contestId
int64 0
1.01k
| index
stringclasses 57
values | name
stringlengths 2
58
| type
stringclasses 2
values | rating
int64 0
3.5k
| tags
listlengths 0
11
| title
stringclasses 522
values | time-limit
stringclasses 8
values | memory-limit
stringclasses 8
values | problem-description
stringlengths 0
7.15k
| input-specification
stringlengths 0
2.05k
| output-specification
stringlengths 0
1.5k
| demo-input
listlengths 0
7
| demo-output
listlengths 0
7
| note
stringlengths 0
5.24k
| points
float64 0
425k
| test_cases
listlengths 0
402
| creationTimeSeconds
int64 1.37B
1.7B
| relativeTimeSeconds
int64 8
2.15B
| programmingLanguage
stringclasses 3
values | verdict
stringclasses 14
values | testset
stringclasses 12
values | passedTestCount
int64 0
1k
| timeConsumedMillis
int64 0
15k
| memoryConsumedBytes
int64 0
805M
| code
stringlengths 3
65.5k
| prompt
stringlengths 262
8.2k
| response
stringlengths 17
65.5k
| score
float64 -1
3.99
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
476
|
E
|
Dreamoon and Strings
|
PROGRAMMING
| 2,200
|
[
"dp",
"strings"
] | null | null |
Dreamoon has a string *s* and a pattern string *p*. He first removes exactly *x* characters from *s* obtaining string *s*' as a result. Then he calculates that is defined as the maximal number of non-overlapping substrings equal to *p* that can be found in *s*'. He wants to make this number as big as possible.
More formally, let's define as maximum value of over all *s*' that can be obtained by removing exactly *x* characters from *s*. Dreamoon wants to know for all *x* from 0 to |*s*| where |*s*| denotes the length of string *s*.
|
The first line of the input contains the string *s* (1<=≤<=|*s*|<=≤<=2<=000).
The second line of the input contains the string *p* (1<=≤<=|*p*|<=≤<=500).
Both strings will only consist of lower case English letters.
|
Print |*s*|<=+<=1 space-separated integers in a single line representing the for all *x* from 0 to |*s*|.
|
[
"aaaaa\naa\n",
"axbaxxb\nab\n"
] |
[
"2 2 1 1 0 0\n",
"0 1 1 2 1 1 0 0\n"
] |
For the first sample, the corresponding optimal values of *s*' after removal 0 through |*s*| = 5 characters from *s* are {"aaaaa", "aaaa", "aaa", "aa", "a", ""}.
For the second sample, possible corresponding optimal values of *s*' are {"axbaxxb", "abaxxb", "axbab", "abab", "aba", "ab", "a", ""}.
| 2,500
|
[
{
"input": "aaaaa\naa",
"output": "2 2 1 1 0 0"
},
{
"input": "axbaxxb\nab",
"output": "0 1 1 2 1 1 0 0"
},
{
"input": "aabb\nab",
"output": "1 1 1 0 0"
},
{
"input": "aaaaaaaaaaaaaaa\na",
"output": "15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0"
},
{
"input": "aaaaaaaaaaa\nb",
"output": "0 0 0 0 0 0 0 0 0 0 0 0"
},
{
"input": "ababababababababa\naba",
"output": "4 4 4 4 4 4 3 3 3 2 2 2 1 1 1 0 0 0"
},
{
"input": "axxbaxxbaxxb\nab",
"output": "0 0 1 1 2 2 3 2 2 1 1 0 0"
},
{
"input": "axaxxbaxabxbaxxbxb\nab",
"output": "1 1 2 2 3 3 3 3 3 3 3 3 3 2 2 1 1 0 0"
},
{
"input": "ababcc\nabc",
"output": "1 1 1 1 0 0 0"
},
{
"input": "a\na",
"output": "1 0"
},
{
"input": "a\nb",
"output": "0 0"
},
{
"input": "a\naa",
"output": "0 0"
},
{
"input": "a\nab",
"output": "0 0"
},
{
"input": "a\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "0 0"
},
{
"input": "a\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0 0"
},
{
"input": "abxxxaxbxaxxxba\naba",
"output": "0 0 1 1 1 1 2 2 2 2 1 1 1 0 0 0"
}
] | 1,698,961,766
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 43
| 171
| 20,172,800
|
s, p = input(), input()
n, m = len(s) + 1, len(p)
d = [[0] * n for t in range(n)]
for x in range(1, n):
i, j = x, m
while i and j:
j -= s[i - 1] == p[j - 1]
i -= 1
if not j:
for y in range(i + 1): d[x][y + x - i - m] = d[i][y] + 1
for y in range(x): d[x][y] = max(d[x][y], d[x - 1][y])
print(*d[-1])
|
Title: Dreamoon and Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dreamoon has a string *s* and a pattern string *p*. He first removes exactly *x* characters from *s* obtaining string *s*' as a result. Then he calculates that is defined as the maximal number of non-overlapping substrings equal to *p* that can be found in *s*'. He wants to make this number as big as possible.
More formally, let's define as maximum value of over all *s*' that can be obtained by removing exactly *x* characters from *s*. Dreamoon wants to know for all *x* from 0 to |*s*| where |*s*| denotes the length of string *s*.
Input Specification:
The first line of the input contains the string *s* (1<=≤<=|*s*|<=≤<=2<=000).
The second line of the input contains the string *p* (1<=≤<=|*p*|<=≤<=500).
Both strings will only consist of lower case English letters.
Output Specification:
Print |*s*|<=+<=1 space-separated integers in a single line representing the for all *x* from 0 to |*s*|.
Demo Input:
['aaaaa\naa\n', 'axbaxxb\nab\n']
Demo Output:
['2 2 1 1 0 0\n', '0 1 1 2 1 1 0 0\n']
Note:
For the first sample, the corresponding optimal values of *s*' after removal 0 through |*s*| = 5 characters from *s* are {"aaaaa", "aaaa", "aaa", "aa", "a", ""}.
For the second sample, possible corresponding optimal values of *s*' are {"axbaxxb", "abaxxb", "axbab", "abab", "aba", "ab", "a", ""}.
|
```python
s, p = input(), input()
n, m = len(s) + 1, len(p)
d = [[0] * n for t in range(n)]
for x in range(1, n):
i, j = x, m
while i and j:
j -= s[i - 1] == p[j - 1]
i -= 1
if not j:
for y in range(i + 1): d[x][y + x - i - m] = d[i][y] + 1
for y in range(x): d[x][y] = max(d[x][y], d[x - 1][y])
print(*d[-1])
```
| 3
|
|
990
|
C
|
Bracket Sequences Concatenation Problem
|
PROGRAMMING
| 1,500
|
[
"implementation"
] | null | null |
A bracket sequence is a string containing only characters "(" and ")".
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You are given $n$ bracket sequences $s_1, s_2, \dots , s_n$. Calculate the number of pairs $i, j \, (1 \le i, j \le n)$ such that the bracket sequence $s_i + s_j$ is a regular bracket sequence. Operation $+$ means concatenation i.e. "()(" + ")()" = "()()()".
If $s_i + s_j$ and $s_j + s_i$ are regular bracket sequences and $i \ne j$, then both pairs $(i, j)$ and $(j, i)$ must be counted in the answer. Also, if $s_i + s_i$ is a regular bracket sequence, the pair $(i, i)$ must be counted in the answer.
|
The first line contains one integer $n \, (1 \le n \le 3 \cdot 10^5)$ — the number of bracket sequences. The following $n$ lines contain bracket sequences — non-empty strings consisting only of characters "(" and ")". The sum of lengths of all bracket sequences does not exceed $3 \cdot 10^5$.
|
In the single line print a single integer — the number of pairs $i, j \, (1 \le i, j \le n)$ such that the bracket sequence $s_i + s_j$ is a regular bracket sequence.
|
[
"3\n)\n()\n(\n",
"2\n()\n()\n"
] |
[
"2\n",
"4\n"
] |
In the first example, suitable pairs are $(3, 1)$ and $(2, 2)$.
In the second example, any pair is suitable, namely $(1, 1), (1, 2), (2, 1), (2, 2)$.
| 0
|
[
{
"input": "3\n)\n()\n(",
"output": "2"
},
{
"input": "2\n()\n()",
"output": "4"
},
{
"input": "7\n()(\n)\n)(\n())\n(((\n()()()\n()",
"output": "6"
},
{
"input": "6\n(\n((\n(((\n))))\n)))))\n))))))",
"output": "0"
},
{
"input": "9\n(()\n((())\n(\n)\n(()()(()())))\n)\n)(()(\n)())(\n)()(",
"output": "9"
},
{
"input": "2\n(((((((((()\n)))))))))",
"output": "1"
},
{
"input": "1\n)(",
"output": "0"
},
{
"input": "1\n()",
"output": "1"
},
{
"input": "2\n(((\n)))",
"output": "1"
},
{
"input": "10\n()()(\n)((\n)()(((()(\n(((()(\n)(()((\n))\n()()()()\n))()))((\n)\n))())(",
"output": "2"
},
{
"input": "3\n)())(\n()(()(\n(((",
"output": "0"
},
{
"input": "2\n((((((((((((((((((((((((\n))))))))))))))))))))))))",
"output": "1"
},
{
"input": "2\n((\n))",
"output": "1"
}
] | 1,528,657,111
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 5
| 2,000
| 6,144,000
|
n = int(input())
s = list()
global stack
stack = list()
for i in range(0,n):
s.append(input())
#---------------------------------stack---------------------
def push(x,):
stack.append(x)
def pop():
if len(stack)!=0:
return(stack.pop())
def check(a,b,count):
c = a+b
for i in range(0,len(c)):
if c[i]=='(':
push(c[i])
if c[i]==')':
y = pop()
if y == None:
count = 0
break
else:count = 1
if len(stack)!=0:
count = 0
return count
#-------------------------------------------------------------
ans = 0
for i in range(0,n):
for j in range(0,n):
count = 0
if check(s[i],s[j],count):
ans = ans + 1
stack = list()
print(ans)
|
Title: Bracket Sequences Concatenation Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A bracket sequence is a string containing only characters "(" and ")".
A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.
You are given $n$ bracket sequences $s_1, s_2, \dots , s_n$. Calculate the number of pairs $i, j \, (1 \le i, j \le n)$ such that the bracket sequence $s_i + s_j$ is a regular bracket sequence. Operation $+$ means concatenation i.e. "()(" + ")()" = "()()()".
If $s_i + s_j$ and $s_j + s_i$ are regular bracket sequences and $i \ne j$, then both pairs $(i, j)$ and $(j, i)$ must be counted in the answer. Also, if $s_i + s_i$ is a regular bracket sequence, the pair $(i, i)$ must be counted in the answer.
Input Specification:
The first line contains one integer $n \, (1 \le n \le 3 \cdot 10^5)$ — the number of bracket sequences. The following $n$ lines contain bracket sequences — non-empty strings consisting only of characters "(" and ")". The sum of lengths of all bracket sequences does not exceed $3 \cdot 10^5$.
Output Specification:
In the single line print a single integer — the number of pairs $i, j \, (1 \le i, j \le n)$ such that the bracket sequence $s_i + s_j$ is a regular bracket sequence.
Demo Input:
['3\n)\n()\n(\n', '2\n()\n()\n']
Demo Output:
['2\n', '4\n']
Note:
In the first example, suitable pairs are $(3, 1)$ and $(2, 2)$.
In the second example, any pair is suitable, namely $(1, 1), (1, 2), (2, 1), (2, 2)$.
|
```python
n = int(input())
s = list()
global stack
stack = list()
for i in range(0,n):
s.append(input())
#---------------------------------stack---------------------
def push(x,):
stack.append(x)
def pop():
if len(stack)!=0:
return(stack.pop())
def check(a,b,count):
c = a+b
for i in range(0,len(c)):
if c[i]=='(':
push(c[i])
if c[i]==')':
y = pop()
if y == None:
count = 0
break
else:count = 1
if len(stack)!=0:
count = 0
return count
#-------------------------------------------------------------
ans = 0
for i in range(0,n):
for j in range(0,n):
count = 0
if check(s[i],s[j],count):
ans = ans + 1
stack = list()
print(ans)
```
| 0
|
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,586,519,908
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 218
| 0
|
import sys
lines = sys.stdin.readlines()
st = lines[0].strip()
lowerC, upperC = 0,0
for l in st:
if ord(l) >= ord("a"): lowerC += 1
else: upperC += 1
if lowerC >= upperC: print(st.lower())
else: print(st.upper())
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
import sys
lines = sys.stdin.readlines()
st = lines[0].strip()
lowerC, upperC = 0,0
for l in st:
if ord(l) >= ord("a"): lowerC += 1
else: upperC += 1
if lowerC >= upperC: print(st.lower())
else: print(st.upper())
```
| 3.9455
|
988
|
D
|
Points and Powers of Two
|
PROGRAMMING
| 1,800
|
[
"brute force",
"math"
] | null | null |
There are $n$ distinct points on a coordinate line, the coordinate of $i$-th point equals to $x_i$. Choose a subset of the given set of points such that the distance between each pair of points in a subset is an integral power of two. It is necessary to consider each pair of points, not only adjacent. Note that any subset containing one element satisfies the condition above. Among all these subsets, choose a subset with maximum possible size.
In other words, you have to choose the maximum possible number of points $x_{i_1}, x_{i_2}, \dots, x_{i_m}$ such that for each pair $x_{i_j}$, $x_{i_k}$ it is true that $|x_{i_j} - x_{i_k}| = 2^d$ where $d$ is some non-negative integer number (not necessarily the same for each pair of points).
|
The first line contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of points.
The second line contains $n$ pairwise distinct integers $x_1, x_2, \dots, x_n$ ($-10^9 \le x_i \le 10^9$) — the coordinates of points.
|
In the first line print $m$ — the maximum possible number of points in a subset that satisfies the conditions described above.
In the second line print $m$ integers — the coordinates of points in the subset you have chosen.
If there are multiple answers, print any of them.
|
[
"6\n3 5 4 7 10 12\n",
"5\n-1 2 5 8 11\n"
] |
[
"3\n7 3 5",
"1\n8\n"
] |
In the first example the answer is $[7, 3, 5]$. Note, that $|7-3|=4=2^2$, $|7-5|=2=2^1$ and $|3-5|=2=2^1$. You can't find a subset having more points satisfying the required property.
| 0
|
[
{
"input": "6\n3 5 4 7 10 12",
"output": "3\n3 4 5 "
},
{
"input": "5\n-1 2 5 8 11",
"output": "1\n-1 "
},
{
"input": "1\n42",
"output": "1\n42 "
},
{
"input": "3\n0 -536870912 536870912",
"output": "3\n-536870912 0 536870912 "
},
{
"input": "2\n536870912 -536870912",
"output": "2\n-536870912 536870912 "
},
{
"input": "3\n1 2 3",
"output": "3\n1 2 3 "
},
{
"input": "4\n1 2 3 8",
"output": "3\n1 2 3 "
},
{
"input": "2\n1 2",
"output": "2\n1 2 "
},
{
"input": "3\n0 1 2",
"output": "3\n0 1 2 "
},
{
"input": "2\n-3 -2",
"output": "2\n-3 -2 "
},
{
"input": "2\n-4 -2",
"output": "2\n-4 -2 "
},
{
"input": "2\n2 1",
"output": "2\n1 2 "
},
{
"input": "1\n1",
"output": "1\n1 "
},
{
"input": "3\n0 2 6",
"output": "2\n0 2 "
},
{
"input": "3\n2 4 8",
"output": "2\n2 4 "
},
{
"input": "2\n1 0",
"output": "2\n0 1 "
},
{
"input": "3\n5 6 7",
"output": "3\n5 6 7 "
},
{
"input": "3\n-1 1 0",
"output": "3\n-1 0 1 "
}
] | 1,620,758,023
| 6,043
|
PyPy 3
|
OK
|
TESTS
| 99
| 920
| 44,953,600
|
from sys import stdin
input = stdin.readline
def solve():
n = int(input())
x = [int(x) for x in input().split()]
s = set(x)
ans = [x[0]]
for i in range(n):
for j in range(0, 32):
if x[i] + 2**j in s:
ans = [x[i], x[i] + 2**j]
if x[i] + (2**j * 2) in s:
ans.append(x[i] + (2**j * 2))
return ans
return ans
ans = solve()
print(len(ans))
print(*ans)
|
Title: Points and Powers of Two
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are $n$ distinct points on a coordinate line, the coordinate of $i$-th point equals to $x_i$. Choose a subset of the given set of points such that the distance between each pair of points in a subset is an integral power of two. It is necessary to consider each pair of points, not only adjacent. Note that any subset containing one element satisfies the condition above. Among all these subsets, choose a subset with maximum possible size.
In other words, you have to choose the maximum possible number of points $x_{i_1}, x_{i_2}, \dots, x_{i_m}$ such that for each pair $x_{i_j}$, $x_{i_k}$ it is true that $|x_{i_j} - x_{i_k}| = 2^d$ where $d$ is some non-negative integer number (not necessarily the same for each pair of points).
Input Specification:
The first line contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of points.
The second line contains $n$ pairwise distinct integers $x_1, x_2, \dots, x_n$ ($-10^9 \le x_i \le 10^9$) — the coordinates of points.
Output Specification:
In the first line print $m$ — the maximum possible number of points in a subset that satisfies the conditions described above.
In the second line print $m$ integers — the coordinates of points in the subset you have chosen.
If there are multiple answers, print any of them.
Demo Input:
['6\n3 5 4 7 10 12\n', '5\n-1 2 5 8 11\n']
Demo Output:
['3\n7 3 5', '1\n8\n']
Note:
In the first example the answer is $[7, 3, 5]$. Note, that $|7-3|=4=2^2$, $|7-5|=2=2^1$ and $|3-5|=2=2^1$. You can't find a subset having more points satisfying the required property.
|
```python
from sys import stdin
input = stdin.readline
def solve():
n = int(input())
x = [int(x) for x in input().split()]
s = set(x)
ans = [x[0]]
for i in range(n):
for j in range(0, 32):
if x[i] + 2**j in s:
ans = [x[i], x[i] + 2**j]
if x[i] + (2**j * 2) in s:
ans.append(x[i] + (2**j * 2))
return ans
return ans
ans = solve()
print(len(ans))
print(*ans)
```
| 3
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,690,043,958
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 15
| 0
|
words = int(input())
for i in range(words):
print(words)
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
words = int(input())
for i in range(words):
print(words)
```
| 0
|
954
|
A
|
Diagonal Walking
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mikhail walks on a 2D plane. He can go either up or right. You are given a sequence of Mikhail's moves. He thinks that this sequence is too long and he wants to make it as short as possible.
In the given sequence moving up is described by character U and moving right is described by character R. Mikhail can replace any pair of consecutive moves RU or UR with a diagonal move (described as character D). After that, he can go on and do some other replacements, until there is no pair of consecutive moves RU or UR left.
Your problem is to print the minimum possible length of the sequence of moves after the replacements.
|
The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains the sequence consisting of *n* characters U and R.
|
Print the minimum possible length of the sequence of moves after all replacements are done.
|
[
"5\nRUURU\n",
"17\nUUURRRRRUUURURUUU\n"
] |
[
"3\n",
"13\n"
] |
In the first test the shortened sequence of moves may be DUD (its length is 3).
In the second test the shortened sequence of moves can be UUDRRRDUDDUUU (its length is 13).
| 0
|
[
{
"input": "5\nRUURU",
"output": "3"
},
{
"input": "17\nUUURRRRRUUURURUUU",
"output": "13"
},
{
"input": "100\nUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU",
"output": "100"
},
{
"input": "100\nRRURRUUUURURRRURRRRURRRRRRURRUURRRUUURUURURRURUURUURRUURUURRURURUUUUURUUUUUURRUUURRRURRURRRUURRUUUUR",
"output": "67"
},
{
"input": "100\nUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUURUUUUUUUUUUUUUUUUUUUUU",
"output": "99"
},
{
"input": "3\nRUR",
"output": "2"
},
{
"input": "1\nR",
"output": "1"
},
{
"input": "5\nRURUU",
"output": "3"
},
{
"input": "1\nU",
"output": "1"
},
{
"input": "2\nUR",
"output": "1"
},
{
"input": "23\nUUUUUUUUUUUUUUUUUUUUUUU",
"output": "23"
}
] | 1,637,473,411
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
n=int(input())
s=input()
count=0
for i in range(n):
if s[i] in 'RU'':
s.replace('RU','D')
elif s[i] in 'UR':
s.replace('UR','D')
print(s)
|
Title: Diagonal Walking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mikhail walks on a 2D plane. He can go either up or right. You are given a sequence of Mikhail's moves. He thinks that this sequence is too long and he wants to make it as short as possible.
In the given sequence moving up is described by character U and moving right is described by character R. Mikhail can replace any pair of consecutive moves RU or UR with a diagonal move (described as character D). After that, he can go on and do some other replacements, until there is no pair of consecutive moves RU or UR left.
Your problem is to print the minimum possible length of the sequence of moves after the replacements.
Input Specification:
The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence. The second line contains the sequence consisting of *n* characters U and R.
Output Specification:
Print the minimum possible length of the sequence of moves after all replacements are done.
Demo Input:
['5\nRUURU\n', '17\nUUURRRRRUUURURUUU\n']
Demo Output:
['3\n', '13\n']
Note:
In the first test the shortened sequence of moves may be DUD (its length is 3).
In the second test the shortened sequence of moves can be UUDRRRDUDDUUU (its length is 13).
|
```python
n=int(input())
s=input()
count=0
for i in range(n):
if s[i] in 'RU'':
s.replace('RU','D')
elif s[i] in 'UR':
s.replace('UR','D')
print(s)
```
| -1
|
|
278
|
A
|
Circle Line
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The circle line of the Berland subway has *n* stations. We know the distances between all pairs of neighboring stations:
- *d*1 is the distance between the 1-st and the 2-nd station;- *d*2 is the distance between the 2-nd and the 3-rd station;...- *d**n*<=-<=1 is the distance between the *n*<=-<=1-th and the *n*-th station;- *d**n* is the distance between the *n*-th and the 1-st station.
The trains go along the circle line in both directions. Find the shortest distance between stations with numbers *s* and *t*.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — the number of stations on the circle line. The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (1<=≤<=*d**i*<=≤<=100) — the distances between pairs of neighboring stations. The third line contains two integers *s* and *t* (1<=≤<=*s*,<=*t*<=≤<=*n*) — the numbers of stations, between which you need to find the shortest distance. These numbers can be the same.
The numbers in the lines are separated by single spaces.
|
Print a single number — the length of the shortest path between stations number *s* and *t*.
|
[
"4\n2 3 4 9\n1 3\n",
"4\n5 8 2 100\n4 1\n",
"3\n1 1 1\n3 1\n",
"3\n31 41 59\n1 1\n"
] |
[
"5\n",
"15\n",
"1\n",
"0\n"
] |
In the first sample the length of path 1 → 2 → 3 equals 5, the length of path 1 → 4 → 3 equals 13.
In the second sample the length of path 4 → 1 is 100, the length of path 4 → 3 → 2 → 1 is 15.
In the third sample the length of path 3 → 1 is 1, the length of path 3 → 2 → 1 is 2.
In the fourth sample the numbers of stations are the same, so the shortest distance equals 0.
| 500
|
[
{
"input": "4\n2 3 4 9\n1 3",
"output": "5"
},
{
"input": "4\n5 8 2 100\n4 1",
"output": "15"
},
{
"input": "3\n1 1 1\n3 1",
"output": "1"
},
{
"input": "3\n31 41 59\n1 1",
"output": "0"
},
{
"input": "5\n16 13 10 30 15\n4 2",
"output": "23"
},
{
"input": "6\n89 82 87 32 67 33\n4 4",
"output": "0"
},
{
"input": "7\n2 3 17 10 2 2 2\n4 2",
"output": "18"
},
{
"input": "3\n4 37 33\n3 3",
"output": "0"
},
{
"input": "8\n87 40 96 7 86 86 72 97\n6 8",
"output": "158"
},
{
"input": "10\n91 94 75 99 100 91 79 86 79 92\n2 8",
"output": "348"
},
{
"input": "19\n1 1 1 1 2 1 1 1 1 1 2 1 3 2 2 1 1 1 2\n7 7",
"output": "0"
},
{
"input": "34\n96 65 24 99 74 76 97 93 99 69 94 82 92 91 98 83 95 97 96 81 90 95 86 87 43 78 88 86 82 62 76 99 83 96\n21 16",
"output": "452"
},
{
"input": "50\n75 98 65 75 99 89 84 65 9 53 62 61 61 53 80 7 6 47 86 1 89 27 67 1 31 39 53 92 19 20 76 41 60 15 29 94 76 82 87 89 93 38 42 6 87 36 100 97 93 71\n2 6",
"output": "337"
},
{
"input": "99\n1 15 72 78 23 22 26 98 7 2 75 58 100 98 45 79 92 69 79 72 33 88 62 9 15 87 17 73 68 54 34 89 51 91 28 44 20 11 74 7 85 61 30 46 95 72 36 18 48 22 42 46 29 46 86 53 96 55 98 34 60 37 75 54 1 81 20 68 84 19 18 18 75 84 86 57 73 34 23 43 81 87 47 96 57 41 69 1 52 44 54 7 85 35 5 1 19 26 7\n4 64",
"output": "1740"
},
{
"input": "100\n33 63 21 27 49 82 86 93 43 55 4 72 89 85 5 34 80 7 23 13 21 49 22 73 89 65 81 25 6 92 82 66 58 88 48 96 1 1 16 48 67 96 84 63 87 76 20 100 36 4 31 41 35 62 55 76 74 70 68 41 4 16 39 81 2 41 34 73 66 57 41 89 78 93 68 96 87 47 92 60 40 58 81 12 19 74 56 83 56 61 83 97 26 92 62 52 39 57 89 95\n71 5",
"output": "2127"
},
{
"input": "100\n95 98 99 81 98 96 100 92 96 90 99 91 98 98 91 78 97 100 96 98 87 93 96 99 91 92 96 92 90 97 85 83 99 95 66 91 87 89 100 95 100 88 99 84 96 79 99 100 94 100 99 99 92 89 99 91 100 94 98 97 91 92 90 87 84 99 97 98 93 100 90 85 75 95 86 71 98 93 91 87 92 95 98 94 95 94 100 98 96 100 97 96 95 95 86 86 94 97 98 96\n67 57",
"output": "932"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 97 100 100 100 100 100 99 100 100 99 99 100 99 100 100 100 100 100 100 100 100 100 97 99 98 98 100 98 98 100 99 100 100 100 100 99 100 98 100 99 98 99 98 98 100 100 100 100 100 100 100 100 100 100 99 100 100 100 100 100 98 100 99 99 100 96 100 96 100 99 100 100 99 100 99 100 100 100 99 100 100 100 100 98 98 97 100 100 99 98\n16 6",
"output": "997"
},
{
"input": "100\n3 6 23 4 23 1 2 14 2 3 3 9 17 8 10 5 1 14 8 5 7 4 13 8 5 6 24 3 12 3 4 9 2 8 2 1 2 1 3 2 1 6 14 23 8 6 3 5 7 8 18 9 2 5 22 6 13 16 2 4 31 20 4 3 3 6 6 1 1 18 5 11 1 14 4 16 6 37 11 1 8 3 7 11 21 14 3 3 12 2 5 1 9 16 3 1 3 4 4 2\n98 24",
"output": "195"
},
{
"input": "100\n1 1 3 1 1 2 1 2 1 1 2 2 2 1 1 1 1 1 1 3 1 1 1 3 1 3 3 1 1 2 1 1 1 1 1 2 1 1 1 4 1 1 3 3 2 1 1 1 1 1 2 2 1 3 1 1 1 2 4 1 1 2 5 2 1 1 2 1 1 1 2 3 1 1 1 1 2 1 1 1 1 2 1 1 1 1 1 2 2 3 1 7 3 1 3 1 2 1 2 1\n49 10",
"output": "60"
},
{
"input": "100\n75 62 31 96 62 76 93 96 72 67 88 35 67 34 60 56 95 86 82 48 64 61 74 100 56 98 76 98 78 55 53 10 12 78 58 45 86 90 93 77 69 73 88 66 92 88 33 50 95 69 89 12 93 57 93 89 59 53 71 86 15 13 61 93 24 100 58 76 46 95 76 82 50 20 79 38 5 72 99 81 55 90 90 65 85 44 63 39 6 34 98 72 88 30 59 73 84 61 25 67\n86 25",
"output": "2523"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100\n1 51",
"output": "5000"
},
{
"input": "4\n1 1 1 1\n2 4",
"output": "2"
},
{
"input": "4\n1 1 2 1\n2 4",
"output": "2"
}
] | 1,591,639,718
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 23
| 218
| 0
|
n=int(input())
l=list(map(int,input().split()))
x,y=map(int,input().split())
s,t=min(x,y),max(x,y)
print(min(sum(l[s-1:t-1]),sum(l[t-1:])+sum(l[:s-1])))
|
Title: Circle Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The circle line of the Berland subway has *n* stations. We know the distances between all pairs of neighboring stations:
- *d*1 is the distance between the 1-st and the 2-nd station;- *d*2 is the distance between the 2-nd and the 3-rd station;...- *d**n*<=-<=1 is the distance between the *n*<=-<=1-th and the *n*-th station;- *d**n* is the distance between the *n*-th and the 1-st station.
The trains go along the circle line in both directions. Find the shortest distance between stations with numbers *s* and *t*.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — the number of stations on the circle line. The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (1<=≤<=*d**i*<=≤<=100) — the distances between pairs of neighboring stations. The third line contains two integers *s* and *t* (1<=≤<=*s*,<=*t*<=≤<=*n*) — the numbers of stations, between which you need to find the shortest distance. These numbers can be the same.
The numbers in the lines are separated by single spaces.
Output Specification:
Print a single number — the length of the shortest path between stations number *s* and *t*.
Demo Input:
['4\n2 3 4 9\n1 3\n', '4\n5 8 2 100\n4 1\n', '3\n1 1 1\n3 1\n', '3\n31 41 59\n1 1\n']
Demo Output:
['5\n', '15\n', '1\n', '0\n']
Note:
In the first sample the length of path 1 → 2 → 3 equals 5, the length of path 1 → 4 → 3 equals 13.
In the second sample the length of path 4 → 1 is 100, the length of path 4 → 3 → 2 → 1 is 15.
In the third sample the length of path 3 → 1 is 1, the length of path 3 → 2 → 1 is 2.
In the fourth sample the numbers of stations are the same, so the shortest distance equals 0.
|
```python
n=int(input())
l=list(map(int,input().split()))
x,y=map(int,input().split())
s,t=min(x,y),max(x,y)
print(min(sum(l[s-1:t-1]),sum(l[t-1:])+sum(l[:s-1])))
```
| 3
|
|
733
|
B
|
Parade
|
PROGRAMMING
| 1,100
|
[
"math"
] | null | null |
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be *n* columns participating in the parade, the *i*-th column consists of *l**i* soldiers, who start to march from left leg, and *r**i* soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if *L* is the total number of soldiers on the parade who start to march from the left leg, and *R* is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |*L*<=-<=*R*|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index *i* and swap values *l**i* and *r**i*.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
|
The first line contains single integer *n* (1<=≤<=*n*<=≤<=105) — the number of columns.
The next *n* lines contain the pairs of integers *l**i* and *r**i* (1<=≤<=*l**i*,<=*r**i*<=≤<=500) — the number of soldiers in the *i*-th column which start to march from the left or the right leg respectively.
|
Print single integer *k* — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to *n* in the order they are given in the input data.
If there are several answers, print any of them.
|
[
"3\n5 6\n8 9\n10 3\n",
"2\n6 5\n5 6\n",
"6\n5 9\n1 3\n4 8\n4 5\n23 54\n12 32\n"
] |
[
"3\n",
"1\n",
"0\n"
] |
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9.
| 1,000
|
[
{
"input": "3\n5 6\n8 9\n10 3",
"output": "3"
},
{
"input": "2\n6 5\n5 6",
"output": "1"
},
{
"input": "6\n5 9\n1 3\n4 8\n4 5\n23 54\n12 32",
"output": "0"
},
{
"input": "2\n500 499\n500 500",
"output": "0"
},
{
"input": "1\n139 252",
"output": "0"
},
{
"input": "10\n18 18\n71 471\n121 362\n467 107\n138 254\n13 337\n499 373\n337 387\n147 417\n76 417",
"output": "4"
},
{
"input": "4\n4 1\n5 3\n7 6\n3 5",
"output": "4"
},
{
"input": "3\n6 5\n9 8\n3 10",
"output": "3"
},
{
"input": "3\n100 9\n1 3\n1 5",
"output": "1"
},
{
"input": "4\n10 1\n10 2\n10 3\n1 10",
"output": "4"
},
{
"input": "5\n25 1\n24 1\n2 3\n2 3\n2 3",
"output": "3"
},
{
"input": "3\n90 1\n1 90\n100 1",
"output": "2"
},
{
"input": "3\n1 123\n22 1\n12 1",
"output": "1"
},
{
"input": "3\n4 5\n7 6\n10 9",
"output": "1"
},
{
"input": "4\n50 1\n50 1\n50 1\n1 49",
"output": "4"
},
{
"input": "7\n2 1\n2 1\n2 1\n2 1\n1 200\n1 200\n1 200",
"output": "1"
},
{
"input": "5\n10 8\n7 6\n2 8\n9 1\n7 1",
"output": "3"
},
{
"input": "3\n22 1\n12 1\n1 123",
"output": "3"
},
{
"input": "3\n10 8\n9 7\n4 5",
"output": "3"
},
{
"input": "5\n1 2\n4 8\n4 8\n32 16\n128 64",
"output": "2"
},
{
"input": "5\n10 1\n10 1\n10 1\n100 1\n1 5",
"output": "5"
}
] | 1,583,872,075
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 40
| 873
| 9,420,800
|
import sys
import math
import bisect
def solve(A, B):
n = len(A)
sum_a = sum(A)
sum_b = sum(B)
max_val = abs(sum_a - sum_b)
sol = 0
for i in range(n):
tmp_a = sum_a - A[i] + B[i]
tmp_b = sum_b - B[i] + A[i]
val = abs(tmp_a - tmp_b)
if max_val < val:
max_val = val
sol = i + 1
return sol
def main():
n = int(input())
A = []
B = []
for i in range(n):
a, b = map(int, input().split())
A.append(a)
B.append(b)
ans = solve(A, B)
print(ans)
if __name__ == "__main__":
main()
|
Title: Parade
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be *n* columns participating in the parade, the *i*-th column consists of *l**i* soldiers, who start to march from left leg, and *r**i* soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if *L* is the total number of soldiers on the parade who start to march from the left leg, and *R* is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |*L*<=-<=*R*|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index *i* and swap values *l**i* and *r**i*.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=105) — the number of columns.
The next *n* lines contain the pairs of integers *l**i* and *r**i* (1<=≤<=*l**i*,<=*r**i*<=≤<=500) — the number of soldiers in the *i*-th column which start to march from the left or the right leg respectively.
Output Specification:
Print single integer *k* — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to *n* in the order they are given in the input data.
If there are several answers, print any of them.
Demo Input:
['3\n5 6\n8 9\n10 3\n', '2\n6 5\n5 6\n', '6\n5 9\n1 3\n4 8\n4 5\n23 54\n12 32\n']
Demo Output:
['3\n', '1\n', '0\n']
Note:
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9.
|
```python
import sys
import math
import bisect
def solve(A, B):
n = len(A)
sum_a = sum(A)
sum_b = sum(B)
max_val = abs(sum_a - sum_b)
sol = 0
for i in range(n):
tmp_a = sum_a - A[i] + B[i]
tmp_b = sum_b - B[i] + A[i]
val = abs(tmp_a - tmp_b)
if max_val < val:
max_val = val
sol = i + 1
return sol
def main():
n = int(input())
A = []
B = []
for i in range(n):
a, b = map(int, input().split())
A.append(a)
B.append(b)
ans = solve(A, B)
print(ans)
if __name__ == "__main__":
main()
```
| 3
|
|
509
|
A
|
Maximum in Table
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.
|
The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.
|
Print a single line containing a positive integer *m* — the maximum value in the table.
|
[
"1\n",
"5\n"
] |
[
"1",
"70"
] |
In the second test the rows of the table look as follows:
| 0
|
[
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "70"
},
{
"input": "2",
"output": "2"
},
{
"input": "3",
"output": "6"
},
{
"input": "4",
"output": "20"
},
{
"input": "6",
"output": "252"
},
{
"input": "7",
"output": "924"
},
{
"input": "9",
"output": "12870"
},
{
"input": "10",
"output": "48620"
},
{
"input": "8",
"output": "3432"
}
] | 1,644,604,108
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 10
| 62
| 0
|
import math
a = int(input()) - 1
print(math.factorial(2*a)//(math.factorial(a)**2))
|
Title: Maximum in Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.
Input Specification:
The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.
Output Specification:
Print a single line containing a positive integer *m* — the maximum value in the table.
Demo Input:
['1\n', '5\n']
Demo Output:
['1', '70']
Note:
In the second test the rows of the table look as follows:
|
```python
import math
a = int(input()) - 1
print(math.factorial(2*a)//(math.factorial(a)**2))
```
| 3
|
|
215
|
A
|
Bicycle Chain
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation.
We know that the *i*-th star on the pedal axle has *a**i* (0<=<<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=<<=*b*1<=<<=*b*2<=<<=...<=<<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value .
Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears.
In the problem, fraction denotes division in real numbers, that is, no rounding is performed.
|
The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing.
The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing.
It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces.
|
Print the number of "integer" gears with the maximum ratio among all "integer" gears.
|
[
"2\n4 5\n3\n12 13 15\n",
"4\n1 2 3 4\n5\n10 11 12 13 14\n"
] |
[
"2\n",
"1\n"
] |
In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15.
| 500
|
[
{
"input": "2\n4 5\n3\n12 13 15",
"output": "2"
},
{
"input": "4\n1 2 3 4\n5\n10 11 12 13 14",
"output": "1"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1 2\n1\n1",
"output": "1"
},
{
"input": "1\n1\n2\n1 2",
"output": "1"
},
{
"input": "4\n3 7 11 13\n4\n51 119 187 221",
"output": "4"
},
{
"input": "4\n2 3 4 5\n3\n1 2 3",
"output": "2"
},
{
"input": "10\n6 12 13 20 48 53 74 92 96 97\n10\n1 21 32 36 47 54 69 75 95 97",
"output": "1"
},
{
"input": "10\n5 9 10 14 15 17 19 22 24 26\n10\n2 11 17 19 21 22 24 25 27 28",
"output": "1"
},
{
"input": "10\n24 53 56 126 354 432 442 740 795 856\n10\n273 438 494 619 689 711 894 947 954 958",
"output": "1"
},
{
"input": "10\n3 4 6 7 8 10 14 16 19 20\n10\n3 4 5 7 8 10 15 16 18 20",
"output": "1"
},
{
"input": "10\n1 6 8 14 15 17 25 27 34 39\n10\n1 8 16 17 19 22 32 39 44 50",
"output": "1"
},
{
"input": "10\n5 21 22 23 25 32 35 36 38 39\n10\n3 7 8 9 18 21 23 24 36 38",
"output": "4"
},
{
"input": "50\n5 8 13 16 19 20 21 22 24 27 28 29 30 32 33 34 35 43 45 48 50 51 54 55 58 59 60 61 62 65 70 71 72 76 78 79 80 81 83 84 85 87 89 91 92 94 97 98 99 100\n50\n2 3 5 6 7 10 15 16 17 20 23 28 29 30 31 34 36 37 40 42 45 46 48 54 55 56 58 59 61 62 69 70 71 72 75 76 78 82 84 85 86 87 88 89 90 91 92 97 99 100",
"output": "1"
},
{
"input": "50\n3 5 6 8 9 11 13 19 21 23 24 32 34 35 42 50 51 52 56 58 59 69 70 72 73 75 76 77 78 80 83 88 90 95 96 100 101 102 108 109 113 119 124 135 138 141 142 143 145 150\n50\n5 8 10 11 18 19 23 30 35 43 51 53 55 58 63 68 69 71 77 78 79 82 83 86 88 89 91 92 93 94 96 102 103 105 109 110 113 114 116 123 124 126 127 132 133 135 136 137 142 149",
"output": "1"
},
{
"input": "50\n6 16 24 25 27 33 36 40 51 60 62 65 71 72 75 77 85 87 91 93 98 102 103 106 117 118 120 121 122 123 125 131 134 136 143 148 155 157 160 161 164 166 170 178 184 187 188 192 194 197\n50\n5 9 17 23 27 34 40 44 47 59 62 70 81 82 87 88 89 90 98 101 102 110 113 114 115 116 119 122 124 128 130 137 138 140 144 150 152 155 159 164 166 169 171 175 185 186 187 189 190 193",
"output": "1"
},
{
"input": "50\n14 22 23 31 32 35 48 63 76 79 88 97 101 102 103 104 106 113 114 115 116 126 136 138 145 152 155 156 162 170 172 173 179 180 182 203 208 210 212 222 226 229 231 232 235 237 245 246 247 248\n50\n2 5 6 16 28 44 45 46 54 55 56 63 72 80 87 93 94 96 97 100 101 103 132 135 140 160 164 165 167 168 173 180 182 185 186 192 194 198 199 202 203 211 213 216 217 227 232 233 236 245",
"output": "1"
},
{
"input": "50\n14 19 33 35 38 41 51 54 69 70 71 73 76 80 84 94 102 104 105 106 107 113 121 128 131 168 180 181 187 191 195 201 205 207 210 216 220 238 249 251 263 271 272 275 281 283 285 286 291 294\n50\n2 3 5 20 21 35 38 40 43 48 49 52 55 64 73 77 82 97 109 113 119 121 125 132 137 139 145 146 149 180 182 197 203 229 234 241 244 251 264 271 274 281 284 285 287 291 292 293 294 298",
"output": "1"
},
{
"input": "50\n2 4 5 16 18 19 22 23 25 26 34 44 48 54 67 79 80 84 92 110 116 133 138 154 163 171 174 202 205 218 228 229 234 245 247 249 250 263 270 272 274 275 277 283 289 310 312 334 339 342\n50\n1 5 17 18 25 37 46 47 48 59 67 75 80 83 84 107 115 122 137 141 159 162 175 180 184 204 221 224 240 243 247 248 249 258 259 260 264 266 269 271 274 293 294 306 329 330 334 335 342 350",
"output": "1"
},
{
"input": "50\n6 9 11 21 28 39 42 56 60 63 81 88 91 95 105 110 117 125 149 165 174 176 185 189 193 196 205 231 233 268 278 279 281 286 289 292 298 303 305 306 334 342 350 353 361 371 372 375 376 378\n50\n6 17 20 43 45 52 58 59 82 83 88 102 111 118 121 131 145 173 190 191 200 216 224 225 232 235 243 256 260 271 290 291 321 322 323 329 331 333 334 341 343 348 351 354 356 360 366 379 387 388",
"output": "1"
},
{
"input": "10\n17 239 443 467 661 1069 1823 2333 3767 4201\n20\n51 83 97 457 593 717 997 1329 1401 1459 1471 1983 2371 2539 3207 3251 3329 5469 6637 6999",
"output": "8"
},
{
"input": "20\n179 359 401 467 521 601 919 941 1103 1279 1709 1913 1949 2003 2099 2143 2179 2213 2399 4673\n20\n151 181 191 251 421 967 1109 1181 1249 1447 1471 1553 1619 2327 2551 2791 3049 3727 6071 7813",
"output": "3"
},
{
"input": "20\n79 113 151 709 809 983 1291 1399 1409 1429 2377 2659 2671 2897 3217 3511 3557 3797 3823 4363\n10\n19 101 659 797 1027 1963 2129 2971 3299 9217",
"output": "3"
},
{
"input": "30\n19 47 109 179 307 331 389 401 461 509 547 569 617 853 883 1249 1361 1381 1511 1723 1741 1783 2459 2531 2621 3533 3821 4091 5557 6217\n20\n401 443 563 941 967 997 1535 1567 1655 1747 1787 1945 1999 2251 2305 2543 2735 4415 6245 7555",
"output": "8"
},
{
"input": "30\n3 43 97 179 257 313 353 359 367 389 397 457 547 599 601 647 1013 1021 1063 1433 1481 1531 1669 3181 3373 3559 3769 4157 4549 5197\n50\n13 15 17 19 29 79 113 193 197 199 215 223 271 293 359 485 487 569 601 683 895 919 941 967 1283 1285 1289 1549 1565 1765 1795 1835 1907 1931 1945 1985 1993 2285 2731 2735 2995 3257 4049 4139 5105 5315 7165 7405 7655 8345",
"output": "20"
},
{
"input": "50\n11 17 23 53 59 109 137 149 173 251 353 379 419 421 439 503 593 607 661 773 821 877 941 997 1061 1117 1153 1229 1289 1297 1321 1609 1747 2311 2389 2543 2693 3041 3083 3137 3181 3209 3331 3373 3617 3767 4201 4409 4931 6379\n50\n55 59 67 73 85 89 101 115 211 263 295 353 545 599 607 685 739 745 997 1031 1255 1493 1523 1667 1709 1895 1949 2161 2195 2965 3019 3035 3305 3361 3373 3673 3739 3865 3881 4231 4253 4385 4985 5305 5585 5765 6145 6445 8045 8735",
"output": "23"
},
{
"input": "5\n33 78 146 3055 4268\n5\n2211 2584 5226 9402 9782",
"output": "3"
},
{
"input": "5\n35 48 52 86 8001\n10\n332 3430 3554 4704 4860 5096 6215 7583 8228 8428",
"output": "4"
},
{
"input": "10\n97 184 207 228 269 2084 4450 6396 7214 9457\n16\n338 1179 1284 1545 1570 2444 3167 3395 3397 5550 6440 7245 7804 7980 9415 9959",
"output": "5"
},
{
"input": "30\n25 30 41 57 58 62 70 72 76 79 84 85 88 91 98 101 104 109 119 129 136 139 148 151 926 1372 3093 3936 5423 7350\n25\n1600 1920 2624 3648 3712 3968 4480 4608 4864 5056 5376 5440 5632 5824 6272 6464 6656 6934 6976 7616 8256 8704 8896 9472 9664",
"output": "24"
},
{
"input": "5\n33 78 146 3055 4268\n5\n2211 2584 5226 9402 9782",
"output": "3"
},
{
"input": "5\n35 48 52 86 8001\n10\n332 3430 3554 4704 4860 5096 6215 7583 8228 8428",
"output": "4"
},
{
"input": "10\n97 184 207 228 269 2084 4450 6396 7214 9457\n16\n338 1179 1284 1545 1570 2444 3167 3395 3397 5550 6440 7245 7804 7980 9415 9959",
"output": "5"
},
{
"input": "30\n25 30 41 57 58 62 70 72 76 79 84 85 88 91 98 101 104 109 119 129 136 139 148 151 926 1372 3093 3936 5423 7350\n25\n1600 1920 2624 3648 3712 3968 4480 4608 4864 5056 5376 5440 5632 5824 6272 6464 6656 6934 6976 7616 8256 8704 8896 9472 9664",
"output": "24"
},
{
"input": "47\n66 262 357 457 513 530 538 540 592 691 707 979 1015 1242 1246 1667 1823 1886 1963 2133 2649 2679 2916 2949 3413 3523 3699 3958 4393 4922 5233 5306 5799 6036 6302 6629 7208 7282 7315 7822 7833 7927 8068 8150 8870 8962 9987\n39\n167 199 360 528 1515 1643 1986 1988 2154 2397 2856 3552 3656 3784 3980 4096 4104 4240 4320 4736 4951 5266 5656 5849 5850 6169 6517 6875 7244 7339 7689 7832 8120 8716 9503 9509 9933 9936 9968",
"output": "12"
},
{
"input": "1\n94\n50\n423 446 485 1214 1468 1507 1853 1930 1999 2258 2271 2285 2425 2543 2715 2743 2992 3196 4074 4108 4448 4475 4652 5057 5250 5312 5356 5375 5731 5986 6298 6501 6521 7146 7255 7276 7332 7481 7998 8141 8413 8665 8908 9221 9336 9491 9504 9677 9693 9706",
"output": "1"
},
{
"input": "50\n51 67 75 186 194 355 512 561 720 876 1077 1221 1503 1820 2153 2385 2568 2608 2937 2969 3271 3311 3481 4081 4093 4171 4255 4256 4829 5020 5192 5636 5817 6156 6712 6717 7153 7436 7608 7612 7866 7988 8264 8293 8867 9311 9879 9882 9889 9908\n1\n5394",
"output": "1"
},
{
"input": "50\n26 367 495 585 675 789 855 1185 1312 1606 2037 2241 2587 2612 2628 2807 2873 2924 3774 4067 4376 4668 4902 5001 5082 5100 5104 5209 5345 5515 5661 5777 5902 5907 6155 6323 6675 6791 7503 8159 8207 8254 8740 8848 8855 8933 9069 9164 9171 9586\n5\n1557 6246 7545 8074 8284",
"output": "1"
},
{
"input": "5\n25 58 91 110 2658\n50\n21 372 909 1172 1517 1554 1797 1802 1843 1977 2006 2025 2137 2225 2317 2507 2645 2754 2919 3024 3202 3212 3267 3852 4374 4487 4553 4668 4883 4911 4916 5016 5021 5068 5104 5162 5683 5856 6374 6871 7333 7531 8099 8135 8173 8215 8462 8776 9433 9790",
"output": "4"
},
{
"input": "45\n37 48 56 59 69 70 79 83 85 86 99 114 131 134 135 145 156 250 1739 1947 2116 2315 2449 3104 3666 4008 4406 4723 4829 5345 5836 6262 6296 6870 7065 7110 7130 7510 7595 8092 8442 8574 9032 9091 9355\n50\n343 846 893 1110 1651 1837 2162 2331 2596 3012 3024 3131 3294 3394 3528 3717 3997 4125 4347 4410 4581 4977 5030 5070 5119 5229 5355 5413 5418 5474 5763 5940 6151 6161 6164 6237 6506 6519 6783 7182 7413 7534 8069 8253 8442 8505 9135 9308 9828 9902",
"output": "17"
},
{
"input": "50\n17 20 22 28 36 38 46 47 48 50 52 57 58 62 63 69 70 74 75 78 79 81 82 86 87 90 93 95 103 202 292 442 1756 1769 2208 2311 2799 2957 3483 4280 4324 4932 5109 5204 6225 6354 6561 7136 8754 9670\n40\n68 214 957 1649 1940 2078 2134 2716 3492 3686 4462 4559 4656 4756 4850 5044 5490 5529 5592 5626 6014 6111 6693 6790 7178 7275 7566 7663 7702 7857 7954 8342 8511 8730 8957 9021 9215 9377 9445 9991",
"output": "28"
},
{
"input": "39\n10 13 21 25 36 38 47 48 58 64 68 69 73 79 86 972 2012 2215 2267 2503 3717 3945 4197 4800 5266 6169 6612 6824 7023 7322 7582 7766 8381 8626 8879 9079 9088 9838 9968\n50\n432 877 970 1152 1202 1223 1261 1435 1454 1578 1843 1907 2003 2037 2183 2195 2215 2425 3065 3492 3615 3637 3686 3946 4189 4415 4559 4656 4665 4707 4886 4887 5626 5703 5955 6208 6521 6581 6596 6693 6985 7013 7081 7343 7663 8332 8342 8637 9207 9862",
"output": "15"
},
{
"input": "50\n7 144 269 339 395 505 625 688 709 950 1102 1152 1350 1381 1641 1830 1977 1999 2093 2180 2718 3308 3574 4168 4232 4259 4393 4689 4982 5154 5476 5581 5635 5721 6159 6302 6741 7010 7152 7315 7417 7482 8116 8239 8640 9347 9395 9614 9661 9822\n20\n84 162 292 1728 1866 2088 3228 3470 4068 5318 5470 6060 6380 6929 7500 8256 8399 8467 8508 9691",
"output": "8"
},
{
"input": "50\n159 880 1070 1139 1358 1608 1691 1841 2073 2171 2213 2597 2692 2759 2879 2931 3173 3217 3441 4201 4878 5106 5129 5253 5395 5647 5968 6019 6130 6276 6286 6330 6409 6728 7488 7713 7765 7828 7899 8064 8264 8457 8483 8685 8900 8946 8965 9133 9187 9638\n45\n57 159 1070 1139 1391 1608 1691 1841 2171 2213 2692 2759 2931 3173 3217 3441 4201 4878 5106 5129 5253 5647 5968 6130 6276 6286 6409 7488 7694 7713 7765 7828 7899 8003 8064 8081 8244 8264 8685 8900 8946 8965 9133 9638 9673",
"output": "38"
},
{
"input": "3\n3 4 5\n3\n6 20 25",
"output": "2"
},
{
"input": "4\n2 3 5 8\n4\n2 6 8 10",
"output": "1"
},
{
"input": "4\n3 5 7 11\n4\n3 5 7 22",
"output": "1"
},
{
"input": "2\n2 3\n3\n20 30 50",
"output": "1"
},
{
"input": "3\n1 2 3\n4\n2 4 6 49",
"output": "1"
},
{
"input": "2\n4 5\n3\n12 15 20",
"output": "1"
},
{
"input": "3\n2 5 7\n3\n4 5 7",
"output": "1"
},
{
"input": "3\n3 5 8\n3\n6 8 10",
"output": "2"
},
{
"input": "2\n2 3\n4\n4 6 9 33",
"output": "1"
},
{
"input": "2\n2 3\n4\n4 6 21 40",
"output": "1"
},
{
"input": "3\n4 9 10\n3\n8 9 10",
"output": "1"
},
{
"input": "5\n1 5 6 9 51\n5\n5 12 18 27 10000",
"output": "1"
},
{
"input": "13\n1 2 3 4 5 6 7 8 9 10 11 12 13\n1\n14",
"output": "1"
}
] | 1,589,525,191
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 57
| 310
| 21,401,600
|
n = int(input())
a = list(map(int, input().split()))
m = int(input())
b = list(map(int, input().split()))
temp = [0]
for i in a:
for j in b:
if j%i==0:
if j/i > max(temp):
temp = []
temp.append(j/i)
elif j/i == max(temp):
temp.append(j/i)
print(len(temp))
|
Title: Bicycle Chain
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation.
