r\n6 5 10 5 5 10 7 8 5 5\r\n10 9 8 7 6 9 7 9 6 5\r\n", "output": "480\r\n"}, {"input": "10\r\n9 10 10 10 9 5 9 7 8 7\r\n11 6 10 4 4 15 7 15 5\r\n3 9 11 12 11 1 13 13 1 5\r\n6 15 9 12 3 2 8 12 11 10\r\n7 1 1 6 10 2 6 1 14 2\r\n8 14 2 3 6 1 14 1 12\r\n6 10 9 3 5\r\n13 12 12 7 13 4 4 8 10\r\n5 6 4 3 14 9 13\r\n8 12 1 5 7 4 13 1\r\n1 9 5 3 5 1 4\r\n", "output": "240\r\n"}, {"input": "10\r\n5 5 5 5 5 5 5 5 5 5\r\n5 5 4 5 4\r\n6 5 7 7 6\r\n5 4 4 5 5\r\n4 4 5 5 5\r\n7 6 4 5 7\r\n4 6 5 4 5\r\n6 6 7 6 6\r\n4 5 4 4 7\r\n7 5 4 4 5\r\n6 6 7 4 4\r\n", "output": "190\r\n"}, {"input": "1\r\n1\r\n100\r\n", "output": "515\r\n"}, {"input": "1\r\n90\r\n90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90\r\n", "output": "41850\r\n"}] | false | stdio | null | true |
408/B | 408 | B | Python 3 | TESTS | 4 | 31 | 0 | 158936970 | a = input()
b = input()
o = {chr(x): 0 for x in range(97, 123)}
e = {chr(x): 0 for x in range(97, 123)}
for item in a:
o[item] += 1
for item in b:
e[item] += 1
new = 0
for key, value in o.items():
new += min(value, e[key])
if new == 0:print(-1)
else:print(new) | 21 | 46 | 0 | 165679408 | n = input()
m = input()
s = set(m)
c= 0
for i in s:
a1 = n.count(i)
a2 = m.count(i)
if a1 == 0:
c = -1
break
else:
c+=min(a1,a2)
print(c) | Codeforces Round 239 (Div. 2) | CF | 2,014 | 1 | 256 | Garland | Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought n colored sheets of paper, the area of each sheet is 1 square meter.
The garland must consist of exactly m pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.
Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of m pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get. | The first line contains a non-empty sequence of n (1 ≤ n ≤ 1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.
The second line contains a non-empty sequence of m (1 ≤ m ≤ 1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make. | Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer. | null | In the first test sample Vasya can make an garland of area 6: he can use both sheets of color b, three (but not four) sheets of color a and cut a single sheet of color c in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.
In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color z. | [{"input": "aaabbac\naabbccac", "output": "6"}, {"input": "a\nz", "output": "-1"}] | 1,200 | ["implementation"] | 21 | [{"input": "aaabbac\r\naabbccac\r\n", "output": "6\r\n"}, {"input": "a\r\nz\r\n", "output": "-1"}, {"input": "r\r\nr\r\n", "output": "1\r\n"}, {"input": "stnsdn\r\nndnndsn\r\n", "output": "4\r\n"}, {"input": "yqfqfp\r\ntttwtqq\r\n", "output": "-1"}, {"input": "zzbbrrtrtzr\r\ntbbtrrrzr\r\n", "output": "9\r\n"}, {"input": "ivvfisvsvii\r\npaihjinno\r\n", "output": "-1"}, {"input": "zbvwnlgkshqerxptyod\r\nz\r\n", "output": "1\r\n"}, {"input": "xlktwjymocqrahnbesf\r\nfoo\r\n", "output": "2\r\n"}, {"input": "bbzmzqazmbambnmzaabznmbabzqnaabmabmnnabbmnzaanzzezebzabqaabzqaemeqqammmbazmmz\r\naznnbbmeebmanbeemzmemqbaeebnqenqzzbanebmnzqqebqmmnmqqzmmeqqqaaezemmazqqmqaqnnqqzbzeeazammmenbbamzbmnaenemenaaaebnmanebqmqnznqbenmqqnnnaeaebqmamennmqqeaaqqbammnzqmnmqnqbbezmemznqmanzmmqzzzzembqnzqbanamezqaqbazenenqqznqaebzaeezbqqbmeeaqnmmbnqbbnmaqqemaeaezaabmbnbzzaae\r\n", "output": "77\r\n"}, {"input": "lccfdfnfflncddlksndcwnfcllnnaswcdnldafcalssfcdnkkaklwnnacsncfwanwnwfadawcsdcfwflnnlfsfclkfknlaldna\r\nuaaldlllhedgugugueahddhedbuddaedhaaeugdubaealbgbagedldbl\r\n", "output": "-1"}, {"input": "hvewdvwdwudrwarwmddwnazmwauzatrmwptwwevavpmwrtruwnpwantewrnwmepdwvtmnveanunrvrtwpvhhnuhnmpptdttzmmndtvudmzhhannmmnahdpzamuvhzaavnhtnumwrwvttdetvuewaaennddwuvzvaptdzrzhtetwwzmzedrwuvrwznprhdvnavrruvvhzuwpdtmpwmzrwvermrhdamv\r\nuvzhwtpuputnahwwarduzddhpnwwvettprwavdmnztdnrddmarmvuevtdezndnezvarhmppwwnmvnrtddzhhnzrwuhvpwmezuurundarwdazwptrpeurrnwautddnhdmhtwhwvvtavdzezumetzezpazndhuentmrhamutrtttpevtuutemdnvwnwnmnvmznatneweuaahdavmaddhnrdenwwztrwh\r\n", "output": "199\r\n"}, {"input": "aaccddff\r\nabcdf\r\n", "output": "-1"}] | false | stdio | null | true |
408/B | 408 | B | Python 3 | TESTS | 4 | 31 | 0 | 161124165 | import sys
input = sys.stdin.readline
from collections import defaultdict
s = input()[:-1]
w = input()[:-1]
d = defaultdict(int)
for i in s:
d[i] += 1
c = 0
for i in w:
if d[i] > 0:
d[i] -= 1
c += 1
if c == 0:
print(-1)
else:
print(c) | 21 | 46 | 0 | 166918276 | x=input()
y=input()
d1={}
d2={}
count=0
for i in x:
d1[i]=x.count(i)
for i in y:
d2[i]=y.count(i)
for i in d2.keys():
if i in d1.keys():
if d1[i]<d2[i]:
count+=d1[i]
else:
count+=d2[i]
else:
count=-1
break
print(count) | Codeforces Round 239 (Div. 2) | CF | 2,014 | 1 | 256 | Garland | Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought n colored sheets of paper, the area of each sheet is 1 square meter.
The garland must consist of exactly m pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.
Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of m pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get. | The first line contains a non-empty sequence of n (1 ≤ n ≤ 1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.
The second line contains a non-empty sequence of m (1 ≤ m ≤ 1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make. | Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer. | null | In the first test sample Vasya can make an garland of area 6: he can use both sheets of color b, three (but not four) sheets of color a and cut a single sheet of color c in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.
In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color z. | [{"input": "aaabbac\naabbccac", "output": "6"}, {"input": "a\nz", "output": "-1"}] | 1,200 | ["implementation"] | 21 | [{"input": "aaabbac\r\naabbccac\r\n", "output": "6\r\n"}, {"input": "a\r\nz\r\n", "output": "-1"}, {"input": "r\r\nr\r\n", "output": "1\r\n"}, {"input": "stnsdn\r\nndnndsn\r\n", "output": "4\r\n"}, {"input": "yqfqfp\r\ntttwtqq\r\n", "output": "-1"}, {"input": "zzbbrrtrtzr\r\ntbbtrrrzr\r\n", "output": "9\r\n"}, {"input": "ivvfisvsvii\r\npaihjinno\r\n", "output": "-1"}, {"input": "zbvwnlgkshqerxptyod\r\nz\r\n", "output": "1\r\n"}, {"input": "xlktwjymocqrahnbesf\r\nfoo\r\n", "output": "2\r\n"}, {"input": "bbzmzqazmbambnmzaabznmbabzqnaabmabmnnabbmnzaanzzezebzabqaabzqaemeqqammmbazmmz\r\naznnbbmeebmanbeemzmemqbaeebnqenqzzbanebmnzqqebqmmnmqqzmmeqqqaaezemmazqqmqaqnnqqzbzeeazammmenbbamzbmnaenemenaaaebnmanebqmqnznqbenmqqnnnaeaebqmamennmqqeaaqqbammnzqmnmqnqbbezmemznqmanzmmqzzzzembqnzqbanamezqaqbazenenqqznqaebzaeezbqqbmeeaqnmmbnqbbnmaqqemaeaezaabmbnbzzaae\r\n", "output": "77\r\n"}, {"input": "lccfdfnfflncddlksndcwnfcllnnaswcdnldafcalssfcdnkkaklwnnacsncfwanwnwfadawcsdcfwflnnlfsfclkfknlaldna\r\nuaaldlllhedgugugueahddhedbuddaedhaaeugdubaealbgbagedldbl\r\n", "output": "-1"}, {"input": "hvewdvwdwudrwarwmddwnazmwauzatrmwptwwevavpmwrtruwnpwantewrnwmepdwvtmnveanunrvrtwpvhhnuhnmpptdttzmmndtvudmzhhannmmnahdpzamuvhzaavnhtnumwrwvttdetvuewaaennddwuvzvaptdzrzhtetwwzmzedrwuvrwznprhdvnavrruvvhzuwpdtmpwmzrwvermrhdamv\r\nuvzhwtpuputnahwwarduzddhpnwwvettprwavdmnztdnrddmarmvuevtdezndnezvarhmppwwnmvnrtddzhhnzrwuhvpwmezuurundarwdazwptrpeurrnwautddnhdmhtwhwvvtavdzezumetzezpazndhuentmrhamutrtttpevtuutemdnvwnwnmnvmznatneweuaahdavmaddhnrdenwwztrwh\r\n", "output": "199\r\n"}, {"input": "aaccddff\r\nabcdf\r\n", "output": "-1"}] | false | stdio | null | true |
599/C | 599 | C | Python 3 | TESTS | 3 | 31 | 0 | 159296371 | n = int(input())
a = [*map(int,input().split())]
s = sorted(a)
window = [set(),set()]
ans = 0
for i in range(n):
window[0].add(a[i]); window[1].add(s[i])
if window[0] == window[1]:
ans +=1
window[0].clear; window[1].clear
print(ans) | 39 | 124 | 17,817,600 | 224633886 | n=int(input())
h=list(map(int,input().split()))
ans,mn,mx=0,[],[]
for i in range(n):
mn.append(float("inf"))
mx.append(float("-inf"))
mn[-1],mx[0]=h[-1],h[0]
for i in range(n-2,-1,-1):
mn[i]=min(h[i],mn[i+1])
for i in range(1,n):
mx[i]=max(mx[i-1],h[i])
for i in range(n-1):
if mx[i]<=mn[i+1]:
ans+=1
print(ans+1) | Codeforces Round 332 (Div. 2) | CF | 2,015 | 2 | 256 | Day at the Beach | One day Squidward, Spongebob and Patrick decided to go to the beach. Unfortunately, the weather was bad, so the friends were unable to ride waves. However, they decided to spent their time building sand castles.
At the end of the day there were n castles built by friends. Castles are numbered from 1 to n, and the height of the i-th castle is equal to hi. When friends were about to leave, Squidward noticed, that castles are not ordered by their height, and this looks ugly. Now friends are going to reorder the castles in a way to obtain that condition hi ≤ hi + 1 holds for all i from 1 to n - 1.
Squidward suggested the following process of sorting castles:
- Castles are split into blocks — groups of consecutive castles. Therefore the block from i to j will include castles i, i + 1, ..., j. A block may consist of a single castle.
- The partitioning is chosen in such a way that every castle is a part of exactly one block.
- Each block is sorted independently from other blocks, that is the sequence hi, hi + 1, ..., hj becomes sorted.
- The partitioning should satisfy the condition that after each block is sorted, the sequence hi becomes sorted too. This may always be achieved by saying that the whole sequence is a single block.
Even Patrick understands that increasing the number of blocks in partitioning will ease the sorting process. Now friends ask you to count the maximum possible number of blocks in a partitioning that satisfies all the above requirements. | The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of castles Spongebob, Patrick and Squidward made from sand during the day.
The next line contains n integers hi (1 ≤ hi ≤ 109). The i-th of these integers corresponds to the height of the i-th castle. | Print the maximum possible number of blocks in a valid partitioning. | null | In the first sample the partitioning looks like that: [1][2][3].
In the second sample the partitioning is: [2, 1][3, 2] | [{"input": "3\n1 2 3", "output": "3"}, {"input": "4\n2 1 3 2", "output": "2"}] | 1,600 | ["sortings"] | 39 | [{"input": "3\r\n1 2 3\r\n", "output": "3\r\n"}, {"input": "4\r\n2 1 3 2\r\n", "output": "2\r\n"}, {"input": "17\r\n1 45 22 39 28 23 23 100 500 778 777 778 1001 1002 1005 1003 1005\r\n", "output": "10\r\n"}, {"input": "101\r\n1 50 170 148 214 153 132 234 181 188 180 225 226 200 197 122 181 168 87 220 223 160 235 94 257 145 199 235 102 146 119 60 109 134 209 260 210 191 180 271 236 195 155 169 166 143 246 102 208 137 278 269 156 251 198 165 111 198 151 213 256 121 276 163 179 285 104 99 139 122 188 184 215 242 244 115 304 259 135 149 104 72 303 291 124 237 112 165 183 168 71 139 85 131 137 107 120 267 235 337 69\r\n", "output": "3\r\n"}, {"input": "10\r\n1 2 2 2 2 2 2 2 2 1\r\n", "output": "2\r\n"}, {"input": "25\r\n1 2 3 4 4 4 4 4 4 4 2 3 5 5 7 9 8 5 10 12 15 12 100500 800600 228228228\r\n", "output": "12\r\n"}, {"input": "10\r\n17 18 19 19 18 17 100 500 100 100\r\n", "output": "4\r\n"}, {"input": "10\r\n1 1 1 1 5 5 1 1 1 1\r\n", "output": "5\r\n"}, {"input": "20\r\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000\r\n", "output": "20\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}, {"input": "5\r\n1 5 3 5 2\r\n", "output": "2\r\n"}, {"input": "10\r\n1 1 1 1 2 2 2 2 4 3\r\n", "output": "9\r\n"}, {"input": "20\r\n1 2 2 2 5 6 6 6 7 7 8 9 15 15 16 16 17 18 19 19\r\n", "output": "20\r\n"}, {"input": "4\r\n2 2 1 1\r\n", "output": "1\r\n"}] | false | stdio | null | true |
408/B | 408 | B | Python 3 | TESTS | 4 | 31 | 0 | 156544851 | we=input("")
need=input("")
l=[]
n=[]
for i in need:
x=0
for j in need:
if i==j:
x=x+1
l.append([i,x])
for i in we:
y=0
for j in we:
if i==j:
y=y+1
n.append([i,y])
fin=[]
las=[]
for i in n:
if i not in fin:
fin.append(i)#we
for i in l:
if i not in las:
las.append(i) #need
num=0
for i in las:
for j in fin:
if i[0]==j[0]:
if i[1]<j[1]:
num=num+i[1]
else:
num=num+j[1]
if num==0:
print(-1)
else:
print(num) | 21 | 46 | 0 | 173273807 | n = input()
m = input()
nCount = [0]*26
mCount = [0]*26
for c in n:
nCount[ord(c)-ord('a')]+=1
for c in m:
mCount[ord(c)-ord('a')]+=1
area = 0
for i in range(26):
if mCount[i] !=0 and nCount[i]==0:
print(-1)
exit()
area+=min(mCount[i],nCount[i])
print(area) | Codeforces Round 239 (Div. 2) | CF | 2,014 | 1 | 256 | Garland | Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought n colored sheets of paper, the area of each sheet is 1 square meter.
The garland must consist of exactly m pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.
Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of m pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get. | The first line contains a non-empty sequence of n (1 ≤ n ≤ 1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.
The second line contains a non-empty sequence of m (1 ≤ m ≤ 1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make. | Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer. | null | In the first test sample Vasya can make an garland of area 6: he can use both sheets of color b, three (but not four) sheets of color a and cut a single sheet of color c in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.
In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color z. | [{"input": "aaabbac\naabbccac", "output": "6"}, {"input": "a\nz", "output": "-1"}] | 1,200 | ["implementation"] | 21 | [{"input": "aaabbac\r\naabbccac\r\n", "output": "6\r\n"}, {"input": "a\r\nz\r\n", "output": "-1"}, {"input": "r\r\nr\r\n", "output": "1\r\n"}, {"input": "stnsdn\r\nndnndsn\r\n", "output": "4\r\n"}, {"input": "yqfqfp\r\ntttwtqq\r\n", "output": "-1"}, {"input": "zzbbrrtrtzr\r\ntbbtrrrzr\r\n", "output": "9\r\n"}, {"input": "ivvfisvsvii\r\npaihjinno\r\n", "output": "-1"}, {"input": "zbvwnlgkshqerxptyod\r\nz\r\n", "output": "1\r\n"}, {"input": "xlktwjymocqrahnbesf\r\nfoo\r\n", "output": "2\r\n"}, {"input": "bbzmzqazmbambnmzaabznmbabzqnaabmabmnnabbmnzaanzzezebzabqaabzqaemeqqammmbazmmz\r\naznnbbmeebmanbeemzmemqbaeebnqenqzzbanebmnzqqebqmmnmqqzmmeqqqaaezemmazqqmqaqnnqqzbzeeazammmenbbamzbmnaenemenaaaebnmanebqmqnznqbenmqqnnnaeaebqmamennmqqeaaqqbammnzqmnmqnqbbezmemznqmanzmmqzzzzembqnzqbanamezqaqbazenenqqznqaebzaeezbqqbmeeaqnmmbnqbbnmaqqemaeaezaabmbnbzzaae\r\n", "output": "77\r\n"}, {"input": "lccfdfnfflncddlksndcwnfcllnnaswcdnldafcalssfcdnkkaklwnnacsncfwanwnwfadawcsdcfwflnnlfsfclkfknlaldna\r\nuaaldlllhedgugugueahddhedbuddaedhaaeugdubaealbgbagedldbl\r\n", "output": "-1"}, {"input": "hvewdvwdwudrwarwmddwnazmwauzatrmwptwwevavpmwrtruwnpwantewrnwmepdwvtmnveanunrvrtwpvhhnuhnmpptdttzmmndtvudmzhhannmmnahdpzamuvhzaavnhtnumwrwvttdetvuewaaennddwuvzvaptdzrzhtetwwzmzedrwuvrwznprhdvnavrruvvhzuwpdtmpwmzrwvermrhdamv\r\nuvzhwtpuputnahwwarduzddhpnwwvettprwavdmnztdnrddmarmvuevtdezndnezvarhmppwwnmvnrtddzhhnzrwuhvpwmezuurundarwdazwptrpeurrnwautddnhdmhtwhwvvtavdzezumetzezpazndhuentmrhamutrtttpevtuutemdnvwnwnmnvmznatneweuaahdavmaddhnrdenwwztrwh\r\n", "output": "199\r\n"}, {"input": "aaccddff\r\nabcdf\r\n", "output": "-1"}] | false | stdio | null | true |
408/B | 408 | B | Python 3 | TESTS | 4 | 31 | 0 | 146593203 | a=[j for j in str(input())]
b=[i for i in str(input())]
c=list(set(b))
ans=0
for k in c:
ans+=min(a.count(k),b.count(k))
print(-1) if ans==0 else print(ans) | 21 | 46 | 0 | 181379203 | have = input()
required = input()
h_d = dict()
r_d = dict()
for i in have:
h_d[i] = h_d.get(i, 0)+1
for i in required:
r_d[i] = r_d.get(i, 0)+1
ans = 0
for i in r_d:
if i not in h_d:
print(-1)
break
else:
ans += min(r_d[i], h_d[i])
else:
print(ans) | Codeforces Round 239 (Div. 2) | CF | 2,014 | 1 | 256 | Garland | Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought n colored sheets of paper, the area of each sheet is 1 square meter.
The garland must consist of exactly m pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.
Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of m pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get. | The first line contains a non-empty sequence of n (1 ≤ n ≤ 1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.
The second line contains a non-empty sequence of m (1 ≤ m ≤ 1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make. | Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer. | null | In the first test sample Vasya can make an garland of area 6: he can use both sheets of color b, three (but not four) sheets of color a and cut a single sheet of color c in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.
In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color z. | [{"input": "aaabbac\naabbccac", "output": "6"}, {"input": "a\nz", "output": "-1"}] | 1,200 | ["implementation"] | 21 | [{"input": "aaabbac\r\naabbccac\r\n", "output": "6\r\n"}, {"input": "a\r\nz\r\n", "output": "-1"}, {"input": "r\r\nr\r\n", "output": "1\r\n"}, {"input": "stnsdn\r\nndnndsn\r\n", "output": "4\r\n"}, {"input": "yqfqfp\r\ntttwtqq\r\n", "output": "-1"}, {"input": "zzbbrrtrtzr\r\ntbbtrrrzr\r\n", "output": "9\r\n"}, {"input": "ivvfisvsvii\r\npaihjinno\r\n", "output": "-1"}, {"input": "zbvwnlgkshqerxptyod\r\nz\r\n", "output": "1\r\n"}, {"input": "xlktwjymocqrahnbesf\r\nfoo\r\n", "output": "2\r\n"}, {"input": "bbzmzqazmbambnmzaabznmbabzqnaabmabmnnabbmnzaanzzezebzabqaabzqaemeqqammmbazmmz\r\naznnbbmeebmanbeemzmemqbaeebnqenqzzbanebmnzqqebqmmnmqqzmmeqqqaaezemmazqqmqaqnnqqzbzeeazammmenbbamzbmnaenemenaaaebnmanebqmqnznqbenmqqnnnaeaebqmamennmqqeaaqqbammnzqmnmqnqbbezmemznqmanzmmqzzzzembqnzqbanamezqaqbazenenqqznqaebzaeezbqqbmeeaqnmmbnqbbnmaqqemaeaezaabmbnbzzaae\r\n", "output": "77\r\n"}, {"input": "lccfdfnfflncddlksndcwnfcllnnaswcdnldafcalssfcdnkkaklwnnacsncfwanwnwfadawcsdcfwflnnlfsfclkfknlaldna\r\nuaaldlllhedgugugueahddhedbuddaedhaaeugdubaealbgbagedldbl\r\n", "output": "-1"}, {"input": "hvewdvwdwudrwarwmddwnazmwauzatrmwptwwevavpmwrtruwnpwantewrnwmepdwvtmnveanunrvrtwpvhhnuhnmpptdttzmmndtvudmzhhannmmnahdpzamuvhzaavnhtnumwrwvttdetvuewaaennddwuvzvaptdzrzhtetwwzmzedrwuvrwznprhdvnavrruvvhzuwpdtmpwmzrwvermrhdamv\r\nuvzhwtpuputnahwwarduzddhpnwwvettprwavdmnztdnrddmarmvuevtdezndnezvarhmppwwnmvnrtddzhhnzrwuhvpwmezuurundarwdazwptrpeurrnwautddnhdmhtwhwvvtavdzezumetzezpazndhuentmrhamutrtttpevtuutemdnvwnwnmnvmznatneweuaahdavmaddhnrdenwwztrwh\r\n", "output": "199\r\n"}, {"input": "aaccddff\r\nabcdf\r\n", "output": "-1"}] | false | stdio | null | true |
12/C | 12 | C | Python 3 | TESTS | 9 | 31 | 0 | 192682492 | a,b=map(int,input().split())
d=[]
c=[int(i) for i in input().split()]
for i in range(b):
e=input()
if d.count(e)==0:
d=d+[e]
c.sort()
k=b-len(d)
print(sum(c[0:len(d)])+k*c[0],sum(c[-len(d):-1])+(k+1)*c[-1]) | 25 | 31 | 0 | 229903267 | n, m = input().split()
prices = sorted(list(map(int, input().split())))
fruits_count = {}
for i in range(int(m)):
fruit = input()
fruits_count[fruit] = 1 if fruit not in fruits_count else fruits_count[fruit] + 1
fruits_prices_max = {x: y for x, y in zip(sorted(fruits_count.keys(), key=lambda x: fruits_count[x], reverse=True), sorted(prices, reverse=True))}
fruits_prices_min = {x: y for x, y in zip(sorted(fruits_count.keys(), key=lambda x: fruits_count[x], reverse=True), sorted(prices))}
print(sum([fruits_count[i] * fruits_prices_min[i] for i in fruits_count.keys()]), sum([fruits_count[i] * fruits_prices_max[i] for i in fruits_count.keys()])) | Codeforces Beta Round 12 (Div 2 Only) | ICPC | 2,010 | 1 | 256 | Fruits | The spring is coming and it means that a lot of fruits appear on the counters. One sunny day little boy Valera decided to go shopping. He made a list of m fruits he wanted to buy. If Valera want to buy more than one fruit of some kind, he includes it into the list several times.
When he came to the fruit stall of Ashot, he saw that the seller hadn't distributed price tags to the goods, but put all price tags on the counter. Later Ashot will attach every price tag to some kind of fruits, and Valera will be able to count the total price of all fruits from his list. But Valera wants to know now what can be the smallest total price (in case of the most «lucky» for him distribution of price tags) and the largest total price (in case of the most «unlucky» for him distribution of price tags). | The first line of the input contains two integer number n and m (1 ≤ n, m ≤ 100) — the number of price tags (which is equal to the number of different kinds of fruits that Ashot sells) and the number of items in Valera's list. The second line contains n space-separated positive integer numbers. Each of them doesn't exceed 100 and stands for the price of one fruit of some kind. The following m lines contain names of the fruits from the list. Each name is a non-empty string of small Latin letters which length doesn't exceed 32. It is guaranteed that the number of distinct fruits from the list is less of equal to n. Also it is known that the seller has in stock all fruits that Valera wants to buy. | Print two numbers a and b (a ≤ b) — the minimum and the maximum possible sum which Valera may need to buy all fruits from his list. | null | null | [{"input": "5 3\n4 2 1 10 5\napple\norange\nmango", "output": "7 19"}, {"input": "6 5\n3 5 1 6 8 1\npeach\ngrapefruit\nbanana\norange\norange", "output": "11 30"}] | 1,100 | ["greedy", "implementation", "sortings"] | 25 | [{"input": "5 3\r\n4 2 1 10 5\r\napple\r\norange\r\nmango\r\n", "output": "7 19\r\n"}, {"input": "6 5\r\n3 5 1 6 8 1\r\npeach\r\ngrapefruit\r\nbanana\r\norange\r\norange\r\n", "output": "11 30\r\n"}, {"input": "2 2\r\n91 82\r\neiiofpfpmemlakcystpun\r\nmcnzeiiofpfpmemlakcystpunfl\r\n", "output": "173 173\r\n"}, {"input": "1 4\r\n1\r\nu\r\nu\r\nu\r\nu\r\n", "output": "4 4\r\n"}, {"input": "3 3\r\n4 2 3\r\nwivujdxzjm\r\nawagljmtc\r\nwivujdxzjm\r\n", "output": "7 11\r\n"}, {"input": "3 4\r\n10 10 10\r\nodchpcsdhldqnkbhwtwnx\r\nldqnkbhwtwnxk\r\nodchpcsdhldqnkbhwtwnx\r\nldqnkbhwtwnxk\r\n", "output": "40 40\r\n"}, {"input": "3 1\r\n14 26 22\r\naag\r\n", "output": "14 26\r\n"}, {"input": "2 2\r\n5 5\r\ndcypj\r\npiyqiagzjlvbhgfndhfu\r\n", "output": "10 10\r\n"}, {"input": "4 3\r\n5 3 10 3\r\nxzjhplrzkbbzkypfazf\r\nxzjhplrzkbbzkypfazf\r\nh\r\n", "output": "9 25\r\n"}, {"input": "5 5\r\n10 10 6 7 9\r\niyerjkvzibxhllkeuagptnoqrzm\r\nvzibxhllkeuag\r\niyerjkvzibxhllkeuagptnoqrzm\r\nnoq\r\nnoq\r\n", "output": "35 49\r\n"}, {"input": "10 8\r\n19 18 20 13 19 13 11 10 19 16\r\nkayangqlsqmcd\r\nqls\r\nqydawlbludrgrjfjrhd\r\nfjrh\r\nqls\r\nqls\r\nrnmmayh\r\nkayangqlsqmcd\r\n", "output": "94 154\r\n"}, {"input": "5 15\r\n61 56 95 42 85\r\noq\r\ndwxivk\r\ntxdxzsfdj\r\noq\r\noq\r\ndwxivk\r\ntxdxzsfdj\r\ndwxivk\r\ntxdxzsfdj\r\nk\r\nk\r\ndwxivk\r\noq\r\nk\r\ntxdxzsfdj\r\n", "output": "891 1132\r\n"}, {"input": "12 18\r\n42 44 69 16 81 64 12 68 70 75 75 67\r\nfm\r\nqamklzfmrjnqgdspwfasjnplg\r\nqamklzfmrjnqgdspwfasjnplg\r\nqamklzfmrjnqgdspwfasjnplg\r\nl\r\nl\r\nl\r\nfm\r\nqamklzfmrjnqgdspwfasjnplg\r\nl\r\nnplgwotfm\r\np\r\nl\r\namklzfm\r\ntkpubqamklzfmrjn\r\npwf\r\nfm\r\np\r\n", "output": "606 1338\r\n"}] | false | stdio | null | true |
755/B | 755 | B | PyPy 3 | TESTS | 20 | 249 | 5,939,200 | 91597448 | # cook your dish here
n, m = map(int, input().split())
Ene = set()
Pol = set()
for i in range(n):
Pol.add(input())
for i in range(m):
Ene.add(input())
comm_ele = 0
for num in Pol:
if num in Ene:
comm_ele += 1
if n>m:
print("YES")
elif n<m:
print("NO")
else:
if comm_ele == n:
print("NO") if n%2 ==0 else print("YES")
elif comm_ele == 0:
print("NO")
else:
print("NO") if comm_ele%2 == n%2 else print("YES") | 33 | 46 | 819,200 | 138663974 | n, m = input().split()
n = int(n)
m = int(m)
words = set()
repeated = 0
p = 0
e = 0
for i in range(n):
x = input()
words.add(x)
p += 1
for i in range(m):
x = input()
if x in words:
p -= 1
repeated += 1
else:
e += 1
if repeated >= 2:
if repeated % 2 == 1:
p += repeated//2 + 1
e += repeated//2
else:
p += repeated//2
e += repeated//2
else:
p += repeated
if p > e:
print('YES')
else:
print('NO') | 8VC Venture Cup 2017 - Elimination Round | CF | 2,017 | 1 | 256 | PolandBall and Game | PolandBall is playing a game with EnemyBall. The rules are simple. Players have to say words in turns. You cannot say a word which was already said. PolandBall starts. The Ball which can't say a new word loses.
You're given two lists of words familiar to PolandBall and EnemyBall. Can you determine who wins the game, if both play optimally? | The first input line contains two integers n and m (1 ≤ n, m ≤ 103) — number of words PolandBall and EnemyBall know, respectively.
Then n strings follow, one per line — words familiar to PolandBall.
Then m strings follow, one per line — words familiar to EnemyBall.
Note that one Ball cannot know a word more than once (strings are unique), but some words can be known by both players.
Each word is non-empty and consists of no more than 500 lowercase English alphabet letters. | In a single line of print the answer — "YES" if PolandBall wins and "NO" otherwise. Both Balls play optimally. | null | In the first example PolandBall knows much more words and wins effortlessly.
In the second example if PolandBall says kremowka first, then EnemyBall cannot use that word anymore. EnemyBall can only say wiedenska. PolandBall says wadowicka and wins. | [{"input": "5 1\npolandball\nis\na\ncool\ncharacter\nnope", "output": "YES"}, {"input": "2 2\nkremowka\nwadowicka\nkremowka\nwiedenska", "output": "YES"}, {"input": "1 2\na\na\nb", "output": "NO"}] | 1,100 | ["binary search", "data structures", "games", "greedy", "sortings", "strings"] | 33 | [{"input": "5 1\r\npolandball\r\nis\r\na\r\ncool\r\ncharacter\r\nnope\r\n", "output": "YES"}, {"input": "2 2\r\nkremowka\r\nwadowicka\r\nkremowka\r\nwiedenska\r\n", "output": "YES"}, {"input": "1 2\r\na\r\na\r\nb\r\n", "output": "NO"}, {"input": "2 2\r\na\r\nb\r\nb\r\nc\r\n", "output": "YES"}, {"input": "2 1\r\nc\r\na\r\na\r\n", "output": "YES"}, {"input": "3 3\r\nab\r\nbc\r\ncd\r\ncd\r\ndf\r\nfg\r\n", "output": "YES"}, {"input": "3 3\r\nc\r\na\r\nb\r\na\r\nd\r\ng\r\n", "output": "YES"}, {"input": "1 1\r\naa\r\naa\r\n", "output": "YES"}, {"input": "2 1\r\na\r\nb\r\na\r\n", "output": "YES"}, {"input": "6 5\r\na\r\nb\r\nc\r\nd\r\ne\r\nf\r\nf\r\ne\r\nd\r\nz\r\ny\r\n", "output": "YES"}, {"input": "3 2\r\na\r\nb\r\nc\r\nd\r\ne\r\n", "output": "YES"}] | false | stdio | null | true |
175/C | 175 | C | PyPy 3-64 | TESTS | 3 | 92 | 0 | 166126786 | '''\https://codeforces.com/contest/175/problem/C
输入 n(≤100) 表示 n 种怪物，然后输入 n 行，每行两个数字表示怪物的数量 (≤1e9)，和怪物的分数(≤1000)。
然后输入 t(≤100) 和一个长为 t 的数组 p，下标从 1 开始，1≤p[1]<p[2]<...<p[t]≤1e12，
表示在你累计击败 p[i] 个怪物之后，得分系数将变为 i+1（初始得分系数为 1）。
击败一只怪物的得分 = 怪物的分数 * 当前得分系数。
你可以按照任意顺序打怪，输出击败所有怪物后的最大得分。
输入
1
5 10
2
3 6
输出 70
解释 前三只怪物得分系数为 1，后两只怪物得分系数为 2，总分=3*10*1+2*10*2=70
输入
2
3 8
5 10
1
20
输出 74
解释 所有怪物的得分系数均为 1，总分=3*8*1+5*10*1=74
'''
n = int(input())
arr = []
for _ in range(n):
a, b = map(int, input().split())
arr.append((a, b))
arr.sort(key=lambda x: x[1])
t = int(input())
nums = list(map(int, input().split()))
res = 0
j = 0
pre = 0
for k, v in arr:
while j<t and pre + k >= nums[j]:
res += (nums[j] - pre) * v * (j + 1)
k -= nums[j] - pre
pre = nums[j]
j += 1
if j>=t:break
res += v * k * (j + 1)
nums[j] -= k
pre += k
print(res) | 90 | 218 | 0 | 119527083 | n=int(input())
a=[list(map(int,input().split()))[::-1] for i in range(n)]
t=int(input())
p=list(map(int,input().split()))
b=0
i=0
a.sort()
c=0
for j in range(n):
while i<t and p[i]-b<=a[j][1]:
c+=(p[i]-b)*(i+1)*a[j][0]
a[j][1]-=p[i]-b
b=p[i]
i+=1
c+=a[j][1]*(i+1)*a[j][0]
b+=a[j][1]
print(c) | Codeforces Round 115 | CF | 2,012 | 2 | 256 | Geometry Horse | Vasya plays the Geometry Horse.
The game goal is to destroy geometric figures of the game world. A certain number of points is given for destroying each figure depending on the figure type and the current factor value.
There are n types of geometric figures. The number of figures of type ki and figure cost ci is known for each figure type. A player gets ci·f points for destroying one figure of type i, where f is the current factor. The factor value can be an integer number from 1 to t + 1, inclusive. At the beginning of the game the factor value is equal to 1. The factor is set to i + 1 after destruction of pi (1 ≤ i ≤ t) figures, so the (pi + 1)-th figure to be destroyed is considered with factor equal to i + 1.
Your task is to determine the maximum number of points Vasya can get after he destroys all figures. Take into account that Vasya is so tough that he can destroy figures in any order chosen by him. | The first line contains the only integer number n (1 ≤ n ≤ 100) — the number of figure types.
Each of the following n lines contains two integer numbers ki and ci (1 ≤ ki ≤ 109, 0 ≤ ci ≤ 1000), separated with space — the number of figures of the i-th type and the cost of one i-type figure, correspondingly.
The next line contains the only integer number t (1 ≤ t ≤ 100) — the number that describe the factor's changes.
The next line contains t integer numbers pi (1 ≤ p1 < p2 < ... < pt ≤ 1012), separated with spaces.
Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator. | Print the only number — the maximum number of points Vasya can get. | null | In the first example Vasya destroys three figures first and gets 3·1·10 = 30 points. Then the factor will become equal to 2 and after destroying the last two figures Vasya will get 2·2·10 = 40 points. As a result Vasya will get 70 points.
In the second example all 8 figures will be destroyed with factor 1, so Vasya will get (3·8 + 5·10)·1 = 74 points. | [{"input": "1\n5 10\n2\n3 6", "output": "70"}, {"input": "2\n3 8\n5 10\n1\n20", "output": "74"}] | 1,600 | ["greedy", "implementation", "sortings", "two pointers"] | 90 | [{"input": "1\r\n5 10\r\n2\r\n3 6\r\n", "output": "70"}, {"input": "2\r\n3 8\r\n5 10\r\n1\r\n20\r\n", "output": "74"}, {"input": "3\r\n10 3\r\n20 2\r\n30 1\r\n3\r\n30 50 60\r\n", "output": "200"}, {"input": "1\r\n100 1000\r\n1\r\n1\r\n", "output": "199000"}, {"input": "1\r\n1 1000\r\n1\r\n1\r\n", "output": "1000"}, {"input": "1\r\n1 1000\r\n1\r\n2\r\n", "output": "1000"}, {"input": "2\r\n1000000000 1000\r\n1 1\r\n1\r\n10\r\n", "output": "1999999991001"}, {"input": "6\r\n5 9\r\n63 3\r\n30 4\r\n25 6\r\n48 2\r\n29 9\r\n8\r\n105 137 172 192 632 722 972 981\r\n", "output": "2251"}, {"input": "7\r\n9902 9\r\n5809 6\r\n2358 0\r\n6868 7\r\n9630 2\r\n8302 10\r\n9422 3\r\n4\r\n2148 4563 8488 9575\r\n", "output": "1481866"}, {"input": "9\r\n60129 6\r\n44235 10\r\n13131 8\r\n2012 2\r\n27536 4\r\n38950 6\r\n39080 2\r\n13892 3\r\n48709 0\r\n1\r\n23853\r\n", "output": "2751752"}, {"input": "10\r\n3466127 4\r\n3477072 1\r\n9690039 9\r\n9885165 6\r\n2559197 4\r\n3448456 3\r\n9169542 1\r\n6915866 2\r\n1702896 10\r\n8934261 5\r\n6\r\n3041416 5811699 5920083 8250213 8694306 8899250\r\n", "output": "1843409345"}, {"input": "4\r\n4059578 5\r\n20774712 1\r\n64867825 7\r\n5606945 8\r\n1\r\n337246111\r\n", "output": "540002937"}, {"input": "1\r\n555 100\r\n10\r\n1 2 3 4 5 6 7 8 9 10\r\n", "output": "605000"}, {"input": "1\r\n1 1\r\n1\r\n100000000000\r\n", "output": "1"}, {"input": "12\r\n1000000000 1\r\n1000000000 2\r\n1000000000 3\r\n1000000000 4\r\n1000000000 5\r\n1000000000 6\r\n1000000000 7\r\n1000000000 8\r\n1000000000 9\r\n1000000000 10\r\n1000000000 11\r\n1000000000 12\r\n1\r\n10000000000\r\n", "output": "101000000000"}, {"input": "11\r\n1000000000 1\r\n1000000000 2\r\n1000000000 3\r\n1000000000 4\r\n1000000000 5\r\n1000000000 6\r\n1000000000 7\r\n1000000000 8\r\n1000000000 9\r\n1000000000 10\r\n1000000000 11\r\n1\r\n10000000000\r\n", "output": "77000000000"}, {"input": "1\r\n10 10\r\n3\r\n1 2 3\r\n", "output": "340"}, {"input": "1\r\n1000000000 1000\r\n2\r\n3 6\r\n", "output": "2999999991000"}, {"input": "1\r\n100 100\r\n3\r\n3 6 9\r\n", "output": "38200"}, {"input": "1\r\n10 1\r\n10\r\n1 2 3 4 5 6 7 8 9 10\r\n", "output": "55"}, {"input": "1\r\n10 10\r\n5\r\n1 2 3 4 5\r\n", "output": "450"}, {"input": "10\r\n10 10\r\n10 10\r\n10 10\r\n10 10\r\n10 10\r\n10 10\r\n10 10\r\n10 10\r\n10 10\r\n10 10\r\n1\r\n1\r\n", "output": "1990"}, {"input": "1\r\n10 10\r\n2\r\n3 6\r\n", "output": "210"}, {"input": "10\r\n1000 1000\r\n1000 1000\r\n1000 1000\r\n1000 1000\r\n1000 1000\r\n1000 1000\r\n1000 1000\r\n1000 1000\r\n1000 1000\r\n1000 1000\r\n1\r\n1000000\r\n", "output": "10000000"}] | false | stdio | null | true |
12/C | 12 | C | Python 3 | TESTS | 1 | 61 | 307,200 | 105277235 | n,k=map(int,input().split())
a=[int(x) for x in input().split()]
a.sort()
dic={}
for _ in range(k):
s=input()
if s in dic:
dic[s]+=1
else:
dic[s]=1
t=0
cnt=0
sorted(dic.items(),reverse=True)
for i in dic:
cnt+=a[t]*dic[i]
t+=1
print(cnt,end=" ")
t=n-1
cnt=0
for i in dic:
cnt+=a[t]*dic[i]
t-=1
print(cnt) | 25 | 46 | 0 | 147141946 | n,m=list(map(int,input().split()))
p=list(map(int,input().split()))
f=[]
k=[]
for i in range(m):
f.append(input())
for i in set(f):
k.append(f.count(i))
k.sort(reverse=True)
p.sort()
h=0
min=0
max=0
for i in k:
min+=i*p[h]
h+=1
max+=i*p[-h]
print(min,max) | Codeforces Beta Round 12 (Div 2 Only) | ICPC | 2,010 | 1 | 256 | Fruits | The spring is coming and it means that a lot of fruits appear on the counters. One sunny day little boy Valera decided to go shopping. He made a list of m fruits he wanted to buy. If Valera want to buy more than one fruit of some kind, he includes it into the list several times.