We know that the *i*-th star on the pedal axle has *a**i* (0<=<<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=<<=*b*1<=<<=*b*2<=<<=...<=<<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value .
Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears.
In the problem, fraction denotes division in real numbers, that is, no rounding is performed.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing.
The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing.
It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces.
Output Specification:
Print the number of "integer" gears with the maximum ratio among all "integer" gears.
Demo Input:
['2\n4 5\n3\n12 13 15\n', '4\n1 2 3 4\n5\n10 11 12 13 14\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15.
|
```python
n = int(input())
a = list(map(int, input().split()))
m = int(input())
b = list(map(int, input().split()))
temp = [0]
for i in a:
for j in b:
if j%i==0:
if j/i > max(temp):
temp = []
temp.append(j/i)
elif j/i == max(temp):
temp.append(j/i)
print(len(temp))
```
| 3
|
|
785
|
A
|
Anton and Polyhedrons
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
|
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
|
[
"4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n",
"3\nDodecahedron\nOctahedron\nOctahedron\n"
] |
[
"42\n",
"28\n"
] |
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
| 500
|
[
{
"input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron",
"output": "42"
},
{
"input": "3\nDodecahedron\nOctahedron\nOctahedron",
"output": "28"
},
{
"input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosahedron\nTetrahedron\nOctahedron\nDodecahedron\nIcosahedron\nOctahedron\nIcosahedron\nTetrahedron\nDodecahedron\nTetrahedron\nOctahedron\nCube\nCube\nDodecahedron\nTetrahedron",
"output": "256"
},
{
"input": "1\nTetrahedron",
"output": "4"
},
{
"input": "1\nCube",
"output": "6"
},
{
"input": "1\nOctahedron",
"output": "8"
},
{
"input": "1\nDodecahedron",
"output": "12"
},
{
"input": "1\nIcosahedron",
"output": "20"
},
{
"input": "28\nOctahedron\nDodecahedron\nOctahedron\nOctahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nDodecahedron\nDodecahedron\nCube\nDodecahedron\nCube\nTetrahedron\nCube\nCube\nTetrahedron\nDodecahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron\nDodecahedron\nIcosahedron\nDodecahedron\nDodecahedron\nIcosahedron\nIcosahedron",
"output": "340"
}
] | 1,694,959,754
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 248
| 16,076,800
|
b=[]
for i in range(int(input())):b.append(input())
print(b.count('Tetrahedron')*4+b.count('Cube')*6+b.count('Octahedron')*8+b.count('Dodecahedron')*12+b.count('Icosahedron')*20)
|
Title: Anton and Polyhedrons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
Output Specification:
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
Demo Input:
['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n']
Demo Output:
['42\n', '28\n']
Note:
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces.
|
```python
b=[]
for i in range(int(input())):b.append(input())
print(b.count('Tetrahedron')*4+b.count('Cube')*6+b.count('Octahedron')*8+b.count('Dodecahedron')*12+b.count('Icosahedron')*20)
```
| 3
|
|
275
|
A
|
Lights Out
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.
|
The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed.
|
Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".
|
[
"1 0 0\n0 0 0\n0 0 1\n",
"1 0 1\n8 8 8\n2 0 3\n"
] |
[
"001\n010\n100\n",
"010\n011\n100\n"
] |
none
| 500
|
[
{
"input": "1 0 0\n0 0 0\n0 0 1",
"output": "001\n010\n100"
},
{
"input": "1 0 1\n8 8 8\n2 0 3",
"output": "010\n011\n100"
},
{
"input": "13 85 77\n25 50 45\n65 79 9",
"output": "000\n010\n000"
},
{
"input": "96 95 5\n8 84 74\n67 31 61",
"output": "011\n011\n101"
},
{
"input": "24 54 37\n60 63 6\n1 84 26",
"output": "110\n101\n011"
},
{
"input": "23 10 40\n15 6 40\n92 80 77",
"output": "101\n100\n000"
},
{
"input": "62 74 80\n95 74 93\n2 47 95",
"output": "010\n001\n110"
},
{
"input": "80 83 48\n26 0 66\n47 76 37",
"output": "000\n000\n010"
},
{
"input": "32 15 65\n7 54 36\n5 51 3",
"output": "111\n101\n001"
},
{
"input": "22 97 12\n71 8 24\n100 21 64",
"output": "100\n001\n100"
},
{
"input": "46 37 13\n87 0 50\n90 8 55",
"output": "111\n011\n000"
},
{
"input": "57 43 58\n20 82 83\n66 16 52",
"output": "111\n010\n110"
},
{
"input": "45 56 93\n47 51 59\n18 51 63",
"output": "101\n011\n100"
},
{
"input": "47 66 67\n14 1 37\n27 81 69",
"output": "001\n001\n110"
},
{
"input": "26 69 69\n85 18 23\n14 22 74",
"output": "110\n001\n010"
},
{
"input": "10 70 65\n94 27 25\n74 66 30",
"output": "111\n010\n100"
},
{
"input": "97 1 74\n15 99 1\n88 68 86",
"output": "001\n011\n000"
},
{
"input": "36 48 42\n45 41 66\n26 64 1",
"output": "001\n111\n010"
},
{
"input": "52 81 97\n29 77 71\n66 11 2",
"output": "100\n100\n111"
},
{
"input": "18 66 33\n19 49 49\n48 46 26",
"output": "011\n100\n000"
},
{
"input": "68 79 52\n51 39 100\n29 14 26",
"output": "110\n000\n111"
},
{
"input": "91 69 77\n91 26 64\n91 88 57",
"output": "001\n011\n110"
},
{
"input": "16 69 64\n48 21 80\n81 51 51",
"output": "010\n101\n111"
},
{
"input": "96 14 2\n100 18 12\n65 34 89",
"output": "111\n010\n010"
},
{
"input": "93 95 90\n8 59 42\n53 13 19",
"output": "100\n001\n111"
},
{
"input": "71 84 18\n100 19 67\n9 76 15",
"output": "010\n010\n001"
},
{
"input": "38 93 85\n21 88 64\n4 96 25",
"output": "111\n011\n000"
},
{
"input": "75 20 20\n60 5 78\n77 4 69",
"output": "011\n001\n000"
},
{
"input": "65 70 96\n19 6 83\n33 37 82",
"output": "100\n000\n011"
},
{
"input": "11 13 60\n17 13 46\n42 21 39",
"output": "000\n011\n101"
},
{
"input": "0 0 0\n0 0 0\n0 0 0",
"output": "111\n111\n111"
},
{
"input": "0 0 0\n0 1 0\n0 0 0",
"output": "101\n000\n101"
},
{
"input": "0 0 0\n0 0 0\n0 0 1",
"output": "111\n110\n100"
}
] | 1,595,269,529
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 33
| 155
| 20,172,800
|
a = [[8]*5 for i in range(5)]
b = [[int(j) for j in input().split()] for i in range(3)]
c = [[1]*5 for i in range(5)]
n, k = 0, 0
for i in range(1,4):
for j in range(1,4):
a[i][j] = b[n][k]
k+=1
n+=1
k = 0
for i in range(1,4):
for j in range(1,4):
if a[i][j] % 2 == 0:
a[i][j] = 0
else:
a[i][j] = 1
for i in range(len(a)-1):
for j in range(len(a[i])-1):
if i>=1 and i<=3 and j>=1 and j<=3:
if a[i][j] % 2 != 0:
c[i][j] = 1 - c[i][j]
c[i+1][j] = 1 - c[i+1][j]
c[i-1][j] = 1 - c[i-1][j]
c[i][j+1] = 1 - c[i][j+1]
c[i][j-1] = 1 - c[i][j-1]
for i in range(1,4):
for j in range(1,4):
print(c[i][j], end = "")
print()
|
Title: Lights Out
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.
Input Specification:
The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed.
Output Specification:
Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".
Demo Input:
['1 0 0\n0 0 0\n0 0 1\n', '1 0 1\n8 8 8\n2 0 3\n']
Demo Output:
['001\n010\n100\n', '010\n011\n100\n']
Note:
none
|
```python
a = [[8]*5 for i in range(5)]
b = [[int(j) for j in input().split()] for i in range(3)]
c = [[1]*5 for i in range(5)]
n, k = 0, 0
for i in range(1,4):
for j in range(1,4):
a[i][j] = b[n][k]
k+=1
n+=1
k = 0
for i in range(1,4):
for j in range(1,4):
if a[i][j] % 2 == 0:
a[i][j] = 0
else:
a[i][j] = 1
for i in range(len(a)-1):
for j in range(len(a[i])-1):
if i>=1 and i<=3 and j>=1 and j<=3:
if a[i][j] % 2 != 0:
c[i][j] = 1 - c[i][j]
c[i+1][j] = 1 - c[i+1][j]
c[i-1][j] = 1 - c[i-1][j]
c[i][j+1] = 1 - c[i][j+1]
c[i][j-1] = 1 - c[i][j-1]
for i in range(1,4):
for j in range(1,4):
print(c[i][j], end = "")
print()
```
| 3
|
|
266
|
B
|
Queue at the School
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"graph matchings",
"implementation",
"shortest paths"
] | null | null |
During the break the schoolchildren, boys and girls, formed a queue of *n* people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to *n*, at that the person in the position number 1 is served first. Then, if at time *x* a boy stands on the *i*-th position and a girl stands on the (*i*<=+<=1)-th position, then at time *x*<=+<=1 the *i*-th position will have a girl and the (*i*<=+<=1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after *t* seconds.
|
The first line contains two integers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string *s*, which represents the schoolchildren's initial arrangement. If the *i*-th position in the queue contains a boy, then the *i*-th character of string *s* equals "B", otherwise the *i*-th character equals "G".
|
Print string *a*, which describes the arrangement after *t* seconds. If the *i*-th position has a boy after the needed time, then the *i*-th character *a* must equal "B", otherwise it must equal "G".
|
[
"5 1\nBGGBG\n",
"5 2\nBGGBG\n",
"4 1\nGGGB\n"
] |
[
"GBGGB\n",
"GGBGB\n",
"GGGB\n"
] |
none
| 500
|
[
{
"input": "5 1\nBGGBG",
"output": "GBGGB"
},
{
"input": "5 2\nBGGBG",
"output": "GGBGB"
},
{
"input": "4 1\nGGGB",
"output": "GGGB"
},
{
"input": "2 1\nBB",
"output": "BB"
},
{
"input": "2 1\nBG",
"output": "GB"
},
{
"input": "6 2\nBBGBBG",
"output": "GBBGBB"
},
{
"input": "8 3\nBBGBGBGB",
"output": "GGBGBBBB"
},
{
"input": "10 3\nBBGBBBBBBG",
"output": "GBBBBBGBBB"
},
{
"input": "22 7\nGBGGBGGGGGBBBGGBGBGBBB",
"output": "GGGGGGGGBGGBGGBBBBBBBB"
},
{
"input": "50 4\nGBBGBBBGGGGGBBGGBBBBGGGBBBGBBBGGBGGBGBBBGGBGGBGGBG",
"output": "GGBGBGBGBGBGGGBBGBGBGBGBBBGBGBGBGBGBGBGBGBGBGGBGBB"
},
{
"input": "50 8\nGGGGBGGBGGGBGBBBGGGGGGGGBBGBGBGBBGGBGGBGGGGGGGGBBG",
"output": "GGGGGGGGGGGGBGGBGBGBGBGBGGGGGGBGBGBGBGBGBGGBGGBGBB"
},
{
"input": "50 30\nBGGGGGGBGGBGBGGGGBGBBGBBBGGBBBGBGBGGGGGBGBBGBGBGGG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "20 20\nBBGGBGGGGBBBGBBGGGBB",
"output": "GGGGGGGGGGBBBBBBBBBB"
},
{
"input": "27 6\nGBGBGBGGGGGGBGGBGGBBGBBBGBB",
"output": "GGGGGGGBGBGBGGGGGBGBBBBBBBB"
},
{
"input": "46 11\nBGGGGGBGBGGBGGGBBGBBGBBGGBBGBBGBGGGGGGGBGBGBGB",
"output": "GGGGGGGGGGGBGGGGGBBGBGBGBGBGBGBGBGBGBGBGBBBBBB"
},
{
"input": "50 6\nBGGBBBBGGBBBBBBGGBGBGBBBBGBBBBBBGBBBBBBBBBBBBBBBBB",
"output": "GGGGBBBBBGBGBGBGBBBGBBBBBBGBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 10\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 8\nGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "50 10\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBGB",
"output": "BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBGBBBBBBBBBBB"
},
{
"input": "50 13\nGGGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG",
"output": "GGGGGGGGGGGGGGGGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "1 1\nB",
"output": "B"
},
{
"input": "1 1\nG",
"output": "G"
},
{
"input": "1 50\nB",
"output": "B"
},
{
"input": "1 50\nG",
"output": "G"
},
{
"input": "50 50\nBBBBBBBBGGBBBBBBGBBBBBBBBBBBGBBBBBBBBBBBBBBGBBBBBB",
"output": "GGGGGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "50 50\nGGBBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBGGGGGGBG",
"output": "GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGBBBBB"
},
{
"input": "6 3\nGGBBBG",
"output": "GGGBBB"
},
{
"input": "26 3\nGBBGBBBBBGGGBGBGGGBGBGGBBG",
"output": "GGBBBBGBGBGBGGGBGBGGGBGBBB"
},
{
"input": "46 3\nGGBBGGGGBBGBGBBBBBGGGBGGGBBGGGBBBGGBGGBBBGBGBB",
"output": "GGGGBGBGGGBBBBBGBGBGBGGGBGGBGBGBGBGBGBGBGBBBBB"
},
{
"input": "44 8\nBGBBBBBBBBBGGBBGBGBGGBBBBBGBBGBBBBBBBBBGBBGB",
"output": "GBBGBGBGBGBGBGBBBBGBBGBBBBBBBBBGBBGBBBBBBBBB"
},
{
"input": "20 20\nBBGGBGGGGBBBGBBGGGBB",
"output": "GGGGGGGGGGBBBBBBBBBB"
},
{
"input": "30 25\nBGGBBGBGGBGBGBBGBGGGGBGBGGBBBB",
"output": "GGGGGGGGGGGGGGGBBBBBBBBBBBBBBB"
},
{
"input": "17 42\nBBGBGBGGGGGGBBGGG",
"output": "GGGGGGGGGGGBBBBBB"
},
{
"input": "30 50\nBGGBBGGGGGGGGBBGGGBBGGBBBGBBGG",
"output": "GGGGGGGGGGGGGGGGGGBBBBBBBBBBBB"
},
{
"input": "31 33\nBBGGBBGGBGBBBBGGBGBBBGBGGBBGBBB",
"output": "GGGGGGGGGGGGBBBBBBBBBBBBBBBBBBB"
},
{
"input": "2 2\nBG",
"output": "GB"
},
{
"input": "8 8\nGGGBGGBB",
"output": "GGGGGBBB"
},
{
"input": "34 34\nBGGBBBBGGBBGBGGGGGGBBGGGGGBGGBGGGB",
"output": "GGGGGGGGGGGGGGGGGGGGGBBBBBBBBBBBBB"
},
{
"input": "34 20\nBBBBGBGGGBGGGBGBGGBBBBGGBBGGGBGBBG",
"output": "GGGGGGGGGGGGGGGGGBBBBBBBBBBBBBBBBB"
},
{
"input": "50 50\nBBGBBBBBBBBBBBBGBBBGBBBBBBBGBBBBBBGBBBGBBGBBGBBBGG",
"output": "GGGGGGGGGGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "10 10\nGGGGGGGGGG",
"output": "GGGGGGGGGG"
},
{
"input": "10 10\nBBBBBBBBBB",
"output": "BBBBBBBBBB"
},
{
"input": "10 10\nBGBGBGBGBG",
"output": "GGGGGBBBBB"
},
{
"input": "1 1\nB",
"output": "B"
}
] | 1,696,771,104
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 92
| 0
|
n,t = map(int,input().split())
que = list(input())
for j in range(t):
check = 0
for i in range(len(que)-1):
if que[i+1] =="G" and que[i] == "B" and i >= check:
que[i],que[i+1] = que[i+1],que[i]
check = i + 2
print("".join(que))
|
Title: Queue at the School
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the break the schoolchildren, boys and girls, formed a queue of *n* people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to *n*, at that the person in the position number 1 is served first. Then, if at time *x* a boy stands on the *i*-th position and a girl stands on the (*i*<=+<=1)-th position, then at time *x*<=+<=1 the *i*-th position will have a girl and the (*i*<=+<=1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after *t* seconds.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*,<=*t*<=≤<=50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string *s*, which represents the schoolchildren's initial arrangement. If the *i*-th position in the queue contains a boy, then the *i*-th character of string *s* equals "B", otherwise the *i*-th character equals "G".
Output Specification:
Print string *a*, which describes the arrangement after *t* seconds. If the *i*-th position has a boy after the needed time, then the *i*-th character *a* must equal "B", otherwise it must equal "G".
Demo Input:
['5 1\nBGGBG\n', '5 2\nBGGBG\n', '4 1\nGGGB\n']
Demo Output:
['GBGGB\n', 'GGBGB\n', 'GGGB\n']
Note:
none
|
```python
n,t = map(int,input().split())
que = list(input())
for j in range(t):
check = 0
for i in range(len(que)-1):
if que[i+1] =="G" and que[i] == "B" and i >= check:
que[i],que[i+1] = que[i+1],que[i]
check = i + 2
print("".join(que))
```
| 3
|
|
17
|
A
|
Noldbach problem
|
PROGRAMMING
| 1,000
|
[
"brute force",
"math",
"number theory"
] |
A. Noldbach problem
|
2
|
64
|
Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1.
Two prime numbers are called neighboring if there are no other prime numbers between them.
You are to help Nick, and find out if he is right or wrong.
|
The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000).
|
Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO.
|
[
"27 2\n",
"45 7\n"
] |
[
"YES",
"NO"
] |
In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form.
| 0
|
[
{
"input": "27 2",
"output": "YES"
},
{
"input": "45 7",
"output": "NO"
},
{
"input": "2 0",
"output": "YES"
},
{
"input": "15 1",
"output": "YES"
},
{
"input": "17 1",
"output": "YES"
},
{
"input": "34 5",
"output": "NO"
},
{
"input": "37 4",
"output": "YES"
},
{
"input": "43 5",
"output": "YES"
},
{
"input": "47 7",
"output": "NO"
},
{
"input": "50 5",
"output": "YES"
},
{
"input": "57 6",
"output": "YES"
},
{
"input": "60 8",
"output": "NO"
},
{
"input": "62 7",
"output": "YES"
},
{
"input": "76 9",
"output": "NO"
},
{
"input": "69 7",
"output": "YES"
},
{
"input": "113 10",
"output": "YES"
},
{
"input": "141 11",
"output": "YES"
},
{
"input": "207 16",
"output": "NO"
},
{
"input": "231 18",
"output": "NO"
},
{
"input": "296 19",
"output": "YES"
},
{
"input": "332 20",
"output": "YES"
},
{
"input": "378 24",
"output": "NO"
},
{
"input": "428 23",
"output": "YES"
},
{
"input": "497 27",
"output": "NO"
},
{
"input": "640 32",
"output": "YES"
},
{
"input": "798 35",
"output": "YES"
},
{
"input": "802 35",
"output": "YES"
},
{
"input": "864 40",
"output": "NO"
},
{
"input": "953 44",
"output": "NO"
},
{
"input": "995 44",
"output": "YES"
},
{
"input": "1000 44",
"output": "YES"
},
{
"input": "1000 0",
"output": "YES"
},
{
"input": "1000 1000",
"output": "NO"
},
{
"input": "2 1000",
"output": "NO"
},
{
"input": "2 0",
"output": "YES"
}
] | 1,667,046,426
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 122
| 0
|
# Hydro submission #635d1c16e2991f80a1014e3d@1667046423273
n,k =map( int,input().split())
a = []
count=0
for i in range(2, n + 1):
for j in range(2, i):
if i % j == 0:
break
else:
a.append(i)
for i in a:
for j in range(1, len(a)):
if a[j-1]+a[j]+1==i:
count+=1
if count < k:
print("NO")
else:
print("YES")
|
Title: Noldbach problem
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Nick is interested in prime numbers. Once he read about Goldbach problem. It states that every even integer greater than 2 can be expressed as the sum of two primes. That got Nick's attention and he decided to invent a problem of his own and call it Noldbach problem. Since Nick is interested only in prime numbers, Noldbach problem states that at least *k* prime numbers from 2 to *n* inclusively can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1.
Two prime numbers are called neighboring if there are no other prime numbers between them.
You are to help Nick, and find out if he is right or wrong.
Input Specification:
The first line of the input contains two integers *n* (2<=≤<=*n*<=≤<=1000) and *k* (0<=≤<=*k*<=≤<=1000).
Output Specification:
Output YES if at least *k* prime numbers from 2 to *n* inclusively can be expressed as it was described above. Otherwise output NO.
Demo Input:
['27 2\n', '45 7\n']
Demo Output:
['YES', 'NO']
Note:
In the first sample the answer is YES since at least two numbers can be expressed as it was described (for example, 13 and 19). In the second sample the answer is NO since it is impossible to express 7 prime numbers from 2 to 45 in the desired form.
|
```python
# Hydro submission #635d1c16e2991f80a1014e3d@1667046423273
n,k =map( int,input().split())
a = []
count=0
for i in range(2, n + 1):
for j in range(2, i):
if i % j == 0:
break
else:
a.append(i)
for i in a:
for j in range(1, len(a)):
if a[j-1]+a[j]+1==i:
count+=1
if count < k:
print("NO")
else:
print("YES")
```
| 3.9695
|
318
|
A
|
Even Odds
|
PROGRAMMING
| 900
|
[
"math"
] | null | null |
Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*.
|
The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
Print the number that will stand at the position number *k* after Volodya's manipulations.
|
[
"10 3\n",
"7 7\n"
] |
[
"5",
"6"
] |
In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5.
| 500
|
[
{
"input": "10 3",
"output": "5"
},
{
"input": "7 7",
"output": "6"
},
{
"input": "7 1",
"output": "1"
},
{
"input": "7 2",
"output": "3"
},
{
"input": "8 5",
"output": "2"
},
{
"input": "8 3",
"output": "5"
},
{
"input": "8 4",
"output": "7"
},
{
"input": "1000000000000 500000000001",
"output": "2"
},
{
"input": "999999999997 499999999999",
"output": "999999999997"
},
{
"input": "999999999999 999999999999",
"output": "999999999998"
},
{
"input": "1000000000000 1",
"output": "1"
},
{
"input": "999999999999 1",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000000 1000000000000",
"output": "1000000000000"
},
{
"input": "1000000000000 500000000000",
"output": "999999999999"
},
{
"input": "1000000000000 499999999999",
"output": "999999999997"
},
{
"input": "999999999997 499999999998",
"output": "999999999995"
},
{
"input": "619234238 556154835",
"output": "493075432"
},
{
"input": "38151981 36650624",
"output": "35149266"
},
{
"input": "680402465 442571217",
"output": "204739968"
},
{
"input": "109135284 9408714",
"output": "18817427"
},
{
"input": "603701841 56038951",
"output": "112077901"
},
{
"input": "356764822 321510177",
"output": "286255532"
},
{
"input": "284911189 142190783",
"output": "284381565"
},
{
"input": "91028405 61435545",
"output": "31842684"
}
] | 1,691,730,759
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 62
| 0
|
n,k=map(int,input().split())
# if n is odd
if n%2:
no_of_odd=(n//2)+1
else:
no_of_odd=n//2
if k<(no_of_odd):
print((k*2)-1)
else:
print((k-no_of_odd)*2)
|
Title: Even Odds
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*.
Input Specification:
The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print the number that will stand at the position number *k* after Volodya's manipulations.
Demo Input:
['10 3\n', '7 7\n']
Demo Output:
['5', '6']
Note:
In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5.
|
```python
n,k=map(int,input().split())
# if n is odd
if n%2:
no_of_odd=(n//2)+1
else:
no_of_odd=n//2
if k<(no_of_odd):
print((k*2)-1)
else:
print((k-no_of_odd)*2)
```
| 0
|
|
740
|
A
|
Alyona and copybooks
|
PROGRAMMING
| 1,300
|
[
"brute force",
"implementation"
] | null | null |
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for *a* rubles, a pack of two copybooks for *b* rubles, and a pack of three copybooks for *c* rubles. Alyona already has *n* copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks *k* that *n*<=+<=*k* is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
|
The only line contains 4 integers *n*, *a*, *b*, *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=109).
|
Print the minimum amount of rubles she should pay to buy such number of copybooks *k* that *n*<=+<=*k* is divisible by 4.
|
[
"1 1 3 4\n",
"6 2 1 1\n",
"4 4 4 4\n",
"999999999 1000000000 1000000000 1000000000\n"
] |
[
"3\n",
"1\n",
"0\n",
"1000000000\n"
] |
In the first example Alyona can buy 3 packs of 1 copybook for 3*a* = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for *b* = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook.
| 500
|
[
{
"input": "1 1 3 4",
"output": "3"
},
{
"input": "6 2 1 1",
"output": "1"
},
{
"input": "4 4 4 4",
"output": "0"
},
{
"input": "999999999 1000000000 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "1016 3 2 1",
"output": "0"
},
{
"input": "17 100 100 1",
"output": "1"
},
{
"input": "17 2 3 100",
"output": "5"
},
{
"input": "18 1 3 3",
"output": "2"
},
{
"input": "19 1 1 1",
"output": "1"
},
{
"input": "999999997 999999990 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "999999998 1000000000 999999990 1000000000",
"output": "999999990"
},
{
"input": "634074578 336470888 481199252 167959139",
"output": "335918278"
},
{
"input": "999999999 1000000000 1000000000 999999990",
"output": "1000000000"
},
{
"input": "804928248 75475634 54748096 641009859",
"output": "0"
},
{
"input": "535590429 374288891 923264237 524125987",
"output": "524125987"
},
{
"input": "561219907 673102149 496813081 702209411",
"output": "673102149"
},
{
"input": "291882089 412106895 365329221 585325539",
"output": "585325539"
},
{
"input": "757703054 5887448 643910770 58376259",
"output": "11774896"
},
{
"input": "783332532 449924898 72235422 941492387",
"output": "0"
},
{
"input": "513994713 43705451 940751563 824608515",
"output": "131116353"
},
{
"input": "539624191 782710197 514300407 2691939",
"output": "8075817"
},
{
"input": "983359971 640274071 598196518 802030518",
"output": "640274071"
},
{
"input": "8989449 379278816 26521171 685146646",
"output": "405799987"
},
{
"input": "34618927 678092074 895037311 863230070",
"output": "678092074"
},
{
"input": "205472596 417096820 468586155 41313494",
"output": "0"
},
{
"input": "19 5 1 2",
"output": "3"
},
{
"input": "17 1 2 2",
"output": "2"
},
{
"input": "18 3 3 1",
"output": "2"
},
{
"input": "19 4 3 1",
"output": "3"
},
{
"input": "936134778 715910077 747167704 219396918",
"output": "438793836"
},
{
"input": "961764255 454914823 615683844 102513046",
"output": "307539138"
},
{
"input": "692426437 48695377 189232688 985629174",
"output": "146086131"
},
{
"input": "863280107 347508634 912524637 458679894",
"output": "347508634"
},
{
"input": "593942288 86513380 486073481 341796022",
"output": "0"
},
{
"input": "914539062 680293934 764655030 519879446",
"output": "764655030"
},
{
"input": "552472140 509061481 586588704 452405440",
"output": "0"
},
{
"input": "723325809 807874739 160137548 335521569",
"output": "335521569"
},
{
"input": "748955287 546879484 733686393 808572289",
"output": "546879484"
},
{
"input": "774584765 845692742 162011045 691688417",
"output": "691688417"
},
{
"input": "505246946 439473295 30527185 869771841",
"output": "30527185"
},
{
"input": "676100616 178478041 604076030 752887969",
"output": "0"
},
{
"input": "701730093 477291299 177624874 930971393",
"output": "654916173"
},
{
"input": "432392275 216296044 751173719 109054817",
"output": "216296044"
},
{
"input": "458021753 810076598 324722563 992170945",
"output": "992170945"
},
{
"input": "188683934 254114048 48014511 170254369",
"output": "48014511"
},
{
"input": "561775796 937657403 280013594 248004555",
"output": "0"
},
{
"input": "1000000000 1000000000 1000000000 1000000000",
"output": "0"
},
{
"input": "3 10000 10000 3",
"output": "9"
},
{
"input": "3 12 3 4",
"output": "7"
},
{
"input": "3 10000 10000 1",
"output": "3"
},
{
"input": "3 1000 1000 1",
"output": "3"
},
{
"input": "3 10 10 1",
"output": "3"
},
{
"input": "3 100 100 1",
"output": "3"
},
{
"input": "3 100000 10000 1",
"output": "3"
},
{
"input": "7 10 2 3",
"output": "5"
},
{
"input": "3 1000 1000 2",
"output": "6"
},
{
"input": "1 100000 1 100000",
"output": "100000"
},
{
"input": "7 4 3 1",
"output": "3"
},
{
"input": "3 1000 1000 3",
"output": "9"
},
{
"input": "3 1000 1 1",
"output": "2"
},
{
"input": "3 10 1 1",
"output": "2"
},
{
"input": "3 100000 1 1",
"output": "2"
},
{
"input": "3 100 1 1",
"output": "2"
},
{
"input": "3 100000 100000 1",
"output": "3"
},
{
"input": "3 1000 1 100",
"output": "101"
},
{
"input": "3 1000000000 1 1000000000",
"output": "1000000000"
},
{
"input": "3 1000 1 10",
"output": "11"
},
{
"input": "3 200 1 100",
"output": "101"
},
{
"input": "7 4 1 1",
"output": "2"
},
{
"input": "7 4 12 1",
"output": "3"
},
{
"input": "3 9 1 1",
"output": "2"
},
{
"input": "3 10000000 1000000 1",
"output": "3"
},
{
"input": "7 1000 1000 1",
"output": "3"
},
{
"input": "3 10000 1 30",
"output": "31"
},
{
"input": "3 1000 1 2",
"output": "3"
},
{
"input": "7 12 6 1",
"output": "3"
},
{
"input": "3 100000 1 1000",
"output": "1001"
},
{
"input": "7 1000 1000 3",
"output": "9"
},
{
"input": "3 4 3 1",
"output": "3"
},
{
"input": "3 3000000 1 100000",
"output": "100001"
},
{
"input": "3 3 1 1",
"output": "2"
},
{
"input": "3 10 1 5",
"output": "6"
},
{
"input": "3 2000 2000 1",
"output": "3"
},
{
"input": "3 10000000 10000000 1",
"output": "3"
},
{
"input": "3 5 1 1",
"output": "2"
},
{
"input": "3 100 1 33",
"output": "34"
},
{
"input": "7 9 2 7",
"output": "9"
},
{
"input": "4448 2 3 6",
"output": "0"
},
{
"input": "2228 1 6 3",
"output": "0"
}
] | 1,486,672,320
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 11
| 61
| 4,608,000
|
import math
def main():
n, a, b, c = map(int, input().split())
n = (4 - (n%4)) % 4
if n == 0:
print(0)
elif n == 1:
print(a)
elif n == 2:
print(min(2*a, b))
elif n == 3:
print(min(3*a, c, a + b))
main()
|
Title: Alyona and copybooks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for *a* rubles, a pack of two copybooks for *b* rubles, and a pack of three copybooks for *c* rubles. Alyona already has *n* copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks *k* that *n*<=+<=*k* is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input Specification:
The only line contains 4 integers *n*, *a*, *b*, *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=109).
Output Specification:
Print the minimum amount of rubles she should pay to buy such number of copybooks *k* that *n*<=+<=*k* is divisible by 4.
Demo Input:
['1 1 3 4\n', '6 2 1 1\n', '4 4 4 4\n', '999999999 1000000000 1000000000 1000000000\n']
Demo Output:
['3\n', '1\n', '0\n', '1000000000\n']
Note:
In the first example Alyona can buy 3 packs of 1 copybook for 3*a* = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for *b* = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook.
|
```python
import math
def main():
n, a, b, c = map(int, input().split())
n = (4 - (n%4)) % 4
if n == 0:
print(0)
elif n == 1:
print(a)
elif n == 2:
print(min(2*a, b))
elif n == 3:
print(min(3*a, c, a + b))
main()
```
| 0
|
|
723
|
A
|
The New Year: Meeting Friends
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"sortings"
] | null | null |
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
|
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
|
Print one integer — the minimum total distance the friends need to travel in order to meet together.
|
[
"7 1 4\n",
"30 20 10\n"
] |
[
"6\n",
"20\n"
] |
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
| 500
|
[
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"input": "71 85 88",
"output": "17"
},
{
"input": "30 38 99",
"output": "69"
},
{
"input": "23 82 95",
"output": "72"
},
{
"input": "22 41 47",
"output": "25"
},
{
"input": "9 94 77",
"output": "85"
},
{
"input": "1 53 51",
"output": "52"
},
{
"input": "25 97 93",
"output": "72"
},
{
"input": "42 53 51",
"output": "11"
},
{
"input": "81 96 94",
"output": "15"
},
{
"input": "21 5 93",
"output": "88"
},
{
"input": "50 13 75",
"output": "62"
},
{
"input": "41 28 98",
"output": "70"
},
{
"input": "69 46 82",
"output": "36"
},
{
"input": "87 28 89",
"output": "61"
},
{
"input": "44 45 40",
"output": "5"
},
{
"input": "86 97 68",
"output": "29"
},
{
"input": "43 92 30",
"output": "62"
},
{
"input": "16 70 1",
"output": "69"
},
{
"input": "40 46 19",
"output": "27"
},
{
"input": "71 38 56",
"output": "33"
},
{
"input": "82 21 80",
"output": "61"
},
{
"input": "75 8 35",
"output": "67"
},
{
"input": "75 24 28",
"output": "51"
},
{
"input": "78 23 56",
"output": "55"
},
{
"input": "85 31 10",
"output": "75"
},
{
"input": "76 50 9",
"output": "67"
},
{
"input": "95 37 34",
"output": "61"
},
{
"input": "84 61 35",
"output": "49"
},
{
"input": "87 85 37",
"output": "50"
},
{
"input": "1 3 2",
"output": "2"
},
{
"input": "4 2 6",
"output": "4"
},
{
"input": "6 9 3",
"output": "6"
},
{
"input": "12 4 8",
"output": "8"
},
{
"input": "15 10 5",
"output": "10"
},
{
"input": "1 50 17",
"output": "49"
},
{
"input": "10 5 15",
"output": "10"
},
{
"input": "8 1 9",
"output": "8"
},
{
"input": "3 5 4",
"output": "2"
},
{
"input": "2 1 3",
"output": "2"
},
{
"input": "1 8 2",
"output": "7"
},
{
"input": "1 100 2",
"output": "99"
},
{
"input": "1 4 6",
"output": "5"
}
] | 1,674,138,062
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
a,b,c=map(int,input().split())
l=[]
l.append(a)
l.append(b)
l.append(c)
l.sort()
print(a[1]-a[0]+a[2]-a[1])
|
Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integer — the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
|
```python
a,b,c=map(int,input().split())
l=[]
l.append(a)
l.append(b)
l.append(c)
l.sort()
print(a[1]-a[0]+a[2]-a[1])
```
| -1
|
|
992
|
A
|
Nastya and an Array
|
PROGRAMMING
| 800
|
[
"implementation",
"sortings"
] | null | null |
Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array.
|
Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero.
|
[
"5\n1 1 1 1 1\n",
"3\n2 0 -1\n",
"4\n5 -6 -5 1\n"
] |
[
"1\n",
"2\n",
"4\n"
] |
In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element.
| 500
|
[
{
"input": "5\n1 1 1 1 1",
"output": "1"
},
{
"input": "3\n2 0 -1",
"output": "2"
},
{
"input": "4\n5 -6 -5 1",
"output": "4"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n21794 -79194",
"output": "2"
},
{
"input": "3\n-63526 95085 -5239",
"output": "3"
},
{
"input": "3\n0 53372 -20572",
"output": "2"
},
{
"input": "13\n-2075 -32242 27034 -37618 -96962 82203 64846 48249 -71761 28908 -21222 -61370 46899",
"output": "13"
},
{
"input": "5\n806 0 1308 1954 683",
"output": "4"
},
{
"input": "8\n-26 0 -249 -289 -126 -206 288 -11",
"output": "7"
},
{
"input": "10\n2 2 2 1 2 -1 0 2 -1 1",
"output": "3"
},
{
"input": "1\n8",
"output": "1"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "10"
},
{
"input": "5\n2 0 -1 0 0",
"output": "2"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "5\n0 0 0 0 0",
"output": "0"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "2\n-1 0",
"output": "1"
},
{
"input": "4\n0 0 0 0",
"output": "0"
},
{
"input": "8\n10 9 -1 0 0 3 2 3",
"output": "5"
},
{
"input": "5\n5 0 1 2 3",
"output": "4"
},
{
"input": "3\n1 1 0",
"output": "1"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "5\n1 2 0 0 0",
"output": "2"
},
{
"input": "5\n1 0 0 0 0",
"output": "1"
},
{
"input": "5\n4 5 6 0 0",
"output": "3"
},
{
"input": "4\n-1 0 0 1",
"output": "2"
},
{
"input": "5\n3 0 0 4 5",
"output": "3"
},
{
"input": "3\n0 0 2",
"output": "1"
},
{
"input": "3\n1 0 0",
"output": "1"
},
{
"input": "4\n0 0 0 4",
"output": "1"
},
{
"input": "5\n-1 0 0 0 0",
"output": "1"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "10\n0 0 0 0 0 1 2 3 0 0",
"output": "3"
},
{
"input": "4\n0 1 2 3",
"output": "3"
},
{
"input": "3\n0 1 2",
"output": "2"
},
{
"input": "4\n2 0 0 -1",
"output": "2"
}
] | 1,641,313,907
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 5
| 1,000
| 7,065,600
|
n=int(input())
q=[]
x=[int(x) for x in input().split()]
for i in range(n):
if x[i] not in q:
if x[i]!=0:
q.append(x[i])
print(len(q))
|
Title: Nastya and an Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array.
Output Specification:
Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero.
Demo Input:
['5\n1 1 1 1 1\n', '3\n2 0 -1\n', '4\n5 -6 -5 1\n']
Demo Output:
['1\n', '2\n', '4\n']
Note:
In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element.
|
```python
n=int(input())
q=[]
x=[int(x) for x in input().split()]
for i in range(n):
if x[i] not in q:
if x[i]!=0:
q.append(x[i])
print(len(q))
```
| 0
|
|
962
|
D
|
Merge Equals
|
PROGRAMMING
| 1,600
|
[
"data structures",
"implementation"
] | null | null |
You are given an array of positive integers. While there are at least two equal elements, we will perform the following operation. We choose the smallest value $x$ that occurs in the array $2$ or more times. Take the first two occurrences of $x$ in this array (the two leftmost occurrences). Remove the left of these two occurrences, and the right one is replaced by the sum of this two values (that is, $2 \cdot x$).
Determine how the array will look after described operations are performed.
For example, consider the given array looks like $[3, 4, 1, 2, 2, 1, 1]$. It will be changed in the following way: $[3, 4, 1, 2, 2, 1, 1]~\rightarrow~[3, 4, 2, 2, 2, 1]~\rightarrow~[3, 4, 4, 2, 1]~\rightarrow~[3, 8, 2, 1]$.
If the given array is look like $[1, 1, 3, 1, 1]$ it will be changed in the following way: $[1, 1, 3, 1, 1]~\rightarrow~[2, 3, 1, 1]~\rightarrow~[2, 3, 2]~\rightarrow~[3, 4]$.
|
The first line contains a single integer $n$ ($2 \le n \le 150\,000$) — the number of elements in the array.
The second line contains a sequence from $n$ elements $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^{9}$) — the elements of the array.
|
In the first line print an integer $k$ — the number of elements in the array after all the performed operations. In the second line print $k$ integers — the elements of the array after all the performed operations.
|
[
"7\n3 4 1 2 2 1 1\n",
"5\n1 1 3 1 1\n",
"5\n10 40 20 50 30\n"
] |
[
"4\n3 8 2 1 \n",
"2\n3 4 \n",
"5\n10 40 20 50 30 \n"
] |
The first two examples were considered in the statement.
In the third example all integers in the given array are distinct, so it will not change.
| 0
|
[
{
"input": "7\n3 4 1 2 2 1 1",
"output": "4\n3 8 2 1 "
},
{
"input": "5\n1 1 3 1 1",
"output": "2\n3 4 "
},
{
"input": "5\n10 40 20 50 30",
"output": "5\n10 40 20 50 30 "
},
{
"input": "100\n10 10 15 12 15 13 15 12 10 10 15 11 13 14 13 14 10 13 12 10 14 12 13 11 14 15 12 11 11 15 12 12 11 14 14 14 15 10 10 15 15 13 13 15 10 12 14 10 12 13 11 15 11 13 14 12 10 12 11 14 13 15 13 15 13 14 14 11 12 13 11 14 10 10 15 10 15 12 15 12 13 10 11 13 15 11 10 12 10 12 14 14 13 12 14 10 12 13 11 13",
"output": "12\n88 240 15 44 160 192 208 224 20 24 11 26 "
},
{
"input": "2\n1000000000 1000000000",
"output": "1\n2000000000 "
},
{
"input": "3\n500000000 500000000 1000000000",
"output": "1\n2000000000 "
},
{
"input": "9\n8 536870913 536870913 536870913 536870913 536870913 536870913 536870913 536870913",
"output": "2\n8 4294967304 "
},
{
"input": "34\n967614464 967614464 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "2\n1935228928 32000000000 "
}
] | 1,696,505,745
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
import heapq
n = int(input())
a = [int(i) for i in input().split()]
a = [[a[i],i] for i in range(n)]
c = []
heapq.heapify(a)
heapq.heapify(c)
while a:
x = heapq.heappop(a)
if not a:
heapq.heappush(c,x)
break
y = heapq.heappop(a)
if x[0]!=y[0]:
heapq.heappush(c,x)
heapq.heappush(a,y)
else:
y[0]*=2
heapq.heappush(a,y)
while c:
heapq.heappush(a,heapq.heappop(c))
c.sort(key = lambda x: x[1])
for i in c:
print(i[0], end = " ")
|
Title: Merge Equals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array of positive integers. While there are at least two equal elements, we will perform the following operation. We choose the smallest value $x$ that occurs in the array $2$ or more times. Take the first two occurrences of $x$ in this array (the two leftmost occurrences). Remove the left of these two occurrences, and the right one is replaced by the sum of this two values (that is, $2 \cdot x$).
Determine how the array will look after described operations are performed.
For example, consider the given array looks like $[3, 4, 1, 2, 2, 1, 1]$. It will be changed in the following way: $[3, 4, 1, 2, 2, 1, 1]~\rightarrow~[3, 4, 2, 2, 2, 1]~\rightarrow~[3, 4, 4, 2, 1]~\rightarrow~[3, 8, 2, 1]$.
If the given array is look like $[1, 1, 3, 1, 1]$ it will be changed in the following way: $[1, 1, 3, 1, 1]~\rightarrow~[2, 3, 1, 1]~\rightarrow~[2, 3, 2]~\rightarrow~[3, 4]$.
Input Specification:
The first line contains a single integer $n$ ($2 \le n \le 150\,000$) — the number of elements in the array.
The second line contains a sequence from $n$ elements $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^{9}$) — the elements of the array.
Output Specification:
In the first line print an integer $k$ — the number of elements in the array after all the performed operations. In the second line print $k$ integers — the elements of the array after all the performed operations.
Demo Input:
['7\n3 4 1 2 2 1 1\n', '5\n1 1 3 1 1\n', '5\n10 40 20 50 30\n']
Demo Output:
['4\n3 8 2 1 \n', '2\n3 4 \n', '5\n10 40 20 50 30 \n']
Note:
The first two examples were considered in the statement.
In the third example all integers in the given array are distinct, so it will not change.
|
```python
import heapq
n = int(input())
a = [int(i) for i in input().split()]
a = [[a[i],i] for i in range(n)]
c = []
heapq.heapify(a)
heapq.heapify(c)
while a:
x = heapq.heappop(a)
if not a:
heapq.heappush(c,x)
break
y = heapq.heappop(a)
if x[0]!=y[0]:
heapq.heappush(c,x)
heapq.heappush(a,y)
else:
y[0]*=2
heapq.heappush(a,y)
while c:
heapq.heappush(a,heapq.heappop(c))
c.sort(key = lambda x: x[1])
for i in c:
print(i[0], end = " ")
```
| 0
|
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,630,059,635
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 78
| 6,758,400
|
def chat_room():
s = input()
res = "hello"
j = 0
count = 0
for i in range(len(s)):
try:
if s[i] == res[j]:
j+=1
count += 1
if count == 5:
return "YES"
except:
pass
return "NO"
print(chat_room())
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
def chat_room():
s = input()
res = "hello"
j = 0
count = 0
for i in range(len(s)):
try:
if s[i] == res[j]:
j+=1
count += 1
if count == 5:
return "YES"
except:
pass
return "NO"
print(chat_room())
```
| 3.948411
|
903
|
C
|
Boxes Packing
|
PROGRAMMING
| 1,200
|
[
"greedy"
] | null | null |
Mishka has got *n* empty boxes. For every *i* (1<=≤<=*i*<=≤<=*n*), *i*-th box is a cube with side length *a**i*.
Mishka can put a box *i* into another box *j* if the following conditions are met:
- *i*-th box is not put into another box; - *j*-th box doesn't contain any other boxes; - box *i* is smaller than box *j* (*a**i*<=<<=*a**j*).
Mishka can put boxes into each other an arbitrary number of times. He wants to minimize the number of visible boxes. A box is called visible iff it is not put into some another box.
Help Mishka to determine the minimum possible number of visible boxes!
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=5000) — the number of boxes Mishka has got.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is the side length of *i*-th box.
|
Print the minimum possible number of visible boxes.
|
[
"3\n1 2 3\n",
"4\n4 2 4 3\n"
] |
[
"1\n",
"2\n"
] |
In the first example it is possible to put box 1 into box 2, and 2 into 3.
In the second example Mishka can put box 2 into box 3, and box 4 into box 1.
| 0
|
[
{
"input": "3\n1 2 3",
"output": "1"
},
{
"input": "4\n4 2 4 3",
"output": "2"
},
{
"input": "10\n58 58 58 58 58 58 58 58 58 58",
"output": "10"
},
{
"input": "10\n86 89 89 86 86 89 86 86 89 89",
"output": "5"
},
{
"input": "100\n981 288 186 186 292 876 341 288 981 360 783 907 292 186 341 292 360 876 360 360 981 398 783 288 292 398 876 981 398 907 783 360 288 981 907 186 360 288 186 981 186 288 907 876 288 907 876 360 341 292 907 783 907 783 292 981 907 292 876 398 783 876 398 341 876 186 288 186 981 341 398 360 907 981 341 186 292 981 292 398 876 783 292 186 360 292 288 292 876 398 288 292 341 288 398 360 360 292 981 360",
"output": "14"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n9",
"output": "1"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n131",
"output": "1"
},
{
"input": "9\n1 1 1 1 1 1 1 1 1",
"output": "9"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "11"
},
{
"input": "8\n1 2 3 4 5 6 7 8",
"output": "1"
},
{
"input": "8\n1 1 1 1 1 1 1 1",
"output": "8"
},
{
"input": "5\n1 1 1 1 1",
"output": "5"
}
] | 1,610,140,339
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 24
| 62
| 409,600
|
n = int(input())
a = input().split()
for i in range(0,n):
a[i] = int(a[i])
a.sort()
a.append("x")
y = 1
i = 0
while i < n-1:
occ = 1
while a[i] == a[i+1]:
occ += 1
i += 1
if occ > y:
y = occ
i += 1
print(y)
|
Title: Boxes Packing
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka has got *n* empty boxes. For every *i* (1<=≤<=*i*<=≤<=*n*), *i*-th box is a cube with side length *a**i*.
Mishka can put a box *i* into another box *j* if the following conditions are met:
- *i*-th box is not put into another box; - *j*-th box doesn't contain any other boxes; - box *i* is smaller than box *j* (*a**i*<=<<=*a**j*).
Mishka can put boxes into each other an arbitrary number of times. He wants to minimize the number of visible boxes. A box is called visible iff it is not put into some another box.
Help Mishka to determine the minimum possible number of visible boxes!
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=5000) — the number of boxes Mishka has got.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=109), where *a**i* is the side length of *i*-th box.
Output Specification:
Print the minimum possible number of visible boxes.
Demo Input:
['3\n1 2 3\n', '4\n4 2 4 3\n']
Demo Output:
['1\n', '2\n']
Note:
In the first example it is possible to put box 1 into box 2, and 2 into 3.