When he came to the fruit stall of Ashot, he saw that the seller hadn't distributed price tags to the goods, but put all price tags on the counter. Later Ashot will attach every price tag to some kind of fruits, and Valera will be able to count the total price of all fruits from his list. But Valera wants to know now what can be the smallest total price (in case of the most «lucky» for him distribution of price tags) and the largest total price (in case of the most «unlucky» for him distribution of price tags). | The first line of the input contains two integer number n and m (1 ≤ n, m ≤ 100) — the number of price tags (which is equal to the number of different kinds of fruits that Ashot sells) and the number of items in Valera's list. The second line contains n space-separated positive integer numbers. Each of them doesn't exceed 100 and stands for the price of one fruit of some kind. The following m lines contain names of the fruits from the list. Each name is a non-empty string of small Latin letters which length doesn't exceed 32. It is guaranteed that the number of distinct fruits from the list is less of equal to n. Also it is known that the seller has in stock all fruits that Valera wants to buy. | Print two numbers a and b (a ≤ b) — the minimum and the maximum possible sum which Valera may need to buy all fruits from his list. | null | null | [{"input": "5 3\n4 2 1 10 5\napple\norange\nmango", "output": "7 19"}, {"input": "6 5\n3 5 1 6 8 1\npeach\ngrapefruit\nbanana\norange\norange", "output": "11 30"}] | 1,100 | ["greedy", "implementation", "sortings"] | 25 | [{"input": "5 3\r\n4 2 1 10 5\r\napple\r\norange\r\nmango\r\n", "output": "7 19\r\n"}, {"input": "6 5\r\n3 5 1 6 8 1\r\npeach\r\ngrapefruit\r\nbanana\r\norange\r\norange\r\n", "output": "11 30\r\n"}, {"input": "2 2\r\n91 82\r\neiiofpfpmemlakcystpun\r\nmcnzeiiofpfpmemlakcystpunfl\r\n", "output": "173 173\r\n"}, {"input": "1 4\r\n1\r\nu\r\nu\r\nu\r\nu\r\n", "output": "4 4\r\n"}, {"input": "3 3\r\n4 2 3\r\nwivujdxzjm\r\nawagljmtc\r\nwivujdxzjm\r\n", "output": "7 11\r\n"}, {"input": "3 4\r\n10 10 10\r\nodchpcsdhldqnkbhwtwnx\r\nldqnkbhwtwnxk\r\nodchpcsdhldqnkbhwtwnx\r\nldqnkbhwtwnxk\r\n", "output": "40 40\r\n"}, {"input": "3 1\r\n14 26 22\r\naag\r\n", "output": "14 26\r\n"}, {"input": "2 2\r\n5 5\r\ndcypj\r\npiyqiagzjlvbhgfndhfu\r\n", "output": "10 10\r\n"}, {"input": "4 3\r\n5 3 10 3\r\nxzjhplrzkbbzkypfazf\r\nxzjhplrzkbbzkypfazf\r\nh\r\n", "output": "9 25\r\n"}, {"input": "5 5\r\n10 10 6 7 9\r\niyerjkvzibxhllkeuagptnoqrzm\r\nvzibxhllkeuag\r\niyerjkvzibxhllkeuagptnoqrzm\r\nnoq\r\nnoq\r\n", "output": "35 49\r\n"}, {"input": "10 8\r\n19 18 20 13 19 13 11 10 19 16\r\nkayangqlsqmcd\r\nqls\r\nqydawlbludrgrjfjrhd\r\nfjrh\r\nqls\r\nqls\r\nrnmmayh\r\nkayangqlsqmcd\r\n", "output": "94 154\r\n"}, {"input": "5 15\r\n61 56 95 42 85\r\noq\r\ndwxivk\r\ntxdxzsfdj\r\noq\r\noq\r\ndwxivk\r\ntxdxzsfdj\r\ndwxivk\r\ntxdxzsfdj\r\nk\r\nk\r\ndwxivk\r\noq\r\nk\r\ntxdxzsfdj\r\n", "output": "891 1132\r\n"}, {"input": "12 18\r\n42 44 69 16 81 64 12 68 70 75 75 67\r\nfm\r\nqamklzfmrjnqgdspwfasjnplg\r\nqamklzfmrjnqgdspwfasjnplg\r\nqamklzfmrjnqgdspwfasjnplg\r\nl\r\nl\r\nl\r\nfm\r\nqamklzfmrjnqgdspwfasjnplg\r\nl\r\nnplgwotfm\r\np\r\nl\r\namklzfm\r\ntkpubqamklzfmrjn\r\npwf\r\nfm\r\np\r\n", "output": "606 1338\r\n"}] | false | stdio | null | true |
455/B | 455 | B | PyPy 3-64 | TESTS | 4 | 62 | 0 | 214539881 | n, k = [int(x) for x in input().split()]
l = []
for i in range(n):
l.append(len(input()))
if min(l) % 2 == 0:
print("Second")
else:
if k % 2 == 0:
print("Second")
else:
print("First") | 75 | 218 | 26,521,600 | 230712465 | # LUOGU_RID: 132709230
# pypy3
from collections import *
from itertools import *
from functools import *
from bisect import *
from heapq import *
import sys
from math import gcd
IN = lambda: sys.stdin.readline().rstrip("\r\n")
PN = lambda x: sys.stdout.write(x)
I = lambda: int(IN())
S = lambda: IN().split()
M = lambda: map(int, IN().split())
L = lambda: list(map(int, IN().split()))
G = lambda: map(lambda x: int(x) - 1, IN().split())
tr = [[0 for i in range(26)] for _ in range(100010)]
son = [0 for i in range(100010)]
idx = 0
def ins(s):
p = 0
global idx
for c in s:
u = ord(c) - ord('a')
if tr[p][u] == 0:
idx += 1
tr[p][u] = idx
son[p] += 1
p = tr[p][u]
n, k = M()
for _ in range(n):
ins(IN())
def win(p):
for i in range(26):
if tr[p][i] != 0 and not win(tr[p][i]):
return True
return False
def lose(p):
for i in range(26):
if tr[p][i] != 0 and not lose(tr[p][i]):
return True
if son[p] == 0:
return True
return False
f, g = win(0), lose(0)
if f:
if g:
print("First")
else:
print("First" if k % 2 == 1 else "Second")
else:
print("Second") | Codeforces Round 260 (Div. 1) | CF | 2,014 | 1 | 256 | A Lot of Games | Andrew, Fedor and Alex are inventive guys. Now they invent the game with strings for two players.
Given a group of n non-empty strings. During the game two players build the word together, initially the word is empty. The players move in turns. On his step player must add a single letter in the end of the word, the resulting word must be prefix of at least one string from the group. A player loses if he cannot move.
Andrew and Alex decided to play this game k times. The player who is the loser of the i-th game makes the first move in the (i + 1)-th game. Guys decided that the winner of all games is the player who wins the last (k-th) game. Andrew and Alex already started the game. Fedor wants to know who wins the game if both players will play optimally. Help him. | The first line contains two integers, n and k (1 ≤ n ≤ 105; 1 ≤ k ≤ 109).
Each of the next n lines contains a single non-empty string from the given group. The total length of all strings from the group doesn't exceed 105. Each string of the group consists only of lowercase English letters. | If the player who moves first wins, print "First", otherwise print "Second" (without the quotes). | null | null | [{"input": "2 3\na\nb", "output": "First"}, {"input": "3 1\na\nb\nc", "output": "First"}, {"input": "1 2\nab", "output": "Second"}] | 1,900 | ["dfs and similar", "dp", "games", "implementation", "strings", "trees"] | 75 | [{"input": "2 3\r\na\r\nb\r\n", "output": "First\r\n"}, {"input": "3 1\r\na\r\nb\r\nc\r\n", "output": "First\r\n"}, {"input": "1 2\r\nab\r\n", "output": "Second\r\n"}, {"input": "5 6\r\nabas\r\ndsfdf\r\nabacaba\r\ndartsidius\r\nkolobok\r\n", "output": "Second\r\n"}, {"input": "4 2\r\naaaa\r\nbbbb\r\nccccc\r\ndumbavumba\r\n", "output": "First\r\n"}, {"input": "3 8\r\nso\r\nbad\r\ntest\r\n", "output": "First\r\n"}, {"input": "5 2\r\nwelcome\r\nto\r\nthe\r\nmatrix\r\nneo\r\n", "output": "First\r\n"}, {"input": "6 4\r\ndog\r\ncat\r\ncow\r\nhot\r\nice\r\nlol\r\n", "output": "Second\r\n"}, {"input": "4 8\r\nla\r\na\r\nz\r\nka\r\n", "output": "First\r\n"}, {"input": "3 2\r\nop\r\nhop\r\ncop\r\n", "output": "First\r\n"}, {"input": "3 3\r\nabasdfabab\r\nabaaasdfdsf\r\nasdfaba\r\n", "output": "Second\r\n"}, {"input": "2 2\r\naba\r\naa\r\n", "output": "Second\r\n"}, {"input": "4 1\r\naa\r\naba\r\nba\r\nbba\r\n", "output": "Second\r\n"}, {"input": "1 3\r\nab\r\n", "output": "Second\r\n"}, {"input": "3 3\r\naa\r\nabb\r\ncc\r\n", "output": "Second\r\n"}] | false | stdio | null | true |
10/A | 10 | A | PyPy 3 | TESTS | 16 | 280 | 0 | 84325696 | n,p1,p2,p3,t1,t2 = map(int,input().split())
a=0
b=0
for c in range(n):
l,r = map(int,input().split())
if(c==0):
b+=(r-l)*p1
a=r
else:
diff = l-a
if(diff > 0):
if(diff <= t1):
b+= diff*p1
else:
if(diff < t1+t2 and diff >= t1):
b+= t1*p1 + (diff-t1)*p2
elif(diff > t1+t2):
b+= t1*p1 + (t2)*p2 + (diff-(t1+t2))*p3
b+=(r-l)*p1
a=r
print(b) | 30 | 62 | 0 | 186159465 | def main():
n, p1, p2, p3, t1, t2 = [int(_) for _ in input().split()]
minutes = [0, 0, 0]
r_prev = 0
for _ in range(n):
l, r = [int(_) for _ in input().split()]
minutes[0] += r - l
if r_prev:
rest = l - r_prev
if rest <= t1:
minutes[0] += rest
else:
minutes[0] += t1
rest -= t1
if rest <= t2:
minutes[1] += rest
else:
minutes[1] += t2
rest -= t2
minutes[2] += rest
r_prev = r
print(minutes[0] * p1 + minutes[1] * p2 + minutes[2] * p3)
if __name__ == "__main__":
main() | 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 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes P3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into n time periods [l1, r1], [l2, r2], ..., [ln, rn]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [l1, rn]. | 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 and the end of the i-th period of work. | 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 41 20 33 43 4\r\n46 465\r\n598 875\r\n967 980\r\n1135 1151\r\n1194 1245\r\n", "output": "46995"}, {"input": "6 88 28 100 53 36\r\n440 445\r\n525 614\r\n644 844\r\n1238 1261\r\n1305 1307\r\n1425 1434\r\n", "output": "85540"}, {"input": "7 46 61 55 28 59\r\n24 26\r\n31 61\r\n66 133\r\n161 612\r\n741 746\r\n771 849\r\n1345 1357\r\n", "output": "67147"}, {"input": "8 83 18 30 28 5\r\n196 249\r\n313 544\r\n585 630\r\n718 843\r\n1040 1194\r\n1207 1246\r\n1268 1370\r\n1414 1422\r\n", "output": "85876"}, {"input": "9 31 65 27 53 54\r\n164 176\r\n194 210\r\n485 538\r\n617 690\r\n875 886\r\n888 902\r\n955 957\r\n1020 1200\r\n1205 1282\r\n", "output": "38570"}] | false | stdio | null | true |
59/E | 59 | E | PyPy 3 | TESTS | 6 | 902 | 51,712,000 | 92545286 | import os, sys
from io import IOBase, BytesIO
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill():
pass
return super(FastIO, self).read()
def readline(self):
while self.newlines == 0:
s = self._fill()
self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# Cout implemented in Python
import sys
class ostream:
def __lshift__(self, a):
sys.stdout.write(str(a))
return self
cout = ostream()
endl = "\n"
import heapq
def solve():
n, m, k_num = map(int, input().split())
edge_list = {x: [] for x in range(1, n + 1)}
k = {}
for i in range(m):
u, v = map(int, input().split())
edge_list[u].append(v)
edge_list[v].append(u)
for i in range(k_num):
a, b, c = map(int, input().split())
k[(a, b, c)] = True
hq = [(0, 1)]
dist = [[float("inf")] * (n + 1) for i in range(n + 1)]
dist[0][1] = 0
path = [[0] * (n + 1) for i in range(n + 1)]
Found = None
while hq:
ele = heapq.heappop(hq)
# visited[ele[1]] = True
for edge in edge_list[ele[1]]:
# print(ele[0], ele[1], edge)
if dist[ele[0]][ele[1]] + 1 < dist[ele[1]][edge] and (ele[0], ele[1], edge) not in k:
dist[ele[1]][edge] = dist[ele[0]][ele[1]] + 1
path[ele[1]][edge] = ele[0]
heapq.heappush(hq, (ele[1], edge))
if edge == n:
Found = ele[1]
break
if not Found:
cout<<-1<<"\n"
return
# print(path)
pt = [n, Found]
st = path[Found][n]
while st:
pt.append(st)
st = path[pt[-1]][pt[-2]]
cout<<len(pt) - 1<<"\n"
cout<< " ".join(map(str, pt[::-1]))<<"\n"
def main():
solve()
if __name__ == "__main__":
main() | 43 | 1,090 | 32,256,000 | 222025354 | import sys
from collections import deque
readline = sys.stdin.readline
N = 0
graph = []
forbbiden = set()
def read_input():
global N
n, m, k = [int(w) for w in readline().split()]
for _ in range(n + 1):
graph.append([])
for _ in range(m):
u, v = [int(w) for w in readline().split()]
graph[u].append(v)
graph[v].append(u)
for _ in range(k):
forbbiden.add(tuple(int(w) for w in readline().split()))
N = n
def solve()->list:
def get_trail(t:tuple) -> list:
buf = []
while t in seen:
buf.append(t)
t = seen[t]
return [tmp[0] for tmp in buf][::-1] + [N]
seen = {}
que = deque() # parent, node, step
que.append((0, 1, 0))
while que:
p, node, step = que.popleft()
for neigh in graph[node]:
if (p, node, neigh) in forbbiden or (node, neigh) in seen:
continue
seen[(node, neigh)] = (p, node)
if neigh == N:
return get_trail((node, neigh))
que.append((node, neigh, step + 1))
return []
def write_output(result: list):
print(len(result) - 1)
if result:
print(*result)
read_input()
write_output(solve()) | Codeforces Beta Round 55 (Div. 2) | CF | 2,011 | 3 | 256 | Shortest Path | In Ancient Berland there were n cities and m two-way roads of equal length. The cities are numbered with integers from 1 to n inclusively. According to an ancient superstition, if a traveller visits three cities ai, bi, ci in row, without visiting other cities between them, a great disaster awaits him. Overall there are k such city triplets. Each triplet is ordered, which means that, for example, you are allowed to visit the cities in the following order: ai, ci, bi. Vasya wants to get from the city 1 to the city n and not fulfil the superstition. Find out which minimal number of roads he should take. Also you are required to find one of his possible path routes. | The first line contains three integers n, m, k (2 ≤ n ≤ 3000, 1 ≤ m ≤ 20000, 0 ≤ k ≤ 105) which are the number of cities, the number of roads and the number of the forbidden triplets correspondingly.
Then follow m lines each containing two integers xi, yi (1 ≤ xi, yi ≤ n) which are the road descriptions. The road is described by the numbers of the cities it joins. No road joins a city with itself, there cannot be more than one road between a pair of cities.
Then follow k lines each containing three integers ai, bi, ci (1 ≤ ai, bi, ci ≤ n) which are the forbidden triplets. Each ordered triplet is listed mo more than one time. All three cities in each triplet are distinct.
City n can be unreachable from city 1 by roads. | If there are no path from 1 to n print -1. Otherwise on the first line print the number of roads d along the shortest path from the city 1 to the city n. On the second line print d + 1 numbers — any of the possible shortest paths for Vasya. The path should start in the city 1 and end in the city n. | null | null | [{"input": "4 4 1\n1 2\n2 3\n3 4\n1 3\n1 4 3", "output": "2\n1 3 4"}, {"input": "3 1 0\n1 2", "output": "-1"}, {"input": "4 4 2\n1 2\n2 3\n3 4\n1 3\n1 2 3\n1 3 4", "output": "4\n1 3 2 3 4"}] | 2,000 | ["graphs", "shortest paths"] | 43 | [{"input": "4 4 1\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 4 3\r\n", "output": "2\r\n1 3 4\r\n"}, {"input": "3 1 0\r\n1 2\r\n", "output": "-1\r\n"}, {"input": "4 4 2\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 2 3\r\n1 3 4\r\n", "output": "4\r\n1 3 2 3 4\r\n"}, {"input": "4 4 1\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 2 3\r\n", "output": "2\r\n1 3 4\r\n"}, {"input": "2 1 0\r\n1 2\r\n", "output": "1\r\n1 2\r\n"}, {"input": "4 4 1\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 3 4\r\n", "output": "3\r\n1 2 3 4\r\n"}, {"input": "3 2 0\r\n1 2\r\n3 2\r\n", "output": "2\r\n1 2 3\r\n"}, {"input": "3 2 1\r\n1 2\r\n3 2\r\n1 2 3\r\n", "output": "-1\r\n"}, {"input": "4 4 4\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 2 3\r\n1 3 4\r\n1 2 4\r\n1 3 2\r\n", "output": "-1\r\n"}] | false | stdio | null | true |
59/E | 59 | E | Python 3 | TESTS | 6 | 342 | 10,649,600 | 177773499 | import sys
import sys, threading
from heapq import heapify, heappop, heappush
from math import log10, floor, pow, gcd, sqrt, inf
from collections import defaultdict, deque, Counter
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
def main():
cities, roads, triplet = inlt()
edges = defaultdict(list)
triplets = set()
for _ in range(roads):
pointA, pointB = inlt()
edges[pointA].append(pointB)
edges[pointB].append(pointA)
for _ in range(triplet):
triplets.add(tuple(inlt()))
# print(edges)
visited = defaultdict(int)
final = int("1"*(cities+1),2)
qu = deque([[(1, 1<<1 ),"1"]])
counts = 0
while qu:
x, count = qu.popleft()
node, comb = x
# print(comb)
if len(count) >= 3:
danger = (int(count[-3]), int(count[-2]), int(count[-1]))
if danger in triplets:
continue
if x[0] == cities:
print(len(count) -1)
for i in count:
print(i, end=" ")
return
if visited[x] == 2:
continue
visited[x] += 1
for i in edges[node]:
tmp = 1<<i
count += str(i)
qu.append([(i,tmp|comb),count])
count = count[:-1]
print(-1)
main()
# threading.stack_size(1 << 27)
# sys.setrecursionlimit(1 << 30)
# main_thread = threading.Thread(target=main)
# main_thread.start()
# main_thread.join() | 43 | 1,090 | 42,393,600 | 154832292 | import sys
input = sys.stdin.readline
from collections import deque
n, m, k = list(map(int, input().split()))
g = [[] for _ in range(n+1)]
for _ in range(m):
i, j = list(map(int, input().split()))
g[i].append(j)
g[j].append(i)
bad_seq = set()
for _ in range(k):
i, j, k = list(map(int, input().split()))
bad_seq.add((i,j,k))
visited_edges = set()
q = deque()
q.append([1, [-1]])
output = [-1]
while q:
j, path = q.popleft()
if j == n:
output = path[1:] + [j]
break
for elem in g[j]:
if (j, elem) not in visited_edges:
if (path[-1], j, elem) not in bad_seq:
visited_edges.add((j, elem))
q.append((elem, path + [j]))
if len(output) > 1:
print(len(output)-1)
print(*output)
else:
print(-1) | Codeforces Beta Round 55 (Div. 2) | CF | 2,011 | 3 | 256 | Shortest Path | In Ancient Berland there were n cities and m two-way roads of equal length. The cities are numbered with integers from 1 to n inclusively. According to an ancient superstition, if a traveller visits three cities ai, bi, ci in row, without visiting other cities between them, a great disaster awaits him. Overall there are k such city triplets. Each triplet is ordered, which means that, for example, you are allowed to visit the cities in the following order: ai, ci, bi. Vasya wants to get from the city 1 to the city n and not fulfil the superstition. Find out which minimal number of roads he should take. Also you are required to find one of his possible path routes. | The first line contains three integers n, m, k (2 ≤ n ≤ 3000, 1 ≤ m ≤ 20000, 0 ≤ k ≤ 105) which are the number of cities, the number of roads and the number of the forbidden triplets correspondingly.
Then follow m lines each containing two integers xi, yi (1 ≤ xi, yi ≤ n) which are the road descriptions. The road is described by the numbers of the cities it joins. No road joins a city with itself, there cannot be more than one road between a pair of cities.
Then follow k lines each containing three integers ai, bi, ci (1 ≤ ai, bi, ci ≤ n) which are the forbidden triplets. Each ordered triplet is listed mo more than one time. All three cities in each triplet are distinct.
City n can be unreachable from city 1 by roads. | If there are no path from 1 to n print -1. Otherwise on the first line print the number of roads d along the shortest path from the city 1 to the city n. On the second line print d + 1 numbers — any of the possible shortest paths for Vasya. The path should start in the city 1 and end in the city n. | null | null | [{"input": "4 4 1\n1 2\n2 3\n3 4\n1 3\n1 4 3", "output": "2\n1 3 4"}, {"input": "3 1 0\n1 2", "output": "-1"}, {"input": "4 4 2\n1 2\n2 3\n3 4\n1 3\n1 2 3\n1 3 4", "output": "4\n1 3 2 3 4"}] | 2,000 | ["graphs", "shortest paths"] | 43 | [{"input": "4 4 1\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 4 3\r\n", "output": "2\r\n1 3 4\r\n"}, {"input": "3 1 0\r\n1 2\r\n", "output": "-1\r\n"}, {"input": "4 4 2\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 2 3\r\n1 3 4\r\n", "output": "4\r\n1 3 2 3 4\r\n"}, {"input": "4 4 1\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 2 3\r\n", "output": "2\r\n1 3 4\r\n"}, {"input": "2 1 0\r\n1 2\r\n", "output": "1\r\n1 2\r\n"}, {"input": "4 4 1\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 3 4\r\n", "output": "3\r\n1 2 3 4\r\n"}, {"input": "3 2 0\r\n1 2\r\n3 2\r\n", "output": "2\r\n1 2 3\r\n"}, {"input": "3 2 1\r\n1 2\r\n3 2\r\n1 2 3\r\n", "output": "-1\r\n"}, {"input": "4 4 4\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 2 3\r\n1 3 4\r\n1 2 4\r\n1 3 2\r\n", "output": "-1\r\n"}] | false | stdio | null | true |
46/C | 46 | C | Python 3 | TESTS | 2 | 92 | 0 | 115054827 | a = int(input())
s = input()
d = s.count('H')
p = []
for i in range(len(s)):
if i+d > len(s):
n = d+i - len(s)
m = d - n
h = s[:m] + s[-n:]
k = h.count("T")
p.append(k)
else:
h = s[i:d+i]
k = h.count("T")
mi = a
for i in range(len(p)):
if p[i] < mi:
mi = p[i]
print(mi) | 27 | 92 | 0 | 196760951 | # LUOGU_RID: 104227169
n = int(input())
s = input()
print(min((s + s)[i:i+s.count('H')].count('T') for i in range(n))) | School Personal Contest #2 (Winter Computer School 2010/11) - Codeforces Beta Round 43 (ACM-ICPC Rules) | ICPC | 2,010 | 2 | 256 | Hamsters and Tigers | Today there is going to be an unusual performance at the circus — hamsters and tigers will perform together! All of them stand in circle along the arena edge and now the trainer faces a difficult task: he wants to swap the animals' positions so that all the hamsters stood together and all the tigers also stood together. The trainer swaps the animals in pairs not to create a mess. He orders two animals to step out of the circle and swap places. As hamsters feel highly uncomfortable when tigers are nearby as well as tigers get nervous when there's so much potential prey around (consisting not only of hamsters but also of yummier spectators), the trainer wants to spend as little time as possible moving the animals, i.e. he wants to achieve it with the minimal number of swaps. Your task is to help him. | The first line contains number n (2 ≤ n ≤ 1000) which indicates the total number of animals in the arena. The second line contains the description of the animals' positions. The line consists of n symbols "H" and "T". The "H"s correspond to hamsters and the "T"s correspond to tigers. It is guaranteed that at least one hamster and one tiger are present on the arena. The animals are given in the order in which they are located circle-wise, in addition, the last animal stands near the first one. | Print the single number which is the minimal number of swaps that let the trainer to achieve his goal. | null | In the first example we shouldn't move anybody because the animals of each species already stand apart from the other species. In the second example you may swap, for example, the tiger in position 2 with the hamster in position 5 and then — the tiger in position 9 with the hamster in position 7. | [{"input": "3\nHTH", "output": "0"}, {"input": "9\nHTHTHTHHT", "output": "2"}] | 1,600 | ["two pointers"] | 27 | [{"input": "3\r\nHTH\r\n", "output": "0\r\n"}, {"input": "9\r\nHTHTHTHHT\r\n", "output": "2\r\n"}, {"input": "2\r\nTH\r\n", "output": "0\r\n"}, {"input": "4\r\nHTTH\r\n", "output": "0\r\n"}, {"input": "4\r\nHTHT\r\n", "output": "1\r\n"}, {"input": "7\r\nTTTHTTT\r\n", "output": "0\r\n"}, {"input": "8\r\nHHTHHTHH\r\n", "output": "1\r\n"}, {"input": "13\r\nHTTTHHHTTTTHH\r\n", "output": "3\r\n"}, {"input": "20\r\nTTHTHTHHTHTTHHTTTHHH\r\n", "output": "4\r\n"}, {"input": "35\r\nTTTTTTHTTHTTTTTHTTTTTTTTTTTHTHTTTTT\r\n", "output": "3\r\n"}, {"input": "120\r\nTTTTTTTHTHTHTTTTTHTHTTTTHTTTTTTTTTTTTTTTTTTTTHTTHTTTTHTTHTTTTTTTTTTTTTTTHTTTTTTHTHTTHTTTTTTHTTTTTTTTTHTTHTTTTHTTTHTTTTTH\r\n", "output": "14\r\n"}, {"input": "19\r\nHHHHHHHHHHHHHTTTHHH\r\n", "output": "0\r\n"}, {"input": "87\r\nHTHHTTHHHHTHHHHHTTTHHTHHHHTTTTHHHTTHHTHTHTHHTTHTHHTHTHTTHHHTTTTTHTTHHHHHHTHHTHHTHTTHTHH\r\n", "output": "17\r\n"}, {"input": "178\r\nTHHHTHTTTHTTHTTHHHHHTTTHTTHHTHTTTHTHTTTTTHHHTHTHHHTHHHTTTTTTTTHHHHTTHHTHHHHTHTTTHHHHHHTHHTHTTHTHTTTTTTTTTHHTTHHTHTTHHTHHHHHTTHHTTHHTTHHHTTHHTTTTHTHHHTHTTHTHTTTHHHHTHHTHHHTHTTTTTT\r\n", "output": "40\r\n"}] | false | stdio | null | true |
10/A | 10 | A | Python 3 | TESTS | 28 | 218 | 307,200 | 73820064 | n,p1,p2,p3,t1,t2=map(int,input().split())
ans=0
l1,r1=0,0
for i in range(n):
l2,r2=map(int,input().split())
ans+=(r2-l2)*p1
if l1==0:
l1,r1=l2,r2
continue
t=l2-r1
if t<=t1:
ans+=t*p1
else:
ans+=t1*p1
t-=t1
if t<=t2:
ans+=t*p2
else:
ans+=t2*p2
t-=t2
ans+=t*p3
l1,r1=l2,r2
print(ans) | 30 | 62 | 0 | 195229808 | n,p1,p2,p3,T1,T2=map(int,input().split())
total,previousTime=0,-1
for k in range(n):
start,finish=map(int,input().split())
if previousTime<0:
previousTime=start
total +=p1*(finish-start)
timeIdle=start-previousTime
if timeIdle>T1+T2:
total +=(timeIdle-T1-T2)*p3
timeIdle=T1+T2
if timeIdle>T1:
total +=(timeIdle-T1)*p2
timeIdle=T1
total +=timeIdle*p1
previousTime=finish
print(total) | 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 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes P3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into n time periods [l1, r1], [l2, r2], ..., [ln, rn]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [l1, rn]. | 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 and the end of the i-th period of work. | 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 41 20 33 43 4\r\n46 465\r\n598 875\r\n967 980\r\n1135 1151\r\n1194 1245\r\n", "output": "46995"}, {"input": "6 88 28 100 53 36\r\n440 445\r\n525 614\r\n644 844\r\n1238 1261\r\n1305 1307\r\n1425 1434\r\n", "output": "85540"}, {"input": "7 46 61 55 28 59\r\n24 26\r\n31 61\r\n66 133\r\n161 612\r\n741 746\r\n771 849\r\n1345 1357\r\n", "output": "67147"}, {"input": "8 83 18 30 28 5\r\n196 249\r\n313 544\r\n585 630\r\n718 843\r\n1040 1194\r\n1207 1246\r\n1268 1370\r\n1414 1422\r\n", "output": "85876"}, {"input": "9 31 65 27 53 54\r\n164 176\r\n194 210\r\n485 538\r\n617 690\r\n875 886\r\n888 902\r\n955 957\r\n1020 1200\r\n1205 1282\r\n", "output": "38570"}] | false | stdio | null | true |
449/B | 449 | B | PyPy 3 | TESTS | 3 | 967 | 51,712,000 | 98030169 | #===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from functools import cmp_to_key
# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt
from copy import deepcopy
from collections import deque
from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect
#==============================================================================================
#fast I/O region
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.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
#===============================================================================================
#some shortcuts
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var)) #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n") #as stdout.write always print sring.
def testcase(t):
for pp in range(t):
solve(pp)
def printlist(a) :
for p in range(0,len(a)):
out(str(a[p]) + ' ')
def google(p):
print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
y%=(p-1) #not so sure about this. used when y>p-1. if p is prime.
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) : # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
if (n <= 1) : return False
if (n <= 3) : return True
if (n % 2 == 0 or n % 3 == 0) : return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
inf = pow(10,20)
mod = 10**9+7
#===============================================================================================
# code here ;))
def djkistra(g,st,dist,lol,vis): #g contains b,dist(a to b) and dist is initiaalised by 10**9 initiallly
pq = []
dist[st] = 0
heappush(pq,(0,st))
while(len(pq) != 0):
curr = heappop(pq)[1]
for i in range(0,len(g[curr])):
b = g[curr][i][0]
w = g[curr][i][1]
if(dist[b] > dist[curr] + w):
dist[b] = dist[curr]+w
heappush(pq,(dist[b],b))
def modif_djkistra(g,dist,usedtrains):
h = []
for i in range(len(g)):
if(dist[i] != inf):
heappush(h,(dist[i],i))
while(len(h) != 0):
d,curr = heappop(h)
if(d != dist[curr]): #dublicate train with larger length
continue
for to,newd in g[curr]:
if(newd+d<=dist[to]):
usedtrains[to] = False
if(dist[to] > newd+d):
heappush(h,(newd+d,to))
dist[to] = newd+d
def solve(case):
n,m,k = sep()
dist = [inf]*n;dist[0] = 0
g = [[] for i in range(n)]
for i in range(m):
a,b,c = sep()
a-=1
b-=1
g[a].append((b,c))
g[b].append((a,c))
have = []
for i in range(k):
a,b = sep()
a-=1
dist[a] = min(dist[a],b)
g[0].append((a,b))
g[a].append((0,b))
have.append(a)
usedtrain = [True]*n
modif_djkistra(g,dist,usedtrain)
cnt = 0
have = list(set(have))
for i in range(n):
if(usedtrain[i]):
cnt+=1
# print(cnt)
print(k - cnt)
testcase(1)
# testcase(int(inp())) | 45 | 1,886 | 133,222,400 | 202495416 | import os
import sys
import threading
from io import BytesIO, IOBase
from heapq import heappush, heappop, heapify
from collections import defaultdict, deque, Counter
from bisect import bisect_left as bl
from bisect import bisect_right as br
# 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(): return list(input().split())
def main():
n,m,k=rl()
g=[[] for i in range(n+1)]
for _ in range(m):
u,v,x=rl()
g[u].append((v,x,1))
g[v].append((u,x,1))
t=[[] for i in range(n+1)]
for _ in range(k):
s,y=rl()
t[s].append(y)
g[1].append((s,y,2))
g[s].append((1,y,2))
dis=[float('inf') for i in range(n+1)]
dis[1]=0
vis=[False for i in range(n+1)]
q=[]
heappush(q,(0,1))
while q:
cd,cn=heappop(q)
if vis[cn]:continue
vis[cn]=True
for (nn,nw,ty) in g[cn]:
nd=nw+dis[cn]
if nd<dis[nn] and not vis[nn]:
dis[nn]=nd
heappush(q,(dis[nn],nn))
res=0
for i in range(2,n+1):
gr=0
gt=0
for (nn,nw,ty) in g[i]:
if dis[nn]+nw==dis[i]:
if ty==2:gt+=1
else:gr+=1
if gr>0:res+=len(t[i])
else:res+=len(t[i])-1
print(res)
pass
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
# threading.Thread(target=main).start() | Codeforces Round 257 (Div. 1) | CF | 2,014 | 2 | 256 | Jzzhu and Cities | Jzzhu is the president of country A. There are n cities numbered from 1 to n in his country. City 1 is the capital of A. Also there are m roads connecting the cities. One can go from city ui to vi (and vise versa) using the i-th road, the length of this road is xi. Finally, there are k train routes in the country. One can use the i-th train route to go from capital of the country to city si (and vise versa), the length of this route is yi.
Jzzhu doesn't want to waste the money of the country, so he is going to close some of the train routes. Please tell Jzzhu the maximum number of the train routes which can be closed under the following condition: the length of the shortest path from every city to the capital mustn't change. | The first line contains three integers n, m, k (2 ≤ n ≤ 105; 1 ≤ m ≤ 3·105; 1 ≤ k ≤ 105).
Each of the next m lines contains three integers ui, vi, xi (1 ≤ ui, vi ≤ n; ui ≠ vi; 1 ≤ xi ≤ 109).
Each of the next k lines contains two integers si and yi (2 ≤ si ≤ n; 1 ≤ yi ≤ 109).
It is guaranteed that there is at least one way from every city to the capital. Note, that there can be multiple roads between two cities. Also, there can be multiple routes going to the same city from the capital. | Output a single integer representing the maximum number of the train routes which can be closed. | null | null | [{"input": "5 5 3\n1 2 1\n2 3 2\n1 3 3\n3 4 4\n1 5 5\n3 5\n4 5\n5 5", "output": "2"}, {"input": "2 2 3\n1 2 2\n2 1 3\n2 1\n2 2\n2 3", "output": "2"}] | 2,000 | ["graphs", "greedy", "shortest paths"] | 45 | [{"input": "5 5 3\r\n1 2 1\r\n2 3 2\r\n1 3 3\r\n3 4 4\r\n1 5 5\r\n3 5\r\n4 5\r\n5 5\r\n", "output": "2\r\n"}, {"input": "2 2 3\r\n1 2 2\r\n2 1 3\r\n2 1\r\n2 2\r\n2 3\r\n", "output": "2\r\n"}, {"input": "5 4 3\r\n1 2 999999999\r\n2 3 1000000000\r\n3 4 529529529\r\n5 1 524524524\r\n5 524444444\r\n5 529999999\r\n2 1000000000\r\n", "output": "2\r\n"}, {"input": "3 2 5\r\n1 2 2\r\n2 3 4\r\n3 5\r\n3 5\r\n3 5\r\n3 6\r\n3 7\r\n", "output": "4\r\n"}, {"input": "5 5 3\r\n1 2 999999999\r\n2 3 1000000000\r\n3 4 529529529\r\n5 1 524524524\r\n5 3 1000000000\r\n5 524444444\r\n5 529999999\r\n2 1000000000\r\n", "output": "2\r\n"}, {"input": "2 1 5\r\n1 2 4\r\n2 3\r\n2 5\r\n2 4\r\n2 4\r\n2 5\r\n", "output": "4\r\n"}, {"input": "3 3 6\r\n1 2 499999999\r\n2 3 500000000\r\n1 3 999999999\r\n2 499999999\r\n2 500000000\r\n2 499999999\r\n3 999999999\r\n3 1000000000\r\n3 1000000000\r\n", "output": "6\r\n"}, {"input": "2 1 1\r\n1 2 1\r\n2 1000000000\r\n", "output": "1\r\n"}, {"input": "3 2 2\r\n1 2 4\r\n2 3 4\r\n2 2\r\n3 6\r\n", "output": "1\r\n"}, {"input": "5 5 2\r\n1 2 100\r\n2 3 100\r\n3 4 100\r\n4 5 20\r\n2 5 5\r\n5 50\r\n4 1\r\n", "output": "1\r\n"}, {"input": "3 2 2\r\n1 2 100\r\n2 3 1\r\n2 1\r\n3 3\r\n", "output": "1\r\n"}] | false | stdio | null | true |
59/E | 59 | E | PyPy 3-64 | TESTS | 15 | 808 | 34,918,400 | 203823090 | # author: birsnot - Nardos
from collections import defaultdict, deque
from heapq import heappop, heappush
from sys import setrecursionlimit, stdin
from threading import Thread, stack_size
# def input(): return stdin.readline()[:-1]
input = stdin.readline
def I(): return int(input())
def II(): return map(int, input().split())
def IL(): return list(map(int, input().split()))
def SIL(): return sorted(map(int, input().split()))
def solve():
N, M, K = II()
adj = defaultdict(list)
for _ in range(M):
u, v = II()
adj[u].append(v)
adj[v].append(u)
ks = set()
for _ in range(K):
u, v, w = II()
ks.add((u, v, w))
pq = [(0, 1, 0)]
parents = defaultdict(list)
i = 10*M
while pq and i:
d, v, p = heappop(pq)
heappush(parents[v], (d, -p))
i -= 1
for u in adj[v]:
if (p, v, u) not in ks:
if u == N:
ans = deque([v, u])
while v != 1:
_, p = heappop(parents[v])
if (-p, v, u) not in ks:
u, v = v, -p
ans.appendleft(v)
print(d + 1)
print(*ans)
return
heappush(pq, (d + 1, u, v))
print(-1)
def main():
# T = I()
T = 1
for ___ in range(T):
solve()
main()
# setrecursionlimit(100050)
# stack_size(1<<27)
# _mt = Thread(target=main)
# _mt.start()
# _mt.join() | 43 | 1,496 | 161,792,000 | 226494619 | import sys
from array import array # noqa: F401
from collections import deque
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n, m, k = map(int, input().split())
adj = [[] for _ in range(n)]
for u, v in (map(int, input().split()) for _ in range(m)):
adj[u - 1].append(v - 1)
adj[v - 1].append(u - 1)
forbidden = [set() for _ in range(n)]
for a, b, c in (map(int, input().split()) for _ in range(k)):
forbidden[c - 1].add((a - 1, b - 1))
inf = 10**9
dp = [[inf] * n for _ in range(n)]
prev = [[-1] * n for _ in range(n)]
dp[0][0] = 0
dq = deque([(0, 0, 0)])
while dq:
v, p, cost = dq.popleft()
if dp[v][p] < cost:
continue
for dest in adj[v]:
if (p, v) in forbidden[dest] or dp[dest][v] <= cost + 1:
continue
dp[dest][v] = cost + 1
prev[dest][v] = p
dq.append((dest, v, cost + 1))
shortest = inf
p, v = -1, -1
for i in range(n):
if shortest > dp[-1][i]:
shortest = dp[-1][i]
p, v = i, n - 1
if shortest == inf:
print(-1)
else:
ans = []
while v != 0:
ans.append(v + 1)
v, p = p, prev[v][p]
ans.append(1)
print(shortest)
print(*reversed(ans)) | Codeforces Beta Round 55 (Div. 2) | CF | 2,011 | 3 | 256 | Shortest Path | In Ancient Berland there were n cities and m two-way roads of equal length. The cities are numbered with integers from 1 to n inclusively. According to an ancient superstition, if a traveller visits three cities ai, bi, ci in row, without visiting other cities between them, a great disaster awaits him. Overall there are k such city triplets. Each triplet is ordered, which means that, for example, you are allowed to visit the cities in the following order: ai, ci, bi. Vasya wants to get from the city 1 to the city n and not fulfil the superstition. Find out which minimal number of roads he should take. Also you are required to find one of his possible path routes. | The first line contains three integers n, m, k (2 ≤ n ≤ 3000, 1 ≤ m ≤ 20000, 0 ≤ k ≤ 105) which are the number of cities, the number of roads and the number of the forbidden triplets correspondingly.
Then follow m lines each containing two integers xi, yi (1 ≤ xi, yi ≤ n) which are the road descriptions. The road is described by the numbers of the cities it joins. No road joins a city with itself, there cannot be more than one road between a pair of cities.
Then follow k lines each containing three integers ai, bi, ci (1 ≤ ai, bi, ci ≤ n) which are the forbidden triplets. Each ordered triplet is listed mo more than one time. All three cities in each triplet are distinct.
City n can be unreachable from city 1 by roads. | If there are no path from 1 to n print -1. Otherwise on the first line print the number of roads d along the shortest path from the city 1 to the city n. On the second line print d + 1 numbers — any of the possible shortest paths for Vasya. The path should start in the city 1 and end in the city n. | null | null | [{"input": "4 4 1\n1 2\n2 3\n3 4\n1 3\n1 4 3", "output": "2\n1 3 4"}, {"input": "3 1 0\n1 2", "output": "-1"}, {"input": "4 4 2\n1 2\n2 3\n3 4\n1 3\n1 2 3\n1 3 4", "output": "4\n1 3 2 3 4"}] | 2,000 | ["graphs", "shortest paths"] | 43 | [{"input": "4 4 1\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 4 3\r\n", "output": "2\r\n1 3 4\r\n"}, {"input": "3 1 0\r\n1 2\r\n", "output": "-1\r\n"}, {"input": "4 4 2\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 2 3\r\n1 3 4\r\n", "output": "4\r\n1 3 2 3 4\r\n"}, {"input": "4 4 1\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 2 3\r\n", "output": "2\r\n1 3 4\r\n"}, {"input": "2 1 0\r\n1 2\r\n", "output": "1\r\n1 2\r\n"}, {"input": "4 4 1\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 3 4\r\n", "output": "3\r\n1 2 3 4\r\n"}, {"input": "3 2 0\r\n1 2\r\n3 2\r\n", "output": "2\r\n1 2 3\r\n"}, {"input": "3 2 1\r\n1 2\r\n3 2\r\n1 2 3\r\n", "output": "-1\r\n"}, {"input": "4 4 4\r\n1 2\r\n2 3\r\n3 4\r\n1 3\r\n1 2 3\r\n1 3 4\r\n1 2 4\r\n1 3 2\r\n", "output": "-1\r\n"}] | false | stdio | null | true |
169/A | 169 | A | Python 3 | TESTS | 6 | 77 | 7,372,800 | 37508191 | n, a, b = map(int,input().split())
h = list(map(int, input().split()))
h.sort()
ans = h[b] - h[a]
if a == b:
ans = h[b] - h[a-1]
print(ans) | 29 | 46 | 0 | 144504780 | n,a,b=map(int, input().split())
h = sorted([int(i) for i in input().split()])
print(h[b]-h[b-1]) | VK Cup 2012 Round 2 (Unofficial Div. 2 Edition) | CF | 2,012 | 2 | 256 | Chores | Petya and Vasya are brothers. Today is a special day for them as their parents left them home alone and commissioned them to do n chores. Each chore is characterized by a single parameter — its complexity. The complexity of the i-th chore equals hi.
As Petya is older, he wants to take the chores with complexity larger than some value x (hi > x) to leave to Vasya the chores with complexity less than or equal to x (hi ≤ x). The brothers have already decided that Petya will do exactly a chores and Vasya will do exactly b chores (a + b = n).
In how many ways can they choose an integer x so that Petya got exactly a chores and Vasya got exactly b chores? | The first input line contains three integers n, a and b (2 ≤ n ≤ 2000; a, b ≥ 1; a + b = n) — the total number of chores, the number of Petya's chores and the number of Vasya's chores.
The next line contains a sequence of integers h1, h2, ..., hn (1 ≤ hi ≤ 109), hi is the complexity of the i-th chore. The numbers in the given sequence are not necessarily different.
All numbers on the lines are separated by single spaces. | Print the required number of ways to choose an integer value of x. If there are no such ways, print 0. | null | In the first sample the possible values of x are 3, 4 or 5.
In the second sample it is impossible to find such x, that Petya got 3 chores and Vasya got 4. | [{"input": "5 2 3\n6 2 3 100 1", "output": "3"}, {"input": "7 3 4\n1 1 9 1 1 1 1", "output": "0"}] | 800 | ["sortings"] | 29 | [{"input": "5 2 3\r\n6 2 3 100 1\r\n", "output": "3\r\n"}, {"input": "7 3 4\r\n1 1 9 1 1 1 1\r\n", "output": "0\r\n"}, {"input": "2 1 1\r\n10 2\r\n", "output": "8\r\n"}, {"input": "2 1 1\r\n7 7\r\n", "output": "0\r\n"}, {"input": "2 1 1\r\n1 1000000000\r\n", "output": "999999999\r\n"}, {"input": "3 1 2\r\n6 5 5\r\n", "output": "1\r\n"}, {"input": "3 2 1\r\n10 10 8\r\n", "output": "2\r\n"}, {"input": "8 3 5\r\n42 55 61 72 83 10 22 33\r\n", "output": "6\r\n"}, {"input": "10 5 5\r\n1 2 3 4 5 999999999 999999998 999999997 999999996 999999995\r\n", "output": "999999990\r\n"}, {"input": "4 1 3\r\n10 8 7 3\r\n", "output": "2\r\n"}, {"input": "4 2 2\r\n402 10 10 402\r\n", "output": "392\r\n"}, {"input": "4 1 3\r\n10 402 402 10\r\n", "output": "0\r\n"}, {"input": "4 3 1\r\n100 100 200 200\r\n", "output": "0\r\n"}, {"input": "102 101 1\r\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\r\n", "output": "0\r\n"}] | false | stdio | null | true |
169/A | 169 | A | Python 3 | TESTS | 6 | 93 | 0 | 78791133 | n , a , b = map(int,input().split())
l = list(map(int,input().split()))
l.sort()
if a!=b:
print(l[b]-l[a])
else:print(l[b]-l[a-1]) | 29 | 46 | 0 | 145054398 | n, older, younger = (list(map(int,input().split())))
chores = (list(map(int,input().split())))
chores.sort(reverse = True)
c_old = chores[older-1]
chores.sort()
c_young = chores[younger-1]
print((c_old)-(c_young)) | VK Cup 2012 Round 2 (Unofficial Div. 2 Edition) | CF | 2,012 | 2 | 256 | Chores | Petya and Vasya are brothers. Today is a special day for them as their parents left them home alone and commissioned them to do n chores. Each chore is characterized by a single parameter — its complexity. The complexity of the i-th chore equals hi.