In the second example Mishka can put box 2 into box 3, and box 4 into box 1.
|
```python
n = int(input())
a = input().split()
for i in range(0,n):
a[i] = int(a[i])
a.sort()
a.append("x")
y = 1
i = 0
while i < n-1:
occ = 1
while a[i] == a[i+1]:
occ += 1
i += 1
if occ > y:
y = occ
i += 1
print(y)
```
| 3
|
|
34
|
B
|
Sale
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] |
B. Sale
|
2
|
256
|
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
|
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
|
[
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] |
[
"8\n",
"7\n"
] |
none
| 1,000
|
[
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "6"
},
{
"input": "5 1\n998 997 985 937 998",
"output": "0"
},
{
"input": "2 2\n-742 -187",
"output": "929"
},
{
"input": "3 3\n522 597 384",
"output": "0"
},
{
"input": "4 2\n-215 -620 192 647",
"output": "835"
},
{
"input": "10 6\n557 605 685 231 910 633 130 838 -564 -85",
"output": "649"
},
{
"input": "20 14\n932 442 960 943 624 624 955 998 631 910 850 517 715 123 1000 155 -10 961 966 59",
"output": "10"
},
{
"input": "30 5\n991 997 996 967 977 999 991 986 1000 965 984 997 998 1000 958 983 974 1000 991 999 1000 978 961 992 990 998 998 978 998 1000",
"output": "0"
},
{
"input": "50 20\n-815 -947 -946 -993 -992 -846 -884 -954 -963 -733 -940 -746 -766 -930 -821 -937 -937 -999 -914 -938 -936 -975 -939 -981 -977 -952 -925 -901 -952 -978 -994 -957 -946 -896 -905 -836 -994 -951 -887 -939 -859 -953 -985 -988 -946 -829 -956 -842 -799 -886",
"output": "19441"
},
{
"input": "88 64\n999 999 1000 1000 999 996 995 1000 1000 999 1000 997 998 1000 999 1000 997 1000 993 998 994 999 998 996 1000 997 1000 1000 1000 997 1000 998 997 1000 1000 998 1000 998 999 1000 996 999 999 999 996 995 999 1000 998 999 1000 999 999 1000 1000 1000 996 1000 1000 1000 997 1000 1000 997 999 1000 1000 1000 1000 1000 999 999 1000 1000 996 999 1000 1000 995 999 1000 996 1000 998 999 999 1000 999",
"output": "0"
},
{
"input": "99 17\n-993 -994 -959 -989 -991 -995 -976 -997 -990 -1000 -996 -994 -999 -995 -1000 -983 -979 -1000 -989 -968 -994 -992 -962 -993 -999 -983 -991 -979 -995 -993 -973 -999 -995 -995 -999 -993 -995 -992 -947 -1000 -999 -998 -982 -988 -979 -993 -963 -988 -980 -990 -979 -976 -995 -999 -981 -988 -998 -999 -970 -1000 -983 -994 -943 -975 -998 -977 -973 -997 -959 -999 -983 -985 -950 -977 -977 -991 -998 -973 -987 -985 -985 -986 -984 -994 -978 -998 -989 -989 -988 -970 -985 -974 -997 -981 -962 -972 -995 -988 -993",
"output": "16984"
},
{
"input": "100 37\n205 19 -501 404 912 -435 -322 -469 -655 880 -804 -470 793 312 -108 586 -642 -928 906 605 -353 -800 745 -440 -207 752 -50 -28 498 -800 -62 -195 602 -833 489 352 536 404 -775 23 145 -512 524 759 651 -461 -427 -557 684 -366 62 592 -563 -811 64 418 -881 -308 591 -318 -145 -261 -321 -216 -18 595 -202 960 -4 219 226 -238 -882 -963 425 970 -434 -160 243 -672 -4 873 8 -633 904 -298 -151 -377 -61 -72 -677 -66 197 -716 3 -870 -30 152 -469 981",
"output": "21743"
},
{
"input": "100 99\n-931 -806 -830 -828 -916 -962 -660 -867 -952 -966 -820 -906 -724 -982 -680 -717 -488 -741 -897 -613 -986 -797 -964 -939 -808 -932 -810 -860 -641 -916 -858 -628 -821 -929 -917 -976 -664 -985 -778 -665 -624 -928 -940 -958 -884 -757 -878 -896 -634 -526 -514 -873 -990 -919 -988 -878 -650 -973 -774 -783 -733 -648 -756 -895 -833 -974 -832 -725 -841 -748 -806 -613 -924 -867 -881 -943 -864 -991 -809 -926 -777 -817 -998 -682 -910 -996 -241 -722 -964 -904 -821 -920 -835 -699 -805 -632 -779 -317 -915 -654",
"output": "81283"
},
{
"input": "100 14\n995 994 745 684 510 737 984 690 979 977 542 933 871 603 758 653 962 997 747 974 773 766 975 770 527 960 841 989 963 865 974 967 950 984 757 685 986 809 982 959 931 880 978 867 805 562 970 900 834 782 616 885 910 608 974 918 576 700 871 980 656 941 978 759 767 840 573 859 841 928 693 853 716 927 976 851 962 962 627 797 707 873 869 988 993 533 665 887 962 880 929 980 877 887 572 790 721 883 848 782",
"output": "0"
},
{
"input": "100 84\n768 946 998 752 931 912 826 1000 991 910 875 962 901 952 958 733 959 908 872 840 923 826 952 980 974 980 947 955 959 822 997 963 966 933 829 923 971 999 926 932 865 984 974 858 994 855 949 941 992 861 951 949 991 711 763 728 935 485 716 907 869 952 960 859 909 963 978 942 968 933 923 909 997 962 687 764 924 774 875 1000 961 951 987 974 848 921 966 859 995 997 974 931 886 941 974 986 906 978 998 823",
"output": "0"
},
{
"input": "100 80\n-795 -994 -833 -930 -974 -980 -950 -940 -788 -927 -583 -956 -945 -949 -809 -974 -957 -736 -967 -908 -975 -961 -986 -983 -963 -771 -952 -847 -751 -741 -982 -959 -925 -931 -839 -937 -880 -914 -858 -998 -812 -911 -862 -965 -943 -984 -738 -920 -950 -998 -909 -998 -781 -901 -677 -940 -985 -951 -675 -952 -967 -949 -882 -641 -969 -937 -975 -993 -913 -941 -807 -851 -832 -960 -939 -943 -895 -929 -528 -880 -823 -930 -888 -862 -948 -966 -962 -857 -799 -969 -833 -998 -952 -878 -946 -971 -976 -974 -723 -992",
"output": "75068"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 1\n555",
"output": "0"
},
{
"input": "1 1\n-1",
"output": "1"
},
{
"input": "1 1\n-24",
"output": "24"
}
] | 1,583,959,112
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 248
| 0
|
n, m = map(int, input().split())
a = list(map(int, input().split()))
a = sorted(a)
suma = 0
for i in range(n):
if (a[i] < 0) and m:
suma += abs(a[i])
m -= 1
print(suma)
|
Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none
|
```python
n, m = map(int, input().split())
a = list(map(int, input().split()))
a = sorted(a)
suma = 0
for i in range(n):
if (a[i] < 0) and m:
suma += abs(a[i])
m -= 1
print(suma)
```
| 3.938
|
869
|
A
|
The Artful Expedient
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
Rock... Paper!
After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up with a new game as a substitute. The game works as follows.
A positive integer *n* is decided first. Both Koyomi and Karen independently choose *n* distinct positive integers, denoted by *x*1,<=*x*2,<=...,<=*x**n* and *y*1,<=*y*2,<=...,<=*y**n* respectively. They reveal their sequences, and repeat until all of 2*n* integers become distinct, which is the only final state to be kept and considered.
Then they count the number of ordered pairs (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) such that the value *x**i* xor *y**j* equals to one of the 2*n* integers. Here xor means the [bitwise exclusive or](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation on two integers, and is denoted by operators ^ and/or xor in most programming languages.
Karen claims a win if the number of such pairs is even, and Koyomi does otherwise. And you're here to help determine the winner of their latest game.
|
The first line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=2<=000) — the length of both sequences.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=2·106) — the integers finally chosen by Koyomi.
The third line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (1<=≤<=*y**i*<=≤<=2·106) — the integers finally chosen by Karen.
Input guarantees that the given 2*n* integers are pairwise distinct, that is, no pair (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) exists such that one of the following holds: *x**i*<==<=*y**j*; *i*<=≠<=*j* and *x**i*<==<=*x**j*; *i*<=≠<=*j* and *y**i*<==<=*y**j*.
|
Output one line — the name of the winner, that is, "Koyomi" or "Karen" (without quotes). Please be aware of the capitalization.
|
[
"3\n1 2 3\n4 5 6\n",
"5\n2 4 6 8 10\n9 7 5 3 1\n"
] |
[
"Karen\n",
"Karen\n"
] |
In the first example, there are 6 pairs satisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3). Thus, Karen wins since 6 is an even number.
In the second example, there are 16 such pairs, and Karen wins again.
| 500
|
[
{
"input": "3\n1 2 3\n4 5 6",
"output": "Karen"
},
{
"input": "5\n2 4 6 8 10\n9 7 5 3 1",
"output": "Karen"
},
{
"input": "1\n1\n2000000",
"output": "Karen"
},
{
"input": "2\n97153 2000000\n1999998 254",
"output": "Karen"
},
{
"input": "15\n31 30 29 28 27 26 25 24 23 22 21 20 19 18 17\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15",
"output": "Karen"
},
{
"input": "30\n79656 68607 871714 1858841 237684 1177337 532141 161161 1111201 527235 323345 1979059 665353 507265 1290761 610606 1238375 743262 106355 1167830 180315 1233029 816465 752968 782570 1499881 1328457 1867240 13948 1302782\n322597 1868510 1958236 1348157 765908 1023636 874300 537124 631783 414906 886318 1931572 1381013 992451 1305644 1525745 716087 83173 303248 1572710 43084 333341 992413 267806 70390 644521 1014900 497068 178940 1920268",
"output": "Karen"
},
{
"input": "30\n1143673 436496 1214486 1315862 148404 724601 1430740 1433008 1654610 1635673 614673 1713408 1270999 1697 1463796 50027 525482 1659078 688200 842647 518551 877506 1017082 1807856 3280 759698 1208220 470180 829800 1960886\n1312613 1965095 967255 1289012 1950383 582960 856825 49684 808824 319418 1968270 190821 344545 211332 1219388 1773751 1876402 132626 541448 1584672 24276 1053225 1823073 1858232 1209173 1035991 1956373 1237148 1973608 848873",
"output": "Karen"
},
{
"input": "1\n2\n3",
"output": "Karen"
},
{
"input": "1\n1048576\n1020000",
"output": "Karen"
},
{
"input": "3\n9 33 69\n71 74 100",
"output": "Karen"
},
{
"input": "3\n1 2 3\n9 5 6",
"output": "Karen"
},
{
"input": "3\n1 7 8\n9 10 20",
"output": "Karen"
},
{
"input": "3\n1 3 2\n4 5 8",
"output": "Karen"
},
{
"input": "3\n2 1 100\n3 4 9",
"output": "Karen"
},
{
"input": "3\n3 1 100\n2 1000 100000",
"output": "Karen"
},
{
"input": "3\n1 2 5\n3 4 6",
"output": "Karen"
},
{
"input": "3\n3 1 8\n2 4 17",
"output": "Karen"
},
{
"input": "3\n1 5 6\n7 8 3",
"output": "Karen"
},
{
"input": "1\n1\n3",
"output": "Karen"
},
{
"input": "3\n1 3 10\n2 4 20",
"output": "Karen"
},
{
"input": "3\n7 8 10\n15 9 11",
"output": "Karen"
},
{
"input": "3\n5 6 8\n3 100 9",
"output": "Karen"
},
{
"input": "3\n1 2 3\n4 5 8",
"output": "Karen"
},
{
"input": "3\n1 2 19\n3 7 30",
"output": "Karen"
},
{
"input": "3\n1 2 3\n6 7 8",
"output": "Karen"
},
{
"input": "3\n1 4 55\n2 3 9",
"output": "Karen"
},
{
"input": "3\n1 100 200\n5 4 500",
"output": "Karen"
},
{
"input": "1\n6\n7",
"output": "Karen"
},
{
"input": "3\n1 3 5\n2 4 13",
"output": "Karen"
},
{
"input": "3\n3 1 100\n2 1000 10000",
"output": "Karen"
},
{
"input": "3\n1 3 9\n2 4 40",
"output": "Karen"
},
{
"input": "2\n1 199999\n1935807 2000000",
"output": "Karen"
},
{
"input": "3\n1 3 8\n2 4 24",
"output": "Karen"
},
{
"input": "2\n4 1\n7 3",
"output": "Karen"
},
{
"input": "3\n1 2 4\n3 7 8",
"output": "Karen"
},
{
"input": "3\n1 6 10000\n2 3 100000",
"output": "Karen"
}
] | 1,605,441,221
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 49
| 171
| 2,150,400
|
import sys
n = int(sys.stdin.readline())
arr1 = list(map(int, sys.stdin.readline().split()))
arr2 = list(map(int, sys.stdin.readline().split()))
cnt = 0
arr = arr1 + arr2
for i in arr:
for j in arr:
temp = i ^ j
if temp == 2 * n:
cnt += 1
print('Karen' if cnt % 2 == 0 else 'Koyomi')
|
Title: The Artful Expedient
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Rock... Paper!
After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up with a new game as a substitute. The game works as follows.
A positive integer *n* is decided first. Both Koyomi and Karen independently choose *n* distinct positive integers, denoted by *x*1,<=*x*2,<=...,<=*x**n* and *y*1,<=*y*2,<=...,<=*y**n* respectively. They reveal their sequences, and repeat until all of 2*n* integers become distinct, which is the only final state to be kept and considered.
Then they count the number of ordered pairs (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) such that the value *x**i* xor *y**j* equals to one of the 2*n* integers. Here xor means the [bitwise exclusive or](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) operation on two integers, and is denoted by operators ^ and/or xor in most programming languages.
Karen claims a win if the number of such pairs is even, and Koyomi does otherwise. And you're here to help determine the winner of their latest game.
Input Specification:
The first line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=2<=000) — the length of both sequences.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=2·106) — the integers finally chosen by Koyomi.
The third line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (1<=≤<=*y**i*<=≤<=2·106) — the integers finally chosen by Karen.
Input guarantees that the given 2*n* integers are pairwise distinct, that is, no pair (*i*,<=*j*) (1<=≤<=*i*,<=*j*<=≤<=*n*) exists such that one of the following holds: *x**i*<==<=*y**j*; *i*<=≠<=*j* and *x**i*<==<=*x**j*; *i*<=≠<=*j* and *y**i*<==<=*y**j*.
Output Specification:
Output one line — the name of the winner, that is, "Koyomi" or "Karen" (without quotes). Please be aware of the capitalization.
Demo Input:
['3\n1 2 3\n4 5 6\n', '5\n2 4 6 8 10\n9 7 5 3 1\n']
Demo Output:
['Karen\n', 'Karen\n']
Note:
In the first example, there are 6 pairs satisfying the constraint: (1, 1), (1, 2), (2, 1), (2, 3), (3, 2) and (3, 3). Thus, Karen wins since 6 is an even number.
In the second example, there are 16 such pairs, and Karen wins again.
|
```python
import sys
n = int(sys.stdin.readline())
arr1 = list(map(int, sys.stdin.readline().split()))
arr2 = list(map(int, sys.stdin.readline().split()))
cnt = 0
arr = arr1 + arr2
for i in arr:
for j in arr:
temp = i ^ j
if temp == 2 * n:
cnt += 1
print('Karen' if cnt % 2 == 0 else 'Koyomi')
```
| 3
|
|
263
|
A
|
Beautiful Matrix
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
|
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
|
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
|
[
"0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n",
"0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n"
] |
[
"3\n",
"1\n"
] |
none
| 500
|
[
{
"input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1",
"output": "4"
},
{
"input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "3"
}
] | 1,694,337,085
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 92
| 0
|
t = 4
y = 3
u = 2
for z in range(3):
n, q, w, e, r = map(int, input().split())
if n == 1:
print(t)
else:
if q == 1:
print(y)
else:
if w == 1:
print(u)
else:
if e == 1:
print(y)
else:
if r == 1:
print(t)
t = t - 1
y = y - 1
u = u - 1
for z in range(2):
t = t + 1
y = y + 1
u = u + 1
n, q, w, e, r = map(int, input().split())
if n == 1:
print(t)
else:
if q == 1:
print(y)
else:
if w == 1:
print(u)
else:
if e == 1:
print(y)
else:
if r == 1:
print(t)
|
Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
Input Specification:
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
Output Specification:
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
Demo Input:
['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
t = 4
y = 3
u = 2
for z in range(3):
n, q, w, e, r = map(int, input().split())
if n == 1:
print(t)
else:
if q == 1:
print(y)
else:
if w == 1:
print(u)
else:
if e == 1:
print(y)
else:
if r == 1:
print(t)
t = t - 1
y = y - 1
u = u - 1
for z in range(2):
t = t + 1
y = y + 1
u = u + 1
n, q, w, e, r = map(int, input().split())
if n == 1:
print(t)
else:
if q == 1:
print(y)
else:
if w == 1:
print(u)
else:
if e == 1:
print(y)
else:
if r == 1:
print(t)
```
| 0
|
|
995
|
B
|
Suit and Tie
|
PROGRAMMING
| 1,400
|
[
"greedy",
"implementation",
"math"
] | null | null |
Allen is hosting a formal dinner party. $2n$ people come to the event in $n$ pairs (couples). After a night of fun, Allen wants to line everyone up for a final picture. The $2n$ people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this makes the picture more aesthetic.
Help Allen find the minimum number of swaps of adjacent positions he must perform to make it so that each couple occupies adjacent positions in the line.
|
The first line contains a single integer $n$ ($1 \le n \le 100$), the number of pairs of people.
The second line contains $2n$ integers $a_1, a_2, \dots, a_{2n}$. For each $i$ with $1 \le i \le n$, $i$ appears exactly twice. If $a_j = a_k = i$, that means that the $j$-th and $k$-th people in the line form a couple.
|
Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions.
|
[
"4\n1 1 2 3 3 2 4 4\n",
"3\n1 1 2 2 3 3\n",
"3\n3 1 2 3 1 2\n"
] |
[
"2\n",
"0\n",
"3\n"
] |
In the first sample case, we can transform $1 1 2 3 3 2 4 4 \rightarrow 1 1 2 3 2 3 4 4 \rightarrow 1 1 2 2 3 3 4 4$ in two steps. Note that the sequence $1 1 2 3 3 2 4 4 \rightarrow 1 1 3 2 3 2 4 4 \rightarrow 1 1 3 3 2 2 4 4$ also works in the same number of steps.
The second sample case already satisfies the constraints; therefore we need $0$ swaps.
| 750
|
[
{
"input": "4\n1 1 2 3 3 2 4 4",
"output": "2"
},
{
"input": "3\n1 1 2 2 3 3",
"output": "0"
},
{
"input": "3\n3 1 2 3 1 2",
"output": "3"
},
{
"input": "8\n7 6 2 1 4 3 3 7 2 6 5 1 8 5 8 4",
"output": "27"
},
{
"input": "2\n1 2 1 2",
"output": "1"
},
{
"input": "3\n1 2 3 3 1 2",
"output": "5"
},
{
"input": "38\n26 28 23 34 33 14 38 15 35 36 30 1 19 17 18 28 22 15 9 27 11 16 17 32 7 21 6 8 32 26 33 23 18 4 2 25 29 3 35 8 38 37 31 37 12 25 3 27 16 24 5 20 12 13 29 11 30 22 9 19 2 24 7 10 34 4 36 21 14 31 13 6 20 10 5 1",
"output": "744"
},
{
"input": "24\n21 21 22 5 8 5 15 11 13 16 17 9 3 18 15 1 12 12 7 2 22 19 20 19 23 14 8 24 4 23 16 17 9 10 1 6 4 2 7 3 18 11 24 10 13 6 20 14",
"output": "259"
},
{
"input": "1\n1 1",
"output": "0"
},
{
"input": "19\n15 19 18 8 12 2 11 7 5 2 1 1 9 9 3 3 16 6 15 17 13 18 4 14 5 8 10 12 6 11 17 13 14 16 19 7 4 10",
"output": "181"
},
{
"input": "8\n3 1 5 2 1 6 3 5 6 2 4 8 8 4 7 7",
"output": "13"
},
{
"input": "2\n2 1 1 2",
"output": "2"
},
{
"input": "81\n48 22 31 24 73 77 79 75 37 78 43 56 20 33 70 34 6 50 51 21 39 29 20 11 73 53 39 61 28 17 55 52 28 57 52 74 35 13 55 2 57 9 46 81 60 47 21 68 1 53 31 64 42 9 79 80 69 30 32 24 15 2 69 10 22 3 71 19 67 66 17 50 62 36 32 65 58 18 25 59 38 10 14 51 23 16 29 81 45 40 18 54 47 12 45 74 41 34 75 44 19 77 71 67 7 16 35 49 15 3 38 4 7 25 76 66 5 65 27 6 1 72 37 42 26 60 12 64 44 41 80 13 49 68 76 48 11 78 40 61 30 43 62 58 5 4 33 26 54 27 36 72 63 63 59 70 23 8 56 8 46 14",
"output": "3186"
},
{
"input": "84\n10 29 12 22 55 3 81 33 64 78 46 44 69 41 34 71 24 12 22 54 63 9 65 40 36 81 32 37 83 50 28 84 53 25 72 77 41 35 50 8 29 78 72 53 21 63 16 1 79 20 66 23 38 18 44 5 27 77 32 52 42 60 67 62 64 52 14 80 4 19 15 45 40 47 42 46 68 18 70 8 3 36 65 38 73 43 59 20 66 6 51 10 58 55 51 13 4 5 43 82 71 21 9 33 47 11 61 30 76 27 24 48 75 15 48 75 2 31 83 67 59 74 56 11 39 13 45 76 26 30 39 17 61 57 68 7 70 62 49 57 49 84 31 26 56 54 74 16 60 1 80 35 82 28 79 73 14 69 6 19 25 34 23 2 58 37 7 17",
"output": "3279"
},
{
"input": "4\n3 4 2 4 1 2 1 3",
"output": "8"
},
{
"input": "75\n28 28 42 3 39 39 73 73 75 75 30 30 21 9 57 41 26 70 15 15 65 65 24 24 4 4 62 62 17 17 29 29 37 37 18 18 1 1 8 8 63 63 49 49 5 5 59 59 19 19 34 34 48 48 10 10 14 42 22 22 38 38 50 50 60 60 64 35 47 31 72 72 41 52 46 46 20 20 21 9 7 7 36 36 2 2 6 6 70 26 69 69 16 16 61 61 66 66 33 33 44 44 11 11 23 23 40 40 12 12 64 35 56 56 27 27 53 53 3 14 43 43 31 47 68 68 13 13 74 74 67 67 71 71 45 45 57 52 32 32 25 25 58 58 55 55 51 51 54 54",
"output": "870"
},
{
"input": "35\n6 32 4 19 9 34 20 29 22 26 19 14 33 11 17 31 30 13 7 12 8 16 5 5 21 15 18 28 34 3 2 10 23 24 35 6 32 4 25 9 1 11 24 20 26 25 2 13 22 17 31 30 33 7 12 8 16 27 27 21 15 18 28 1 3 14 10 23 29 35",
"output": "673"
},
{
"input": "86\n33 6 22 8 54 43 57 85 70 41 20 17 35 12 66 25 45 78 67 55 50 19 31 75 77 29 58 78 34 15 40 48 14 82 6 37 44 53 62 23 56 22 34 18 71 83 21 80 47 38 3 42 60 9 73 49 84 7 76 30 5 4 11 28 69 16 26 10 59 48 64 46 32 68 24 63 79 36 13 1 27 61 39 74 2 51 51 2 74 39 61 27 1 13 36 79 86 24 68 32 46 64 63 59 10 26 16 69 28 11 4 5 30 76 7 84 49 73 9 60 42 3 38 47 80 21 83 72 18 52 65 56 23 62 53 44 37 81 82 14 86 40 15 52 72 58 29 77 85 31 19 50 55 67 71 45 25 66 12 35 17 20 41 70 75 57 43 54 8 65 81 33",
"output": "6194"
}
] | 1,650,765,689
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 61
| 0
|
n = int(input())
p = input().split(" ")
p = [int(p[i]) for i in range (2*n)]
c = 0
# slides the value at index ind to i+1, returns number of spots slid
def slide(i, ind):
temp = p[ind]
for j in range (ind-1, i, -1):
p[j+1] = p[j]
print(p)
p[i+1] = temp
return ind-i-1
for i in range (0, 2*n, 2):
ind = p.index(p[i], i+1)
c += slide(i, ind)
print(c)
|
Title: Suit and Tie
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Allen is hosting a formal dinner party. $2n$ people come to the event in $n$ pairs (couples). After a night of fun, Allen wants to line everyone up for a final picture. The $2n$ people line up, but Allen doesn't like the ordering. Allen prefers if each pair occupies adjacent positions in the line, as this makes the picture more aesthetic.
Help Allen find the minimum number of swaps of adjacent positions he must perform to make it so that each couple occupies adjacent positions in the line.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 100$), the number of pairs of people.
The second line contains $2n$ integers $a_1, a_2, \dots, a_{2n}$. For each $i$ with $1 \le i \le n$, $i$ appears exactly twice. If $a_j = a_k = i$, that means that the $j$-th and $k$-th people in the line form a couple.
Output Specification:
Output a single integer, representing the minimum number of adjacent swaps needed to line the people up so that each pair occupies adjacent positions.
Demo Input:
['4\n1 1 2 3 3 2 4 4\n', '3\n1 1 2 2 3 3\n', '3\n3 1 2 3 1 2\n']
Demo Output:
['2\n', '0\n', '3\n']
Note:
In the first sample case, we can transform $1 1 2 3 3 2 4 4 \rightarrow 1 1 2 3 2 3 4 4 \rightarrow 1 1 2 2 3 3 4 4$ in two steps. Note that the sequence $1 1 2 3 3 2 4 4 \rightarrow 1 1 3 2 3 2 4 4 \rightarrow 1 1 3 3 2 2 4 4$ also works in the same number of steps.
The second sample case already satisfies the constraints; therefore we need $0$ swaps.
|
```python
n = int(input())
p = input().split(" ")
p = [int(p[i]) for i in range (2*n)]
c = 0
# slides the value at index ind to i+1, returns number of spots slid
def slide(i, ind):
temp = p[ind]
for j in range (ind-1, i, -1):
p[j+1] = p[j]
print(p)
p[i+1] = temp
return ind-i-1
for i in range (0, 2*n, 2):
ind = p.index(p[i], i+1)
c += slide(i, ind)
print(c)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of *n* gemstones, the *i*-th of which has color *c**i*. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
|
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=500) — the number of gemstones.
The second line contains *n* space-separated integers, the *i*-th of which is *c**i* (1<=≤<=*c**i*<=≤<=*n*) — the color of the *i*-th gemstone in a line.
|
Print a single integer — the minimum number of seconds needed to destroy the entire line.
|
[
"3\n1 2 1\n",
"3\n1 2 3\n",
"7\n1 4 4 2 3 2 1\n"
] |
[
"1\n",
"3\n",
"2\n"
] |
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
| 0
|
[
{
"input": "3\n1 2 1",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "7\n1 4 4 2 3 2 1",
"output": "2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "8\n1 2 1 3 4 1 2 1",
"output": "2"
},
{
"input": "50\n5 7 5 10 7 9 1 9 10 2 8 3 5 7 3 10 2 3 7 6 2 7 1 2 2 2 4 7 3 5 8 3 4 4 1 6 7 10 5 4 8 1 9 5 5 3 4 4 8 3",
"output": "21"
},
{
"input": "50\n13 17 20 5 14 19 4 17 9 13 10 19 16 13 17 2 18 3 1 9 19 4 19 10 17 12 16 20 10 11 15 10 3 19 8 6 2 8 9 15 13 7 8 8 5 8 15 18 9 4",
"output": "28"
},
{
"input": "50\n22 19 14 22 20 11 16 28 23 15 3 23 6 16 30 15 15 10 24 28 19 19 22 30 28 1 27 12 12 14 17 30 17 26 21 26 27 1 11 23 9 30 18 19 17 29 11 20 29 24",
"output": "25"
},
{
"input": "50\n30 17 31 15 10 3 39 36 5 29 16 11 31 2 38 1 32 40 7 15 39 34 24 11 4 23 9 35 39 32 4 5 14 37 10 34 11 33 30 14 4 34 23 10 34 34 26 34 26 16",
"output": "36"
},
{
"input": "50\n19 25 46 17 1 41 50 19 7 1 43 8 19 38 42 32 38 22 8 5 5 31 29 35 43 12 23 48 40 29 30 9 46 3 39 24 36 36 32 22 21 29 43 33 36 49 48 22 47 37",
"output": "36"
},
{
"input": "6\n1 2 1 1 3 1",
"output": "2"
}
] | 1,689,171,604
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
print("_RANDOM_GUESS_1689171603.416308")# 1689171603.416328
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of *n* gemstones, the *i*-th of which has color *c**i*. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input Specification:
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=500) — the number of gemstones.
The second line contains *n* space-separated integers, the *i*-th of which is *c**i* (1<=≤<=*c**i*<=≤<=*n*) — the color of the *i*-th gemstone in a line.
Output Specification:
Print a single integer — the minimum number of seconds needed to destroy the entire line.
Demo Input:
['3\n1 2 1\n', '3\n1 2 3\n', '7\n1 4 4 2 3 2 1\n']
Demo Output:
['1\n', '3\n', '2\n']
Note:
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
|
```python
print("_RANDOM_GUESS_1689171603.416308")# 1689171603.416328
```
| 0
|
|
876
|
A
|
Trip For Meal
|
PROGRAMMING
| 900
|
[
"math"
] | null | null |
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is *a* meters, between Rabbit's and Eeyore's house is *b* meters, between Owl's and Eeyore's house is *c* meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal *n* times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal *n* times, traveling minimum possible distance. Help him to find this distance.
|
First line contains an integer *n* (1<=≤<=*n*<=≤<=100) — number of visits.
Second line contains an integer *a* (1<=≤<=*a*<=≤<=100) — distance between Rabbit's and Owl's houses.
Third line contains an integer *b* (1<=≤<=*b*<=≤<=100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer *c* (1<=≤<=*c*<=≤<=100) — distance between Owl's and Eeyore's houses.
|
Output one number — minimum distance in meters Winnie must go through to have a meal *n* times.
|
[
"3\n2\n3\n1\n",
"1\n2\n3\n5\n"
] |
[
"3\n",
"0\n"
] |
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
| 500
|
[
{
"input": "3\n2\n3\n1",
"output": "3"
},
{
"input": "1\n2\n3\n5",
"output": "0"
},
{
"input": "10\n1\n8\n3",
"output": "9"
},
{
"input": "7\n10\n5\n6",
"output": "30"
},
{
"input": "9\n9\n7\n5",
"output": "42"
},
{
"input": "9\n37\n85\n76",
"output": "296"
},
{
"input": "76\n46\n77\n11",
"output": "860"
},
{
"input": "80\n42\n1\n37",
"output": "79"
},
{
"input": "8\n80\n55\n1",
"output": "61"
},
{
"input": "10\n13\n72\n17",
"output": "117"
},
{
"input": "9\n24\n1\n63",
"output": "8"
},
{
"input": "65\n5\n8\n7",
"output": "320"
},
{
"input": "56\n8\n9\n3",
"output": "170"
},
{
"input": "59\n8\n1\n2",
"output": "58"
},
{
"input": "75\n50\n50\n5",
"output": "415"
},
{
"input": "75\n54\n76\n66",
"output": "3996"
},
{
"input": "73\n71\n69\n66",
"output": "4755"
},
{
"input": "83\n58\n88\n16",
"output": "1354"
},
{
"input": "74\n31\n11\n79",
"output": "803"
},
{
"input": "62\n27\n16\n72",
"output": "976"
},
{
"input": "72\n95\n27\n9",
"output": "657"
},
{
"input": "1\n2\n2\n1",
"output": "0"
},
{
"input": "1\n1\n1\n1",
"output": "0"
},
{
"input": "1\n1\n1\n99",
"output": "0"
},
{
"input": "100\n100\n100\n100",
"output": "9900"
},
{
"input": "2\n1\n1\n3",
"output": "1"
},
{
"input": "1\n3\n2\n1",
"output": "0"
},
{
"input": "1\n5\n6\n1",
"output": "0"
},
{
"input": "1\n2\n6\n1",
"output": "0"
},
{
"input": "1\n30\n20\n1",
"output": "0"
},
{
"input": "1\n15\n13\n11",
"output": "0"
},
{
"input": "1\n100\n50\n1",
"output": "0"
},
{
"input": "1\n2\n3\n1",
"output": "0"
},
{
"input": "5\n5\n5\n5",
"output": "20"
},
{
"input": "1\n100\n50\n5",
"output": "0"
},
{
"input": "1\n5\n7\n3",
"output": "0"
},
{
"input": "1\n3\n4\n2",
"output": "0"
},
{
"input": "3\n3\n3\n3",
"output": "6"
},
{
"input": "1\n99\n98\n1",
"output": "0"
},
{
"input": "5\n3\n3\n1",
"output": "6"
},
{
"input": "2\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1\n7\n8",
"output": "1"
},
{
"input": "2\n2\n2\n1",
"output": "2"
},
{
"input": "2\n2\n3\n1",
"output": "2"
},
{
"input": "1\n5\n6\n2",
"output": "0"
},
{
"input": "100\n1\n1\n100",
"output": "99"
},
{
"input": "2\n3\n2\n1",
"output": "2"
},
{
"input": "1\n99\n98\n97",
"output": "0"
}
] | 1,508,408,815
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 78
| 5,939,200
|
import math
import re
def ria():
return [int(i) for i in input().split()]
def ri():
return int(input())
def rfa():
return [float(i) for i in input().split()]
n = ria()[0]
a = ria()[0]
b = ria()[0]
c = ria()[0]
n -= 1
if n == 0:
print(0)
exit()
print(min(a + c * (n - 1),b + c * (n - 1),a*n,b*n))
|
Title: Trip For Meal
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is *a* meters, between Rabbit's and Eeyore's house is *b* meters, between Owl's and Eeyore's house is *c* meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal *n* times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal *n* times, traveling minimum possible distance. Help him to find this distance.
Input Specification:
First line contains an integer *n* (1<=≤<=*n*<=≤<=100) — number of visits.
Second line contains an integer *a* (1<=≤<=*a*<=≤<=100) — distance between Rabbit's and Owl's houses.
Third line contains an integer *b* (1<=≤<=*b*<=≤<=100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer *c* (1<=≤<=*c*<=≤<=100) — distance between Owl's and Eeyore's houses.
Output Specification:
Output one number — minimum distance in meters Winnie must go through to have a meal *n* times.
Demo Input:
['3\n2\n3\n1\n', '1\n2\n3\n5\n']
Demo Output:
['3\n', '0\n']
Note:
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
|
```python
import math
import re
def ria():
return [int(i) for i in input().split()]
def ri():
return int(input())
def rfa():
return [float(i) for i in input().split()]
n = ria()[0]
a = ria()[0]
b = ria()[0]
c = ria()[0]
n -= 1
if n == 0:
print(0)
exit()
print(min(a + c * (n - 1),b + c * (n - 1),a*n,b*n))
```
| 3
|
|
645
|
C
|
Enduring Exodus
|
PROGRAMMING
| 1,600
|
[
"binary search",
"two pointers"
] | null | null |
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his *k* cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of *n* rooms located in a row, some of which are occupied.
Farmer John wants to book a set of *k*<=+<=1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms *i* and *j* is defined as |*j*<=-<=*i*|. Help Farmer John protect his cows by calculating this minimum possible distance.
|
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=<<=*n*<=≤<=100<=000) — the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length *n* describing the rooms. The *i*-th character of the string will be '0' if the *i*-th room is free, and '1' if the *i*-th room is occupied. It is guaranteed that at least *k*<=+<=1 characters of this string are '0', so there exists at least one possible choice of *k*<=+<=1 rooms for Farmer John and his cows to stay in.
|
Print the minimum possible distance between Farmer John's room and his farthest cow.
|
[
"7 2\n0100100\n",
"5 1\n01010\n",
"3 2\n000\n"
] |
[
"2\n",
"2\n",
"1\n"
] |
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1.
| 1,500
|
[
{
"input": "7 2\n0100100",
"output": "2"
},
{
"input": "5 1\n01010",
"output": "2"
},
{
"input": "3 2\n000",
"output": "1"
},
{
"input": "10 1\n1101111101",
"output": "6"
},
{
"input": "2 1\n00",
"output": "1"
},
{
"input": "3 1\n010",
"output": "2"
},
{
"input": "8 7\n00000000",
"output": "4"
},
{
"input": "7 6\n0000000",
"output": "3"
},
{
"input": "112 12\n0110101000000010101110010111100101011010011110100111111100011101011111000111101101110100111011110001100110110010",
"output": "10"
},
{
"input": "9 8\n000000000",
"output": "4"
},
{
"input": "9 3\n010001000",
"output": "2"
},
{
"input": "5 3\n00000",
"output": "2"
},
{
"input": "8 7\n00000000",
"output": "4"
},
{
"input": "6 1\n000011",
"output": "1"
},
{
"input": "100 40\n0010010100000100011100010100110001101100110000110010000000001010000111100000100100100101010010001100",
"output": "30"
},
{
"input": "93 79\n000000000000000000011000000000000000000000000000000000000000000000010000000000100000100000000",
"output": "42"
},
{
"input": "31 11\n0000001011011100010000000110001",
"output": "7"
},
{
"input": "47 46\n00000000000000000000000000000000000000000000000",
"output": "23"
},
{
"input": "100 96\n0000000000000010000010000000000000000000000000000000000000000000000000000010000000000000000000000000",
"output": "50"
},
{
"input": "491 89\n01111101111111100000111010110001010001110111000010101111101000100010010111011101110110111101101010111000111000011100011010010010111111000011011010100110001000011100111000001011100010001111101111101000111001100110010100101000001110010100100100100101001100010101001000010000111110011000000100000100101000100101000001001101011011100000110101111110101001001000100110010000010110101011000101011001001011001000110000011111001110101011000000110101000000100110001101111000101001001001100001001110101",
"output": "73"
},
{
"input": "308 17\n01000000100000000000000001000001000010000000000000000001001110000001010001000110000000000000100101000000010000001000000000001100000110000000000000000001000000000000000100000001000010001000000001000000000000000100010000000000000000000000000000000000001000000000001001101100000000000010000000000000000000000000",
"output": "9"
},
{
"input": "8 4\n00111000",
"output": "5"
},
{
"input": "18 2\n010111110111011110",
"output": "5"
},
{
"input": "29 3\n01110011111111111111110110110",
"output": "17"
}
] | 1,691,544,707
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 6
| 46
| 0
|
n,k = list(map(int,input().split()))
rooms = input()
R = []
for i in range(len(rooms)):
if rooms[i] == "0":
R.append(i)
l = 0
m = 1
r = k
minDist = 100000
while r <= len(R) - 1:
mid = (R[l] + R[r]) / 2
if R[m] >= mid:
longDist = R[m] - R[l]
else:
longDist = R[r] - R[m]
if longDist < minDist:
minDist = longDist
l += 1
m += 1
r += 1
print(minDist)
|
Title: Enduring Exodus
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In an attempt to escape the Mischievous Mess Makers' antics, Farmer John has abandoned his farm and is traveling to the other side of Bovinia. During the journey, he and his *k* cows have decided to stay at the luxurious Grand Moo-dapest Hotel. The hotel consists of *n* rooms located in a row, some of which are occupied.
Farmer John wants to book a set of *k*<=+<=1 currently unoccupied rooms for him and his cows. He wants his cows to stay as safe as possible, so he wishes to minimize the maximum distance from his room to the room of his cow. The distance between rooms *i* and *j* is defined as |*j*<=-<=*i*|. Help Farmer John protect his cows by calculating this minimum possible distance.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=<<=*n*<=≤<=100<=000) — the number of rooms in the hotel and the number of cows travelling with Farmer John.
The second line contains a string of length *n* describing the rooms. The *i*-th character of the string will be '0' if the *i*-th room is free, and '1' if the *i*-th room is occupied. It is guaranteed that at least *k*<=+<=1 characters of this string are '0', so there exists at least one possible choice of *k*<=+<=1 rooms for Farmer John and his cows to stay in.
Output Specification:
Print the minimum possible distance between Farmer John's room and his farthest cow.
Demo Input:
['7 2\n0100100\n', '5 1\n01010\n', '3 2\n000\n']
Demo Output:
['2\n', '2\n', '1\n']
Note:
In the first sample, Farmer John can book room 3 for himself, and rooms 1 and 4 for his cows. The distance to the farthest cow is 2. Note that it is impossible to make this distance 1, as there is no block of three consecutive unoccupied rooms.
In the second sample, Farmer John can book room 1 for himself and room 3 for his single cow. The distance between him and his cow is 2.
In the third sample, Farmer John books all three available rooms, taking the middle room for himself so that both cows are next to him. His distance from the farthest cow is 1.
|
```python
n,k = list(map(int,input().split()))
rooms = input()
R = []
for i in range(len(rooms)):
if rooms[i] == "0":
R.append(i)
l = 0
m = 1
r = k
minDist = 100000
while r <= len(R) - 1:
mid = (R[l] + R[r]) / 2
if R[m] >= mid:
longDist = R[m] - R[l]
else:
longDist = R[r] - R[m]
if longDist < minDist:
minDist = longDist
l += 1
m += 1
r += 1
print(minDist)
```
| 0
|
|
523
|
C
|
Name Quest
|
PROGRAMMING
| 1,400
|
[
"*special",
"greedy"
] | null | null |
A Martian boy is named *s* — he has got this name quite recently from his parents for his coming of age birthday. Now he enjoys looking for his name everywhere. If he sees that he can obtain his name from some string by removing zero or more letters (at that, the remaining letters remain in the same order), he gets happy. For example, if *s*=«aba», then strings «baobab», «aabbaa», «helloabahello» make him very happy and strings «aab», «baaa» and «helloabhello» do not.
However rather than being happy once, he loves twice as much being happy twice! So, when he got string *t* as a present, he wanted to cut it in two parts (the left part and the right part) so that each part made him happy.
Help *s* determine the number of distinct ways to cut the given string *t* into two parts in the required manner.
|
The first line contains string *s*, consisting of lowercase English letters. The length of string *s* is from 1 to 1000 letters.
The second line contains string *t*, that also consists of lowercase English letters. The length of string *t* is from 1 to 106 letters.
|
Print the sought number of ways to cut string *t* in two so that each part made *s* happy.
|
[
"aba\nbaobababbah\n",
"mars\nsunvenusearthmarsjupitersaturnuranusneptune\n"
] |
[
"2\n",
"0\n"
] |
none
| 1,500
|
[
{
"input": "aba\nbaobababbah",
"output": "2"
},
{
"input": "mars\nsunvenusearthmarsjupitersaturnuranusneptune",
"output": "0"
},
{
"input": "rry\nsorrymercuryismissedabove",
"output": "3"
},
{
"input": "a\naaaaaaaaaaaaaa",
"output": "13"
},
{
"input": "abcaba\nabcabaabcabaabcabaabcaba",
"output": "13"
},
{
"input": "lol\nlol",
"output": "0"
},
{
"input": "aaaaaaaaaaaa\naaaaaa",
"output": "0"
},
{
"input": "a\na",
"output": "0"
}
] | 1,426,362,235
| 17,035
|
Python 3
|
MEMORY_LIMIT_EXCEEDED
|
PRETESTS
| 6
| 265
| 268,390,400
|
s = str(input())
t = str(input())
def get_lowest(start, l):
for el in l:
if el > start:
return el
return False
def find_name(v, s):
letters = []
for l in s:
letter = []
for x in range(len(v)):
if l == v[x]:
letter.append(x)
letters.append(letter)
return letters
def check_letters(letters):
ind = ''
tmp = -1
for l in letters:
low = get_lowest(tmp, l)
tmp = low
ind += str(low)
return ind
variations = []
for x in range(len(s), len(t) - len(s) + 1):
variations.append([t[:x], t[x:]])
e = 0
for variation in variations:
c = 0
for v in variation:
if 'False' not in check_letters(find_name(v, s)):
c += 1
if c == 2:
e += 1
print(e)
|
Title: Name Quest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A Martian boy is named *s* — he has got this name quite recently from his parents for his coming of age birthday. Now he enjoys looking for his name everywhere. If he sees that he can obtain his name from some string by removing zero or more letters (at that, the remaining letters remain in the same order), he gets happy. For example, if *s*=«aba», then strings «baobab», «aabbaa», «helloabahello» make him very happy and strings «aab», «baaa» and «helloabhello» do not.
However rather than being happy once, he loves twice as much being happy twice! So, when he got string *t* as a present, he wanted to cut it in two parts (the left part and the right part) so that each part made him happy.
Help *s* determine the number of distinct ways to cut the given string *t* into two parts in the required manner.
Input Specification:
The first line contains string *s*, consisting of lowercase English letters. The length of string *s* is from 1 to 1000 letters.
The second line contains string *t*, that also consists of lowercase English letters. The length of string *t* is from 1 to 106 letters.
Output Specification:
Print the sought number of ways to cut string *t* in two so that each part made *s* happy.
Demo Input:
['aba\nbaobababbah\n', 'mars\nsunvenusearthmarsjupitersaturnuranusneptune\n']
Demo Output:
['2\n', '0\n']
Note:
none
|
```python
s = str(input())
t = str(input())
def get_lowest(start, l):
for el in l:
if el > start:
return el
return False
def find_name(v, s):
letters = []
for l in s:
letter = []
for x in range(len(v)):
if l == v[x]:
letter.append(x)
letters.append(letter)
return letters
def check_letters(letters):
ind = ''
tmp = -1
for l in letters:
low = get_lowest(tmp, l)
tmp = low
ind += str(low)
return ind
variations = []
for x in range(len(s), len(t) - len(s) + 1):
variations.append([t[:x], t[x:]])
e = 0
for variation in variations:
c = 0
for v in variation:
if 'False' not in check_letters(find_name(v, s)):
c += 1
if c == 2:
e += 1
print(e)
```
| 0
|
|
369
|
B
|
Valera and Contest
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"implementation",
"math"
] | null | null |
Valera loves to participate in competitions. Especially in programming contests. Today he has participated in the contest with his team, consisting of *n* students (including Valera). This contest was an individual competition, so each student in the team solved problems individually.
After the contest was over, Valera was interested in results. He found out that:
- each student in the team scored at least *l* points and at most *r* points; - in total, all members of the team scored exactly *s**all* points; - the total score of the *k* members of the team who scored the most points is equal to exactly *s**k*; more formally, if *a*1,<=*a*2,<=...,<=*a**n* is the sequence of points earned by the team of students in the non-increasing order (*a*1<=≥<=*a*2<=≥<=...<=≥<=*a**n*), then *s**k*<==<=*a*1<=+<=*a*2<=+<=...<=+<=*a**k*.
However, Valera did not find out exactly how many points each of *n* students scored. Valera asked you to recover any distribution of scores between the students of the team, such that all the conditions above are met.
|
The first line of the input contains exactly six integers *n*,<=*k*,<=*l*,<=*r*,<=*s**all*,<=*s**k* (1<=≤<=*n*,<=*k*,<=*l*,<=*r*<=≤<=1000; *l*<=≤<=*r*; *k*<=≤<=*n*; 1<=≤<=*s**k*<=≤<=*s**all*<=≤<=106).
It's guaranteed that the input is such that the answer exists.
|
Print exactly *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the number of points each student scored. If there are multiple solutions, you can print any of them. You can print the distribution of points in any order.
|
[
"5 3 1 3 13 9\n",
"5 3 1 3 15 9\n"
] |
[
"2 3 2 3 3 ",
"3 3 3 3 3 "
] |
none
| 1,000
|
[
{
"input": "5 3 1 3 13 9",
"output": "2 3 2 3 3 "
},
{
"input": "5 3 1 3 15 9",
"output": "3 3 3 3 3 "
},
{
"input": "50 25 1 1 50 25",
"output": "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 "
},
{
"input": "1000 700 782 1000 892330 648282",
"output": "926 926 926 814 926 926 926 926 926 813 926 814 926 813 813 926 814 926 926 926 814 926 814 926 813 927 814 926 813 926 926 813 926 926 926 927 926 813 926 813 926 926 926 926 813 926 926 926 813 813 926 926 814 926 926 926 814 926 813 927 926 926 927 926 926 926 926 926 926 926 927 813 926 814 926 926 926 926 813 813 814 926 927 814 926 926 813 926 813 926 926 814 926 926 926 926 926 926 814 926 926 927 926 926 926 926 926 814 926 926 813 926 926 926 926 927 813 926 926 927 926 926 926 926 926 926 926 926..."
},
{
"input": "1000 999 500 503 501513 501013",
"output": "501 502 502 501 501 502 502 502 501 501 502 501 502 501 501 501 501 502 502 502 501 502 501 502 501 502 501 502 501 501 502 501 501 502 502 502 501 501 502 501 502 501 502 502 501 501 502 502 501 501 502 502 501 502 502 501 501 502 501 502 501 502 502 502 502 502 502 501 502 502 502 501 502 501 502 502 501 502 501 501 501 501 502 501 502 502 501 502 501 501 502 501 502 502 501 502 502 501 501 502 502 502 501 501 502 502 502 501 502 502 501 501 501 501 502 502 500 501 502 502 502 502 502 502 501 502 501 502..."
},
{
"input": "999 998 500 501 500009 499509",
"output": "500 501 501 500 500 501 501 501 500 500 501 500 501 500 500 500 500 501 501 501 500 501 500 501 500 501 500 501 500 500 501 500 500 501 501 501 500 500 501 500 501 500 501 501 500 500 501 501 500 500 501 501 500 501 501 500 500 501 500 501 500 501 501 501 501 501 501 500 501 501 501 500 501 500 501 501 500 501 500 500 500 500 501 500 501 501 500 501 500 500 501 500 501 501 500 501 501 500 500 501 501 501 500 500 501 501 501 500 501 501 500 500 500 500 501 501 501 500 501 501 501 501 501 501 500 501 500 501..."
},
{
"input": "999 998 500 500 499500 499000",
"output": "500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500..."
},
{
"input": "999 997 500 502 500516 499516",
"output": "501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 502 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 501 502 501 501 501 501 501 501 501 501 500 501 501 501 501 501 502 501 501 501 501 501 501 501 501 501 501 501..."
},
{
"input": "1000 300 50 500 269795 127658",
"output": "203 204 203 203 203 203 203 426 203 203 204 203 425 203 203 203 203 204 203 203 203 425 203 204 203 426 203 203 203 203 203 203 203 426 203 426 203 203 426 203 203 203 203 203 203 203 203 204 203 203 425 203 203 425 425 203 203 425 203 426 203 204 426 426 425 426 203 203 425 203 426 203 425 203 425 425 203 425 203 203 203 203 426 203 425 203 203 425 203 203 203 203 203 203 203 203 426 203 203 203 425 426 203 203 203 203 203 203 425 425 203 203 203 203 203 426 203 203 203 426 204 203 203 203 203 203 203 426..."
},
{
"input": "50 25 1000 1000 50000 25000",
"output": "1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 "
},
{
"input": "100 100 500 1000 75589 75589",
"output": "756 756 756 755 755 756 756 756 756 755 756 756 756 756 756 756 755 756 755 756 756 756 756 756 756 756 756 756 756 755 756 756 756 756 756 756 756 756 756 756 756 756 756 756 756 756 755 756 756 756 756 756 756 756 756 756 756 756 756 756 756 756 756 756 756 755 756 756 756 756 756 755 756 756 756 755 756 756 756 756 756 756 756 756 756 756 756 756 756 756 756 756 755 756 756 756 756 756 756 756 "
},
{
"input": "1 1 1000 1000 1000 1000",
"output": "1000 "
},
{
"input": "2 2 500 1000 1000 1000",
"output": "500 500 "
},
{
"input": "1000 500 1 1000 500500 500000",
"output": "1 1000 1000 1 1 1000 1000 1000 1 1 1000 1 1000 1 1 1 1 1000 1000 1000 1 1000 1 1000 1 1000 1 1000 1 1 1000 1 1 1000 1000 1000 1 1 1000 1 1000 1 1000 1000 1 1 1 1000 1 1 1000 1000 1 1000 1000 1 1 1000 1 1000 1 1000 1000 1000 1000 1000 1000 1 1000 1000 1000 1 1000 1 1000 1000 1 1000 1 1 1 1 1000 1 1000 1000 1 1000 1 1 1000 1 1000 1000 1 1000 1000 1 1 1000 1000 1000 1 1 1000 1000 1000 1 1000 1000 1 1 1 1 1000 1000 1 1 1000 1000 1000 1000 1000 1000 1 1000 1 1000 1 1 1 1 1000 1 1 1000 1000 1000 1000 1000 1 1 1 ..."
},
{
"input": "1000 500 500 1000 750000 375000",
"output": "750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750 750..."
},
{
"input": "300 100 1 3 600 200",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "300 100 1 3 900 300",
"output": "3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3..."
},
{
"input": "300 100 1 3 300 100",
"output": "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 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 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..."
},
{
"input": "1 1 1 1 1 1",
"output": "1 "
},
{
"input": "432 32 13 123 28942 3824",
"output": "63 63 63 63 63 63 62 63 63 63 63 63 63 63 62 63 63 63 62 62 63 63 62 63 62 120 62 62 63 63 62 63 63 63 63 63 119 120 63 63 63 62 63 62 63 63 63 63 63 63 63 63 63 63 63 120 63 63 63 63 63 63 63 63 63 63 63 63 63 62 63 63 63 63 63 63 62 63 63 63 63 63 63 63 63 62 63 63 63 63 63 63 63 62 63 63 63 63 63 63 63 119 63 62 62 62 63 63 63 63 63 120 63 63 62 63 120 63 62 63 63 62 62 63 63 62 63 63 63 120 63 120 63 63 63 62 63 63 63 63 62 63 62 62 63 63 63 62 63 63 63 63 63 63 63 62 63 62 63 63 63 62 63 119 62 63 62 ..."