As Petya is older, he wants to take the chores with complexity larger than some value x (hi > x) to leave to Vasya the chores with complexity less than or equal to x (hi ≤ x). The brothers have already decided that Petya will do exactly a chores and Vasya will do exactly b chores (a + b = n).
In how many ways can they choose an integer x so that Petya got exactly a chores and Vasya got exactly b chores? | The first input line contains three integers n, a and b (2 ≤ n ≤ 2000; a, b ≥ 1; a + b = n) — the total number of chores, the number of Petya's chores and the number of Vasya's chores.
The next line contains a sequence of integers h1, h2, ..., hn (1 ≤ hi ≤ 109), hi is the complexity of the i-th chore. The numbers in the given sequence are not necessarily different.
All numbers on the lines are separated by single spaces. | Print the required number of ways to choose an integer value of x. If there are no such ways, print 0. | null | In the first sample the possible values of x are 3, 4 or 5.
In the second sample it is impossible to find such x, that Petya got 3 chores and Vasya got 4. | [{"input": "5 2 3\n6 2 3 100 1", "output": "3"}, {"input": "7 3 4\n1 1 9 1 1 1 1", "output": "0"}] | 800 | ["sortings"] | 29 | [{"input": "5 2 3\r\n6 2 3 100 1\r\n", "output": "3\r\n"}, {"input": "7 3 4\r\n1 1 9 1 1 1 1\r\n", "output": "0\r\n"}, {"input": "2 1 1\r\n10 2\r\n", "output": "8\r\n"}, {"input": "2 1 1\r\n7 7\r\n", "output": "0\r\n"}, {"input": "2 1 1\r\n1 1000000000\r\n", "output": "999999999\r\n"}, {"input": "3 1 2\r\n6 5 5\r\n", "output": "1\r\n"}, {"input": "3 2 1\r\n10 10 8\r\n", "output": "2\r\n"}, {"input": "8 3 5\r\n42 55 61 72 83 10 22 33\r\n", "output": "6\r\n"}, {"input": "10 5 5\r\n1 2 3 4 5 999999999 999999998 999999997 999999996 999999995\r\n", "output": "999999990\r\n"}, {"input": "4 1 3\r\n10 8 7 3\r\n", "output": "2\r\n"}, {"input": "4 2 2\r\n402 10 10 402\r\n", "output": "392\r\n"}, {"input": "4 1 3\r\n10 402 402 10\r\n", "output": "0\r\n"}, {"input": "4 3 1\r\n100 100 200 200\r\n", "output": "0\r\n"}, {"input": "102 101 1\r\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\r\n", "output": "0\r\n"}] | false | stdio | null | true |
754/D | 754 | D | PyPy 3-64 | TESTS | 4 | 77 | 2,150,400 | 174838101 | import sys
from bisect import bisect_left
from collections import *
from itertools import *
from math import *
from array import *
from functools import lru_cache
import heapq
import bisect
import random
import io, os
if sys.hexversion == 50923504:
sys.stdin = open('cfinput.txt')
RI = lambda: map(int, sys.stdin.buffer.readline().split())
RS = lambda: map(bytes.decode, sys.stdin.buffer.readline().strip().split())
RILST = lambda: list(RI())
MOD = 10 ** 9 + 7
"""https://codeforces.com/problemset/problem/754/D
输入 n, k (1≤k≤n≤3e5) 和 n 个闭区间，区间左右端点在 [-1e9,1e9] 内，区间的编号从 1 开始。
请你选择 k 个区间，使得这 k 个区间的交集的大小尽量大（只考虑整数），输出这个最大值，以及对应的区间的编号。
思考题：如果改成并集呢？
输入
4 2
1 100
40 70
120 130
125 180
输出
31
1 2
输入
3 2
1 12
15 20
25 30
输出
0
1 2
输入
5 2
1 10
5 15
14 50
30 70
99 100
输出
21
3 4
"""
# ms
def solve(n, k, rs, ps):
rs.sort()
ans = 0
out = []
left = 0
h = []
for i, (l, r) in enumerate(rs):
while len(h) > k - 1:
heapq.heappop(h)
if len(h) == k - 1:
d = min(h[0][0], r) - l + 1
if d > ans:
ans = d
out = h[:]
left = i + 1
heapq.heappush(h, (r, i + 1))
print(ans)
if ans == 0:
return print(*range(1, k + 1))
print(' '.join(map(lambda x: str(x[1]), out)), left)
if __name__ == '__main__':
n, k = RI()
rs = []
ps = []
for _ in range(n):
l, r = RI()
ps.append(l)
ps.append(r)
rs.append((l, r))
solve(n, k, rs, ps) | 77 | 3,353 | 112,844,800 | 174818794 | import os, sys
from io import BytesIO, IOBase
from collections import defaultdict, deque, Counter
from bisect import bisect_left, bisect_right
from heapq import heappush, heappop
from functools import lru_cache
from itertools import accumulate
import math
import sys
from sys import stdout
# sys.setrecursionlimit(10 ** 6)
# Fast IO Region
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.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# sys.setrecursionlimit(2000)
ii = lambda: int(input())
mii = lambda: map(int, input().split())
lmii = lambda: list(map(int, input().split()))
mod = 998244353
c2i = lambda c: ord(c) - ord('a')
i2c = lambda i: chr(ord('a') + i)
def solve():
n, m = mii()
itv = []
for i in range(n):
l, r = mii()
itv.append([l, r, i])
itv.sort()
length = -1
u = -1
hr = []
for i in range(len(itv)):
l, r, j = itv[i]
heappush(hr, [r, l, j])
if len(hr) > m:
heappop(hr)
v = hr[0][0] - l
if v > length and len(hr) == m:
length = v
u = i
if u == -1:
print(0)
print(*list(range(1, m + 1)))
else:
hr = []
for i in range(len(itv)):
l, r, j = itv[i]
heappush(hr, [r, l, j])
if len(hr) > m:
heappop(hr)
if i == u:
print(length + 1)
print(*[hr[k][2] + 1 for k in range(len(hr))])
for _ in range(1):
solve() | Codeforces Round 390 (Div. 2) | CF | 2,017 | 4 | 256 | Fedor and coupons | All our characters have hobbies. The same is true for Fedor. He enjoys shopping in the neighboring supermarket.
The goods in the supermarket have unique integer ids. Also, for every integer there is a product with id equal to this integer. Fedor has n discount coupons, the i-th of them can be used with products with ids ranging from li to ri, inclusive. Today Fedor wants to take exactly k coupons with him.
Fedor wants to choose the k coupons in such a way that the number of such products x that all coupons can be used with this product x is as large as possible (for better understanding, see examples). Fedor wants to save his time as well, so he asks you to choose coupons for him. Help Fedor! | The first line contains two integers n and k (1 ≤ k ≤ n ≤ 3·105) — the number of coupons Fedor has, and the number of coupons he wants to choose.
Each of the next n lines contains two integers li and ri ( - 109 ≤ li ≤ ri ≤ 109) — the description of the i-th coupon. The coupons can be equal. | In the first line print single integer — the maximum number of products with which all the chosen coupons can be used. The products with which at least one coupon cannot be used shouldn't be counted.
In the second line print k distinct integers p1, p2, ..., pk (1 ≤ pi ≤ n) — the ids of the coupons which Fedor should choose.
If there are multiple answers, print any of them. | null | In the first example if we take the first two coupons then all the products with ids in range [40, 70] can be bought with both coupons. There are 31 products in total.
In the second example, no product can be bought with two coupons, that is why the answer is 0. Fedor can choose any two coupons in this example. | [{"input": "4 2\n1 100\n40 70\n120 130\n125 180", "output": "31\n1 2"}, {"input": "3 2\n1 12\n15 20\n25 30", "output": "0\n1 2"}, {"input": "5 2\n1 10\n5 15\n14 50\n30 70\n99 100", "output": "21\n3 4"}] | 2,100 | ["binary search", "data structures", "greedy", "sortings"] | 77 | [{"input": "4 2\r\n1 100\r\n40 70\r\n120 130\r\n125 180\r\n", "output": "31\r\n1 2 \r\n"}, {"input": "3 2\r\n1 12\r\n15 20\r\n25 30\r\n", "output": "0\r\n1 2 \r\n"}, {"input": "5 2\r\n1 10\r\n5 15\r\n14 50\r\n30 70\r\n99 100\r\n", "output": "21\r\n3 4 \r\n"}, {"input": "7 6\r\n-8 6\r\n7 9\r\n-10 -5\r\n-6 10\r\n-7 -3\r\n5 8\r\n4 10\r\n", "output": "0\r\n1 2 3 4 5 6 \r\n"}, {"input": "9 6\r\n-7 -3\r\n-3 10\r\n-6 1\r\n-1 8\r\n-9 4\r\n-7 -6\r\n-5 -3\r\n-10 -2\r\n3 4\r\n", "output": "1\r\n1 2 3 5 7 8 \r\n"}, {"input": "7 7\r\n9 10\r\n-5 3\r\n-6 2\r\n1 6\r\n-9 6\r\n-10 7\r\n-7 -5\r\n", "output": "0\r\n1 2 3 4 5 6 7 \r\n"}, {"input": "23 2\r\n-629722518 -626148345\r\n739975524 825702590\r\n-360913153 -208398929\r\n76588954 101603025\r\n-723230356 -650106339\r\n-117490984 -101920679\r\n-39187628 -2520915\r\n717852164 720343632\r\n-611281114 -579708833\r\n-141791522 -122348148\r\n605078929 699430996\r\n-873386085 -820238799\r\n-922404067 -873522961\r\n7572046 13337057\r\n975081176 977171682\r\n901338407 964254238\r\n325388219 346712972\r\n505189756 516497863\r\n-425326983 -422098946\r\n520670681 522544433\r\n-410872616 -367919621\r\n359488350 447471156\r\n-566203447 -488202136\r\n", "output": "0\r\n1 2 \r\n"}, {"input": "24 21\r\n240694945 246896662\r\n240694930 246896647\r\n240695065 246896782\r\n240695050 246896767\r\n240695080 246896797\r\n240694960 246896677\r\n240694975 246896692\r\n240694825 246896542\r\n240694900 246896617\r\n240694915 246896632\r\n240694885 246896602\r\n240694855 246896572\r\n240694870 246896587\r\n240694795 246896512\r\n240695095 246896812\r\n240695125 246896842\r\n240695005 246896722\r\n240694990 246896707\r\n240695140 246896857\r\n240695020 246896737\r\n240695035 246896752\r\n240694840 246896557\r\n240694810 246896527\r\n240695110 246896827\r\n", "output": "6201418\r\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 17 18 20 21 22 23 \r\n"}, {"input": "1 1\r\n2 2\r\n", "output": "1\r\n1 \r\n"}, {"input": "1 1\r\n-1000000000 1000000000\r\n", "output": "2000000001\r\n1 \r\n"}, {"input": "2 1\r\n-1000000000 -1000000000\r\n1000000000 1000000000\r\n", "output": "1\r\n1 \r\n"}, {"input": "7 3\r\n3 3\r\n-6 -1\r\n6 7\r\n2 8\r\n3 10\r\n-8 0\r\n-3 10\r\n", "output": "6\r\n4 5 7 \r\n"}, {"input": "5 4\r\n4 7\r\n-4 2\r\n-7 -7\r\n-5 -2\r\n-8 -8\r\n", "output": "0\r\n1 2 3 4 \r\n"}, {"input": "7 7\r\n0 7\r\n9 9\r\n-10 -7\r\n5 8\r\n-10 4\r\n-7 0\r\n-3 5\r\n", "output": "0\r\n1 2 3 4 5 6 7 \r\n"}, {"input": "9 2\r\n5 10\r\n-10 -10\r\n0 10\r\n-6 3\r\n-8 7\r\n6 10\r\n-8 1\r\n5 7\r\n2 2\r\n", "output": "10\r\n5 7 \r\n"}, {"input": "9 5\r\n-2 1\r\n-6 9\r\n-7 -2\r\n5 7\r\n-10 -7\r\n-9 -2\r\n1 4\r\n-1 10\r\n4 8\r\n", "output": "0\r\n1 2 3 4 5 \r\n"}, {"input": "54 7\r\n-98 -39\r\n14 60\r\n-23 -5\r\n58 75\r\n14 16\r\n-40 20\r\n-6 10\r\n11 60\r\n-47 54\r\n-71 -17\r\n-48 -25\r\n-87 -46\r\n-10 99\r\n-97 -88\r\n-14 94\r\n-25 29\r\n-96 -92\r\n68 75\r\n-75 2\r\n12 84\r\n-47 3\r\n-88 49\r\n-37 88\r\n-61 -25\r\n36 67\r\n30 54\r\n12 31\r\n-71 60\r\n-18 -15\r\n-61 -47\r\n-51 -41\r\n-67 51\r\n26 37\r\n18 94\r\n-67 52\r\n-16 56\r\n-5 26\r\n27 57\r\n36 91\r\n-61 61\r\n71 86\r\n27 73\r\n-57 -39\r\n54 71\r\n-16 14\r\n-97 81\r\n-32 49\r\n-18 50\r\n-63 93\r\n51 70\r\n8 66\r\n43 45\r\n-2 99\r\n11 98\r\n", "output": "111\r\n22 28 32 35 40 46 49 \r\n"}, {"input": "52 18\r\n-50 54\r\n35 65\r\n67 82\r\n-87 -10\r\n-39 4\r\n-55 -18\r\n-27 90\r\n-42 73\r\n18 43\r\n70 85\r\n-85 -22\r\n-1 60\r\n-89 23\r\n-78 -75\r\n-14 69\r\n-69 50\r\n-93 74\r\n-10 45\r\n-81 -72\r\n-24 86\r\n-89 100\r\n25 70\r\n-65 -61\r\n-45 100\r\n-49 -23\r\n-74 -59\r\n-81 -15\r\n-58 47\r\n-65 -58\r\n-47 16\r\n-22 91\r\n-85 19\r\n-81 77\r\n79 87\r\n-31 88\r\n26 32\r\n11 90\r\n7 46\r\n64 83\r\n-51 -20\r\n-76 44\r\n-22 75\r\n45 84\r\n-98 46\r\n-20 78\r\n-88 -47\r\n-41 65\r\n2 93\r\n-66 69\r\n-73 94\r\n-85 -44\r\n-65 -23\r\n", "output": "67\r\n1 7 8 16 17 20 21 24 28 31 33 35 41 42 44 47 49 50 \r\n"}] | false | stdio | import sys
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, k = map(int, f.readline().split())
coupons = []
for _ in range(n):
l, r = map(int, f.readline().split())
coupons.append((l, r))
# Read reference output (ans_ref)
with open(output_path) as f:
ans_ref = int(f.readline().strip())
# Read submission output
with open(submission_path) as f:
try:
ans_sub = int(f.readline().strip())
if ans_sub != ans_ref:
print(0)
return
selected = list(map(int, f.readline().strip().split()))
except:
print(0)
return
# Check selected coupons
if len(selected) != k:
print(0)
return
seen = set()
for p in selected:
if p < 1 or p > n or p in seen:
print(0)
return
seen.add(p)
# Compute intersection
max_l = -float('inf')
min_r = float('inf')
for idx in selected:
l, r = coupons[idx - 1]
max_l = max(max_l, l)
min_r = min(min_r, r)
intersection = max(0, min_r - max_l + 1)
print(1 if intersection == ans_ref else 0)
if __name__ == "__main__":
main() | true |
489/A | 489 | A | PyPy 3-64 | TESTS | 5 | 46 | 0 | 192591472 | n=int(input())
lis=list(map(int,input().split()))
l=[]
for i in lis:
l.append(i)
l.sort()
index1=[]
index2=[]
count=0
pointer=0
for i in range(len(lis)):
a=l[pointer]
b=lis.index(a)
if lis[i]>a:
index1.append(b)
index2.append(i)
lis[i],lis[b]=lis[b],lis[i]
count+=1
pointer+=1
print(count)
for i in range(len(index1)):
print(index1[i],index2[i]) | 22 | 77 | 819,200 | 8726157 | from sys import stdin
lines = list(filter(None, stdin.read().split('\n')))
def parseline(line):
return list(map(int, line.split()))
lines = list(map(parseline, lines))
n, = lines[0]
ai = lines[1];
assert(len(ai) == n)
sai = sorted((x, i) for i, x in enumerate(ai))
pos = list(range(n))
iai = list(range(n))
print(n)
for j, e in enumerate(sai):
x, i = e
print(j, pos[i])
alpha = j
beta = pos[i]
assert pos[iai[j]] == j
pos[i], pos[iai[j]] = j, pos[i]
iai[alpha], iai[beta] = iai[beta], iai[alpha] | Codeforces Round 277.5 (Div. 2) | CF | 2,014 | 1 | 256 | SwapSort | In this problem your goal is to sort an array consisting of n integers in at most n swaps. For the given array find the sequence of swaps that makes the array sorted in the non-descending order. Swaps are performed consecutively, one after another.
Note that in this problem you do not have to minimize the number of swaps — your task is to find any sequence that is no longer than n. | The first line of the input contains integer n (1 ≤ n ≤ 3000) — the number of array elements. The second line contains elements of array: a0, a1, ..., an - 1 ( - 109 ≤ ai ≤ 109), where ai is the i-th element of the array. The elements are numerated from 0 to n - 1 from left to right. Some integers may appear in the array more than once. | In the first line print k (0 ≤ k ≤ n) — the number of swaps. Next k lines must contain the descriptions of the k swaps, one per line. Each swap should be printed as a pair of integers i, j (0 ≤ i, j ≤ n - 1), representing the swap of elements ai and aj. You can print indices in the pairs in any order. The swaps are performed in the order they appear in the output, from the first to the last. It is allowed to print i = j and swap the same pair of elements multiple times.
If there are multiple answers, print any of them. It is guaranteed that at least one answer exists. | null | null | [{"input": "5\n5 2 5 1 4", "output": "2\n0 3\n4 2"}, {"input": "6\n10 20 20 40 60 60", "output": "0"}, {"input": "2\n101 100", "output": "1\n0 1"}] | 1,200 | ["greedy", "implementation", "sortings"] | 22 | [{"input": "5\r\n5 2 5 1 4\r\n", "output": "2\r\n0 3\r\n4 2\r\n"}, {"input": "6\r\n10 20 20 40 60 60\r\n", "output": "0\r\n"}, {"input": "2\r\n101 100\r\n", "output": "1\r\n0 1\r\n"}, {"input": "1\r\n1000\r\n", "output": "0\r\n"}, {"input": "2\r\n1000000000 -1000000000\r\n", "output": "1\r\n0 1\r\n"}, {"input": "8\r\n5 2 6 8 3 1 6 8\r\n", "output": "4\r\n0 5\r\n4 2\r\n5 3\r\n6 5\r\n"}, {"input": "2\r\n200000000 199999999\r\n", "output": "1\r\n0 1\r\n"}, {"input": "3\r\n100000000 100000002 100000001\r\n", "output": "1\r\n1 2\r\n"}, {"input": "5\r\n1000000000 -10000000 0 8888888 7777777\r\n", "output": "3\r\n0 1\r\n2 1\r\n4 2\r\n"}, {"input": "5\r\n10 30 20 50 40\r\n", "output": "2\r\n1 2\r\n4 3\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) as f:
n = int(f.readline().strip())
arr = list(map(int, f.readline().split()))
with open(sub_path) as f:
lines = f.readlines()
if not lines:
print(0)
return
try:
k = int(lines[0].strip())
swaps = []
for line in lines[1:1 + k]:
i, j = map(int, line.strip().split())
swaps.append((i, j))
except (ValueError, IndexError):
print(0)
return
if not (0 <= k <= n):
print(0)
return
for i, j in swaps:
if not (0 <= i < n and 0 <= j < n):
print(0)
return
current = arr.copy()
for i, j in swaps:
current[i], current[j] = current[j], current[i]
sorted_arr = sorted(arr)
if current == sorted_arr:
print(1)
else:
print(0)
if __name__ == "__main__":
main()
| true |
479/D | 479 | D | PyPy 3-64 | TESTS | 2 | 46 | 0 | 221741730 | import sys
input = lambda: sys.stdin.readline().rstrip()
N,L,x,y = map(int, input().split())
A = list(map(int, input().split()))
seen = set()
cx,cy=0,0
for a in A:
if str(a-x) in seen:
cx=1
if str(a-y) in seen:
cy=1
seen.add(str(a))
if cx==cy==1:
exit(print(0))
if cx==1:
print(1)
print(y)
exit(0)
if cy==1:
print(1)
print(x)
exit(0)
seen=set()
t1,t2 = y-x,y+x
for a in A:
if str(a-t1) in seen:
if a>=y:
print(1)
exit(print(a-y))
if a+x<=l:
print(1)
exit(print(a+x))
if str(a-t2) in seen:
print(1)
print(a-x) | 111 | 202 | 10,854,400 | 8309555 | import itertools
import math
def can_measure(a, d):
return any(i + d in a for i in a)
def main():
n, l, x, y = map(int, input().split())
a = set(map(int, input().split()))
can_x = can_measure(a, x)
can_y = can_measure(a, y)
if can_x and can_y:
print(0)
elif can_x:
print(1)
print(y)
elif can_y:
print(1)
print(x)
else:
for i in a:
if i + x + y in a:
print(1)
print(i + x)
break
else:
t = i + x - y in a
if 0 <= i + x <= l and t:
print(1)
print(i + x)
break;
if 0 <= i - y <= l and t:
print(1)
print(i - y)
break;
else:
print(2)
print(x, y)
if __name__ == "__main__":
main() | Codeforces Round 274 (Div. 2) | CF | 2,014 | 1 | 256 | Long Jumps | Valery is a PE teacher at a school in Berland. Soon the students are going to take a test in long jumps, and Valery has lost his favorite ruler!
However, there is no reason for disappointment, as Valery has found another ruler, its length is l centimeters. The ruler already has n marks, with which he can make measurements. We assume that the marks are numbered from 1 to n in the order they appear from the beginning of the ruler to its end. The first point coincides with the beginning of the ruler and represents the origin. The last mark coincides with the end of the ruler, at distance l from the origin. This ruler can be repesented by an increasing sequence a1, a2, ..., an, where ai denotes the distance of the i-th mark from the origin (a1 = 0, an = l).
Valery believes that with a ruler he can measure the distance of d centimeters, if there is a pair of integers i and j (1 ≤ i ≤ j ≤ n), such that the distance between the i-th and the j-th mark is exactly equal to d (in other words, aj - ai = d).
Under the rules, the girls should be able to jump at least x centimeters, and the boys should be able to jump at least y (x < y) centimeters. To test the children's abilities, Valery needs a ruler to measure each of the distances x and y.
Your task is to determine what is the minimum number of additional marks you need to add on the ruler so that they can be used to measure the distances x and y. Valery can add the marks at any integer non-negative distance from the origin not exceeding the length of the ruler. | The first line contains four positive space-separated integers n, l, x, y (2 ≤ n ≤ 105, 2 ≤ l ≤ 109, 1 ≤ x < y ≤ l) — the number of marks, the length of the ruler and the jump norms for girls and boys, correspondingly.
The second line contains a sequence of n integers a1, a2, ..., an (0 = a1 < a2 < ... < an = l), where ai shows the distance from the i-th mark to the origin. | In the first line print a single non-negative integer v — the minimum number of marks that you need to add on the ruler.
In the second line print v space-separated integers p1, p2, ..., pv (0 ≤ pi ≤ l). Number pi means that the i-th mark should be at the distance of pi centimeters from the origin. Print the marks in any order. If there are multiple solutions, print any of them. | null | In the first sample it is impossible to initially measure the distance of 230 centimeters. For that it is enough to add a 20 centimeter mark or a 230 centimeter mark.
In the second sample you already can use the ruler to measure the distances of 185 and 230 centimeters, so you don't have to add new marks.
In the third sample the ruler only contains the initial and the final marks. We will need to add two marks to be able to test the children's skills. | [{"input": "3 250 185 230\n0 185 250", "output": "1\n230"}, {"input": "4 250 185 230\n0 20 185 250", "output": "0"}, {"input": "2 300 185 230\n0 300", "output": "2\n185 230"}] | 1,700 | ["binary search", "greedy", "implementation"] | 111 | [{"input": "3 250 185 230\r\n0 185 250\r\n", "output": "1\r\n230\r\n"}, {"input": "4 250 185 230\r\n0 20 185 250\r\n", "output": "0\r\n"}, {"input": "2 300 185 230\r\n0 300\r\n", "output": "2\r\n185 230\r\n"}, {"input": "4 300 4 5\r\n0 6 7 300\r\n", "output": "1\r\n11\r\n"}, {"input": "2 100 30 70\r\n0 100\r\n", "output": "1\r\n30\r\n"}, {"input": "2 300 140 160\r\n0 300\r\n", "output": "1\r\n140\r\n"}, {"input": "4 300 1 2\r\n0 298 299 300\r\n", "output": "0\r\n"}, {"input": "3 350 150 160\r\n0 310 350\r\n", "output": "1\r\n150\r\n"}, {"input": "4 300 4 5\r\n0 298 299 300\r\n", "output": "1\r\n294\r\n"}, {"input": "19 180 117 148\r\n0 1 19 20 21 28 57 65 68 70 78 88 100 116 154 157 173 179 180\r\n", "output": "2\r\n117 148\r\n"}, {"input": "14 134 99 114\r\n0 6 8 19 50 61 69 83 84 96 111 114 125 134\r\n", "output": "1\r\n99\r\n"}, {"input": "18 187 27 157\r\n0 17 18 31 36 37 40 53 73 86 96 107 119 150 167 181 184 187\r\n", "output": "1\r\n27\r\n"}, {"input": "20 179 69 120\r\n0 6 8 11 21 24 55 61 83 84 96 111 114 116 125 140 147 154 176 179\r\n", "output": "1\r\n27\r\n"}, {"input": "16 115 62 112\r\n0 5 24 32 38 43 44 57 62 72 74 92 103 105 113 115\r\n", "output": "1\r\n112\r\n"}, {"input": "112 1867 1261 1606\r\n0 7 17 43 67 70 87 112 129 141 148 162 179 180 189 202 211 220 231 247 250 277 308 311 327 376 400 406 409 417 418 444 480 512 514 515 518 547 572 575 578 587 612 617 654 684 701 742 757 761 788 821 825 835 841 843 850 858 869 872 881 936 939 969 970 971 997 1026 1040 1045 1068 1070 1073 1076 1095 1110 1115 1154 1166 1178 1179 1203 1204 1225 1237 1241 1246 1275 1302 1305 1311 1312 1315 1338 1340 1419 1428 1560 1561 1576 1591 1594 1618 1643 1658 1660 1664 1689 1803 1822 1835 1867\r\n", "output": "1\r\n1808\r\n"}, {"input": "2 2 1 2\r\n0 2\r\n", "output": "1\r\n1\r\n"}, {"input": "3 2 1 2\r\n0 1 2\r\n", "output": "0\r\n"}, {"input": "3 10 2 3\r\n0 1 10\r\n", "output": "1\r\n3\r\n"}, {"input": "4 10 3 5\r\n0 1 9 10\r\n", "output": "1\r\n4\r\n"}, {"input": "5 1000 777 778\r\n0 1 500 501 1000\r\n", "output": "1\r\n778\r\n"}, {"input": "3 10 1 3\r\n0 2 10\r\n", "output": "1\r\n3\r\n"}, {"input": "4 300 120 150\r\n0 110 140 300\r\n", "output": "1\r\n260\r\n"}, {"input": "5 401 300 400\r\n0 100 250 350 401\r\n", "output": "1\r\n400\r\n"}, {"input": "3 10 1 8\r\n0 7 10\r\n", "output": "1\r\n8\r\n"}, {"input": "4 1000 2 3\r\n0 400 405 1000\r\n", "output": "1\r\n402\r\n"}, {"input": "6 12 7 10\r\n0 1 3 4 6 12\r\n", "output": "1\r\n10\r\n"}, {"input": "4 1000 10 20\r\n0 500 530 1000\r\n", "output": "1\r\n510\r\n"}, {"input": "3 8 2 3\r\n0 7 8\r\n", "output": "1\r\n5\r\n"}, {"input": "4 10 8 9\r\n0 4 5 10\r\n", "output": "2\r\n8 9\r\n"}, {"input": "4 10 7 8\r\n0 5 6 10\r\n", "output": "2\r\n7 8\r\n"}, {"input": "6 35 29 30\r\n0 10 11 31 32 35\r\n", "output": "1\r\n2\r\n"}, {"input": "5 200000 1 100029\r\n0 100000 100009 100010 200000\r\n", "output": "1\r\n100029\r\n"}, {"input": "4 1000 900 901\r\n0 950 951 1000\r\n", "output": "1\r\n50\r\n"}, {"input": "6 504 400 500\r\n0 3 5 103 105 504\r\n", "output": "1\r\n503\r\n"}, {"input": "5 550 300 400\r\n0 151 251 450 550\r\n", "output": "1\r\n150\r\n"}, {"input": "4 300 40 50\r\n0 280 290 300\r\n", "output": "1\r\n240\r\n"}, {"input": "2 1000000000 100000000 500000000\r\n0 1000000000\r\n", "output": "2\r\n100000000 500000000\r\n"}, {"input": "4 600 100 400\r\n0 50 350 600\r\n", "output": "1\r\n450\r\n"}, {"input": "4 100 7 8\r\n0 3 4 100\r\n", "output": "1\r\n11\r\n"}, {"input": "4 100 80 81\r\n0 2 3 100\r\n", "output": "1\r\n83\r\n"}, {"input": "3 13 8 10\r\n0 2 13\r\n", "output": "1\r\n10\r\n"}, {"input": "4 10 7 8\r\n0 4 5 10\r\n", "output": "2\r\n7 8\r\n"}, {"input": "3 450 100 400\r\n0 150 450\r\n", "output": "1\r\n50\r\n"}, {"input": "4 500 30 50\r\n0 20 40 500\r\n", "output": "1\r\n50\r\n"}, {"input": "4 100 10 11\r\n0 4 5 100\r\n", "output": "1\r\n15\r\n"}, {"input": "2 10 5 7\r\n0 10\r\n", "output": "2\r\n5 7\r\n"}, {"input": "6 100 70 71\r\n0 50 51 90 91 100\r\n", "output": "1\r\n20\r\n"}, {"input": "4 9 6 7\r\n0 4 5 9\r\n", "output": "2\r\n6 7\r\n"}, {"input": "3 10 1 8\r\n0 3 10\r\n", "output": "1\r\n2\r\n"}, {"input": "3 12 1 2\r\n0 10 12\r\n", "output": "1\r\n1\r\n"}, {"input": "4 100 3 5\r\n0 40 48 100\r\n", "output": "1\r\n43\r\n"}, {"input": "3 20 17 18\r\n0 19 20\r\n", "output": "1\r\n2\r\n"}, {"input": "4 1000 45 46\r\n0 2 3 1000\r\n", "output": "1\r\n48\r\n"}, {"input": "4 10 5 7\r\n0 4 6 10\r\n", "output": "2\r\n5 7\r\n"}, {"input": "3 12 1 3\r\n0 10 12\r\n", "output": "1\r\n9\r\n"}, {"input": "4 20 6 7\r\n0 1 15 20\r\n", "output": "1\r\n7\r\n"}, {"input": "3 11 3 5\r\n0 9 11\r\n", "output": "1\r\n6\r\n"}, {"input": "3 100 9 10\r\n0 99 100\r\n", "output": "1\r\n90\r\n"}, {"input": "3 10 7 8\r\n0 1 10\r\n", "output": "1\r\n8\r\n"}, {"input": "3 10 5 6\r\n0 9 10\r\n", "output": "1\r\n4\r\n"}, {"input": "3 10 7 8\r\n0 9 10\r\n", "output": "1\r\n2\r\n"}, {"input": "3 10 6 7\r\n0 9 10\r\n", "output": "1\r\n3\r\n"}, {"input": "3 9 6 7\r\n0 1 9\r\n", "output": "1\r\n7\r\n"}, {"input": "3 1000000000 99 100\r\n0 1 1000000000\r\n", "output": "1\r\n100\r\n"}, {"input": "4 10 3 5\r\n0 2 4 10\r\n", "output": "1\r\n5\r\n"}, {"input": "4 100 90 91\r\n0 7 8 100\r\n", "output": "1\r\n98\r\n"}, {"input": "4 100 80 81\r\n0 98 99 100\r\n", "output": "1\r\n18\r\n"}] | false | stdio | import sys
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:
lines = f.readlines()
n, l, x, y = map(int, lines[0].split())
a = list(map(int, lines[1].split()))
a_set = set(a)
# Read reference output (v_ref)
with open(output_path) as f:
ref_lines = f.readlines()
v_ref = int(ref_lines[0].strip())
# Read submission output
with open(submission_path) as f:
sub_lines = [line.strip() for line in f.readlines() if line.strip()]
if not sub_lines:
print(0)
return
try:
v_sub = int(sub_lines[0])
except:
print(0)
return
marks_sub = []
if len(sub_lines) > 1:
try:
marks_sub = list(map(int, sub_lines[1].split()))
except:
print(0)
return
# Check v_sub matches v_ref
if v_sub != v_ref:
print(0)
return
# Check number of marks
if len(marks_sub) != v_sub:
print(0)
return
# Validate each mark
seen = set()
for p in marks_sub:
if not (0 <= p <= l):
print(0)
return
if p in a_set:
print(0)
return
if p in seen:
print(0)
return
seen.add(p)
# Check x and y can be measured
combined = a_set.union(seen)
combined_set = set(combined)
# Check x
x_ok = any((m + x) in combined_set for m in combined_set)
if not x_ok:
print(0)
return
# Check y
y_ok = any((m + y) in combined_set for m in combined_set)
if not y_ok:
print(0)
return
print(1)
if __name__ == "__main__":
main() | true |
427/B | 427 | B | PyPy 3-64 | TESTS | 5 | 124 | 25,292,800 | 208332398 | n,t,c=map(int,input().strip().split())
li=list(map(int,input().strip().split()))
di={}
i=0
fli=[]
while i<n:
ct=0
si=i
maxi=li[i]
while ct<c and i<n:
if li[i]>maxi:
maxi=li[i]
ct+=1
i+=1
lii=[maxi]*c
fli=fli+lii
ans=0
si=0
for i in range(0,n-c+1):
if fli[i]<=t:
ans+=1
print(ans) | 80 | 93 | 21,913,600 | 226506538 | def ways(n, t, c, prisoners):
valid = 0
ways = 0
if n < c:
return 0
for p in prisoners:
if p <= t:
valid += 1
else:
valid = 0
if c <= valid:
ways += 1
return ways
n, t, c = map(int, input().split())
prisoners = map(int, input().split())
print(ways(n, t, c, prisoners)) | Codeforces Round 244 (Div. 2) | CF | 2,014 | 1 | 256 | Prison Transfer | The prison of your city has n prisoners. As the prison can't accommodate all of them, the city mayor has decided to transfer c of the prisoners to a prison located in another city.
For this reason, he made the n prisoners to stand in a line, with a number written on their chests. The number is the severity of the crime he/she has committed. The greater the number, the more severe his/her crime was.
Then, the mayor told you to choose the c prisoners, who will be transferred to the other prison. He also imposed two conditions. They are,
- The chosen c prisoners has to form a contiguous segment of prisoners.
- Any of the chosen prisoner's crime level should not be greater then t. Because, that will make the prisoner a severe criminal and the mayor doesn't want to take the risk of his running away during the transfer.