},
{
"input": "504 32 13 123 33704 3791",
"output": "64 63 63 63 63 63 63 64 63 63 63 64 63 63 63 64 63 63 63 63 64 63 63 63 63 119 63 63 64 64 63 63 63 64 63 64 118 119 64 64 63 63 63 63 63 64 63 63 64 64 64 63 63 63 64 118 63 64 63 64 63 63 64 64 63 64 63 64 64 63 64 64 64 64 63 63 63 64 63 64 64 64 64 64 63 63 63 64 64 64 63 63 63 63 64 63 64 63 63 63 63 118 63 63 63 63 63 63 64 63 63 119 63 63 63 64 119 63 63 64 63 63 63 63 64 63 64 64 64 119 63 119 64 64 63 63 63 64 63 63 63 64 63 63 63 63 64 63 64 63 64 64 63 64 63 63 63 63 64 64 64 63 64 118 63 63 63 ..."
},
{
"input": "999 32 13 123 68122 3876",
"output": "66 67 67 66 66 67 67 67 66 66 67 66 67 66 66 66 66 67 67 66 66 67 66 67 66 121 66 66 66 66 67 66 66 67 67 67 66 66 67 66 66 66 66 67 66 66 66 67 66 66 67 67 66 67 67 66 66 67 66 67 66 67 67 67 67 67 67 66 67 67 67 66 67 66 67 67 66 67 66 66 66 66 67 66 67 67 66 67 66 66 66 66 66 66 66 67 67 66 66 66 67 121 66 66 67 67 67 66 67 67 66 66 66 66 67 67 122 66 67 67 67 67 67 66 66 67 66 67 66 66 66 66 67 66 66 67 67 67 67 67 66 66 66 67 67 66 67 67 66 66 66 67 66 67 66 66 66 66 67 67 66 66 66 121 67 66 66 67 66 ..."
},
{
"input": "489 32 13 123 33009 3885",
"output": "64 64 64 64 63 64 63 64 64 64 64 64 64 64 63 64 64 64 63 63 64 64 63 64 63 122 63 63 64 64 63 64 64 64 63 64 121 122 64 64 63 63 63 63 64 64 64 64 64 64 64 64 64 64 64 121 63 64 64 64 64 64 64 64 64 64 63 64 64 63 64 64 64 64 64 64 63 64 63 64 64 64 64 64 64 63 64 64 64 64 63 63 63 63 64 63 64 64 64 63 64 121 64 63 63 64 63 64 64 64 64 122 64 64 63 64 122 64 64 64 64 63 64 64 64 64 64 64 64 122 64 122 64 64 64 63 64 64 64 64 63 64 63 64 64 64 64 63 64 63 64 64 64 64 64 63 64 63 64 64 64 64 64 121 63 64 63 ..."
},
{
"input": "234 32 13 123 16337 3715",
"output": "62 63 62 62 63 62 62 62 62 63 62 63 63 62 63 62 63 63 63 116 62 63 63 63 63 116 62 63 62 63 63 63 62 62 63 63 116 116 62 63 63 63 62 62 116 62 63 116 62 62 62 63 116 62 62 116 63 62 62 63 63 63 63 63 63 63 62 62 62 116 63 62 62 62 62 62 62 62 63 63 62 63 63 63 63 62 62 62 62 62 63 62 63 62 62 63 63 63 63 116 63 116 63 63 116 62 63 63 62 62 62 116 116 63 62 63 117 63 62 63 63 62 63 62 62 116 62 63 63 117 62 116 62 63 62 63 62 62 116 63 62 63 63 62 62 62 63 63 63 62 63 63 62 62 62 62 63 63 63 62 63 62 63 116..."
},
{
"input": "998 997 13 13 12974 12961",
"output": "13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 1..."
},
{
"input": "999 999 13 333 169609 169609",
"output": "170 170 170 170 170 170 170 170 170 169 170 170 170 169 169 170 169 170 170 170 170 170 169 170 169 170 169 170 169 170 170 169 170 170 170 170 170 169 170 169 170 170 170 170 169 170 170 170 169 169 170 170 169 170 170 170 170 170 169 170 170 170 170 170 170 170 170 170 170 170 170 169 170 170 170 170 170 170 169 169 170 170 170 170 170 170 169 170 169 170 170 169 170 170 170 170 170 170 169 170 170 170 170 170 170 170 170 169 170 170 169 170 170 170 170 170 170 170 170 170 170 170 170 170 170 170 170 170..."
},
{
"input": "999 998 13 533 270345 270332",
"output": "271 271 271 271 271 271 271 271 271 270 271 271 271 270 270 271 271 271 271 271 271 271 271 271 270 271 271 271 270 271 271 270 271 271 271 271 271 271 271 270 271 271 271 271 270 271 271 271 270 270 271 271 271 271 271 271 271 271 270 271 271 271 271 271 271 271 271 271 271 271 271 270 271 271 271 271 271 271 270 270 271 271 271 271 271 271 270 271 270 271 271 271 271 271 271 271 271 271 271 271 271 271 271 271 271 271 271 271 271 271 13 271 271 271 271 271 271 271 271 271 271 271 271 271 271 271 271 271 ..."
},
{
"input": "998 123 13 293 151330 33752",
"output": "134 135 135 134 134 135 135 135 134 134 135 134 135 134 134 134 134 135 135 134 134 135 134 135 134 275 134 134 134 134 135 134 134 135 135 275 134 134 135 134 134 134 134 135 134 134 134 135 134 134 135 135 134 135 135 134 134 135 134 275 134 135 275 274 135 274 135 134 135 135 274 134 135 134 135 135 134 135 134 134 134 134 274 134 135 135 134 135 134 134 134 134 134 134 134 135 274 134 134 134 135 275 134 134 135 135 135 134 135 135 135 134 134 134 135 274 275 134 135 275 135 135 135 134 134 135 134 274..."
},
{
"input": "995 993 123 743 437780 437534",
"output": "440 441 441 440 440 441 441 441 440 440 441 440 441 440 440 440 440 441 441 441 440 441 440 441 440 441 440 441 440 440 441 440 440 441 441 441 440 440 441 440 441 441 441 441 440 440 441 441 440 440 441 441 440 441 441 441 440 441 440 441 440 441 441 441 441 441 441 441 441 441 441 440 441 440 441 441 440 441 440 440 440 441 441 440 441 441 440 441 440 440 441 440 441 441 441 441 441 440 440 441 441 441 441 440 441 441 441 440 441 441 441 440 441 441 441 441 441 440 441 441 441 441 441 441 440 441 441 441..."
},
{
"input": "999 999 111 111 110889 110889",
"output": "111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111..."
},
{
"input": "1000 1000 111 111 111000 111000",
"output": "111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111..."
},
{
"input": "1000 567 999 1000 999489 566922",
"output": "999 1000 1000 999 999 1000 1000 1000 999 999 1000 999 1000 999 999 999 999 1000 1000 1000 999 1000 999 1000 999 1000 999 1000 999 999 1000 999 999 1000 1000 1000 999 999 1000 999 1000 999 1000 1000 999 999 999 1000 999 999 1000 1000 999 1000 1000 999 999 1000 999 1000 999 1000 1000 1000 1000 1000 1000 999 1000 1000 1000 999 1000 999 1000 1000 999 1000 999 999 999 999 1000 999 1000 1000 999 1000 999 999 1000 999 1000 1000 999 1000 1000 999 999 1000 1000 1000 999 999 1000 1000 1000 999 1000 1000 999 999 999 ..."
},
{
"input": "1000 567 998 1000 998981 566754",
"output": "998 1000 999 998 999 999 999 1000 999 998 1000 998 1000 998 998 999 998 999 999 999 998 1000 998 1000 998 1000 998 999 998 998 999 998 999 1000 999 1000 998 998 1000 998 999 999 999 999 998 998 999 1000 998 998 1000 999 998 1000 1000 999 998 1000 998 1000 998 1000 1000 1000 1000 1000 999 999 1000 999 1000 998 1000 998 1000 1000 999 1000 998 998 998 999 1000 998 1000 999 998 1000 998 998 999 998 999 999 999 999 1000 998 998 999 1000 1000 999 999 999 999 999 998 1000 1000 998 998 999 999 999 1000 998 998 999..."
},
{
"input": "1000 567 996 1000 997986 566445",
"output": "997 999 999 997 997 999 999 999 997 996 999 997 999 996 996 997 997 999 999 999 997 999 997 999 996 1000 997 999 996 997 999 996 997 999 999 999 997 996 999 996 999 999 999 999 996 997 999 999 996 996 999 999 997 999 999 997 997 999 996 999 997 999 999 999 999 999 999 999 999 999 999 996 999 997 999 999 997 999 996 996 997 997 999 997 999 999 996 999 996 997 999 997 999 999 999 999 999 997 997 999 999 999 999 997 999 999 999 997 999 999 996 997 997 999 999 999 996 997 999 999 999 999 999 999 997 999 999 99..."
},
{
"input": "1000 567 996 1000 997986 566445",
"output": "997 999 999 997 997 999 999 999 997 996 999 997 999 996 996 997 997 999 999 999 997 999 997 999 996 1000 997 999 996 997 999 996 997 999 999 999 997 996 999 996 999 999 999 999 996 997 999 999 996 996 999 999 997 999 999 997 997 999 996 999 997 999 999 999 999 999 999 999 999 999 999 996 999 997 999 999 997 999 996 996 997 997 999 997 999 999 996 999 996 997 999 997 999 999 999 999 999 997 997 999 999 999 999 997 999 999 999 997 999 999 996 997 997 999 999 999 996 997 999 999 999 999 999 999 997 999 999 99..."
},
{
"input": "1 1 1 1000 656 656",
"output": "656 "
},
{
"input": "2 1 1 1000 683 550",
"output": "550 133 "
},
{
"input": "3 2 1 1000 1816 1652",
"output": "826 826 164 "
},
{
"input": "5 5 1 1000 3288 3288",
"output": "657 658 657 658 658 "
},
{
"input": "5 3 1 1000 2732 2055",
"output": "338 685 339 685 685 "
},
{
"input": "1 1 10 100 50 50",
"output": "50 "
},
{
"input": "3 3 1 5 14 14",
"output": "5 5 4 "
},
{
"input": "1 1 1 100 50 50",
"output": "50 "
},
{
"input": "5 5 1 5 5 5",
"output": "1 1 1 1 1 "
},
{
"input": "1 1 1 10 10 10",
"output": "10 "
},
{
"input": "5 5 1 3 5 5",
"output": "1 1 1 1 1 "
},
{
"input": "1000 1000 1 1000 1000000 1000000",
"output": "1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1..."
},
{
"input": "3 3 1 3 9 9",
"output": "3 3 3 "
},
{
"input": "1000 1000 389 999 686847 686847",
"output": "687 687 687 687 687 687 687 687 687 686 687 687 687 686 686 687 687 687 687 687 687 687 687 687 686 687 687 687 686 687 687 686 687 687 687 687 687 686 687 686 687 687 687 687 686 687 687 687 686 686 687 687 687 687 687 687 687 687 686 687 687 687 687 687 687 687 687 687 687 687 687 686 687 687 687 687 687 687 686 686 687 687 687 687 687 687 686 687 686 687 687 687 687 687 687 687 687 687 687 687 687 687 687 687 687 687 687 687 687 687 686 687 687 687 687 687 686 687 687 687 687 687 687 687 687 687 687 687..."
},
{
"input": "2 2 2 2 4 4",
"output": "2 2 "
}
] | 1,587,213,667
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 124
| 0
|
n,k,l,r,s,s1=[int(i) for i in input().split()]
l2=[]
l1=[]
c=int(s1//k)
while(len(l1)<k):
l1.append(c)
d=sum(l1)
if(d==s1):
pass
else:
while(d<s1):
for i in range(0,len(l1)):
l1[i]=l1[i]+1
d=d+1
d=n-k
while(len(l2)<d):
l2.append(l)
e=sum(l2)
if(e==(s-s1)):
pass
else:
while(e<(s-s1)):
for i in range(0,len(l2)):
l2[i]=l2[i]+1
e=e+1
print(l1+l2)
|
Title: Valera and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera loves to participate in competitions. Especially in programming contests. Today he has participated in the contest with his team, consisting of *n* students (including Valera). This contest was an individual competition, so each student in the team solved problems individually.
After the contest was over, Valera was interested in results. He found out that:
- each student in the team scored at least *l* points and at most *r* points; - in total, all members of the team scored exactly *s**all* points; - the total score of the *k* members of the team who scored the most points is equal to exactly *s**k*; more formally, if *a*1,<=*a*2,<=...,<=*a**n* is the sequence of points earned by the team of students in the non-increasing order (*a*1<=≥<=*a*2<=≥<=...<=≥<=*a**n*), then *s**k*<==<=*a*1<=+<=*a*2<=+<=...<=+<=*a**k*.
However, Valera did not find out exactly how many points each of *n* students scored. Valera asked you to recover any distribution of scores between the students of the team, such that all the conditions above are met.
Input Specification:
The first line of the input contains exactly six integers *n*,<=*k*,<=*l*,<=*r*,<=*s**all*,<=*s**k* (1<=≤<=*n*,<=*k*,<=*l*,<=*r*<=≤<=1000; *l*<=≤<=*r*; *k*<=≤<=*n*; 1<=≤<=*s**k*<=≤<=*s**all*<=≤<=106).
It's guaranteed that the input is such that the answer exists.
Output Specification:
Print exactly *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the number of points each student scored. If there are multiple solutions, you can print any of them. You can print the distribution of points in any order.
Demo Input:
['5 3 1 3 13 9\n', '5 3 1 3 15 9\n']
Demo Output:
['2 3 2 3 3 ', '3 3 3 3 3 ']
Note:
none
|
```python
n,k,l,r,s,s1=[int(i) for i in input().split()]
l2=[]
l1=[]
c=int(s1//k)
while(len(l1)<k):
l1.append(c)
d=sum(l1)
if(d==s1):
pass
else:
while(d<s1):
for i in range(0,len(l1)):
l1[i]=l1[i]+1
d=d+1
d=n-k
while(len(l2)<d):
l2.append(l)
e=sum(l2)
if(e==(s-s1)):
pass
else:
while(e<(s-s1)):
for i in range(0,len(l2)):
l2[i]=l2[i]+1
e=e+1
print(l1+l2)
```
| 0
|
|
408
|
A
|
Line to Cashier
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
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 *k**i* people. The *j*-th person standing in the queue to the *i*-th cashier has *m**i*,<=*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: *k*1,<=*k*2,<=...,<=*k**n* (1<=≤<=*k**i*<=≤<=100), where *k**i* is the number of people in the queue to the *i*-th cashier.
The *i*-th of the next *n* lines contains *k**i* space-separated integers: *m**i*,<=1,<=*m**i*,<=2,<=...,<=*m**i*,<=*k**i* (1<=≤<=*m**i*,<=*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.
|
[
"1\n1\n1\n",
"4\n1 4 3 2\n100\n1 2 2 3\n1 9 1\n7 8\n"
] |
[
"20\n",
"100\n"
] |
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.
| 500
|
[
{
"input": "1\n1\n1",
"output": "20"
},
{
"input": "4\n1 4 3 2\n100\n1 2 2 3\n1 9 1\n7 8",
"output": "100"
},
{
"input": "4\n5 4 5 5\n3 1 3 1 2\n3 1 1 3\n1 1 1 2 2\n2 2 1 1 3",
"output": "100"
},
{
"input": "5\n5 3 6 6 4\n7 5 3 3 9\n6 8 2\n1 10 8 5 9 2\n9 7 8 5 9 10\n9 8 3 3",
"output": "125"
},
{
"input": "5\n10 10 10 10 10\n6 7 8 6 8 5 9 8 10 5\n9 6 9 8 7 8 8 10 8 5\n8 7 7 8 7 5 6 8 9 5\n6 5 10 5 5 10 7 8 5 5\n10 9 8 7 6 9 7 9 6 5",
"output": "480"
},
{
"input": "10\n9 10 10 10 9 5 9 7 8 7\n11 6 10 4 4 15 7 15 5\n3 9 11 12 11 1 13 13 1 5\n6 15 9 12 3 2 8 12 11 10\n7 1 1 6 10 2 6 1 14 2\n8 14 2 3 6 1 14 1 12\n6 10 9 3 5\n13 12 12 7 13 4 4 8 10\n5 6 4 3 14 9 13\n8 12 1 5 7 4 13 1\n1 9 5 3 5 1 4",
"output": "240"
},
{
"input": "10\n5 5 5 5 5 5 5 5 5 5\n5 5 4 5 4\n6 5 7 7 6\n5 4 4 5 5\n4 4 5 5 5\n7 6 4 5 7\n4 6 5 4 5\n6 6 7 6 6\n4 5 4 4 7\n7 5 4 4 5\n6 6 7 4 4",
"output": "190"
},
{
"input": "1\n1\n100",
"output": "515"
},
{
"input": "1\n90\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",
"output": "41850"
}
] | 1,587,757,966
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 124
| 0
|
n=int(input())
l=list(map(int,input().split()))
t=[]
for i in range(n):
k=list(map(int,input().split()))
p=0
for i in k:
p +=i*5
else:
p +=15*(len(k))
t +=[p]
else:
print(min(t))
|
Title: Line to Cashier
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
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 *k**i* people. The *j*-th person standing in the queue to the *i*-th cashier has *m**i*,<=*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.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of cashes in the shop. The second line contains *n* space-separated integers: *k*1,<=*k*2,<=...,<=*k**n* (1<=≤<=*k**i*<=≤<=100), where *k**i* is the number of people in the queue to the *i*-th cashier.
The *i*-th of the next *n* lines contains *k**i* space-separated integers: *m**i*,<=1,<=*m**i*,<=2,<=...,<=*m**i*,<=*k**i* (1<=≤<=*m**i*,<=*j*<=≤<=100) — the number of products the *j*-th person in the queue for the *i*-th cash has.
Output Specification:
Print a single integer — the minimum number of seconds Vasya needs to get to the cashier.
Demo Input:
['1\n1\n1\n', '4\n1 4 3 2\n100\n1 2 2 3\n1 9 1\n7 8\n']
Demo Output:
['20\n', '100\n']
Note:
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.
|
```python
n=int(input())
l=list(map(int,input().split()))
t=[]
for i in range(n):
k=list(map(int,input().split()))
p=0
for i in k:
p +=i*5
else:
p +=15*(len(k))
t +=[p]
else:
print(min(t))
```
| 3
|
|
789
|
B
|
Masha and geometric depression
|
PROGRAMMING
| 1,700
|
[
"brute force",
"implementation",
"math"
] | null | null |
Masha really loves algebra. On the last lesson, her strict teacher Dvastan gave she new exercise.
You are given geometric progression *b* defined by two integers *b*1 and *q*. Remind that a geometric progression is a sequence of integers *b*1,<=*b*2,<=*b*3,<=..., where for each *i*<=><=1 the respective term satisfies the condition *b**i*<==<=*b**i*<=-<=1·*q*, where *q* is called the common ratio of the progression. Progressions in Uzhlyandia are unusual: both *b*1 and *q* can equal 0. Also, Dvastan gave Masha *m* "bad" integers *a*1,<=*a*2,<=...,<=*a**m*, and an integer *l*.
Masha writes all progression terms one by one onto the board (including repetitive) while condition |*b**i*|<=≤<=*l* is satisfied (|*x*| means absolute value of *x*). There is an exception: if a term equals one of the "bad" integers, Masha skips it (doesn't write onto the board) and moves forward to the next term.
But the lesson is going to end soon, so Masha has to calculate how many integers will be written on the board. In order not to get into depression, Masha asked you for help: help her calculate how many numbers she will write, or print "inf" in case she needs to write infinitely many integers.
|
The first line of input contains four integers *b*1, *q*, *l*, *m* (-109<=≤<=*b*1,<=*q*<=≤<=109, 1<=≤<=*l*<=≤<=109, 1<=≤<=*m*<=≤<=105) — the initial term and the common ratio of progression, absolute value of maximal number that can be written on the board and the number of "bad" integers, respectively.
The second line contains *m* distinct integers *a*1,<=*a*2,<=...,<=*a**m* (-109<=≤<=*a**i*<=≤<=109) — numbers that will never be written on the board.
|
Print the only integer, meaning the number of progression terms that will be written on the board if it is finite, or "inf" (without quotes) otherwise.
|
[
"3 2 30 4\n6 14 25 48\n",
"123 1 2143435 4\n123 11 -5453 141245\n",
"123 1 2143435 4\n54343 -13 6 124\n"
] |
[
"3",
"0",
"inf"
] |
In the first sample case, Masha will write integers 3, 12, 24. Progression term 6 will be skipped because it is a "bad" integer. Terms bigger than 24 won't be written because they exceed *l* by absolute value.
In the second case, Masha won't write any number because all terms are equal 123 and this is a "bad" integer.
In the third case, Masha will write infinitely integers 123.
| 1,000
|
[
{
"input": "3 2 30 4\n6 14 25 48",
"output": "3"
},
{
"input": "123 1 2143435 4\n123 11 -5453 141245",
"output": "0"
},
{
"input": "123 1 2143435 4\n54343 -13 6 124",
"output": "inf"
},
{
"input": "3 2 25 2\n379195692 -69874783",
"output": "4"
},
{
"input": "3 2 30 3\n-691070108 -934106649 -220744807",
"output": "4"
},
{
"input": "3 3 104 17\n9 -73896485 -290898562 5254410 409659728 -916522518 -435516126 94354167 262981034 -375897180 -80186684 -173062070 -288705544 -699097793 -11447747 320434295 503414250",
"output": "3"
},
{
"input": "-1000000000 -1000000000 1 1\n232512888",
"output": "0"
},
{
"input": "11 0 228 5\n-1 0 1 5 -11245",
"output": "1"
},
{
"input": "11 0 228 5\n-1 -17 1 5 -11245",
"output": "inf"
},
{
"input": "0 0 2143435 5\n-1 -153 1 5 -11245",
"output": "inf"
},
{
"input": "123 0 2143435 4\n5433 0 123 -645",
"output": "0"
},
{
"input": "123 -1 2143435 5\n-123 0 12 5 -11245",
"output": "inf"
},
{
"input": "123 0 21 4\n543453 -123 6 1424",
"output": "0"
},
{
"input": "3 2 115 16\n24 48 12 96 3 720031148 -367712651 -838596957 558177735 -963046495 -313322487 -465018432 -618984128 -607173835 144854086 178041956",
"output": "1"
},
{
"input": "-3 0 92055 36\n-92974174 -486557474 -663622151 695596393 177960746 -563227474 -364263320 -676254242 -614140218 71456762 -764104225 705056581 -106398436 332755134 -199942822 -732751692 658942664 677739866 886535704 183687802 -784248291 -22550621 -938674499 637055091 -704750213 780395802 778342470 -999059668 -794361783 796469192 215667969 354336794 -60195289 -885080928 -290279020 201221317",
"output": "inf"
},
{
"input": "0 -3 2143435 5\n-1 0 1 5 -11245",
"output": "0"
},
{
"input": "123 -1 2143435 5\n-123 0 123 -5453 141245",
"output": "0"
},
{
"input": "123 0 2143435 4\n5433 0 -123 -645",
"output": "1"
},
{
"input": "11 0 2 5\n-1 0 1 5 -11245",
"output": "0"
},
{
"input": "2 2 4 1\n2",
"output": "1"
},
{
"input": "1 -2 1000000000 1\n0",
"output": "30"
},
{
"input": "0 8 10 1\n5",
"output": "inf"
},
{
"input": "-1000 0 10 1\n5",
"output": "0"
},
{
"input": "0 2 2143435 4\n54343 -13 6 124",
"output": "inf"
},
{
"input": "0 8 5 1\n9",
"output": "inf"
},
{
"input": "-10 1 5 1\n100",
"output": "0"
},
{
"input": "123 -1 2143435 4\n54343 -13 6 123",
"output": "inf"
},
{
"input": "-5 -1 10 1\n-5",
"output": "inf"
},
{
"input": "2 0 1 1\n2",
"output": "0"
},
{
"input": "0 5 8 1\n10",
"output": "inf"
},
{
"input": "0 5 100 2\n34 56",
"output": "inf"
},
{
"input": "15 -1 15 4\n15 -15 1 2",
"output": "0"
},
{
"input": "10 -1 2 1\n1",
"output": "0"
},
{
"input": "2 0 2 1\n2",
"output": "inf"
},
{
"input": "4 0 4 1\n0",
"output": "1"
},
{
"input": "10 10 10 1\n123",
"output": "1"
},
{
"input": "2 2 4 1\n3",
"output": "2"
},
{
"input": "0 1 1 1\n0",
"output": "0"
},
{
"input": "3 2 30 1\n3",
"output": "3"
},
{
"input": "1000000000 100000 1000000000 4\n5433 13 6 0",
"output": "1"
},
{
"input": "-2 0 1 1\n1",
"output": "0"
},
{
"input": "2 -1 10 1\n2",
"output": "inf"
},
{
"input": "1 -1 2 1\n1",
"output": "inf"
},
{
"input": "0 10 10 1\n2",
"output": "inf"
},
{
"input": "0 35 2 1\n3",
"output": "inf"
},
{
"input": "3 1 3 1\n5",
"output": "inf"
},
{
"input": "3 2 3 4\n6 14 25 48",
"output": "1"
},
{
"input": "0 69 12 1\n1",
"output": "inf"
},
{
"input": "100 0 100000 1\n100",
"output": "inf"
},
{
"input": "0 4 1000 3\n5 6 7",
"output": "inf"
},
{
"input": "0 2 100 1\n5",
"output": "inf"
},
{
"input": "3 2 24 4\n6 14 25 48",
"output": "3"
},
{
"input": "0 4 1 1\n2",
"output": "inf"
},
{
"input": "1 5 10000 1\n125",
"output": "5"
},
{
"input": "2 -1 1 1\n1",
"output": "0"
},
{
"input": "0 3 100 1\n5",
"output": "inf"
},
{
"input": "0 3 3 1\n1",
"output": "inf"
},
{
"input": "0 2 5 1\n1",
"output": "inf"
},
{
"input": "5 -1 100 1\n5",
"output": "inf"
},
{
"input": "-20 0 10 1\n0",
"output": "0"
},
{
"input": "3 0 1 1\n3",
"output": "0"
},
{
"input": "2 -1 3 1\n2",
"output": "inf"
},
{
"input": "1 1 1000000000 1\n100",
"output": "inf"
},
{
"input": "5 -1 3 1\n0",
"output": "0"
},
{
"input": "0 5 10 1\n2",
"output": "inf"
},
{
"input": "123 0 125 1\n123",
"output": "inf"
},
{
"input": "2 -1 100 1\n2",
"output": "inf"
},
{
"input": "5 2 100 1\n5",
"output": "4"
},
{
"input": "-5 0 1 1\n1",
"output": "0"
},
{
"input": "-3 0 1 1\n-3",
"output": "0"
},
{
"input": "2 -2 10 1\n1",
"output": "3"
},
{
"input": "0 2 30 4\n6 14 25 48",
"output": "inf"
},
{
"input": "1 -1 1 1\n1",
"output": "inf"
},
{
"input": "2 -1 6 1\n2",
"output": "inf"
},
{
"input": "-3 1 100 1\n-3",
"output": "0"
},
{
"input": "1 0 2 1\n1",
"output": "inf"
},
{
"input": "1000000000 999999998 1000000000 1\n0",
"output": "1"
},
{
"input": "1 0 2143435 4\n1 -123 -5453 141245",
"output": "inf"
},
{
"input": "-1000 0 100 1\n-1000",
"output": "0"
},
{
"input": "100 10 2 1\n100",
"output": "0"
},
{
"input": "-3 1 100 1\n3",
"output": "inf"
},
{
"input": "123 -1 10000 1\n123",
"output": "inf"
},
{
"input": "1 -1 2143435 4\n1 -123 -5453 141245",
"output": "inf"
},
{
"input": "5 1 5 5\n1 2 3 4 0",
"output": "inf"
},
{
"input": "-100 -1 1 1\n1",
"output": "0"
},
{
"input": "10 -1 3 2\n10 8",
"output": "0"
},
{
"input": "-10 0 5 1\n0",
"output": "0"
},
{
"input": "3 0 3 1\n0",
"output": "1"
},
{
"input": "2 0 2 1\n-1",
"output": "inf"
},
{
"input": "5 0 20 1\n5",
"output": "inf"
},
{
"input": "-4 1 1 1\n0",
"output": "0"
},
{
"input": "11 0 1111 1\n11",
"output": "inf"
},
{
"input": "2 0 3 1\n2",
"output": "inf"
},
{
"input": "-1 -1 2143435 4\n-1 -123 -5453 141245",
"output": "inf"
},
{
"input": "-100 0 50 1\n0",
"output": "0"
},
{
"input": "5 1 2 1\n2",
"output": "0"
},
{
"input": "3 0 3 1\n4",
"output": "inf"
},
{
"input": "0 23 3 1\n3",
"output": "inf"
},
{
"input": "-1000 0 100 1\n2",
"output": "0"
},
{
"input": "1 -1 10 1\n1",
"output": "inf"
}
] | 1,500,815,816
| 3,716
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 1
| 78
| 23,142,400
|
def ans(n):
print(n)
sys.exit()
b1, q, l, m = map(int, input().split())
w = set(map(int, input().split()))
inf = 'inf'
#possiblity for answering inf
if b1 == 0:
if (0 in w):
ans(inf)
else:
ans(0)
# b1 != 0
if q == 0:
if 0 not in w:
ans(inf)
elif abs(b1) <= l and (b1 not in w):
ans(1)
else:
ans(0)
# q != 0
if q == 1:
if (b1 not in w) and (abs(b1)<= l):
ans(inf)
else:
ans(0)
if q == -1:
if (abs(b1) <=l):
if ((b1 not in w) or (-b1 not in w)):
ans(inf)
else:
ans(0)
else:
ans(0)
#no more inf
# |q| > 1
ans = 0
while abs(b1) <= 10**9 + 10:
if abs(b1) <= l and (b1 not in w):
ans += 1
b1 *= q
print( ans)
|
Title: Masha and geometric depression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Masha really loves algebra. On the last lesson, her strict teacher Dvastan gave she new exercise.
You are given geometric progression *b* defined by two integers *b*1 and *q*. Remind that a geometric progression is a sequence of integers *b*1,<=*b*2,<=*b*3,<=..., where for each *i*<=><=1 the respective term satisfies the condition *b**i*<==<=*b**i*<=-<=1·*q*, where *q* is called the common ratio of the progression. Progressions in Uzhlyandia are unusual: both *b*1 and *q* can equal 0. Also, Dvastan gave Masha *m* "bad" integers *a*1,<=*a*2,<=...,<=*a**m*, and an integer *l*.
Masha writes all progression terms one by one onto the board (including repetitive) while condition |*b**i*|<=≤<=*l* is satisfied (|*x*| means absolute value of *x*). There is an exception: if a term equals one of the "bad" integers, Masha skips it (doesn't write onto the board) and moves forward to the next term.
But the lesson is going to end soon, so Masha has to calculate how many integers will be written on the board. In order not to get into depression, Masha asked you for help: help her calculate how many numbers she will write, or print "inf" in case she needs to write infinitely many integers.
Input Specification:
The first line of input contains four integers *b*1, *q*, *l*, *m* (-109<=≤<=*b*1,<=*q*<=≤<=109, 1<=≤<=*l*<=≤<=109, 1<=≤<=*m*<=≤<=105) — the initial term and the common ratio of progression, absolute value of maximal number that can be written on the board and the number of "bad" integers, respectively.
The second line contains *m* distinct integers *a*1,<=*a*2,<=...,<=*a**m* (-109<=≤<=*a**i*<=≤<=109) — numbers that will never be written on the board.
Output Specification:
Print the only integer, meaning the number of progression terms that will be written on the board if it is finite, or "inf" (without quotes) otherwise.
Demo Input:
['3 2 30 4\n6 14 25 48\n', '123 1 2143435 4\n123 11 -5453 141245\n', '123 1 2143435 4\n54343 -13 6 124\n']
Demo Output:
['3', '0', 'inf']
Note:
In the first sample case, Masha will write integers 3, 12, 24. Progression term 6 will be skipped because it is a "bad" integer. Terms bigger than 24 won't be written because they exceed *l* by absolute value.
In the second case, Masha won't write any number because all terms are equal 123 and this is a "bad" integer.
In the third case, Masha will write infinitely integers 123.
|
```python
def ans(n):
print(n)
sys.exit()
b1, q, l, m = map(int, input().split())
w = set(map(int, input().split()))
inf = 'inf'
#possiblity for answering inf
if b1 == 0:
if (0 in w):
ans(inf)
else:
ans(0)
# b1 != 0
if q == 0:
if 0 not in w:
ans(inf)
elif abs(b1) <= l and (b1 not in w):
ans(1)
else:
ans(0)
# q != 0
if q == 1:
if (b1 not in w) and (abs(b1)<= l):
ans(inf)
else:
ans(0)
if q == -1:
if (abs(b1) <=l):
if ((b1 not in w) or (-b1 not in w)):
ans(inf)
else:
ans(0)
else:
ans(0)
#no more inf
# |q| > 1
ans = 0
while abs(b1) <= 10**9 + 10:
if abs(b1) <= l and (b1 not in w):
ans += 1
b1 *= q
print( ans)
```
| -1
|
|
895
|
A
|
Pizza Separation
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360.
|
Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya.
|
[
"4\n90 90 90 90\n",
"3\n100 100 160\n",
"1\n360\n",
"4\n170 30 150 10\n"
] |
[
"0\n",
"40\n",
"360\n",
"0\n"
] |
In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector.
| 500
|
[
{
"input": "4\n90 90 90 90",
"output": "0"
},
{
"input": "3\n100 100 160",
"output": "40"
},
{
"input": "1\n360",
"output": "360"
},
{
"input": "4\n170 30 150 10",
"output": "0"
},
{
"input": "5\n10 10 10 10 320",
"output": "280"
},
{
"input": "8\n45 45 45 45 45 45 45 45",
"output": "0"
},
{
"input": "3\n120 120 120",
"output": "120"
},
{
"input": "5\n110 90 70 50 40",
"output": "40"
},
{
"input": "2\n170 190",
"output": "20"
},
{
"input": "15\n25 25 25 25 25 25 25 25 25 25 25 25 25 25 10",
"output": "10"
},
{
"input": "5\n30 60 180 60 30",
"output": "0"
},
{
"input": "2\n359 1",
"output": "358"
},
{
"input": "5\n100 100 30 100 30",
"output": "40"
},
{
"input": "5\n36 34 35 11 244",
"output": "128"
},
{
"input": "5\n96 94 95 71 4",
"output": "18"
},
{
"input": "2\n85 275",
"output": "190"
},
{
"input": "3\n281 67 12",
"output": "202"
},
{
"input": "5\n211 113 25 9 2",
"output": "62"
},
{
"input": "13\n286 58 6 1 1 1 1 1 1 1 1 1 1",
"output": "212"
},
{
"input": "15\n172 69 41 67 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "20\n226 96 2 20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "92"
},
{
"input": "50\n148 53 32 11 4 56 8 2 5 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",
"output": "0"
},
{
"input": "3\n1 1 358",
"output": "356"
},
{
"input": "20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 341",
"output": "322"
},
{
"input": "33\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 1 328",
"output": "296"
},
{
"input": "70\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 1 291",
"output": "222"
},
{
"input": "130\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 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 1 231",
"output": "102"
},
{
"input": "200\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 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 1 161",
"output": "0"
},
{
"input": "222\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 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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 139",
"output": "0"
},
{
"input": "10\n8 3 11 4 1 10 10 1 8 304",
"output": "248"
},
{
"input": "12\n8 7 7 3 11 2 10 1 10 8 10 283",
"output": "206"
},
{
"input": "13\n10 8 9 10 5 9 4 1 10 11 1 7 275",
"output": "190"
},
{
"input": "14\n1 6 3 11 9 5 9 8 5 6 7 3 7 280",
"output": "200"
},
{
"input": "15\n10 11 5 4 11 5 4 1 5 4 5 5 9 6 275",
"output": "190"
},
{
"input": "30\n8 7 5 8 3 7 2 4 3 8 11 3 9 11 2 4 1 4 5 6 11 5 8 3 6 3 11 2 11 189",
"output": "18"
},
{
"input": "70\n5 3 6 8 9 2 8 9 11 5 2 8 9 11 7 6 6 9 7 11 7 6 3 8 2 4 4 8 4 3 2 2 3 5 6 5 11 2 7 7 5 8 10 5 2 1 10 9 4 10 7 1 8 10 9 1 5 1 1 1 2 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "29\n2 10 1 5 7 2 9 11 9 9 10 8 4 11 2 5 4 1 4 9 6 10 8 3 1 3 8 9 189",
"output": "18"
},
{
"input": "35\n3 4 11 4 4 2 3 4 3 9 7 10 2 7 8 3 11 3 6 4 6 7 11 10 8 7 6 7 2 8 5 3 2 2 168",
"output": "0"
},
{
"input": "60\n4 10 3 10 6 3 11 8 11 9 3 5 9 2 6 5 6 9 4 10 1 1 3 7 2 10 5 5 3 10 5 2 1 2 9 11 11 9 11 4 11 7 5 6 10 9 3 4 7 8 7 3 6 7 8 5 1 1 1 5",
"output": "0"
},
{
"input": "71\n3 11 8 1 10 1 7 9 6 4 11 10 11 2 4 1 11 7 9 10 11 4 8 7 11 3 8 4 1 8 4 2 9 9 7 10 10 9 5 7 9 7 2 1 7 6 5 11 5 9 4 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "63\n2 11 5 8 7 9 9 8 10 5 9 10 11 8 10 2 3 5 3 7 5 10 2 9 4 8 1 8 5 9 7 7 1 8 7 7 9 10 10 10 8 7 7 2 2 8 9 7 10 8 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "81\n5 8 7 11 2 7 1 1 5 8 7 2 3 11 4 9 7 6 4 4 2 1 1 7 9 4 1 8 3 1 4 10 7 9 9 8 11 3 4 3 10 8 6 4 7 2 4 3 6 11 11 10 7 10 2 10 8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "47\n5 3 7 4 2 7 8 1 9 10 5 11 10 7 7 5 1 3 2 11 3 8 6 1 6 10 8 3 2 10 5 6 8 6 9 7 10 9 7 4 8 11 10 1 5 11 68",
"output": "0"
},
{
"input": "100\n5 8 9 3 2 3 9 8 11 10 4 8 1 1 1 1 6 5 10 9 5 3 7 7 2 11 10 2 3 2 2 8 7 3 5 5 10 9 2 5 10 6 7 7 4 7 7 8 2 8 9 9 2 4 1 1 3 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",
"output": "0"
},
{
"input": "120\n9 11 3 7 3 7 9 1 10 7 11 4 1 5 3 5 6 3 1 11 8 8 11 7 3 5 1 9 1 7 10 10 10 10 9 5 4 8 2 8 2 1 4 5 3 11 3 5 1 1 2 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",
"output": "0"
},
{
"input": "200\n7 7 9 8 2 8 5 8 3 9 7 10 2 9 11 8 11 7 5 2 6 3 11 9 5 1 10 2 1 2 2 2 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 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",
"output": "0"
},
{
"input": "220\n3 2 8 1 3 5 5 11 1 5 2 6 9 2 2 6 8 10 7 1 3 2 10 9 10 10 4 10 9 5 1 1 2 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 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",
"output": "0"
},
{
"input": "6\n27 15 28 34 41 215",
"output": "70"
},
{
"input": "7\n41 38 41 31 22 41 146",
"output": "14"
},
{
"input": "8\n24 27 34 23 29 23 30 170",
"output": "20"
},
{
"input": "9\n11 11 20 20 33 32 35 26 172",
"output": "6"
},
{
"input": "10\n36 13 28 13 33 34 23 25 34 121",
"output": "0"
},
{
"input": "11\n19 37 13 41 37 15 32 12 19 35 100",
"output": "10"
},
{
"input": "12\n37 25 34 38 21 24 34 38 11 29 28 41",
"output": "2"
},
{
"input": "13\n24 40 20 26 25 29 39 29 35 28 19 18 28",
"output": "2"
},
{
"input": "14\n11 21 40 19 28 34 13 16 23 30 34 22 25 44",
"output": "4"
},
{
"input": "3\n95 91 174",
"output": "12"
},
{
"input": "4\n82 75 78 125",
"output": "46"
},
{
"input": "6\n87 75 88 94 15 1",
"output": "4"
},
{
"input": "10\n27 52 58 64 45 64 1 19 2 28",
"output": "12"
},
{
"input": "50\n14 12 11 8 1 6 11 6 7 8 4 11 4 5 7 3 5 4 7 24 10 2 3 4 6 13 2 1 8 7 5 13 10 8 5 20 1 2 23 7 14 3 4 4 2 8 8 2 6 1",
"output": "0"
},
{
"input": "100\n3 3 4 3 3 6 3 2 8 2 13 3 1 1 2 1 3 4 1 7 1 2 2 6 3 2 10 3 1 2 5 6 2 3 3 2 3 11 8 3 2 6 1 3 3 4 7 7 2 2 1 2 6 3 3 2 3 1 3 8 2 6 4 2 1 12 2 2 2 1 4 1 4 1 3 1 3 1 5 2 6 6 7 1 2 3 2 4 4 2 5 9 8 2 4 6 5 1 1 3",
"output": "0"
},
{
"input": "150\n1 5 1 2 2 2 1 4 2 2 2 3 1 2 1 2 2 2 2 1 2 2 2 1 5 3 4 1 3 4 5 2 4 2 1 2 2 1 1 2 3 2 4 2 2 3 3 1 1 5 2 3 2 1 9 2 1 1 2 1 4 1 1 3 2 2 2 1 2 2 2 1 3 3 4 2 2 1 3 3 3 1 4 3 4 1 2 2 1 1 1 2 2 5 4 1 1 1 2 1 2 3 2 2 6 3 3 3 1 2 1 1 2 8 2 2 4 3 4 5 3 1 4 2 2 2 2 1 4 4 1 1 2 2 4 9 6 3 1 1 2 1 3 4 1 3 2 2 2 1",
"output": "0"
},
{
"input": "200\n1 2 1 3 1 3 1 2 1 4 6 1 2 2 2 2 1 1 1 1 3 2 1 2 2 2 1 2 2 2 2 1 1 1 3 2 3 1 1 2 1 1 2 1 1 1 1 1 1 2 1 2 2 4 1 3 1 2 1 2 2 1 2 1 3 1 1 2 2 1 1 1 1 2 4 1 2 1 1 1 2 1 3 1 1 3 1 2 2 4 1 1 2 1 2 1 2 2 2 2 1 1 2 1 2 1 3 3 1 1 1 2 1 3 3 1 2 1 3 1 3 3 1 2 2 1 4 1 2 2 1 2 2 4 2 5 1 2 2 1 2 1 2 1 5 2 1 2 2 1 2 4 1 2 2 4 2 3 2 3 1 2 1 1 2 2 2 1 1 2 1 4 1 2 1 1 2 1 2 3 1 1 1 2 2 3 1 3 2 2 3 1 2 1 2 1 1 2 1 2",
"output": "0"
},
{
"input": "5\n35 80 45 100 100",
"output": "40"
},
{
"input": "4\n90 179 90 1",
"output": "2"
},
{
"input": "5\n50 50 20 160 80",
"output": "0"
},
{
"input": "5\n30 175 30 5 120",
"output": "10"
},
{
"input": "4\n170 30 10 150",
"output": "20"
},
{
"input": "6\n90 30 90 30 90 30",
"output": "60"
},
{
"input": "4\n70 80 110 100",
"output": "20"
},
{
"input": "7\n35 45 70 100 10 10 90",
"output": "0"
},
{
"input": "6\n50 90 10 90 20 100",
"output": "20"
},
{
"input": "6\n10 155 162 1 26 6",
"output": "18"
},
{
"input": "7\n80 90 80 45 10 10 45",
"output": "20"
},
{
"input": "4\n18 36 162 144",
"output": "36"
},
{
"input": "5\n20 50 50 160 80",
"output": "40"
},
{
"input": "5\n10 30 140 20 160",
"output": "0"
},
{
"input": "6\n90 80 60 50 40 40",
"output": "20"
},
{
"input": "9\n40 20 20 20 20 20 20 40 160",
"output": "40"
},
{
"input": "4\n90 54 90 126",
"output": "72"
},
{
"input": "4\n150 170 30 10",
"output": "20"
},
{
"input": "8\n130 12 13 85 41 67 5 7",
"output": "26"
},
{
"input": "7\n70 170 20 10 30 30 30",
"output": "20"
},
{
"input": "8\n100 100 50 50 15 15 15 15",
"output": "40"
},
{
"input": "4\n100 70 80 110",
"output": "20"
},
{
"input": "5\n160 130 40 20 10",
"output": "20"
},
{
"input": "4\n20 149 151 40",
"output": "22"
},
{
"input": "4\n100 10 100 150",
"output": "60"
},
{
"input": "6\n19 64 105 168 1 3",
"output": "16"
},
{
"input": "8\n10 10 70 70 90 90 10 10",
"output": "0"
}
] | 1,511,712,779
| 479
|
PyPy 3
|
OK
|
TESTS
| 93
| 139
| 23,859,200
|
n = int(input())
best = 360
tab = list(map(int, input().split()))
pre_sum = [0]
for i in range(1, len(tab) + 1):
pre_sum += [tab[i - 1] + pre_sum[-1]]
for i in range(n + 1):
for j in range(i, n + 1):
best = min([best, 2 * abs(180 - (pre_sum[j] - pre_sum[i - 1]))])
print(best);
|
Title: Pizza Separation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Students Vasya and Petya are studying at the BSU (Byteland State University). At one of the breaks they decided to order a pizza. In this problem pizza is a circle of some radius. The pizza was delivered already cut into *n* pieces. The *i*-th piece is a sector of angle equal to *a**i*. Vasya and Petya want to divide all pieces of pizza into two continuous sectors in such way that the difference between angles of these sectors is minimal. Sector angle is sum of angles of all pieces in it. Pay attention, that one of sectors can be empty.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=360) — the number of pieces into which the delivered pizza was cut.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=360) — the angles of the sectors into which the pizza was cut. The sum of all *a**i* is 360.
Output Specification:
Print one integer — the minimal difference between angles of sectors that will go to Vasya and Petya.
Demo Input:
['4\n90 90 90 90\n', '3\n100 100 160\n', '1\n360\n', '4\n170 30 150 10\n']
Demo Output:
['0\n', '40\n', '360\n', '0\n']
Note:
In first sample Vasya can take 1 and 2 pieces, Petya can take 3 and 4 pieces. Then the answer is |(90 + 90) - (90 + 90)| = 0.
In third sample there is only one piece of pizza that can be taken by only one from Vasya and Petya. So the answer is |360 - 0| = 360.
In fourth sample Vasya can take 1 and 4 pieces, then Petya will take 2 and 3 pieces. So the answer is |(170 + 10) - (30 + 150)| = 0.
Picture explaning fourth sample:
<img class="tex-graphics" src="https://espresso.codeforces.com/4bb3450aca241f92fedcba5479bf1b6d22cf813d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Both red and green sectors consist of two adjacent pieces of pizza. So Vasya can take green sector, then Petya will take red sector.
|
```python
n = int(input())
best = 360
tab = list(map(int, input().split()))
pre_sum = [0]
for i in range(1, len(tab) + 1):
pre_sum += [tab[i - 1] + pre_sum[-1]]
for i in range(n + 1):
for j in range(i, n + 1):
best = min([best, 2 * abs(180 - (pre_sum[j] - pre_sum[i - 1]))])
print(best);
```
| 3
|
|
450
|
B
|
Jzzhu and Sequences
|
PROGRAMMING
| 1,300
|
[
"implementation",
"math"
] | null | null |
Jzzhu has invented a kind of sequences, they meet the following property:
You are given *x* and *y*, please calculate *f**n* modulo 1000000007 (109<=+<=7).
|
The first line contains two integers *x* and *y* (|*x*|,<=|*y*|<=≤<=109). The second line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
|
Output a single integer representing *f**n* modulo 1000000007 (109<=+<=7).
|
[
"2 3\n3\n",
"0 -1\n2\n"
] |
[
"1\n",
"1000000006\n"
] |
In the first sample, *f*<sub class="lower-index">2</sub> = *f*<sub class="lower-index">1</sub> + *f*<sub class="lower-index">3</sub>, 3 = 2 + *f*<sub class="lower-index">3</sub>, *f*<sub class="lower-index">3</sub> = 1.
In the second sample, *f*<sub class="lower-index">2</sub> = - 1; - 1 modulo (10<sup class="upper-index">9</sup> + 7) equals (10<sup class="upper-index">9</sup> + 6).
| 1,000
|
[
{
"input": "2 3\n3",
"output": "1"
},
{
"input": "0 -1\n2",
"output": "1000000006"
},
{
"input": "-9 -11\n12345",
"output": "1000000005"
},
{
"input": "0 0\n1000000000",
"output": "0"
},
{
"input": "-1000000000 1000000000\n2000000000",
"output": "1000000000"
},
{
"input": "-12345678 12345678\n1912345678",
"output": "12345678"
},
{
"input": "728374857 678374857\n1928374839",
"output": "950000007"
},
{
"input": "278374837 992837483\n1000000000",
"output": "721625170"
},
{
"input": "-693849384 502938493\n982838498",
"output": "502938493"
},
{
"input": "-783928374 983738273\n992837483",
"output": "16261734"
},
{
"input": "-872837483 -682738473\n999999999",
"output": "190099010"
},
{
"input": "-892837483 -998273847\n999283948",
"output": "892837483"
},
{
"input": "-283938494 738473848\n1999999999",
"output": "716061513"
},
{
"input": "-278374857 819283838\n1",
"output": "721625150"
},
{
"input": "-1000000000 123456789\n1",
"output": "7"
},
{
"input": "-529529529 -524524524\n2",
"output": "475475483"
},
{
"input": "1 2\n2000000000",
"output": "2"
},
{
"input": "-1 -2\n2000000000",
"output": "1000000005"
},
{
"input": "1 2\n1999999999",
"output": "1"
},
{
"input": "1 2\n1999999998",
"output": "1000000006"
},
{
"input": "1 2\n1999999997",
"output": "1000000005"
},
{
"input": "1 2\n1999999996",
"output": "1000000006"
},
{
"input": "69975122 366233206\n1189460676",
"output": "703741923"
},
{
"input": "812229413 904420051\n806905621",
"output": "812229413"
},
{
"input": "872099024 962697902\n1505821695",
"output": "90598878"
},
{
"input": "887387283 909670917\n754835014",
"output": "112612724"
},
{
"input": "37759824 131342932\n854621399",
"output": "868657075"
},
{
"input": "-246822123 800496170\n626323615",
"output": "753177884"
},
{
"input": "-861439463 974126967\n349411083",
"output": "835566423"
},
{
"input": "-69811049 258093841\n1412447",
"output": "741906166"
},
{
"input": "844509330 -887335829\n123329059",
"output": "844509330"
},
{
"input": "83712471 -876177148\n1213284777",
"output": "40110388"
},
{
"input": "598730524 -718984219\n1282749880",
"output": "401269483"
},
{
"input": "-474244697 -745885656\n1517883612",
"output": "271640959"
},
{
"input": "-502583588 -894906953\n1154189557",
"output": "497416419"
},
{
"input": "-636523651 -873305815\n154879215",
"output": "763217843"
},
{
"input": "721765550 594845720\n78862386",
"output": "126919830"
},
{
"input": "364141461 158854993\n1337196589",
"output": "364141461"
},
{
"input": "878985260 677031952\n394707801",
"output": "798046699"
},
{
"input": "439527072 -24854079\n1129147002",
"output": "464381151"
},
{
"input": "840435009 -612103127\n565968986",
"output": "387896880"
},
{
"input": "875035447 -826471373\n561914518",
"output": "124964560"
},
{
"input": "-342526698 305357084\n70776744",
"output": "352116225"
},
{
"input": "-903244186 899202229\n1527859274",
"output": "899202229"
},
{
"input": "-839482546 815166320\n1127472130",
"output": "839482546"
},
{
"input": "-976992569 -958313041\n1686580818",
"output": "981320479"
},
{
"input": "-497338894 -51069176\n737081851",
"output": "502661113"
},
{
"input": "-697962643 -143148799\n1287886520",
"output": "856851208"
},
{
"input": "-982572938 -482658433\n1259858332",
"output": "982572938"
},
{
"input": "123123 78817\n2000000000",
"output": "78817"
},
{
"input": "1000000000 -1000000000\n3",
"output": "14"
},
{
"input": "-1000000000 1000000000\n6",
"output": "14"
},
{
"input": "2 3\n6",
"output": "1000000006"
},
{
"input": "0 -1\n6",
"output": "1"
},
{
"input": "500000000 -1000000000\n600000003",
"output": "500000014"
},
{
"input": "-1000000000 1000000000\n3",
"output": "999999993"
},
{
"input": "1 3\n6",
"output": "1000000005"
},
{
"input": "1 2\n12",
"output": "1000000006"
},
{
"input": "7 -1000000000\n3",
"output": "0"
},
{
"input": "-999999997 999999997\n6",
"output": "20"
},
{
"input": "3 4\n6",
"output": "1000000006"
},
{
"input": "-1 2\n6",
"output": "1000000004"
},
{
"input": "2 3\n12",
"output": "1000000006"
},
{
"input": "4 18\n6",
"output": "999999993"
},
{
"input": "1 2\n6",
"output": "1000000006"
},
{
"input": "1000000000 -1000000000\n6",
"output": "999999993"
},
{
"input": "999999999 -999999999\n3",
"output": "16"
},
{
"input": "-1 0\n1",
"output": "1000000006"
},
{
"input": "1000000000 -1000000000\n9",
"output": "14"
},
{
"input": "999999999 -1000000000\n12",
"output": "999999992"
},
{
"input": "1000000000 -7\n3",
"output": "0"
},
{
"input": "-5 5\n6",
"output": "999999997"
},
{
"input": "5 9\n6",
"output": "1000000003"
},
{
"input": "-15 -10\n1",
"output": "999999992"
}
] | 1,646,111,788
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 74
| 62
| 0
|
f1,f2 = map(int, input().split())
n = int(input())
md = 10**9+7
if ((n-1)//3)%2==0:
ans = ([f1, f2, f2-f1][(n-1)%3])%md
else:
ans = ([-f1, -f2, f1-f2][(n-1)%3])%md
print(ans)
|
Title: Jzzhu and Sequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jzzhu has invented a kind of sequences, they meet the following property:
You are given *x* and *y*, please calculate *f**n* modulo 1000000007 (109<=+<=7).