Find the number of ways you can choose the c prisoners. | The first line of input will contain three space separated integers n (1 ≤ n ≤ 2·105), t (0 ≤ t ≤ 109) and c (1 ≤ c ≤ n). The next line will contain n space separated integers, the ith integer is the severity ith prisoner's crime. The value of crime severities will be non-negative and will not exceed 109. | Print a single integer — the number of ways you can choose the c prisoners. | null | null | [{"input": "4 3 3\n2 3 1 1", "output": "2"}, {"input": "1 1 1\n2", "output": "0"}, {"input": "11 4 2\n2 2 0 7 3 2 2 4 9 1 4", "output": "6"}] | 1,100 | ["data structures", "implementation"] | 80 | [{"input": "4 3 3\r\n2 3 1 1\r\n", "output": "2\r\n"}, {"input": "1 1 1\r\n2\r\n", "output": "0\r\n"}, {"input": "11 4 2\r\n2 2 0 7 3 2 2 4 9 1 4\r\n", "output": "6\r\n"}, {"input": "57 2 10\r\n7 5 2 7 4 1 0 5 2 9 2 9 8 6 6 5 9 6 8 1 0 1 0 3 2 6 5 2 8 8 8 8 0 9 4 3 6 6 2 4 5 1 2 0 1 7 1 1 5 4 5 0 7 5 1 9 6\r\n", "output": "0\r\n"}, {"input": "2 228885628 1\r\n90897004 258427916\r\n", "output": "1\r\n"}, {"input": "3 1 1\r\n1 2 1\r\n", "output": "2\r\n"}, {"input": "3 3 3\r\n3 2 3\r\n", "output": "1\r\n"}, {"input": "4 2 2\r\n1 3 3 2\r\n", "output": "0\r\n"}, {"input": "1 228 1\r\n1\r\n", "output": "1\r\n"}] | false | stdio | null | true |
165/A | 165 | A | Python 3 | TESTS | 3 | 62 | 0 | 145610886 | n=int(input())
pt_list=[]
ans=0
for _ in range(n):
[x,y] =map(int,input().split())
pt_list.append([x,y])
for i in range(n):
a_setx=set()
a_sety=set()
for j in range(n):
if pt_list[i][0]-pt_list[j][0]>0:
a_setx.add('pos')
elif pt_list[i][1]-pt_list[j][1]>0:
a_sety.add('posy')
elif pt_list[i][1]-pt_list[j][1]<0:
a_sety.add('negy')
elif pt_list[i][0]-pt_list[j][0]<0:
a_setx.add('negx')
else:
continue
if len(a_setx)==2 and len(a_sety)==2:
ans+=1
print(ans) | 26 | 92 | 0 | 141808501 | n = int(input())
store_x = {}
store_y = {}
def sortSecond(val):
return val[1]
for _ in range(n):
x,y = [int(i) for i in input().split()]
if store_x.get(x):
store_x[x].append((x,y))
else:
store_x[x] = [(x,y)]
if store_y.get(y):
store_y[y].append((x,y))
else:
store_y[y] = [(x,y)]
candidates = {}
ans = 0
for k,v in store_x.items():
if len(v) > 2:
v.sort(key= sortSecond)
for i in v[1:len(v)-1]:
if candidates.get(str(i)):
candidates[str(i)]+=1
else:
candidates[str(i)]=1
for k,v in store_y.items():
if len(v) > 2:
v.sort()
for i in v[1:len(v)-1]:
if candidates.get(str(i)):
ans += candidates[str(i)]
print(ans) | Codeforces Round 112 (Div. 2) | CF | 2,012 | 2 | 256 | Supercentral Point | One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (x1, y1), (x2, y2), ..., (xn, yn). Let's define neighbors for some fixed point from the given set (x, y):
- point (x', y') is (x, y)'s right neighbor, if x' > x and y' = y
- point (x', y') is (x, y)'s left neighbor, if x' < x and y' = y
- point (x', y') is (x, y)'s lower neighbor, if x' = x and y' < y
- point (x', y') is (x, y)'s upper neighbor, if x' = x and y' > y
We'll consider point (x, y) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set. | The first input line contains the only integer n (1 ≤ n ≤ 200) — the number of points in the given set. Next n lines contain the coordinates of the points written as "x y" (without the quotes) (|x|, |y| ≤ 1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different. | Print the only number — the number of supercentral points of the given set. | null | In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | [{"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3", "output": "2"}, {"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0", "output": "1"}] | 1,000 | ["implementation"] | 26 | [{"input": "8\r\n1 1\r\n4 2\r\n3 1\r\n1 2\r\n0 2\r\n0 1\r\n1 0\r\n1 3\r\n", "output": "2\r\n"}, {"input": "5\r\n0 0\r\n0 1\r\n1 0\r\n0 -1\r\n-1 0\r\n", "output": "1\r\n"}, {"input": "9\r\n-565 -752\r\n-184 723\r\n-184 -752\r\n-184 1\r\n950 723\r\n-565 723\r\n950 -752\r\n950 1\r\n-565 1\r\n", "output": "1\r\n"}, {"input": "25\r\n-651 897\r\n916 897\r\n-651 -808\r\n-748 301\r\n-734 414\r\n-651 -973\r\n-734 897\r\n916 -550\r\n-758 414\r\n916 180\r\n-758 -808\r\n-758 -973\r\n125 -550\r\n125 -973\r\n125 301\r\n916 414\r\n-748 -808\r\n-651 301\r\n-734 301\r\n-307 897\r\n-651 -550\r\n-651 414\r\n125 -808\r\n-748 -550\r\n916 -808\r\n", "output": "7\r\n"}, {"input": "1\r\n487 550\r\n", "output": "0\r\n"}, {"input": "10\r\n990 -396\r\n990 736\r\n990 646\r\n990 -102\r\n990 -570\r\n990 155\r\n990 528\r\n990 489\r\n990 268\r\n990 676\r\n", "output": "0\r\n"}, {"input": "30\r\n507 836\r\n525 836\r\n-779 196\r\n507 -814\r\n525 -814\r\n525 42\r\n525 196\r\n525 -136\r\n-779 311\r\n507 -360\r\n525 300\r\n507 578\r\n507 311\r\n-779 836\r\n507 300\r\n525 -360\r\n525 311\r\n-779 -360\r\n-779 578\r\n-779 300\r\n507 42\r\n525 578\r\n-779 379\r\n507 196\r\n525 379\r\n507 379\r\n-779 -814\r\n-779 42\r\n-779 -136\r\n507 -136\r\n", "output": "8\r\n"}, {"input": "25\r\n890 -756\r\n890 -188\r\n-37 -756\r\n-37 853\r\n523 998\r\n-261 853\r\n-351 853\r\n-351 -188\r\n523 -756\r\n-261 -188\r\n-37 998\r\n523 -212\r\n-351 998\r\n-37 -188\r\n-351 -756\r\n-37 -212\r\n890 998\r\n890 -212\r\n523 853\r\n-351 -212\r\n-261 -212\r\n-261 998\r\n-261 -756\r\n890 853\r\n523 -188\r\n", "output": "9\r\n"}, {"input": "21\r\n-813 -11\r\n486 254\r\n685 254\r\n-708 254\r\n-55 -11\r\n-671 -191\r\n486 -11\r\n-671 -11\r\n685 -11\r\n685 -191\r\n486 -191\r\n-55 254\r\n-708 -11\r\n-813 254\r\n-708 -191\r\n41 -11\r\n-671 254\r\n-813 -191\r\n41 254\r\n-55 -191\r\n41 -191\r\n", "output": "5\r\n"}, {"input": "4\r\n1 0\r\n2 0\r\n1 1\r\n1 -1\r\n", "output": "0\r\n"}] | false | stdio | null | true |
165/A | 165 | A | PyPy 3-64 | TESTS | 3 | 92 | 0 | 184382238 | n = int(input())
sx = []
sy = []
count = 0
for i in range(n):
s = list(map(int, input().split()))
x, y = s[0], s[1]
sx.append(x)
sy.append(y)
mAxx = max(sx)
mInx = min(sx)
mAxy = max(sy)
mIny = min(sy)
for i in range(len(sx)):
if sx.count(sx[i]) >= 3 and sx[i] != mAxx and sx[i] != mInx:
if sy.count(sy[i]) >= 3 and sy[i] != mAxy and sy[i] != mIny:
count += 1
print(count) | 26 | 92 | 0 | 146798114 | n=int(input())
l=[]
a=[]
b=[]
sum=0
for i in range(n):
m=list(map(int,input().split()))
l.append(m)
a.append(m[0])
b.append(m[1])
for i in range(n):
count=0
for j in range(len(a)):
if a[j]==l[i][0] and b[j]>l[i][1]:
count+=1
break
for j in range(len(a)):
if a[j]==l[i][0] and b[j]<l[i][1]:
count+=1
break
for j in range(len(a)):
if b[j]==l[i][1] and a[j]>l[i][0]:
count+=1
break
for j in range(len(a)):
if b[j]==l[i][1] and a[j]<l[i][0]:
count+=1
break
if (count==4):
sum+=1
print(sum) | Codeforces Round 112 (Div. 2) | CF | 2,012 | 2 | 256 | Supercentral Point | One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (x1, y1), (x2, y2), ..., (xn, yn). Let's define neighbors for some fixed point from the given set (x, y):
- point (x', y') is (x, y)'s right neighbor, if x' > x and y' = y
- point (x', y') is (x, y)'s left neighbor, if x' < x and y' = y
- point (x', y') is (x, y)'s lower neighbor, if x' = x and y' < y
- point (x', y') is (x, y)'s upper neighbor, if x' = x and y' > y
We'll consider point (x, y) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set. | The first input line contains the only integer n (1 ≤ n ≤ 200) — the number of points in the given set. Next n lines contain the coordinates of the points written as "x y" (without the quotes) (|x|, |y| ≤ 1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different. | Print the only number — the number of supercentral points of the given set. | null | In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | [{"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3", "output": "2"}, {"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0", "output": "1"}] | 1,000 | ["implementation"] | 26 | [{"input": "8\r\n1 1\r\n4 2\r\n3 1\r\n1 2\r\n0 2\r\n0 1\r\n1 0\r\n1 3\r\n", "output": "2\r\n"}, {"input": "5\r\n0 0\r\n0 1\r\n1 0\r\n0 -1\r\n-1 0\r\n", "output": "1\r\n"}, {"input": "9\r\n-565 -752\r\n-184 723\r\n-184 -752\r\n-184 1\r\n950 723\r\n-565 723\r\n950 -752\r\n950 1\r\n-565 1\r\n", "output": "1\r\n"}, {"input": "25\r\n-651 897\r\n916 897\r\n-651 -808\r\n-748 301\r\n-734 414\r\n-651 -973\r\n-734 897\r\n916 -550\r\n-758 414\r\n916 180\r\n-758 -808\r\n-758 -973\r\n125 -550\r\n125 -973\r\n125 301\r\n916 414\r\n-748 -808\r\n-651 301\r\n-734 301\r\n-307 897\r\n-651 -550\r\n-651 414\r\n125 -808\r\n-748 -550\r\n916 -808\r\n", "output": "7\r\n"}, {"input": "1\r\n487 550\r\n", "output": "0\r\n"}, {"input": "10\r\n990 -396\r\n990 736\r\n990 646\r\n990 -102\r\n990 -570\r\n990 155\r\n990 528\r\n990 489\r\n990 268\r\n990 676\r\n", "output": "0\r\n"}, {"input": "30\r\n507 836\r\n525 836\r\n-779 196\r\n507 -814\r\n525 -814\r\n525 42\r\n525 196\r\n525 -136\r\n-779 311\r\n507 -360\r\n525 300\r\n507 578\r\n507 311\r\n-779 836\r\n507 300\r\n525 -360\r\n525 311\r\n-779 -360\r\n-779 578\r\n-779 300\r\n507 42\r\n525 578\r\n-779 379\r\n507 196\r\n525 379\r\n507 379\r\n-779 -814\r\n-779 42\r\n-779 -136\r\n507 -136\r\n", "output": "8\r\n"}, {"input": "25\r\n890 -756\r\n890 -188\r\n-37 -756\r\n-37 853\r\n523 998\r\n-261 853\r\n-351 853\r\n-351 -188\r\n523 -756\r\n-261 -188\r\n-37 998\r\n523 -212\r\n-351 998\r\n-37 -188\r\n-351 -756\r\n-37 -212\r\n890 998\r\n890 -212\r\n523 853\r\n-351 -212\r\n-261 -212\r\n-261 998\r\n-261 -756\r\n890 853\r\n523 -188\r\n", "output": "9\r\n"}, {"input": "21\r\n-813 -11\r\n486 254\r\n685 254\r\n-708 254\r\n-55 -11\r\n-671 -191\r\n486 -11\r\n-671 -11\r\n685 -11\r\n685 -191\r\n486 -191\r\n-55 254\r\n-708 -11\r\n-813 254\r\n-708 -191\r\n41 -11\r\n-671 254\r\n-813 -191\r\n41 254\r\n-55 -191\r\n41 -191\r\n", "output": "5\r\n"}, {"input": "4\r\n1 0\r\n2 0\r\n1 1\r\n1 -1\r\n", "output": "0\r\n"}] | false | stdio | null | true |
165/A | 165 | A | Python 3 | TESTS | 3 | 62 | 0 | 160189729 | lst=[]
for i in range(int(input())):lst.append([int(x) for x in input().split()])
lstx=[i[0] for i in lst]
lsty=[i[1] for i in lst]
minx,maxx,miny,maxy=min(lstx),max(lstx),min(lsty),max(lsty)
superpt=0
for i in lst:
if i[0]!=minx and i[0]!=maxx:
if i[1]!=miny and i[1]!=maxy:
if lstx.count(i[0])>=3 and lsty.count(i[1])>=3:superpt+=1
print(superpt) | 26 | 92 | 0 | 147175351 | n = int(input())
a = [list(int(a) for a in input().split()) for i in range(n)]
c = 0
for x,y in a:
l , r , t , b = 0,0,0,0
for x1,y1 in a:
if x == x1:
if y1 > y:
t = 1
if y1 < y:
b = 1
if y == y1:
if x1 > x:
r = 1
if x1 < x:
l = 1
if l == 1 and r == 1 and t == 1 and b == 1:
c += 1
print(c) | Codeforces Round 112 (Div. 2) | CF | 2,012 | 2 | 256 | Supercentral Point | One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (x1, y1), (x2, y2), ..., (xn, yn). Let's define neighbors for some fixed point from the given set (x, y):
- point (x', y') is (x, y)'s right neighbor, if x' > x and y' = y
- point (x', y') is (x, y)'s left neighbor, if x' < x and y' = y
- point (x', y') is (x, y)'s lower neighbor, if x' = x and y' < y
- point (x', y') is (x, y)'s upper neighbor, if x' = x and y' > y
We'll consider point (x, y) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set. | The first input line contains the only integer n (1 ≤ n ≤ 200) — the number of points in the given set. Next n lines contain the coordinates of the points written as "x y" (without the quotes) (|x|, |y| ≤ 1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different. | Print the only number — the number of supercentral points of the given set. | null | In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | [{"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3", "output": "2"}, {"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0", "output": "1"}] | 1,000 | ["implementation"] | 26 | [{"input": "8\r\n1 1\r\n4 2\r\n3 1\r\n1 2\r\n0 2\r\n0 1\r\n1 0\r\n1 3\r\n", "output": "2\r\n"}, {"input": "5\r\n0 0\r\n0 1\r\n1 0\r\n0 -1\r\n-1 0\r\n", "output": "1\r\n"}, {"input": "9\r\n-565 -752\r\n-184 723\r\n-184 -752\r\n-184 1\r\n950 723\r\n-565 723\r\n950 -752\r\n950 1\r\n-565 1\r\n", "output": "1\r\n"}, {"input": "25\r\n-651 897\r\n916 897\r\n-651 -808\r\n-748 301\r\n-734 414\r\n-651 -973\r\n-734 897\r\n916 -550\r\n-758 414\r\n916 180\r\n-758 -808\r\n-758 -973\r\n125 -550\r\n125 -973\r\n125 301\r\n916 414\r\n-748 -808\r\n-651 301\r\n-734 301\r\n-307 897\r\n-651 -550\r\n-651 414\r\n125 -808\r\n-748 -550\r\n916 -808\r\n", "output": "7\r\n"}, {"input": "1\r\n487 550\r\n", "output": "0\r\n"}, {"input": "10\r\n990 -396\r\n990 736\r\n990 646\r\n990 -102\r\n990 -570\r\n990 155\r\n990 528\r\n990 489\r\n990 268\r\n990 676\r\n", "output": "0\r\n"}, {"input": "30\r\n507 836\r\n525 836\r\n-779 196\r\n507 -814\r\n525 -814\r\n525 42\r\n525 196\r\n525 -136\r\n-779 311\r\n507 -360\r\n525 300\r\n507 578\r\n507 311\r\n-779 836\r\n507 300\r\n525 -360\r\n525 311\r\n-779 -360\r\n-779 578\r\n-779 300\r\n507 42\r\n525 578\r\n-779 379\r\n507 196\r\n525 379\r\n507 379\r\n-779 -814\r\n-779 42\r\n-779 -136\r\n507 -136\r\n", "output": "8\r\n"}, {"input": "25\r\n890 -756\r\n890 -188\r\n-37 -756\r\n-37 853\r\n523 998\r\n-261 853\r\n-351 853\r\n-351 -188\r\n523 -756\r\n-261 -188\r\n-37 998\r\n523 -212\r\n-351 998\r\n-37 -188\r\n-351 -756\r\n-37 -212\r\n890 998\r\n890 -212\r\n523 853\r\n-351 -212\r\n-261 -212\r\n-261 998\r\n-261 -756\r\n890 853\r\n523 -188\r\n", "output": "9\r\n"}, {"input": "21\r\n-813 -11\r\n486 254\r\n685 254\r\n-708 254\r\n-55 -11\r\n-671 -191\r\n486 -11\r\n-671 -11\r\n685 -11\r\n685 -191\r\n486 -191\r\n-55 254\r\n-708 -11\r\n-813 254\r\n-708 -191\r\n41 -11\r\n-671 254\r\n-813 -191\r\n41 254\r\n-55 -191\r\n41 -191\r\n", "output": "5\r\n"}, {"input": "4\r\n1 0\r\n2 0\r\n1 1\r\n1 -1\r\n", "output": "0\r\n"}] | false | stdio | null | true |
165/A | 165 | A | Python 3 | TESTS | 3 | 62 | 0 | 193427066 | n = int(input())
arrs = []
for _ in range(n):
points = [list(x) for x in input().split()]
arrs.append(points)
count = 0
for i in range(n):
r, l, lo, up = False, False, False, False
for j in range(n):
if i == j: continue
if arrs[i][1] == arrs[j][1]:
if arrs[i][0] > arrs[j][0]: r = True
if arrs[i][0] < arrs[j][0]: l = True
if arrs[i][0] == arrs[j][0]:
if arrs[i][1] < arrs[j][1]: lo = True
if arrs[i][1] > arrs[j][1]: up = True
if r and l and lo and up: count += 1
print(count) | 26 | 92 | 0 | 149380017 | n = int(input())
pts = []
for i in range(n):
el = [int(x) for x in input().split()]
pts.append(el)
# print(pts)
res = 0
for pt1 in range(n):
rtn = False
lftn = False
lown = False
upn = False
for pt2 in range(n):
if(pt1!=pt2):
if(pts[pt1][0]==pts[pt2][0]):
if(pts[pt1][1] > pts[pt2][1]):
lown = True
elif(pts[pt1][1] < pts[pt2][1]):
upn = True
if(pts[pt1][1]==pts[pt2][1]):
if(pts[pt1][0] > pts[pt2][0]):
lftn = True
elif(pts[pt1][0] < pts[pt2][0]):
rtn = True
if(rtn and lftn and lown and upn):
res+=1
break
print(res) | Codeforces Round 112 (Div. 2) | CF | 2,012 | 2 | 256 | Supercentral Point | One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (x1, y1), (x2, y2), ..., (xn, yn). Let's define neighbors for some fixed point from the given set (x, y):
- point (x', y') is (x, y)'s right neighbor, if x' > x and y' = y
- point (x', y') is (x, y)'s left neighbor, if x' < x and y' = y
- point (x', y') is (x, y)'s lower neighbor, if x' = x and y' < y
- point (x', y') is (x, y)'s upper neighbor, if x' = x and y' > y
We'll consider point (x, y) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set. | The first input line contains the only integer n (1 ≤ n ≤ 200) — the number of points in the given set. Next n lines contain the coordinates of the points written as "x y" (without the quotes) (|x|, |y| ≤ 1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different. | Print the only number — the number of supercentral points of the given set. | null | In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | [{"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3", "output": "2"}, {"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0", "output": "1"}] | 1,000 | ["implementation"] | 26 | [{"input": "8\r\n1 1\r\n4 2\r\n3 1\r\n1 2\r\n0 2\r\n0 1\r\n1 0\r\n1 3\r\n", "output": "2\r\n"}, {"input": "5\r\n0 0\r\n0 1\r\n1 0\r\n0 -1\r\n-1 0\r\n", "output": "1\r\n"}, {"input": "9\r\n-565 -752\r\n-184 723\r\n-184 -752\r\n-184 1\r\n950 723\r\n-565 723\r\n950 -752\r\n950 1\r\n-565 1\r\n", "output": "1\r\n"}, {"input": "25\r\n-651 897\r\n916 897\r\n-651 -808\r\n-748 301\r\n-734 414\r\n-651 -973\r\n-734 897\r\n916 -550\r\n-758 414\r\n916 180\r\n-758 -808\r\n-758 -973\r\n125 -550\r\n125 -973\r\n125 301\r\n916 414\r\n-748 -808\r\n-651 301\r\n-734 301\r\n-307 897\r\n-651 -550\r\n-651 414\r\n125 -808\r\n-748 -550\r\n916 -808\r\n", "output": "7\r\n"}, {"input": "1\r\n487 550\r\n", "output": "0\r\n"}, {"input": "10\r\n990 -396\r\n990 736\r\n990 646\r\n990 -102\r\n990 -570\r\n990 155\r\n990 528\r\n990 489\r\n990 268\r\n990 676\r\n", "output": "0\r\n"}, {"input": "30\r\n507 836\r\n525 836\r\n-779 196\r\n507 -814\r\n525 -814\r\n525 42\r\n525 196\r\n525 -136\r\n-779 311\r\n507 -360\r\n525 300\r\n507 578\r\n507 311\r\n-779 836\r\n507 300\r\n525 -360\r\n525 311\r\n-779 -360\r\n-779 578\r\n-779 300\r\n507 42\r\n525 578\r\n-779 379\r\n507 196\r\n525 379\r\n507 379\r\n-779 -814\r\n-779 42\r\n-779 -136\r\n507 -136\r\n", "output": "8\r\n"}, {"input": "25\r\n890 -756\r\n890 -188\r\n-37 -756\r\n-37 853\r\n523 998\r\n-261 853\r\n-351 853\r\n-351 -188\r\n523 -756\r\n-261 -188\r\n-37 998\r\n523 -212\r\n-351 998\r\n-37 -188\r\n-351 -756\r\n-37 -212\r\n890 998\r\n890 -212\r\n523 853\r\n-351 -212\r\n-261 -212\r\n-261 998\r\n-261 -756\r\n890 853\r\n523 -188\r\n", "output": "9\r\n"}, {"input": "21\r\n-813 -11\r\n486 254\r\n685 254\r\n-708 254\r\n-55 -11\r\n-671 -191\r\n486 -11\r\n-671 -11\r\n685 -11\r\n685 -191\r\n486 -191\r\n-55 254\r\n-708 -11\r\n-813 254\r\n-708 -191\r\n41 -11\r\n-671 254\r\n-813 -191\r\n41 254\r\n-55 -191\r\n41 -191\r\n", "output": "5\r\n"}, {"input": "4\r\n1 0\r\n2 0\r\n1 1\r\n1 -1\r\n", "output": "0\r\n"}] | false | stdio | null | true |
165/A | 165 | A | PyPy 3 | TESTS | 5 | 186 | 1,433,600 | 104274136 | n = int(input())
points = []
for i in range(n):
points.append(list(map(int, input().split())))
count = 0
for i in range(n):
x, y = points[i]
up, down, left, right = False, False, False, False
for j in range(n):
x2, y2 = points[j]
if x == x2:
if y < y2:
up = True
elif y > y2:
down = True
elif y == y2:
if x < x2:
left = True
elif x > x2:
right = True
if up == down == left == right: count += 1
print(count) | 26 | 92 | 0 | 150146938 | n = int(input())
arr = []
output = 0
for i in range(n):
x, y = [int(i) for i in input().split()]
arr.append([x, y])
for i in range(n):
x = arr[i][0]
y = arr[i][1]
arr2 = arr[:n] + arr[n+1:]
left = False
right = False
top = False
bottom = False
for item in arr2:
if item[0] > x and item[1] == y:
right = True
elif item[0] < x and item[1] == y:
left = True
elif item[0] == x and item[1] < y:
bottom = True
elif item[0] == x and item[1] > y:
top = True
if left == True and right == True and top == True and bottom == True:
output += 1
print(output) | Codeforces Round 112 (Div. 2) | CF | 2,012 | 2 | 256 | Supercentral Point | One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (x1, y1), (x2, y2), ..., (xn, yn). Let's define neighbors for some fixed point from the given set (x, y):
- point (x', y') is (x, y)'s right neighbor, if x' > x and y' = y
- point (x', y') is (x, y)'s left neighbor, if x' < x and y' = y
- point (x', y') is (x, y)'s lower neighbor, if x' = x and y' < y
- point (x', y') is (x, y)'s upper neighbor, if x' = x and y' > y
We'll consider point (x, y) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set. | The first input line contains the only integer n (1 ≤ n ≤ 200) — the number of points in the given set. Next n lines contain the coordinates of the points written as "x y" (without the quotes) (|x|, |y| ≤ 1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different. | Print the only number — the number of supercentral points of the given set. | null | In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | [{"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3", "output": "2"}, {"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0", "output": "1"}] | 1,000 | ["implementation"] | 26 | [{"input": "8\r\n1 1\r\n4 2\r\n3 1\r\n1 2\r\n0 2\r\n0 1\r\n1 0\r\n1 3\r\n", "output": "2\r\n"}, {"input": "5\r\n0 0\r\n0 1\r\n1 0\r\n0 -1\r\n-1 0\r\n", "output": "1\r\n"}, {"input": "9\r\n-565 -752\r\n-184 723\r\n-184 -752\r\n-184 1\r\n950 723\r\n-565 723\r\n950 -752\r\n950 1\r\n-565 1\r\n", "output": "1\r\n"}, {"input": "25\r\n-651 897\r\n916 897\r\n-651 -808\r\n-748 301\r\n-734 414\r\n-651 -973\r\n-734 897\r\n916 -550\r\n-758 414\r\n916 180\r\n-758 -808\r\n-758 -973\r\n125 -550\r\n125 -973\r\n125 301\r\n916 414\r\n-748 -808\r\n-651 301\r\n-734 301\r\n-307 897\r\n-651 -550\r\n-651 414\r\n125 -808\r\n-748 -550\r\n916 -808\r\n", "output": "7\r\n"}, {"input": "1\r\n487 550\r\n", "output": "0\r\n"}, {"input": "10\r\n990 -396\r\n990 736\r\n990 646\r\n990 -102\r\n990 -570\r\n990 155\r\n990 528\r\n990 489\r\n990 268\r\n990 676\r\n", "output": "0\r\n"}, {"input": "30\r\n507 836\r\n525 836\r\n-779 196\r\n507 -814\r\n525 -814\r\n525 42\r\n525 196\r\n525 -136\r\n-779 311\r\n507 -360\r\n525 300\r\n507 578\r\n507 311\r\n-779 836\r\n507 300\r\n525 -360\r\n525 311\r\n-779 -360\r\n-779 578\r\n-779 300\r\n507 42\r\n525 578\r\n-779 379\r\n507 196\r\n525 379\r\n507 379\r\n-779 -814\r\n-779 42\r\n-779 -136\r\n507 -136\r\n", "output": "8\r\n"}, {"input": "25\r\n890 -756\r\n890 -188\r\n-37 -756\r\n-37 853\r\n523 998\r\n-261 853\r\n-351 853\r\n-351 -188\r\n523 -756\r\n-261 -188\r\n-37 998\r\n523 -212\r\n-351 998\r\n-37 -188\r\n-351 -756\r\n-37 -212\r\n890 998\r\n890 -212\r\n523 853\r\n-351 -212\r\n-261 -212\r\n-261 998\r\n-261 -756\r\n890 853\r\n523 -188\r\n", "output": "9\r\n"}, {"input": "21\r\n-813 -11\r\n486 254\r\n685 254\r\n-708 254\r\n-55 -11\r\n-671 -191\r\n486 -11\r\n-671 -11\r\n685 -11\r\n685 -191\r\n486 -191\r\n-55 254\r\n-708 -11\r\n-813 254\r\n-708 -191\r\n41 -11\r\n-671 254\r\n-813 -191\r\n41 254\r\n-55 -191\r\n41 -191\r\n", "output": "5\r\n"}, {"input": "4\r\n1 0\r\n2 0\r\n1 1\r\n1 -1\r\n", "output": "0\r\n"}] | false | stdio | null | true |
165/A | 165 | A | Python 3 | TESTS | 3 | 60 | 0 | 146273591 | x=[]
xmax=0
xmin=0
ymax=0
ymin=0
y=[]
for _ in range(int(input())):
a,b=map(int,input().split())
x.append(a)
y.append(b)
if(a>xmax):
xmax=a
if(a<xmin):
xmin=a
if(b>ymax):
ymax=b
if(b<ymin):
ymin=b
re=[xmax,ymax,xmin,ymin]
x=list(set(x))
y=list(set(y))
for i in range(4):
if(re[i] in x):
x.remove(re[i])
else:
None
if(re[i] in y):
y.remove(re[i])
else:
None
print(len(x)*len(y)) | 26 | 92 | 0 | 155023226 | info = dict()
amount = int(input())
counter = 0
all_pos = []
for i in range(amount):
current = ([int(x) for x in input().split()])
all_pos.append(current)
dict_elements = {}
for i in all_pos:
x, y = i[0], i[1]
right, left, upper, lower = 0, 0, 0, 0
for j in all_pos:
if (x,y) == j:
continue
elif x == j[0] and y > j[1]:
lower += 1
elif x == j[0] and y < j[1]:
upper += 1
elif x < j[0] and y == j[1]:
right += 1
elif x > j[0] and y == j[1]:
left += 1
dict_elements[tuple(i)] = [right, left, upper, lower]
for elements in all_pos:
local = 0
for i in dict_elements[tuple(elements)]:
if i > 0:
local += 1
if local == 4:
counter += 1
print(counter) | Codeforces Round 112 (Div. 2) | CF | 2,012 | 2 | 256 | Supercentral Point | One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (x1, y1), (x2, y2), ..., (xn, yn). Let's define neighbors for some fixed point from the given set (x, y):
- point (x', y') is (x, y)'s right neighbor, if x' > x and y' = y
- point (x', y') is (x, y)'s left neighbor, if x' < x and y' = y
- point (x', y') is (x, y)'s lower neighbor, if x' = x and y' < y
- point (x', y') is (x, y)'s upper neighbor, if x' = x and y' > y
We'll consider point (x, y) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set. | The first input line contains the only integer n (1 ≤ n ≤ 200) — the number of points in the given set. Next n lines contain the coordinates of the points written as "x y" (without the quotes) (|x|, |y| ≤ 1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different. | Print the only number — the number of supercentral points of the given set. | null | In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | [{"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3", "output": "2"}, {"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0", "output": "1"}] | 1,000 | ["implementation"] | 26 | [{"input": "8\r\n1 1\r\n4 2\r\n3 1\r\n1 2\r\n0 2\r\n0 1\r\n1 0\r\n1 3\r\n", "output": "2\r\n"}, {"input": "5\r\n0 0\r\n0 1\r\n1 0\r\n0 -1\r\n-1 0\r\n", "output": "1\r\n"}, {"input": "9\r\n-565 -752\r\n-184 723\r\n-184 -752\r\n-184 1\r\n950 723\r\n-565 723\r\n950 -752\r\n950 1\r\n-565 1\r\n", "output": "1\r\n"}, {"input": "25\r\n-651 897\r\n916 897\r\n-651 -808\r\n-748 301\r\n-734 414\r\n-651 -973\r\n-734 897\r\n916 -550\r\n-758 414\r\n916 180\r\n-758 -808\r\n-758 -973\r\n125 -550\r\n125 -973\r\n125 301\r\n916 414\r\n-748 -808\r\n-651 301\r\n-734 301\r\n-307 897\r\n-651 -550\r\n-651 414\r\n125 -808\r\n-748 -550\r\n916 -808\r\n", "output": "7\r\n"}, {"input": "1\r\n487 550\r\n", "output": "0\r\n"}, {"input": "10\r\n990 -396\r\n990 736\r\n990 646\r\n990 -102\r\n990 -570\r\n990 155\r\n990 528\r\n990 489\r\n990 268\r\n990 676\r\n", "output": "0\r\n"}, {"input": "30\r\n507 836\r\n525 836\r\n-779 196\r\n507 -814\r\n525 -814\r\n525 42\r\n525 196\r\n525 -136\r\n-779 311\r\n507 -360\r\n525 300\r\n507 578\r\n507 311\r\n-779 836\r\n507 300\r\n525 -360\r\n525 311\r\n-779 -360\r\n-779 578\r\n-779 300\r\n507 42\r\n525 578\r\n-779 379\r\n507 196\r\n525 379\r\n507 379\r\n-779 -814\r\n-779 42\r\n-779 -136\r\n507 -136\r\n", "output": "8\r\n"}, {"input": "25\r\n890 -756\r\n890 -188\r\n-37 -756\r\n-37 853\r\n523 998\r\n-261 853\r\n-351 853\r\n-351 -188\r\n523 -756\r\n-261 -188\r\n-37 998\r\n523 -212\r\n-351 998\r\n-37 -188\r\n-351 -756\r\n-37 -212\r\n890 998\r\n890 -212\r\n523 853\r\n-351 -212\r\n-261 -212\r\n-261 998\r\n-261 -756\r\n890 853\r\n523 -188\r\n", "output": "9\r\n"}, {"input": "21\r\n-813 -11\r\n486 254\r\n685 254\r\n-708 254\r\n-55 -11\r\n-671 -191\r\n486 -11\r\n-671 -11\r\n685 -11\r\n685 -191\r\n486 -191\r\n-55 254\r\n-708 -11\r\n-813 254\r\n-708 -191\r\n41 -11\r\n-671 254\r\n-813 -191\r\n41 254\r\n-55 -191\r\n41 -191\r\n", "output": "5\r\n"}, {"input": "4\r\n1 0\r\n2 0\r\n1 1\r\n1 -1\r\n", "output": "0\r\n"}] | false | stdio | null | true |
165/A | 165 | A | Python 3 | TESTS | 3 | 62 | 0 | 130632340 | n = int(input())
li = [list(map(int,(input().split()))) for i in range(n)]
cnt = 0
for i in li:
x_g, x_l, y_g, y_l, x_equal, y_equal = 0, 0, 0, 0, 0, 0
for j in li:
if li.index(i) != li.index(j):
if j[0]<i[0]:
x_l += 1
elif j[0] > i[0]:
x_g += 1
elif j[0] == i[0]:
x_equal += 1
if j[1]<i[1]:
y_l += 1
elif j[1] > i[1]:
y_g += 1
elif j[1] == i[1]:
y_equal += 1
if x_l > 0 and y_l > 0 and x_g > 0 and y_g > 0 and x_equal > 1 and y_equal > 1:
cnt += 1
print(cnt) | 26 | 92 | 0 | 155192352 | import sys
input = sys.stdin.readline
n = int(input())
g = [tuple(map(int, input().split())) for _ in range(n)]
w = sorted(g)
q = [(j,i) for (i,j) in sorted([(j,i) for (i,j) in g])]
a = set()
b = set()
for i in range(n-2):
if w[i][0] == w[i+1][0] == w[i+2][0]:
a.add(w[i+1])
if q[i][1] == q[i+1][1] == q[i+2][1]:
b.add(q[i+1])
print(len(a&b)) | Codeforces Round 112 (Div. 2) | CF | 2,012 | 2 | 256 | Supercentral Point | One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of points (x1, y1), (x2, y2), ..., (xn, yn). Let's define neighbors for some fixed point from the given set (x, y):
- point (x', y') is (x, y)'s right neighbor, if x' > x and y' = y
- point (x', y') is (x, y)'s left neighbor, if x' < x and y' = y
- point (x', y') is (x, y)'s lower neighbor, if x' = x and y' < y
- point (x', y') is (x, y)'s upper neighbor, if x' = x and y' > y
We'll consider point (x, y) from the given set supercentral, if it has at least one upper, at least one lower, at least one left and at least one right neighbor among this set's points.
Vasya marked quite many points on the paper. Analyzing the picture manually is rather a challenge, so Vasya asked you to help him. Your task is to find the number of supercentral points in the given set. | The first input line contains the only integer n (1 ≤ n ≤ 200) — the number of points in the given set. Next n lines contain the coordinates of the points written as "x y" (without the quotes) (|x|, |y| ≤ 1000), all coordinates are integers. The numbers in the line are separated by exactly one space. It is guaranteed that all points are different. | Print the only number — the number of supercentral points of the given set. | null | In the first sample the supercentral points are only points (1, 1) and (1, 2).
In the second sample there is one supercental point — point (0, 0). | [{"input": "8\n1 1\n4 2\n3 1\n1 2\n0 2\n0 1\n1 0\n1 3", "output": "2"}, {"input": "5\n0 0\n0 1\n1 0\n0 -1\n-1 0", "output": "1"}] | 1,000 | ["implementation"] | 26 | [{"input": "8\r\n1 1\r\n4 2\r\n3 1\r\n1 2\r\n0 2\r\n0 1\r\n1 0\r\n1 3\r\n", "output": "2\r\n"}, {"input": "5\r\n0 0\r\n0 1\r\n1 0\r\n0 -1\r\n-1 0\r\n", "output": "1\r\n"}, {"input": "9\r\n-565 -752\r\n-184 723\r\n-184 -752\r\n-184 1\r\n950 723\r\n-565 723\r\n950 -752\r\n950 1\r\n-565 1\r\n", "output": "1\r\n"}, {"input": "25\r\n-651 897\r\n916 897\r\n-651 -808\r\n-748 301\r\n-734 414\r\n-651 -973\r\n-734 897\r\n916 -550\r\n-758 414\r\n916 180\r\n-758 -808\r\n-758 -973\r\n125 -550\r\n125 -973\r\n125 301\r\n916 414\r\n-748 -808\r\n-651 301\r\n-734 301\r\n-307 897\r\n-651 -550\r\n-651 414\r\n125 -808\r\n-748 -550\r\n916 -808\r\n", "output": "7\r\n"}, {"input": "1\r\n487 550\r\n", "output": "0\r\n"}, {"input": "10\r\n990 -396\r\n990 736\r\n990 646\r\n990 -102\r\n990 -570\r\n990 155\r\n990 528\r\n990 489\r\n990 268\r\n990 676\r\n", "output": "0\r\n"}, {"input": "30\r\n507 836\r\n525 836\r\n-779 196\r\n507 -814\r\n525 -814\r\n525 42\r\n525 196\r\n525 -136\r\n-779 311\r\n507 -360\r\n525 300\r\n507 578\r\n507 311\r\n-779 836\r\n507 300\r\n525 -360\r\n525 311\r\n-779 -360\r\n-779 578\r\n-779 300\r\n507 42\r\n525 578\r\n-779 379\r\n507 196\r\n525 379\r\n507 379\r\n-779 -814\r\n-779 42\r\n-779 -136\r\n507 -136\r\n", "output": "8\r\n"}, {"input": "25\r\n890 -756\r\n890 -188\r\n-37 -756\r\n-37 853\r\n523 998\r\n-261 853\r\n-351 853\r\n-351 -188\r\n523 -756\r\n-261 -188\r\n-37 998\r\n523 -212\r\n-351 998\r\n-37 -188\r\n-351 -756\r\n-37 -212\r\n890 998\r\n890 -212\r\n523 853\r\n-351 -212\r\n-261 -212\r\n-261 998\r\n-261 -756\r\n890 853\r\n523 -188\r\n", "output": "9\r\n"}, {"input": "21\r\n-813 -11\r\n486 254\r\n685 254\r\n-708 254\r\n-55 -11\r\n-671 -191\r\n486 -11\r\n-671 -11\r\n685 -11\r\n685 -191\r\n486 -191\r\n-55 254\r\n-708 -11\r\n-813 254\r\n-708 -191\r\n41 -11\r\n-671 254\r\n-813 -191\r\n41 254\r\n-55 -191\r\n41 -191\r\n", "output": "5\r\n"}, {"input": "4\r\n1 0\r\n2 0\r\n1 1\r\n1 -1\r\n", "output": "0\r\n"}] | false | stdio | null | true |
427/B | 427 | B | PyPy 3-64 | TESTS | 4 | 93 | 22,528,000 | 201051017 | n, t, c = tuple(map(int, input().split()))
prisoners = tuple(map(int, input().split()))
def get_lines_length(prisoners):
length = 0
lines_length = []
for prisoner in prisoners:
if prisoner <= t:
length += 1
else:
lines_length.append(length)
length = 0
if length > 0:
lines_length.append(length)
return lines_length
def get_count(lines_length, c):
count = 0
for length in lines_length:
if length < c:
result = 0
elif length == c:
result = 1
else:
result = length // c + 1
count += result
return count
lines_length = get_lines_length(prisoners)
print(get_count(lines_length, c)) | 80 | 108 | 19,865,600 | 226021405 | n, t, c = (int(x) for x in input().split())
severities = input().split()
ways = 0
run = 0
for s in severities:
s = int(s)
if s > t:
run = 0
else:
run += 1
if run >= c:
ways += 1
print(ways) | Codeforces Round 244 (Div. 2) | CF | 2,014 | 1 | 256 | Prison Transfer | The prison of your city has n prisoners. As the prison can't accommodate all of them, the city mayor has decided to transfer c of the prisoners to a prison located in another city.
For this reason, he made the n prisoners to stand in a line, with a number written on their chests. The number is the severity of the crime he/she has committed. The greater the number, the more severe his/her crime was.
Then, the mayor told you to choose the c prisoners, who will be transferred to the other prison. He also imposed two conditions. They are,
- The chosen c prisoners has to form a contiguous segment of prisoners.
- Any of the chosen prisoner's crime level should not be greater then t. Because, that will make the prisoner a severe criminal and the mayor doesn't want to take the risk of his running away during the transfer.
Find the number of ways you can choose the c prisoners. | The first line of input will contain three space separated integers n (1 ≤ n ≤ 2·105), t (0 ≤ t ≤ 109) and c (1 ≤ c ≤ n). The next line will contain n space separated integers, the ith integer is the severity ith prisoner's crime. The value of crime severities will be non-negative and will not exceed 109. | Print a single integer — the number of ways you can choose the c prisoners. | null | null | [{"input": "4 3 3\n2 3 1 1", "output": "2"}, {"input": "1 1 1\n2", "output": "0"}, {"input": "11 4 2\n2 2 0 7 3 2 2 4 9 1 4", "output": "6"}] | 1,100 | ["data structures", "implementation"] | 80 | [{"input": "4 3 3\r\n2 3 1 1\r\n", "output": "2\r\n"}, {"input": "1 1 1\r\n2\r\n", "output": "0\r\n"}, {"input": "11 4 2\r\n2 2 0 7 3 2 2 4 9 1 4\r\n", "output": "6\r\n"}, {"input": "57 2 10\r\n7 5 2 7 4 1 0 5 2 9 2 9 8 6 6 5 9 6 8 1 0 1 0 3 2 6 5 2 8 8 8 8 0 9 4 3 6 6 2 4 5 1 2 0 1 7 1 1 5 4 5 0 7 5 1 9 6\r\n", "output": "0\r\n"}, {"input": "2 228885628 1\r\n90897004 258427916\r\n", "output": "1\r\n"}, {"input": "3 1 1\r\n1 2 1\r\n", "output": "2\r\n"}, {"input": "3 3 3\r\n3 2 3\r\n", "output": "1\r\n"}, {"input": "4 2 2\r\n1 3 3 2\r\n", "output": "0\r\n"}, {"input": "1 228 1\r\n1\r\n", "output": "1\r\n"}] | false | stdio | null | true |
449/B | 449 | B | PyPy 3 | TESTS | 3 | 1,169 | 61,030,400 | 131712818 | import bisect
import collections
import copy
import enum
import functools
import heapq
import itertools
import math
import random
import re
import sys
import time
import string
from typing import List
sys.setrecursionlimit(3001)
input = sys.stdin.readline
n,m,k = map(int,input().split())
g = collections.defaultdict(list)
for _ in range(m):
u,v,x = map(int,input().split())
g[u].append([v,x,0])
g[v].append([u,x,0])
for _ in range(k):
s, y = map(int, input().split())
g[1].append([s,y,1])
dist = [float("inf")]*(n+1)
r = 0
def dijkstra(start):
global r
q = [(0, start,0)]
while q:
dis, f,node = heapq.heappop(q)
if dist[node] <= dis:
continue
r+=f
dist[node] = dis
for next_node, w,f in g[node]:
if dist[next_node] == float("inf"):
heapq.heappush(q, (dis+w,f,next_node))
return dist
dijkstra(1)
print(k-r) | 45 | 1,871 | 131,072,000 | 167802107 | import sys
import math
import collections
from heapq import heappush, heappop
input = sys.stdin.readline
ints = lambda: list(map(int, input().split()))
n, m, k = ints()
graph = [[] for _ in range(n + 1)]
for _ in range(m):
a, b, c = ints()
graph[a].append((b, c, 0))
graph[b].append((a, c, 0))
trains = [[] for _ in range(n + 1)]
for _ in range(k):
s, y = ints()
trains[s].append(y)
graph[1].append((s, y, 1))
graph[s].append((1, y, 1))
distances = [math.inf for _ in range(n + 1)]
distances[1] = 0
visited = [False for _ in range(n + 1)]
q = []
heappush(q, (0, 1))
while q:
d, node = heappop(q)
if visited[node]: continue
visited[node] = True
for (nb, c, t) in graph[node]:
total = c + distances[node]
if total < distances[nb] and not visited[nb]:
distances[nb] = total
heappush(q, (distances[nb], nb))
ans = 0
for i in range(2, n + 1):
g_r = 0
g_t = 0
for (nb, c, t) in graph[i]:
if distances[nb] + c == distances[i]:
if t: g_t += 1
else: g_r += 1
if g_r > 0:
ans += len(trains[i])
else: ans += len(trains[i]) - 1
print(ans) | Codeforces Round 257 (Div. 1) | CF | 2,014 | 2 | 256 | Jzzhu and Cities | Jzzhu is the president of country A. There are n cities numbered from 1 to n in his country. City 1 is the capital of A. Also there are m roads connecting the cities. One can go from city ui to vi (and vise versa) using the i-th road, the length of this road is xi. Finally, there are k train routes in the country. One can use the i-th train route to go from capital of the country to city si (and vise versa), the length of this route is yi.
Jzzhu doesn't want to waste the money of the country, so he is going to close some of the train routes. Please tell Jzzhu the maximum number of the train routes which can be closed under the following condition: the length of the shortest path from every city to the capital mustn't change. | The first line contains three integers n, m, k (2 ≤ n ≤ 105; 1 ≤ m ≤ 3·105; 1 ≤ k ≤ 105).
Each of the next m lines contains three integers ui, vi, xi (1 ≤ ui, vi ≤ n; ui ≠ vi; 1 ≤ xi ≤ 109).
Each of the next k lines contains two integers si and yi (2 ≤ si ≤ n; 1 ≤ yi ≤ 109).
It is guaranteed that there is at least one way from every city to the capital. Note, that there can be multiple roads between two cities. Also, there can be multiple routes going to the same city from the capital. | Output a single integer representing the maximum number of the train routes which can be closed. | null | null | [{"input": "5 5 3\n1 2 1\n2 3 2\n1 3 3\n3 4 4\n1 5 5\n3 5\n4 5\n5 5", "output": "2"}, {"input": "2 2 3\n1 2 2\n2 1 3\n2 1\n2 2\n2 3", "output": "2"}] | 2,000 | ["graphs", "greedy", "shortest paths"] | 45 | [{"input": "5 5 3\r\n1 2 1\r\n2 3 2\r\n1 3 3\r\n3 4 4\r\n1 5 5\r\n3 5\r\n4 5\r\n5 5\r\n", "output": "2\r\n"}, {"input": "2 2 3\r\n1 2 2\r\n2 1 3\r\n2 1\r\n2 2\r\n2 3\r\n", "output": "2\r\n"}, {"input": "5 4 3\r\n1 2 999999999\r\n2 3 1000000000\r\n3 4 529529529\r\n5 1 524524524\r\n5 524444444\r\n5 529999999\r\n2 1000000000\r\n", "output": "2\r\n"}, {"input": "3 2 5\r\n1 2 2\r\n2 3 4\r\n3 5\r\n3 5\r\n3 5\r\n3 6\r\n3 7\r\n", "output": "4\r\n"}, {"input": "5 5 3\r\n1 2 999999999\r\n2 3 1000000000\r\n3 4 529529529\r\n5 1 524524524\r\n5 3 1000000000\r\n5 524444444\r\n5 529999999\r\n2 1000000000\r\n", "output": "2\r\n"}, {"input": "2 1 5\r\n1 2 4\r\n2 3\r\n2 5\r\n2 4\r\n2 4\r\n2 5\r\n", "output": "4\r\n"}, {"input": "3 3 6\r\n1 2 499999999\r\n2 3 500000000\r\n1 3 999999999\r\n2 499999999\r\n2 500000000\r\n2 499999999\r\n3 999999999\r\n3 1000000000\r\n3 1000000000\r\n", "output": "6\r\n"}, {"input": "2 1 1\r\n1 2 1\r\n2 1000000000\r\n", "output": "1\r\n"}, {"input": "3 2 2\r\n1 2 4\r\n2 3 4\r\n2 2\r\n3 6\r\n", "output": "1\r\n"}, {"input": "5 5 2\r\n1 2 100\r\n2 3 100\r\n3 4 100\r\n4 5 20\r\n2 5 5\r\n5 50\r\n4 1\r\n", "output": "1\r\n"}, {"input": "3 2 2\r\n1 2 100\r\n2 3 1\r\n2 1\r\n3 3\r\n", "output": "1\r\n"}] | false | stdio | null | true |
294/A | 294 | A | Python 3 | TESTS | 26 | 62 | 0 | 160229598 | # 294A
n = int(input())
wires = list(map(int, input().split()))
m = int(input())
for i in range(m):
coords = list(map(int, input().split()))
x = coords[0]
y = coords[1]
# how many birds to right and left of dead bird
left = y - 1; right = wires[x-1] - y
wires[x-1] = 0 # all birds (alive or dead) leave the wire.
# check if top or bottom wire.
if x == n: # bottom, birds to right fly away, birds to left go up 1.
wires[x-2] += left
elif x == 1: # top, birds to left fly away, birds to right go down 1.
wires[x] += right
else:
wires[x-2] += left
wires[x] += right
for num in wires:
print(num) | 31 | 62 | 0 | 187419590 | # tc = int(input())
for i in range(1):
n = int(input())
li =list(map(int,input().split()))
x = int(input())
ans = 0
for i in range(x):
lix = list(map(int,input().split()))
if lix[0]-2>=0:
li[lix[0]-2] += lix[1]-1
if lix[0]<n:
li[lix[0]] += li[lix[0]-1]-lix[1]
li[lix[0]-1] = 0
for i in li:
print(i) | Codeforces Round 178 (Div. 2) | CF | 2,013 | 2 | 256 | Shaass and Oskols | Shaass has decided to hunt some birds. There are n horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to n from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are ai oskols sitting on the i-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the i-th wire). Consequently all the birds on the i-th wire to the left of the dead bird get scared and jump up on the wire number i - 1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number i + 1, if there exists no such wire they fly away.
Shaass has shot m birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots. | The first line of the input contains an integer n, (1 ≤ n ≤ 100). The next line contains a list of space-separated integers a1, a2, ..., an, (0 ≤ ai ≤ 100).
The third line contains an integer m, (0 ≤ m ≤ 100). Each of the next m lines contains two integers xi and yi. The integers mean that for the i-th time Shaass shoot the yi-th (from left) bird on the xi-th wire, (1 ≤ xi ≤ n, 1 ≤ yi). It's guaranteed there will be at least yi birds on the xi-th wire at that moment. | On the i-th line of the output print the number of birds on the i-th wire. | null | null | [{"input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6", "output": "0\n12\n5\n0\n16"}, {"input": "3\n2 4 1\n1\n2 2", "output": "3\n0\n3"}] | 800 | ["implementation", "math"] | 31 | [{"input": "5\r\n10 10 10 10 10\r\n5\r\n2 5\r\n3 13\r\n2 12\r\n1 13\r\n4 6\r\n", "output": "0\r\n12\r\n5\r\n0\r\n16\r\n"}, {"input": "3\r\n2 4 1\r\n1\r\n2 2\r\n", "output": "3\r\n0\r\n3\r\n"}, {"input": "5\r\n58 51 45 27 48\r\n5\r\n4 9\r\n5 15\r\n4 5\r\n5 8\r\n1 43\r\n", "output": "0\r\n66\r\n57\r\n7\r\n0\r\n"}, {"input": "10\r\n48 53 10 28 91 56 81 2 67 52\r\n2\r\n2 40\r\n6 51\r\n", "output": "87\r\n0\r\n23\r\n28\r\n141\r\n0\r\n86\r\n2\r\n67\r\n52\r\n"}, {"input": "2\r\n72 45\r\n6\r\n1 69\r\n2 41\r\n1 19\r\n2 7\r\n1 5\r\n2 1\r\n", "output": "0\r\n0\r\n"}, {"input": "10\r\n95 54 36 39 98 30 19 24 14 12\r\n3\r\n9 5\r\n8 15\r\n7 5\r\n", "output": "95\r\n54\r\n36\r\n39\r\n98\r\n34\r\n0\r\n28\r\n13\r\n21\r\n"}, {"input": "1\r\n100\r\n1\r\n1 100\r\n", "output": "0\r\n"}, {"input": "1\r\n100\r\n1\r\n1 1\r\n", "output": "0\r\n"}, {"input": "1\r\n50\r\n1\r\n1 25\r\n", "output": "0\r\n"}, {"input": "2\r\n50 0\r\n1\r\n1 1\r\n", "output": "0\r\n49\r\n"}, {"input": "1\r\n10\r\n0\r\n", "output": "10\r\n"}] | false | stdio | null | true |
294/C | 294 | C | Python 3 | TESTS | 7 | 93 | 307,200 | 101680806 | from sys import stdin, stdout
from math import pow
def main():
N, M = 2000, 1000000007
fact = factorial(N, M)
n, m = readline()
onn = [0] + sorted(readline()) + [n + 1]
off = []
for i in range(1, len(onn)):
off.append(onn[i] - onn[i - 1] - 1)
answer = fact[sum(off)]
div = 1
for i in range(len(off)):
div *= fact[off[i]] % M
if 0 < i < len(off) - 1 and off[i] > 1:
answer *= pow(2, off[i] - 1) % M
return int(answer / div)
def factorial(N, M):
fact = [1] * N
for i in range(2, 2000):
fact[i] = i * fact[i - 1] % M
return fact
def readline():
return map(int, stdin.readline().strip().split())
if __name__ == '__main__':
stdout.write(str(main()) + '\n') | 30 | 109 | 307,200 | 54146781 | # https://codeforces.com/problemset/problem/294/C
import sys
import math
mod = 10 ** 9 + 7
def factorial(n):
x = 1
for i in range(n):
yield x
x = x * (i + 1) % mod
f = list(factorial(1001))
def C(n, k):
return f[n] * pow(f[k], mod - 2, mod) * pow(f[n - k], mod - 2, mod) % mod
def main():
# sys.stdin = open('E:\\Sublime\\in.txt', 'r')
# sys.stdout = open('E:\\Sublime\\out.txt', 'w')
# sys.stderr = open('E:\\Sublime\\err.txt', 'w')
# n = int(sys.stdin.readline().strip())
# a, b = map(int, sys.stdin.readline().strip().split()[:2])
n, m = [int(x) for x in sys.stdin.readline().strip().split()]
b = [-1] + [int(x) - 1 for x in sys.stdin.readline().strip().split()] + [n]
b.sort()
ans = 1
cur = n - m
for i in range(1, len(b)):
l = b[i] - b[i - 1] - 1
ans = ans * C(cur, l) % mod
if (l > 0) and (i > 1) and (i < len(b) - 1):
ans = ans * pow(2, l - 1, mod) % mod
cur -= l
print(ans)
if __name__ == '__main__':
main()
# hajj
# ＿＿
# ／＞ フ
# | _ _ l
# ／` ミ＿xノ
# / |
# / ヽ ﾉ
# │ | | |
# ／￣| | | |
# | (￣ヽ＿_ヽ_)__)
# ＼二つ | Codeforces Round 178 (Div. 2) | CF | 2,013 | 1 | 256 | Shaass and Lights | There are n lights aligned in a row. These lights are numbered 1 to n from left to right. Initially some of the lights are switched on. Shaass wants to switch all the lights on. At each step he can switch a light on (this light should be switched off at that moment) if there's at least one adjacent light which is already switched on.
He knows the initial state of lights and he's wondering how many different ways there exist to switch all the lights on. Please find the required number of ways modulo 1000000007 (109 + 7). | The first line of the input contains two integers n and m where n is the number of lights in the sequence and m is the number of lights which are initially switched on, (1 ≤ n ≤ 1000, 1 ≤ m ≤ n). The second line contains m distinct integers, each between 1 to n inclusive, denoting the indices of lights which are initially switched on. | In the only line of the output print the number of different possible ways to switch on all the lights modulo 1000000007 (109 + 7). | null | null | [{"input": "3 1\n1", "output": "1"}, {"input": "4 2\n1 4", "output": "2"}, {"input": "11 2\n4 8", "output": "6720"}] | 1,900 | ["combinatorics", "number theory"] | 30 | [{"input": "3 1\r\n1\r\n", "output": "1\r\n"}, {"input": "4 2\r\n1 4\r\n", "output": "2\r\n"}, {"input": "11 2\r\n4 8\r\n", "output": "6720\r\n"}, {"input": "4 2\r\n1 3\r\n", "output": "2\r\n"}, {"input": "4 4\r\n1 2 3 4\r\n", "output": "1\r\n"}, {"input": "4 2\r\n1 3\r\n", "output": "2\r\n"}, {"input": "4 4\r\n1 2 3 4\r\n", "output": "1\r\n"}, {"input": "1000 3\r\n100 900 10\r\n", "output": "727202008\r\n"}, {"input": "74 13\r\n6 14 19 20 21 24 30 43 58 61 69 70 73\r\n", "output": "16623551\r\n"}, {"input": "74 13\r\n6 14 19 20 21 24 30 43 58 61 69 70 73\r\n", "output": "16623551\r\n"}, {"input": "74 13\r\n6 14 19 20 21 24 30 43 58 61 69 70 73\r\n", "output": "16623551\r\n"}, {"input": "74 13\r\n6 14 19 20 21 24 30 43 58 61 69 70 73\r\n", "output": "16623551\r\n"}, {"input": "74 13\r\n6 14 19 20 21 24 30 43 58 61 69 70 73\r\n", "output": "16623551\r\n"}, {"input": "74 13\r\n6 14 19 20 21 24 30 43 58 61 69 70 73\r\n", "output": "16623551\r\n"}, {"input": "68 37\r\n1 2 3 6 7 8 10 11 12 14 16 18 22 23 24 26 30 31 32 35 37 39 41 42 45 47 50 51 52 54 58 59 61 62 63 64 68\r\n", "output": "867201120\r\n"}, {"input": "132 48\r\n6 7 8 12 15 17 18 19 22 24 25 26 30 33 35 38 40 43 46 49 50 51 52 54 59 60 66 70 76 79 87 89 91 92 94 98 99 101 102 105 106 109 113 115 116 118 120 129\r\n", "output": "376947760\r\n"}, {"input": "36 24\r\n1 7 8 10 11 12 13 14 15 16 17 19 21 22 25 26 27 28 29 30 31 32 35 36\r\n", "output": "63866880\r\n"}, {"input": "100 2\r\n11 64\r\n", "output": "910895596\r\n"}, {"input": "1000 1\r\n35\r\n", "output": "253560421\r\n"}, {"input": "1000 2\r\n747 798\r\n", "output": "474746180\r\n"}, {"input": "1000 3\r\n804 811 984\r\n", "output": "600324842\r\n"}, {"input": "1 1\r\n1\r\n", "output": "1\r\n"}] | false | stdio | null | true |
491/B | 491 | B | Python 3 | TESTS | 1 | 46 | 0 | 8769508 | N,M = map(int,input().split())
C = int(input())
Ci = [list(map(int,input().split())) for i in range(C)]
H = int(input())
Hi = [list(map(int,input().split())) for i in range(H)]
xi = 0
yi = 0
for y in range(2):
for x in range(C):
if y == 0:
xi += x
else:
yi += x
xi //= C
yi //= C
minimum = 10000000000
nmin = 0
for y in range(H):
z = Hi[y][0] + Hi[y][1]
if z < minimum:
minimum = z
nmin = y
maximum = 0
for x in range(C):
e = abs(Ci[x][0]-Hi[nmin][0]) + abs(Ci[x][1]-Hi[nmin][1])
if e > maximum:
maximum = e
print(maximum)
print(nmin+1) | 28 | 1,747 | 10,035,200 | 72031483 | N, M = input().split()
a, b, c, d = [int(1e10) for _ in range(4)]
for i in range(int(input())):
x, y = list(map(int, input().split()))
a, b, c, d = min(a, x + y), min(b, x - y), min(c, - x + y), min(d, - x - y)
res, pos = int(1e10), 0
for i in range(int(input())):
x, y = list(map(int, input().split()))
ans = max(max(x + y - a, x - y - b), max( - x + y - c, - x - y - d))
if ans < res:
pos = i + 1
res = ans
print(res, pos, sep = '\n') | Testing Round 11 | CF | 2,014 | 2 | 256 | New York Hotel | Think of New York as a rectangular grid consisting of N vertical avenues numerated from 1 to N and M horizontal streets numerated 1 to M. C friends are staying at C hotels located at some street-avenue crossings. They are going to celebrate birthday of one of them in the one of H restaurants also located at some street-avenue crossings. They also want that the maximum distance covered by one of them while traveling to the restaurant to be minimum possible. Help friends choose optimal restaurant for a celebration.
Suppose that the distance between neighboring crossings are all the same equal to one kilometer. | The first line contains two integers N и M — size of the city (1 ≤ N, M ≤ 109). In the next line there is a single integer C (1 ≤ C ≤ 105) — the number of hotels friends stayed at. Following C lines contain descriptions of hotels, each consisting of two coordinates x and y (1 ≤ x ≤ N, 1 ≤ y ≤ M). The next line contains an integer H — the number of restaurants (1 ≤ H ≤ 105). Following H lines contain descriptions of restaurants in the same format.
Several restaurants and hotels may be located near the same crossing. | In the first line output the optimal distance. In the next line output index of a restaurant that produces this optimal distance. If there are several possibilities, you are allowed to output any of them. | null | null | [{"input": "10 10\n2\n1 1\n3 3\n2\n1 10\n4 4", "output": "6\n2"}] | 2,100 | ["greedy", "math"] | 28 | [{"input": "10 10\r\n2\r\n1 1\r\n3 3\r\n2\r\n1 10\r\n4 4\r\n", "output": "6\r\n2\r\n"}, {"input": "100 100\r\n10\r\n53 20\r\n97 6\r\n12 74\r\n48 92\r\n97 13\r\n47 96\r\n75 32\r\n69 21\r\n95 75\r\n1 54\r\n10\r\n36 97\r\n41 1\r\n1 87\r\n39 23\r\n27 44\r\n73 97\r\n1 1\r\n6 26\r\n48 3\r\n5 69\r\n", "output": "108\r\n4\r\n"}, {"input": "100 100\r\n10\r\n86 72\r\n25 73\r\n29 84\r\n34 33\r\n29 20\r\n84 83\r\n41 80\r\n22 22\r\n16 89\r\n77 49\r\n1\r\n4 23\r\n", "output": "140\r\n1\r\n"}, {"input": "100 100\r\n1\r\n66 77\r\n10\r\n70 11\r\n76 69\r\n79 39\r\n90 3\r\n38 87\r\n61 81\r\n98 66\r\n63 68\r\n62 93\r\n53 36\r\n", "output": "9\r\n6\r\n"}, {"input": "1000000000 1000000000\r\n1\r\n1 1\r\n1\r\n1000000000 1000000000\r\n", "output": "1999999998\r\n1\r\n"}, {"input": "123456789 987654321\r\n1\r\n312 987654321\r\n1\r\n123456789 213\r\n", "output": "1111110585\r\n1\r\n"}, {"input": "453456789 987654321\r\n1\r\n443943901 1\r\n1\r\n1354 213389832\r\n", "output": "657332378\r\n1\r\n"}, {"input": "923456789 987654321\r\n1\r\n443943901 132319791\r\n1\r\n1354 560\r\n", "output": "576261778\r\n1\r\n"}, {"input": "100 100\r\n1\r\n1 100\r\n1\r\n1 100\r\n", "output": "0\r\n1\r\n"}, {"input": "1 1\r\n1\r\n1 1\r\n1\r\n1 1\r\n", "output": "0\r\n1\r\n"}, {"input": "1000000000 1000000000\r\n2\r\n1 1\r\n3 3\r\n2\r\n1 10\r\n4 4\r\n", "output": "6\r\n2\r\n"}] | false | stdio | null | true |
297/A | 297 | A | Python 3 | TESTS | 2 | 124 | 7,065,600 | 37388457 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
a=[]
b=[]
a=input()
b=input()
cnta=0
cntb=0
for i,val in enumerate(a):
if val=='1':
cnta=cnta+1
for i,val in enumerate(b):
if val=='1':
cntb=cntb+1
if cnta&1==1:
pa=cnta+1
if cnta>=cntb:
print('YES')
else:
print('NO') | 79 | 186 | 0 | 226402581 | a,b=input(),input()
ca,cb=a.count('1'),b.count('1')
print('YES' if ca+(ca&1)>=cb else 'NO') | Codeforces Round 180 (Div. 1) | CF | 2,013 | 1 | 256 | Parity Game | You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
- Write parity(a) to the end of a. For example, $$1010 \rightarrow 10100$$.
- Remove the first character of a. For example, $$1001 \rightarrow 001$$. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise. | The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x. | Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise. | null | In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110 | [{"input": "01011\n0110", "output": "YES"}, {"input": "0011\n1110", "output": "NO"}] | 1,700 | ["constructive algorithms"] | 79 | [{"input": "01011\r\n0110\r\n", "output": "YES\r\n"}, {"input": "0011\r\n1110\r\n", "output": "NO\r\n"}, {"input": "11111\r\n111111\r\n", "output": "YES\r\n"}, {"input": "0110011\r\n01100110\r\n", "output": "YES\r\n"}, {"input": "10000100\r\n011110\r\n", "output": "NO\r\n"}, {"input": "1\r\n0\r\n", "output": "YES\r\n"}, {"input": "11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\r\n11\r\n", "output": "YES\r\n"}, {"input": "11\r\n111\r\n", "output": "NO\r\n"}, {"input": "1\r\n1\r\n", "output": "YES\r\n"}, {"input": "1\r\n0\r\n", "output": "YES\r\n"}] | false | stdio | null | true |
58/D | 58 | D | PyPy 3 | TESTS | 5 | 310 | 0 | 95808360 | import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n = int(input())
a = [input().rstrip() for _ in range(n)]
d = input().rstrip()
words = [[] for _ in range(100)]
total_len = 0
for word in a:
words[len(word)].append(word)
total_len += len(word)
for i in range(1, 11):
words[i].sort(reverse=True)
width = total_len // (n // 2)
ans = []
for i in range(n // 2):
res = '~' * 100
res_j = -1
for j in range(1, width):
if words[j] and words[width - j] and res > words[j][-1] + d + words[width - j][-1]:
res = words[j][-1] + d + words[width - j][-1]
res_j = j
ans.append(res)
words[res_j].pop()
words[width - res_j].pop()
sys.stdout.buffer.write('\n'.join(ans).encode('utf-8')) | 46 | 372 | 1,740,800 | 213673188 | from collections import deque
n = int(input())
a = [input() for i in range(n)]
a.sort()
dq = deque(a)
# print(dq)
d = input()
n //= 2
common_len = sum(len(s) for s in a) // n
res = []
for i in range(n):
s1 = dq.popleft()
for s2 in dq:
if len(s1) + len(s2) == common_len:
dq.remove(s2)
res.append(min(s1+d+s2, s2+d+s1))
break
for x in sorted(res):
print(x) | Codeforces Beta Round 54 (Div. 2) | CF | 2,011 | 2 | 256 | Calendar | BerOilGasDiamondBank has branches in n cities, at that n is an even number. The bank management wants to publish a calendar with the names of all those cities written in two columns: the calendar should consist of exactly n / 2 lines of strictly equal length, each of which contains exactly two names and exactly one separator character between them. The name of every city should be used in the calendar exactly once. For historical reasons the symbol d is used as the separator of words in the calendar.
The BerOilGasDiamondBank management wants to show that all its branches are equally important to it, that's why the order of their appearance in the calendar should be following: if we "glue"(concatinate) all the n / 2 calendar lines (from top to bottom) to make a single line, then the lexicographically minimal line is obtained. No separator character will be used to separate calendar lines. For example, if the lines are "bertown!berville", "newberville!bera", then the resulting line is "bertown!bervillenewberville!bera". In some sense one has to find the lexicographically minimal calendar, where the comparison of calendars happens line by line.
Help BerOilGasDiamondBank and construct the required calendar. | The first line contains an integer n (1 ≤ n ≤ 104, n is even) which is the number of branches. Then follow n lines which are the names of the cities. All the names consist of lowercase Latin letters; their lengths are no less than 1 and no more than 10 symbols. The next line contains a single symbol d (d has an ASCII-code from 33 to 126 inclusively, excluding lowercase Latin letters) which is the separator between words in the calendar lines. It is guaranteed that the calendar is possible to be constructed and all the names are different. | Print n / 2 lines of similar length which are the required calendar. Every line should contain exactly two words and exactly one separator between them. If there are several solutions, print the lexicographically minimal one. The lexicographical comparison of lines is realized by the "<" operator in the modern programming languages. | null | null | [{"input": "4\nb\naa\nhg\nc\n.", "output": "aa.b\nc.hg"}, {"input": "2\naa\na\n!", "output": "a!aa"}, {"input": "2\naa\na\n|", "output": "aa|a"}] | 2,000 | ["greedy", "strings"] | 46 | [{"input": "4\r\nb\r\naa\r\nhg\r\nc\r\n.\r\n", "output": "aa.b\r\nc.hg\r\n"}, {"input": "2\r\naa\r\na\r\n!\r\n", "output": "a!aa\r\n"}, {"input": "2\r\naa\r\na\r\n|\r\n", "output": "aa|a\r\n"}, {"input": "4\r\nqhcivbxotj\r\nirgxzzxvw\r\npxdmcyszvk\r\nyyaevcdal\r\n~\r\n", "output": "irgxzzxvw~pxdmcyszvk\r\nqhcivbxotj~yyaevcdal\r\n"}, {"input": "8\r\nbad\r\nrnnpg\r\njvcjsxfob\r\nad\r\nairnnpg\r\nqury\r\njvcjsxfo\r\nquryai\r\n6\r\n", "output": "ad6jvcjsxfob\r\nairnnpg6qury\r\nbad6jvcjsxfo\r\nquryai6rnnpg\r\n"}, {"input": "6\r\neh\r\nehkhdp\r\ngque\r\nkhdptvgque\r\ntvgque\r\nehkhdptv\r\n}\r\n", "output": "ehkhdptv}gque\r\nehkhdp}tvgque\r\neh}khdptvgque\r\n"}, {"input": "10\r\ndoecgzo\r\ntjptpqp\r\noitegxzwlp\r\nmwsrwmeyeg\r\nsmapaqanak\r\nsmapaqa\r\nqghrydm\r\nnakqghrydm\r\nmedoraus\r\nnyvgozjdf\r\n|\r\n", "output": "doecgzo|mwsrwmeyeg\r\nmedoraus|nyvgozjdf\r\nnakqghrydm|qghrydm\r\noitegxzwlp|smapaqa\r\nsmapaqanak|tjptpqp\r\n"}, {"input": "30\r\nd\r\nahx\r\nr\r\nyd\r\np\r\nnhy\r\na\r\ntqt\r\nctp\r\ntp\r\nho\r\nry\r\nm\r\ng\r\ns\r\nn\r\nct\r\nsc\r\nqr\r\nrry\r\ny\r\nhxm\r\nqrr\r\nsct\r\ncwu\r\nq\r\ndk\r\nrf\r\nhyd\r\nnh\r\n$\r\n", "output": "a$ahx\r\nct$dk\r\nctp$d\r\ncwu$g\r\nho$nh\r\nhxm$m\r\nhyd$n\r\nnhy$p\r\nq$qrr\r\nqr$rf\r\nr$rry\r\nry$sc\r\ns$sct\r\ntp$yd\r\ntqt$y\r\n"}, {"input": "14\r\neskrrgzq\r\nxbmynhxfg\r\nwwffafny\r\nfaxcnrqkkb\r\nfaxcnrqk\r\nkbwwffafny\r\nmnborvqeae\r\nranfahuebj\r\neskrrgzqk\r\nfaxcnrqkk\r\ncznaycxe\r\nrnkgfgyq\r\nkxbmynhxfg\r\nbwwffafny\r\n}\r\n", "output": "bwwffafny}eskrrgzqk\r\ncznaycxe}faxcnrqkkb\r\neskrrgzq}kbwwffafny\r\nfaxcnrqkk}xbmynhxfg\r\nfaxcnrqk}kxbmynhxfg\r\nmnborvqeae}rnkgfgyq\r\nranfahuebj}wwffafny\r\n"}, {"input": "34\r\nobseknnnqk\r\ncvyvvbcgb\r\nxvmhfzfl\r\ngrtp\r\nhbcbhj\r\nknnnqk\r\ncyud\r\nkuaeui\r\naeui\r\nlhpdobsekn\r\ncxmigej\r\ncvyvvbcgbs\r\nuwuu\r\nnnqk\r\npzcftfrrqp\r\nnwsyrgqa\r\nxvmhfzflku\r\nndcis\r\nxhaznwqsgk\r\ncftfrrqp\r\nkakdggegew\r\njjzvokhh\r\nlhpdobse\r\nxjjzvokhh\r\nlhpd\r\nsuwuu\r\ntuatbwof\r\nvpsuday\r\nndcisx\r\nfggxici\r\nbfnipz\r\nknzjio\r\noirksxb\r\nbfni\r\n~\r\n", "output": "aeui~cvyvvbcgbs\r\nbfnipz~cftfrrqp\r\nbfni~kakdggegew\r\ncvyvvbcgb~ndcis\r\ncxmigej~fggxici\r\ncyud~lhpdobsekn\r\ngrtp~obseknnnqk\r\nhbcbhj~jjzvokhh\r\nknnnqk~lhpdobse\r\nknzjio~nwsyrgqa\r\nkuaeui~tuatbwof\r\nlhpd~pzcftfrrqp\r\nndcisx~xvmhfzfl\r\nnnqk~xhaznwqsgk\r\noirksxb~vpsuday\r\nsuwuu~xjjzvokhh\r\nuwuu~xvmhfzflku\r\n"}, {"input": "58\r\nesgdfmf\r\nxfkluadj\r\nqhvh\r\njwhuyhm\r\nmgi\r\nysgc\r\nvhhenku\r\npb\r\ntr\r\nu\r\njyrpjnpd\r\nkluadjo\r\nopb\r\ncopb\r\ngcyhceo\r\nr\r\ndjo\r\nxfklu\r\neo\r\nadjo\r\nfkluadjo\r\nybe\r\nwljwh\r\nqhvhh\r\nrhgotp\r\nyhceo\r\nuyhm\r\nvdd\r\nyhm\r\nysgcyhc\r\nvddrhg\r\nril\r\nwljwhu\r\nx\r\nqh\r\nhceo\r\ntfcopb\r\nmgitfc\r\nvddrh\r\nmgitfco\r\nxf\r\nmgitf\r\ncyoybe\r\notp\r\no\r\nljwhuyhm\r\nysgcy\r\nhhenku\r\nwl\r\ngotp\r\nqhv\r\nw\r\nhenku\r\nenku\r\nys\r\nrilcyo\r\nxfklua\r\nqhvhhenk\r\n|\r\n", "output": "adjo|henku\r\ncopb|mgitf\r\ncyoybe|djo\r\nenku|qhvhh\r\neo|esgdfmf\r\nfkluadjo|o\r\ngcyhceo|pb\r\ngotp|vddrh\r\nhceo|wljwh\r\nhhenku|mgi\r\njwhuyhm|qh\r\njyrpjnpd|r\r\nkluadjo|tr\r\nljwhuyhm|u\r\nmgitfco|wl\r\nmgitfc|opb\r\notp|rhgotp\r\nqhvhhenk|w\r\nqhvh|xfklu\r\nqhv|rilcyo\r\nril|tfcopb\r\nuyhm|yhceo\r\nvddrhg|vdd\r\nvhhenku|xf\r\nwljwhu|ybe\r\nxfkluadj|x\r\nxfklua|yhm\r\nysgcyhc|ys\r\nysgcy|ysgc\r\n"}, {"input": "76\r\nsd\r\nwhx\r\nk\r\nce\r\nthm\r\nbyfi\r\npju\r\nbn\r\ndz\r\non\r\nizr\r\niswh\r\nl\r\nwig\r\ns\r\nju\r\nsr\r\nie\r\nx\r\nbth\r\nzvi\r\nlxth\r\ndmzz\r\nbnqq\r\nan\r\ny\r\ng\r\nvlj\r\nc\r\nhdu\r\nlx\r\nwkyd\r\ndb\r\nrmr\r\nrv\r\nis\r\ngv\r\nu\r\nbyf\r\nm\r\nqqb\r\nwe\r\nb\r\ne\r\nnioo\r\niek\r\no\r\nymk\r\nifpw\r\nisw\r\nammm\r\ncgk\r\ncq\r\nhhv\r\nq\r\nat\r\nd\r\ney\r\nn\r\nrhq\r\ncecg\r\nqsh\r\nak\r\nhx\r\nrve\r\nlaly\r\ni\r\nbnsa\r\nioou\r\nsk\r\nkg\r\nqshs\r\nwzmn\r\nupt\r\nvwvr\r\nyjj\r\nN\r\n", "output": "akNbth\r\nammmNb\r\nanNbyf\r\natNcgk\r\nbnNhdu\r\nbnqqNc\r\nbnsaNd\r\nbyfiNe\r\nceNhhv\r\ncecgNg\r\ncqNiek\r\ndbNisw\r\ndmzzNi\r\ndzNizr\r\neyNpju\r\ngvNqqb\r\nhxNqsh\r\nieNrhq\r\nifpwNk\r\nioouNl\r\nisNrmr\r\niswhNm\r\njuNrve\r\nkgNthm\r\nlalyNn\r\nlxNupt\r\nlxthNo\r\nniooNq\r\nonNvlj\r\nqshsNs\r\nrvNwhx\r\nsdNwig\r\nskNyjj\r\nsrNymk\r\nuNvwvr\r\nweNzvi\r\nwkydNx\r\nwzmnNy\r\n"}, {"input": "10\r\npo\r\negf\r\ne\r\ngfuzaqsi\r\nsi\r\nhpo\r\nuldiig\r\negfuzaq\r\nuldiigh\r\nuzaqsi\r\n{\r\n", "output": "egfuzaq{po\r\negf{uldiig\r\ne{gfuzaqsi\r\nhpo{uzaqsi\r\nsi{uldiigh\r\n"}, {"input": "4\r\na\r\nf\r\nz\r\nh\r\n!\r\n", "output": "a!f\r\nh!z\r\n"}] | false | stdio | null | true |
580/C | 580 | C | PyPy 3-64 | TESTS | 34 | 343 | 163,225,600 | 226967627 | import sys, threading
input = sys.stdin.readline
from collections import defaultdict
input = sys.stdin.readline
# returns the first number where key becomes true for a given delegate type key
def bs(low=1, high=1, key = lambda x: True):
while low <= high:
mid = (low + high)//2
if key(mid):
high = mid-1
else:
low = mid+1
return low
def main():
n, m = list(map(int, input().split()))
cats = list(map(int, input().split()))
graph = [[] for _ in range(n+1)]
for _ in range(n-1):
u, v = list(map(int, input().split()))
graph[u].append(v)
graph[v].append(u)
seen = [0]*n
def dfs(node, seen, m, n_cats):
count = 0
seen[node-1] = 1
isleaf = 1
if cats[node-1]:
n_cats+=1
else:
n_cats = 0
if n_cats > m:
return 0
for nbr in graph[node]:
if seen[nbr-1] == 0:
isleaf = 0
count+=dfs(nbr,seen, m, n_cats)
count+=isleaf
return count
print(dfs(1, seen, m, 0))
# Set the stack size
threading.stack_size(1 << 27)
# Create and start the main thread
main_thread = threading.Thread(target=main)
main_thread.start()
# Wait for the main thread to complete
main_thread.join() | 40 | 233 | 29,286,400 | 225884260 | import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return int(input())
def inlt():
return list(map(int, input().split()))
def insr():
s = input()
return list(s[: len(s) - 1])
def invr():
return map(int, input().split())
n, m = invr()
cat = inlt()
adj = [[] for i in range(n)]
for _ in range(n - 1):
a, b = invr()
adj[a - 1].append(b - 1)
adj[b - 1].append(a - 1)
ans = [0]
stack = [(-1, 0, 0)]
while stack:
prev, c, cnt = stack.pop()
if cat[c] == 1:
cnt += 1
else:
cnt = 0
if cnt > m:
continue
if adj[c] == [prev]:
ans[0] += 1
continue
for i in adj[c]:
if i == prev:
continue
else:
stack.append((c, i, cnt))
print(ans[0]) | Codeforces Round 321 (Div. 2) | CF | 2,015 | 2 | 256 | Kefa and Park | Kefa decided to celebrate his first big salary by going to the restaurant.
He lives by an unusual park. The park is a rooted tree consisting of n vertices with the root at vertex 1. Vertex 1 also contains Kefa's house. Unfortunaely for our hero, the park also contains cats. Kefa has already found out what are the vertices with cats in them.
The leaf vertices of the park contain restaurants. Kefa wants to choose a restaurant where he will go, but unfortunately he is very afraid of cats, so there is no way he will go to the restaurant if the path from the restaurant to his house contains more than m consecutive vertices with cats.
Your task is to help Kefa count the number of restaurants where he can go. | The first line contains two integers, n and m (2 ≤ n ≤ 105, 1 ≤ m ≤ n) — the number of vertices of the tree and the maximum number of consecutive vertices with cats that is still ok for Kefa.
The second line contains n integers a1, a2, ..., an, where each ai either equals to 0 (then vertex i has no cat), or equals to 1 (then vertex i has a cat).
Next n - 1 lines contains the edges of the tree in the format "xi yi" (without the quotes) (1 ≤ xi, yi ≤ n, xi ≠ yi), where xi and yi are the vertices of the tree, connected by an edge.
It is guaranteed that the given set of edges specifies a tree. | A single integer — the number of distinct leaves of a tree the path to which from Kefa's home contains at most m consecutive vertices with cats. | null | Let us remind you that a tree is a connected graph on n vertices and n - 1 edge. A rooted tree is a tree with a special vertex called root. In a rooted tree among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child. A vertex is called a leaf, if it has no children.
Note to the first sample test: The vertices containing cats are marked red. The restaurants are at vertices 2, 3, 4. Kefa can't go only to the restaurant located at vertex 2.
Note to the second sample test: The restaurants are located at vertices 4, 5, 6, 7. Kefa can't go to restaurants 6, 7. | [{"input": "4 1\n1 1 0 0\n1 2\n1 3\n1 4", "output": "2"}, {"input": "7 1\n1 0 1 1 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7", "output": "2"}] | 1,500 | ["dfs and similar", "graphs", "trees"] | 40 | [{"input": "4 1\r\n1 1 0 0\r\n1 2\r\n1 3\r\n1 4\r\n", "output": "2\r\n"}, {"input": "7 1\r\n1 0 1 1 0 0 0\r\n1 2\r\n1 3\r\n2 4\r\n2 5\r\n3 6\r\n3 7\r\n", "output": "2\r\n"}, {"input": "3 2\r\n1 1 1\r\n1 2\r\n2 3\r\n", "output": "0\r\n"}, {"input": "5 2\r\n1 1 0 1 1\r\n1 2\r\n2 3\r\n3 4\r\n4 5\r\n", "output": "1\r\n"}, {"input": "6 1\r\n1 0 1 1 0 0\r\n1 2\r\n1 3\r\n1 4\r\n1 5\r\n1 6\r\n", "output": "3\r\n"}, {"input": "7 3\r\n1 1 1 1 1 0 1\r\n1 2\r\n1 3\r\n2 4\r\n3 5\r\n5 6\r\n6 7\r\n", "output": "2\r\n"}, {"input": "15 2\r\n1 0 1 0 1 0 0 0 0 0 0 0 0 0 0\r\n1 2\r\n1 3\r\n2 4\r\n2 5\r\n3 6\r\n3 7\r\n4 8\r\n4 9\r\n5 10\r\n5 11\r\n6 12\r\n6 13\r\n7 14\r\n7 15\r\n", "output": "8\r\n"}, {"input": "2 1\r\n1 1\r\n2 1\r\n", "output": "0\r\n"}, {"input": "12 3\r\n1 0 1 0 1 1 1 1 0 0 0 0\r\n6 7\r\n12 1\r\n9 7\r\n1 4\r\n10 7\r\n7 1\r\n11 8\r\n5 1\r\n3 7\r\n5 8\r\n4 2\r\n", "output": "7\r\n"}] | false | stdio | null | true |
3/B | 3 | B | Python 3 | TESTS | 14 | 1,310 | 10,547,200 | 22710047 | def print_v(v):
print(v, eval(v))
n,v = map(int, input().split())
#n - number of vehicles
#v - volume of bus
vehicle1 = []
vehicle2 = []
for i in range(n):
t,p = map(int, input().split())
if t == 1:
vehicle1.append((p,i+1))
else:
vehicle2.append((p,i+1))
vehicle1.sort(key=lambda x: x[0],reverse=True)
vehicle2.sort(key=lambda x: x[0],reverse=True)
index1 = 0
index2 = 0
if n == 9875 and v == 13641:
f = len(vehicle1)//100 + 1
for i in range(f):
print(vehicle1[i*100:(i+1)*100])
current_v = 0 #current space in the car
current_result = 0 #current maximal value
numbers = []
while True:
if(v - current_v >= 2):
if(index2 < len(vehicle2) or index1 + 1 < len(vehicle1)):
if(index2 < len(vehicle2) and index1 + 1 < len(vehicle1)):
if(vehicle2[index2][0] > vehicle1[index1][0] + vehicle1[index1+1][0]):
numbers.append((vehicle2[index2][0], vehicle2[index2][1], 2))
index2 += 1
current_v += 2
else:
numbers.append((vehicle1[index1][0], vehicle1[index1][1], 1))
numbers.append((vehicle1[index1+1][0], vehicle1[index1+1][1], 1))
index1 += 2
current_v += 2
else:
if(index2 < len(vehicle2)):
numbers.append((vehicle2[index2][0], vehicle2[index2][1], 2))
index2 += 1
current_v += 2
if(index1 + 1 < len(vehicle1)):
numbers.append((vehicle1[index1][0], vehicle1[index1][1], 1))
numbers.append((vehicle1[index1+1][0], vehicle1[index1+1][1], 1))
index1 += 2
current_v += 2
else:
break
else:
break
def sum_numbers(numbers):
if(len(numbers) == 0):
return 0
result = 0
for item in numbers:
result += item[0]
return result
if(len(vehicle1)%2 == 1):
if(v - current_v >= 1):
numbers.append((vehicle1[index1][0], vehicle1[index1][1], 1))
current_v += 1
is_1_in_numbers = False
is_1_in_numbers_twice = False
counter = 0
for item in numbers:
if item[-1] == 1:
is_1_in_numbers = True
counter += 1
if(counter == 2):
is_1_in_numbers_twice = True
break
if(v - current_v == 1 and index2 < len(vehicle2) and is_1_in_numbers):
temp_numbers = numbers.copy()
for i in range(len(temp_numbers) - 1,-1,-1):
if temp_numbers[i][-1] == 1:
del(temp_numbers[i])
break
temp_numbers.append((vehicle2[index2][0], vehicle2[index2][1], 2))
if (sum_numbers(temp_numbers) > sum_numbers(numbers)):
numbers = temp_numbers
elif (v - current_v == 0 and index2 < len(vehicle2) and is_1_in_numbers_twice):
temp_numbers = numbers.copy()
counter = 0
for i in range(len(temp_numbers) - 1,-1,-1):
if temp_numbers[i][-1] == 1:
del(temp_numbers[i])
counter += 1
if(counter == 2):
break
temp_numbers.append((vehicle2[index2][0], vehicle2[index2][1], 2))
if (sum_numbers(temp_numbers) > sum_numbers(numbers)):
numbers = temp_numbers
numbers.sort(key=lambda x:x[1])
print(sum_numbers(numbers))
for item in numbers:
print(item[1]) | 32 | 466 | 16,076,800 | 185738569 | from heapq import nlargest
from itertools import accumulate, chain
import sys
n, v = map(int, sys.stdin.readline().split())
a1, a2 = [], []
i1, i2 = [], []
for i in range(n):
t, p = map(int, sys.stdin.readline().split())
if t == 1:
a1.append(p)
i1.append(i)
else:
a2.append(p)
i2.append(i)
acc1 = list(accumulate(sorted(a1, reverse=True), initial=0))
acc2 = list(accumulate(sorted(a2, reverse=True), initial=0))
best = (0, 0, 0)
for n2 in range(min(v // 2, len(a2)) + 1):
n1 = min(v - 2 * n2, len(a1))
best = max(best, (acc1[n1] + acc2[n2], n1, n2))
ans, n1, n2 = best
sys.stdout.write(str(ans) + "\n")
ii1 = sorted(range(len(a1)), key=a1.__getitem__)[len(a1) - n1 :]
ii2 = sorted(range(len(a2)), key=a2.__getitem__)[len(a2) - n2 :]
sys.stdout.write(" ".join(str(i1[i] + 1) for i in ii1))
sys.stdout.write(" ")
sys.stdout.write(" ".join(str(i2[i] + 1) for i in ii2))
sys.stdout.write("\n") | Codeforces Beta Round 3 | ICPC | 2,010 | 2 | 64 | Lorry | A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres).
Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body. | The first line contains a pair of integer numbers n and v (1 ≤ n ≤ 105; 1 ≤ v ≤ 109), where n is the number of waterborne vehicles in the boat depot, and v is the truck body volume of the lorry in cubic metres. The following n lines contain the information about the waterborne vehicles, that is a pair of numbers ti, pi (1 ≤ ti ≤ 2; 1 ≤ pi ≤ 104), where ti is the vehicle type (1 – a kayak, 2 – a catamaran), and pi is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file. | In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them. | null | null | [{"input": "3 2\n1 2\n2 7\n1 3", "output": "7\n2"}] | 1,900 | ["greedy", "sortings"] | 32 | [{"input": "3 2\r\n1 2\r\n2 7\r\n1 3\r\n", "output": "7\r\n2\r\n"}, {"input": "5 3\r\n1 9\r\n2 9\r\n1 9\r\n2 10\r\n1 6\r\n", "output": "24\r\n3 1 5\r\n"}, {"input": "10 10\r\n1 14\r\n2 15\r\n2 11\r\n2 12\r\n2 9\r\n1 14\r\n2 15\r\n1 9\r\n2 11\r\n2 6\r\n", "output": "81\r\n6 1 7 2 4 9\r\n"}, {"input": "20 19\r\n2 47\r\n1 37\r\n1 48\r\n2 42\r\n2 48\r\n1 38\r\n2 47\r\n1 48\r\n2 47\r\n1 41\r\n2 46\r\n1 28\r\n1 49\r\n1 45\r\n2 34\r\n1 43\r\n2 29\r\n1 46\r\n2 45\r\n2 18\r\n", "output": "630\r\n13 8 3 18 14 16 10 6 2 5 9 7 1 11\r\n"}, {"input": "50 27\r\n2 93\r\n1 98\r\n2 62\r\n1 56\r\n1 86\r\n1 42\r\n2 67\r\n2 97\r\n2 59\r\n1 73\r\n1 83\r\n2 96\r\n1 20\r\n1 66\r\n1 84\r\n1 83\r\n1 91\r\n2 97\r\n1 81\r\n2 88\r\n2 63\r\n1 99\r\n2 57\r\n1 39\r\n1 74\r\n2 88\r\n1 30\r\n2 68\r\n1 100\r\n2 57\r\n1 87\r\n1 93\r\n1 83\r\n1 100\r\n1 91\r\n1 14\r\n1 38\r\n2 98\r\n2 85\r\n2 61\r\n1 44\r\n2 93\r\n2 66\r\n2 55\r\n2 74\r\n1 67\r\n2 67\r\n1 85\r\n2 59\r\n1 83\r\n", "output": "2055\r\n34 29 22 2 32 35 17 31 5 48 15 50 33 16 11 19 25 10 46 14 4 38 18 8\r\n"}, {"input": "1 1\r\n1 600\r\n", "output": "600\r\n1\r\n"}, {"input": "10 14\r\n2 230\r\n2 516\r\n2 527\r\n2 172\r\n2 854\r\n2 61\r\n1 52\r\n2 154\r\n2 832\r\n2 774\r\n", "output": "3905\r\n5 9 10 3 2 1 4\r\n"}, {"input": "8 8\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n2 100\r\n2 100\r\n2 100\r\n2 100\r\n", "output": "400\r\n8 7 6 5\r\n"}, {"input": "8 4\r\n1 100\r\n1 100\r\n1 100\r\n1 100\r\n2 1\r\n2 1\r\n2 1\r\n2 1\r\n", "output": "400\r\n4 3 2 1\r\n"}] | false | stdio | import sys
def read_lines(path):
with open(path, 'r') as f:
return [line.strip() for line in f.readlines() if line.strip()]
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input
input_lines = read_lines(input_path)
if not input_lines:
print(0)
return
n, v = map(int, input_lines[0].split())
vehicles = []
for line in input_lines[1:1+n]:
ti, pi = map(int, line.split())
vehicles.append( (ti, pi) )
# Read judge output
judge_lines = read_lines(output_path)
if not judge_lines:
print(0)
return
try:
judge_sum = int(judge_lines[0])
except:
print(0)
return
judge_vehicles = []
if len(judge_lines) >=2:
judge_vehicles = list(map(int, ' '.join(judge_lines[1:]).split()))
# Read submission output
sub_lines = read_lines(submission_path)
if len(sub_lines) <1:
print(0)
return
try:
sub_sum = int(sub_lines[0])
except:
print(0)
return
sub_vehicles = []
if len(sub_lines) >=2:
sub_vehicles = list(map(int, ' '.join(sub_lines[1:]).split()))
# Check sum matches
if sub_sum != judge_sum:
print(0)
return
# Check vehicle indices are valid and unique
seen = set()
for idx in sub_vehicles:
if idx <1 or idx >n or idx in seen:
print(0)
return
seen.add(idx)
# Compute sum_t and sum_p
sum_t = 0
sum_p =0
for idx in sub_vehicles:
ti, pi = vehicles[idx-1]
sum_t += ti
sum_p += pi
if sum_t >v:
print(0)
return
if sum_p != sub_sum:
print(0)
return
# All checks passed
print(1)
if __name__ == '__main__':
main() | true |
301/C | 301 | C | PyPy 3 | TESTS | 1 | 154 | 20,172,800 | 125260456 | res = '''
0??<>1
1??<>2
2??<>3
3??<>4
4??<>5
5??<>6
6??<>7
7??<>8
8??<>9
9??>>??0
?0>>0?
?1>>1?
?2>>2?
?3>>3?
?4>>4?
?5>>5?
?6>>6?
?7>>7?
?8>>8?
?9>>9?
?>>??
0>>0?
1>>1?
2>>2?
3>>3?
4>>4?
5>>5?
6>>6?
7>>7?
8>>8?
9>>9?
'''
print(res.strip()) | 23 | 125 | 0 | 3818363 | print('9??>>??0')
for i in range(9):
print('{}??<>{}'.format(i, i + 1))
print('??<>1')
for i in range(10):
print('?{0}>>{0}?'.format(i))
print('?>>??')
print('>>?') | Codeforces Round 182 (Div. 1) | CF | 2,013 | 2 | 256 | Yaroslav and Algorithm | Yaroslav likes algorithms. We'll describe one of his favorite algorithms.
1. The algorithm receives a string as the input. We denote this input string as a.
2. The algorithm consists of some number of command. Сommand number i looks either as si >> wi, or as si <> wi, where si and wi are some possibly empty strings of length at most 7, consisting of digits and characters "?".
3. At each iteration, the algorithm looks for a command with the minimum index i, such that si occurs in a as a substring. If this command is not found the algorithm terminates.
4. Let's denote the number of the found command as k. In string a the first occurrence of the string sk is replaced by string wk. If the found command at that had form sk >> wk, then the algorithm continues its execution and proceeds to the next iteration. Otherwise, the algorithm terminates.
5. The value of string a after algorithm termination is considered to be the output of the algorithm.
Yaroslav has a set of n positive integers, he needs to come up with his favorite algorithm that will increase each of the given numbers by one. More formally, if we consider each number as a string representing the decimal representation of the number, then being run on each of these strings separately, the algorithm should receive the output string that is a recording of the corresponding number increased by one.
Help Yaroslav. | The first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the set. The next n lines contains one positive integer each. All the given numbers are less than 1025. | Print the algorithm which can individually increase each number of the set. In the i-th line print the command number i without spaces.
Your algorithm will be launched for each of these numbers. The answer will be considered correct if:
- Each line will a correct algorithm command (see the description in the problem statement).
- The number of commands should not exceed 50.
- The algorithm will increase each of the given numbers by one.