Input Specification:
The first line contains two integers *x* and *y* (|*x*|,<=|*y*|<=≤<=109). The second line contains a single integer *n* (1<=≤<=*n*<=≤<=2·109).
Output Specification:
Output a single integer representing *f**n* modulo 1000000007 (109<=+<=7).
Demo Input:
['2 3\n3\n', '0 -1\n2\n']
Demo Output:
['1\n', '1000000006\n']
Note:
In the first sample, *f*<sub class="lower-index">2</sub> = *f*<sub class="lower-index">1</sub> + *f*<sub class="lower-index">3</sub>, 3 = 2 + *f*<sub class="lower-index">3</sub>, *f*<sub class="lower-index">3</sub> = 1.
In the second sample, *f*<sub class="lower-index">2</sub> = - 1; - 1 modulo (10<sup class="upper-index">9</sup> + 7) equals (10<sup class="upper-index">9</sup> + 6).
|
```python
f1,f2 = map(int, input().split())
n = int(input())
md = 10**9+7
if ((n-1)//3)%2==0:
ans = ([f1, f2, f2-f1][(n-1)%3])%md
else:
ans = ([-f1, -f2, f1-f2][(n-1)%3])%md
print(ans)
```
| 3
|
|
215
|
A
|
Bicycle Chain
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation.
We know that the *i*-th star on the pedal axle has *a**i* (0<=<<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=<<=*b*1<=<<=*b*2<=<<=...<=<<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value .
Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears.
In the problem, fraction denotes division in real numbers, that is, no rounding is performed.
|
The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing.
The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing.
It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces.
|
Print the number of "integer" gears with the maximum ratio among all "integer" gears.
|
[
"2\n4 5\n3\n12 13 15\n",
"4\n1 2 3 4\n5\n10 11 12 13 14\n"
] |
[
"2\n",
"1\n"
] |
In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15.
| 500
|
[
{
"input": "2\n4 5\n3\n12 13 15",
"output": "2"
},
{
"input": "4\n1 2 3 4\n5\n10 11 12 13 14",
"output": "1"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1 2\n1\n1",
"output": "1"
},
{
"input": "1\n1\n2\n1 2",
"output": "1"
},
{
"input": "4\n3 7 11 13\n4\n51 119 187 221",
"output": "4"
},
{
"input": "4\n2 3 4 5\n3\n1 2 3",
"output": "2"
},
{
"input": "10\n6 12 13 20 48 53 74 92 96 97\n10\n1 21 32 36 47 54 69 75 95 97",
"output": "1"
},
{
"input": "10\n5 9 10 14 15 17 19 22 24 26\n10\n2 11 17 19 21 22 24 25 27 28",
"output": "1"
},
{
"input": "10\n24 53 56 126 354 432 442 740 795 856\n10\n273 438 494 619 689 711 894 947 954 958",
"output": "1"
},
{
"input": "10\n3 4 6 7 8 10 14 16 19 20\n10\n3 4 5 7 8 10 15 16 18 20",
"output": "1"
},
{
"input": "10\n1 6 8 14 15 17 25 27 34 39\n10\n1 8 16 17 19 22 32 39 44 50",
"output": "1"
},
{
"input": "10\n5 21 22 23 25 32 35 36 38 39\n10\n3 7 8 9 18 21 23 24 36 38",
"output": "4"
},
{
"input": "50\n5 8 13 16 19 20 21 22 24 27 28 29 30 32 33 34 35 43 45 48 50 51 54 55 58 59 60 61 62 65 70 71 72 76 78 79 80 81 83 84 85 87 89 91 92 94 97 98 99 100\n50\n2 3 5 6 7 10 15 16 17 20 23 28 29 30 31 34 36 37 40 42 45 46 48 54 55 56 58 59 61 62 69 70 71 72 75 76 78 82 84 85 86 87 88 89 90 91 92 97 99 100",
"output": "1"
},
{
"input": "50\n3 5 6 8 9 11 13 19 21 23 24 32 34 35 42 50 51 52 56 58 59 69 70 72 73 75 76 77 78 80 83 88 90 95 96 100 101 102 108 109 113 119 124 135 138 141 142 143 145 150\n50\n5 8 10 11 18 19 23 30 35 43 51 53 55 58 63 68 69 71 77 78 79 82 83 86 88 89 91 92 93 94 96 102 103 105 109 110 113 114 116 123 124 126 127 132 133 135 136 137 142 149",
"output": "1"
},
{
"input": "50\n6 16 24 25 27 33 36 40 51 60 62 65 71 72 75 77 85 87 91 93 98 102 103 106 117 118 120 121 122 123 125 131 134 136 143 148 155 157 160 161 164 166 170 178 184 187 188 192 194 197\n50\n5 9 17 23 27 34 40 44 47 59 62 70 81 82 87 88 89 90 98 101 102 110 113 114 115 116 119 122 124 128 130 137 138 140 144 150 152 155 159 164 166 169 171 175 185 186 187 189 190 193",
"output": "1"
},
{
"input": "50\n14 22 23 31 32 35 48 63 76 79 88 97 101 102 103 104 106 113 114 115 116 126 136 138 145 152 155 156 162 170 172 173 179 180 182 203 208 210 212 222 226 229 231 232 235 237 245 246 247 248\n50\n2 5 6 16 28 44 45 46 54 55 56 63 72 80 87 93 94 96 97 100 101 103 132 135 140 160 164 165 167 168 173 180 182 185 186 192 194 198 199 202 203 211 213 216 217 227 232 233 236 245",
"output": "1"
},
{
"input": "50\n14 19 33 35 38 41 51 54 69 70 71 73 76 80 84 94 102 104 105 106 107 113 121 128 131 168 180 181 187 191 195 201 205 207 210 216 220 238 249 251 263 271 272 275 281 283 285 286 291 294\n50\n2 3 5 20 21 35 38 40 43 48 49 52 55 64 73 77 82 97 109 113 119 121 125 132 137 139 145 146 149 180 182 197 203 229 234 241 244 251 264 271 274 281 284 285 287 291 292 293 294 298",
"output": "1"
},
{
"input": "50\n2 4 5 16 18 19 22 23 25 26 34 44 48 54 67 79 80 84 92 110 116 133 138 154 163 171 174 202 205 218 228 229 234 245 247 249 250 263 270 272 274 275 277 283 289 310 312 334 339 342\n50\n1 5 17 18 25 37 46 47 48 59 67 75 80 83 84 107 115 122 137 141 159 162 175 180 184 204 221 224 240 243 247 248 249 258 259 260 264 266 269 271 274 293 294 306 329 330 334 335 342 350",
"output": "1"
},
{
"input": "50\n6 9 11 21 28 39 42 56 60 63 81 88 91 95 105 110 117 125 149 165 174 176 185 189 193 196 205 231 233 268 278 279 281 286 289 292 298 303 305 306 334 342 350 353 361 371 372 375 376 378\n50\n6 17 20 43 45 52 58 59 82 83 88 102 111 118 121 131 145 173 190 191 200 216 224 225 232 235 243 256 260 271 290 291 321 322 323 329 331 333 334 341 343 348 351 354 356 360 366 379 387 388",
"output": "1"
},
{
"input": "10\n17 239 443 467 661 1069 1823 2333 3767 4201\n20\n51 83 97 457 593 717 997 1329 1401 1459 1471 1983 2371 2539 3207 3251 3329 5469 6637 6999",
"output": "8"
},
{
"input": "20\n179 359 401 467 521 601 919 941 1103 1279 1709 1913 1949 2003 2099 2143 2179 2213 2399 4673\n20\n151 181 191 251 421 967 1109 1181 1249 1447 1471 1553 1619 2327 2551 2791 3049 3727 6071 7813",
"output": "3"
},
{
"input": "20\n79 113 151 709 809 983 1291 1399 1409 1429 2377 2659 2671 2897 3217 3511 3557 3797 3823 4363\n10\n19 101 659 797 1027 1963 2129 2971 3299 9217",
"output": "3"
},
{
"input": "30\n19 47 109 179 307 331 389 401 461 509 547 569 617 853 883 1249 1361 1381 1511 1723 1741 1783 2459 2531 2621 3533 3821 4091 5557 6217\n20\n401 443 563 941 967 997 1535 1567 1655 1747 1787 1945 1999 2251 2305 2543 2735 4415 6245 7555",
"output": "8"
},
{
"input": "30\n3 43 97 179 257 313 353 359 367 389 397 457 547 599 601 647 1013 1021 1063 1433 1481 1531 1669 3181 3373 3559 3769 4157 4549 5197\n50\n13 15 17 19 29 79 113 193 197 199 215 223 271 293 359 485 487 569 601 683 895 919 941 967 1283 1285 1289 1549 1565 1765 1795 1835 1907 1931 1945 1985 1993 2285 2731 2735 2995 3257 4049 4139 5105 5315 7165 7405 7655 8345",
"output": "20"
},
{
"input": "50\n11 17 23 53 59 109 137 149 173 251 353 379 419 421 439 503 593 607 661 773 821 877 941 997 1061 1117 1153 1229 1289 1297 1321 1609 1747 2311 2389 2543 2693 3041 3083 3137 3181 3209 3331 3373 3617 3767 4201 4409 4931 6379\n50\n55 59 67 73 85 89 101 115 211 263 295 353 545 599 607 685 739 745 997 1031 1255 1493 1523 1667 1709 1895 1949 2161 2195 2965 3019 3035 3305 3361 3373 3673 3739 3865 3881 4231 4253 4385 4985 5305 5585 5765 6145 6445 8045 8735",
"output": "23"
},
{
"input": "5\n33 78 146 3055 4268\n5\n2211 2584 5226 9402 9782",
"output": "3"
},
{
"input": "5\n35 48 52 86 8001\n10\n332 3430 3554 4704 4860 5096 6215 7583 8228 8428",
"output": "4"
},
{
"input": "10\n97 184 207 228 269 2084 4450 6396 7214 9457\n16\n338 1179 1284 1545 1570 2444 3167 3395 3397 5550 6440 7245 7804 7980 9415 9959",
"output": "5"
},
{
"input": "30\n25 30 41 57 58 62 70 72 76 79 84 85 88 91 98 101 104 109 119 129 136 139 148 151 926 1372 3093 3936 5423 7350\n25\n1600 1920 2624 3648 3712 3968 4480 4608 4864 5056 5376 5440 5632 5824 6272 6464 6656 6934 6976 7616 8256 8704 8896 9472 9664",
"output": "24"
},
{
"input": "5\n33 78 146 3055 4268\n5\n2211 2584 5226 9402 9782",
"output": "3"
},
{
"input": "5\n35 48 52 86 8001\n10\n332 3430 3554 4704 4860 5096 6215 7583 8228 8428",
"output": "4"
},
{
"input": "10\n97 184 207 228 269 2084 4450 6396 7214 9457\n16\n338 1179 1284 1545 1570 2444 3167 3395 3397 5550 6440 7245 7804 7980 9415 9959",
"output": "5"
},
{
"input": "30\n25 30 41 57 58 62 70 72 76 79 84 85 88 91 98 101 104 109 119 129 136 139 148 151 926 1372 3093 3936 5423 7350\n25\n1600 1920 2624 3648 3712 3968 4480 4608 4864 5056 5376 5440 5632 5824 6272 6464 6656 6934 6976 7616 8256 8704 8896 9472 9664",
"output": "24"
},
{
"input": "47\n66 262 357 457 513 530 538 540 592 691 707 979 1015 1242 1246 1667 1823 1886 1963 2133 2649 2679 2916 2949 3413 3523 3699 3958 4393 4922 5233 5306 5799 6036 6302 6629 7208 7282 7315 7822 7833 7927 8068 8150 8870 8962 9987\n39\n167 199 360 528 1515 1643 1986 1988 2154 2397 2856 3552 3656 3784 3980 4096 4104 4240 4320 4736 4951 5266 5656 5849 5850 6169 6517 6875 7244 7339 7689 7832 8120 8716 9503 9509 9933 9936 9968",
"output": "12"
},
{
"input": "1\n94\n50\n423 446 485 1214 1468 1507 1853 1930 1999 2258 2271 2285 2425 2543 2715 2743 2992 3196 4074 4108 4448 4475 4652 5057 5250 5312 5356 5375 5731 5986 6298 6501 6521 7146 7255 7276 7332 7481 7998 8141 8413 8665 8908 9221 9336 9491 9504 9677 9693 9706",
"output": "1"
},
{
"input": "50\n51 67 75 186 194 355 512 561 720 876 1077 1221 1503 1820 2153 2385 2568 2608 2937 2969 3271 3311 3481 4081 4093 4171 4255 4256 4829 5020 5192 5636 5817 6156 6712 6717 7153 7436 7608 7612 7866 7988 8264 8293 8867 9311 9879 9882 9889 9908\n1\n5394",
"output": "1"
},
{
"input": "50\n26 367 495 585 675 789 855 1185 1312 1606 2037 2241 2587 2612 2628 2807 2873 2924 3774 4067 4376 4668 4902 5001 5082 5100 5104 5209 5345 5515 5661 5777 5902 5907 6155 6323 6675 6791 7503 8159 8207 8254 8740 8848 8855 8933 9069 9164 9171 9586\n5\n1557 6246 7545 8074 8284",
"output": "1"
},
{
"input": "5\n25 58 91 110 2658\n50\n21 372 909 1172 1517 1554 1797 1802 1843 1977 2006 2025 2137 2225 2317 2507 2645 2754 2919 3024 3202 3212 3267 3852 4374 4487 4553 4668 4883 4911 4916 5016 5021 5068 5104 5162 5683 5856 6374 6871 7333 7531 8099 8135 8173 8215 8462 8776 9433 9790",
"output": "4"
},
{
"input": "45\n37 48 56 59 69 70 79 83 85 86 99 114 131 134 135 145 156 250 1739 1947 2116 2315 2449 3104 3666 4008 4406 4723 4829 5345 5836 6262 6296 6870 7065 7110 7130 7510 7595 8092 8442 8574 9032 9091 9355\n50\n343 846 893 1110 1651 1837 2162 2331 2596 3012 3024 3131 3294 3394 3528 3717 3997 4125 4347 4410 4581 4977 5030 5070 5119 5229 5355 5413 5418 5474 5763 5940 6151 6161 6164 6237 6506 6519 6783 7182 7413 7534 8069 8253 8442 8505 9135 9308 9828 9902",
"output": "17"
},
{
"input": "50\n17 20 22 28 36 38 46 47 48 50 52 57 58 62 63 69 70 74 75 78 79 81 82 86 87 90 93 95 103 202 292 442 1756 1769 2208 2311 2799 2957 3483 4280 4324 4932 5109 5204 6225 6354 6561 7136 8754 9670\n40\n68 214 957 1649 1940 2078 2134 2716 3492 3686 4462 4559 4656 4756 4850 5044 5490 5529 5592 5626 6014 6111 6693 6790 7178 7275 7566 7663 7702 7857 7954 8342 8511 8730 8957 9021 9215 9377 9445 9991",
"output": "28"
},
{
"input": "39\n10 13 21 25 36 38 47 48 58 64 68 69 73 79 86 972 2012 2215 2267 2503 3717 3945 4197 4800 5266 6169 6612 6824 7023 7322 7582 7766 8381 8626 8879 9079 9088 9838 9968\n50\n432 877 970 1152 1202 1223 1261 1435 1454 1578 1843 1907 2003 2037 2183 2195 2215 2425 3065 3492 3615 3637 3686 3946 4189 4415 4559 4656 4665 4707 4886 4887 5626 5703 5955 6208 6521 6581 6596 6693 6985 7013 7081 7343 7663 8332 8342 8637 9207 9862",
"output": "15"
},
{
"input": "50\n7 144 269 339 395 505 625 688 709 950 1102 1152 1350 1381 1641 1830 1977 1999 2093 2180 2718 3308 3574 4168 4232 4259 4393 4689 4982 5154 5476 5581 5635 5721 6159 6302 6741 7010 7152 7315 7417 7482 8116 8239 8640 9347 9395 9614 9661 9822\n20\n84 162 292 1728 1866 2088 3228 3470 4068 5318 5470 6060 6380 6929 7500 8256 8399 8467 8508 9691",
"output": "8"
},
{
"input": "50\n159 880 1070 1139 1358 1608 1691 1841 2073 2171 2213 2597 2692 2759 2879 2931 3173 3217 3441 4201 4878 5106 5129 5253 5395 5647 5968 6019 6130 6276 6286 6330 6409 6728 7488 7713 7765 7828 7899 8064 8264 8457 8483 8685 8900 8946 8965 9133 9187 9638\n45\n57 159 1070 1139 1391 1608 1691 1841 2171 2213 2692 2759 2931 3173 3217 3441 4201 4878 5106 5129 5253 5647 5968 6130 6276 6286 6409 7488 7694 7713 7765 7828 7899 8003 8064 8081 8244 8264 8685 8900 8946 8965 9133 9638 9673",
"output": "38"
},
{
"input": "3\n3 4 5\n3\n6 20 25",
"output": "2"
},
{
"input": "4\n2 3 5 8\n4\n2 6 8 10",
"output": "1"
},
{
"input": "4\n3 5 7 11\n4\n3 5 7 22",
"output": "1"
},
{
"input": "2\n2 3\n3\n20 30 50",
"output": "1"
},
{
"input": "3\n1 2 3\n4\n2 4 6 49",
"output": "1"
},
{
"input": "2\n4 5\n3\n12 15 20",
"output": "1"
},
{
"input": "3\n2 5 7\n3\n4 5 7",
"output": "1"
},
{
"input": "3\n3 5 8\n3\n6 8 10",
"output": "2"
},
{
"input": "2\n2 3\n4\n4 6 9 33",
"output": "1"
},
{
"input": "2\n2 3\n4\n4 6 21 40",
"output": "1"
},
{
"input": "3\n4 9 10\n3\n8 9 10",
"output": "1"
},
{
"input": "5\n1 5 6 9 51\n5\n5 12 18 27 10000",
"output": "1"
},
{
"input": "13\n1 2 3 4 5 6 7 8 9 10 11 12 13\n1\n14",
"output": "1"
}
] | 1,591,781,582
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 57
| 280
| 1,331,200
|
# cook your dish here
n = int(input())
a = list(map(int, input().split()))
m = int(input())
b = list(map(int, input().split()))
count=0
m =0
mylist = []
for i in a:
for j in b:
if j%i==0:
c = j/i
mylist.append(c)
m = max(mylist)
count = mylist.count(m)
print(count)
|
Title: Bicycle Chain
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya's bicycle chain drive consists of two parts: *n* stars are attached to the pedal axle, *m* stars are attached to the rear wheel axle. The chain helps to rotate the rear wheel by transmitting the pedal rotation.
We know that the *i*-th star on the pedal axle has *a**i* (0<=<<=*a*1<=<<=*a*2<=<<=...<=<<=*a**n*) teeth, and the *j*-th star on the rear wheel axle has *b**j* (0<=<<=*b*1<=<<=*b*2<=<<=...<=<<=*b**m*) teeth. Any pair (*i*,<=*j*) (1<=≤<=*i*<=≤<=*n*; 1<=≤<=*j*<=≤<=*m*) is called a gear and sets the indexes of stars to which the chain is currently attached. Gear (*i*,<=*j*) has a gear ratio, equal to the value .
Since Vasya likes integers, he wants to find such gears (*i*,<=*j*), that their ratios are integers. On the other hand, Vasya likes fast driving, so among all "integer" gears (*i*,<=*j*) he wants to choose a gear with the maximum ratio. Help him to find the number of such gears.
In the problem, fraction denotes division in real numbers, that is, no rounding is performed.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stars on the bicycle's pedal axle. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) in the order of strict increasing.
The third input line contains integer *m* (1<=≤<=*m*<=≤<=50) — the number of stars on the rear wheel axle. The fourth line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=104) in the order of strict increasing.
It is guaranteed that there exists at least one gear (*i*,<=*j*), that its gear ratio is an integer. The numbers on the lines are separated by spaces.
Output Specification:
Print the number of "integer" gears with the maximum ratio among all "integer" gears.
Demo Input:
['2\n4 5\n3\n12 13 15\n', '4\n1 2 3 4\n5\n10 11 12 13 14\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample the maximum "integer" gear ratio equals 3. There are two gears that have such gear ratio. For one of them *a*<sub class="lower-index">1</sub> = 4, *b*<sub class="lower-index">1</sub> = 12, and for the other *a*<sub class="lower-index">2</sub> = 5, *b*<sub class="lower-index">3</sub> = 15.
|
```python
# cook your dish here
n = int(input())
a = list(map(int, input().split()))
m = int(input())
b = list(map(int, input().split()))
count=0
m =0
mylist = []
for i in a:
for j in b:
if j%i==0:
c = j/i
mylist.append(c)
m = max(mylist)
count = mylist.count(m)
print(count)
```
| 3
|
|
426
|
B
|
Sereja and Mirroring
|
PROGRAMMING
| 1,300
|
[
"implementation"
] | null | null |
Let's assume that we are given a matrix *b* of size *x*<=×<=*y*, let's determine the operation of mirroring matrix *b*. The mirroring of matrix *b* is a 2*x*<=×<=*y* matrix *c* which has the following properties:
- the upper half of matrix *c* (rows with numbers from 1 to *x*) exactly matches *b*; - the lower half of matrix *c* (rows with numbers from *x*<=+<=1 to 2*x*) is symmetric to the upper one; the symmetry line is the line that separates two halves (the line that goes in the middle, between rows *x* and *x*<=+<=1).
Sereja has an *n*<=×<=*m* matrix *a*. He wants to find such matrix *b*, that it can be transformed into matrix *a*, if we'll perform on it several (possibly zero) mirrorings. What minimum number of rows can such matrix contain?
|
The first line contains two integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). Each of the next *n* lines contains *m* integers — the elements of matrix *a*. The *i*-th line contains integers *a**i*1,<=*a**i*2,<=...,<=*a**im* (0<=≤<=*a**ij*<=≤<=1) — the *i*-th row of the matrix *a*.
|
In the single line, print the answer to the problem — the minimum number of rows of matrix *b*.
|
[
"4 3\n0 0 1\n1 1 0\n1 1 0\n0 0 1\n",
"3 3\n0 0 0\n0 0 0\n0 0 0\n",
"8 1\n0\n1\n1\n0\n0\n1\n1\n0\n"
] |
[
"2\n",
"3\n",
"2\n"
] |
In the first test sample the answer is a 2 × 3 matrix *b*:
If we perform a mirroring operation with this matrix, we get the matrix *a* that is given in the input:
| 1,000
|
[
{
"input": "4 3\n0 0 1\n1 1 0\n1 1 0\n0 0 1",
"output": "2"
},
{
"input": "3 3\n0 0 0\n0 0 0\n0 0 0",
"output": "3"
},
{
"input": "8 1\n0\n1\n1\n0\n0\n1\n1\n0",
"output": "2"
},
{
"input": "10 4\n0 0 1 0\n0 0 1 0\n1 1 0 1\n0 0 1 1\n1 0 1 0\n1 0 1 0\n0 0 1 1\n1 1 0 1\n0 0 1 0\n0 0 1 0",
"output": "5"
},
{
"input": "10 3\n0 0 0\n1 1 1\n1 1 0\n0 0 0\n0 1 1\n0 1 1\n0 0 0\n1 1 0\n1 1 1\n0 0 0",
"output": "5"
},
{
"input": "8 4\n1 0 0 0\n1 1 0 0\n1 0 0 1\n1 1 1 1\n0 0 1 1\n0 1 0 1\n0 1 1 1\n1 0 0 0",
"output": "8"
},
{
"input": "2 9\n1 0 0 1 1 1 0 1 0\n1 0 0 1 0 0 0 1 1",
"output": "2"
},
{
"input": "10 3\n0 1 0\n1 1 1\n1 0 1\n0 0 1\n1 0 1\n1 0 0\n1 1 0\n1 1 1\n1 0 1\n0 0 1",
"output": "10"
},
{
"input": "8 4\n1 1 0 1\n0 0 0 0\n0 0 0 0\n1 1 0 1\n1 1 0 1\n0 0 0 0\n0 0 0 0\n1 1 0 1",
"output": "2"
},
{
"input": "8 7\n1 1 0 0 1 1 0\n1 1 0 0 1 1 0\n1 1 0 0 1 1 0\n1 1 0 0 1 1 0\n1 1 0 0 1 1 0\n1 1 0 0 1 1 0\n1 1 0 0 1 1 0\n1 1 0 0 1 1 0",
"output": "1"
},
{
"input": "6 5\n0 0 1 0 1\n1 0 0 1 0\n1 1 1 0 0\n1 0 1 1 0\n0 0 0 0 0\n1 0 1 0 0",
"output": "6"
},
{
"input": "1 69\n0 0 1 1 1 1 0 0 1 1 1 0 0 0 1 1 1 1 1 1 1 0 0 1 0 0 1 1 1 0 0 1 1 0 1 1 1 1 0 0 0 0 0 0 1 1 1 0 1 1 0 1 0 0 1 0 0 0 1 1 1 1 1 1 1 1 0 1 0",
"output": "1"
},
{
"input": "8 20\n0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 1 1 0\n1 0 1 0 1 0 0 0 0 1 0 1 0 1 1 0 1 1 1 1\n1 0 1 0 1 0 0 0 0 1 0 1 0 1 1 0 1 1 1 1\n0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 1 1 0\n0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 1 1 0\n1 0 1 0 1 0 0 0 0 1 0 1 0 1 1 0 1 1 1 1\n1 0 1 0 1 0 0 0 0 1 0 1 0 1 1 0 1 1 1 1\n0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 1 1 0",
"output": "2"
},
{
"input": "1 1\n0",
"output": "1"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "2 2\n1 0\n0 1",
"output": "2"
},
{
"input": "2 2\n0 1\n0 1",
"output": "1"
},
{
"input": "1 2\n0 1",
"output": "1"
},
{
"input": "1 100\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",
"output": "1"
},
{
"input": "1 100\n1 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 1",
"output": "1"
},
{
"input": "1 100\n0 0 0 0 0 0 0 0 1 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",
"output": "1"
},
{
"input": "100 1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "25"
},
{
"input": "100 1\n1\n1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1\n1",
"output": "50"
},
{
"input": "100 1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1\n0\n0\n0",
"output": "100"
},
{
"input": "8 1\n1\n0\n0\n1\n1\n0\n1\n1",
"output": "8"
},
{
"input": "6 1\n0\n0\n0\n0\n0\n0",
"output": "3"
},
{
"input": "10 2\n1 1\n0 0\n0 0\n1 1\n0 0\n0 0\n1 1\n0 0\n0 0\n1 1",
"output": "5"
},
{
"input": "4 2\n1 1\n0 0\n0 0\n0 0",
"output": "4"
},
{
"input": "6 3\n1 1 1\n0 0 0\n1 1 1\n1 1 1\n0 0 0\n1 1 1",
"output": "3"
},
{
"input": "6 3\n1 1 1\n1 0 1\n1 1 1\n1 1 1\n1 0 1\n1 1 1",
"output": "3"
},
{
"input": "6 3\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "3"
},
{
"input": "4 3\n1 0 1\n0 1 0\n1 0 1\n1 0 1",
"output": "4"
},
{
"input": "6 1\n1\n1\n1\n1\n1\n1",
"output": "3"
},
{
"input": "10 1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "5"
},
{
"input": "3 1\n1\n1\n1",
"output": "3"
},
{
"input": "6 3\n0 0 0\n0 0 0\n0 0 0\n0 0 0\n0 0 0\n0 0 0",
"output": "3"
},
{
"input": "6 1\n1\n0\n1\n1\n0\n1",
"output": "3"
},
{
"input": "6 6\n0 0 0 0 0 0\n0 0 0 0 0 0\n0 0 0 0 0 0\n0 0 0 0 0 0\n0 0 0 0 0 0\n0 0 0 0 0 0",
"output": "3"
},
{
"input": "3 1\n1\n0\n1",
"output": "3"
},
{
"input": "12 3\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "3"
},
{
"input": "12 1\n0\n1\n0\n0\n1\n0\n0\n1\n0\n0\n1\n0",
"output": "3"
},
{
"input": "2 3\n0 0 0\n0 0 0",
"output": "1"
},
{
"input": "3 3\n1 1 1\n1 1 1\n0 0 0",
"output": "3"
},
{
"input": "10 1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "5"
},
{
"input": "12 1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "3"
},
{
"input": "6 2\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "3"
},
{
"input": "6 3\n1 1 0\n0 0 0\n1 1 0\n1 1 0\n0 0 0\n1 1 0",
"output": "3"
},
{
"input": "6 2\n0 1\n0 1\n0 1\n0 1\n0 1\n0 1",
"output": "3"
},
{
"input": "12 1\n1\n1\n0\n0\n1\n1\n1\n1\n0\n0\n1\n1",
"output": "3"
},
{
"input": "6 2\n1 0\n1 0\n1 0\n1 0\n1 0\n1 0",
"output": "3"
}
] | 1,620,296,595
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 65
| 124
| 21,606,400
|
def issymmetric(n, mat):
if n%2: return False
else:
return mat[:n//2] == mat[n//2:][::-1]
n, m = map(int, input().split())
mat = []
for i in range(n):
mat.append(list(map(int, input().split())))
#print(n, mat)
while issymmetric(n, mat):
mat = mat[:n//2]
n //= 2
print(n)
|
Title: Sereja and Mirroring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's assume that we are given a matrix *b* of size *x*<=×<=*y*, let's determine the operation of mirroring matrix *b*. The mirroring of matrix *b* is a 2*x*<=×<=*y* matrix *c* which has the following properties:
- the upper half of matrix *c* (rows with numbers from 1 to *x*) exactly matches *b*; - the lower half of matrix *c* (rows with numbers from *x*<=+<=1 to 2*x*) is symmetric to the upper one; the symmetry line is the line that separates two halves (the line that goes in the middle, between rows *x* and *x*<=+<=1).
Sereja has an *n*<=×<=*m* matrix *a*. He wants to find such matrix *b*, that it can be transformed into matrix *a*, if we'll perform on it several (possibly zero) mirrorings. What minimum number of rows can such matrix contain?
Input Specification:
The first line contains two integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100). Each of the next *n* lines contains *m* integers — the elements of matrix *a*. The *i*-th line contains integers *a**i*1,<=*a**i*2,<=...,<=*a**im* (0<=≤<=*a**ij*<=≤<=1) — the *i*-th row of the matrix *a*.
Output Specification:
In the single line, print the answer to the problem — the minimum number of rows of matrix *b*.
Demo Input:
['4 3\n0 0 1\n1 1 0\n1 1 0\n0 0 1\n', '3 3\n0 0 0\n0 0 0\n0 0 0\n', '8 1\n0\n1\n1\n0\n0\n1\n1\n0\n']
Demo Output:
['2\n', '3\n', '2\n']
Note:
In the first test sample the answer is a 2 × 3 matrix *b*:
If we perform a mirroring operation with this matrix, we get the matrix *a* that is given in the input:
|
```python
def issymmetric(n, mat):
if n%2: return False
else:
return mat[:n//2] == mat[n//2:][::-1]
n, m = map(int, input().split())
mat = []
for i in range(n):
mat.append(list(map(int, input().split())))
#print(n, mat)
while issymmetric(n, mat):
mat = mat[:n//2]
n //= 2
print(n)
```
| 3
|
|
916
|
A
|
Jamie and Alarm Snooze
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"math"
] | null | null |
Jamie loves sleeping. One day, he decides that he needs to wake up at exactly *hh*:<=*mm*. However, he hates waking up, so he wants to make waking up less painful by setting the alarm at a lucky time. He will then press the snooze button every *x* minutes until *hh*:<=*mm* is reached, and only then he will wake up. He wants to know what is the smallest number of times he needs to press the snooze button.
A time is considered lucky if it contains a digit '7'. For example, 13:<=07 and 17:<=27 are lucky, while 00:<=48 and 21:<=34 are not lucky.
Note that it is not necessary that the time set for the alarm and the wake-up time are on the same day. It is guaranteed that there is a lucky time Jamie can set so that he can wake at *hh*:<=*mm*.
Formally, find the smallest possible non-negative integer *y* such that the time representation of the time *x*·*y* minutes before *hh*:<=*mm* contains the digit '7'.
Jamie uses 24-hours clock, so after 23:<=59 comes 00:<=00.
|
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=60).
The second line contains two two-digit integers, *hh* and *mm* (00<=≤<=*hh*<=≤<=23,<=00<=≤<=*mm*<=≤<=59).
|
Print the minimum number of times he needs to press the button.
|
[
"3\n11 23\n",
"5\n01 07\n"
] |
[
"2\n",
"0\n"
] |
In the first sample, Jamie needs to wake up at 11:23. So, he can set his alarm at 11:17. He would press the snooze button when the alarm rings at 11:17 and at 11:20.
In the second sample, Jamie can set his alarm at exactly at 01:07 which is lucky.
| 500
|
[
{
"input": "3\n11 23",
"output": "2"
},
{
"input": "5\n01 07",
"output": "0"
},
{
"input": "34\n09 24",
"output": "3"
},
{
"input": "2\n14 37",
"output": "0"
},
{
"input": "14\n19 54",
"output": "9"
},
{
"input": "42\n15 44",
"output": "12"
},
{
"input": "46\n02 43",
"output": "1"
},
{
"input": "14\n06 41",
"output": "1"
},
{
"input": "26\n04 58",
"output": "26"
},
{
"input": "54\n16 47",
"output": "0"
},
{
"input": "38\n20 01",
"output": "3"
},
{
"input": "11\n02 05",
"output": "8"
},
{
"input": "55\n22 10",
"output": "5"
},
{
"input": "23\n10 08",
"output": "6"
},
{
"input": "23\n23 14",
"output": "9"
},
{
"input": "51\n03 27",
"output": "0"
},
{
"input": "35\n15 25",
"output": "13"
},
{
"input": "3\n12 15",
"output": "6"
},
{
"input": "47\n00 28",
"output": "3"
},
{
"input": "31\n13 34",
"output": "7"
},
{
"input": "59\n17 32",
"output": "0"
},
{
"input": "25\n11 03",
"output": "8"
},
{
"input": "9\n16 53",
"output": "4"
},
{
"input": "53\n04 06",
"output": "3"
},
{
"input": "37\n00 12",
"output": "5"
},
{
"input": "5\n13 10",
"output": "63"
},
{
"input": "50\n01 59",
"output": "10"
},
{
"input": "34\n06 13",
"output": "4"
},
{
"input": "2\n18 19",
"output": "1"
},
{
"input": "46\n06 16",
"output": "17"
},
{
"input": "14\n03 30",
"output": "41"
},
{
"input": "40\n13 37",
"output": "0"
},
{
"input": "24\n17 51",
"output": "0"
},
{
"input": "8\n14 57",
"output": "0"
},
{
"input": "52\n18 54",
"output": "2"
},
{
"input": "20\n15 52",
"output": "24"
},
{
"input": "20\n03 58",
"output": "30"
},
{
"input": "48\n07 11",
"output": "0"
},
{
"input": "32\n04 01",
"output": "2"
},
{
"input": "60\n08 15",
"output": "1"
},
{
"input": "44\n20 20",
"output": "4"
},
{
"input": "55\n15 35",
"output": "9"
},
{
"input": "55\n03 49",
"output": "11"
},
{
"input": "23\n16 39",
"output": "4"
},
{
"input": "7\n20 36",
"output": "7"
},
{
"input": "35\n16 42",
"output": "1"
},
{
"input": "35\n05 56",
"output": "21"
},
{
"input": "3\n17 45",
"output": "0"
},
{
"input": "47\n05 59",
"output": "6"
},
{
"input": "15\n10 13",
"output": "9"
},
{
"input": "59\n06 18",
"output": "9"
},
{
"input": "34\n17 18",
"output": "0"
},
{
"input": "18\n05 23",
"output": "2"
},
{
"input": "46\n17 21",
"output": "0"
},
{
"input": "30\n06 27",
"output": "0"
},
{
"input": "14\n18 40",
"output": "3"
},
{
"input": "58\n22 54",
"output": "6"
},
{
"input": "26\n19 44",
"output": "5"
},
{
"input": "10\n15 57",
"output": "0"
},
{
"input": "54\n20 47",
"output": "0"
},
{
"input": "22\n08 45",
"output": "3"
},
{
"input": "48\n18 08",
"output": "1"
},
{
"input": "32\n07 06",
"output": "0"
},
{
"input": "60\n19 19",
"output": "2"
},
{
"input": "45\n07 25",
"output": "0"
},
{
"input": "29\n12 39",
"output": "8"
},
{
"input": "13\n08 28",
"output": "3"
},
{
"input": "41\n21 42",
"output": "5"
},
{
"input": "41\n09 32",
"output": "3"
},
{
"input": "9\n21 45",
"output": "2"
},
{
"input": "37\n10 43",
"output": "5"
},
{
"input": "3\n20 50",
"output": "1"
},
{
"input": "47\n00 04",
"output": "1"
},
{
"input": "15\n13 10",
"output": "21"
},
{
"input": "15\n17 23",
"output": "0"
},
{
"input": "43\n22 13",
"output": "2"
},
{
"input": "27\n10 26",
"output": "6"
},
{
"input": "55\n22 24",
"output": "5"
},
{
"input": "55\n03 30",
"output": "11"
},
{
"input": "24\n23 27",
"output": "0"
},
{
"input": "52\n11 33",
"output": "3"
},
{
"input": "18\n22 48",
"output": "17"
},
{
"input": "1\n12 55",
"output": "8"
},
{
"input": "1\n04 27",
"output": "0"
},
{
"input": "1\n12 52",
"output": "5"
},
{
"input": "1\n20 16",
"output": "9"
},
{
"input": "1\n04 41",
"output": "4"
},
{
"input": "1\n20 21",
"output": "4"
},
{
"input": "1\n04 45",
"output": "8"
},
{
"input": "1\n12 18",
"output": "1"
},
{
"input": "1\n04 42",
"output": "5"
},
{
"input": "1\n02 59",
"output": "2"
},
{
"input": "1\n18 24",
"output": "7"
},
{
"input": "1\n02 04",
"output": "7"
},
{
"input": "1\n18 28",
"output": "1"
},
{
"input": "1\n18 01",
"output": "2"
},
{
"input": "1\n10 25",
"output": "8"
},
{
"input": "1\n02 49",
"output": "2"
},
{
"input": "1\n02 30",
"output": "3"
},
{
"input": "1\n18 54",
"output": "7"
},
{
"input": "1\n02 19",
"output": "2"
},
{
"input": "1\n05 25",
"output": "8"
},
{
"input": "60\n23 55",
"output": "6"
},
{
"input": "60\n08 19",
"output": "1"
},
{
"input": "60\n00 00",
"output": "7"
},
{
"input": "60\n08 24",
"output": "1"
},
{
"input": "60\n16 13",
"output": "9"
},
{
"input": "60\n08 21",
"output": "1"
},
{
"input": "60\n16 45",
"output": "9"
},
{
"input": "60\n08 26",
"output": "1"
},
{
"input": "60\n08 50",
"output": "1"
},
{
"input": "60\n05 21",
"output": "12"
},
{
"input": "60\n13 29",
"output": "6"
},
{
"input": "60\n05 18",
"output": "12"
},
{
"input": "60\n13 42",
"output": "6"
},
{
"input": "60\n05 07",
"output": "0"
},
{
"input": "60\n05 47",
"output": "0"
},
{
"input": "60\n21 55",
"output": "4"
},
{
"input": "60\n05 36",
"output": "12"
},
{
"input": "60\n21 08",
"output": "4"
},
{
"input": "60\n21 32",
"output": "4"
},
{
"input": "60\n16 31",
"output": "9"
},
{
"input": "5\n00 00",
"output": "73"
},
{
"input": "2\n06 58",
"output": "390"
},
{
"input": "60\n00 00",
"output": "7"
},
{
"input": "2\n00 00",
"output": "181"
},
{
"input": "10\n00 00",
"output": "37"
},
{
"input": "60\n01 00",
"output": "8"
},
{
"input": "12\n00 06",
"output": "31"
},
{
"input": "1\n00 01",
"output": "4"
},
{
"input": "5\n00 05",
"output": "74"
},
{
"input": "60\n01 01",
"output": "8"
},
{
"input": "11\n18 11",
"output": "2"
},
{
"input": "60\n01 15",
"output": "8"
},
{
"input": "10\n00 16",
"output": "38"
},
{
"input": "60\n00 59",
"output": "7"
},
{
"input": "30\n00 00",
"output": "13"
},
{
"input": "60\n01 05",
"output": "8"
},
{
"input": "4\n00 03",
"output": "4"
},
{
"input": "4\n00 00",
"output": "91"
},
{
"input": "60\n00 01",
"output": "7"
},
{
"input": "6\n00 03",
"output": "1"
},
{
"input": "13\n00 00",
"output": "1"
},
{
"input": "1\n18 01",
"output": "2"
},
{
"input": "5\n06 00",
"output": "145"
},
{
"input": "60\n04 08",
"output": "11"
},
{
"input": "5\n01 55",
"output": "96"
},
{
"input": "8\n00 08",
"output": "47"
},
{
"input": "23\n18 23",
"output": "2"
},
{
"input": "6\n00 06",
"output": "62"
},
{
"input": "59\n18 59",
"output": "2"
},
{
"input": "11\n00 10",
"output": "3"
},
{
"input": "10\n00 01",
"output": "37"
},
{
"input": "59\n00 00",
"output": "7"
},
{
"input": "10\n18 10",
"output": "2"
},
{
"input": "5\n00 01",
"output": "73"
},
{
"input": "1\n00 00",
"output": "3"
},
{
"input": "8\n00 14",
"output": "47"
},
{
"input": "60\n03 00",
"output": "10"
},
{
"input": "60\n00 10",
"output": "7"
},
{
"input": "5\n01 13",
"output": "87"
},
{
"input": "30\n02 43",
"output": "18"
},
{
"input": "17\n00 08",
"output": "3"
},
{
"input": "3\n00 00",
"output": "1"
},
{
"input": "60\n00 05",
"output": "7"
},
{
"input": "5\n18 05",
"output": "2"
},
{
"input": "30\n00 30",
"output": "14"
},
{
"input": "1\n00 06",
"output": "9"
},
{
"input": "55\n00 00",
"output": "7"
},
{
"input": "8\n02 08",
"output": "62"
},
{
"input": "7\n00 00",
"output": "9"
},
{
"input": "6\n08 06",
"output": "2"
},
{
"input": "48\n06 24",
"output": "16"
},
{
"input": "8\n06 58",
"output": "98"
},
{
"input": "3\n12 00",
"output": "1"
},
{
"input": "5\n01 06",
"output": "86"
},
{
"input": "2\n00 08",
"output": "185"
},
{
"input": "3\n18 03",
"output": "2"
},
{
"input": "1\n17 00",
"output": "0"
},
{
"input": "59\n00 48",
"output": "7"
},
{
"input": "5\n12 01",
"output": "49"
},
{
"input": "55\n01 25",
"output": "9"
},
{
"input": "2\n07 23",
"output": "0"
},
{
"input": "10\n01 10",
"output": "44"
},
{
"input": "2\n00 01",
"output": "2"
},
{
"input": "59\n00 01",
"output": "6"
},
{
"input": "5\n00 02",
"output": "1"
},
{
"input": "4\n01 02",
"output": "106"
},
{
"input": "5\n00 06",
"output": "74"
},
{
"input": "42\n00 08",
"output": "9"
},
{
"input": "60\n01 20",
"output": "8"
},
{
"input": "3\n06 00",
"output": "1"
},
{
"input": "4\n00 01",
"output": "1"
},
{
"input": "2\n00 06",
"output": "184"
},
{
"input": "1\n00 57",
"output": "0"
},
{
"input": "6\n00 00",
"output": "61"
},
{
"input": "5\n08 40",
"output": "9"
},
{
"input": "58\n00 55",
"output": "1"
},
{
"input": "2\n00 02",
"output": "182"
},
{
"input": "1\n08 01",
"output": "2"
},
{
"input": "10\n10 10",
"output": "14"
},
{
"input": "60\n01 11",
"output": "8"
},
{
"input": "2\n07 00",
"output": "0"
},
{
"input": "15\n00 03",
"output": "25"
},
{
"input": "6\n04 34",
"output": "106"
},
{
"input": "16\n00 16",
"output": "24"
},
{
"input": "2\n00 59",
"output": "1"
},
{
"input": "59\n00 08",
"output": "7"
},
{
"input": "10\n03 10",
"output": "56"
},
{
"input": "3\n08 03",
"output": "2"
},
{
"input": "20\n06 11",
"output": "37"
},
{
"input": "4\n01 00",
"output": "106"
},
{
"input": "38\n01 08",
"output": "12"
},
{
"input": "60\n00 06",
"output": "7"
},
{
"input": "5\n12 00",
"output": "49"
},
{
"input": "6\n01 42",
"output": "78"
},
{
"input": "4\n00 04",
"output": "92"
},
{
"input": "60\n04 05",
"output": "11"
},
{
"input": "1\n00 53",
"output": "6"
},
{
"input": "5\n08 05",
"output": "2"
},
{
"input": "60\n18 45",
"output": "1"
},
{
"input": "60\n06 23",
"output": "13"
},
{
"input": "6\n00 15",
"output": "3"
},
{
"input": "58\n00 06",
"output": "7"
},
{
"input": "2\n06 44",
"output": "383"
},
{
"input": "1\n08 00",
"output": "1"
},
{
"input": "10\n06 58",
"output": "78"
},
{
"input": "59\n00 58",
"output": "8"
},
{
"input": "1\n18 00",
"output": "1"
},
{
"input": "50\n00 42",
"output": "9"
},
{
"input": "30\n18 30",
"output": "2"
},
{
"input": "60\n21 59",
"output": "4"
},
{
"input": "2\n10 52",
"output": "87"
},
{
"input": "56\n00 00",
"output": "7"
},
{
"input": "16\n18 16",
"output": "2"
},
{
"input": "5\n01 05",
"output": "86"
},
{
"input": "5\n05 00",
"output": "133"
},
{
"input": "5\n23 59",
"output": "72"
},
{
"input": "7\n17 13",
"output": "0"
},
{
"input": "58\n00 00",
"output": "7"
},
{
"input": "15\n00 07",
"output": "0"
},
{
"input": "59\n08 00",
"output": "1"
},
{
"input": "46\n00 00",
"output": "8"
},
{
"input": "59\n01 05",
"output": "2"
},
{
"input": "2\n01 00",
"output": "211"
},
{
"input": "60\n00 24",
"output": "7"
},
{
"input": "10\n00 08",
"output": "37"
},
{
"input": "10\n00 06",
"output": "37"
},
{
"input": "60\n01 24",
"output": "8"
},
{
"input": "50\n00 10",
"output": "8"
},
{
"input": "2\n03 00",
"output": "271"
},
{
"input": "4\n19 04",
"output": "17"
},
{
"input": "25\n00 23",
"output": "16"
},
{
"input": "10\n01 01",
"output": "43"
}
] | 1,632,191,461
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 10
| 140
| 20,172,800
|
n = int(input())
time = input().split()
if '7' in time[0] or '7' in time[1]:
ans = 0
else:
x, ans, y = int(time[1]), 0, int(time[0])
while '7' not in str(x) and '7' not in str(y):
if x > n:
x -= n
ans += 1
if x <= n:
y -= 1
diff = x - n
x = 60 + diff
ans += 1
print(ans)
|
Title: Jamie and Alarm Snooze
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jamie loves sleeping. One day, he decides that he needs to wake up at exactly *hh*:<=*mm*. However, he hates waking up, so he wants to make waking up less painful by setting the alarm at a lucky time. He will then press the snooze button every *x* minutes until *hh*:<=*mm* is reached, and only then he will wake up. He wants to know what is the smallest number of times he needs to press the snooze button.
A time is considered lucky if it contains a digit '7'. For example, 13:<=07 and 17:<=27 are lucky, while 00:<=48 and 21:<=34 are not lucky.
Note that it is not necessary that the time set for the alarm and the wake-up time are on the same day. It is guaranteed that there is a lucky time Jamie can set so that he can wake at *hh*:<=*mm*.
Formally, find the smallest possible non-negative integer *y* such that the time representation of the time *x*·*y* minutes before *hh*:<=*mm* contains the digit '7'.
Jamie uses 24-hours clock, so after 23:<=59 comes 00:<=00.
Input Specification:
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=60).
The second line contains two two-digit integers, *hh* and *mm* (00<=≤<=*hh*<=≤<=23,<=00<=≤<=*mm*<=≤<=59).
Output Specification:
Print the minimum number of times he needs to press the button.
Demo Input:
['3\n11 23\n', '5\n01 07\n']
Demo Output:
['2\n', '0\n']
Note:
In the first sample, Jamie needs to wake up at 11:23. So, he can set his alarm at 11:17. He would press the snooze button when the alarm rings at 11:17 and at 11:20.