- To get a respond, the algorithm will perform no more than 200 iterations for each number. | null | null | [{"input": "2\n10\n79", "output": "10<>11\n79<>80"}] | 2,500 | ["constructive algorithms"] | 23 | [{"input": "2\r\n10\r\n79\r\n", "output": "10<>11\r\n79<>80\r\n"}, {"input": "5\r\n9\r\n99\r\n999\r\n9999\r\n99999\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "5\r\n99999\r\n9999\r\n999\r\n99\r\n9\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n392\r\n605\r\n903\r\n154\r\n293\r\n383\r\n422\r\n717\r\n719\r\n896\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n448\r\n727\r\n772\r\n539\r\n870\r\n913\r\n668\r\n300\r\n36\r\n895\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n704\r\n812\r\n323\r\n334\r\n674\r\n665\r\n142\r\n712\r\n254\r\n869\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n548\r\n645\r\n663\r\n758\r\n38\r\n860\r\n724\r\n742\r\n530\r\n779\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n317\r\n36\r\n191\r\n843\r\n289\r\n107\r\n41\r\n943\r\n265\r\n649\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n447\r\n806\r\n891\r\n730\r\n371\r\n351\r\n7\r\n102\r\n394\r\n549\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n630\r\n624\r\n85\r\n955\r\n757\r\n841\r\n967\r\n377\r\n932\r\n309\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n1\r\n2\r\n3\r\n4\r\n5\r\n6\r\n7\r\n8\r\n9\r\n10\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "1\r\n9999999999999999999999999\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}] | false | stdio | import sys
def main(input_path, output_path, submission_output_path):
with open(input_path) as f:
n = int(f.readline())
numbers = [int(line.strip()) for line in f]
commands = []
with open(submission_output_path) as f:
for line in f:
line = line.strip()
if '>>' in line:
parts = line.split('>>', 1)
op = '>>'
elif '<>' in line:
parts = line.split('<>', 1)
op = '<>'
else:
print(0)
return
s, w = parts[0], parts[1]
if len(s) > 7 or len(w) > 7:
print(0)
return
valid_chars = set('0123456789?')
for c in s + w:
if c not in valid_chars:
print(0)
return
commands.append((s, op, w))
if len(commands) > 50 or not commands:
print(0)
return
for num in numbers:
current = str(num)
expected = str(num + 1)
iterations = 0
while True:
found = False
applied_op = None
for cmd in commands:
s, op, w = cmd
if s in current:
idx = current.find(s)
current = current[:idx] + w + current[idx+len(s):]
iterations += 1
found = True
applied_op = op
break
if not found:
break
if iterations > 200:
break
if applied_op == '<>':
break
if current != expected or iterations > 200:
print(0)
return
print(1)
if __name__ == "__main__":
main(sys.argv[1], sys.argv[2], sys.argv[3]) | true |
732/D | 732 | D | PyPy 3-64 | TESTS | 48 | 93 | 18,534,400 | 217602631 | n,m=map(int,input().split())
days=list(map(int,input().split()))
sub=list(map(int,input().split()))
def binarysearch():
l=sum(sub)+m
r=n
ans=-1
while l<=r:
mid=l+(r-l)//2
count=[0]*(n)
# print(mid)
for i in range(mid):
count[days[i]-1]+=1
zeroz=0
t=True
# print(count)
for i in range(mid):
f=days[i]
if f==0:
zeroz+=1
else:
if count[f-1]==1:
if zeroz<sub[f-1]:
t=False
break
else:
zeroz-=sub[f-1]
else:
zeroz+=1
count[f-1]-=1
if t:
ans=mid
r=mid-1
else:
l=mid+1
return ans
print(binarysearch()) | 53 | 139 | 12,492,800 | 192753286 | import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def binary_search(c1, c2):
c = (c1 + c2 + 1) // 2
while abs(c1 - c2) > 1:
c = (c1 + c2 + 1) // 2
if ok(c):
c2 = c
else:
c1 = c
c = max(c - 1, 0)
while not ok(c):
c += 1
return c
def ok(c):
if c > n:
return True
x, y = [], [0] * (m + 1)
for i in range(c, 0, -1):
di = d[i]
if not di:
continue
if not y[di]:
x.append(di)
y[di] = i
if len(x) < m:
return False
s = 0
for i in x[::-1]:
s += a[i] + 1
if s > y[i]:
return False
return True
n, m = map(int, input().split())
d = [0] + list(map(int, input().split()))
a = [0] + list(map(int, input().split()))
if not ok(n):
ans = -1
print(ans)
exit()
ans = binary_search(0, n + 1)
print(ans) | Codeforces Round 377 (Div. 2) | CF | 2,016 | 1 | 256 | Exams | Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time. | The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i. | Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1. | null | In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it. | [{"input": "7 2\n0 1 0 2 1 0 2\n2 1", "output": "5"}, {"input": "10 3\n0 0 1 2 3 0 2 0 1 2\n1 1 4", "output": "9"}, {"input": "5 1\n1 1 1 1 1\n5", "output": "-1"}] | 1,700 | ["binary search", "greedy", "sortings"] | 53 | [{"input": "7 2\r\n0 1 0 2 1 0 2\r\n2 1\r\n", "output": "5\r\n"}, {"input": "10 3\r\n0 0 1 2 3 0 2 0 1 2\r\n1 1 4\r\n", "output": "9\r\n"}, {"input": "5 1\r\n1 1 1 1 1\r\n5\r\n", "output": "-1\r\n"}, {"input": "100 10\r\n1 1 6 6 6 2 5 7 6 5 3 7 10 10 8 9 7 6 9 2 6 7 8 6 7 5 2 5 10 1 10 1 8 10 2 9 7 1 6 8 3 10 9 4 4 8 8 6 6 1 5 5 6 5 6 6 6 9 4 7 5 4 6 6 1 1 2 1 8 10 6 2 1 7 2 1 8 10 9 2 7 3 1 5 10 2 8 10 10 10 8 9 5 4 6 10 8 9 6 6\r\n2 4 10 11 5 2 6 7 2 15\r\n", "output": "74\r\n"}, {"input": "1 1\r\n1\r\n1\r\n", "output": "-1\r\n"}, {"input": "3 2\r\n0 0 0\r\n2 1\r\n", "output": "-1\r\n"}, {"input": "4 2\r\n0 1 0 2\r\n1 1\r\n", "output": "4\r\n"}, {"input": "10 1\r\n0 1 0 0 0 0 0 0 0 1\r\n1\r\n", "output": "2\r\n"}, {"input": "5 1\r\n0 0 0 0 1\r\n1\r\n", "output": "5\r\n"}, {"input": "7 2\r\n0 0 0 0 0 1 2\r\n1 1\r\n", "output": "7\r\n"}, {"input": "10 3\r\n0 0 1 2 2 0 2 0 1 3\r\n1 1 4\r\n", "output": "10\r\n"}, {"input": "6 2\r\n1 1 1 1 1 2\r\n1 1\r\n", "output": "6\r\n"}, {"input": "6 2\r\n1 0 0 0 0 2\r\n1 1\r\n", "output": "-1\r\n"}] | false | stdio | null | true |
3/B | 3 | B | Python 3 | TESTS | 12 | 842 | 14,233,600 | 91114214 | import sys
n, v = sys.stdin.readline().strip().split()
n = int(n)
v = int(v)
lines = []
for i in range(0, n):
lines.append(sys.stdin.readline().strip())
vehicles = []
for index, vehicle in enumerate(lines):
typ, capacity = vehicle.split()
typ = int(typ)
capacity = int(capacity)
eff = capacity / typ
vehicles.append([eff, typ, capacity, index])
ori_list = vehicles.copy()
vehicles.sort()
lorry_cap = 0
count = 0
caps = 0
indices = []
rev_vehicles = vehicles[::-1]
while lorry_cap < v and count < len(rev_vehicles):
caps += rev_vehicles[count][2]
lorry_cap += rev_vehicles[count][1]
indices.append(rev_vehicles[count][3])
count += 1
if lorry_cap > v:
caps -= rev_vehicles[count-1][2]
lorry_cap -= rev_vehicles[count-1][1]
indices.pop(-1)
for j in range(count, len(rev_vehicles)):
if rev_vehicles[count][1] == 1:
caps += rev_vehicles[count][2]
lorry_cap += rev_vehicles[count][1]
indices.append(rev_vehicles[count][3])
break
if lorry_cap < v:
for l in range(count-1, len(rev_vehicles)):
for a in indices[::-1]:
if ori_list[a][1] == 1 and ori_list[a][2] < rev_vehicles[l][2]:
caps -= ori_list[a][2]
caps += rev_vehicles[l][2]
indices.remove(a)
indices.append(rev_vehicles[l][3])
break
else:
continue
break
print(caps)
num_veh = ''
for m in indices:
num_veh += str(m+1)
num_veh += ' '
print(num_veh[:-1]) | 32 | 778 | 12,083,200 | 78422702 | import sys
n, v = map(int, input().split())
kayak = []
catam = []
for i in range(0,n):
nums = list(map(int, sys.stdin.readline().split()))
if nums[0] == 1:
kayak.append((nums[1], i + 1))
else:
catam.append((nums[1], i + 1))
# sorts by [0] elementwise
kayak.sort(key=lambda pair: pair[0])
catam.sort(key=lambda pair: pair[0])
# Greedily take kayaks and catamarans
truck = []
while v > 0:
if v == 1:
if kayak:
truck.append(kayak.pop())
v -= 1
else:
if not kayak and not catam:
v = 0
elif not kayak:
truck.append(catam.pop())
v -= 2
elif not catam:
truck.append(kayak.pop())
v -= 1
else:
if len(kayak) == 1:
if kayak[0][0] > catam[-1][0]:
truck.append(kayak.pop())
v -= 1
else:
truck.append(catam.pop())
v -= 2
else:
if kayak[-1][0]+kayak[-2][0] > catam[-1][0]:
truck.append(kayak.pop())
v -= 1
else:
truck.append(catam.pop())
v -= 2
ans = 0
for i in truck:
ans += i[0]
print(ans)
for i in truck:
sys.stdout.write(str(i[1]) + " ") | Codeforces Beta Round 3 | ICPC | 2,010 | 2 | 64 | Lorry | A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres).
Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body. | The first line contains a pair of integer numbers n and v (1 ≤ n ≤ 105; 1 ≤ v ≤ 109), where n is the number of waterborne vehicles in the boat depot, and v is the truck body volume of the lorry in cubic metres. The following n lines contain the information about the waterborne vehicles, that is a pair of numbers ti, pi (1 ≤ ti ≤ 2; 1 ≤ pi ≤ 104), where ti is the vehicle type (1 – a kayak, 2 – a catamaran), and pi is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file. | In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them. | null | null | [{"input": "3 2\n1 2\n2 7\n1 3", "output": "7\n2"}] | 1,900 | ["greedy", "sortings"] | 32 | [{"input": "3 2\r\n1 2\r\n2 7\r\n1 3\r\n", "output": "7\r\n2\r\n"}, {"input": "5 3\r\n1 9\r\n2 9\r\n1 9\r\n2 10\r\n1 6\r\n", "output": "24\r\n3 1 5\r\n"}, {"input": "10 10\r\n1 14\r\n2 15\r\n2 11\r\n2 12\r\n2 9\r\n1 14\r\n2 15\r\n1 9\r\n2 11\r\n2 6\r\n", "output": "81\r\n6 1 7 2 4 9\r\n"}, {"input": "20 19\r\n2 47\r\n1 37\r\n1 48\r\n2 42\r\n2 48\r\n1 38\r\n2 47\r\n1 48\r\n2 47\r\n1 41\r\n2 46\r\n1 28\r\n1 49\r\n1 45\r\n2 34\r\n1 43\r\n2 29\r\n1 46\r\n2 45\r\n2 18\r\n", "output": "630\r\n13 8 3 18 14 16 10 6 2 5 9 7 1 11\r\n"}, {"input": "50 27\r\n2 93\r\n1 98\r\n2 62\r\n1 56\r\n1 86\r\n1 42\r\n2 67\r\n2 97\r\n2 59\r\n1 73\r\n1 83\r\n2 96\r\n1 20\r\n1 66\r\n1 84\r\n1 83\r\n1 91\r\n2 97\r\n1 81\r\n2 88\r\n2 63\r\n1 99\r\n2 57\r\n1 39\r\n1 74\r\n2 88\r\n1 30\r\n2 68\r\n1 100\r\n2 57\r\n1 87\r\n1 93\r\n1 83\r\n1 100\r\n1 91\r\n1 14\r\n1 38\r\n2 98\r\n2 85\r\n2 61\r\n1 44\r\n2 93\r\n2 66\r\n2 55\r\n2 74\r\n1 67\r\n2 67\r\n1 85\r\n2 59\r\n1 83\r\n", "output": "2055\r\n34 29 22 2 32 35 17 31 5 48 15 50 33 16 11 19 25 10 46 14 4 38 18 8\r\n"}, {"input": "1 1\r\n1 600\r\n", "output": "600\r\n1\r\n"}, {"input": "10 14\r\n2 230\r\n2 516\r\n2 527\r\n2 172\r\n2 854\r\n2 61\r\n1 52\r\n2 154\r\n2 832\r\n2 774\r\n", "output": "3905\r\n5 9 10 3 2 1 4\r\n"}, {"input": "8 8\r\n1 1\r\n1 1\r\n1 1\r\n1 1\r\n2 100\r\n2 100\r\n2 100\r\n2 100\r\n", "output": "400\r\n8 7 6 5\r\n"}, {"input": "8 4\r\n1 100\r\n1 100\r\n1 100\r\n1 100\r\n2 1\r\n2 1\r\n2 1\r\n2 1\r\n", "output": "400\r\n4 3 2 1\r\n"}] | false | stdio | import sys
def read_lines(path):
with open(path, 'r') as f:
return [line.strip() for line in f.readlines() if line.strip()]
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input
input_lines = read_lines(input_path)
if not input_lines:
print(0)
return
n, v = map(int, input_lines[0].split())
vehicles = []
for line in input_lines[1:1+n]:
ti, pi = map(int, line.split())
vehicles.append( (ti, pi) )
# Read judge output
judge_lines = read_lines(output_path)
if not judge_lines:
print(0)
return
try:
judge_sum = int(judge_lines[0])
except:
print(0)
return
judge_vehicles = []
if len(judge_lines) >=2:
judge_vehicles = list(map(int, ' '.join(judge_lines[1:]).split()))
# Read submission output
sub_lines = read_lines(submission_path)
if len(sub_lines) <1:
print(0)
return
try:
sub_sum = int(sub_lines[0])
except:
print(0)
return
sub_vehicles = []
if len(sub_lines) >=2:
sub_vehicles = list(map(int, ' '.join(sub_lines[1:]).split()))
# Check sum matches
if sub_sum != judge_sum:
print(0)
return
# Check vehicle indices are valid and unique
seen = set()
for idx in sub_vehicles:
if idx <1 or idx >n or idx in seen:
print(0)
return
seen.add(idx)
# Compute sum_t and sum_p
sum_t = 0
sum_p =0
for idx in sub_vehicles:
ti, pi = vehicles[idx-1]
sum_t += ti
sum_p += pi
if sum_t >v:
print(0)
return
if sum_p != sub_sum:
print(0)
return
# All checks passed
print(1)
if __name__ == '__main__':
main() | true |
301/C | 301 | C | PyPy 3 | TESTS | 0 | 154 | 20,172,800 | 125260427 | res = '''
0??<>1
1??<>2
2??<>3
3??<>4
4??<>5
5??<>6
6??<>7
7??<>8
8??<>9
9??>>??0
?0>>0?
?1>>1?
?2>>2?
?3>>3?
?4>>4?
?5>>5?
?6>>6?
?7>>7?
?8>>8?
?9>>9?
?>>??
0>>0?
1>>1?
2>>2?
3>>3?
4>>4?
5>>5?
6>>6?
7>>7?
8>>8?
9>>9?
'''
print(res) | 23 | 248 | 0 | 56067152 | print("0??<>1");
print("1??<>2");
print("2??<>3");
print("3??<>4");
print("4??<>5");
print("5??<>6");
print("6??<>7");
print("7??<>8");
print("8??<>9");
print("9??>>??0");
print("??<>1");
print("?9>>9?");
print("?8>>8?");
print("?7>>7?");
print("?6>>6?");
print("?5>>5?");
print("?4>>4?");
print("?3>>3?");
print("?2>>2?");
print("?1>>1?");
print("?0>>0?");
print("9>>9?");
print("8>>8?");
print("7>>7?");
print("6>>6?");
print("5>>5?");
print("4>>4?");
print("3>>3?");
print("2>>2?");
print("1>>1?");
print("0>>0?"); | Codeforces Round 182 (Div. 1) | CF | 2,013 | 2 | 256 | Yaroslav and Algorithm | Yaroslav likes algorithms. We'll describe one of his favorite algorithms.
1. The algorithm receives a string as the input. We denote this input string as a.
2. The algorithm consists of some number of command. Сommand number i looks either as si >> wi, or as si <> wi, where si and wi are some possibly empty strings of length at most 7, consisting of digits and characters "?".
3. At each iteration, the algorithm looks for a command with the minimum index i, such that si occurs in a as a substring. If this command is not found the algorithm terminates.
4. Let's denote the number of the found command as k. In string a the first occurrence of the string sk is replaced by string wk. If the found command at that had form sk >> wk, then the algorithm continues its execution and proceeds to the next iteration. Otherwise, the algorithm terminates.
5. The value of string a after algorithm termination is considered to be the output of the algorithm.
Yaroslav has a set of n positive integers, he needs to come up with his favorite algorithm that will increase each of the given numbers by one. More formally, if we consider each number as a string representing the decimal representation of the number, then being run on each of these strings separately, the algorithm should receive the output string that is a recording of the corresponding number increased by one.
Help Yaroslav. | The first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the set. The next n lines contains one positive integer each. All the given numbers are less than 1025. | Print the algorithm which can individually increase each number of the set. In the i-th line print the command number i without spaces.
Your algorithm will be launched for each of these numbers. The answer will be considered correct if:
- Each line will a correct algorithm command (see the description in the problem statement).
- The number of commands should not exceed 50.
- The algorithm will increase each of the given numbers by one.
- To get a respond, the algorithm will perform no more than 200 iterations for each number. | null | null | [{"input": "2\n10\n79", "output": "10<>11\n79<>80"}] | 2,500 | ["constructive algorithms"] | 23 | [{"input": "2\r\n10\r\n79\r\n", "output": "10<>11\r\n79<>80\r\n"}, {"input": "5\r\n9\r\n99\r\n999\r\n9999\r\n99999\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "5\r\n99999\r\n9999\r\n999\r\n99\r\n9\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n392\r\n605\r\n903\r\n154\r\n293\r\n383\r\n422\r\n717\r\n719\r\n896\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n448\r\n727\r\n772\r\n539\r\n870\r\n913\r\n668\r\n300\r\n36\r\n895\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n704\r\n812\r\n323\r\n334\r\n674\r\n665\r\n142\r\n712\r\n254\r\n869\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n548\r\n645\r\n663\r\n758\r\n38\r\n860\r\n724\r\n742\r\n530\r\n779\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n317\r\n36\r\n191\r\n843\r\n289\r\n107\r\n41\r\n943\r\n265\r\n649\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n447\r\n806\r\n891\r\n730\r\n371\r\n351\r\n7\r\n102\r\n394\r\n549\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n630\r\n624\r\n85\r\n955\r\n757\r\n841\r\n967\r\n377\r\n932\r\n309\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n1\r\n2\r\n3\r\n4\r\n5\r\n6\r\n7\r\n8\r\n9\r\n10\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "10\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n1\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}, {"input": "1\r\n9999999999999999999999999\r\n", "output": "0??<>1\r\n1??<>2\r\n2??<>3\r\n3??<>4\r\n4??<>5\r\n5??<>6\r\n6??<>7\r\n7??<>8\r\n8??<>9\r\n9??>>??0\r\n??<>1\r\n?0>>0?\r\n?1>>1?\r\n?2>>2?\r\n?3>>3?\r\n?4>>4?\r\n?5>>5?\r\n?6>>6?\r\n?7>>7?\r\n?8>>8?\r\n?9>>9?\r\n?>>??\r\n>>?\r\n"}] | false | stdio | import sys
def main(input_path, output_path, submission_output_path):
with open(input_path) as f:
n = int(f.readline())
numbers = [int(line.strip()) for line in f]
commands = []
with open(submission_output_path) as f:
for line in f:
line = line.strip()
if '>>' in line:
parts = line.split('>>', 1)
op = '>>'
elif '<>' in line:
parts = line.split('<>', 1)
op = '<>'
else:
print(0)
return
s, w = parts[0], parts[1]
if len(s) > 7 or len(w) > 7:
print(0)
return
valid_chars = set('0123456789?')
for c in s + w:
if c not in valid_chars:
print(0)
return
commands.append((s, op, w))
if len(commands) > 50 or not commands:
print(0)
return
for num in numbers:
current = str(num)
expected = str(num + 1)
iterations = 0
while True:
found = False
applied_op = None
for cmd in commands:
s, op, w = cmd
if s in current:
idx = current.find(s)
current = current[:idx] + w + current[idx+len(s):]
iterations += 1
found = True
applied_op = op
break
if not found:
break
if iterations > 200:
break
if applied_op == '<>':
break
if current != expected or iterations > 200:
print(0)
return
print(1)
if __name__ == "__main__":
main(sys.argv[1], sys.argv[2], sys.argv[3]) | true |
128/B | 128 | B | Python 3 | TESTS | 9 | 842 | 14,745,600 | 17777690 | from heapq import *
l=input()
k=int(input())
n=len(l)
if k>n*(n+1)/2:
print("No such line.")
quit()
ss=[(l[i],i) for i in range(n)]
while k:
k-=1
t=heappop(ss)
if k==0:
print(t[0])
else:
if t[1]<n-1:
heappush(ss,(t[0]+l[t[1]+1],t[1]+1)) | 63 | 404 | 13,824,000 | 219201798 | def get_input():
s = input()
l = len(s)
k = int(input())
return s, l, k
def find_kth_sum(s, l, k):
if l * (l + 1) / 2 < k:
return 'No such line.'
a = [i for i in range(l)]
p = 0
ans = ''
while True:
t = [(s[i + p], i) for i in a]
t = sorted(t)
i = j = 0
prod = 0
while i < len(t):
pre = prod
while j < len(t) and t[i][0] == t[j][0]:
prod += l - t[j][1] - p
j += 1
if prod >= k:
break
i = j
ans += t[i][0]
if pre + j - i >= k:
break
k -= pre + j - i
p += 1
a = [t[q][1] for q in range(i, j) if t[q][1] + p < l]
return ans
def main():
s, l, k = get_input()
result = find_kth_sum(s, l, k)
print(result)
if __name__ == "__main__":
main() | Codeforces Beta Round 94 (Div. 1 Only) | CF | 2,011 | 2 | 256 | String | One day in the IT lesson Anna and Maria learned about the lexicographic order.
String x is lexicographically less than string y, if either x is a prefix of y (and x ≠ y), or there exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi, and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of the string a. The lexicographic comparison of strings is implemented by operator < in modern programming languages.
The teacher gave Anna and Maria homework. She gave them a string of length n. They should write out all substrings of the given string, including the whole initial string, and the equal substrings (for example, one should write out the following substrings from the string "aab": "a", "a", "aa", "ab", "aab", "b"). The resulting strings should be sorted in the lexicographical order. The cunning teacher doesn't want to check all these strings. That's why she said to find only the k-th string from the list. Help Anna and Maria do the homework. | The first line contains a non-empty string that only consists of small Latin letters ("a"-"z"), whose length does not exceed 105. The second line contains the only integer k (1 ≤ k ≤ 105). | Print the string Anna and Maria need — the k-th (in the lexicographical order) substring of the given string. If the total number of substrings is less than k, print a string saying "No such line." (without the quotes). | null | In the second sample before string "bc" follow strings "a", "ab", "abc", "b". | [{"input": "aa\n2", "output": "a"}, {"input": "abc\n5", "output": "bc"}, {"input": "abab\n7", "output": "b"}] | 2,100 | ["brute force", "constructive algorithms", "hashing", "implementation", "string suffix structures", "strings"] | 63 | [{"input": "aa\r\n2\r\n", "output": "a\r\n"}, {"input": "abc\r\n5\r\n", "output": "bc\r\n"}, {"input": "abab\r\n7\r\n", "output": "b\r\n"}, {"input": "codeforces\r\n1\r\n", "output": "c\r\n"}, {"input": "cccc\r\n8\r\n", "output": "ccc\r\n"}, {"input": "abcdefghijklmnopqrstuvwxyz\r\n27\r\n", "output": "b\r\n"}, {"input": "cba\r\n6\r\n", "output": "cba\r\n"}, {"input": "z\r\n100000\r\n", "output": "No such line.\r\n"}, {"input": "aaaaaaaaaa\r\n90\r\n", "output": "No such line.\r\n"}] | false | stdio | null | true |
58/C | 58 | C | Python 3 | TESTS | 3 | 248 | 0 | 82570103 | n=int(input())
l=list(map(int,input().split()))
i=0
j=n-1
x=0
a=l[0]
c=0
while i<=j:
if x+a!=l[j]:
c+=1
j-=1
i+=1
x+=1
print(c) | 40 | 342 | 14,438,400 | 86234698 | n = int(input())
a = [0] * 100005
l = list(map(int, input().split()))
for i in range(n):
a[l[i] - min(i, n - i - 1)] += 1
print(n - max(a[1::])) | Codeforces Beta Round 54 (Div. 2) | CF | 2,011 | 2 | 256 | Trees | On Bertown's main street n trees are growing, the tree number i has the height of ai meters (1 ≤ i ≤ n). By the arrival of the President of Berland these trees were decided to be changed so that their heights formed a beautiful sequence. This means that the heights of trees on ends (the 1st one and the n-th one) should be equal to each other, the heights of the 2-nd and the (n - 1)-th tree must also be equal to each other, at that the height of the 2-nd tree should be larger than the height of the first tree by 1, and so on. In other words, the heights of the trees, standing at equal distance from the edge (of one end of the sequence) must be equal to each other, and with the increasing of the distance from the edge by 1 the tree height must also increase by 1. For example, the sequences "2 3 4 5 5 4 3 2" and "1 2 3 2 1" are beautiful, and '1 3 3 1" and "1 2 3 1" are not.
Changing the height of a tree is a very expensive operation, using advanced technologies invented by Berland scientists. In one operation you can choose any tree and change its height to any number, either increase or decrease. Note that even after the change the height should remain a positive integer, i. e, it can't be less than or equal to zero. Identify the smallest number of changes of the trees' height needed for the sequence of their heights to become beautiful. | The first line contains integer n (1 ≤ n ≤ 105) which is the number of trees. The second line contains integers ai (1 ≤ ai ≤ 105) which are the heights of the trees. | Print a single number which is the minimal number of trees whose heights will have to be changed for the sequence to become beautiful. | null | null | [{"input": "3\n2 2 2", "output": "1"}, {"input": "4\n1 2 2 1", "output": "0"}] | 1,800 | ["brute force"] | 40 | [{"input": "3\r\n2 2 2\r\n", "output": "1\r\n"}, {"input": "4\r\n1 2 2 1\r\n", "output": "0\r\n"}, {"input": "3\r\n61452 50974 73849\r\n", "output": "2\r\n"}, {"input": "4\r\n86002 1199 86003 86002\r\n", "output": "1\r\n"}, {"input": "5\r\n92605 92606 41969 98774 92605\r\n", "output": "2\r\n"}, {"input": "10\r\n1 1 2 3 4 4 3 2 1 10\r\n", "output": "9\r\n"}, {"input": "10\r\n73905 73906 73907 85732 73909 73909 73908 73907 73906 73905\r\n", "output": "1\r\n"}, {"input": "10\r\n60718 99414 65042 65043 65044 38495 95782 65042 65041 65040\r\n", "output": "4\r\n"}, {"input": "7\r\n1 2 3 7 6 5 4\r\n", "output": "3\r\n"}, {"input": "5\r\n4 5 6 5 1\r\n", "output": "1\r\n"}, {"input": "1\r\n100000\r\n", "output": "0\r\n"}, {"input": "5\r\n1 1 1 1 1\r\n", "output": "3\r\n"}, {"input": "7\r\n1 2 3 7 6 5 4\r\n", "output": "3\r\n"}, {"input": "1\r\n2727\r\n", "output": "0\r\n"}, {"input": "5\r\n1 2 3 2 4\r\n", "output": "1\r\n"}, {"input": "9\r\n100 12 13 14 15 14 13 12 11\r\n", "output": "1\r\n"}, {"input": "5\r\n2 4 5 4 2\r\n", "output": "2\r\n"}, {"input": "5\r\n3 3 4 3 2\r\n", "output": "1\r\n"}] | false | stdio | null | true |
547/D | 547 | D | Python 3 | TESTS | 0 | 30 | 0 | 227211410 | n = int(input())
points = []
# Read and sort the points based on x-coordinates
for _ in range(n):
x, y = map(int, input().split())
points.append((x, y))
points.sort()
# Initialize variables to keep track of red and blue fish count
red_count = 0
blue_count = 0
# Assign colors to points while maintaining the balance
colors = []
for x, y in points:
if x % 2 == 0:
if y % 2 == 0:
# Even x, even y
if red_count <= blue_count:
colors.append('r')
red_count += 1
else:
colors.append('b')
blue_count += 1
else:
# Even x, odd y
if red_count >= blue_count:
colors.append('r')
red_count += 1
else:
colors.append('b')
blue_count += 1
else:
if y % 2 == 0:
# Odd x, even y
if red_count >= blue_count:
colors.append('r')
red_count += 1
else:
colors.append('b')
blue_count += 1
else:
# Odd x, odd y
if red_count <= blue_count:
colors.append('r')
red_count += 1
else:
colors.append('b')
blue_count += 1
# Output the assigned colors
print("".join(colors)) | 50 | 889 | 112,640,000 | 224152629 | from sys import stdin, stdout
from collections import defaultdict
time = 0
c = 2*10**5
n = 4*10**5+2
col = 0
finished= [0]*n
for_node = [0]*n
f_range=range
f_len=len
def dfs_euler_tour(node: int, graph):
stack = []
global colour
global col
global time
stack.append((node, -1))
while stack:
s, ind = stack.pop()
if ind > -1:
if ind < f_len(colour):
colour[ind] = col
col = col ^ 1
index = for_node[s]
while(index < f_len(graph[s])):
v,i=graph[s][index]
if not visited[i]:
stack.append((v, i))
visited[i] = True
index += 1
for_node[s] = index
break
index += 1
finished[s]+1
m = int(stdin.readline())
graph =defaultdict(list)
edges=[]
debug=[]
for i in f_range(m):
u, v = stdin.readline().split()
u, v = (int(u)-1, int(v)-1 + c+1)
edges.append((u, v))
graph[u].append((v, i))
graph[v].append((u, i))
if u == 30630-1 and m == 199809:
debug.append(i)
colour = [-1]*m
odds= [i for i in graph.keys() if f_len(graph[i]) % 2]
if odds:
for i in f_range(f_len(odds)):
u = odds[i]
ind=f_len(edges)
if u<c:
v = n-1
else:
v = c
edges.append((u, v))
graph[u].append((v, ind))
graph[v].append((u, ind))
if f_len(graph[n-1]) % 2:
u=n-1
v=c
ind = f_len(edges)
edges.append((u, v))
graph[u].append((v, ind))
graph[v].append((u, ind))
visited = [False]*f_len(edges)
for i in graph.keys():
if not finished[i]:
dfs_euler_tour(i, graph)
sol = ''.join(['b' if i == 1 else 'r' for i in colour])
sol += '\n'
stdout.write(sol) | Codeforces Round 305 (Div. 1) | CF | 2,015 | 3 | 256 | Mike and Fish | As everyone knows, bears love fish. But Mike is a strange bear; He hates fish! The even more strange thing about him is he has an infinite number of blue and red fish.
He has marked n distinct points in the plane. i-th point is point (xi, yi). He wants to put exactly one fish in each of these points such that the difference between the number of red fish and the blue fish on each horizontal or vertical line is at most 1.
He can't find a way to perform that! Please help him. | The first line of input contains integer n (1 ≤ n ≤ 2 × 105).
The next n lines contain the information about the points, i-th line contains two integers xi and yi (1 ≤ xi, yi ≤ 2 × 105), the i-th point coordinates.
It is guaranteed that there is at least one valid answer. | Print the answer as a sequence of n characters 'r' (for red) or 'b' (for blue) where i-th character denotes the color of the fish in the i-th point. | null | null | [{"input": "4\n1 1\n1 2\n2 1\n2 2", "output": "brrb"}, {"input": "3\n1 1\n1 2\n2 1", "output": "brr"}] | 2,600 | ["constructive algorithms", "dfs and similar", "graphs"] | 50 | [{"input": "4\r\n1 1\r\n1 2\r\n2 1\r\n2 2\r\n", "output": "brrb\r\n"}, {"input": "3\r\n1 1\r\n1 2\r\n2 1\r\n", "output": "brr\r\n"}, {"input": "3\r\n157210 22861\r\n175396 39466\r\n40933 17093\r\n", "output": "rrr\r\n"}, {"input": "5\r\n55599 84144\r\n169207 98421\r\n1909 186625\r\n31525 147710\r\n7781 82078\r\n", "output": "rrrrr\r\n"}, {"input": "15\r\n44249 54630\r\n165741 91307\r\n49455 83026\r\n52521 88269\r\n39286 65158\r\n38282 73821\r\n96608 30032\r\n155832 122920\r\n22021 13546\r\n161556 192797\r\n168062 8224\r\n161221 155335\r\n5670 180269\r\n89163 128733\r\n151226 75254\r\n", "output": "rrrrrrrrrrrrrrr\r\n"}, {"input": "9\r\n95316 68815\r\n95316 23738\r\n60801 169893\r\n84639 68815\r\n109462 87456\r\n22940 37614\r\n172202 151462\r\n84639 23738\r\n109462 151462\r\n", "output": "rbrbrrrrb\r\n"}, {"input": "2\r\n196356 153892\r\n134153 153892\r\n", "output": "rb\r\n"}, {"input": "10\r\n4126 18194\r\n143965 18194\r\n32687 18194\r\n118527 18194\r\n186573 18194\r\n97223 18194\r\n179697 18194\r\n175536 18194\r\n107767 18194\r\n127019 18194\r\n", "output": "bbbbrbrrrr\r\n"}, {"input": "1\r\n1 1\r\n", "output": "r\r\n"}, {"input": "1\r\n1000 3434\r\n", "output": "r\r\n"}, {"input": "1\r\n200000 200000\r\n", "output": "r\r\n"}, {"input": "2\r\n1 2\r\n2 3\r\n", "output": "rr\r\n"}, {"input": "2\r\n1 2\r\n1 3\r\n", "output": "br\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:
n = int(f.readline())
points = []
for _ in range(n):
x, y = map(int, f.readline().split())
points.append((x, y))
with open(submission_path, 'r') as f:
s = f.read().strip()
if len(s) != n:
print(0)
return
for c in s:
if c not in ('r', 'b'):
print(0)
return
x_counts = {}
y_counts = {}
for (x, y), color in zip(points, s):
if x not in x_counts:
x_counts[x] = {'r': 0, 'b': 0}
x_counts[x][color] += 1
if y not in y_counts:
y_counts[y] = {'r': 0, 'b': 0}
y_counts[y][color] += 1
for x in x_counts:
r = x_counts[x]['r']
b = x_counts[x]['b']
if abs(r - b) > 1:
print(0)
return
for y in y_counts:
r = y_counts[y]['r']
b = y_counts[y]['b']
if abs(r - b) > 1:
print(0)
return
print(1)
if __name__ == "__main__":
main() | true |
545/D | 545 | D | Python 3 | TESTS | 13 | 77 | 10,342,400 | 229507916 | n = int(input())
queue =map(int, input().split())
queue2=set()
number_ones=0
for i in queue:
if i == 1:
number_ones+=1
else:
queue2.add(i)
queue=sorted(queue2)
# print(queue)
if number_ones > 1:
queue=[1]*2+queue
# print(queue)
n=len(queue)
sum_ = 0
i = 0
while True:
if i == n:
break
x = queue[i]
if x >= sum_:
sum_ += x
i += 1
else:
queue.pop(i)
n -= 1
print(n)
# print(queue)
# print(number_ones) | 61 | 78 | 13,107,200 | 225590587 | def solve(n):
n.sort()
count = 0
curr = 0
for num in n:
if num < curr:
continue
count += 1
curr += num
return count
def main():
t = int(input())
print(solve([int(x) for x in input().split()]))
if __name__ == "__main__":
main() | Codeforces Round 303 (Div. 2) | CF | 2,015 | 1 | 256 | Queue | Little girl Susie went shopping with her mom and she wondered how to improve service quality.
There are n people in the queue. For each person we know time ti needed to serve him. A person will be disappointed if the time he waits is more than the time needed to serve him. The time a person waits is the total time when all the people who stand in the queue in front of him are served. Susie thought that if we swap some people in the queue, then we can decrease the number of people who are disappointed.
Help Susie find out what is the maximum number of not disappointed people can be achieved by swapping people in the queue. | The first line contains integer n (1 ≤ n ≤ 105).
The next line contains n integers ti (1 ≤ ti ≤ 109), separated by spaces. | Print a single number — the maximum number of not disappointed people in the queue. | null | Value 4 is achieved at such an arrangement, for example: 1, 2, 3, 5, 15. Thus, you can make everything feel not disappointed except for the person with time 5. | [{"input": "5\n15 2 1 5 3", "output": "4"}] | 1,300 | ["greedy", "implementation", "sortings"] | 61 | [{"input": "5\r\n15 2 1 5 3\r\n", "output": "4\r\n"}, {"input": "15\r\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\r\n", "output": "2\r\n"}, {"input": "10\r\n13 2 5 55 21 34 1 8 1 3\r\n", "output": "6\r\n"}, {"input": "10\r\n8 256 16 1 2 1 64 4 128 32\r\n", "output": "10\r\n"}, {"input": "10\r\n10000 40000 10000 50000 20000 100000 10000 100 30000 500\r\n", "output": "6\r\n"}, {"input": "1\r\n1000000000\r\n", "output": "1\r\n"}, {"input": "15\r\n9 11 45 86 52 65 35 3 93 7 21 45 15 11 39\r\n", "output": "6\r\n"}, {"input": "20\r\n16839799 17525904 91276752 42650694 60106463 12243176 54892123 25142243 16015971 41250998 11150057 6994983 67700784 16562412 82163675 46178521 33914268 91966607 93976858 84100064\r\n", "output": "5\r\n"}, {"input": "26\r\n1000 4110030 64221 131521030 942374833 1003 2055015 32110 513757 16440130 263042057 32880256 128439 557559573 16051 8220066 469240078 65760513 256878 790176315 4012 2005 1027508 928528684 8030 805074697\r\n", "output": "21\r\n"}] | false | stdio | null | true |
546/B | 546 | B | PyPy 3-64 | TESTS | 9 | 61 | 1,740,800 | 232331284 | c = int(input())
a = str(input()).split()
b = [int(x) for x in a]
counter = 0
arr = []
dups = []
for x in b:
if x in arr:
dups.append(x)
else:
arr.append(x)
for x in dups:
while x in arr:
x = x + 1
counter += 1
else:
continue
print(counter) | 49 | 62 | 307,200 | 11211283 | import sys
n = int(input())
if n == 1:
print(0)
sys.exit(0)
l = list(map(int, input().split(" ")))
l.sort()
min = l[0]
val = 0
for i in range(1, len(l)):
if l[i] <= l[i-1]:
val += 1+l[i-1]-l[i]
l[i]+= 1+l[i-1]-l[i]
print(val) | Codeforces Round 304 (Div. 2) | CF | 2,015 | 3 | 256 | Soldier and Badges | Colonel has n badges. He wants to give one badge to every of his n soldiers. Each badge has a coolness factor, which shows how much it's owner reached. Coolness factor can be increased by one for the cost of one coin.
For every pair of soldiers one of them should get a badge with strictly higher factor than the second one. Exact values of their factors aren't important, they just need to have distinct factors.
Colonel knows, which soldier is supposed to get which badge initially, but there is a problem. Some of badges may have the same factor of coolness. Help him and calculate how much money has to be paid for making all badges have different factors of coolness. | First line of input consists of one integer n (1 ≤ n ≤ 3000).
Next line consists of n integers ai (1 ≤ ai ≤ n), which stand for coolness factor of each badge. | Output single integer — minimum amount of coins the colonel has to pay. | null | In first sample test we can increase factor of first badge by 1.
In second sample test we can increase factors of the second and the third badge by 1. | [{"input": "4\n1 3 1 4", "output": "1"}, {"input": "5\n1 2 3 2 5", "output": "2"}] | 1,200 | ["brute force", "greedy", "implementation", "sortings"] | 49 | [{"input": "4\r\n1 3 1 4\r\n", "output": "1"}, {"input": "5\r\n1 2 3 2 5\r\n", "output": "2"}, {"input": "5\r\n1 5 3 2 4\r\n", "output": "0"}, {"input": "10\r\n1 1 2 3 4 5 6 7 8 9\r\n", "output": "9"}, {"input": "11\r\n9 2 10 3 1 5 7 1 4 8 6\r\n", "output": "10"}, {"input": "4\r\n4 3 2 2\r\n", "output": "3"}, {"input": "1\r\n1\r\n", "output": "0"}, {"input": "50\r\n49 37 30 2 18 48 14 48 50 27 1 43 46 5 21 28 44 2 24 17 41 38 25 18 43 28 25 21 28 23 26 27 4 31 50 18 23 11 13 28 44 47 1 26 43 25 22 46 32 45\r\n", "output": "170"}, {"input": "50\r\n37 31 19 46 45 1 9 37 15 19 15 10 17 16 38 13 26 25 36 13 7 21 12 41 46 19 3 50 14 49 49 40 29 41 47 29 3 42 13 21 10 21 9 33 38 30 24 40 5 26\r\n", "output": "135"}, {"input": "50\r\n18 13 50 12 23 29 31 44 28 29 33 31 17 38 27 37 36 34 40 4 27 2 8 27 50 27 21 28 11 13 47 25 15 26 9 15 22 3 22 45 9 12 5 5 46 44 23 34 12 25\r\n", "output": "138"}, {"input": "50\r\n24 44 39 44 11 20 6 43 4 21 43 12 41 3 25 25 24 7 16 36 32 2 2 29 34 30 33 9 18 3 14 28 26 49 29 5 5 36 44 21 36 37 1 25 46 10 10 24 10 39\r\n", "output": "128"}, {"input": "50\r\n7 5 18 2 7 12 8 20 41 4 7 3 7 10 22 1 19 9 20 10 23 3 6 3 30 13 6 18 3 3 18 38 9 7 2 1 2 5 25 10 13 1 8 34 1 26 13 8 13 2\r\n", "output": "699"}, {"input": "50\r\n2 19 24 3 12 4 14 9 10 19 6 1 26 6 11 1 4 34 17 1 3 35 17 2 17 17 5 5 12 1 24 35 2 5 43 23 21 4 18 3 11 5 1 21 3 3 3 1 10 10\r\n", "output": "692"}, {"input": "50\r\n2 2 4 19 5 7 2 35 3 12 1 18 17 16 40 4 15 36 1 11 13 3 14 1 4 10 1 12 43 7 9 9 4 3 28 9 12 12 1 33 3 23 11 24 20 20 2 4 26 4\r\n", "output": "660"}, {"input": "50\r\n5 3 25 6 30 6 39 15 3 19 1 38 1 3 17 3 8 13 4 10 14 3 2 3 20 1 21 21 27 31 6 6 14 28 3 13 49 8 12 6 17 13 45 1 6 18 12 7 31 14\r\n", "output": "574"}, {"input": "50\r\n10 25 27 13 28 35 40 39 3 6 18 29 44 1 26 2 45 36 9 46 41 12 33 19 8 22 15 48 34 20 11 32 1 47 43 23 7 5 14 30 31 21 38 42 24 49 4 37 16 17\r\n", "output": "49"}, {"input": "50\r\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\r\n", "output": "1225"}, {"input": "50\r\n50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50\r\n", "output": "1225"}, {"input": "3\r\n1 3 3\r\n", "output": "1"}, {"input": "10\r\n4 4 4 4 4 4 5 5 5 5\r\n", "output": "41"}, {"input": "4\r\n1 4 4 4\r\n", "output": "3"}, {"input": "3\r\n1 1 1\r\n", "output": "3"}, {"input": "3\r\n3 3 3\r\n", "output": "3"}] | false | stdio | null | true |
158/C | 158 | C | PyPy 3-64 | TESTS | 6 | 124 | 0 | 231104758 | a = int(input(""))
d = []
for i in range(a):
l = input("")
if l == "pwd":
[print("/", x, sep="", end = "") for x in d]
print("/")
else:
ll="".join(l[3:])
ll = ll.split("/")
#print(ll)
for w in ll:
if w == "..":
d.pop()
elif w != "":
d.append(w) | 29 | 124 | 0 | 18858108 | class Shell:
def __init__(self):
self.wd = ['']
def cd(self, where: str):
glob = where.startswith('/')
directories = where.split('/')
if directories and not directories[0]:
directories.pop(0)
if glob:
self._reset()
for d in directories:
self._cd_one_dir(d)
def pwd(self):
if len(self.wd) == 1:
print('/')
else:
print('/', '/'.join(self.wd[1:]), '/', sep='')
def _reset(self):
self.wd = ['']
def _cd_one_dir(self, d):
if d == '..':
self.wd.pop()
else:
self.wd.append(d)
n = int(input())
sh = Shell()
for _ in range(n):
cmd = input()
if cmd.startswith('cd'):
sh.cd(cmd.split()[-1])
else:
sh.pwd() | VK Cup 2012 Qualification Round 1 | CF | 2,012 | 3 | 256 | Cd and pwd commands | Vasya is writing an operating system shell, and it should have commands for working with directories. To begin with, he decided to go with just two commands: cd (change the current directory) and pwd (display the current directory).