In the second sample, Jamie can set his alarm at exactly at 01:07 which is lucky.
|
```python
n = int(input())
time = input().split()
if '7' in time[0] or '7' in time[1]:
ans = 0
else:
x, ans, y = int(time[1]), 0, int(time[0])
while '7' not in str(x) and '7' not in str(y):
if x > n:
x -= n
ans += 1
if x <= n:
y -= 1
diff = x - n
x = 60 + diff
ans += 1
print(ans)
```
| 0
|
|
439
|
A
|
Devu, the Singer and Churu, the Joker
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation"
] | null | null |
Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
|
The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).
|
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
|
[
"3 30\n2 2 1\n",
"3 20\n2 1 1\n"
] |
[
"5\n",
"-1\n"
] |
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.
| 500
|
[
{
"input": "3 30\n2 2 1",
"output": "5"
},
{
"input": "3 20\n2 1 1",
"output": "-1"
},
{
"input": "50 10000\n5 4 10 9 9 6 7 7 7 3 3 7 7 4 7 4 10 10 1 7 10 3 1 4 5 7 2 10 10 10 2 3 4 7 6 1 8 4 7 3 8 8 4 10 1 1 9 2 6 1",
"output": "1943"
},
{
"input": "50 10000\n4 7 15 9 11 12 20 9 14 14 10 13 6 13 14 17 6 8 20 12 10 15 13 17 5 12 13 11 7 5 5 2 3 15 13 7 14 14 19 2 13 14 5 15 3 19 15 16 4 1",
"output": "1891"
},
{
"input": "100 9000\n5 2 3 1 1 3 4 9 9 6 7 10 10 10 2 10 6 8 8 6 7 9 9 5 6 2 1 10 10 9 4 5 9 2 4 3 8 5 6 1 1 5 3 6 2 6 6 6 5 8 3 6 7 3 1 10 9 1 8 3 10 9 5 6 3 4 1 1 10 10 2 3 4 8 10 10 5 1 5 3 6 8 10 6 10 2 1 8 10 1 7 6 9 10 5 2 3 5 3 2",
"output": "1688"
},
{
"input": "100 8007\n5 19 14 18 9 6 15 8 1 14 11 20 3 17 7 12 2 6 3 17 7 20 1 14 20 17 2 10 13 7 18 18 9 10 16 8 1 11 11 9 13 18 9 20 12 12 7 15 12 17 11 5 11 15 9 2 15 1 18 3 18 16 15 4 10 5 18 13 13 12 3 8 17 2 12 2 13 3 1 13 2 4 9 10 18 10 14 4 4 17 12 19 2 9 6 5 5 20 18 12",
"output": "1391"
},
{
"input": "39 2412\n1 1 1 1 1 1 26 1 1 1 99 1 1 1 1 1 1 1 1 1 1 88 7 1 1 1 1 76 1 1 1 93 40 1 13 1 68 1 32",
"output": "368"
},
{
"input": "39 2617\n47 1 1 1 63 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 70 1 99 63 1 1 1 1 1 1 1 1 64 1 1",
"output": "435"
},
{
"input": "39 3681\n83 77 1 94 85 47 1 98 29 16 1 1 1 71 96 85 31 97 96 93 40 50 98 1 60 51 1 96 100 72 1 1 1 89 1 93 1 92 100",
"output": "326"
},
{
"input": "45 894\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 28 28 1 1 1 1 1 1 1 1 1 1 1 1 1 1 99 3 1 1",
"output": "139"
},
{
"input": "45 4534\n1 99 65 99 4 46 54 80 51 30 96 1 28 30 44 70 78 1 1 100 1 62 1 1 1 85 1 1 1 61 1 46 75 1 61 77 97 26 67 1 1 63 81 85 86",
"output": "514"
},
{
"input": "72 3538\n52 1 8 1 1 1 7 1 1 1 1 48 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 40 1 1 38 1 1 1 1 1 1 1 1 1 1 1 35 1 93 79 1 1 1 1 1 1 1 1 1 51 1 1 1 1 1 1 1 1 1 1 1 1 96 1",
"output": "586"
},
{
"input": "81 2200\n1 59 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 93 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 50 1 1 1 1 1 1 1 1 1 1 1",
"output": "384"
},
{
"input": "81 2577\n85 91 1 1 2 1 1 100 1 80 1 1 17 86 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 37 1 66 24 1 1 96 49 1 66 1 44 1 1 1 1 98 1 1 1 1 35 1 37 3 35 1 1 87 64 1 24 1 58 1 1 42 83 5 1 1 1 1 1 95 1 94 1 50 1 1",
"output": "174"
},
{
"input": "81 4131\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 16 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",
"output": "807"
},
{
"input": "81 6315\n1 1 67 100 1 99 36 1 92 5 1 96 42 12 1 57 91 1 1 66 41 30 74 95 1 37 1 39 91 69 1 52 77 47 65 1 1 93 96 74 90 35 85 76 71 92 92 1 1 67 92 74 1 1 86 76 35 1 56 16 27 57 37 95 1 40 20 100 51 1 80 60 45 79 95 1 46 1 25 100 96",
"output": "490"
},
{
"input": "96 1688\n1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 45 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 25 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 71 1 1 1 30 1 1 1",
"output": "284"
},
{
"input": "96 8889\n1 1 18 1 1 1 1 1 1 1 1 1 99 1 1 1 1 88 1 45 1 1 1 1 1 1 1 1 1 1 1 1 1 1 96 1 1 1 1 21 1 1 1 1 1 1 1 73 1 1 1 1 1 10 1 1 1 1 1 1 1 46 43 1 1 1 1 1 98 1 1 1 1 1 1 6 1 1 1 1 1 74 1 25 1 55 1 1 1 13 1 1 54 1 1 1",
"output": "1589"
},
{
"input": "10 100\n1 1 1 1 1 1 1 1 1 1",
"output": "18"
},
{
"input": "100 10000\n54 46 72 94 79 83 91 54 73 3 24 55 54 31 28 20 19 6 25 19 47 23 1 70 15 87 51 39 54 77 55 5 60 3 15 99 56 88 22 78 79 21 38 27 28 86 7 88 12 59 55 70 25 1 70 49 1 45 69 72 50 17 4 56 8 100 90 34 35 20 61 76 88 79 4 74 65 68 75 26 40 72 59 94 10 67 96 85 29 90 47 24 44 1 66 93 55 36 1 99",
"output": "1017"
},
{
"input": "100 6000\n41 31 23 17 24 78 26 96 93 48 46 2 49 33 35 9 73 100 34 48 83 36 33 69 43 24 3 74 8 81 27 33 94 38 77 9 76 90 62 90 21 67 22 22 12 2 17 27 61 18 72 85 59 65 71 38 90 75 74 66 60 47 58 50 90 95 75 10 5 100 97 29 83 88 65 26 93 90 22 98 36 55 70 38 50 92 88 72 99 96 25 14 74 16 25 92 67 94 77 96",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 6\n1",
"output": "1"
},
{
"input": "1 5\n1",
"output": "0"
},
{
"input": "1 3\n4",
"output": "-1"
},
{
"input": "3 24\n2 1 2",
"output": "-1"
}
] | 1,579,578,805
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 108
| 307,200
|
n,d=[int(x) for x in input().split()]
t=[int(x) for x in input().split()]
summ=sum(t)+(n-1)*10
if summ>d:
print(-1)
else:
print((n-1)*2+(d-summ)//5)
|
Title: Devu, the Singer and Churu, the Joker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
Input Specification:
The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).
Output Specification:
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
Demo Input:
['3 30\n2 2 1\n', '3 20\n2 1 1\n']
Demo Output:
['5\n', '-1\n']
Note:
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1.
|
```python
n,d=[int(x) for x in input().split()]
t=[int(x) for x in input().split()]
summ=sum(t)+(n-1)*10
if summ>d:
print(-1)
else:
print((n-1)*2+(d-summ)//5)
```
| 3
|
|
919
|
B
|
Perfect Number
|
PROGRAMMING
| 1,100
|
[
"binary search",
"brute force",
"dp",
"implementation",
"number theory"
] | null | null |
We consider a positive integer perfect, if and only if the sum of its digits is exactly $10$. Given a positive integer $k$, your task is to find the $k$-th smallest perfect positive integer.
|
A single line with a positive integer $k$ ($1 \leq k \leq 10\,000$).
|
A single number, denoting the $k$-th smallest perfect integer.
|
[
"1\n",
"2\n"
] |
[
"19\n",
"28\n"
] |
The first perfect integer is $19$ and the second one is $28$.
| 750
|
[
{
"input": "1",
"output": "19"
},
{
"input": "2",
"output": "28"
},
{
"input": "13",
"output": "136"
},
{
"input": "101",
"output": "1432"
},
{
"input": "1023",
"output": "100270"
},
{
"input": "9999",
"output": "10800010"
},
{
"input": "10000",
"output": "10800100"
},
{
"input": "2333",
"output": "310060"
},
{
"input": "9139",
"output": "10134010"
},
{
"input": "9859",
"output": "10422001"
},
{
"input": "5987",
"output": "2221201"
},
{
"input": "3",
"output": "37"
},
{
"input": "4",
"output": "46"
},
{
"input": "5",
"output": "55"
},
{
"input": "7",
"output": "73"
},
{
"input": "9",
"output": "91"
},
{
"input": "10",
"output": "109"
},
{
"input": "15",
"output": "154"
},
{
"input": "51",
"output": "613"
},
{
"input": "99",
"output": "1414"
},
{
"input": "138",
"output": "2224"
},
{
"input": "233",
"output": "5050"
},
{
"input": "666",
"output": "27100"
},
{
"input": "1234",
"output": "110206"
},
{
"input": "3567",
"output": "1033003"
},
{
"input": "4445",
"output": "1221301"
},
{
"input": "5689",
"output": "2114002"
},
{
"input": "6666",
"output": "3102004"
},
{
"input": "7777",
"output": "5300200"
},
{
"input": "8888",
"output": "10110061"
},
{
"input": "9411",
"output": "10214200"
},
{
"input": "5539",
"output": "2101114"
},
{
"input": "6259",
"output": "2511100"
},
{
"input": "2387",
"output": "312220"
},
{
"input": "8515",
"output": "10030114"
}
] | 1,669,771,069
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 5
| 2,000
| 0
|
def checkPerf(k):
ds=[int(ch) for ch in str(k)]
if(sum(ds)==10):
return True
return False
cnt=0
i=0
n=int(input())
while True:
i+=1
if(checkPerf(i)):
cnt+=1
if(cnt==n):
print(i)
break
|
Title: Perfect Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We consider a positive integer perfect, if and only if the sum of its digits is exactly $10$. Given a positive integer $k$, your task is to find the $k$-th smallest perfect positive integer.
Input Specification:
A single line with a positive integer $k$ ($1 \leq k \leq 10\,000$).
Output Specification:
A single number, denoting the $k$-th smallest perfect integer.
Demo Input:
['1\n', '2\n']
Demo Output:
['19\n', '28\n']
Note:
The first perfect integer is $19$ and the second one is $28$.
|
```python
def checkPerf(k):
ds=[int(ch) for ch in str(k)]
if(sum(ds)==10):
return True
return False
cnt=0
i=0
n=int(input())
while True:
i+=1
if(checkPerf(i)):
cnt+=1
if(cnt==n):
print(i)
break
```
| 0
|
|
812
|
A
|
Sagheer and Crossroads
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (*l* — left, *s* — straight, *r* — right) and a light *p* for a pedestrian crossing.
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
|
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers *l*, *s*, *r*, *p* — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
|
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
|
[
"1 0 0 1\n0 1 0 0\n0 0 1 0\n0 0 0 1\n",
"0 1 1 0\n1 0 1 0\n1 1 0 0\n0 0 0 1\n",
"1 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 1 0\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.
| 500
|
[
{
"input": "1 0 0 1\n0 1 0 0\n0 0 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 1 1 0\n1 0 1 0\n1 1 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "1 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 1 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "1 1 1 0\n0 1 0 1\n1 1 1 0\n1 1 1 1",
"output": "YES"
},
{
"input": "0 1 1 0\n0 1 0 0\n1 0 0 1\n1 0 0 0",
"output": "YES"
},
{
"input": "1 0 0 0\n0 1 0 0\n1 1 0 0\n0 1 1 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 1 0 1\n1 0 1 1\n1 1 1 0",
"output": "YES"
},
{
"input": "1 1 0 0\n0 1 0 1\n1 1 1 0\n0 0 1 1",
"output": "YES"
},
{
"input": "0 1 0 0\n0 0 0 0\n1 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 1 0\n0 0 0 0\n1 1 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 1 0\n0 1 0 1\n1 0 1 0\n0 0 1 0",
"output": "YES"
},
{
"input": "1 1 1 0\n0 1 0 1\n1 1 1 1\n0 0 0 1",
"output": "YES"
},
{
"input": "0 0 1 0\n0 0 0 0\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 1 0 1\n1 0 1 1\n0 0 0 1",
"output": "YES"
},
{
"input": "1 1 0 0\n0 1 0 0\n1 1 1 0\n1 0 1 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "1 0 1 0\n1 1 0 0\n1 1 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 1 0\n1 1 0 0\n1 0 1 0\n1 0 0 0",
"output": "NO"
},
{
"input": "0 0 1 0\n1 0 0 0\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "0 1 1 0\n1 1 0 1\n1 0 0 1\n1 1 1 0",
"output": "YES"
},
{
"input": "1 0 0 0\n1 1 0 0\n1 1 0 1\n0 0 1 0",
"output": "YES"
},
{
"input": "0 0 0 0\n1 1 0 0\n0 0 0 1\n0 0 1 0",
"output": "NO"
},
{
"input": "0 1 0 0\n0 0 0 1\n0 1 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 1 0 0\n1 1 0 1\n1 0 0 1\n1 1 0 1",
"output": "YES"
},
{
"input": "1 0 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 1 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 1 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n1 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 1 0 0\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 1 0\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n1 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 1 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 0 1 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 0 0 0\n1 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 0 0 0\n0 1 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 0 0 0\n0 0 1 0",
"output": "YES"
},
{
"input": "1 0 0 0\n0 0 0 1\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 1 0 0\n0 0 0 1\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 1 0\n0 0 0 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n1 0 0 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 1 0 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 1 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 1\n1 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 1 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 1 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 0\n0 1 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 0\n0 0 1 0",
"output": "NO"
},
{
"input": "1 0 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "NO"
},
{
"input": "0 1 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 1 0\n0 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n1 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 1 0 0\n0 0 0 1\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 1 0\n0 0 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n1 0 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 1 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 1 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n1 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n0 1 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n0 0 1 0",
"output": "NO"
},
{
"input": "1 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 1 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 1 0\n0 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n1 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 1 0 0\n0 0 0 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 1 0\n0 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n1 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 1 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n1 0 0 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n0 1 0 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 1 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "1 1 1 1\n1 1 1 1\n1 1 1 1\n1 1 1 1",
"output": "YES"
},
{
"input": "1 0 0 0\n0 1 0 0\n0 0 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "1 1 1 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "1 0 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 1 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 1 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n1 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 1 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 1 1 0\n1 0 1 0\n1 1 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "1 1 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "1 1 1 0\n1 1 1 0\n1 1 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "1 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 1 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 1 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 1 1 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 1 0 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 0 1\n0 0 0 0\n0 1 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 0 1\n1 0 0 0\n0 0 0 0",
"output": "YES"
}
] | 1,592,865,393
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 93
| 6,963,200
|
class block:
def __init__(self, l,s,r,p,no):
self.id=no
self.l = l
self.s = s
self.r = r
self.p =p
self.opposite=(no+2)%4
self.right=(no+1)%4
self.left=(no-1)%4
self.attacking=[]
if(l):
self.attacking.append(self.left)
if(r):
self.attacking.append(self.right)
if(s):
self.attacking.append(self.opposite)
def hits(self,a):
if a.p and a.id in self.attacking:
return True
A=[]
for i in range(4):
l,s,r,p=map(int,input().split())
a=block(l,s,r,p,i)
A.append(a)
for i in range(4):
for j in range(i+1,4):
if A[i].hits(A[j]):
print("YES")
exit()
print("NO")
|
Title: Sagheer and Crossroads
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (*l* — left, *s* — straight, *r* — right) and a light *p* for a pedestrian crossing.
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input Specification:
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers *l*, *s*, *r*, *p* — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output Specification:
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Demo Input:
['1 0 0 1\n0 1 0 0\n0 0 1 0\n0 0 0 1\n', '0 1 1 0\n1 0 1 0\n1 1 0 0\n0 0 0 1\n', '1 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 1 0\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.
|
```python
class block:
def __init__(self, l,s,r,p,no):
self.id=no
self.l = l
self.s = s
self.r = r
self.p =p
self.opposite=(no+2)%4
self.right=(no+1)%4
self.left=(no-1)%4
self.attacking=[]
if(l):
self.attacking.append(self.left)
if(r):
self.attacking.append(self.right)
if(s):
self.attacking.append(self.opposite)
def hits(self,a):
if a.p and a.id in self.attacking:
return True
A=[]
for i in range(4):
l,s,r,p=map(int,input().split())
a=block(l,s,r,p,i)
A.append(a)
for i in range(4):
for j in range(i+1,4):
if A[i].hits(A[j]):
print("YES")
exit()
print("NO")
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2,<=100]. Your task is to say if the hidden number is prime or composite.
Integer *x*<=><=1 is called prime if it has exactly two distinct divisors, 1 and *x*. If integer *x*<=><=1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2,<=100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2,<=100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
|
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
|
Up to 20 times you can ask a query — print an integer from interval [2,<=100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
- fflush(stdout) in C++; - System.out.flush() in Java; - stdout.flush() in Python; - flush(output) in Pascal; - See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2,<=100]. Of course, his/her solution won't be able to read the hidden number from the input.
|
[
"yes\nno\nyes\n",
"no\nyes\nno\nno\nno\n"
] |
[
"2\n80\n5\ncomposite\n",
"58\n59\n78\n78\n2\nprime\n"
] |
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ea790051c34ea7d2761cd9b096412ca7c647a173.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/35c6952617fa94ec3e0ea8e63aa1c3c49b3ba420.png" style="max-width: 100.0%;max-height: 100.0%;"/>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries).
| 0
|
[
{
"input": "30",
"output": "composite 4"
},
{
"input": "59",
"output": "prime 15"
},
{
"input": "2",
"output": "prime 16"
},
{
"input": "7",
"output": "prime 16"
},
{
"input": "9",
"output": "composite 3"
},
{
"input": "13",
"output": "prime 15"
},
{
"input": "55",
"output": "composite 6"
},
{
"input": "89",
"output": "prime 15"
},
{
"input": "3",
"output": "prime 16"
},
{
"input": "4",
"output": "composite 2"
},
{
"input": "6",
"output": "composite 4"
},
{
"input": "8",
"output": "composite 2"
},
{
"input": "11",
"output": "prime 15"
},
{
"input": "12",
"output": "composite 2"
},
{
"input": "25",
"output": "composite 4"
},
{
"input": "36",
"output": "composite 2"
},
{
"input": "46",
"output": "composite 10"
},
{
"input": "47",
"output": "prime 15"
},
{
"input": "49",
"output": "composite 5"
},
{
"input": "51",
"output": "composite 8"
},
{
"input": "53",
"output": "prime 15"
},
{
"input": "59",
"output": "prime 15"
},
{
"input": "64",
"output": "composite 2"
},
{
"input": "81",
"output": "composite 3"
},
{
"input": "91",
"output": "composite 7"
},
{
"input": "93",
"output": "composite 12"
},
{
"input": "94",
"output": "composite 16"
},
{
"input": "95",
"output": "composite 9"
},
{
"input": "96",
"output": "composite 2"
},
{
"input": "97",
"output": "prime 15"
},
{
"input": "99",
"output": "composite 3"
},
{
"input": "100",
"output": "composite 2"
},
{
"input": "26",
"output": "composite 7"
},
{
"input": "94",
"output": "composite 16"
},
{
"input": "25",
"output": "composite 4"
},
{
"input": "46",
"output": "composite 10"
},
{
"input": "49",
"output": "composite 5"
},
{
"input": "82",
"output": "composite 14"
},
{
"input": "87",
"output": "composite 11"
},
{
"input": "85",
"output": "composite 8"
},
{
"input": "74",
"output": "composite 13"
},
{
"input": "86",
"output": "composite 15"
},
{
"input": "93",
"output": "composite 12"
},
{
"input": "58",
"output": "composite 11"
},
{
"input": "47",
"output": "prime 15"
},
{
"input": "22",
"output": "composite 6"
},
{
"input": "23",
"output": "prime 15"
},
{
"input": "62",
"output": "composite 12"
},
{
"input": "38",
"output": "composite 9"
},
{
"input": "91",
"output": "composite 7"
},
{
"input": "34",
"output": "composite 8"
},
{
"input": "95",
"output": "composite 9"
},
{
"input": "4",
"output": "composite 2"
},
{
"input": "16",
"output": "composite 2"
},
{
"input": "96",
"output": "composite 2"
},
{
"input": "69",
"output": "composite 10"
},
{
"input": "65",
"output": "composite 7"
},
{
"input": "6",
"output": "composite 4"
},
{
"input": "18",
"output": "composite 4"
}
] | 1,686,442,617
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 6
| 46
| 0
|
v=[]
c=0
flag=False
for i in range(2,10):
v+=[i]
for j in v:
print(j,flush=True)
if input()=="yes":
c+=1
if c >=2:
print("composite")
else:
print("prime")
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This is an interactive problem. In the output section below you will see the information about flushing the output.
Bear Limak thinks of some hidden number — an integer from interval [2,<=100]. Your task is to say if the hidden number is prime or composite.
Integer *x*<=><=1 is called prime if it has exactly two distinct divisors, 1 and *x*. If integer *x*<=><=1 is not prime, it's called composite.
You can ask up to 20 queries about divisors of the hidden number. In each query you should print an integer from interval [2,<=100]. The system will answer "yes" if your integer is a divisor of the hidden number. Otherwise, the answer will be "no".
For example, if the hidden number is 14 then the system will answer "yes" only if you print 2, 7 or 14.
When you are done asking queries, print "prime" or "composite" and terminate your program.
You will get the Wrong Answer verdict if you ask more than 20 queries, or if you print an integer not from the range [2,<=100]. Also, you will get the Wrong Answer verdict if the printed answer isn't correct.
You will get the Idleness Limit Exceeded verdict if you don't print anything (but you should) or if you forget about flushing the output (more info below).
Input Specification:
After each query you should read one string from the input. It will be "yes" if the printed integer is a divisor of the hidden number, and "no" otherwise.
Output Specification:
Up to 20 times you can ask a query — print an integer from interval [2,<=100] in one line. You have to both print the end-of-line character and flush the output. After flushing you should read a response from the input.
In any moment you can print the answer "prime" or "composite" (without the quotes). After that, flush the output and terminate your program.
To flush you can use (just after printing an integer and end-of-line):
- fflush(stdout) in C++; - System.out.flush() in Java; - stdout.flush() in Python; - flush(output) in Pascal; - See the documentation for other languages.
Hacking. To hack someone, as the input you should print the hidden number — one integer from the interval [2,<=100]. Of course, his/her solution won't be able to read the hidden number from the input.
Demo Input:
['yes\nno\nyes\n', 'no\nyes\nno\nno\nno\n']
Demo Output:
['2\n80\n5\ncomposite\n', '58\n59\n78\n78\n2\nprime\n']
Note:
The hidden number in the first query is 30. In a table below you can see a better form of the provided example of the communication process.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ea790051c34ea7d2761cd9b096412ca7c647a173.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The hidden number is divisible by both 2 and 5. Thus, it must be composite. Note that it isn't necessary to know the exact value of the hidden number. In this test, the hidden number is 30.
<img align="middle" class="tex-formula" src="https://espresso.codeforces.com/35c6952617fa94ec3e0ea8e63aa1c3c49b3ba420.png" style="max-width: 100.0%;max-height: 100.0%;"/>
59 is a divisor of the hidden number. In the interval [2, 100] there is only one number with this divisor. The hidden number must be 59, which is prime. Note that the answer is known even after the second query and you could print it then and terminate. Though, it isn't forbidden to ask unnecessary queries (unless you exceed the limit of 20 queries).
|
```python
v=[]
c=0
flag=False
for i in range(2,10):
v+=[i]
for j in v:
print(j,flush=True)
if input()=="yes":
c+=1
if c >=2:
print("composite")
else:
print("prime")
```
| 0
|
|
886
|
A
|
ACM ICPC
|
PROGRAMMING
| 1,000
|
[
"brute force"
] | null | null |
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
|
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
|
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
|
[
"1 3 2 1 2 1\n",
"1 1 1 1 1 99\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater.
| 500
|
[
{
"input": "1 3 2 1 2 1",
"output": "YES"
},
{
"input": "1 1 1 1 1 99",
"output": "NO"
},
{
"input": "1000 1000 1000 1000 1000 1000",
"output": "YES"
},
{
"input": "0 0 0 0 0 0",
"output": "YES"
},
{
"input": "633 609 369 704 573 416",
"output": "NO"
},
{
"input": "353 313 327 470 597 31",
"output": "NO"
},
{
"input": "835 638 673 624 232 266",
"output": "NO"
},
{
"input": "936 342 19 398 247 874",
"output": "NO"
},
{
"input": "417 666 978 553 271 488",
"output": "NO"
},
{
"input": "71 66 124 199 67 147",
"output": "YES"
},
{
"input": "54 26 0 171 239 12",
"output": "YES"
},
{
"input": "72 8 186 92 267 69",
"output": "YES"
},
{
"input": "180 179 188 50 75 214",
"output": "YES"
},
{
"input": "16 169 110 136 404 277",
"output": "YES"
},
{
"input": "101 400 9 200 300 10",
"output": "YES"
},
{
"input": "101 400 200 9 300 10",
"output": "YES"
},
{
"input": "101 200 400 9 300 10",
"output": "YES"
},
{
"input": "101 400 200 300 9 10",
"output": "YES"
},
{
"input": "101 200 400 300 9 10",
"output": "YES"
},
{
"input": "4 4 4 4 5 4",
"output": "NO"
},
{
"input": "2 2 2 2 2 1",
"output": "NO"
},
{
"input": "1000 1000 999 1000 1000 1000",
"output": "NO"
},
{
"input": "129 1 10 29 8 111",
"output": "NO"
},
{
"input": "1000 1000 1000 999 999 1000",
"output": "YES"
},
{
"input": "101 200 300 400 9 10",
"output": "YES"
},
{
"input": "101 400 200 300 10 9",
"output": "YES"
},
{
"input": "101 200 400 300 10 9",
"output": "YES"
},
{
"input": "101 200 300 400 10 9",
"output": "YES"
},
{
"input": "101 200 300 10 400 9",
"output": "YES"
},
{
"input": "1 1 1 1 1 5",
"output": "NO"
},
{
"input": "8 1 1 3 3 0",
"output": "NO"
},
{
"input": "1 1 2 2 3 3",
"output": "YES"
},
{
"input": "1 2 2 5 2 5",
"output": "NO"
},
{
"input": "1 2 3 6 6 6",
"output": "NO"
},
{
"input": "36 91 7 86 51 89",
"output": "NO"
},
{
"input": "10 1 1 1 23 24",
"output": "NO"
},
{
"input": "1 1 1 10 23 24",
"output": "NO"
},
{
"input": "20 10 1 2 3 44",
"output": "NO"
},
{
"input": "7 0 14 11 8 6",
"output": "NO"
},
{
"input": "100 496 1 1 1 1",
"output": "NO"
},
{
"input": "5 4 2 5 11 3",
"output": "NO"
},
{
"input": "1 3 7 8 8 9",
"output": "YES"
},
{
"input": "1 3 4 5 18 19",
"output": "YES"
},
{
"input": "5 5 1 2 2 15",
"output": "NO"
},
{
"input": "2 1 0 0 0 5",
"output": "NO"
},
{
"input": "1 6 6 1 20 2",
"output": "NO"
},
{
"input": "2 10 0 0 0 0",
"output": "NO"
},
{
"input": "1 1 3 1 1 11",
"output": "NO"
},
{
"input": "10 10 1 1 1 37",
"output": "NO"
},
{
"input": "1 1 0 0 0 4",
"output": "NO"
},
{
"input": "1 1 10 1 1 28",
"output": "NO"
},
{
"input": "1 5 5 5 6 8",
"output": "YES"
},
{
"input": "0 2 3 4 4 5",
"output": "YES"
}
] | 1,653,604,300
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 46
| 0
|
from sys import exit
a=list(map(int,input().split()))
a=sorted(a)
sum=0
for i in a:
sum=sum+i
half=sum/2
half=half-a[-1]
a.pop(-1)
for i in a:
x=a.pop(i)
rest=half-x
if rest in a:
print("YES")
exit()
a.insert(i, x)
print("NO")
|
Title: ACM ICPC
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
Input Specification:
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
Output Specification:
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['1 3 2 1 2 1\n', '1 1 1 1 1 99\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater.
|
```python
from sys import exit
a=list(map(int,input().split()))
a=sorted(a)
sum=0
for i in a:
sum=sum+i
half=sum/2
half=half-a[-1]
a.pop(-1)
for i in a:
x=a.pop(i)
rest=half-x
if rest in a:
print("YES")
exit()
a.insert(i, x)
print("NO")
```
| -1
|
|
1,008
|
A
|
Romaji
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese.
|
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
|
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower).
|
[
"sumimasen\n",
"ninja\n",
"codeforces\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.
| 500
|
[
{
"input": "sumimasen",
"output": "YES"
},
{
"input": "ninja",
"output": "YES"
},
{
"input": "codeforces",
"output": "NO"
},
{
"input": "auuaoonntanonnuewannnnpuuinniwoonennyolonnnvienonpoujinndinunnenannmuveoiuuhikucuziuhunnnmunzancenen",
"output": "YES"
},
{
"input": "n",
"output": "YES"
},
{
"input": "necnei",
"output": "NO"
},
{
"input": "nternn",
"output": "NO"
},
{
"input": "aucunuohja",
"output": "NO"
},
{
"input": "a",
"output": "YES"
},
{
"input": "b",
"output": "NO"
},
{
"input": "nn",
"output": "YES"
},
{
"input": "nnnzaaa",
"output": "YES"
},
{
"input": "zn",
"output": "NO"
},
{
"input": "ab",
"output": "NO"
},
{
"input": "aaaaaaaaaa",
"output": "YES"
},
{
"input": "aaaaaaaaab",
"output": "NO"
},
{
"input": "aaaaaaaaan",
"output": "YES"
},
{
"input": "baaaaaaaaa",
"output": "YES"
},
{
"input": "naaaaaaaaa",
"output": "YES"
},
{
"input": "nbaaaaaaaa",
"output": "YES"
},
{
"input": "bbaaaaaaaa",
"output": "NO"
},
{
"input": "bnaaaaaaaa",
"output": "NO"
},
{
"input": "eonwonojannonnufimiiniewuqaienokacevecinfuqihatenhunliquuyebayiaenifuexuanenuaounnboancaeowonu",
"output": "YES"
},
{
"input": "uixinnepnlinqaingieianndeakuniooudidonnnqeaituioeneiroionxuowudiooonayenfeonuino",
"output": "NO"
},
{
"input": "nnnnnyigaveteononnnnxaalenxuiiwannntoxonyoqonlejuoxuoconnnentoinnul",
"output": "NO"
},
{
"input": "ndonneasoiunhomuunnhuitonnntunntoanerekonoupunanuauenu",
"output": "YES"
},
{
"input": "anujemogawautiedoneobninnibonuunaoennnyoorufonxionntinimiboonununnnnnleenqunminzayoutanlalo",
"output": "NO"
},
{
"input": "y",
"output": "NO"
},
{
"input": "by",
"output": "NO"
},
{
"input": "yy",
"output": "NO"
},
{
"input": "nbn",
"output": "NO"
},
{
"input": "nz",
"output": "NO"
},
{
"input": "king",
"output": "NO"
},
{
"input": "g",
"output": "NO"
},
{
"input": "az",
"output": "NO"
},
{
"input": "x",
"output": "NO"
},
{
"input": "z",
"output": "NO"
},
{
"input": "yn",
"output": "NO"
},
{
"input": "aeo",
"output": "YES"
},
{
"input": "nb",
"output": "NO"
},
{
"input": "npn",
"output": "NO"
},
{
"input": "kini",
"output": "YES"
},
{
"input": "pya",
"output": "NO"
},
{
"input": "m",
"output": "NO"
},
{
"input": "p",
"output": "NO"
},
{
"input": "aaaaaaaak",
"output": "NO"
},
{
"input": "aab",
"output": "NO"
},
{
"input": "d",
"output": "NO"
},
{
"input": "at",
"output": "NO"
},
{
"input": "aaaaaak",
"output": "NO"
},
{
"input": "aaz",
"output": "NO"
},
{
"input": "aaab",
"output": "NO"
},
{
"input": "s",
"output": "NO"
},
{
"input": "nzzen",
"output": "NO"
},
{
"input": "aeionnhhhn",
"output": "NO"
},
{
"input": "h",
"output": "NO"
},
{
"input": "ny",
"output": "NO"
}
] | 1,637,775,221
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
n=input()
l=len(n)
for i in range(l):
if n(i) not in "a,e,i,o,u,n" and n(i+1) not in "a,e,i,o,u":
print(no)
else:
print(yes)
|
Title: Romaji
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese.
Input Specification:
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
Output Specification:
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower).
Demo Input:
['sumimasen\n', 'ninja\n', 'codeforces\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.
|
```python
n=input()
l=len(n)
for i in range(l):
if n(i) not in "a,e,i,o,u,n" and n(i+1) not in "a,e,i,o,u":
print(no)
else:
print(yes)
```
| -1
|
|
812
|
A
|
Sagheer and Crossroads
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (*l* — left, *s* — straight, *r* — right) and a light *p* for a pedestrian crossing.
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
|
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers *l*, *s*, *r*, *p* — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
|
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
|
[
"1 0 0 1\n0 1 0 0\n0 0 1 0\n0 0 0 1\n",
"0 1 1 0\n1 0 1 0\n1 1 0 0\n0 0 0 1\n",
"1 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 1 0\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.
| 500
|
[
{
"input": "1 0 0 1\n0 1 0 0\n0 0 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 1 1 0\n1 0 1 0\n1 1 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "1 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 1 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "1 1 1 0\n0 1 0 1\n1 1 1 0\n1 1 1 1",
"output": "YES"
},
{
"input": "0 1 1 0\n0 1 0 0\n1 0 0 1\n1 0 0 0",
"output": "YES"
},
{
"input": "1 0 0 0\n0 1 0 0\n1 1 0 0\n0 1 1 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 1 0 1\n1 0 1 1\n1 1 1 0",
"output": "YES"
},
{
"input": "1 1 0 0\n0 1 0 1\n1 1 1 0\n0 0 1 1",
"output": "YES"
},
{
"input": "0 1 0 0\n0 0 0 0\n1 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 1 0\n0 0 0 0\n1 1 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 1 0\n0 1 0 1\n1 0 1 0\n0 0 1 0",
"output": "YES"
},
{
"input": "1 1 1 0\n0 1 0 1\n1 1 1 1\n0 0 0 1",
"output": "YES"
},
{
"input": "0 0 1 0\n0 0 0 0\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 1 0 1\n1 0 1 1\n0 0 0 1",
"output": "YES"
},
{
"input": "1 1 0 0\n0 1 0 0\n1 1 1 0\n1 0 1 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "1 0 1 0\n1 1 0 0\n1 1 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 1 0\n1 1 0 0\n1 0 1 0\n1 0 0 0",
"output": "NO"
},
{
"input": "0 0 1 0\n1 0 0 0\n0 0 0 1\n0 0 0 1",
"output": "NO"
},
{
"input": "0 1 1 0\n1 1 0 1\n1 0 0 1\n1 1 1 0",
"output": "YES"
},
{
"input": "1 0 0 0\n1 1 0 0\n1 1 0 1\n0 0 1 0",
"output": "YES"
},
{
"input": "0 0 0 0\n1 1 0 0\n0 0 0 1\n0 0 1 0",
"output": "NO"
},
{
"input": "0 1 0 0\n0 0 0 1\n0 1 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 1 0 0\n1 1 0 1\n1 0 0 1\n1 1 0 1",
"output": "YES"
},
{
"input": "1 0 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 1 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 1 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n1 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 1 0 0\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 1 0\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n1 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 1 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 0 1 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 0 0 0\n1 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 0 0 0\n0 1 0 0",
"output": "NO"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 0 0 0\n0 0 1 0",
"output": "YES"
},
{
"input": "1 0 0 0\n0 0 0 1\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 1 0 0\n0 0 0 1\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 1 0\n0 0 0 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n1 0 0 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 1 0 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 1 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 1\n1 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 1 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 1 0\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 0\n0 1 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 1\n0 0 0 0\n0 0 1 0",
"output": "NO"
},
{
"input": "1 0 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "NO"
},
{
"input": "0 1 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 1 0\n0 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n1 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 1 0 0\n0 0 0 1\n0 0 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 1 0\n0 0 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n1 0 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 1 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 1 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n1 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n0 1 0 0",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 1\n0 0 1 0",
"output": "NO"
},
{
"input": "1 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 1 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 1 0\n0 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n1 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 1 0 0\n0 0 0 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 1 0\n0 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n1 0 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 1 0 0\n0 0 0 1",
"output": "NO"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n1 0 0 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n0 1 0 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 1 1",
"output": "YES"
},
{
"input": "0 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "NO"
},
{
"input": "1 1 1 1\n1 1 1 1\n1 1 1 1\n1 1 1 1",
"output": "YES"
},
{
"input": "1 0 0 0\n0 1 0 0\n0 0 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "1 1 1 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "1 0 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 1 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 1 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n1 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 1 0 0\n0 0 0 0\n0 0 0 1\n0 0 0 0",
"output": "YES"
},
{
"input": "0 1 1 0\n1 0 1 0\n1 1 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "1 1 0 1\n0 0 0 0\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "1 1 1 0\n1 1 1 0\n1 1 1 0\n0 0 0 1",
"output": "YES"
},
{
"input": "1 0 0 0\n0 0 0 0\n0 0 0 0\n0 0 0 1",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 0 0\n0 1 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 1 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 1 1 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 1 0 1\n0 0 0 0\n0 0 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 0 1\n0 0 0 0\n0 1 0 0",
"output": "YES"
},
{
"input": "0 0 0 1\n0 0 0 1\n1 0 0 0\n0 0 0 0",
"output": "YES"
}
] | 1,622,270,113
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 124
| 2,355,200
|
import sys
import string
from collections import Counter, defaultdict
from math import fsum, sqrt, gcd, ceil, factorial
from operator import *
from itertools import accumulate
inf = float("inf")
# input = sys.stdin.readline
flush = lambda: sys.stdout.flush
comb = lambda x, y: (factorial(x) // factorial(y)) // factorial(x - y)
en = lambda x: list(enumerate(x))
# inputs
# ip = lambda : input().rstrip()
ip = lambda: input()
ii = lambda: int(input())
r = lambda: map(int, input().split())
rr = lambda: list(r())
arr = [rr() for _ in range(4)]
for i in range(4):
if arr[i][3] and (
arr[i][0]
or arr[i][1]
or arr[i][2]
or arr[(i + 2) % 4][1]
or arr[(i + 1) % 4][0]
or arr[(i + 3) % 4][3]
):
exit(print("YES"))
print("NO")
|
Title: Sagheer and Crossroads
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sagheer is walking in the street when he comes to an intersection of two roads. Each road can be represented as two parts where each part has 3 lanes getting into the intersection (one for each direction) and 3 lanes getting out of the intersection, so we have 4 parts in total. Each part has 4 lights, one for each lane getting into the intersection (*l* — left, *s* — straight, *r* — right) and a light *p* for a pedestrian crossing.
An accident is possible if a car can hit a pedestrian. This can happen if the light of a pedestrian crossing of some part and the light of a lane that can get to or from that same part are green at the same time.
Now, Sagheer is monitoring the configuration of the traffic lights. Your task is to help him detect whether an accident is possible.
Input Specification:
The input consists of four lines with each line describing a road part given in a counter-clockwise order.
Each line contains four integers *l*, *s*, *r*, *p* — for the left, straight, right and pedestrian lights, respectively. The possible values are 0 for red light and 1 for green light.
Output Specification:
On a single line, print "YES" if an accident is possible, and "NO" otherwise.
Demo Input:
['1 0 0 1\n0 1 0 0\n0 0 1 0\n0 0 0 1\n', '0 1 1 0\n1 0 1 0\n1 1 0 0\n0 0 0 1\n', '1 0 0 0\n0 0 0 1\n0 0 0 0\n1 0 1 0\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
In the first example, some accidents are possible because cars of part 1 can hit pedestrians of parts 1 and 4. Also, cars of parts 2 and 3 can hit pedestrians of part 4.
In the second example, no car can pass the pedestrian crossing of part 4 which is the only green pedestrian light. So, no accident can occur.
|
```python
import sys
import string
from collections import Counter, defaultdict
from math import fsum, sqrt, gcd, ceil, factorial
from operator import *
from itertools import accumulate
inf = float("inf")
# input = sys.stdin.readline
flush = lambda: sys.stdout.flush
comb = lambda x, y: (factorial(x) // factorial(y)) // factorial(x - y)
en = lambda x: list(enumerate(x))
# inputs
# ip = lambda : input().rstrip()
ip = lambda: input()
ii = lambda: int(input())
r = lambda: map(int, input().split())
rr = lambda: list(r())
arr = [rr() for _ in range(4)]
for i in range(4):
if arr[i][3] and (
arr[i][0]
or arr[i][1]
or arr[i][2]
or arr[(i + 2) % 4][1]
or arr[(i + 1) % 4][0]
or arr[(i + 3) % 4][3]
):
exit(print("YES"))
print("NO")
```
| 0
|
|
690
|
E1
|
Photographs (I)
|
PROGRAMMING
| 1,800
|
[] | null | null |
The Human-Cow Confederation (*HC*2), led by Heidi, has built a base where people and cows can hide, guarded from zombie attacks. The entrance to the base is protected by an automated gate which performs a kind of a Turing test: it shows the entering creature a photograph and asks them whether the top and bottom halves of this photograph have been swapped or not. A person (or even a cow) will have no problem answering such questions; on the other hand, a zombie would just randomly smash one of the two buttons.
The creature is asked a series of such questions. If at least 75% of them are answered correctly, the gate is unlocked; otherwise, a side door opens, beneath which a huge fan is spinning...
Heidi is now building a robot army to fight the zombies, and she wants the robots to also be able to enter the base. You are tasked with programming them to distinguish the images.
The first two images from the test set. The first picture has been rearranged, but not the second.
|
The first line of the input contains the number *q* of questions (1<=≤<=*q*<=≤<=220). After that, *q* questions follow, each of which in the format described below.
The first line of every question contains two space-separated integers *h* and *w* (1<=≤<=*h*,<=*w*<=≤<=600) – the height (number of rows) and width (number of columns) of the photograph. (Most photographs are roughly 200<=×<=300.) After this, *h* lines follow, each describing a single row of the picture. The picture is monochrome (in shades of grey). Its *i*-th row is described by *w* space-separated integers *a**ij* (*j*<==<=1,<=...,<=*w*), where *a**ij* is the brightness of the corresponding pixel (0<=≤<=*a**ij*<=<<=256, where 0 is black and 255 is white).
Each picture will be either a real-life photograph, or a real-life photograph which has been broken up into two pieces and rearranged. More precisely, in the latter case, the topmost rows have been moved to the bottom of the picture. It is guaranteed that *h* is even.
There is only a single input file to be processed, called all.in, and it is downloadable from the online judge. You are also a given another input file, called sample.in, which contains the first 20 pictures from all.in; you are provided the correct answers for sample.in in sample.out. You are also given a directory easy_bmp, which contains the first 50 input photographs in the form of .bmp image files, as well as a directory easy_sample_original_bmp, which contains the first 20 images before rearrangement. Check the notes for the download links.
|
Your program should print *q* lines. The *i*-th line should contain your answer for the *i*-th question: YES if the photograph has been rearranged and NO otherwise. Your answers will be accepted if they all conform to this format and if at least 75% of them are correct.
Because the input is rather huge, feel free to process it locally and submit just your precomputed answers (i.e., a program which just prints your output for the input file all.in).
|
[] |
[] |
The link to download all necessary files is http://assets.codeforces.com/files/690/easy_contestant_package.zip
| 0
|
[] | 1,490,350,123
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 1
| 46
| 4,915,200
|
for x in ['YES', 'NO', 'NO', 'YES', 'NO', 'NO', 'NO', 'NO', 'NO', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'NO', 'YES', 'NO', 'YES', 'NO', 'NO', 'YES', 'NO', 'NO', 'YES', 'NO', 'NO', 'YES', 'YES', 'NO', 'YES', 'NO', 'YES', 'NO', 'YES', 'YES', 'YES', 'NO', 'YES', 'YES', 'NO', 'YES', 'YES', 'NO', 'YES', 'YES', 'YES', 'NO', 'YES', 'NO', 'NO', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'NO', 'NO', 'YES', 'NO', 'NO', 'NO', 'NO', 'YES', 'NO', 'NO', 'NO', 'NO', 'YES', 'NO', 'YES', 'NO', 'NO', 'NO', 'YES', 'YES', 'NO', 'YES', 'YES', 'NO', 'NO', 'NO', 'NO', 'NO', 'NO', 'YES', 'NO', 'NO', 'YES', 'NO', 'YES', 'NO', 'NO', 'NO', 'NO', 'YES', 'YES', 'YES', 'YES', 'YES', 'NO', 'NO', 'NO', 'YES', 'NO', 'YES', 'NO', 'YES', 'NO', 'YES', 'NO', 'NO', 'YES', 'YES', 'YES', 'YES', 'NO', 'NO', 'NO', 'YES', 'NO', 'YES', 'YES', 'YES', 'NO', 'YES', 'NO', 'NO', 'NO', 'YES', 'NO', 'NO', 'NO', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'NO', 'YES', 'NO', 'NO', 'NO', 'NO', 'NO', 'YES', 'NO', 'NO', 'YES', 'NO', 'YES', 'NO', 'YES', 'NO', 'YES', 'NO', 'NO', 'YES', 'YES', 'NO', 'NO', 'YES', 'NO', 'NO', 'NO', 'NO', 'YES', 'NO', 'YES', 'NO', 'NO', 'YES', 'NO', 'YES', 'YES', 'YES', 'NO', 'NO', 'NO', 'YES', 'YES', 'NO', 'YES', 'NO', 'YES', 'YES', 'YES', 'NO', 'YES', 'YES', 'NO', 'YES', 'NO', 'YES', 'YES', 'NO', 'YES', 'YES', 'NO', 'NO', 'YES', 'NO', 'NO', 'NO', 'YES', 'YES', 'NO', 'YES']:
print(x)
|
Title: Photographs (I)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Human-Cow Confederation (*HC*2), led by Heidi, has built a base where people and cows can hide, guarded from zombie attacks. The entrance to the base is protected by an automated gate which performs a kind of a Turing test: it shows the entering creature a photograph and asks them whether the top and bottom halves of this photograph have been swapped or not. A person (or even a cow) will have no problem answering such questions; on the other hand, a zombie would just randomly smash one of the two buttons.
The creature is asked a series of such questions. If at least 75% of them are answered correctly, the gate is unlocked; otherwise, a side door opens, beneath which a huge fan is spinning...
Heidi is now building a robot army to fight the zombies, and she wants the robots to also be able to enter the base. You are tasked with programming them to distinguish the images.
The first two images from the test set. The first picture has been rearranged, but not the second.
Input Specification:
The first line of the input contains the number *q* of questions (1<=≤<=*q*<=≤<=220). After that, *q* questions follow, each of which in the format described below.
The first line of every question contains two space-separated integers *h* and *w* (1<=≤<=*h*,<=*w*<=≤<=600) – the height (number of rows) and width (number of columns) of the photograph. (Most photographs are roughly 200<=×<=300.) After this, *h* lines follow, each describing a single row of the picture. The picture is monochrome (in shades of grey). Its *i*-th row is described by *w* space-separated integers *a**ij* (*j*<==<=1,<=...,<=*w*), where *a**ij* is the brightness of the corresponding pixel (0<=≤<=*a**ij*<=<<=256, where 0 is black and 255 is white).
Each picture will be either a real-life photograph, or a real-life photograph which has been broken up into two pieces and rearranged. More precisely, in the latter case, the topmost rows have been moved to the bottom of the picture. It is guaranteed that *h* is even.
There is only a single input file to be processed, called all.in, and it is downloadable from the online judge. You are also a given another input file, called sample.in, which contains the first 20 pictures from all.in; you are provided the correct answers for sample.in in sample.out. You are also given a directory easy_bmp, which contains the first 50 input photographs in the form of .bmp image files, as well as a directory easy_sample_original_bmp, which contains the first 20 images before rearrangement. Check the notes for the download links.
Output Specification:
Your program should print *q* lines. The *i*-th line should contain your answer for the *i*-th question: YES if the photograph has been rearranged and NO otherwise. Your answers will be accepted if they all conform to this format and if at least 75% of them are correct.
Because the input is rather huge, feel free to process it locally and submit just your precomputed answers (i.e., a program which just prints your output for the input file all.in).
Note:
The link to download all necessary files is http://assets.codeforces.com/files/690/easy_contestant_package.zip
|
```python
for x in ['YES', 'NO', 'NO', 'YES', 'NO', 'NO', 'NO', 'NO', 'NO', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'NO', 'YES', 'NO', 'YES', 'NO', 'NO', 'YES', 'NO', 'NO', 'YES', 'NO', 'NO', 'YES', 'YES', 'NO', 'YES', 'NO', 'YES', 'NO', 'YES', 'YES', 'YES', 'NO', 'YES', 'YES', 'NO', 'YES', 'YES', 'NO', 'YES', 'YES', 'YES', 'NO', 'YES', 'NO', 'NO', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'NO', 'NO', 'YES', 'NO', 'NO', 'NO', 'NO', 'YES', 'NO', 'NO', 'NO', 'NO', 'YES', 'NO', 'YES', 'NO', 'NO', 'NO', 'YES', 'YES', 'NO', 'YES', 'YES', 'NO', 'NO', 'NO', 'NO', 'NO', 'NO', 'YES', 'NO', 'NO', 'YES', 'NO', 'YES', 'NO', 'NO', 'NO', 'NO', 'YES', 'YES', 'YES', 'YES', 'YES', 'NO', 'NO', 'NO', 'YES', 'NO', 'YES', 'NO', 'YES', 'NO', 'YES', 'NO', 'NO', 'YES', 'YES', 'YES', 'YES', 'NO', 'NO', 'NO', 'YES', 'NO', 'YES', 'YES', 'YES', 'NO', 'YES', 'NO', 'NO', 'NO', 'YES', 'NO', 'NO', 'NO', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'YES', 'NO', 'YES', 'NO', 'NO', 'NO', 'NO', 'NO', 'YES', 'NO', 'NO', 'YES', 'NO', 'YES', 'NO', 'YES', 'NO', 'YES', 'NO', 'NO', 'YES', 'YES', 'NO', 'NO', 'YES', 'NO', 'NO', 'NO', 'NO', 'YES', 'NO', 'YES', 'NO', 'NO', 'YES', 'NO', 'YES', 'YES', 'YES', 'NO', 'NO', 'NO', 'YES', 'YES', 'NO', 'YES', 'NO', 'YES', 'YES', 'YES', 'NO', 'YES', 'YES', 'NO', 'YES', 'NO', 'YES', 'YES', 'NO', 'YES', 'YES', 'NO', 'NO', 'YES', 'NO', 'NO', 'NO', 'YES', 'YES', 'NO', 'YES']:
print(x)
```
| 3
|
|
822
|
A
|
I'm bored with life
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"number theory"
] | null | null |
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
|
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
|
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
|
[
"4 3\n"
] |
[
"6\n"
] |
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
| 500
|
[
{
"input": "4 3",
"output": "6"
},
{
"input": "10 399603090",
"output": "3628800"
},
{
"input": "6 973151934",
"output": "720"
},
{
"input": "2 841668075",
"output": "2"
},
{
"input": "7 415216919",
"output": "5040"
},
{
"input": "3 283733059",
"output": "6"
},
{
"input": "11 562314608",
"output": "39916800"
},
{
"input": "3 990639260",
"output": "6"
},
{
"input": "11 859155400",
"output": "39916800"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "5 3",
"output": "6"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "5 4",
"output": "24"
},
{
"input": "1 12",
"output": "1"
},
{
"input": "9 7",
"output": "5040"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "6 11",
"output": "720"
},
{
"input": "6 7",
"output": "720"
},
{
"input": "11 11",
"output": "39916800"
},
{
"input": "4 999832660",
"output": "24"
},
{
"input": "7 999228288",
"output": "5040"
},
{
"input": "11 999257105",
"output": "39916800"
},
{
"input": "11 999286606",
"output": "39916800"
},
{
"input": "3 999279109",
"output": "6"
},
{
"input": "999632727 11",
"output": "39916800"
},
{
"input": "999625230 7",
"output": "5040"
},
{
"input": "999617047 3",
"output": "6"
},
{
"input": "999646548 7",
"output": "5040"
},
{
"input": "999639051 3",
"output": "6"
},
{
"input": "12 12",
"output": "479001600"
},
{
"input": "12 1",
"output": "1"
},
{
"input": "1213 5",
"output": "120"
},
{
"input": "8 9",
"output": "40320"
},
{
"input": "12 9",
"output": "362880"
},
{
"input": "12 1000000000",
"output": "479001600"
},
{
"input": "1000000000 1",
"output": "1"
},
{
"input": "12 13",
"output": "479001600"
},
{
"input": "2 29845",
"output": "2"
},
{
"input": "10 21",
"output": "3628800"
},
{
"input": "12 20",
"output": "479001600"
},
{
"input": "15 12",
"output": "479001600"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000 12",
"output": "479001600"
},
{
"input": "11 30",
"output": "39916800"
},
{
"input": "17 12",
"output": "479001600"
},
{
"input": "4 19",
"output": "24"
},
{
"input": "12 15",
"output": "479001600"
},
{
"input": "20 6",
"output": "720"
},
{
"input": "10 20",
"output": "3628800"
},
{
"input": "10 10",
"output": "3628800"
},
{
"input": "22 12",
"output": "479001600"
},
{
"input": "20 12",
"output": "479001600"
},
{
"input": "12 23",
"output": "479001600"
},
{
"input": "12 22",
"output": "479001600"
},
{
"input": "18 3",
"output": "6"
},
{
"input": "14 10",
"output": "3628800"
},
{
"input": "14 12",
"output": "479001600"
},
{
"input": "8 3",
"output": "6"
},
{
"input": "5 5",
"output": "120"
}
] | 1,680,878,603
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
def binary_gcd(a, b):
if a == 0:
return b
elif b == 0:
return a
elif a == b:
return a
elif a % 2 == 0 and b % 2 == 0:
return 2 * binary_gcd(a // 2, b // 2)
elif a % 2 == 0:
return binary_gcd(a // 2, b)
elif b % 2 == 0:
return binary_gcd(a, b // 2)
elif a > b:
return binary_gcd((a - b) // 2, b)
else:
return binary_gcd((b - a) // 2, a)
def gcd_factorial(a, b):
m = min(a, b)
result = 1
for i in range(1, m+1):
result *= i
if i == m:
return binary_gcd(result, result)
return result
print(input())
a,b= str(input()).split(' ')
print(gcd_factorial(a, b))
|
Title: I'm bored with life
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input Specification:
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
Output Specification:
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
Demo Input:
['4 3\n']
Demo Output:
['6\n']
Note:
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
|
```python
def binary_gcd(a, b):
if a == 0:
return b
elif b == 0:
return a
elif a == b:
return a
elif a % 2 == 0 and b % 2 == 0:
return 2 * binary_gcd(a // 2, b // 2)
elif a % 2 == 0:
return binary_gcd(a // 2, b)
elif b % 2 == 0:
return binary_gcd(a, b // 2)
elif a > b:
return binary_gcd((a - b) // 2, b)
else:
return binary_gcd((b - a) // 2, a)
def gcd_factorial(a, b):
m = min(a, b)
result = 1
for i in range(1, m+1):
result *= i
if i == m:
return binary_gcd(result, result)
return result
print(input())
a,b= str(input()).split(' ')
print(gcd_factorial(a, b))
```
| -1
|
|
74
|
A
|
Room Leader
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Room Leader
|
2
|
256
|
Let us remind you part of the rules of Codeforces. The given rules slightly simplified, use the problem statement as a formal document.