Directories in Vasya's operating system form a traditional hierarchical tree structure. There is a single root directory, denoted by the slash character "/". Every other directory has a name — a non-empty string consisting of lowercase Latin letters. Each directory (except for the root) has a parent directory — the one that contains the given directory. It is denoted as "..".
The command cd takes a single parameter, which is a path in the file system. The command changes the current directory to the directory specified by the path. The path consists of the names of directories separated by slashes. The name of the directory can be "..", which means a step up to the parent directory. «..» can be used in any place of the path, maybe several times. If the path begins with a slash, it is considered to be an absolute path, that is, the directory changes to the specified one, starting from the root. If the parameter begins with a directory name (or ".."), it is considered to be a relative path, that is, the directory changes to the specified directory, starting from the current one.
The command pwd should display the absolute path to the current directory. This path must not contain "..".
Initially, the current directory is the root. All directories mentioned explicitly or passed indirectly within any command cd are considered to exist. It is guaranteed that there is no attempt of transition to the parent directory of the root directory. | The first line of the input data contains the single integer n (1 ≤ n ≤ 50) — the number of commands.
Then follow n lines, each contains one command. Each of these lines contains either command pwd, or command cd, followed by a space-separated non-empty parameter.
The command parameter cd only contains lower case Latin letters, slashes and dots, two slashes cannot go consecutively, dots occur only as the name of a parent pseudo-directory. The command parameter cd does not end with a slash, except when it is the only symbol that points to the root directory. The command parameter has a length from 1 to 200 characters, inclusive.
Directories in the file system can have the same names. | For each command pwd you should print the full absolute path of the given directory, ending with a slash. It should start with a slash and contain the list of slash-separated directories in the order of being nested from the root to the current folder. It should contain no dots. | null | null | [{"input": "7\npwd\ncd /home/vasya\npwd\ncd ..\npwd\ncd vasya/../petya\npwd", "output": "/\n/home/vasya/\n/home/\n/home/petya/"}, {"input": "4\ncd /a/b\npwd\ncd ../a/b\npwd", "output": "/a/b/\n/a/a/b/"}] | 1,400 | ["*special", "data structures", "implementation"] | 29 | [{"input": "7\r\npwd\r\ncd /home/vasya\r\npwd\r\ncd ..\r\npwd\r\ncd vasya/../petya\r\npwd\r\n", "output": "/\r\n/home/vasya/\r\n/home/\r\n/home/petya/\r\n"}, {"input": "4\r\ncd /a/b\r\npwd\r\ncd ../a/b\r\npwd\r\n", "output": "/a/b/\r\n/a/a/b/\r\n"}, {"input": "1\r\npwd\r\n", "output": "/\r\n"}, {"input": "2\r\ncd /test/../test/../test/../test/../a/b/c/..\r\npwd\r\n", "output": "/a/b/\r\n"}, {"input": "9\r\ncd test\r\npwd\r\ncd ..\r\ncd /test\r\npwd\r\ncd ..\r\npwd\r\ncd test/test\r\npwd\r\n", "output": "/test/\r\n/test/\r\n/\r\n/test/test/\r\n"}, {"input": "6\r\ncd a/a/b/b\r\npwd\r\ncd ../..\r\npwd\r\ncd ..\r\npwd\r\n", "output": "/a/a/b/b/\r\n/a/a/\r\n/a/\r\n"}, {"input": "5\r\npwd\r\ncd /xgztbykka\r\npwd\r\ncd /gia/kxfls\r\npwd\r\n", "output": "/\r\n/xgztbykka/\r\n/gia/kxfls/\r\n"}, {"input": "17\r\npwd\r\ncd denwxe/../jhj/rxit/ie\r\npwd\r\ncd /tmyuylvul/qev/ezqit\r\npwd\r\ncd ../gxsfgyuspg/irleht\r\npwd\r\ncd wq/pqyz/tjotsmdzja\r\npwd\r\ncd ia/hs/../u/nemol/ffhf\r\npwd\r\ncd /lrdm/mvwxwb/llib\r\npwd\r\ncd /lmhu/wloover/rqd\r\npwd\r\ncd lkwabdw/../wrqn/x/../ien\r\npwd\r\n", "output": "/\r\n/jhj/rxit/ie/\r\n/tmyuylvul/qev/ezqit/\r\n/tmyuylvul/qev/gxsfgyuspg/irleht/\r\n/tmyuylvul/qev/gxsfgyuspg/irleht/wq/pqyz/tjotsmdzja/\r\n/tmyuylvul/qev/gxsfgyuspg/irleht/wq/pqyz/tjotsmdzja/ia/u/nemol/ffhf/\r\n/lrdm/mvwxwb/llib/\r\n/lmhu/wloover/rqd/\r\n/lmhu/wloover/rqd/wrqn/ien/\r\n"}, {"input": "5\r\ncd /xgztbykka\r\ncd /gia/kxfls\r\ncd /kiaxt/hcx\r\ncd /ufzoiv\r\npwd\r\n", "output": "/ufzoiv/\r\n"}, {"input": "17\r\ncd denwxe/../jhj/rxit/ie\r\ncd /tmyuylvul/qev/ezqit\r\ncd ../gxsfgyuspg/irleht\r\ncd wq/pqyz/tjotsmdzja\r\ncd ia/hs/../u/nemol/ffhf\r\ncd /lrdm/mvwxwb/llib\r\ncd /lmhu/wloover/rqd\r\ncd lkwabdw/../wrqn/x/../ien\r\ncd /rqljh/qyovqhiry/q\r\ncd /d/aargbeotxm/ovv\r\ncd /jaagwy/../xry/w/zdvx\r\ncd /nblqgcua/s/s/c/dgg\r\ncd /jktwitbkgj/ee/../../q\r\ncd wkx/jyphtd/h/../ygwc\r\ncd areyd/regf/ogvklan\r\ncd /wrbi/vbxefrd/jimis\r\npwd\r\n", "output": "/wrbi/vbxefrd/jimis/\r\n"}, {"input": "5\r\npwd\r\ncd ztb/kag\r\npwd\r\npwd\r\npwd\r\n", "output": "/\r\n/ztb/kag/\r\n/ztb/kag/\r\n/ztb/kag/\r\n"}, {"input": "17\r\ncd en/ebhjhjzlrx/pmieg\r\ncd uylvulohqe/wezq/oarx\r\npwd\r\ncd yus/fsi/ehtrs/../vjpq\r\ncd tjotsmdzja/diand/dqb\r\ncd emolqs/hff/rdmy/vw\r\ncd ../llibd/mhuos/oove\r\ncd /rqdqj/kwabd/nj/qng\r\npwd\r\ncd /yie/lrq/hmxq/vqhi\r\ncd qma/../aargbeotxm/ov\r\ncd /jaagwy/../xry/w/zdvx\r\npwd\r\ncd ../gcuagas/s/c/dggmz\r\npwd\r\npwd\r\ncd bkgjifee/../../../vfwkxjoj\r\n", "output": "/en/ebhjhjzlrx/pmieg/uylvulohqe/wezq/oarx/\r\n/rqdqj/kwabd/nj/qng/\r\n/xry/w/zdvx/\r\n/xry/w/gcuagas/s/c/dggmz/\r\n/xry/w/gcuagas/s/c/dggmz/\r\n"}, {"input": "50\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\npwd\r\n", "output": "/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n/\r\n"}, {"input": "1\r\ncd /\r\n", "output": ""}, {"input": "11\r\npwd\r\ncd /home/vasya\r\npwd\r\ncd ..\r\npwd\r\ncd vasya/../../petya\r\npwd\r\ncd /a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a\r\npwd\r\ncd ..\r\npwd\r\n", "output": "/\r\n/home/vasya/\r\n/home/\r\n/petya/\r\n/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/\r\n/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/\r\n"}] | false | stdio | null | true |
432/D | 432 | D | PyPy 3-64 | TESTS | 9 | 249 | 25,804,800 | 199431396 | import bisect
import heapq
import sys
from types import GeneratorType
from functools import cmp_to_key
from collections import defaultdict, Counter, deque
import math
from functools import lru_cache
from heapq import nlargest
from functools import reduce
import random
from itertools import combinations
from itertools import accumulate
from operator import xor, add
from operator import mul
from typing import List
inf = float("inf")
PLATFORM = "CF"
if PLATFORM == "LUOGU":
import numpy as np
sys.setrecursionlimit(1000000)
class FastIO:
def __init__(self):
return
@staticmethod
def _read():
return sys.stdin.readline().strip()
def read_int(self):
return int(self._read())
def read_float(self):
return float(self._read())
def read_ints(self):
return map(int, self._read().split())
def read_floats(self):
return map(float, self._read().split())
def read_ints_minus_one(self):
return map(lambda x: int(x) - 1, self._read().split())
def read_list_ints(self):
return list(map(int, self._read().split()))
def read_list_floats(self):
return list(map(float, self._read().split()))
def read_list_ints_minus_one(self):
return list(map(lambda x: int(x) - 1, self._read().split()))
def read_str(self):
return self._read()
def read_list_strs(self):
return self._read().split()
def read_list_str(self):
return list(self._read())
@staticmethod
def st(x):
return sys.stdout.write(str(x) + '\n')
@staticmethod
def lst(x):
return sys.stdout.write(" ".join(str(w) for w in x) + '\n')
@staticmethod
def round_5(f):
res = int(f)
if f - res >= 0.5:
res += 1
return res
@staticmethod
def max(a, b):
return a if a > b else b
@staticmethod
def min(a, b):
return a if a < b else b
@staticmethod
def bootstrap(f, queue=[]):
def wrappedfunc(*args, **kwargs):
if queue:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if isinstance(to, GeneratorType):
queue.append(to)
to = next(to)
else:
queue.pop()
if not queue:
break
to = queue[-1].send(to)
return to
return wrappedfunc
class NumberTheory:
def __init__(self):
return
@staticmethod
def euler_phi(n):
# 欧拉函数返回小于等于n的与n互质的个数
# 注意1和1互质，而大于1的质数与1不互质
ans = n
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
ans = ans // i * (i - 1)
while n % i == 0:
n = n // i
if n > 1:
ans = ans // n * (n - 1)
return int(ans)
class KMP:
def __init__(self):
return
@staticmethod
def prefix_function(s):
# 计算s[:i]与s[:i]的最长公共真前缀与真后缀
n = len(s)
pi = [0] * n
for i in range(1, n):
j = pi[i - 1]
while j > 0 and s[i] != s[j]:
j = pi[j - 1]
if s[i] == s[j]:
j += 1
pi[i] = j
# pi[0] = 0
return pi
@staticmethod
def z_function(s):
# 计算 s[i:] 与 s 的最长公共前缀
n = len(s)
z = [0] * n
left, r = 0, 0
for i in range(1, n):
if i <= r and z[i - left] < r - i + 1:
z[i] = z[i - left]
else:
z[i] = max(0, r - i + 1)
while i + z[i] < n and s[z[i]] == s[i + z[i]]:
z[i] += 1
if i + z[i] - 1 > r:
left = i
r = i + z[i] - 1
# z[0] = 0
return z
class Solution:
def __init__(self):
return
@staticmethod
def main(ac=FastIO()):
s = ac.read_str()
n = len(s)
z = KMP().z_function(s)
z[0] = n
cnt = Counter(z)
ans = []
for i in range(n-1, -1, -1):
if z[i] == n-i:
ans.append([n-i, cnt[n-i]])
m = len(ans)
for i in range(m-2, -1, -1):
ans[i][1] += ans[i+1][1]
ac.st(m)
for a in ans:
ac.lst(a)
return
Solution().main() | 30 | 108 | 17,100,800 | 219090643 | s = ' ' + input()
n = len(s)
r, c = [-1] * n, [1] * n
for i in range(1, n):
r[i] = r[i - 1] + 1
while r[i] and s[r[i]] != s[i]:
r[i] = r[r[i] - 1] + 1
d, n = [], n - 1
for i in range(n, 1, -1): c[r[i]] += c[i]
while n > 0:
d.append(str(n) + ' ' + str(c[n]))
n = r[n]
print(len(d))
d.reverse()
print('\n'.join(d)) | Codeforces Round 246 (Div. 2) | CF | 2,014 | 1 | 256 | Prefixes and Suffixes | You have a string s = s1s2...s|s|, where |s| is the length of string s, and si its i-th character.
Let's introduce several definitions:
- A substring s[i..j] (1 ≤ i ≤ j ≤ |s|) of string s is string sisi + 1...sj.
- The prefix of string s of length l (1 ≤ l ≤ |s|) is string s[1..l].
- The suffix of string s of length l (1 ≤ l ≤ |s|) is string s[|s| - l + 1..|s|].
Your task is, for any prefix of string s which matches a suffix of string s, print the number of times it occurs in string s as a substring. | The single line contains a sequence of characters s1s2...s|s| (1 ≤ |s| ≤ 105) — string s. The string only consists of uppercase English letters. | In the first line, print integer k (0 ≤ k ≤ |s|) — the number of prefixes that match a suffix of string s. Next print k lines, in each line print two integers li ci. Numbers li ci mean that the prefix of the length li matches the suffix of length li and occurs in string s as a substring ci times. Print pairs li ci in the order of increasing li. | null | null | [{"input": "ABACABA", "output": "3\n1 4\n3 2\n7 1"}, {"input": "AAA", "output": "3\n1 3\n2 2\n3 1"}] | 2,000 | ["dp", "string suffix structures", "strings", "two pointers"] | 30 | [{"input": "ABACABA\r\n", "output": "3\r\n1 4\r\n3 2\r\n7 1\r\n"}, {"input": "AAA\r\n", "output": "3\r\n1 3\r\n2 2\r\n3 1\r\n"}, {"input": "A\r\n", "output": "1\r\n1 1\r\n"}, {"input": "AAAAAAAAAAAAAAAAXAAAAAAAAAAAAAAAAAAAAAAA\r\n", "output": "17\r\n1 39\r\n2 37\r\n3 35\r\n4 33\r\n5 31\r\n6 29\r\n7 27\r\n8 25\r\n9 23\r\n10 21\r\n11 19\r\n12 17\r\n13 15\r\n14 13\r\n15 11\r\n16 9\r\n40 1\r\n"}, {"input": "AB\r\n", "output": "1\r\n2 1\r\n"}, {"input": "AXAXA\r\n", "output": "3\r\n1 3\r\n3 2\r\n5 1\r\n"}, {"input": "CODEFORCES\r\n", "output": "1\r\n10 1\r\n"}, {"input": "GERALDPAVELGERALDPAVEL\r\n", "output": "2\r\n11 2\r\n22 1\r\n"}, {"input": "ZZ\r\n", "output": "2\r\n1 2\r\n2 1\r\n"}] | false | stdio | null | true |
439/B | 439 | B | PyPy 3 | TESTS | 3 | 140 | 0 | 66242144 | a,b=map(int,input().split())
l=list(map(int,input().split()))
l=sorted(l)
c=int(b)
tot=0
d=0
for i in range(a):
if c>0:
tot+=l[i]*c
elif c<1:
break
c-=1
if a<=b:
print(tot)
else:
d=(l[(a-b):])
print(sum(d)+tot) | 31 | 93 | 11,468,800 | 228681113 | def solve(n, x, c):
c.sort()
s = 0
for t in c:
s += t*x
x = max(1,x-1)
return s
n, x = [int(_) for _ in input().split()]
c = [int(_) for _ in input().split()]
print(solve(n, x, c)) | Codeforces Round 251 (Div. 2) | CF | 2,014 | 1 | 256 | Devu, the Dumb Guy | Devu is a dumb guy, his learning curve is very slow. You are supposed to teach him n subjects, the ith subject has ci chapters. When you teach him, you are supposed to teach all the chapters of a subject continuously.
Let us say that his initial per chapter learning power of a subject is x hours. In other words he can learn a chapter of a particular subject in x hours.
Well Devu is not complete dumb, there is a good thing about him too. If you teach him a subject, then time required to teach any chapter of the next subject will require exactly 1 hour less than previously required (see the examples to understand it more clearly). Note that his per chapter learning power can not be less than 1 hour.
You can teach him the n subjects in any possible order. Find out minimum amount of time (in hours) Devu will take to understand all the subjects and you will be free to do some enjoying task rather than teaching a dumb guy.
Please be careful that answer might not fit in 32 bit data type. | The first line will contain two space separated integers n, x (1 ≤ n, x ≤ 105). The next line will contain n space separated integers: c1, c2, ..., cn (1 ≤ ci ≤ 105). | Output a single integer representing the answer to the problem. | null | Look at the first example. Consider the order of subjects: 1, 2. When you teach Devu the first subject, it will take him 3 hours per chapter, so it will take 12 hours to teach first subject. After teaching first subject, his per chapter learning time will be 2 hours. Now teaching him second subject will take 2 × 1 = 2 hours. Hence you will need to spend 12 + 2 = 14 hours.
Consider the order of subjects: 2, 1. When you teach Devu the second subject, then it will take him 3 hours per chapter, so it will take 3 × 1 = 3 hours to teach the second subject. After teaching the second subject, his per chapter learning time will be 2 hours. Now teaching him the first subject will take 2 × 4 = 8 hours. Hence you will need to spend 11 hours.
So overall, minimum of both the cases is 11 hours.
Look at the third example. The order in this example doesn't matter. When you teach Devu the first subject, it will take him 3 hours per chapter. When you teach Devu the second subject, it will take him 2 hours per chapter. When you teach Devu the third subject, it will take him 1 hours per chapter. In total it takes 6 hours. | [{"input": "2 3\n4 1", "output": "11"}, {"input": "4 2\n5 1 2 1", "output": "10"}, {"input": "3 3\n1 1 1", "output": "6"}] | 1,200 | ["implementation", "sortings"] | 31 | [{"input": "2 3\r\n4 1\r\n", "output": "11\r\n"}, {"input": "4 2\r\n5 1 2 1\r\n", "output": "10\r\n"}, {"input": "3 3\r\n1 1 1\r\n", "output": "6\r\n"}, {"input": "20 4\r\n1 1 3 5 5 1 3 4 2 5 2 4 3 1 3 3 3 3 4 3\r\n", "output": "65\r\n"}, {"input": "20 10\r\n6 6 1 2 6 4 5 3 6 5 4 5 6 5 4 6 6 2 3 3\r\n", "output": "196\r\n"}, {"input": "1 1\r\n9273\r\n", "output": "9273\r\n"}, {"input": "1 1\r\n1\r\n", "output": "1\r\n"}, {"input": "1 2\r\n1\r\n", "output": "2\r\n"}, {"input": "1 2\r\n2\r\n", "output": "4\r\n"}, {"input": "2 1\r\n1 2\r\n", "output": "3\r\n"}] | false | stdio | null | true |
489/A | 489 | A | Python 3 | TESTS | 5 | 30 | 0 | 227549872 | elements = int(input())
arr = [int(i) for i in input().split()]
swaps_num = 0
swaps = []
for i in range(elements):
for j in range(i, elements):
if i == j:
continue
if arr[i] > arr[j]:
swaps_num += 1
temp = arr[i]
arr[i] = arr[j]
arr[j] = temp
swaps.append([i, j])
print(swaps_num)
for i in range(len(swaps)):
print(*swaps[i]) | 22 | 78 | 1,228,800 | 8721277 | __author__ = 'alexandrun'
import sys
#sys.stdin = open("p1.in", "r")
n = int(input())
words = input().split()
i = 0
ini = []
for w in words:
ini.append((int(w), i))
i += 1
fin = ini[:]
fin.sort()
#print(ini)
#print(fin)
pos = {}
i = 0
for pair in ini:
pos[pair] = i
i += 1
swaps = []
ifin = 0
for pair in fin:
swaps.append((ifin, pos[pair]))
aux = ini[ifin]
ini[ifin] = ini[pos[pair]]
ini[pos[pair]] = aux
p1 = ini[ifin]
p2 = ini[pos[pair]]
aux = pos[p1]
pos[p1] = pos[p2]
pos[p2] = aux
ifin += 1
#print()
#print(swaps)
#print(ini)
print(len(swaps))
for s in swaps:
print(s[0], s[1]) | Codeforces Round 277.5 (Div. 2) | CF | 2,014 | 1 | 256 | SwapSort | In this problem your goal is to sort an array consisting of n integers in at most n swaps. For the given array find the sequence of swaps that makes the array sorted in the non-descending order. Swaps are performed consecutively, one after another.
Note that in this problem you do not have to minimize the number of swaps — your task is to find any sequence that is no longer than n. | The first line of the input contains integer n (1 ≤ n ≤ 3000) — the number of array elements. The second line contains elements of array: a0, a1, ..., an - 1 ( - 109 ≤ ai ≤ 109), where ai is the i-th element of the array. The elements are numerated from 0 to n - 1 from left to right. Some integers may appear in the array more than once. | In the first line print k (0 ≤ k ≤ n) — the number of swaps. Next k lines must contain the descriptions of the k swaps, one per line. Each swap should be printed as a pair of integers i, j (0 ≤ i, j ≤ n - 1), representing the swap of elements ai and aj. You can print indices in the pairs in any order. The swaps are performed in the order they appear in the output, from the first to the last. It is allowed to print i = j and swap the same pair of elements multiple times.
If there are multiple answers, print any of them. It is guaranteed that at least one answer exists. | null | null | [{"input": "5\n5 2 5 1 4", "output": "2\n0 3\n4 2"}, {"input": "6\n10 20 20 40 60 60", "output": "0"}, {"input": "2\n101 100", "output": "1\n0 1"}] | 1,200 | ["greedy", "implementation", "sortings"] | 22 | [{"input": "5\r\n5 2 5 1 4\r\n", "output": "2\r\n0 3\r\n4 2\r\n"}, {"input": "6\r\n10 20 20 40 60 60\r\n", "output": "0\r\n"}, {"input": "2\r\n101 100\r\n", "output": "1\r\n0 1\r\n"}, {"input": "1\r\n1000\r\n", "output": "0\r\n"}, {"input": "2\r\n1000000000 -1000000000\r\n", "output": "1\r\n0 1\r\n"}, {"input": "8\r\n5 2 6 8 3 1 6 8\r\n", "output": "4\r\n0 5\r\n4 2\r\n5 3\r\n6 5\r\n"}, {"input": "2\r\n200000000 199999999\r\n", "output": "1\r\n0 1\r\n"}, {"input": "3\r\n100000000 100000002 100000001\r\n", "output": "1\r\n1 2\r\n"}, {"input": "5\r\n1000000000 -10000000 0 8888888 7777777\r\n", "output": "3\r\n0 1\r\n2 1\r\n4 2\r\n"}, {"input": "5\r\n10 30 20 50 40\r\n", "output": "2\r\n1 2\r\n4 3\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) as f:
n = int(f.readline().strip())
arr = list(map(int, f.readline().split()))
with open(sub_path) as f:
lines = f.readlines()
if not lines:
print(0)
return
try:
k = int(lines[0].strip())
swaps = []
for line in lines[1:1 + k]:
i, j = map(int, line.strip().split())
swaps.append((i, j))
except (ValueError, IndexError):
print(0)
return
if not (0 <= k <= n):
print(0)
return
for i, j in swaps:
if not (0 <= i < n and 0 <= j < n):
print(0)
return
current = arr.copy()
for i, j in swaps:
current[i], current[j] = current[j], current[i]
sorted_arr = sorted(arr)
if current == sorted_arr:
print(1)
else:
print(0)
if __name__ == "__main__":
main()
| true |
154/A | 154 | A | PyPy 3 | TESTS | 4 | 124 | 17,715,200 | 134807402 | # Author Name: Ajay Meena
# Codeforce : https://codeforces.com/profile/majay1638
import sys
import math
import bisect
import heapq
from bisect import bisect_right
from sys import stdin, stdout
# -------------- INPUT FUNCTIONS ------------------
def get_ints_in_variables(): return map(
int, sys.stdin.readline().strip().split())
def get_int(): return int(sys.stdin.readline())
def get_ints_in_list(): return list(
map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
map(int, sys.stdin.readline().strip().split())) for _ in range(n)]
def get_string(): return sys.stdin.readline().strip()
# -------- SOME CUSTOMIZED FUNCTIONS-----------
def myceil(x, y): return (x + y - 1) // y
# -------------- SOLUTION FUNCTION ------------------
def Solution():
# Write Your Code Here
s = [c for c in get_string()]
n = get_int()
arr = [get_string() for _ in range(n)]
ans = 0
for i in range(len(s)-1):
c = s[i]+s[i+1]
for v in arr:
if v[0] in c and v[1] in c:
s[i+1] = "X"
ans += 1
break
print(ans)
def main():
# Take input Here and Call solution function
Solution()
# calling main Function
if __name__ == '__main__':
main() | 42 | 278 | 2,764,800 | 186980434 | from collections import defaultdict
from sys import stdin
input = stdin.readline
s = input().strip()
k = int(input())
ans = 0
for _ in range(k):
t = input().strip()
a = b = 0
for c in s:
if c == t[0]: a += 1
elif c == t[1]: b += 1
else:
ans += min(a, b)
a = b = 0
ans += min(a, b)
print(ans)
# https://blog.51cto.com/u_15303184/3098436 | Codeforces Round 109 (Div. 1) | CF | 2,012 | 2 | 256 | Hometask | Sergey attends lessons of the N-ish language. Each lesson he receives a hometask. This time the task is to translate some sentence to the N-ish language. Sentences of the N-ish language can be represented as strings consisting of lowercase Latin letters without spaces or punctuation marks.
Sergey totally forgot about the task until half an hour before the next lesson and hastily scribbled something down. But then he recollected that in the last lesson he learned the grammar of N-ish. The spelling rules state that N-ish contains some "forbidden" pairs of letters: such letters can never occur in a sentence next to each other. Also, the order of the letters doesn't matter (for example, if the pair of letters "ab" is forbidden, then any occurrences of substrings "ab" and "ba" are also forbidden). Also, each pair has different letters and each letter occurs in no more than one forbidden pair.
Now Sergey wants to correct his sentence so that it doesn't contain any "forbidden" pairs of letters that stand next to each other. However, he is running out of time, so he decided to simply cross out some letters from the sentence. What smallest number of letters will he have to cross out? When a letter is crossed out, it is "removed" so that the letters to its left and right (if they existed), become neighboring. For example, if we cross out the first letter from the string "aba", we get the string "ba", and if we cross out the second letter, we get "aa". | The first line contains a non-empty string s, consisting of lowercase Latin letters — that's the initial sentence in N-ish, written by Sergey. The length of string s doesn't exceed 105.
The next line contains integer k (0 ≤ k ≤ 13) — the number of forbidden pairs of letters.
Next k lines contain descriptions of forbidden pairs of letters. Each line contains exactly two different lowercase Latin letters without separators that represent the forbidden pairs. It is guaranteed that each letter is included in no more than one pair. | Print the single number — the smallest number of letters that need to be removed to get a string without any forbidden pairs of neighboring letters. Please note that the answer always exists as it is always possible to remove all letters. | null | In the first sample you should remove two letters b.
In the second sample you should remove the second or the third letter. The second restriction doesn't influence the solution. | [{"input": "ababa\n1\nab", "output": "2"}, {"input": "codeforces\n2\ndo\ncs", "output": "1"}] | 1,600 | ["greedy"] | 42 | [{"input": "ababa\r\n1\r\nab\r\n", "output": "2\r\n"}, {"input": "codeforces\r\n2\r\ndo\r\ncs\r\n", "output": "1\r\n"}, {"input": "nllnrlrnll\r\n1\r\nrl\r\n", "output": "1\r\n"}, {"input": "aludfbjtwnkgnfl\r\n1\r\noy\r\n", "output": "0\r\n"}, {"input": "pgpgppgggpbbnnn\r\n2\r\npg\r\nnb\r\n", "output": "7\r\n"}, {"input": "eepeeeeppppppeepeppe\r\n1\r\npe\r\n", "output": "10\r\n"}, {"input": "vefneyamdzoytemupniw\r\n13\r\nve\r\nfg\r\noi\r\nan\r\nck\r\nwx\r\npq\r\nml\r\nbt\r\nud\r\nrz\r\nsj\r\nhy\r\n", "output": "1\r\n"}, {"input": "drvwfaacccwnncfwuptsorrrvvvrgdzytrwweeexzyyyxuwuuk\r\n13\r\nld\r\nac\r\nnp\r\nrv\r\nmo\r\njh\r\ngb\r\nuw\r\nfq\r\nst\r\nkx\r\nzy\r\nei\r\n", "output": "11\r\n"}, {"input": "pninnihzipirpbdggrdglzdpbldtzihgbzdnrgznbpdanhnlag\r\n4\r\nli\r\nqh\r\nad\r\nbp\r\n", "output": "4\r\n"}, {"input": "mbmxuuuuxuuuuhhooooxxxuxxxuxuuxuuuxxjvjvjjjjvvvjjjjjvvjvjjjvvvjjvjjvvvjjjvjvvjvjjjjjmmbmbbbbbmbbbbmm\r\n5\r\nmb\r\nho\r\nxu\r\njv\r\nyp\r\n", "output": "37\r\n"}, {"input": "z\r\n0\r\n", "output": "0\r\n"}, {"input": "t\r\n13\r\nzx\r\nig\r\nyq\r\nbd\r\nph\r\nar\r\nne\r\nwo\r\ntk\r\njl\r\ncv\r\nfs\r\nmu\r\n", "output": "0\r\n"}, {"input": "rbnxovfcwkdjctdjfskaozjzthlcntuaoiavnbsfpuzxyvhfbxetvryvwrqetokdshadxpxijtpkrqvghsrejgnqntwiypiglzmp\r\n13\r\njr\r\nnf\r\nyk\r\ntq\r\nwe\r\nil\r\ngu\r\npb\r\naz\r\nxm\r\nhc\r\nvd\r\nso\r\n", "output": "0\r\n"}, {"input": "yynynnyyyiynyniiiinyynniiyiyyyniyniyynyyyynyynnniiiniyyniiyyiynyiynnnnyiiyiyniyyininiyiiiyynnnyiyinnnnyiinnnnnyninyinyynynyiynyyyiyinyynnyyinynyinininyniniynniiyyiiyy\r\n1\r\nni\r\n", "output": "28\r\n"}, {"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\r\n1\r\nab\r\n", "output": "75\r\n"}] | false | stdio | null | true |
549/D | 549 | D | Python 3 | TESTS | 5 | 46 | 0 | 11490679 | import sys
def solve():
n, m = map(int, input().split())
res = 0
tab = [list(input()) for _ in range(n)]
for row in range(n):
for col in range(m):
tab[row][col] = 1 if tab[row][col] == 'W' else 0
for row in range(n - 1, -1, -1):
for col in range(m - 1, -1, -1):
if tab[row][col] != 0:
diff = tab[row][col]
res+=1
for i in range(row + 1):
for j in range(col + 1):
tab[i][j] -= diff
return res
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
print(solve()) | 47 | 155 | 102,400 | 11568044 | n, m = map(int, input().split(' '))
p = [input() for i in range(n)]
p1 = [[1 if p[i][j] == 'B' else -1 for j in range(m)]
for i in range(n)]
tm = [0 for i in range(m)]
r = 0
for i in range(n-1, -1, -1):
for j in range(m-1, -1, -1):
if tm[j] != p1[i][j]:
r = r + 1
tp = p1[i][j] - tm[j]
for l in range(j + 1):
tm[l] = tm[l] + tp
print(r) | Looksery Cup 2015 | CF | 2,015 | 1 | 256 | Haar Features | The first algorithm for detecting a face on the image working in realtime was developed by Paul Viola and Michael Jones in 2001. A part of the algorithm is a procedure that computes Haar features. As part of this task, we consider a simplified model of this concept.
Let's consider a rectangular image that is represented with a table of size n × m. The table elements are integers that specify the brightness of each pixel in the image.
A feature also is a rectangular table of size n × m. Each cell of a feature is painted black or white.
To calculate the value of the given feature at the given image, you must perform the following steps. First the table of the feature is put over the table of the image (without rotations or reflections), thus each pixel is entirely covered with either black or white cell. The value of a feature in the image is the value of W - B, where W is the total brightness of the pixels in the image, covered with white feature cells, and B is the total brightness of the pixels covered with black feature cells.
Some examples of the most popular Haar features are given below.
Your task is to determine the number of operations that are required to calculate the feature by using the so-called prefix rectangles.
A prefix rectangle is any rectangle on the image, the upper left corner of which coincides with the upper left corner of the image.
You have a variable value, whose value is initially zero. In one operation you can count the sum of pixel values at any prefix rectangle, multiply it by any integer and add to variable value.
You are given a feature. It is necessary to calculate the minimum number of operations required to calculate the values of this attribute at an arbitrary image. For a better understanding of the statement, read the explanation of the first sample. | The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100) — the number of rows and columns in the feature.
Next n lines contain the description of the feature. Each line consists of m characters, the j-th character of the i-th line equals to "W", if this element of the feature is white and "B" if it is black. | Print a single number — the minimum number of operations that you need to make to calculate the value of the feature. | null | The first sample corresponds to feature B, the one shown in the picture. The value of this feature in an image of size 6 × 8 equals to the difference of the total brightness of the pixels in the lower and upper half of the image. To calculate its value, perform the following two operations:
1. add the sum of pixels in the prefix rectangle with the lower right corner in the 6-th row and 8-th column with coefficient 1 to the variable value (the rectangle is indicated by a red frame);
2. add the number of pixels in the prefix rectangle with the lower right corner in the 3-rd row and 8-th column with coefficient - 2 and variable value.
Thus, all the pixels in the lower three rows of the image will be included with factor 1, and all pixels in the upper three rows of the image will be included with factor 1 - 2 = - 1, as required. | [{"input": "6 8\nBBBBBBBB\nBBBBBBBB\nBBBBBBBB\nWWWWWWWW\nWWWWWWWW\nWWWWWWWW", "output": "2"}, {"input": "3 3\nWBW\nBWW\nWWW", "output": "4"}, {"input": "3 6\nWWBBWW\nWWBBWW\nWWBBWW", "output": "3"}, {"input": "4 4\nBBBB\nBBBB\nBBBB\nBBBW", "output": "4"}] | 1,900 | ["greedy", "implementation"] | 47 | [{"input": "6 8\r\nBBBBBBBB\r\nBBBBBBBB\r\nBBBBBBBB\r\nWWWWWWWW\r\nWWWWWWWW\r\nWWWWWWWW\r\n", "output": "2\r\n"}, {"input": "3 3\r\nWBW\r\nBWW\r\nWWW\r\n", "output": "4\r\n"}, {"input": "3 6\r\nWWBBWW\r\nWWBBWW\r\nWWBBWW\r\n", "output": "3\r\n"}, {"input": "4 4\r\nBBBB\r\nBBBB\r\nBBBB\r\nBBBW\r\n", "output": "4\r\n"}, {"input": "10 9\r\nBWWWBWWBB\r\nBBWWBWBBW\r\nBBWBWBWBB\r\nBWBWBBBBB\r\nBBWBWBWBW\r\nBWWBWWBBW\r\nWBWWWBWWW\r\nWBBWBWBWW\r\nBBWWBWWBB\r\nBBWWBWWBW\r\n", "output": "61\r\n"}, {"input": "4 1\r\nW\r\nW\r\nB\r\nB\r\n", "output": "2\r\n"}, {"input": "2 10\r\nBBWBWWBWBB\r\nBBBBBBBBBW\r\n", "output": "10\r\n"}, {"input": "100 1\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nW\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\nB\r\n", "output": "2\r\n"}, {"input": "1 100\r\nWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\r\n", "output": "2\r\n"}, {"input": "4 5\r\nBWWBB\r\nBWBBW\r\nWBWWW\r\nWBWWB\r\n", "output": "13\r\n"}, {"input": "2 9\r\nWBBBWBBBW\r\nBWWBBBBBB\r\n", "output": "9\r\n"}, {"input": "6 6\r\nBBWWWB\r\nWBBBWB\r\nBBBBBW\r\nWWWWWW\r\nBBBBBW\r\nBWWBBB\r\n", "output": "16\r\n"}, {"input": "1 1\r\nW\r\n", "output": "1\r\n"}, {"input": "1 1\r\nB\r\n", "output": "1\r\n"}, {"input": "1 8\r\nWWBWWWWW\r\n", "output": "3\r\n"}, {"input": "2 8\r\nBBBBBBBB\r\nBBBBBBBB\r\n", "output": "1\r\n"}, {"input": "1 52\r\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\r\n", "output": "1\r\n"}, {"input": "11 8\r\nWWWWWWWW\r\nWWWWWWWW\r\nWWWWWWWW\r\nWWWWWWWW\r\nWWWWWWWW\r\nWWWBWWWW\r\nWWWWWWWW\r\nWBWWWWWW\r\nWWWWWWWW\r\nWWWWWWWW\r\nWWWWWWWW\r\n", "output": "9\r\n"}] | false | stdio | null | true |
925/B | 925 | B | PyPy 3-64 | TESTS | 0 | 46 | 512,000 | 144816962 | from math import ceil
from bisect import bisect_left as bl
n,m1,m2=map(int,input().strip().split())
d={m1:0,m2:1}
a=[*map(int,input().strip().split())]
a=list(zip(a,range(len(a))))
a.sort(key=lambda s:s[0])
val={m1:1111,m2:1111}
for i in range(1,n+1):
req=ceil(m1/i)
ll = bl(a, (req, -1))
if n - ll >= i:
val[m1]=i
break
for i in range(1,n+1):
req=ceil(m2/i)
ll = bl(a, (req, -1))
if n - ll >= i:
val[m2]=i
break
m1,m2=sorted([m1,m2],key=lambda s:val[s],reverse=True)
a1=[]
for i in range(1,n+1):
req=ceil(m2/i)
ll=bl(a,(req,-1))
if n-ll>=i:
a1+=a[ll:ll+i]
a=a[:ll]+a[ll+i:]
break
# print(a)
n=len(a)
a2=[]
for i in range(1,n+1):
req=ceil(m1/i)
ll=bl(a,(req,-1))
if n-ll>=i:
a2+=a[ll:ll+i]
break
ans=[None,None]
if m1!=m2:
ans[d[m2]]=a1
ans[d[m1]]=a2
else:
ans[0]=a1
ans[1]=a2
# print(a1,a2)
if len(a1)==0 or len(a2)==0:
print("No")
exit()
print("Yes")
print(len(ans[0]),len(ans[1]))
for val,id in ans[0]:
print(id+1,end=" ")
print('')
for val,id in ans[1]:
print(id+1,end=" ") | 40 | 1,216 | 39,219,200 | 42132645 | # python3
def readline(): return tuple(map(int, input().split()))
def ceil_div(num, den): return (num - 1) // den + 1
def main():
n, x1, x2 = readline()
c = readline()
xx = (x1, x2)
servers = sorted(enumerate(c, start=1), key=lambda p: p[1])
for (i, a) in enumerate(servers):
for (j, x) in enumerate(xx):
kj = ceil_div(x, a[1])
if i + kj < n and (n - i - kj) * servers[i + kj][1] >= sum(xx) - x:
print("Yes")
l1 = servers[i:i+kj]
l2 = servers[i+kj:]
if j: l1, l2 = l2, l1
print(len(l1), len(l2))
print(" ".join(str(d[0]) for d in l1))
print(" ".join(str(d[0]) for d in l2))
return
print("No")
main()
# Made By Mostafa_Khaled | VK Cup 2018 - Round 3 | CF | 2,018 | 2 | 256 | Resource Distribution | One department of some software company has $$$n$$$ servers of different specifications. Servers are indexed with consecutive integers from $$$1$$$ to $$$n$$$. Suppose that the specifications of the $$$j$$$-th server may be expressed with a single integer number $$$c_j$$$ of artificial resource units.
In order for production to work, it is needed to deploy two services $$$S_1$$$ and $$$S_2$$$ to process incoming requests using the servers of the department. Processing of incoming requests of service $$$S_i$$$ takes $$$x_i$$$ resource units.
The described situation happens in an advanced company, that is why each service may be deployed using not only one server, but several servers simultaneously. If service $$$S_i$$$ is deployed using $$$k_i$$$ servers, then the load is divided equally between these servers and each server requires only $$$x_i / k_i$$$ (that may be a fractional number) resource units.
Each server may be left unused at all, or be used for deploying exactly one of the services (but not for two of them simultaneously). The service should not use more resources than the server provides.
Determine if it is possible to deploy both services using the given servers, and if yes, determine which servers should be used for deploying each of the services. | The first line contains three integers $$$n$$$, $$$x_1$$$, $$$x_2$$$ ($$$2 \leq n \leq 300\,000$$$, $$$1 \leq x_1, x_2 \leq 10^9$$$) — the number of servers that the department may use, and resource units requirements for each of the services.
The second line contains $$$n$$$ space-separated integers $$$c_1, c_2, \ldots, c_n$$$ ($$$1 \leq c_i \leq 10^9$$$) — the number of resource units provided by each of the servers. | If it is impossible to deploy both services using the given servers, print the only word "No" (without the quotes).
Otherwise print the word "Yes" (without the quotes).
In the second line print two integers $$$k_1$$$ and $$$k_2$$$ ($$$1 \leq k_1, k_2 \leq n$$$) — the number of servers used for each of the services.
In the third line print $$$k_1$$$ integers, the indices of the servers that will be used for the first service.
In the fourth line print $$$k_2$$$ integers, the indices of the servers that will be used for the second service.
No index may appear twice among the indices you print in the last two lines. If there are several possible answers, it is allowed to print any of them. | null | In the first sample test each of the servers 1, 2 and 6 will will provide $$$8 / 3 = 2.(6)$$$ resource units and each of the servers 5, 4 will provide $$$16 / 2 = 8$$$ resource units.
In the second sample test the first server will provide $$$20$$$ resource units and each of the remaining servers will provide $$$32 / 3 = 10.(6)$$$ resource units. | [{"input": "6 8 16\n3 5 2 9 8 7", "output": "Yes\n3 2\n1 2 6\n5 4"}, {"input": "4 20 32\n21 11 11 12", "output": "Yes\n1 3\n1\n2 3 4"}, {"input": "4 11 32\n5 5 16 16", "output": "No"}, {"input": "5 12 20\n7 8 4 11 9", "output": "No"}] | 1,700 | ["binary search", "implementation", "sortings"] | 40 | [{"input": "6 8 16\r\n3 5 2 9 8 7\r\n", "output": "Yes\r\n4 2\r\n3 1 2 6\r\n5 4\r\n"}, {"input": "4 20 32\r\n21 11 11 12\r\n", "output": "Yes\r\n1 3\r\n1\r\n2 3 4\r\n"}, {"input": "4 11 32\r\n5 5 16 16\r\n", "output": "No\r\n"}, {"input": "5 12 20\r\n7 8 4 11 9\r\n", "output": "No\r\n"}, {"input": "2 1 1\r\n1 1\r\n", "output": "Yes\r\n1 1\r\n1\r\n2\r\n"}, {"input": "2 1 1\r\n1 1000000\r\n", "output": "Yes\r\n1 1\r\n1\r\n2\r\n"}, {"input": "2 1 1\r\n1000000000 1000000000\r\n", "output": "Yes\r\n1 1\r\n1\r\n2\r\n"}, {"input": "2 1 2\r\n1 1\r\n", "output": "No\r\n"}, {"input": "15 250 200\r\n71 2 77 69 100 53 54 40 73 32 82 58 24 82 41\r\n", "output": "Yes\r\n11 3\r\n13 10 8 15 6 7 12 4 1 9 3\r\n11 14 5\r\n"}, {"input": "4 12 11\r\n4 4 6 11\r\n", "output": "Yes\r\n3 1\r\n1 2 3\r\n4\r\n"}] | false | stdio | import sys
def main(input_path, output_path, submission_path):
with open(input_path) as f:
n, x1, x2 = map(int, f.readline().split())
c = list(map(int, f.readline().split()))
with open(output_path) as f:
ref_lines = [line.strip() for line in f]
with open(submission_path) as f:
sub_lines = [line.strip() for line in f]
non_empty_ref = [line for line in ref_lines if line]
non_empty_sub = [line for line in sub_lines if line]
if not non_empty_ref:
print(0)
return
ref_first = non_empty_ref[0]
if ref_first == 'No':
if len(non_empty_sub) == 1 and non_empty_sub[0] == 'No':
print(1)
else:
print(0)
return
else:
if len(non_empty_sub) != 4 or non_empty_sub[0] != 'Yes':
print(0)
return
try:
k1, k2 = map(int, non_empty_sub[1].split())
except:
print(0)
return
if k1 <= 0 or k2 <= 0 or k1 + k2 > n:
print(0)
return
try:
s1 = list(map(int, non_empty_sub[2].split()))
s2 = list(map(int, non_empty_sub[3].split()))
except:
print(0)
return
if len(s1) != k1 or len(s2) != k2:
print(0)
return
all_indices = set()
for s in s1:
if not (1 <= s <= n) or s in all_indices:
print(0)
return
all_indices.add(s)
for s in s2:
if not (1 <= s <= n) or s in all_indices:
print(0)
return
all_indices.add(s)
for s in s1:
if c[s-1] * k1 < x1:
print(0)
return
for s in s2:
if c[s-1] * k2 < x2:
print(0)
return
print(1)
return
if __name__ == "__main__":
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
main(input_path, output_path, submission_path) | true |
154/A | 154 | A | PyPy 3 | TESTS | 4 | 248 | 0 | 57094042 | def is_valid(word, rules):
for rule in rules:
if rule in word:
return False
return True
class CodeforcesTask154ASolution:
def __init__(self):
self.result = ''
self.word = ''
self.rules_count = 0
self.forbidden = []
def read_input(self):
self.word = input()
self.rules_count = int(input())
for x in range(self.rules_count):
self.forbidden.append(input())
def process_task(self):
forbidden_rev = [x[::-1] for x in self.forbidden]
forbs = forbidden_rev + self.forbidden
#print(forbs)
cross = 0
while not is_valid(self.word, forbs):
for rule in forbs:
index = self.word.find(rule)
cross += 1
if index == 0:
self.word = self.word.replace(rule, rule[1], 1)
elif index > 0:
if self.word[index - 1] + self.word[index + 1] in forbs:
self.word = self.word.replace(rule, rule[0], 1)
else:
self.word = self.word.replace(rule, rule[1], 1)
else:
cross -= 1
self.result = str(cross)
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask154ASolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result()) | 42 | 342 | 3,072,000 | 101687730 | import sys
#input = sys.stdin.readline
for _ in range(1):
s=input()
n=len(s)
d={}
for _ in range(int(input())):
a=input()
d[a[0]]=a[1]
d[a[1]]=a[0]
i=0
ans=0
while i<n:
if s[i] in d:
a,b=0,0
while True:
start=i
while i<n and s[i]==s[start]:
a+=1
i+=1
if i<n and s[i]==d[s[start]]:
#b+=1
start=i
while i<n and s[i]==s[start]:
b+=1
i+=1
if i<n and s[i]==d[s[start]]:
continue
else:
ans+=min(a,b)
break
else:
i+=1
print(ans) | Codeforces Round 109 (Div. 1) | CF | 2,012 | 2 | 256 | Hometask | Sergey attends lessons of the N-ish language. Each lesson he receives a hometask. This time the task is to translate some sentence to the N-ish language. Sentences of the N-ish language can be represented as strings consisting of lowercase Latin letters without spaces or punctuation marks.