In the beginning of the round the contestants are divided into rooms. Each room contains exactly *n* participants. During the contest the participants are suggested to solve five problems, *A*, *B*, *C*, *D* and *E*. For each of these problem, depending on when the given problem was solved and whether it was solved at all, the participants receive some points. Besides, a contestant can perform hacks on other contestants. For each successful hack a contestant earns 100 points, for each unsuccessful hack a contestant loses 50 points. The number of points for every contestant is represented by the sum of points he has received from all his problems, including hacks.
You are suggested to determine the leader for some room; the leader is a participant who has maximum points.
|
The first line contains an integer *n*, which is the number of contestants in the room (1<=≤<=*n*<=≤<=50). The next *n* lines contain the participants of a given room. The *i*-th line has the format of "*handle**i* *plus**i* *minus**i* *a**i* *b**i* *c**i* *d**i* *e**i*" — it is the handle of a contestant, the number of successful hacks, the number of unsuccessful hacks and the number of points he has received from problems *A*, *B*, *C*, *D*, *E* correspondingly. The handle of each participant consists of Latin letters, digits and underscores and has the length from 1 to 20 characters. There are the following limitations imposed upon the numbers:
- 0<=≤<=*plus**i*,<=*minus**i*<=≤<=50; - 150<=≤<=*a**i*<=≤<=500 or *a**i*<==<=0, if problem *A* is not solved; - 300<=≤<=*b**i*<=≤<=1000 or *b**i*<==<=0, if problem *B* is not solved; - 450<=≤<=*c**i*<=≤<=1500 or *c**i*<==<=0, if problem *C* is not solved; - 600<=≤<=*d**i*<=≤<=2000 or *d**i*<==<=0, if problem *D* is not solved; - 750<=≤<=*e**i*<=≤<=2500 or *e**i*<==<=0, if problem *E* is not solved.
All the numbers are integer. All the participants have different handles. It is guaranteed that there is exactly one leader in the room (i.e. there are no two participants with the maximal number of points).
|
Print on the single line the handle of the room leader.
|
[
"5\nPetr 3 1 490 920 1000 1200 0\ntourist 2 0 490 950 1100 1400 0\nEgor 7 0 480 900 950 0 1000\nc00lH4x0R 0 10 150 0 0 0 0\nsome_participant 2 1 450 720 900 0 0\n"
] |
[
"tourist"
] |
The number of points that each participant from the example earns, are as follows:
- Petr — 3860 - tourist — 4140 - Egor — 4030 - c00lH4x0R — - 350 - some_participant — 2220
Thus, the leader of the room is tourist.
| 500
|
[
{
"input": "5\nPetr 3 1 490 920 1000 1200 0\ntourist 2 0 490 950 1100 1400 0\nEgor 7 0 480 900 950 0 1000\nc00lH4x0R 0 10 150 0 0 0 0\nsome_participant 2 1 450 720 900 0 0",
"output": "tourist"
},
{
"input": "1\nA 0 0 200 0 0 0 0",
"output": "A"
},
{
"input": "2\n12345678901234567890 1 0 200 0 0 0 0\n_ 1 0 201 0 0 0 0",
"output": "_"
},
{
"input": "5\nAb 0 0 481 900 1200 1600 2000\nCd 0 0 480 899 1200 1600 2000\nEf 0 0 480 900 1200 1600 2000\ngH 0 0 480 900 1200 1599 2000\nij 0 0 480 900 1199 1600 2001",
"output": "Ab"
},
{
"input": "4\nF1 0 0 150 0 0 0 0\nF2 0 1 0 0 0 0 0\nF3 0 2 0 0 0 0 0\nF4 0 3 0 0 0 0 0",
"output": "F1"
},
{
"input": "2\nA87h 5 0 199 0 0 0 0\nBcfg 7 0 0 0 0 0 0",
"output": "Bcfg"
},
{
"input": "10\nKh 40 26 0 0 0 0 1243\nn 46 50 500 0 910 1912 0\nU 18 1 182 0 457 0 0\nFth6A0uT6i 38 30 0 787 0 1121 0\nC5l 24 38 0 689 1082 0 0\nN 47 25 0 0 1065 0 1538\nznyL 9 24 0 315 0 0 0\nJ0kU 27 47 445 0 0 0 0\nlT0rwiD2pg 46 13 0 818 0 0 0\nuJzr 29 14 0 0 0 0 2387",
"output": "N"
},
{
"input": "2\nminus_one 0 4 199 0 0 0 0\nminus_two 0 4 198 0 0 0 0",
"output": "minus_one"
},
{
"input": "10\nW22kb1L1 0 39 0 465 0 1961 865\n1MCXiVYmu5ys0afl 0 38 0 0 0 1982 1241\nCxg706kUJtQ 0 23 211 0 0 1785 1056\nmzEY 0 16 0 0 0 1988 1404\nv8JUjmam5SFP 0 48 0 788 1199 1426 0\n7giq 0 21 0 780 1437 1363 1930\nsXsUGbAulj6Lbiq 0 32 205 0 0 603 0\nRepIrY1Er4PgK 0 13 381 872 927 1488 0\nleKBdKHLnLFz 0 29 220 0 0 1006 889\nD 0 26 497 0 0 0 1815",
"output": "7giq"
},
{
"input": "1\nZ 0 0 0 0 0 0 0",
"output": "Z"
},
{
"input": "3\nAbcd 0 4 189 0 0 0 0\nDefg 0 5 248 0 0 0 0\nGhh 1 3 0 0 0 0 0",
"output": "Defg"
},
{
"input": "3\ndf 0 6 0 0 0 0 0\njnm 1 8 300 0 0 0 0\n_ub_ 3 20 300 310 0 0 0",
"output": "jnm"
},
{
"input": "1\njhgcyt 0 50 0 0 0 0 0",
"output": "jhgcyt"
},
{
"input": "2\njhv 0 50 500 1000 1500 2000 2500\nPetr 2 1 489 910 1100 1300 1000",
"output": "jhv"
},
{
"input": "3\nufu 0 50 0 0 0 0 0\nhzEr65f 1 50 0 0 0 0 0\nytdttjfhfd 0 50 150 0 0 0 0",
"output": "ytdttjfhfd"
},
{
"input": "5\nufuf 0 50 0 0 0 0 0\nyfycy 50 0 500 1000 1500 2000 2500\n__u77 6 7 490 999 1456 1976 1356\n0 1 2 0 0 0 0 2452\ngu7fF 50 0 500 1000 1500 2000 2499",
"output": "yfycy"
},
{
"input": "2\nhfy 0 50 0 0 0 0 2500\nugug 0 50 0 0 0 0 2499",
"output": "hfy"
},
{
"input": "8\nA 0 0 0 0 0 0 0\nb 0 0 0 0 0 0 0\nc 0 0 0 0 0 0 0\nD 0 0 0 0 0 0 0\nE 1 0 0 0 0 0 0\nF 0 0 0 0 0 0 0\ng 0 0 0 0 0 0 0\nH 0 0 0 0 0 0 0",
"output": "E"
},
{
"input": "2\nyyyc 50 50 0 0 0 0 0\nydd 0 0 0 0 0 0 2499",
"output": "yyyc"
},
{
"input": "2\ntom 0 2 0 0 0 0 0\nmac 0 1 0 0 0 0 0",
"output": "mac"
},
{
"input": "1\ncool 0 10 0 0 0 0 0",
"output": "cool"
}
] | 1,616,882,421
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 124
| 307,200
|
total_contestants = int(input())
final_scores = {}
for _ in range(total_contestants):
handle, successful_hacks, unsuccessful_hacks, *problem_values = input().split()
successful_hacks = int(successful_hacks)
unsuccessful_hacks = int(unsuccessful_hacks)
final_scores[handle] = (successful_hacks * 100) - (unsuccessful_hacks * 50) + sum([int(x) for x in problem_values])
highest_score = max(list(final_scores.values()))
handle_with_highest_score = {handle : score_value for (handle, score_value) in final_scores.items() if score_value == highest_score}
print(list(handle_with_highest_score.keys())[0])
|
Title: Room Leader
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Let us remind you part of the rules of Codeforces. The given rules slightly simplified, use the problem statement as a formal document.
In the beginning of the round the contestants are divided into rooms. Each room contains exactly *n* participants. During the contest the participants are suggested to solve five problems, *A*, *B*, *C*, *D* and *E*. For each of these problem, depending on when the given problem was solved and whether it was solved at all, the participants receive some points. Besides, a contestant can perform hacks on other contestants. For each successful hack a contestant earns 100 points, for each unsuccessful hack a contestant loses 50 points. The number of points for every contestant is represented by the sum of points he has received from all his problems, including hacks.
You are suggested to determine the leader for some room; the leader is a participant who has maximum points.
Input Specification:
The first line contains an integer *n*, which is the number of contestants in the room (1<=≤<=*n*<=≤<=50). The next *n* lines contain the participants of a given room. The *i*-th line has the format of "*handle**i* *plus**i* *minus**i* *a**i* *b**i* *c**i* *d**i* *e**i*" — it is the handle of a contestant, the number of successful hacks, the number of unsuccessful hacks and the number of points he has received from problems *A*, *B*, *C*, *D*, *E* correspondingly. The handle of each participant consists of Latin letters, digits and underscores and has the length from 1 to 20 characters. There are the following limitations imposed upon the numbers:
- 0<=≤<=*plus**i*,<=*minus**i*<=≤<=50; - 150<=≤<=*a**i*<=≤<=500 or *a**i*<==<=0, if problem *A* is not solved; - 300<=≤<=*b**i*<=≤<=1000 or *b**i*<==<=0, if problem *B* is not solved; - 450<=≤<=*c**i*<=≤<=1500 or *c**i*<==<=0, if problem *C* is not solved; - 600<=≤<=*d**i*<=≤<=2000 or *d**i*<==<=0, if problem *D* is not solved; - 750<=≤<=*e**i*<=≤<=2500 or *e**i*<==<=0, if problem *E* is not solved.
All the numbers are integer. All the participants have different handles. It is guaranteed that there is exactly one leader in the room (i.e. there are no two participants with the maximal number of points).
Output Specification:
Print on the single line the handle of the room leader.
Demo Input:
['5\nPetr 3 1 490 920 1000 1200 0\ntourist 2 0 490 950 1100 1400 0\nEgor 7 0 480 900 950 0 1000\nc00lH4x0R 0 10 150 0 0 0 0\nsome_participant 2 1 450 720 900 0 0\n']
Demo Output:
['tourist']
Note:
The number of points that each participant from the example earns, are as follows:
- Petr — 3860 - tourist — 4140 - Egor — 4030 - c00lH4x0R — - 350 - some_participant — 2220
Thus, the leader of the room is tourist.
|
```python
total_contestants = int(input())
final_scores = {}
for _ in range(total_contestants):
handle, successful_hacks, unsuccessful_hacks, *problem_values = input().split()
successful_hacks = int(successful_hacks)
unsuccessful_hacks = int(unsuccessful_hacks)
final_scores[handle] = (successful_hacks * 100) - (unsuccessful_hacks * 50) + sum([int(x) for x in problem_values])
highest_score = max(list(final_scores.values()))
handle_with_highest_score = {handle : score_value for (handle, score_value) in final_scores.items() if score_value == highest_score}
print(list(handle_with_highest_score.keys())[0])
```
| 3.968428
|
47
|
B
|
Coins
|
PROGRAMMING
| 1,200
|
[
"implementation"
] |
B. Coins
|
2
|
256
|
One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal.
|
The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B.
|
It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights.
|
[
"A>B\nC<B\nA>C\n",
"A<B\nB>C\nC>A\n"
] |
[
"CBA",
"ACB"
] |
none
| 1,000
|
[
{
"input": "A>B\nC<B\nA>C",
"output": "CBA"
},
{
"input": "A<B\nB>C\nC>A",
"output": "ACB"
},
{
"input": "A<C\nB<A\nB>C",
"output": "Impossible"
},
{
"input": "A<B\nA<C\nB>C",
"output": "ACB"
},
{
"input": "B>A\nC<B\nC>A",
"output": "ACB"
},
{
"input": "A>B\nB>C\nC<A",
"output": "CBA"
},
{
"input": "A>C\nA>B\nB<C",
"output": "BCA"
},
{
"input": "C<B\nB>A\nA<C",
"output": "ACB"
},
{
"input": "C<B\nA>B\nC<A",
"output": "CBA"
},
{
"input": "C>B\nB>A\nA<C",
"output": "ABC"
},
{
"input": "C<B\nB<A\nC>A",
"output": "Impossible"
},
{
"input": "B<C\nC<A\nA>B",
"output": "BCA"
},
{
"input": "A>B\nC<B\nC<A",
"output": "CBA"
},
{
"input": "B>A\nC>B\nA>C",
"output": "Impossible"
},
{
"input": "B<A\nC>B\nC>A",
"output": "BAC"
},
{
"input": "A<B\nC>B\nA<C",
"output": "ABC"
},
{
"input": "A<B\nC<A\nB<C",
"output": "Impossible"
},
{
"input": "A>C\nC<B\nB>A",
"output": "CAB"
},
{
"input": "C>A\nA<B\nB>C",
"output": "ACB"
},
{
"input": "C>A\nC<B\nB>A",
"output": "ACB"
},
{
"input": "B>C\nB>A\nA<C",
"output": "ACB"
},
{
"input": "C<B\nC<A\nB<A",
"output": "CBA"
},
{
"input": "A<C\nA<B\nB>C",
"output": "ACB"
},
{
"input": "B>A\nA>C\nB>C",
"output": "CAB"
},
{
"input": "B<A\nA<C\nC<B",
"output": "Impossible"
},
{
"input": "A<C\nB>C\nA>B",
"output": "Impossible"
},
{
"input": "B>A\nC<A\nC>B",
"output": "Impossible"
},
{
"input": "A>C\nC>B\nB<A",
"output": "BCA"
},
{
"input": "B<C\nB<A\nA>C",
"output": "BCA"
},
{
"input": "A>B\nC>B\nA<C",
"output": "BAC"
},
{
"input": "C<B\nC<A\nB<A",
"output": "CBA"
},
{
"input": "A<C\nA>B\nB>C",
"output": "Impossible"
},
{
"input": "B>A\nB>C\nA<C",
"output": "ACB"
},
{
"input": "B>C\nC<A\nB<A",
"output": "CBA"
},
{
"input": "C>A\nB>A\nB>C",
"output": "ACB"
},
{
"input": "B<A\nB>C\nA<C",
"output": "Impossible"
},
{
"input": "B<C\nA<B\nC>A",
"output": "ABC"
},
{
"input": "C<B\nB>A\nA>C",
"output": "CAB"
},
{
"input": "A>B\nC>A\nB<C",
"output": "BAC"
},
{
"input": "A>B\nB>C\nC<A",
"output": "CBA"
},
{
"input": "B>C\nC>A\nA>B",
"output": "Impossible"
},
{
"input": "B<A\nB>C\nC<A",
"output": "CBA"
},
{
"input": "C>B\nB>A\nC>A",
"output": "ABC"
},
{
"input": "A>C\nA>B\nB>C",
"output": "CBA"
},
{
"input": "B<C\nA>B\nA<C",
"output": "BAC"
},
{
"input": "C>A\nC<B\nB>A",
"output": "ACB"
},
{
"input": "C>B\nA<B\nC<A",
"output": "Impossible"
},
{
"input": "A<C\nC<B\nA>B",
"output": "Impossible"
},
{
"input": "B>C\nA<B\nA<C",
"output": "ACB"
},
{
"input": "C>A\nA<B\nC>B",
"output": "ABC"
}
] | 1,585,102,482
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 216
| 307,200
|
L = list()
for i in range(3):
L.append(input())
if L[0][1] == L[1][1] and L[1][1] == L[2][1]:
print("Impossible")
exit(0)
L1 = list()
for ch in L:
if ch[1] == ">":
L1.append(ch[2]+"<"+ch[0])
else:
L1.append(ch)
result = ""
if L1[0][0] == L1[1][0] or L1[0][0] == L1[2][0]:
result += L1[0][0]
elif L1[1][0] == L1[2][0]:
result += L1[1][0]
if L1[0][2] == L1[1][2] or L1[0][2] == L1[2][2]:
result += L1[0][2]
elif L1[1][2] == L1[2][2]:
result += L1[1][2]
res = ""
if "A" not in result:
res = result[0]+"A"+result[1]
elif "B" not in result:
res = result[0]+"B"+result[1]
else:
res = result[0]+"C"+result[1]
print(res)
|
Title: Coins
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal.
Input Specification:
The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B.
Output Specification:
It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights.
Demo Input:
['A>B\nC<B\nA>C\n', 'A<B\nB>C\nC>A\n']
Demo Output:
['CBA', 'ACB']
Note:
none
|
```python
L = list()
for i in range(3):
L.append(input())
if L[0][1] == L[1][1] and L[1][1] == L[2][1]:
print("Impossible")
exit(0)
L1 = list()
for ch in L:
if ch[1] == ">":
L1.append(ch[2]+"<"+ch[0])
else:
L1.append(ch)
result = ""
if L1[0][0] == L1[1][0] or L1[0][0] == L1[2][0]:
result += L1[0][0]
elif L1[1][0] == L1[2][0]:
result += L1[1][0]
if L1[0][2] == L1[1][2] or L1[0][2] == L1[2][2]:
result += L1[0][2]
elif L1[1][2] == L1[2][2]:
result += L1[1][2]
res = ""
if "A" not in result:
res = result[0]+"A"+result[1]
elif "B" not in result:
res = result[0]+"B"+result[1]
else:
res = result[0]+"C"+result[1]
print(res)
```
| -1
|
337
|
A
|
Puzzles
|
PROGRAMMING
| 900
|
[
"greedy"
] | null | null |
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
|
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
|
Print a single integer — the least possible difference the teacher can obtain.
|
[
"4 6\n10 12 10 7 5 22\n"
] |
[
"5\n"
] |
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
| 500
|
[
{
"input": "4 6\n10 12 10 7 5 22",
"output": "5"
},
{
"input": "2 2\n4 4",
"output": "0"
},
{
"input": "2 10\n4 5 6 7 8 9 10 11 12 12",
"output": "0"
},
{
"input": "4 5\n818 136 713 59 946",
"output": "759"
},
{
"input": "3 20\n446 852 783 313 549 965 40 88 86 617 479 118 768 34 47 826 366 957 463 903",
"output": "13"
},
{
"input": "2 25\n782 633 152 416 432 825 115 97 386 357 836 310 530 413 354 373 847 882 913 682 729 582 671 674 94",
"output": "3"
},
{
"input": "4 25\n226 790 628 528 114 64 239 279 619 39 894 763 763 847 525 93 882 697 999 643 650 244 159 884 190",
"output": "31"
},
{
"input": "2 50\n971 889 628 39 253 157 925 694 129 516 660 272 738 319 611 816 142 717 514 392 41 105 132 676 958 118 306 768 600 685 103 857 704 346 857 309 23 718 618 161 176 379 846 834 640 468 952 878 164 997",
"output": "0"
},
{
"input": "25 50\n582 146 750 905 313 509 402 21 488 512 32 898 282 64 579 869 37 996 377 929 975 697 666 837 311 205 116 992 533 298 648 268 54 479 792 595 152 69 267 417 184 433 894 603 988 712 24 414 301 176",
"output": "412"
},
{
"input": "49 50\n58 820 826 960 271 294 473 102 925 318 729 672 244 914 796 646 868 6 893 882 726 203 528 498 271 195 355 459 721 680 547 147 631 116 169 804 145 996 133 559 110 257 771 476 576 251 607 314 427 886",
"output": "938"
},
{
"input": "50 50\n374 573 323 744 190 806 485 247 628 336 491 606 702 321 991 678 337 579 86 240 993 208 668 686 855 205 363 177 719 249 896 919 782 434 59 647 787 996 286 216 636 212 546 903 958 559 544 126 608 993",
"output": "937"
},
{
"input": "6 50\n6 8 7 8 5 4 4 5 7 8 6 5 7 4 7 7 7 8 6 4 6 6 8 8 7 7 8 7 5 8 5 4 4 7 8 4 4 6 6 6 8 7 4 7 6 6 5 8 4 7",
"output": "0"
},
{
"input": "37 50\n14 5 11 17 8 20 19 16 20 11 17 20 16 9 14 14 13 18 11 20 8 8 8 5 19 17 6 18 10 20 9 7 12 6 14 17 4 4 10 13 7 4 11 6 20 19 12 12 15 19",
"output": "12"
},
{
"input": "40 50\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "0"
},
{
"input": "40 50\n17 20 43 26 41 37 14 8 30 35 30 24 43 8 42 9 41 50 41 35 27 32 35 43 28 36 31 16 5 7 23 16 14 29 8 39 12 16 36 18 49 39 33 37 38 6 6 27 23 17",
"output": "31"
},
{
"input": "2 2\n1000 4",
"output": "996"
},
{
"input": "2 3\n4 502 1000",
"output": "498"
},
{
"input": "3 3\n4 1000 4",
"output": "996"
}
] | 1,680,260,776
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 0
|
n = input().split()
k = input().split()
a = []
l = []
for i in range(len(k)-int(n[0])+1):
for j in range(int(n[0])):
a.append(int(k[i+j]))
l.append(max(a)-min(a))
a = []
print(min(l))
|
Title: Puzzles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The end of the school year is near and Ms. Manana, the teacher, will soon have to say goodbye to a yet another class. She decided to prepare a goodbye present for her *n* students and give each of them a jigsaw puzzle (which, as wikipedia states, is a tiling puzzle that requires the assembly of numerous small, often oddly shaped, interlocking and tessellating pieces).
The shop assistant told the teacher that there are *m* puzzles in the shop, but they might differ in difficulty and size. Specifically, the first jigsaw puzzle consists of *f*1 pieces, the second one consists of *f*2 pieces and so on.
Ms. Manana doesn't want to upset the children, so she decided that the difference between the numbers of pieces in her presents must be as small as possible. Let *A* be the number of pieces in the largest puzzle that the teacher buys and *B* be the number of pieces in the smallest such puzzle. She wants to choose such *n* puzzles that *A*<=-<=*B* is minimum possible. Help the teacher and find the least possible value of *A*<=-<=*B*.
Input Specification:
The first line contains space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=*m*<=≤<=50). The second line contains *m* space-separated integers *f*1,<=*f*2,<=...,<=*f**m* (4<=≤<=*f**i*<=≤<=1000) — the quantities of pieces in the puzzles sold in the shop.
Output Specification:
Print a single integer — the least possible difference the teacher can obtain.
Demo Input:
['4 6\n10 12 10 7 5 22\n']
Demo Output:
['5\n']
Note:
Sample 1. The class has 4 students. The shop sells 6 puzzles. If Ms. Manana buys the first four puzzles consisting of 10, 12, 10 and 7 pieces correspondingly, then the difference between the sizes of the largest and the smallest puzzle will be equal to 5. It is impossible to obtain a smaller difference. Note that the teacher can also buy puzzles 1, 3, 4 and 5 to obtain the difference 5.
|
```python
n = input().split()
k = input().split()
a = []
l = []
for i in range(len(k)-int(n[0])+1):
for j in range(int(n[0])):
a.append(int(k[i+j]))
l.append(max(a)-min(a))
a = []
print(min(l))
```
| 0
|
|
811
|
A
|
Vladik and Courtesy
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.
|
Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively.
|
Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.
|
[
"1 1\n",
"7 6\n"
] |
[
"Valera\n",
"Vladik\n"
] |
Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 500
|
[
{
"input": "1 1",
"output": "Valera"
},
{
"input": "7 6",
"output": "Vladik"
},
{
"input": "25 38",
"output": "Vladik"
},
{
"input": "8311 2468",
"output": "Valera"
},
{
"input": "250708 857756",
"output": "Vladik"
},
{
"input": "957985574 24997558",
"output": "Valera"
},
{
"input": "999963734 999994456",
"output": "Vladik"
},
{
"input": "1000000000 1000000000",
"output": "Vladik"
},
{
"input": "946 879",
"output": "Valera"
},
{
"input": "10819 45238",
"output": "Vladik"
},
{
"input": "101357 236928",
"output": "Vladik"
},
{
"input": "1033090 7376359",
"output": "Vladik"
},
{
"input": "9754309 9525494",
"output": "Valera"
},
{
"input": "90706344 99960537",
"output": "Vladik"
},
{
"input": "965161805 908862070",
"output": "Valera"
},
{
"input": "9 11",
"output": "Valera"
},
{
"input": "3 2",
"output": "Vladik"
},
{
"input": "6 6",
"output": "Vladik"
},
{
"input": "4 4",
"output": "Valera"
},
{
"input": "5 5",
"output": "Valera"
},
{
"input": "5 4",
"output": "Valera"
},
{
"input": "12345680 1",
"output": "Valera"
},
{
"input": "9 10",
"output": "Valera"
},
{
"input": "678 76687",
"output": "Vladik"
},
{
"input": "1 678",
"output": "Vladik"
},
{
"input": "45 1678",
"output": "Vladik"
},
{
"input": "3 3",
"output": "Vladik"
},
{
"input": "10 11",
"output": "Valera"
},
{
"input": "2 1",
"output": "Valera"
},
{
"input": "1 2",
"output": "Vladik"
},
{
"input": "2 2",
"output": "Vladik"
},
{
"input": "4 5",
"output": "Valera"
},
{
"input": "9 6",
"output": "Valera"
},
{
"input": "1 5",
"output": "Vladik"
},
{
"input": "7 8",
"output": "Vladik"
},
{
"input": "1000000000 999982505",
"output": "Valera"
},
{
"input": "12 12",
"output": "Vladik"
},
{
"input": "1000 950",
"output": "Valera"
},
{
"input": "10 9",
"output": "Valera"
},
{
"input": "100 9",
"output": "Valera"
},
{
"input": "1000 996",
"output": "Vladik"
},
{
"input": "9 5",
"output": "Valera"
}
] | 1,495,879,503
| 1,803
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 4
| 62
| 0
|
def winner(a, b):
a_c = 1
b_c = 2
for i in range(a + b):
a -= a_c
b -= b_c
if a<0 or b<0:
break
a_c += 2
b_c += 2
if a<0 and b<0:
return "Vladik"
else:
return "Valera"
if __name__ == '__main__':
a, b = map(int, input().split())
print(winner(a, b))
|
Title: Vladik and Courtesy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.
Input Specification:
Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively.
Output Specification:
Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.
Demo Input:
['1 1\n', '7 6\n']
Demo Output:
['Valera\n', 'Vladik\n']
Note:
Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
def winner(a, b):
a_c = 1
b_c = 2
for i in range(a + b):
a -= a_c
b -= b_c
if a<0 or b<0:
break
a_c += 2
b_c += 2
if a<0 and b<0:
return "Vladik"
else:
return "Valera"
if __name__ == '__main__':
a, b = map(int, input().split())
print(winner(a, b))
```
| 0
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,679,183,576
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
n = input()
n = int(n)
temp = 0
arr = []
result = []
i = 0
while i < n:
temp = input()
arr.insert(i, temp)
i = i + 1
i = 0
while i < n:
temp = arr[i]
if len(temp) > 10:
newStr = temp[0] + str(len(temp)) + temp[len(temp)-1]
else:
newStr = temp
result.insert(i, newStr)
i = i + 1
i = 0
while i < n:
print(result[i])
i = i + 1
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
n = input()
n = int(n)
temp = 0
arr = []
result = []
i = 0
while i < n:
temp = input()
arr.insert(i, temp)
i = i + 1
i = 0
while i < n:
temp = arr[i]
if len(temp) > 10:
newStr = temp[0] + str(len(temp)) + temp[len(temp)-1]
else:
newStr = temp
result.insert(i, newStr)
i = i + 1
i = 0
while i < n:
print(result[i])
i = i + 1
```
| 0
|
599
|
C
|
Day at the Beach
|
PROGRAMMING
| 1,600
|
[
"sortings"
] | null | null |
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 *h**i*. 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 *h**i*<=≤<=*h**i*<=+<=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 *h**i*,<=*h**i*<=+<=1,<=...,<=*h**j* becomes sorted. - The partitioning should satisfy the condition that after each block is sorted, the sequence *h**i* 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 *h**i* (1<=≤<=*h**i*<=≤<=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.
|
[
"3\n1 2 3\n",
"4\n2 1 3 2\n"
] |
[
"3\n",
"2\n"
] |
In the first sample the partitioning looks like that: [1][2][3].
In the second sample the partitioning is: [2, 1][3, 2]
| 1,500
|
[
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "4\n2 1 3 2",
"output": "2"
},
{
"input": "17\n1 45 22 39 28 23 23 100 500 778 777 778 1001 1002 1005 1003 1005",
"output": "10"
},
{
"input": "101\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",
"output": "3"
},
{
"input": "10\n1 2 2 2 2 2 2 2 2 1",
"output": "2"
},
{
"input": "25\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",
"output": "12"
},
{
"input": "10\n17 18 19 19 18 17 100 500 100 100",
"output": "4"
},
{
"input": "10\n1 1 1 1 5 5 1 1 1 1",
"output": "5"
},
{
"input": "20\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "20"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "5\n1 5 3 5 2",
"output": "2"
},
{
"input": "10\n1 1 1 1 2 2 2 2 4 3",
"output": "9"
},
{
"input": "20\n1 2 2 2 5 6 6 6 7 7 8 9 15 15 16 16 17 18 19 19",
"output": "20"
},
{
"input": "4\n2 2 1 1",
"output": "1"
}
] | 1,658,894,851
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 39
| 171
| 20,992,000
|
from collections import Counter
n=int(input())
arr=list(map(int,input().split()))
p=sorted(arr)
ans=0
c=Counter()
seen=set()
for a,b in zip(arr,p):
c[a]+=1
c[b]-=1
if c[a]>0:
seen.add(a)
if c[a]==0 and a in seen:
seen.remove(a)
if c[b]>0:
seen.add(b)
if c[b]==0 and b in seen:
seen.remove(b)
if len(seen)==0:
ans+=1
print(ans)
|
Title: Day at the Beach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
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 *h**i*. 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 *h**i*<=≤<=*h**i*<=+<=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 *h**i*,<=*h**i*<=+<=1,<=...,<=*h**j* becomes sorted. - The partitioning should satisfy the condition that after each block is sorted, the sequence *h**i* 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.
Input Specification:
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 *h**i* (1<=≤<=*h**i*<=≤<=109). The *i*-th of these integers corresponds to the height of the *i*-th castle.
Output Specification:
Print the maximum possible number of blocks in a valid partitioning.
Demo Input:
['3\n1 2 3\n', '4\n2 1 3 2\n']
Demo Output:
['3\n', '2\n']
Note:
In the first sample the partitioning looks like that: [1][2][3].
In the second sample the partitioning is: [2, 1][3, 2]
|
```python
from collections import Counter
n=int(input())
arr=list(map(int,input().split()))
p=sorted(arr)
ans=0
c=Counter()
seen=set()
for a,b in zip(arr,p):
c[a]+=1
c[b]-=1
if c[a]>0:
seen.add(a)
if c[a]==0 and a in seen:
seen.remove(a)
if c[b]>0:
seen.add(b)
if c[b]==0 and b in seen:
seen.remove(b)
if len(seen)==0:
ans+=1
print(ans)
```
| 3
|
|
1,005
|
B
|
Delete from the Left
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"strings"
] | null | null |
You are given two strings $s$ and $t$. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by $1$. You can't choose a string if it is empty.
For example:
- by applying a move to the string "where", the result is the string "here", - by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings $s$ and $t$ equal.
|
The first line of the input contains $s$. In the second line of the input contains $t$. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and $2\cdot10^5$, inclusive.
|
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
|
[
"test\nwest\n",
"codeforces\nyes\n",
"test\nyes\n",
"b\nab\n"
] |
[
"2\n",
"9\n",
"7\n",
"1\n"
] |
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" $8$ times. As a result, the string becomes "codeforces" $\to$ "es". The move should be applied to the string "yes" once. The result is the same string "yes" $\to$ "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
| 0
|
[
{
"input": "test\nwest",
"output": "2"
},
{
"input": "codeforces\nyes",
"output": "9"
},
{
"input": "test\nyes",
"output": "7"
},
{
"input": "b\nab",
"output": "1"
},
{
"input": "z\nz",
"output": "0"
},
{
"input": "abacabadabacaba\nabacabadacaba",
"output": "18"
},
{
"input": "z\na",
"output": "2"
},
{
"input": "ama\nsama",
"output": "1"
},
{
"input": "aaabc\nbc",
"output": "3"
},
{
"input": "lxuyspmieeuyxluv\ndfwxqvavzxctblaa",
"output": "32"
},
{
"input": "ahmedaly\naly",
"output": "5"
},
{
"input": "aa\na",
"output": "1"
},
{
"input": "aa\nba",
"output": "2"
},
{
"input": "cccba\nba",
"output": "3"
},
{
"input": "a\nab",
"output": "3"
},
{
"input": "dadc\ndddc",
"output": "4"
},
{
"input": "aacaaab\nb",
"output": "6"
},
{
"input": "dddc\ncdcde",
"output": "9"
},
{
"input": "bbdab\ndab",
"output": "2"
},
{
"input": "a\naa",
"output": "1"
},
{
"input": "aa\nab",
"output": "4"
},
{
"input": "codeforces\nces",
"output": "7"
}
] | 1,693,933,375
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 1
| 31
| 0
|
first_stroke = input()[::-1]
second_stroke = input()[::-1]
sums_stoke = sum([len(first_stroke), len(second_stroke)])
for i in range(max(len(first_stroke), len(second_stroke))):
if first_stroke[i] == second_stroke[i]:
if i == min(len(first_stroke), len(second_stroke)) - 1:
print(sums_stoke - (i + 1) * 2)
break
else: continue
else:
print(sums_stoke - i * 2)
# Tue Sep 05 2023 20:02:53 GMT+0300 (Moscow Standard Time)
|
Title: Delete from the Left
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two strings $s$ and $t$. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by $1$. You can't choose a string if it is empty.
For example:
- by applying a move to the string "where", the result is the string "here", - by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings $s$ and $t$ equal.
Input Specification:
The first line of the input contains $s$. In the second line of the input contains $t$. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and $2\cdot10^5$, inclusive.
Output Specification:
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Demo Input:
['test\nwest\n', 'codeforces\nyes\n', 'test\nyes\n', 'b\nab\n']
Demo Output:
['2\n', '9\n', '7\n', '1\n']
Note:
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" $8$ times. As a result, the string becomes "codeforces" $\to$ "es". The move should be applied to the string "yes" once. The result is the same string "yes" $\to$ "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted.
|
```python
first_stroke = input()[::-1]
second_stroke = input()[::-1]
sums_stoke = sum([len(first_stroke), len(second_stroke)])
for i in range(max(len(first_stroke), len(second_stroke))):
if first_stroke[i] == second_stroke[i]:
if i == min(len(first_stroke), len(second_stroke)) - 1:
print(sums_stoke - (i + 1) * 2)
break
else: continue
else:
print(sums_stoke - i * 2)
# Tue Sep 05 2023 20:02:53 GMT+0300 (Moscow Standard Time)
```
| -1
|
|
510
|
A
|
Fox And Snake
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
|
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
|
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
|
[
"3 3\n",
"3 4\n",
"5 3\n",
"9 9\n"
] |
[
"###\n..#\n###\n",
"####\n...#\n####\n",
"###\n..#\n###\n#..\n###\n",
"#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n"
] |
none
| 500
|
[
{
"input": "3 3",
"output": "###\n..#\n###"
},
{
"input": "3 4",
"output": "####\n...#\n####"
},
{
"input": "5 3",
"output": "###\n..#\n###\n#..\n###"
},
{
"input": "9 9",
"output": "#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########"
},
{
"input": "3 5",
"output": "#####\n....#\n#####"
},
{
"input": "3 6",
"output": "######\n.....#\n######"
},
{
"input": "7 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "7 4",
"output": "####\n...#\n####\n#...\n####\n...#\n####"
},
{
"input": "49 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "43 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "43 27",
"output": "###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n....................."
},
{
"input": "11 15",
"output": "###############\n..............#\n###############\n#..............\n###############\n..............#\n###############\n#..............\n###############\n..............#\n###############"
},
{
"input": "11 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "19 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "23 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "49 49",
"output": "#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#..."
},
{
"input": "33 43",
"output": "###########################################\n..........................................#\n###########################################\n#..........................................\n###########################################\n..........................................#\n###########################################\n#..........................................\n###########################################\n..........................................#\n###########################################\n#.................."
},
{
"input": "33 44",
"output": "############################################\n...........................................#\n############################################\n#...........................................\n############################################\n...........................................#\n############################################\n#...........................................\n############################################\n...........................................#\n############################################\n#......."
},
{
"input": "45 45",
"output": "#############################################\n............................................#\n#############################################\n#............................................\n#############################################\n............................................#\n#############################################\n#............................................\n#############################################\n............................................#\n#########################################..."
},
{
"input": "45 49",
"output": "#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#..."
},
{
"input": "45 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###"
},
{
"input": "45 4",
"output": "####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####"
},
{
"input": "5 4",
"output": "####\n...#\n####\n#...\n####"
},
{
"input": "5 5",
"output": "#####\n....#\n#####\n#....\n#####"
},
{
"input": "5 6",
"output": "######\n.....#\n######\n#.....\n######"
},
{
"input": "7 5",
"output": "#####\n....#\n#####\n#....\n#####\n....#\n#####"
},
{
"input": "7 6",
"output": "######\n.....#\n######\n#.....\n######\n.....#\n######"
},
{
"input": "7 7",
"output": "#######\n......#\n#######\n#......\n#######\n......#\n#######"
}
] | 1,696,872,419
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 46
| 102,400
|
n,m=map(int,input().split())
a=[]
for i in range(n):
b=[]
for j in range(m):
b.append('.')
a.append(b)
for i in range(m):
a[0][i]='#'
def right():
global row
a[row][m-1]='#'
row+=1
for i in range(m):
a[row][i]='#'
row+=1
def left():
global row
a[row][0]='#'
row+=1
for i in range(m):
a[row][i]='#'
row+=1
row=1
k=1
while row<n:
if k%2!=0:
right()
else:
left()
k+=1
for i in range(len(a)):
for j in range(len(a[i])):
print(a[i][j],end='')
print()
|
Title: Fox And Snake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
Input Specification:
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
Output Specification:
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
Demo Input:
['3 3\n', '3 4\n', '5 3\n', '9 9\n']
Demo Output:
['###\n..#\n###\n', '####\n...#\n####\n', '###\n..#\n###\n#..\n###\n', '#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n']
Note:
none
|
```python
n,m=map(int,input().split())
a=[]
for i in range(n):
b=[]
for j in range(m):
b.append('.')
a.append(b)
for i in range(m):
a[0][i]='#'
def right():
global row
a[row][m-1]='#'
row+=1
for i in range(m):
a[row][i]='#'
row+=1
def left():
global row
a[row][0]='#'
row+=1
for i in range(m):
a[row][i]='#'
row+=1
row=1
k=1
while row<n:
if k%2!=0:
right()
else:
left()
k+=1
for i in range(len(a)):
for j in range(len(a[i])):
print(a[i][j],end='')
print()
```
| 3
|
|
27
|
E
|
Number With The Given Amount Of Divisors
|
PROGRAMMING
| 2,000
|
[
"brute force",
"dp",
"number theory"
] |
E. Number With The Given Amount Of Divisors
|
2
|
256
|
Given the number *n*, find the smallest positive integer which has exactly *n* divisors. It is guaranteed that for the given *n* the answer will not exceed 1018.
|
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000).
|
Output the smallest positive integer with exactly *n* divisors.
|
[
"4\n",
"6\n"
] |
[
"6\n",
"12\n"
] |
none
| 2,500
|
[
{
"input": "1",
"output": "1"
},
{
"input": "7",
"output": "64"
},
{
"input": "8",
"output": "24"
},
{
"input": "9",
"output": "36"
},
{
"input": "10",
"output": "48"
},
{
"input": "15",
"output": "144"
},
{
"input": "20",
"output": "240"
},
{
"input": "47",
"output": "70368744177664"
},
{
"input": "59",
"output": "288230376151711744"
},
{
"input": "100",
"output": "45360"
},
{
"input": "159",
"output": "40532396646334464"
},
{
"input": "265",
"output": "364791569817010176"
},
{
"input": "312",
"output": "14192640"
},
{
"input": "473",
"output": "259700248434180096"
},
{
"input": "637",
"output": "46656000000"
},
{
"input": "500",
"output": "62370000"
},
{
"input": "720",
"output": "61261200"
},
{
"input": "902",
"output": "324625310542725120"
},
{
"input": "940",
"output": "199495389743677440"
},
{
"input": "1000",
"output": "810810000"
},
{
"input": "999",
"output": "757632231014400"
},
{
"input": "118",
"output": "864691128455135232"
}
] | 1,691,314,239
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include<bits/stdc++.h>
#define int long long
#define mod 1000000007
#define test freopen("1.txt","r",stdin);
#define endl "\n"
#define ff first
#define ss second
using namespace std;
int MAXV=2e18;
vector<int>primes={2,3,5,7,11,13,17,19,23,29,31,37,41};
int f(int num,int factors,int i,int n){
if(factors==n){
return num;
}
if(i==primes.size())return MAXV;
if(n<0)return MAXV;
if(num>MAXV)return MAXV;
int ret=MAXV;
for(int power=0;power<=63;power++){
if(factors*(power+1)>n or num>MAXV)break;
ret=min(ret,f(num*pow(primes[i],power),factors*(power+1),i+1,n));
}
return ret;
}
signed main(){
int n;cin>>n;
cout<<f(1,1,0,n);
return 0;
}
|
Title: Number With The Given Amount Of Divisors
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Given the number *n*, find the smallest positive integer which has exactly *n* divisors. It is guaranteed that for the given *n* the answer will not exceed 1018.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000).
Output Specification:
Output the smallest positive integer with exactly *n* divisors.
Demo Input:
['4\n', '6\n']
Demo Output:
['6\n', '12\n']
Note:
none
|
```python
#include<bits/stdc++.h>
#define int long long
#define mod 1000000007
#define test freopen("1.txt","r",stdin);
#define endl "\n"
#define ff first
#define ss second
using namespace std;
int MAXV=2e18;
vector<int>primes={2,3,5,7,11,13,17,19,23,29,31,37,41};
int f(int num,int factors,int i,int n){
if(factors==n){
return num;
}
if(i==primes.size())return MAXV;
if(n<0)return MAXV;
if(num>MAXV)return MAXV;
int ret=MAXV;
for(int power=0;power<=63;power++){
if(factors*(power+1)>n or num>MAXV)break;
ret=min(ret,f(num*pow(primes[i],power),factors*(power+1),i+1,n));
}
return ret;
}
signed main(){
int n;cin>>n;
cout<<f(1,1,0,n);
return 0;
}
```
| -1
|
474
|
A
|
Keyboard
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Our good friend Mole is trying to code a big message. He is typing on an unusual keyboard with characters arranged in following way:
Unfortunately Mole is blind, so sometimes it is problem for him to put his hands accurately. He accidentally moved both his hands with one position to the left or to the right. That means that now he presses not a button he wants, but one neighboring button (left or right, as specified in input).
We have a sequence of characters he has typed and we want to find the original message.
|
First line of the input contains one letter describing direction of shifting ('L' or 'R' respectively for left or right).
Second line contains a sequence of characters written by Mole. The size of this sequence will be no more than 100. Sequence contains only symbols that appear on Mole's keyboard. It doesn't contain spaces as there is no space on Mole's keyboard.
It is guaranteed that even though Mole hands are moved, he is still pressing buttons on keyboard and not hitting outside it.
|
Print a line that contains the original message.
|
[
"R\ns;;upimrrfod;pbr\n"
] |
[
"allyouneedislove\n"
] |
none
| 500
|
[
{
"input": "R\ns;;upimrrfod;pbr",
"output": "allyouneedislove"
},
{
"input": "R\nwertyuiop;lkjhgfdsxcvbnm,.",
"output": "qwertyuiolkjhgfdsazxcvbnm,"
},
{
"input": "L\nzxcvbnm,kjhgfdsaqwertyuio",
"output": "xcvbnm,.lkjhgfdswertyuiop"
},
{
"input": "R\nbubbuduppudup",
"output": "vyvvysyooysyo"
},
{
"input": "L\ngggggggggggggggggggggggggggggggggggggggggg",
"output": "hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh"
},
{
"input": "R\ngggggggggggggggggggggggggggggggggggggggggg",
"output": "ffffffffffffffffffffffffffffffffffffffffff"
},
{
"input": "L\nggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg",
"output": "hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh"
},
{
"input": "R\nggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg",
"output": "fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff"
},
{
"input": "L\nxgwurenkxkiau,c,vonei.zltazmnkhqtwuogkgvgckvja,z.rhanuy.ybebmzcfwozkwvuuiolaqlgvvvewnbuinrncgjwjdsfw",
"output": "cheitrmlclosi.v.bpmro/x;ysx,mljwyeiphlhbhvlbks.x/tjsmiu/unrn,xvgepxlebiiop;sw;hbbbremniomtmvhkekfdge"
},
{
"input": "L\nuoz.vmks,wxrb,nwcvdzh.m,hwsios.lvu,ktes,,ythddhm.sh,d,c,cfj.wqam,bowofbyx,jathqayhreqvixvbmgdokofmym",
"output": "ipx/b,ld.ectn.mevbfxj/,.jedopd/;bi.lyrd..uyjffj,/dj.f.v.vgk/ews,.npepgnuc.ksyjwsujtrwbocbn,hfplpg,u,"
},
{
"input": "R\noedjyrvuw/rn.v.hdwndbiposiewgsn.pnyf;/tsdohp,hrtd/mx,;coj./billd..mwbneohcikrdes/ucjr,wspthleyp,..f,",
"output": "iwshtecyq.eb,c,gsqbsvuoiauwqfab,obtdl.rasigomgers.nzmlxih,.vukks,,nqvbwigxujeswa.yxhemqaorgkwtom,,dm"
},
{
"input": "R\nvgj;o;ijrtfyck,dthccioltcx,crub;oceooognsuvfx/kgo.fbsudv,yod.erdrxhbeiyltxhnrobbb;ydrgroefcr/f;uvdjd",
"output": "cfhliluherdtxjmsrgxxuikrxzmxeyvlixwiiifbaycdz.jfi,dvayscmtis,wesezgvwutkrzgbeivvvltsefeiwdxe.dlycshs"
},
{
"input": "L\nqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq",
"output": "wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww"
},
{
"input": "L\noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo",
"output": "pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp"
},
{
"input": "L\n,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,",
"output": "...................................................................................................."
},
{
"input": "L\nzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
},
{
"input": "R\noooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo",
"output": "iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii"
},
{
"input": "R\nwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww",
"output": "qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq"
},
{
"input": "R\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
"output": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz"
},
{
"input": "L\nq",
"output": "w"
},
{
"input": "L\no",
"output": "p"
},
{
"input": "L\n,",
"output": "."
},
{
"input": "L\nz",
"output": "x"
},
{
"input": "R\n.",
"output": ","
},
{
"input": "R\no",
"output": "i"
},
{
"input": "R\nw",
"output": "q"
},
{
"input": "R\nx",
"output": "z"
},
{
"input": "R\n,./",
"output": "m,."
},
{
"input": "R\nwertuk;;/",
"output": "qweryjll."
},
{
"input": "L\n..",
"output": "//"
}
] | 1,680,886,088
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 31
| 77
| 0
|
l = "qwertyuiopasdfghjkl;zxcvbnm,./"
x = input()
s = input()
y = ""
for i in range(len(s)):
if x == 'L':
if s[i] not in ('p', ';', '/'):
y = y + l[l.index(s[i]) + 1]
else:
y = y + s[i]
else:
if s[i] not in ('q', 'a', 'z'):
y = y + l[l.index(s[i]) - 1]
else:
y = y + s[i]
print(y)
|
Title: Keyboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our good friend Mole is trying to code a big message. He is typing on an unusual keyboard with characters arranged in following way:
Unfortunately Mole is blind, so sometimes it is problem for him to put his hands accurately. He accidentally moved both his hands with one position to the left or to the right. That means that now he presses not a button he wants, but one neighboring button (left or right, as specified in input).
We have a sequence of characters he has typed and we want to find the original message.
Input Specification:
First line of the input contains one letter describing direction of shifting ('L' or 'R' respectively for left or right).
Second line contains a sequence of characters written by Mole. The size of this sequence will be no more than 100. Sequence contains only symbols that appear on Mole's keyboard. It doesn't contain spaces as there is no space on Mole's keyboard.
It is guaranteed that even though Mole hands are moved, he is still pressing buttons on keyboard and not hitting outside it.
Output Specification:
Print a line that contains the original message.
Demo Input:
['R\ns;;upimrrfod;pbr\n']
Demo Output:
['allyouneedislove\n']
Note:
none
|
```python
l = "qwertyuiopasdfghjkl;zxcvbnm,./"
x = input()
s = input()
y = ""
for i in range(len(s)):
if x == 'L':
if s[i] not in ('p', ';', '/'):
y = y + l[l.index(s[i]) + 1]
else:
y = y + s[i]
else:
if s[i] not in ('q', 'a', 'z'):
y = y + l[l.index(s[i]) - 1]
else:
y = y + s[i]
print(y)
```
| 3
|
|
228
|
A
|
Is your horseshoe on the other hoof?