Sergey totally forgot about the task until half an hour before the next lesson and hastily scribbled something down. But then he recollected that in the last lesson he learned the grammar of N-ish. The spelling rules state that N-ish contains some "forbidden" pairs of letters: such letters can never occur in a sentence next to each other. Also, the order of the letters doesn't matter (for example, if the pair of letters "ab" is forbidden, then any occurrences of substrings "ab" and "ba" are also forbidden). Also, each pair has different letters and each letter occurs in no more than one forbidden pair.
Now Sergey wants to correct his sentence so that it doesn't contain any "forbidden" pairs of letters that stand next to each other. However, he is running out of time, so he decided to simply cross out some letters from the sentence. What smallest number of letters will he have to cross out? When a letter is crossed out, it is "removed" so that the letters to its left and right (if they existed), become neighboring. For example, if we cross out the first letter from the string "aba", we get the string "ba", and if we cross out the second letter, we get "aa". | The first line contains a non-empty string s, consisting of lowercase Latin letters — that's the initial sentence in N-ish, written by Sergey. The length of string s doesn't exceed 105.
The next line contains integer k (0 ≤ k ≤ 13) — the number of forbidden pairs of letters.
Next k lines contain descriptions of forbidden pairs of letters. Each line contains exactly two different lowercase Latin letters without separators that represent the forbidden pairs. It is guaranteed that each letter is included in no more than one pair. | Print the single number — the smallest number of letters that need to be removed to get a string without any forbidden pairs of neighboring letters. Please note that the answer always exists as it is always possible to remove all letters. | null | In the first sample you should remove two letters b.
In the second sample you should remove the second or the third letter. The second restriction doesn't influence the solution. | [{"input": "ababa\n1\nab", "output": "2"}, {"input": "codeforces\n2\ndo\ncs", "output": "1"}] | 1,600 | ["greedy"] | 42 | [{"input": "ababa\r\n1\r\nab\r\n", "output": "2\r\n"}, {"input": "codeforces\r\n2\r\ndo\r\ncs\r\n", "output": "1\r\n"}, {"input": "nllnrlrnll\r\n1\r\nrl\r\n", "output": "1\r\n"}, {"input": "aludfbjtwnkgnfl\r\n1\r\noy\r\n", "output": "0\r\n"}, {"input": "pgpgppgggpbbnnn\r\n2\r\npg\r\nnb\r\n", "output": "7\r\n"}, {"input": "eepeeeeppppppeepeppe\r\n1\r\npe\r\n", "output": "10\r\n"}, {"input": "vefneyamdzoytemupniw\r\n13\r\nve\r\nfg\r\noi\r\nan\r\nck\r\nwx\r\npq\r\nml\r\nbt\r\nud\r\nrz\r\nsj\r\nhy\r\n", "output": "1\r\n"}, {"input": "drvwfaacccwnncfwuptsorrrvvvrgdzytrwweeexzyyyxuwuuk\r\n13\r\nld\r\nac\r\nnp\r\nrv\r\nmo\r\njh\r\ngb\r\nuw\r\nfq\r\nst\r\nkx\r\nzy\r\nei\r\n", "output": "11\r\n"}, {"input": "pninnihzipirpbdggrdglzdpbldtzihgbzdnrgznbpdanhnlag\r\n4\r\nli\r\nqh\r\nad\r\nbp\r\n", "output": "4\r\n"}, {"input": "mbmxuuuuxuuuuhhooooxxxuxxxuxuuxuuuxxjvjvjjjjvvvjjjjjvvjvjjjvvvjjvjjvvvjjjvjvvjvjjjjjmmbmbbbbbmbbbbmm\r\n5\r\nmb\r\nho\r\nxu\r\njv\r\nyp\r\n", "output": "37\r\n"}, {"input": "z\r\n0\r\n", "output": "0\r\n"}, {"input": "t\r\n13\r\nzx\r\nig\r\nyq\r\nbd\r\nph\r\nar\r\nne\r\nwo\r\ntk\r\njl\r\ncv\r\nfs\r\nmu\r\n", "output": "0\r\n"}, {"input": "rbnxovfcwkdjctdjfskaozjzthlcntuaoiavnbsfpuzxyvhfbxetvryvwrqetokdshadxpxijtpkrqvghsrejgnqntwiypiglzmp\r\n13\r\njr\r\nnf\r\nyk\r\ntq\r\nwe\r\nil\r\ngu\r\npb\r\naz\r\nxm\r\nhc\r\nvd\r\nso\r\n", "output": "0\r\n"}, {"input": "yynynnyyyiynyniiiinyynniiyiyyyniyniyynyyyynyynnniiiniyyniiyyiynyiynnnnyiiyiyniyyininiyiiiyynnnyiyinnnnyiinnnnnyninyinyynynyiynyyyiyinyynnyyinynyinininyniniynniiyyiiyy\r\n1\r\nni\r\n", "output": "28\r\n"}, {"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\r\n1\r\nab\r\n", "output": "75\r\n"}] | false | stdio | null | true |
560/A | 560 | A | PyPy 3 | TESTS | 11 | 124 | 0 | 62326847 | n=input()
x=input()
if(" 1 " in x):
print(-1)
elif(x[0]=="1"):
print(-1)
elif(x[len(x)-1]=="1"):
print(-1)
else:
print(1) | 16 | 31 | 0 | 146537537 | input()
a=list(map(int,input().split()))
print(-1 if 1 in a else 1) | Codeforces Round 313 (Div. 2) | CF | 2,015 | 2 | 256 | Currency System in Geraldion | A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum? | The first line contains number n (1 ≤ n ≤ 1000) — the number of values of the banknotes that used in Geraldion.
The second line contains n distinct space-separated numbers a1, a2, ..., an (1 ≤ ai ≤ 106) — the values of the banknotes. | Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print - 1. | null | null | [{"input": "5\n1 2 3 4 5", "output": "-1"}] | 1,000 | ["implementation", "sortings"] | 16 | [{"input": "5\r\n1 2 3 4 5\r\n", "output": "-1\r\n"}, {"input": "1\r\n2\r\n", "output": "1\r\n"}, {"input": "10\r\n371054 506438 397130 1 766759 208409 769264 549213 641270 771837\r\n", "output": "-1\r\n"}, {"input": "10\r\n635370 154890 909382 220996 276501 716105 538714 140162 171960 271264\r\n", "output": "1\r\n"}, {"input": "50\r\n110876 835020 859879 999908 712969 788264 287153 921820 330355 499311 209594 484829 296329 940051 174081 931503 1 780512 390075 97866 124255 950067 697612 244256 782385 789882 37608 82153 399889 598867 416717 377988 535636 511221 792568 683271 131077 290194 496712 330720 587436 563481 645817 942562 654093 980561 382937 48293 582608 116156\r\n", "output": "-1\r\n"}, {"input": "50\r\n474421 421097 217233 156339 27075 733996 281778 863492 184707 956857 288561 70997 393786 337382 663642 131184 637 273801 799870 295017 392338 842567 161819 297705 102013 930684 375703 838048 154915 138503 629056 256591 893619 19263 787927 684541 320265 841090 421423 490879 394582 493952 619247 633202 612928 50907 276653 407819 489945 153173\r\n", "output": "1\r\n"}, {"input": "1\r\n1\r\n", "output": "-1\r\n"}, {"input": "1\r\n1000000\r\n", "output": "1\r\n"}, {"input": "2\r\n3 2\r\n", "output": "1\r\n"}, {"input": "2\r\n2 3\r\n", "output": "1\r\n"}] | false | stdio | null | true |
154/A | 154 | A | Python 3 | TESTS | 4 | 92 | 307,200 | 5544384 | t = input()
p, n = {}, len(t)
for i in range(int(input())):
q = input()
p[q[0]], p[q[1]] = q[1], q[0]
if n == 1: print(0)
elif n == 2: print(int(p[t[0]] == t[1]))
else:
a, b, c = t[0], t[1], t[2]
i, s = 3, 0
while True:
if a in p and p[a] == b:
s += 1
if i > n - 2: break
a, b, c = c, t[i], t[i + 1]
i += 2
else:
if i > n - 1:
if b in p and p[b] == c: s += 1
break
a, b, c = b, c, t[i]
i += 1
print(s) | 42 | 372 | 2,662,400 | 101616018 | s=input()
ans=0
for _ in range(int(input())):
a,b=0,0
p=input()
for x in s:
if(x==p[0]):a+=1
elif(x==p[1]):b+=1
else:
ans+=min(a,b);
a,b=0,0
ans+=min(a,b)
print(ans) | Codeforces Round 109 (Div. 1) | CF | 2,012 | 2 | 256 | Hometask | Sergey attends lessons of the N-ish language. Each lesson he receives a hometask. This time the task is to translate some sentence to the N-ish language. Sentences of the N-ish language can be represented as strings consisting of lowercase Latin letters without spaces or punctuation marks.
Sergey totally forgot about the task until half an hour before the next lesson and hastily scribbled something down. But then he recollected that in the last lesson he learned the grammar of N-ish. The spelling rules state that N-ish contains some "forbidden" pairs of letters: such letters can never occur in a sentence next to each other. Also, the order of the letters doesn't matter (for example, if the pair of letters "ab" is forbidden, then any occurrences of substrings "ab" and "ba" are also forbidden). Also, each pair has different letters and each letter occurs in no more than one forbidden pair.
Now Sergey wants to correct his sentence so that it doesn't contain any "forbidden" pairs of letters that stand next to each other. However, he is running out of time, so he decided to simply cross out some letters from the sentence. What smallest number of letters will he have to cross out? When a letter is crossed out, it is "removed" so that the letters to its left and right (if they existed), become neighboring. For example, if we cross out the first letter from the string "aba", we get the string "ba", and if we cross out the second letter, we get "aa". | The first line contains a non-empty string s, consisting of lowercase Latin letters — that's the initial sentence in N-ish, written by Sergey. The length of string s doesn't exceed 105.
The next line contains integer k (0 ≤ k ≤ 13) — the number of forbidden pairs of letters.
Next k lines contain descriptions of forbidden pairs of letters. Each line contains exactly two different lowercase Latin letters without separators that represent the forbidden pairs. It is guaranteed that each letter is included in no more than one pair. | Print the single number — the smallest number of letters that need to be removed to get a string without any forbidden pairs of neighboring letters. Please note that the answer always exists as it is always possible to remove all letters. | null | In the first sample you should remove two letters b.
In the second sample you should remove the second or the third letter. The second restriction doesn't influence the solution. | [{"input": "ababa\n1\nab", "output": "2"}, {"input": "codeforces\n2\ndo\ncs", "output": "1"}] | 1,600 | ["greedy"] | 42 | [{"input": "ababa\r\n1\r\nab\r\n", "output": "2\r\n"}, {"input": "codeforces\r\n2\r\ndo\r\ncs\r\n", "output": "1\r\n"}, {"input": "nllnrlrnll\r\n1\r\nrl\r\n", "output": "1\r\n"}, {"input": "aludfbjtwnkgnfl\r\n1\r\noy\r\n", "output": "0\r\n"}, {"input": "pgpgppgggpbbnnn\r\n2\r\npg\r\nnb\r\n", "output": "7\r\n"}, {"input": "eepeeeeppppppeepeppe\r\n1\r\npe\r\n", "output": "10\r\n"}, {"input": "vefneyamdzoytemupniw\r\n13\r\nve\r\nfg\r\noi\r\nan\r\nck\r\nwx\r\npq\r\nml\r\nbt\r\nud\r\nrz\r\nsj\r\nhy\r\n", "output": "1\r\n"}, {"input": "drvwfaacccwnncfwuptsorrrvvvrgdzytrwweeexzyyyxuwuuk\r\n13\r\nld\r\nac\r\nnp\r\nrv\r\nmo\r\njh\r\ngb\r\nuw\r\nfq\r\nst\r\nkx\r\nzy\r\nei\r\n", "output": "11\r\n"}, {"input": "pninnihzipirpbdggrdglzdpbldtzihgbzdnrgznbpdanhnlag\r\n4\r\nli\r\nqh\r\nad\r\nbp\r\n", "output": "4\r\n"}, {"input": "mbmxuuuuxuuuuhhooooxxxuxxxuxuuxuuuxxjvjvjjjjvvvjjjjjvvjvjjjvvvjjvjjvvvjjjvjvvjvjjjjjmmbmbbbbbmbbbbmm\r\n5\r\nmb\r\nho\r\nxu\r\njv\r\nyp\r\n", "output": "37\r\n"}, {"input": "z\r\n0\r\n", "output": "0\r\n"}, {"input": "t\r\n13\r\nzx\r\nig\r\nyq\r\nbd\r\nph\r\nar\r\nne\r\nwo\r\ntk\r\njl\r\ncv\r\nfs\r\nmu\r\n", "output": "0\r\n"}, {"input": "rbnxovfcwkdjctdjfskaozjzthlcntuaoiavnbsfpuzxyvhfbxetvryvwrqetokdshadxpxijtpkrqvghsrejgnqntwiypiglzmp\r\n13\r\njr\r\nnf\r\nyk\r\ntq\r\nwe\r\nil\r\ngu\r\npb\r\naz\r\nxm\r\nhc\r\nvd\r\nso\r\n", "output": "0\r\n"}, {"input": "yynynnyyyiynyniiiinyynniiyiyyyniyniyynyyyynyynnniiiniyyniiyyiynyiynnnnyiiyiyniyyininiyiiiyynnnyiyinnnnyiinnnnnyninyinyynynyiynyyyiyinyynnyyinynyinininyniniynniiyyiiyy\r\n1\r\nni\r\n", "output": "28\r\n"}, {"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb\r\n1\r\nab\r\n", "output": "75\r\n"}] | false | stdio | null | true |
455/B | 455 | B | PyPy 3-64 | TESTS | 4 | 62 | 0 | 216139921 | def is_winner(words, k):
# Check if all the words have the same parity of lengths (either all even or all odd).
is_all_even = all(len(word) % 2 == 0 for word in words)
is_all_odd = all(len(word) % 2 == 1 for word in words)
# If all the words have the same parity, the winner of the last game depends on the number of games (k).
# If the number of games (k) is odd, the winner of the last game is the same as the first game.
# If the number of games (k) is even, the winner of the last game is the opposite of the first game.
if is_all_even or is_all_odd:
return "First" if k % 2 == 1 else "Second"
else:
return "Second"
# Read input
n, k = map(int, input().split())
words = [input().strip() for _ in range(n)]
# Output the result
print(is_winner(words, k)) | 75 | 701 | 41,984,000 | 229069437 | max_alpha = 26
class TrieNode:
def __init__(self):
self.children = [0] * max_alpha
def add_trie(root, word):
v = root
for char in word:
c = ord(char) - ord('a')
if not v.children[c]:
v.children[c] = TrieNode()
v = v.children[c]
def dfs(v):
v.win = False
v.lose = False
is_leaf = True
for i in range(max_alpha):
if v.children[i]:
is_leaf = False
to = v.children[i]
dfs(to)
v.win |= not to.win
v.lose |= not to.lose
if is_leaf:
v.lose = True
class TrieNode:
def __init__(self):
self.children = [0] * max_alpha
self.win = False
self.lose = False
def answer(res):
print("First" if res else "Second")
exit(0)
n, k = map(int, input().split())
root = TrieNode()
for _ in range(n):
word = input().strip()
add_trie(root, word)
dfs(root)
if k == 1:
answer(root.win)
elif not root.win:
answer(root.win)
elif root.lose:
answer(root.win)
elif k % 2 == 1:
answer(root.win)
else:
answer(not root.win) | Codeforces Round 260 (Div. 1) | CF | 2,014 | 1 | 256 | A Lot of Games | Andrew, Fedor and Alex are inventive guys. Now they invent the game with strings for two players.
Given a group of n non-empty strings. During the game two players build the word together, initially the word is empty. The players move in turns. On his step player must add a single letter in the end of the word, the resulting word must be prefix of at least one string from the group. A player loses if he cannot move.
Andrew and Alex decided to play this game k times. The player who is the loser of the i-th game makes the first move in the (i + 1)-th game. Guys decided that the winner of all games is the player who wins the last (k-th) game. Andrew and Alex already started the game. Fedor wants to know who wins the game if both players will play optimally. Help him. | The first line contains two integers, n and k (1 ≤ n ≤ 105; 1 ≤ k ≤ 109).
Each of the next n lines contains a single non-empty string from the given group. The total length of all strings from the group doesn't exceed 105. Each string of the group consists only of lowercase English letters. | If the player who moves first wins, print "First", otherwise print "Second" (without the quotes). | null | null | [{"input": "2 3\na\nb", "output": "First"}, {"input": "3 1\na\nb\nc", "output": "First"}, {"input": "1 2\nab", "output": "Second"}] | 1,900 | ["dfs and similar", "dp", "games", "implementation", "strings", "trees"] | 75 | [{"input": "2 3\r\na\r\nb\r\n", "output": "First\r\n"}, {"input": "3 1\r\na\r\nb\r\nc\r\n", "output": "First\r\n"}, {"input": "1 2\r\nab\r\n", "output": "Second\r\n"}, {"input": "5 6\r\nabas\r\ndsfdf\r\nabacaba\r\ndartsidius\r\nkolobok\r\n", "output": "Second\r\n"}, {"input": "4 2\r\naaaa\r\nbbbb\r\nccccc\r\ndumbavumba\r\n", "output": "First\r\n"}, {"input": "3 8\r\nso\r\nbad\r\ntest\r\n", "output": "First\r\n"}, {"input": "5 2\r\nwelcome\r\nto\r\nthe\r\nmatrix\r\nneo\r\n", "output": "First\r\n"}, {"input": "6 4\r\ndog\r\ncat\r\ncow\r\nhot\r\nice\r\nlol\r\n", "output": "Second\r\n"}, {"input": "4 8\r\nla\r\na\r\nz\r\nka\r\n", "output": "First\r\n"}, {"input": "3 2\r\nop\r\nhop\r\ncop\r\n", "output": "First\r\n"}, {"input": "3 3\r\nabasdfabab\r\nabaaasdfdsf\r\nasdfaba\r\n", "output": "Second\r\n"}, {"input": "2 2\r\naba\r\naa\r\n", "output": "Second\r\n"}, {"input": "4 1\r\naa\r\naba\r\nba\r\nbba\r\n", "output": "Second\r\n"}, {"input": "1 3\r\nab\r\n", "output": "Second\r\n"}, {"input": "3 3\r\naa\r\nabb\r\ncc\r\n", "output": "Second\r\n"}] | false | stdio | null | true |
562/F | 566 | A | PyPy 3 | TESTS | 0 | 93 | 0 | 12331028 | SIGMA = 26
class Node:
def __init__(self):
self.ch = [None] * SIGMA
self.a = []
self.b = []
def add(self, s, i):
t = self
for c in s:
v = ord(c) - ord('a')
if not t.ch[v]:
t.ch[v] = Node()
t = t.ch[v]
t.a += [i]
def inc(self, s, i):
t = self
for c in s:
v = ord(c) - ord('a')
if not t.ch[v]:
break
t = t.ch[v]
t.b += [i]
def solve(self, d):
if not self:
return (0, [], [], [])
res = 0
pairs = []
for i in range(SIGMA):
if self.ch[i]:
t, a, b, p = self.ch[i].solve(d + 1)
res += t
self.a += a
self.b += b
pairs += p
k = min(len(self.a), len(self.b))
res += d * k
pairs += [(self.a[i], self.b[i]) for i in range(k)]
self.a = self.a[k:]
self.b = self.b[k:]
return (res, self.a, self.b, pairs)
N = int(input())
A = [input() for i in range(N)]
B = [input() for i in range(N)]
T = Node()
for i, s in enumerate(B):
T.add(s, i + 1)
for i, s in enumerate(A):
T.inc(s, i + 1)
res = T.solve(0)
print(res[0])
for i, j in res[3]:
print('%d %d' % (i, j)) | 38 | 1,372 | 262,553,600 | 12334711 | import sys
class Node:
def __init__(self, d):
global nodes
self.ch = {}
self.a = [[], []]
self.d = d
nodes += [self]
nodes = []
pairs = []
res = 0
N = int(sys.stdin.readline())
_input = sys.stdin.readlines()
_input = [s[:-1] for s in _input]
A = [_input[:N], _input[N:]]
T = Node(0)
for i, l in enumerate(A):
for j, s in enumerate(l):
t = T
for c in s:
v = ord(c) - ord('a')
if not v in t.ch:
t.ch[v] = Node(t.d + 1)
t = t.ch[v]
t.a[i] += [j + 1]
for n in reversed(nodes):
for i in n.ch:
n.a[0] += n.ch[i].a[0]
n.a[1] += n.ch[i].a[1]
k = min(len(n.a[0]), len(n.a[1]))
for i in range(k):
pairs += [str(n.a[0][-1]) + ' ' + str(n.a[1][-1])]
n.a[0].pop()
n.a[1].pop()
res += n.d
print(res)
print('\n'.join(pairs)) | VK Cup 2015 - Finals | CF | 2,015 | 2 | 256 | Matching Names | Teachers of one programming summer school decided to make a surprise for the students by giving them names in the style of the "Hobbit" movie. Each student must get a pseudonym maximally similar to his own name. The pseudonym must be a name of some character of the popular saga and now the teachers are busy matching pseudonyms to student names.
There are n students in a summer school. Teachers chose exactly n pseudonyms for them. Each student must get exactly one pseudonym corresponding to him. Let us determine the relevance of a pseudonym b to a student with name a as the length of the largest common prefix a and b. We will represent such value as $$\operatorname{lcp}(a,b)$$. Then we can determine the quality of matching of the pseudonyms to students as a sum of relevances of all pseudonyms to the corresponding students.
Find the matching between students and pseudonyms with the maximum quality. | The first line contains number n (1 ≤ n ≤ 100 000) — the number of students in the summer school.
Next n lines contain the name of the students. Each name is a non-empty word consisting of lowercase English letters. Some names can be repeating.
The last n lines contain the given pseudonyms. Each pseudonym is a non-empty word consisting of small English letters. Some pseudonyms can be repeating.
The total length of all the names and pseudonyms doesn't exceed 800 000 characters. | In the first line print the maximum possible quality of matching pseudonyms to students.
In the next n lines describe the optimal matching. Each line must have the form a b (1 ≤ a, b ≤ n), that means that the student who was number a in the input, must match to the pseudonym number b in the input.
The matching should be a one-to-one correspondence, that is, each student and each pseudonym should occur exactly once in your output. If there are several optimal answers, output any. | null | The first test from the statement the match looks as follows:
- bill → bilbo (lcp = 3)
- galya → galadriel (lcp = 3)
- gennady → gendalf (lcp = 3)
- toshik → torin (lcp = 2)
- boris → smaug (lcp = 0) | [{"input": "5\ngennady\ngalya\nboris\nbill\ntoshik\nbilbo\ntorin\ngendalf\nsmaug\ngaladriel", "output": "11\n4 1\n2 5\n1 3\n5 2\n3 4"}] | 2,300 | [] | 38 | [{"input": "5\r\ngennady\r\ngalya\r\nboris\r\nbill\r\ntoshik\r\nbilbo\r\ntorin\r\ngendalf\r\nsmaug\r\ngaladriel\r\n", "output": "11\r\n4 1\r\n2 5\r\n1 3\r\n5 2\r\n3 4\r\n"}, {"input": "1\r\na\r\na\r\n", "output": "1\r\n1 1\r\n"}, {"input": "2\r\na\r\na\r\na\r\na\r\n", "output": "2\r\n1 1\r\n2 2\r\n"}, {"input": "2\r\na\r\nb\r\na\r\na\r\n", "output": "1\r\n1 1\r\n2 2\r\n"}, {"input": "2\r\nb\r\nb\r\na\r\na\r\n", "output": "0\r\n1 1\r\n2 2\r\n"}, {"input": "2\r\na\r\nb\r\na\r\nb\r\n", "output": "2\r\n1 1\r\n2 2\r\n"}, {"input": "10\r\nbaa\r\na\r\nba\r\naabab\r\naa\r\nbaab\r\nbb\r\nabbbb\r\na\r\na\r\na\r\nba\r\nba\r\nbaabbb\r\nba\r\na\r\naabb\r\nbaa\r\nab\r\nb\r\n", "output": "17\r\n4 7\r\n8 9\r\n2 1\r\n9 6\r\n6 4\r\n1 8\r\n3 2\r\n7 10\r\n10 3\r\n5 5\r\n"}, {"input": "10\r\nabaabbaaa\r\nacccccaacabc\r\nacbaabaaabbca\r\naaccca\r\ncbbba\r\naaba\r\nacab\r\nac\r\ncbac\r\nca\r\nbbbbc\r\nbacbcbcaac\r\nc\r\ncba\r\na\r\nabba\r\nbcabc\r\nabcccaa\r\nab\r\na\r\n", "output": "10\r\n1 9\r\n6 5\r\n4 10\r\n8 6\r\n7 8\r\n9 4\r\n10 3\r\n3 2\r\n2 1\r\n5 7\r\n"}, {"input": "1\r\nzzzz\r\nyyx\r\n", "output": "0\r\n1 1\r\n"}, {"input": "1\r\naa\r\naaa\r\n", "output": "2\r\n1 1\r\n"}, {"input": "1\r\naaa\r\naa\r\n", "output": "2\r\n1 1\r\n"}, {"input": "10\r\nb\r\nb\r\na\r\na\r\na\r\na\r\nb\r\nb\r\na\r\nb\r\nb\r\na\r\na\r\na\r\nb\r\nb\r\nb\r\na\r\nb\r\nb\r\n", "output": "9\r\n3 2\r\n4 3\r\n5 4\r\n6 8\r\n1 1\r\n2 5\r\n7 6\r\n8 7\r\n10 9\r\n9 10\r\n"}, {"input": "10\r\na\r\nb\r\na\r\na\r\nc\r\na\r\na\r\na\r\na\r\na\r\nb\r\nc\r\nc\r\na\r\nc\r\nb\r\na\r\na\r\na\r\nc\r\n", "output": "6\r\n1 4\r\n3 7\r\n4 8\r\n6 9\r\n2 1\r\n5 2\r\n7 6\r\n8 3\r\n9 5\r\n10 10\r\n"}, {"input": "10\r\nw\r\nr\r\na\r\nc\r\nx\r\ne\r\nb\r\nx\r\nw\r\nx\r\nz\r\ng\r\nd\r\ny\r\ns\r\ny\r\nj\r\nh\r\nl\r\nu\r\n", "output": "0\r\n3 3\r\n7 2\r\n4 8\r\n6 7\r\n2 9\r\n1 5\r\n9 10\r\n5 4\r\n8 6\r\n10 1\r\n"}] | false | stdio | import sys
def lcp(s, t):
min_len = min(len(s), len(t))
cnt = 0
for i in range(min_len):
if s[i] == t[i]:
cnt += 1
else:
break
return cnt
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
# Read input
with open(input_path, 'r') as f:
n = int(f.readline())
students = [f.readline().strip() for _ in range(n)]
pseudonyms = [f.readline().strip() for _ in range(n)]
# Read reference sum
with open(output_path, 'r') as f:
ref_sum_line = f.readline().strip()
if not ref_sum_line:
print(0)
return
try:
reference_sum = int(ref_sum_line)
except ValueError:
print(0)
return
# Read submission
with open(submission_path, 'r') as f:
# Read submitted sum
sum_line = f.readline().strip()
if not sum_line:
print(0)
return
try:
submitted_sum = int(sum_line)
except ValueError:
print(0)
return
if submitted_sum != reference_sum:
print(0)
return
# Read the pairs
a_list = []
b_list = []
a_set = set()
b_set = set()
for _ in range(n):
line = f.readline().strip()
parts = line.split()
if len(parts) != 2:
print(0)
return
try:
a = int(parts[0])
b = int(parts[1])
except ValueError:
print(0)
return
if a < 1 or a > n or b < 1 or b > n:
print(0)
return
if a in a_set or b in b_set:
print(0)
return
a_set.add(a)
b_set.add(b)
a_list.append(a)
b_list.append(b)
# Check if all a and b are present
if len(a_set) != n or len(b_set) != n:
print(0)
return
# Compute sum
computed_sum = 0
for a, b in zip(a_list, b_list):
s = students[a-1]
p = pseudonyms[b-1]
computed_sum += lcp(s, p)
if computed_sum != submitted_sum:
print(0)
return
# All checks passed
print(1)
if __name__ == "__main__":
main() | true |
436/C | 436 | C | PyPy 3 | TESTS | 8 | 171 | 23,142,400 | 87968019 | def put():
return map(int, input().split())
def diff(x,y):
ans = 0
for i in range(n*m):
if s[x][i]!= s[y][i]:
ans+=1
return ans
def find(i):
if i==p[i]:
return i
p[i] = find(p[i])
return p[i]
def union(i,j):
if rank[i]>rank[j]:
i,j = j,i
elif rank[i]==rank[j]:
rank[j]+=1
p[i]= j
def dfs(i,p):
print(i+1,p+1)
for j in tree[i]:
if j!=p:
dfs(j,i)
n,m,k,w = put()
s = ['']*k
for i in range(k):
for j in range(n):
s[i]+=input()
edge = []
rank = [0]*k
p = list(range(k))
cost = n*m
tree = [[] for i in range(k)]
for i in range(k):
for j in range(i+1,k):
z = diff(i,j)
edge.append((z,i,j))
edge.sort()
for z,i,j in edge:
u = find(i)
v = find(j)
if u!=v:
union(u,v)
cost+= z*w
tree[i].append(j)
tree[j].append(i)
print(cost)
dfs(0,-1) | 30 | 1,466 | 70,041,600 | 87968659 | def put():
return map(int, input().split())
def diff(x,y):
ans = 0
for i in range(n*m):
if s[x][i]!= s[y][i]:
ans+=1
return ans
def find(i):
if i==p[i]:
return i
p[i] = find(p[i])
return p[i]
def union(i,j):
if rank[i]>rank[j]:
i,j = j,i
elif rank[i]==rank[j]:
rank[j]+=1
p[i]= j
def dfs(i,p):
if i!=0:
print(i,p)
for j in tree[i]:
if j!=p:
dfs(j,i)
n,m,k,w = put()
s = ['']*k
for i in range(k):
for j in range(n):
s[i]+=input()
edge = []
k+=1
rank = [0]*(k)
p = list(range(k))
cost = 0
tree = [[] for i in range(k)]
for i in range(k):
for j in range(i+1,k):
if i==0:
z=n*m
else:
z = diff(i-1,j-1)*w
edge.append((z,i,j))
edge.sort()
for z,i,j in edge:
u = find(i)
v = find(j)
if u!=v:
union(u,v)
cost+= z
tree[i].append(j)
tree[j].append(i)
print(cost)
dfs(0,-1) | Zepto Code Rush 2014 | CF | 2,014 | 2 | 256 | Dungeons and Candies | During the loading of the game "Dungeons and Candies" you are required to get descriptions of k levels from the server. Each description is a map of an n × m checkered rectangular field. Some cells of the field contain candies (each cell has at most one candy). An empty cell is denoted as "." on the map, but if a cell has a candy, it is denoted as a letter of the English alphabet. A level may contain identical candies, in this case the letters in the corresponding cells of the map will be the same.
When you transmit information via a network, you want to minimize traffic — the total size of the transferred data. The levels can be transmitted in any order. There are two ways to transmit the current level A:
1. You can transmit the whole level A. Then you need to transmit n·m bytes via the network.
2. You can transmit the difference between level A and some previously transmitted level B (if it exists); this operation requires to transmit dA, B·w bytes, where dA, B is the number of cells of the field that are different for A and B, and w is a constant. Note, that you should compare only the corresponding cells of levels A and B to calculate dA, B. You cannot transform the maps of levels, i.e. rotate or shift them relatively to each other.
Your task is to find a way to transfer all the k levels and minimize the traffic. | The first line contains four integers n, m, k, w (1 ≤ n, m ≤ 10; 1 ≤ k, w ≤ 1000). Then follows the description of k levels. Each level is described by n lines, each line contains m characters. Each character is either a letter of the English alphabet or a dot ("."). Please note that the case of the letters matters. | In the first line print the required minimum number of transferred bytes.
Then print k pairs of integers x1, y1, x2, y2, ..., xk, yk, describing the way to transfer levels. Pair xi, yi means that level xi needs to be transferred by way yi. If yi equals 0, that means that the level must be transferred using the first way, otherwise yi must be equal to the number of a previously transferred level. It means that you will transfer the difference between levels yi and xi to transfer level xi. Print the pairs in the order of transferring levels. The levels are numbered 1 through k in the order they follow in the input.
If there are multiple optimal solutions, you can print any of them. | null | null | [{"input": "2 3 3 2\nA.A\n...\nA.a\n..C\nX.Y\n...", "output": "14\n1 0\n2 1\n3 1"}, {"input": "1 1 4 1\nA\n.\nB\n.", "output": "3\n1 0\n2 0\n4 2\n3 0"}, {"input": "1 3 5 2\nABA\nBBB\nBBA\nBAB\nABB", "output": "11\n1 0\n3 1\n2 3\n4 2\n5 1"}] | 1,800 | ["dsu", "graphs", "greedy", "trees"] | 30 | [{"input": "2 3 3 2\r\nA.A\r\n...\r\nA.a\r\n..C\r\nX.Y\r\n...\r\n", "output": "14\r\n1 0\r\n2 1\r\n3 1\r\n"}, {"input": "1 1 4 1\r\nA\r\n.\r\nB\r\n.\r\n", "output": "3\r\n1 0\r\n2 0\r\n4 2\r\n3 0\r\n"}, {"input": "1 3 5 2\r\nABA\r\nBBB\r\nBBA\r\nBAB\r\nABB\r\n", "output": "11\r\n1 0\r\n3 1\r\n2 3\r\n4 2\r\n5 1\r\n"}, {"input": "2 2 5 1\r\n..\r\nBA\r\n.A\r\nB.\r\n..\r\nA.\r\nAB\r\n.B\r\n..\r\n..\r\n", "output": "12\r\n1 0\r\n2 1\r\n3 1\r\n5 3\r\n4 5\r\n"}, {"input": "3 3 10 2\r\nBA.\r\n..A\r\n.BB\r\nB..\r\n..B\r\n.AA\r\nB..\r\nAB.\r\n..A\r\nBAB\r\n.A.\r\n.B.\r\n..B\r\nA..\r\n...\r\n...\r\n.B.\r\nBA.\r\n..B\r\n.AB\r\n.B.\r\nB.A\r\n.A.\r\n.BA\r\n..B\r\n...\r\n.A.\r\n.AA\r\n..A\r\n.B.\r\n", "output": "67\r\n1 0\r\n10 1\r\n2 1\r\n3 2\r\n4 1\r\n7 4\r\n9 7\r\n5 9\r\n6 9\r\n8 4\r\n"}, {"input": "3 1 5 1\r\nB\r\nA\r\nB\r\nA\r\nA\r\nB\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\nA\r\n", "output": "5\r\n1 0\r\n2 1\r\n3 2\r\n4 3\r\n5 3\r\n"}, {"input": "3 2 10 1\r\nAB\r\nBA\r\nAB\r\nAA\r\nAA\r\nBA\r\nAA\r\nAA\r\nAB\r\nAB\r\nAB\r\nBA\r\nBA\r\nAB\r\nAA\r\nBB\r\nAB\r\nBA\r\nBB\r\nBB\r\nBA\r\nAA\r\nAA\r\nAB\r\nAB\r\nAB\r\nBA\r\nBB\r\nAB\r\nAA\r\n", "output": "16\r\n1 0\r\n3 1\r\n8 3\r\n2 3\r\n4 2\r\n9 4\r\n6 4\r\n7 6\r\n10 6\r\n5 10\r\n"}, {"input": "2 3 10 2\r\nABB\r\nABA\r\nAAB\r\nBAB\r\nAAA\r\nBBA\r\nBBB\r\nBAA\r\nBBB\r\nABB\r\nABA\r\nBBA\r\nBBB\r\nAAB\r\nABA\r\nABB\r\nBBA\r\nBAB\r\nBBB\r\nBBB\r\n", "output": "38\r\n1 0\r\n5 1\r\n7 5\r\n4 7\r\n9 4\r\n10 5\r\n6 1\r\n3 6\r\n8 1\r\n2 0\r\n"}, {"input": "1 1 1 1\r\n.\r\n", "output": "1\r\n1 0\r\n"}] | false | stdio | import sys
def read_levels(input_path):
with open(input_path) as f:
lines = [line.strip() for line in f.readlines() if line.strip()]
first_line = lines[0].split()
n, m, k, w = map(int, first_line)
levels = [None] # 1-based indexing
idx = 1
for _ in range(k):
level = []
for _ in range(n):
if idx >= len(lines):
return None, None, None, None, None
level.append(lines[idx].strip())
idx += 1
levels.append(level)
return n, m, k, w, levels
def main(input_path, output_path, submission_path):
n, m, k, w, levels = read_levels(input_path)
if levels is None:
return 0
# Read reference output's first line
with open(output_path) as f_ref:
ref_lines = f_ref.read().splitlines()
if not ref_lines:
return 0
ref_total = int(ref_lines[0])
# Read submission output
with open(submission_path) as f_sub:
sub_lines = f_sub.read().splitlines()
if not sub_lines:
return 0
try:
sub_total = int(sub_lines[0])
except:
return 0
# Check sub_total matches reference
if sub_total != ref_total:
return 0
# Parse pairs
tokens = []
for line in sub_lines[1:]:
tokens.extend(line.strip().split())
if len(tokens) != 2 * k:
return 0
pairs = []
try:
for i in range(0, len(tokens), 2):
xi = int(tokens[i])
yi = int(tokens[i+1])
pairs.append((xi, yi))
except:
return 0
# Check all xi are unique and 1..k
seen_xi = set()
for xi, _ in pairs:
if xi < 1 or xi > k or xi in seen_xi:
return 0
seen_xi.add(xi)
if len(seen_xi) != k:
return 0
# Check pairs order and calculate total
transmitted = set()
total = 0
for xi, yi in pairs:
if yi != 0 and yi not in transmitted:
return 0
if yi == 0:
cost = n * m
else:
# Compute d between xi and yi
d = 0
for row in range(n):
for col in range(m):
if levels[xi][row][col] != levels[yi][row][col]:
d += 1
cost = d * w
total += cost
transmitted.add(xi)
if total != sub_total:
return 0
return 1
if __name__ == "__main__":
input_path, output_path, submission_path = sys.argv[1:]
score = main(input_path, output_path, submission_path)
print(score)
| true |
985/C | 985 | C | Python 3 | TESTS | 7 | 202 | 8,192,000 | 38517262 | from bisect import bisect_right
def bad():
print(0)
exit()
n, k, l_ = map(int, input().split())
a = sorted(list(map(int, input().split())))
if n == 1:
print(min(a))
exit()
if a[1] - a[0] > l_:
bad()
b = bisect_right(a, a[0] + l_)
if b < n:
bad()
to_sum = [0]
if n == 2:
print(a[0] + a[n // 2])
exit()
to_sum += [min(k * i, b - n + i) for i in range(1, n)]
print(sum(a[i] for i in to_sum)) | 50 | 93 | 14,131,200 | 215825613 | import sys
input = sys.stdin.readline
n, k, l = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
ans = 0
c = 0
for i in range(n * k - 1, -1, -1 ):
c += 1
if(a[i] - a[0] <= l and c >= k):
ans += a[i]
c -= k
print((ans, 0)[c > 0]) | Educational Codeforces Round 44 (Rated for Div. 2) | ICPC | 2,018 | 2 | 256 | Liebig's Barrels | You have m = n·k wooden staves. The i-th stave has length ai. You have to assemble n barrels consisting of k staves each, you can use any k staves to construct a barrel. Each stave must belong to exactly one barrel.
Let volume vj of barrel j be equal to the length of the minimal stave in it.
You want to assemble exactly n barrels with the maximal total sum of volumes. But you have to make them equal enough, so a difference between volumes of any pair of the resulting barrels must not exceed l, i.e. |vx - vy| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n.
Print maximal total sum of volumes of equal enough barrels or 0 if it's impossible to satisfy the condition above. | The first line contains three space-separated integers n, k and l (1 ≤ n, k ≤ 105, 1 ≤ n·k ≤ 105, 0 ≤ l ≤ 109).
The second line contains m = n·k space-separated integers a1, a2, ..., am (1 ≤ ai ≤ 109) — lengths of staves. | Print single integer — maximal total sum of the volumes of barrels or 0 if it's impossible to construct exactly n barrels satisfying the condition |vx - vy| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n. | null | In the first example you can form the following barrels: [1, 2], [2, 2], [2, 3], [2, 3].
In the second example you can form the following barrels: [10], [10].
In the third example you can form the following barrels: [2, 5].
In the fourth example difference between volumes of barrels in any partition is at least 2 so it is impossible to make barrels equal enough. | [{"input": "4 2 1\n2 2 1 2 3 2 2 3", "output": "7"}, {"input": "2 1 0\n10 10", "output": "20"}, {"input": "1 2 1\n5 2", "output": "2"}, {"input": "3 2 1\n1 2 3 4 5 6", "output": "0"}] | 1,500 | ["greedy"] | 50 | [{"input": "4 2 1\r\n2 2 1 2 3 2 2 3\r\n", "output": "7\r\n"}, {"input": "2 1 0\r\n10 10\r\n", "output": "20\r\n"}, {"input": "1 2 1\r\n5 2\r\n", "output": "2\r\n"}, {"input": "3 2 1\r\n1 2 3 4 5 6\r\n", "output": "0\r\n"}, {"input": "10 3 189\r\n267 697 667 4 52 128 85 616 142 344 413 660 962 194 618 329 266 593 558 447 89 983 964 716 32 890 267 164 654 71\r\n", "output": "0\r\n"}, {"input": "10 3 453\r\n277 706 727 812 692 686 196 507 911 40 498 704 573 381 463 759 704 381 693 640 326 405 47 834 962 521 463 740 520 494\r\n", "output": "2979\r\n"}, {"input": "10 3 795\r\n398 962 417 307 760 534 536 450 421 280 608 111 687 726 941 903 630 900 555 403 795 122 814 188 234 976 679 539 525 104\r\n", "output": "5045\r\n"}, {"input": "6 2 29\r\n1 2 3 3 4 5 5 6 7 7 8 9\r\n", "output": "28\r\n"}, {"input": "2 1 2\r\n1 2\r\n", "output": "3\r\n"}] | false | stdio | null | true |
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