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
|
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
|
Print a single integer — the minimum number of horseshoes Valera needs to buy.
|
[
"1 7 3 3\n",
"7 7 7 7\n"
] |
[
"1\n",
"3\n"
] |
none
| 500
|
[
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 586975278",
"output": "1"
},
{
"input": "259504825 377489979 588153796 377489979",
"output": "1"
},
{
"input": "652588203 931100304 931100304 652588203",
"output": "2"
},
{
"input": "391958720 651507265 391958720 651507265",
"output": "2"
},
{
"input": "90793237 90793237 90793237 90793237",
"output": "3"
},
{
"input": "551651653 551651653 551651653 551651653",
"output": "3"
},
{
"input": "156630260 609654355 668943582 973622757",
"output": "0"
},
{
"input": "17061017 110313588 434481173 796661222",
"output": "0"
},
{
"input": "24975422 256716298 337790533 690960249",
"output": "0"
},
{
"input": "255635360 732742923 798648949 883146723",
"output": "0"
},
{
"input": "133315691 265159773 734556507 265159773",
"output": "1"
},
{
"input": "28442865 741657755 978106882 978106882",
"output": "1"
},
{
"input": "131245479 174845575 497483467 131245479",
"output": "1"
},
{
"input": "139159884 616215581 958341883 616215581",
"output": "1"
},
{
"input": "147784432 947653080 947653080 947653080",
"output": "2"
},
{
"input": "94055790 756126496 756126496 94055790",
"output": "2"
},
{
"input": "240458500 511952208 240458500 511952208",
"output": "2"
},
{
"input": "681828506 972810624 972810624 681828506",
"output": "2"
},
{
"input": "454961014 454961014 454961014 454961014",
"output": "3"
},
{
"input": "915819430 915819430 915819430 915819430",
"output": "3"
},
{
"input": "671645142 671645142 671645142 671645142",
"output": "3"
},
{
"input": "132503558 132503558 132503558 132503558",
"output": "3"
},
{
"input": "5 5 999999 6",
"output": "1"
},
{
"input": "1 1 2 5",
"output": "1"
},
{
"input": "2 1 2 3",
"output": "1"
},
{
"input": "1 1 3 5",
"output": "1"
},
{
"input": "1 1 3 3",
"output": "2"
},
{
"input": "2 2 2 1",
"output": "2"
},
{
"input": "3 1 1 1",
"output": "2"
},
{
"input": "1 2 2 2",
"output": "2"
}
] | 1,685,195,036
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 4
| 92
| 0
|
def horseshoe(data):
y = data.split()
count = 0
for i in range(len(y) - 1):
if y[i + 1] == y[i]:
count += 1
return count
print(horseshoe(input()))
|
Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none
|
```python
def horseshoe(data):
y = data.split()
count = 0
for i in range(len(y) - 1):
if y[i + 1] == y[i]:
count += 1
return count
print(horseshoe(input()))
```
| 0
|
|
336
|
A
|
Vasily the Bear and Triangle
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Vasily the bear has a favorite rectangle, it has one vertex at point (0,<=0), and the opposite vertex at point (*x*,<=*y*). Of course, the sides of Vasya's favorite rectangle are parallel to the coordinate axes.
Vasya also loves triangles, if the triangles have one vertex at point *B*<==<=(0,<=0). That's why today he asks you to find two points *A*<==<=(*x*1,<=*y*1) and *C*<==<=(*x*2,<=*y*2), such that the following conditions hold:
- the coordinates of points: *x*1, *x*2, *y*1, *y*2 are integers. Besides, the following inequation holds: *x*1<=<<=*x*2; - the triangle formed by point *A*, *B* and *C* is rectangular and isosceles ( is right); - all points of the favorite rectangle are located inside or on the border of triangle *ABC*; - the area of triangle *ABC* is as small as possible.
Help the bear, find the required points. It is not so hard to proof that these points are unique.
|
The first line contains two integers *x*,<=*y* (<=-<=109<=≤<=*x*,<=*y*<=≤<=109,<=*x*<=≠<=0,<=*y*<=≠<=0).
|
Print in the single line four integers *x*1,<=*y*1,<=*x*2,<=*y*2 — the coordinates of the required points.
|
[
"10 5\n",
"-10 5\n"
] |
[
"0 15 15 0\n",
"-15 0 0 15\n"
] |
<img class="tex-graphics" src="https://espresso.codeforces.com/a9ea2088c4294ce8f23801562fda36b830df2c3f.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Figure to the first sample
| 500
|
[
{
"input": "10 5",
"output": "0 15 15 0"
},
{
"input": "-10 5",
"output": "-15 0 0 15"
},
{
"input": "20 -10",
"output": "0 -30 30 0"
},
{
"input": "-10 -1000000000",
"output": "-1000000010 0 0 -1000000010"
},
{
"input": "-1000000000 -1000000000",
"output": "-2000000000 0 0 -2000000000"
},
{
"input": "1000000000 1000000000",
"output": "0 2000000000 2000000000 0"
},
{
"input": "-123131 3123141",
"output": "-3246272 0 0 3246272"
},
{
"input": "-23423 -243242423",
"output": "-243265846 0 0 -243265846"
},
{
"input": "123112 4560954",
"output": "0 4684066 4684066 0"
},
{
"input": "1321 -23131",
"output": "0 -24452 24452 0"
},
{
"input": "1000000000 999999999",
"output": "0 1999999999 1999999999 0"
},
{
"input": "54543 432423",
"output": "0 486966 486966 0"
},
{
"input": "1 1",
"output": "0 2 2 0"
},
{
"input": "-1 -1",
"output": "-2 0 0 -2"
},
{
"input": "-1 1",
"output": "-2 0 0 2"
},
{
"input": "1 -1",
"output": "0 -2 2 0"
},
{
"input": "42 -2",
"output": "0 -44 44 0"
},
{
"input": "2 -435",
"output": "0 -437 437 0"
},
{
"input": "76 -76",
"output": "0 -152 152 0"
},
{
"input": "1000000000 1",
"output": "0 1000000001 1000000001 0"
},
{
"input": "1000000000 -1",
"output": "0 -1000000001 1000000001 0"
},
{
"input": "-1000000000 1",
"output": "-1000000001 0 0 1000000001"
},
{
"input": "-1000000000 -1",
"output": "-1000000001 0 0 -1000000001"
},
{
"input": "1000000000 -999999999",
"output": "0 -1999999999 1999999999 0"
},
{
"input": "-1000000000 999999999",
"output": "-1999999999 0 0 1999999999"
},
{
"input": "-1000000000 -999999999",
"output": "-1999999999 0 0 -1999999999"
},
{
"input": "999999999 1000000000",
"output": "0 1999999999 1999999999 0"
},
{
"input": "-999999999 1000000000",
"output": "-1999999999 0 0 1999999999"
},
{
"input": "999999999 -1000000000",
"output": "0 -1999999999 1999999999 0"
},
{
"input": "-999999999 -1000000000",
"output": "-1999999999 0 0 -1999999999"
}
] | 1,593,990,750
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 248
| 20,172,800
|
from sys import stdin, stdout
def stdinhack():
for line in stdin: yield line
next_line = lambda: next(stdinhack()).split()
write = lambda x : stdout.write(x)
def solve ():
x, y = map(int, next_line())
if x>0 and y>0: print(0, x+y, x+y,0)
if x<0 and y>0: print(x-y, 0, 0, -x+y)
if x>0 and y<0: print(-y+x, 0, 0, y-x)
if x<0 and y<0: print(0, x+y, x+y,0)
return
if __name__ == '__main__':
solve()
|
Title: Vasily the Bear and Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the bear has a favorite rectangle, it has one vertex at point (0,<=0), and the opposite vertex at point (*x*,<=*y*). Of course, the sides of Vasya's favorite rectangle are parallel to the coordinate axes.
Vasya also loves triangles, if the triangles have one vertex at point *B*<==<=(0,<=0). That's why today he asks you to find two points *A*<==<=(*x*1,<=*y*1) and *C*<==<=(*x*2,<=*y*2), such that the following conditions hold:
- the coordinates of points: *x*1, *x*2, *y*1, *y*2 are integers. Besides, the following inequation holds: *x*1<=<<=*x*2; - the triangle formed by point *A*, *B* and *C* is rectangular and isosceles ( is right); - all points of the favorite rectangle are located inside or on the border of triangle *ABC*; - the area of triangle *ABC* is as small as possible.
Help the bear, find the required points. It is not so hard to proof that these points are unique.
Input Specification:
The first line contains two integers *x*,<=*y* (<=-<=109<=≤<=*x*,<=*y*<=≤<=109,<=*x*<=≠<=0,<=*y*<=≠<=0).
Output Specification:
Print in the single line four integers *x*1,<=*y*1,<=*x*2,<=*y*2 — the coordinates of the required points.
Demo Input:
['10 5\n', '-10 5\n']
Demo Output:
['0 15 15 0\n', '-15 0 0 15\n']
Note:
<img class="tex-graphics" src="https://espresso.codeforces.com/a9ea2088c4294ce8f23801562fda36b830df2c3f.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Figure to the first sample
|
```python
from sys import stdin, stdout
def stdinhack():
for line in stdin: yield line
next_line = lambda: next(stdinhack()).split()
write = lambda x : stdout.write(x)
def solve ():
x, y = map(int, next_line())
if x>0 and y>0: print(0, x+y, x+y,0)
if x<0 and y>0: print(x-y, 0, 0, -x+y)
if x>0 and y<0: print(-y+x, 0, 0, y-x)
if x<0 and y<0: print(0, x+y, x+y,0)
return
if __name__ == '__main__':
solve()
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
An atom of element X can exist in *n* distinct states with energies *E*1<=<<=*E*2<=<<=...<=<<=*E**n*. Arkady wants to build a laser on this element, using a three-level scheme. Here is a simplified description of the scheme.
Three distinct states *i*, *j* and *k* are selected, where *i*<=<<=*j*<=<<=*k*. After that the following process happens:
1. initially the atom is in the state *i*,1. we spend *E**k*<=-<=*E**i* energy to put the atom in the state *k*,1. the atom emits a photon with useful energy *E**k*<=-<=*E**j* and changes its state to the state *j*,1. the atom spontaneously changes its state to the state *i*, losing energy *E**j*<=-<=*E**i*,1. the process repeats from step 1.
Let's define the energy conversion efficiency as , i. e. the ration between the useful energy of the photon and spent energy.
Due to some limitations, Arkady can only choose such three states that *E**k*<=-<=*E**i*<=≤<=*U*.
Help Arkady to find such the maximum possible energy conversion efficiency within the above constraints.
|
The first line contains two integers *n* and *U* (3<=≤<=*n*<=≤<=105, 1<=≤<=*U*<=≤<=109) — the number of states and the maximum possible difference between *E**k* and *E**i*.
The second line contains a sequence of integers *E*1,<=*E*2,<=...,<=*E**n* (1<=≤<=*E*1<=<<=*E*2...<=<<=*E**n*<=≤<=109). It is guaranteed that all *E**i* are given in increasing order.
|
If it is not possible to choose three states that satisfy all constraints, print -1.
Otherwise, print one real number η — the maximum possible energy conversion efficiency. Your answer is considered correct its absolute or relative error does not exceed 10<=-<=9.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .
|
[
"4 4\n1 3 5 7\n",
"10 8\n10 13 15 16 17 19 20 22 24 25\n",
"3 1\n2 5 10\n"
] |
[
"0.5\n",
"0.875\n",
"-1\n"
] |
In the first example choose states 1, 2 and 3, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/147ae7a830722917b0aa37d064df8eb74cfefb97.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second example choose states 4, 5 and 9, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f68f268de4eb2242167e6ec64e6b8aa60a5703ae.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 0
|
[
{
"input": "4 4\n1 3 5 7",
"output": "0.5"
},
{
"input": "10 8\n10 13 15 16 17 19 20 22 24 25",
"output": "0.875"
},
{
"input": "3 1\n2 5 10",
"output": "-1"
},
{
"input": "5 3\n4 6 8 9 10",
"output": "0.5"
},
{
"input": "10 128\n110 121 140 158 174 188 251 271 272 277",
"output": "0.86554621848739499157"
},
{
"input": "20 17\n104 107 121 131 138 140 143 144 178 192 193 198 201 206 238 242 245 248 255 265",
"output": "0.92857142857142860315"
},
{
"input": "30 23\n102 104 105 107 108 109 110 111 116 118 119 122 127 139 140 142 145 157 166 171 173 174 175 181 187 190 191 193 195 196",
"output": "0.95652173913043481157"
},
{
"input": "50 64\n257 258 350 375 1014 1017 1051 1097 1169 1177 1223 1836 1942 1983 2111 2131 2341 2418 2593 2902 2948 3157 3243 3523 3566 4079 4499 4754 5060 5624 6279 6976 7011 7071 7278 7366 7408 7466 7526 7837 7934 8532 8577 8680 9221 9271 9327 9411 9590 9794",
"output": "0.91891891891891896993"
},
{
"input": "5 2\n4 6 8 9 10",
"output": "0.5"
},
{
"input": "10 2\n110 121 140 158 174 188 251 271 272 277",
"output": "-1"
},
{
"input": "30 5\n102 104 105 107 108 109 110 111 116 118 119 122 127 139 140 142 145 157 166 171 173 174 175 181 187 190 191 193 195 196",
"output": "0.80000000000000004441"
},
{
"input": "10 6\n110 121 140 158 174 188 251 271 272 277",
"output": "0.83333333333333337034"
},
{
"input": "20 4\n104 107 121 131 138 140 143 144 178 192 193 198 201 206 238 242 245 248 255 265",
"output": "0.25"
},
{
"input": "3 1000000000\n1 2 1000000000",
"output": "0.99999999900000002828"
},
{
"input": "3 1\n1 2 3",
"output": "-1"
},
{
"input": "5 1000000000\n1 2 3 999999999 1000000000",
"output": "0.99999999900000002828"
},
{
"input": "10 199\n1 3 190 191 193 195 196 197 199 200",
"output": "0.98994974874371854945"
},
{
"input": "10 300\n80 100 103 140 146 159 392 393 396 398",
"output": "0.98993288590604022747"
},
{
"input": "10 92\n44 119 252 281 303 323 351 363 377 392",
"output": "0.77528089887640450062"
},
{
"input": "4 2\n1 3 5 7",
"output": "-1"
},
{
"input": "8 2\n1 3 7 9 15 17 23 25",
"output": "-1"
},
{
"input": "3 5\n1 2 10",
"output": "-1"
},
{
"input": "4 7\n1 5 8 9",
"output": "0.42857142857142854764"
}
] | 1,553,631,047
| 3,347
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 21
| 187
| 11,264,000
|
n,u=map(int,input().split())
arr=list(map(int,input().split()))
arr.sort()
j,i=1,0
maxi=-1
while(i<n-1):
while(1):
if j>=n or arr[j]-arr[i]>u:
j-=1
break
j+=1
if arr[j]-arr[i]<=u and j!=i:
maxi=max(maxi,(arr[j]-arr[i+1])/(arr[j]-arr[i]))
if i==j:
j+=1
i+=1
print(maxi)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An atom of element X can exist in *n* distinct states with energies *E*1<=<<=*E*2<=<<=...<=<<=*E**n*. Arkady wants to build a laser on this element, using a three-level scheme. Here is a simplified description of the scheme.
Three distinct states *i*, *j* and *k* are selected, where *i*<=<<=*j*<=<<=*k*. After that the following process happens:
1. initially the atom is in the state *i*,1. we spend *E**k*<=-<=*E**i* energy to put the atom in the state *k*,1. the atom emits a photon with useful energy *E**k*<=-<=*E**j* and changes its state to the state *j*,1. the atom spontaneously changes its state to the state *i*, losing energy *E**j*<=-<=*E**i*,1. the process repeats from step 1.
Let's define the energy conversion efficiency as , i. e. the ration between the useful energy of the photon and spent energy.
Due to some limitations, Arkady can only choose such three states that *E**k*<=-<=*E**i*<=≤<=*U*.
Help Arkady to find such the maximum possible energy conversion efficiency within the above constraints.
Input Specification:
The first line contains two integers *n* and *U* (3<=≤<=*n*<=≤<=105, 1<=≤<=*U*<=≤<=109) — the number of states and the maximum possible difference between *E**k* and *E**i*.
The second line contains a sequence of integers *E*1,<=*E*2,<=...,<=*E**n* (1<=≤<=*E*1<=<<=*E*2...<=<<=*E**n*<=≤<=109). It is guaranteed that all *E**i* are given in increasing order.
Output Specification:
If it is not possible to choose three states that satisfy all constraints, print -1.
Otherwise, print one real number η — the maximum possible energy conversion efficiency. Your answer is considered correct its absolute or relative error does not exceed 10<=-<=9.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .
Demo Input:
['4 4\n1 3 5 7\n', '10 8\n10 13 15 16 17 19 20 22 24 25\n', '3 1\n2 5 10\n']
Demo Output:
['0.5\n', '0.875\n', '-1\n']
Note:
In the first example choose states 1, 2 and 3, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/147ae7a830722917b0aa37d064df8eb74cfefb97.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second example choose states 4, 5 and 9, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f68f268de4eb2242167e6ec64e6b8aa60a5703ae.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
n,u=map(int,input().split())
arr=list(map(int,input().split()))
arr.sort()
j,i=1,0
maxi=-1
while(i<n-1):
while(1):
if j>=n or arr[j]-arr[i]>u:
j-=1
break
j+=1
if arr[j]-arr[i]<=u and j!=i:
maxi=max(maxi,(arr[j]-arr[i+1])/(arr[j]-arr[i]))
if i==j:
j+=1
i+=1
print(maxi)
```
| 0
|
|
845
|
B
|
Luba And The Ticket
|
PROGRAMMING
| 1,600
|
[
"brute force",
"greedy",
"implementation"
] | null | null |
Luba has a ticket consisting of 6 digits. In one move she can choose digit in any position and replace it with arbitrary digit. She wants to know the minimum number of digits she needs to replace in order to make the ticket lucky.
The ticket is considered lucky if the sum of first three digits equals to the sum of last three digits.
|
You are given a string consisting of 6 characters (all characters are digits from 0 to 9) — this string denotes Luba's ticket. The ticket can start with the digit 0.
|
Print one number — the minimum possible number of digits Luba needs to replace to make the ticket lucky.
|
[
"000000\n",
"123456\n",
"111000\n"
] |
[
"0\n",
"2\n",
"1\n"
] |
In the first example the ticket is already lucky, so the answer is 0.
In the second example Luba can replace 4 and 5 with zeroes, and the ticket will become lucky. It's easy to see that at least two replacements are required.
In the third example Luba can replace any zero with 3. It's easy to see that at least one replacement is required.
| 0
|
[
{
"input": "000000",
"output": "0"
},
{
"input": "123456",
"output": "2"
},
{
"input": "111000",
"output": "1"
},
{
"input": "120111",
"output": "0"
},
{
"input": "999999",
"output": "0"
},
{
"input": "199880",
"output": "1"
},
{
"input": "899889",
"output": "1"
},
{
"input": "899888",
"output": "1"
},
{
"input": "505777",
"output": "2"
},
{
"input": "999000",
"output": "3"
},
{
"input": "989010",
"output": "3"
},
{
"input": "651894",
"output": "1"
},
{
"input": "858022",
"output": "2"
},
{
"input": "103452",
"output": "1"
},
{
"input": "999801",
"output": "2"
},
{
"input": "999990",
"output": "1"
},
{
"input": "697742",
"output": "1"
},
{
"input": "242367",
"output": "2"
},
{
"input": "099999",
"output": "1"
},
{
"input": "198999",
"output": "1"
},
{
"input": "023680",
"output": "1"
},
{
"input": "999911",
"output": "2"
},
{
"input": "000990",
"output": "2"
},
{
"input": "117099",
"output": "1"
},
{
"input": "990999",
"output": "1"
},
{
"input": "000111",
"output": "1"
},
{
"input": "000444",
"output": "2"
},
{
"input": "202597",
"output": "2"
},
{
"input": "000333",
"output": "1"
},
{
"input": "030039",
"output": "1"
},
{
"input": "000009",
"output": "1"
},
{
"input": "006456",
"output": "1"
},
{
"input": "022995",
"output": "3"
},
{
"input": "999198",
"output": "1"
},
{
"input": "223456",
"output": "2"
},
{
"input": "333665",
"output": "2"
},
{
"input": "123986",
"output": "2"
},
{
"input": "599257",
"output": "1"
},
{
"input": "101488",
"output": "3"
},
{
"input": "111399",
"output": "2"
},
{
"input": "369009",
"output": "1"
},
{
"input": "024887",
"output": "2"
},
{
"input": "314347",
"output": "1"
},
{
"input": "145892",
"output": "1"
},
{
"input": "321933",
"output": "1"
},
{
"input": "100172",
"output": "1"
},
{
"input": "222455",
"output": "2"
},
{
"input": "317596",
"output": "1"
},
{
"input": "979245",
"output": "2"
},
{
"input": "000018",
"output": "1"
},
{
"input": "101389",
"output": "2"
},
{
"input": "123985",
"output": "2"
},
{
"input": "900000",
"output": "1"
},
{
"input": "132069",
"output": "1"
},
{
"input": "949256",
"output": "1"
},
{
"input": "123996",
"output": "2"
},
{
"input": "034988",
"output": "2"
},
{
"input": "320869",
"output": "2"
},
{
"input": "089753",
"output": "1"
},
{
"input": "335667",
"output": "2"
},
{
"input": "868580",
"output": "1"
},
{
"input": "958031",
"output": "2"
},
{
"input": "117999",
"output": "2"
},
{
"input": "000001",
"output": "1"
},
{
"input": "213986",
"output": "2"
},
{
"input": "123987",
"output": "3"
},
{
"input": "111993",
"output": "2"
},
{
"input": "642479",
"output": "1"
},
{
"input": "033788",
"output": "2"
},
{
"input": "766100",
"output": "2"
},
{
"input": "012561",
"output": "1"
},
{
"input": "111695",
"output": "2"
},
{
"input": "123689",
"output": "2"
},
{
"input": "944234",
"output": "1"
},
{
"input": "154999",
"output": "2"
},
{
"input": "333945",
"output": "1"
},
{
"input": "371130",
"output": "1"
},
{
"input": "977330",
"output": "2"
},
{
"input": "777544",
"output": "2"
},
{
"input": "111965",
"output": "2"
},
{
"input": "988430",
"output": "2"
},
{
"input": "123789",
"output": "3"
},
{
"input": "111956",
"output": "2"
},
{
"input": "444776",
"output": "2"
},
{
"input": "001019",
"output": "1"
},
{
"input": "011299",
"output": "2"
},
{
"input": "011389",
"output": "2"
},
{
"input": "999333",
"output": "2"
},
{
"input": "126999",
"output": "2"
},
{
"input": "744438",
"output": "0"
},
{
"input": "588121",
"output": "3"
},
{
"input": "698213",
"output": "2"
},
{
"input": "652858",
"output": "1"
},
{
"input": "989304",
"output": "3"
},
{
"input": "888213",
"output": "3"
},
{
"input": "969503",
"output": "2"
},
{
"input": "988034",
"output": "2"
},
{
"input": "889444",
"output": "2"
},
{
"input": "990900",
"output": "1"
},
{
"input": "301679",
"output": "2"
},
{
"input": "434946",
"output": "1"
},
{
"input": "191578",
"output": "2"
},
{
"input": "118000",
"output": "2"
},
{
"input": "636915",
"output": "0"
},
{
"input": "811010",
"output": "1"
},
{
"input": "822569",
"output": "1"
},
{
"input": "122669",
"output": "2"
},
{
"input": "010339",
"output": "2"
},
{
"input": "213698",
"output": "2"
},
{
"input": "895130",
"output": "2"
},
{
"input": "000900",
"output": "1"
},
{
"input": "191000",
"output": "2"
},
{
"input": "001000",
"output": "1"
},
{
"input": "080189",
"output": "2"
},
{
"input": "990000",
"output": "2"
},
{
"input": "201984",
"output": "2"
},
{
"input": "002667",
"output": "2"
},
{
"input": "877542",
"output": "2"
},
{
"input": "301697",
"output": "2"
},
{
"input": "211597",
"output": "2"
},
{
"input": "420337",
"output": "1"
},
{
"input": "024768",
"output": "2"
},
{
"input": "878033",
"output": "2"
},
{
"input": "788024",
"output": "2"
},
{
"input": "023869",
"output": "2"
},
{
"input": "466341",
"output": "1"
},
{
"input": "696327",
"output": "1"
},
{
"input": "779114",
"output": "2"
},
{
"input": "858643",
"output": "1"
},
{
"input": "011488",
"output": "3"
},
{
"input": "003669",
"output": "2"
},
{
"input": "202877",
"output": "3"
},
{
"input": "738000",
"output": "2"
},
{
"input": "567235",
"output": "2"
},
{
"input": "887321",
"output": "3"
},
{
"input": "401779",
"output": "2"
},
{
"input": "989473",
"output": "2"
},
{
"input": "004977",
"output": "3"
},
{
"input": "023778",
"output": "2"
},
{
"input": "809116",
"output": "1"
},
{
"input": "042762",
"output": "1"
},
{
"input": "777445",
"output": "2"
},
{
"input": "769302",
"output": "2"
},
{
"input": "023977",
"output": "2"
},
{
"input": "990131",
"output": "2"
}
] | 1,503,373,427
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 46
| 0
|
wow = [0]*6
j = 0
rub = input()
for i in rub:
wow[j] = int(i)
j += 1
a1 = [wow[0],wow[1],wow[2]]
a2 = [wow[3],wow[4],wow[5]]
sum1 = sum(a1)
sum2 = sum(a2)
if(sum1 < sum2):
noisud = min(a1)
maksud = max(a2)
pontang = sum2-sum1
if(sum2 < sum1):
noisud = min(a2)
maksud = max(a1)
pontang = sum1-sum2
if(sum1 == sum2):
print(0)
noisud = 9-noisud
mx1 = max(noisud,maksud)
mx2 = max(noisud,maksud)
if(mx1>=pontang):
print(1)
elif(mx1+mx2>=pontang):
print(2)
else:
print(3)
|
Title: Luba And The Ticket
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luba has a ticket consisting of 6 digits. In one move she can choose digit in any position and replace it with arbitrary digit. She wants to know the minimum number of digits she needs to replace in order to make the ticket lucky.
The ticket is considered lucky if the sum of first three digits equals to the sum of last three digits.
Input Specification:
You are given a string consisting of 6 characters (all characters are digits from 0 to 9) — this string denotes Luba's ticket. The ticket can start with the digit 0.
Output Specification:
Print one number — the minimum possible number of digits Luba needs to replace to make the ticket lucky.
Demo Input:
['000000\n', '123456\n', '111000\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first example the ticket is already lucky, so the answer is 0.
In the second example Luba can replace 4 and 5 with zeroes, and the ticket will become lucky. It's easy to see that at least two replacements are required.
In the third example Luba can replace any zero with 3. It's easy to see that at least one replacement is required.
|
```python
wow = [0]*6
j = 0
rub = input()
for i in rub:
wow[j] = int(i)
j += 1
a1 = [wow[0],wow[1],wow[2]]
a2 = [wow[3],wow[4],wow[5]]
sum1 = sum(a1)
sum2 = sum(a2)
if(sum1 < sum2):
noisud = min(a1)
maksud = max(a2)
pontang = sum2-sum1
if(sum2 < sum1):
noisud = min(a2)
maksud = max(a1)
pontang = sum1-sum2
if(sum1 == sum2):
print(0)
noisud = 9-noisud
mx1 = max(noisud,maksud)
mx2 = max(noisud,maksud)
if(mx1>=pontang):
print(1)
elif(mx1+mx2>=pontang):
print(2)
else:
print(3)
```
| -1
|
|
448
|
B
|
Suffix Structures
|
PROGRAMMING
| 1,400
|
[
"implementation",
"strings"
] | null | null |
Bizon the Champion isn't just a bison. He also is a favorite of the "Bizons" team.
At a competition the "Bizons" got the following problem: "You are given two distinct words (strings of English letters), *s* and *t*. You need to transform word *s* into word *t*". The task looked simple to the guys because they know the suffix data structures well. Bizon Senior loves suffix automaton. By applying it once to a string, he can remove from this string any single character. Bizon Middle knows suffix array well. By applying it once to a string, he can swap any two characters of this string. The guys do not know anything about the suffix tree, but it can help them do much more.
Bizon the Champion wonders whether the "Bizons" can solve the problem. Perhaps, the solution do not require both data structures. Find out whether the guys can solve the problem and if they can, how do they do it? Can they solve it either only with use of suffix automaton or only with use of suffix array or they need both structures? Note that any structure may be used an unlimited number of times, the structures may be used in any order.
|
The first line contains a non-empty word *s*. The second line contains a non-empty word *t*. Words *s* and *t* are different. Each word consists only of lowercase English letters. Each word contains at most 100 letters.
|
In the single line print the answer to the problem. Print "need tree" (without the quotes) if word *s* cannot be transformed into word *t* even with use of both suffix array and suffix automaton. Print "automaton" (without the quotes) if you need only the suffix automaton to solve the problem. Print "array" (without the quotes) if you need only the suffix array to solve the problem. Print "both" (without the quotes), if you need both data structures to solve the problem.
It's guaranteed that if you can solve the problem only with use of suffix array, then it is impossible to solve it only with use of suffix automaton. This is also true for suffix automaton.
|
[
"automaton\ntomat\n",
"array\narary\n",
"both\nhot\n",
"need\ntree\n"
] |
[
"automaton\n",
"array\n",
"both\n",
"need tree\n"
] |
In the third sample you can act like that: first transform "both" into "oth" by removing the first character using the suffix automaton and then make two swaps of the string using the suffix array and get "hot".
| 1,000
|
[
{
"input": "automaton\ntomat",
"output": "automaton"
},
{
"input": "array\narary",
"output": "array"
},
{
"input": "both\nhot",
"output": "both"
},
{
"input": "need\ntree",
"output": "need tree"
},
{
"input": "abacaba\naaaa",
"output": "automaton"
},
{
"input": "z\nzz",
"output": "need tree"
},
{
"input": "itwtyhhsdjjffmmoqkkhxjouypznewstyorotxhozlytndehmaxogrohccnqcgkrjrdmnuaogiwmnmsbdaizqkxnkqxxiihbwepc\nsnixfywvcntitcefsgqxjcodwtumurcglfmnamnowzbjzmfzspbfuldraiepeeiyasmrsneekydsbvazoqszyjxkjiotushsddet",
"output": "need tree"
},
{
"input": "y\nu",
"output": "need tree"
},
{
"input": "nbjigpsbammkuuqrxfnmhtimwpflrflehffykbylmnxgadldchdbqklqbremcmzlpxieozgpfgrhegmdcxxfyehzzelcwgkierrj\nbjbakuqrnhimwhffykylmngadhbqkqbrcziefredxxezcgkerj",
"output": "automaton"
},
{
"input": "gzvvawianfysfuxhruarhverinqsbrfxvkcsermuzowahevgskmpvfdljtcztnbkzftfhvnarvkfkqjgrzbrcfthqmspvpqcva\nwnm",
"output": "automaton"
},
{
"input": "dvzohfzgzdjavqwhjcrdphpdqjwtqijabbrhformstqaonlhbglmxugkwviigqaohwvqfhdwwcvdkjrcgxblhvtashhcxssbvpo\nzgvqhpjhforlugkwfwrchvhp",
"output": "automaton"
},
{
"input": "wkfoyetcjivofxaktmauapzeuhcpzjloszzxwydgavebgniiuzrscytsokjkjfkpylvxtlqlquzduywbhqdzmtwprfdohmwgmysy\ny",
"output": "automaton"
},
{
"input": "npeidcoiulxdxzjozsonkdwnoazsbntfclnpubgweaynuhfmrtybqtkuihxxfhwlnquslnhzvqznyofzcbdewnrisqzdhsiyhkxf\nnpeidcoiulxdxzjozsonkdwnoazsbntfclnpubgeaynuhfmrtybqtkuihxxfhwlnquslnhzvqznyofzcbdewnrisqzdhsiyhkxf",
"output": "automaton"
},
{
"input": "gahcqpgmypeahjcwkzahnhmsmxosnikucqwyzklbfwtujjlzvwklqzxakcrcqalhsvsgvknpxsoqkjnyjkypfsiogbcaxjyugeet\ngahcqpgmypeahjwwkzahnhmsmxopnikucacyzklbfwtujjlzvwkoqzxakcrcqqlhsvsgvknpxslgkjnyjkysfoisqbcaxjyuteeg",
"output": "array"
},
{
"input": "vwesbxsifsjqapwridrenumrukgemlldpbtdhxivsrmzbgprtkqgaryniudkjgpjndluwxuohwwysmyuxyrulwsodgunzirudgtx\nugeabdszfshqsksddireguvsukieqlluhngdpxjvwwnzdrtrtrdjiuxgadtgjpxrmlynspyyryngxuiibrmurwpmoxwwuklbwumo",
"output": "array"
},
{
"input": "kjnohlseyntrslfssrshjxclzlsbkfzfwwwgyxsysvmfkxugdwjodfyxhdsveruoioutwmtcbaljomaorvzjsbmglqckmsyieeiu\netihhycsjgdysowuljmaoksoecxawsgsljofkrjftuweidrkwtymyswdlilsozsxevfbformnbsumlxzqzykjvsnrlxufvgbmshc",
"output": "array"
},
{
"input": "ezbpsylkfztypqrefinexshtgglmkoinrktkloitqhfkivoabrfrivvqrcxkjckzvcozpchhiodrbbxuhnwcjigftnrjfiqyxakh\niacxghqffzdbsiqunhxbiooqvfohzticjpvrzykcrlrxklgknyrkrhjxcetmfocierekatfvkbslkkrbhftwngoijpipvqyznthi",
"output": "array"
},
{
"input": "smywwqeolrsytkthfgacnbufzaulgszikbhluzcdbafjclkqueepxbhoamrwswxherzhhuqqcttokbljfbppdinzqgdupkfevmke\nsmywwqeolrsytkthfgacnbufzaulgszikbhluzcdbafjclkqueepxbhoamrwswxherzhhufqcttokbljfbppdinzqgdupkqevmke",
"output": "array"
},
{
"input": "hxsvvydmzhxrswvhkvrbjrfqkazbkjabnrdghposgyfeslzumaovfkallszzumztftgpcilwfrzpvhhbgdzdvnmseqywlzmhhoxh\ndbelhtzgkssyfrqgzuurdjhwvmdbhylhmvphjgxpzhxbb",
"output": "both"
},
{
"input": "nppjzscfgcvdcnsjtiaudvutmgswqbewejlzibczzowgkdrjgxrpirfdaekvngcsonroheepdoeoeevaullbfwprcnhlxextbxpd\nifilrvacohnwcgzuleicucebrfxphosrgwnglxxkqrcorsxegjoppbb",
"output": "both"
},
{
"input": "ggzmtrhkpdswwqgcbtviahqrgzhyhzddtdekchrpjgngupitzyyuipwstgzewktcqpwezidwvvxgjixnflpjhfznokmpbyzczrzk\ngpgwhtzrcytstezmhettkppgmvxlxqnkjzibiqdtceczkbfhdziuajwjqzgwnhnkdzizprgzwud",
"output": "both"
},
{
"input": "iypjqiiqxhtinlmywpetgqqsdopxhghthjopgbodkwrdxzaaxmtaqcfuiarhrvasusanklzcqaytdyzndakcpljqupowompjjved\nhxeatriypptbhnokarhgqdrkqkypqzdttixphngmpqjodzjqlmcztyjfgoswjelwwdaqdjayavsdocuhqsluxaaopniviaumxip",
"output": "both"
},
{
"input": "ypyhyabmljukejpltkgunwuanhxblhiouyltdiczttndrhdprqtlpfanmzlyzbqanfwfyurxhepuzspdvehxnblhajczqcxlqebx\nlladxuucky",
"output": "both"
},
{
"input": "ddmgoarkuhknbtjggnomyxvvavobmylixwuxnnsdrrbibitoteaiydptnvtfblathihflefuggfnyayniragbtkommycpdyhft\ntejwybmyrhmalraptqwhghsckvnnaagtmzhnpwbhzzgfgritqwqqamgssllnicjqdkivrwaqyxngsqopwieljfxcdywjaal",
"output": "need tree"
},
{
"input": "kipjuscf\nkbwfqfwuvkyhmvnaznzsgdgdnpipikbicmlcwehjirmhgwpxwpgfztqjwfqfaapmsgskr",
"output": "need tree"
},
{
"input": "kobhhrqgwbgqkzcoacrhpkegyepzfds\nhlwcgbvvlegoyrcrjhsjywpdnccxtzgmeujxciuwjlnefllwldidlnjswmetkarxqjigokfvmpxpzfxarhkpdcia",
"output": "need tree"
},
{
"input": "lllolloloolllloolollololololollllooololoooloooolololloolloollllolloolloooooooololllolllolllloolllool\nlollollololololooooloooooooooolloolllololooollllol",
"output": "automaton"
},
{
"input": "lloloooolooollololloooloololooollooloollolllloolllllllloollollllolooloollloololollllooloooololooolol\nlooooollooolllololloollooooololollollloloollollolo",
"output": "both"
},
{
"input": "yyyyxxxxyxyyxxxyxxyxxxyyxxxxxyyxxxyxxyxxyyyxxxyxxxyxyxyyxyyxyxxyyyxyxxyxxyxxyyxyyyyxyyyyxxxyyxyxxyyx\nyyyyxxxxyxyyxxxyxxyxxxyyxxxxxyyxxxyxxyxxyyyxxxyxxxxxyxyyxyyxyxxyyyxyxxyxxyxxyyxyyyyxyyyyxxxyyxyxxyyx",
"output": "need tree"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "automaton"
},
{
"input": "abracadabra\nlol",
"output": "need tree"
},
{
"input": "abc\naa",
"output": "need tree"
},
{
"input": "ba\naa",
"output": "need tree"
},
{
"input": "abbb\naaab",
"output": "need tree"
},
{
"input": "baaa\nbb",
"output": "need tree"
},
{
"input": "boosss\nosos",
"output": "both"
}
] | 1,647,207,074
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 4
| 46
| 0
|
s = input()
t = input()
if t in s:
print("automaton")
else:
s_freqs = {}
for c in s:
if c not in s_freqs:
s_freqs[c] = 0
s_freqs[c] += 1
t_freqs = {}
for c in t:
if c not in t_freqs:
t_freqs[c] = 0
t_freqs[c] += 1
if s_freqs == t_freqs:
print("array")
else:
for k, v in t_freqs.items():
if k not in s or s_freqs[k] < t_freqs[k]:
print("need tree")
break
else:
print("both")
|
Title: Suffix Structures
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bizon the Champion isn't just a bison. He also is a favorite of the "Bizons" team.
At a competition the "Bizons" got the following problem: "You are given two distinct words (strings of English letters), *s* and *t*. You need to transform word *s* into word *t*". The task looked simple to the guys because they know the suffix data structures well. Bizon Senior loves suffix automaton. By applying it once to a string, he can remove from this string any single character. Bizon Middle knows suffix array well. By applying it once to a string, he can swap any two characters of this string. The guys do not know anything about the suffix tree, but it can help them do much more.
Bizon the Champion wonders whether the "Bizons" can solve the problem. Perhaps, the solution do not require both data structures. Find out whether the guys can solve the problem and if they can, how do they do it? Can they solve it either only with use of suffix automaton or only with use of suffix array or they need both structures? Note that any structure may be used an unlimited number of times, the structures may be used in any order.
Input Specification:
The first line contains a non-empty word *s*. The second line contains a non-empty word *t*. Words *s* and *t* are different. Each word consists only of lowercase English letters. Each word contains at most 100 letters.
Output Specification:
In the single line print the answer to the problem. Print "need tree" (without the quotes) if word *s* cannot be transformed into word *t* even with use of both suffix array and suffix automaton. Print "automaton" (without the quotes) if you need only the suffix automaton to solve the problem. Print "array" (without the quotes) if you need only the suffix array to solve the problem. Print "both" (without the quotes), if you need both data structures to solve the problem.
It's guaranteed that if you can solve the problem only with use of suffix array, then it is impossible to solve it only with use of suffix automaton. This is also true for suffix automaton.
Demo Input:
['automaton\ntomat\n', 'array\narary\n', 'both\nhot\n', 'need\ntree\n']
Demo Output:
['automaton\n', 'array\n', 'both\n', 'need tree\n']
Note:
In the third sample you can act like that: first transform "both" into "oth" by removing the first character using the suffix automaton and then make two swaps of the string using the suffix array and get "hot".
|
```python
s = input()
t = input()
if t in s:
print("automaton")
else:
s_freqs = {}
for c in s:
if c not in s_freqs:
s_freqs[c] = 0
s_freqs[c] += 1
t_freqs = {}
for c in t:
if c not in t_freqs:
t_freqs[c] = 0
t_freqs[c] += 1
if s_freqs == t_freqs:
print("array")
else:
for k, v in t_freqs.items():
if k not in s or s_freqs[k] < t_freqs[k]:
print("need tree")
break
else:
print("both")
```
| 0
|
|
963
|
A
|
Alternating Sum
|
PROGRAMMING
| 1,800
|
[
"math",
"number theory"
] | null | null |
You are given two integers $a$ and $b$. Moreover, you are given a sequence $s_0, s_1, \dots, s_{n}$. All values in $s$ are integers $1$ or $-1$. It's known that sequence is $k$-periodic and $k$ divides $n+1$. In other words, for each $k \leq i \leq n$ it's satisfied that $s_{i} = s_{i - k}$.
Find out the non-negative remainder of division of $\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}$ by $10^{9} + 9$.
Note that the modulo is unusual!
|
The first line contains four integers $n, a, b$ and $k$ $(1 \leq n \leq 10^{9}, 1 \leq a, b \leq 10^{9}, 1 \leq k \leq 10^{5})$.
The second line contains a sequence of length $k$ consisting of characters '+' and '-'.
If the $i$-th character (0-indexed) is '+', then $s_{i} = 1$, otherwise $s_{i} = -1$.
Note that only the first $k$ members of the sequence are given, the rest can be obtained using the periodicity property.
|
Output a single integer — value of given expression modulo $10^{9} + 9$.
|
[
"2 2 3 3\n+-+\n",
"4 1 5 1\n-\n"
] |
[
"7\n",
"999999228\n"
] |
In the first example:
$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i})$ = $2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2}$ = 7
In the second example:
$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 \equiv 999999228 \pmod{10^{9} + 9}$.
| 500
|
[
{
"input": "2 2 3 3\n+-+",
"output": "7"
},
{
"input": "4 1 5 1\n-",
"output": "999999228"
},
{
"input": "1 1 4 2\n-+",
"output": "3"
},
{
"input": "3 1 4 4\n+--+",
"output": "45"
},
{
"input": "5 1 1 6\n++---+",
"output": "0"
},
{
"input": "5 2 2 6\n+--++-",
"output": "0"
},
{
"input": "686653196 115381398 884618610 3\n+-+",
"output": "542231211"
},
{
"input": "608663287 430477711 172252358 8\n-+--+-+-",
"output": "594681696"
},
{
"input": "904132655 827386249 118827660 334\n+++-+++++--+++----+-+-+-+-+--+-+---++--++--++--+-+-+++-+++--+-+-+----+-+-++++-----+--++++------+++-+-+-++-++++++++-+-++-+++--+--++------+--+-+++--++--+---++-++-+-+-++---++-++--+-+-++-+------+-+----+++-+++--+-+-+--+--+--+------+--+---+--+-++--+++---+-+-++--------+-++--++-+-+-+-+-+-+--+-++++-+++--+--++----+--+-++-++--+--+-+-++-+-++++-",
"output": "188208979"
},
{
"input": "234179195 430477711 115381398 12\n++++-+-+-+++",
"output": "549793323"
},
{
"input": "75952547 967294208 907708706 252\n++--++--+++-+-+--++--++++++---+++-++-+-----++++--++-+-++------+-+-+-++-+-+-++++------++---+-++++---+-+-++++--++++++--+-+++-++--+--+---++++---+-+++-+++--+-+--+++++---+--++-++++--++++-+-++-+++-++-----+-+++++----++--+++-+-+++++-+--++-++-+--+-++++--+-+-+-+",
"output": "605712499"
},
{
"input": "74709071 801809249 753674746 18\n++++++-+-+---+-+--",
"output": "13414893"
},
{
"input": "743329 973758 92942 82\n++----+-++++----+--+++---+--++++-+-+---+++++--+--+++++++--++-+++----+--+++++-+--+-",
"output": "299311566"
},
{
"input": "18111 291387 518587 2\n++",
"output": "724471355"
},
{
"input": "996144 218286 837447 1\n-",
"output": "549104837"
},
{
"input": "179358 828426 548710 67\n++++---+--++----+-+-++++----+--+---+------++-+-++++--+----+---+-+--",
"output": "759716474"
},
{
"input": "397521 174985 279760 1\n+",
"output": "25679493"
},
{
"input": "613632 812232 482342 1\n-",
"output": "891965141"
},
{
"input": "936810 183454 647048 1\n+",
"output": "523548992"
},
{
"input": "231531 250371 921383 28\n++-+------+--+--++++--+-+++-",
"output": "134450934"
},
{
"input": "947301 87242 360762 97\n--+++--+++-++--++-++--++--+++---+++--++++--+++++--+-++-++-----+-++-+--++-----+-++-+--++-++-+-----",
"output": "405016159"
},
{
"input": "425583346 814209084 570987274 1\n+",
"output": "63271171"
},
{
"input": "354062556 688076879 786825319 1\n+",
"output": "545304776"
},
{
"input": "206671954 13571766 192250278 1\n+",
"output": "717117421"
},
{
"input": "23047921 621656196 160244047 1\n-",
"output": "101533009"
},
{
"input": "806038018 740585177 987616107 293\n-+++++--++++---++-+--+-+---+-++++--+--+++--++---++++++++--+++++-+-++-+--+----+--+++-+-++-+++-+-+-+----------++-+-+++++++-+-+-+-++---+++-+-+-------+-+-++--++-++-++-++-+---+--++-++--+++--+++-+-+----++--+-+-++-+---+---+-+-+++------+-+++-+---++-+--+++----+++++---++-++--+----+++-+--+++-+------+-++",
"output": "441468166"
},
{
"input": "262060935 184120408 148332034 148\n+--+-------+-+-+--++-+++--++-+-++++++--++-+++-+++--+-------+-+--+++-+-+-+---++-++-+-++---+--+-+-+--+------+++--+--+-+-+---+---+-+-++++---+++--+++---",
"output": "700325386"
},
{
"input": "919350941 654611542 217223605 186\n++-++-+++++-+++--+---+++++++-++-+----+-++--+-++--++--+++-+++---+--+--++-+-+++-+-+++-++---+--+++-+-+++--+-+-------+-++------++---+-+---++-++-++---+-+--+-+--+++++---+--+--++++-++-++--+--++",
"output": "116291420"
},
{
"input": "289455627 906207104 512692624 154\n-------++--+++---++-++------++----------+--+++-+-+++---+---+++--++++++--+-+-+--+---+-+-++-++--+-++--++++---+-+---+-----+--+-+---------+++-++---++-+-+-----",
"output": "48198216"
},
{
"input": "258833760 515657142 791267045 1\n-",
"output": "935800888"
},
{
"input": "691617927 66917103 843055237 8\n--+++---",
"output": "147768186"
},
{
"input": "379582849 362892355 986900829 50\n++-++---+-+++++--++++--+--++--++-----+------++--+-",
"output": "927469713"
},
{
"input": "176799169 363368399 841293419 1\n+",
"output": "746494802"
},
{
"input": "144808247 203038656 166324035 4\n-+-+",
"output": "909066471"
},
{
"input": "477607531 177367565 20080950 2\n++",
"output": "928662830"
},
{
"input": "682074525 289438443 917164266 1\n+",
"output": "28048785"
},
{
"input": "938449224 59852396 219719125 1\n-",
"output": "648647459"
},
{
"input": "395171426 872478622 193568600 147\n+---++---+-+--+++++--+---+-++++-+-++---++++--+--+-+-++-+-++--------++---+++-+---++---+---+-+--+-++++-+++-+-+-++-+--+++-++-+-+-+-++++++-+---+---++--",
"output": "460881399"
},
{
"input": "403493428 317461491 556701240 1\n-",
"output": "936516261"
},
{
"input": "917751169 330191895 532837377 70\n-+-+++++++--++---++-+++++-+++-----+-+++---+--+-+-++-++-+-+-++-++-+----",
"output": "908035409"
},
{
"input": "252089413 552678586 938424519 1\n-",
"output": "627032736"
},
{
"input": "649316142 320010793 200197645 1\n-",
"output": "323650777"
},
{
"input": "116399299 784781190 299072480 5\n++++-",
"output": "754650814"
}
] | 1,589,705,075
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 6
| 1,000
| 23,040,000
|
q=10**9+9
n,a,b,k=map(int,input().split())
s=input()
result=0
for i in range(n+1):
c=(a**(n-i))%q
d=(b**i)%q
e=[1,-1][s[i%k]=='-']
result+=(c*d*e)
print(result%q)
|
Title: Alternating Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two integers $a$ and $b$. Moreover, you are given a sequence $s_0, s_1, \dots, s_{n}$. All values in $s$ are integers $1$ or $-1$. It's known that sequence is $k$-periodic and $k$ divides $n+1$. In other words, for each $k \leq i \leq n$ it's satisfied that $s_{i} = s_{i - k}$.
Find out the non-negative remainder of division of $\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}$ by $10^{9} + 9$.
Note that the modulo is unusual!
Input Specification:
The first line contains four integers $n, a, b$ and $k$ $(1 \leq n \leq 10^{9}, 1 \leq a, b \leq 10^{9}, 1 \leq k \leq 10^{5})$.
The second line contains a sequence of length $k$ consisting of characters '+' and '-'.
If the $i$-th character (0-indexed) is '+', then $s_{i} = 1$, otherwise $s_{i} = -1$.
Note that only the first $k$ members of the sequence are given, the rest can be obtained using the periodicity property.
Output Specification:
Output a single integer — value of given expression modulo $10^{9} + 9$.
Demo Input:
['2 2 3 3\n+-+\n', '4 1 5 1\n-\n']
Demo Output:
['7\n', '999999228\n']
Note:
In the first example:
$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i})$ = $2^{2} 3^{0} - 2^{1} 3^{1} + 2^{0} 3^{2}$ = 7
In the second example:
$(\sum \limits_{i=0}^{n} s_{i} a^{n - i} b^{i}) = -1^{4} 5^{0} - 1^{3} 5^{1} - 1^{2} 5^{2} - 1^{1} 5^{3} - 1^{0} 5^{4} = -781 \equiv 999999228 \pmod{10^{9} + 9}$.
|
```python
q=10**9+9
n,a,b,k=map(int,input().split())
s=input()
result=0
for i in range(n+1):
c=(a**(n-i))%q
d=(b**i)%q
e=[1,-1][s[i%k]=='-']
result+=(c*d*e)
print(result%q)
```
| 0
|
|
721
|
A
|
One-dimensional Japanese Crossword
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
|
The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
|
[
"3\nBBW\n",
"5\nBWBWB\n",
"4\nWWWW\n",
"4\nBBBB\n",
"13\nWBBBBWWBWBBBW\n"
] |
[
"1\n2 ",
"3\n1 1 1 ",
"0\n",
"1\n4 ",
"3\n4 1 3 "
] |
The last sample case correspond to the picture in the statement.
| 500
|
[
{
"input": "3\nBBW",
"output": "1\n2 "
},
{
"input": "5\nBWBWB",
"output": "3\n1 1 1 "
},
{
"input": "4\nWWWW",
"output": "0"
},
{
"input": "4\nBBBB",
"output": "1\n4 "
},
{
"input": "13\nWBBBBWWBWBBBW",
"output": "3\n4 1 3 "
},
{
"input": "1\nB",
"output": "1\n1 "
},
{
"input": "2\nBB",
"output": "1\n2 "
},
{
"input": "100\nWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWBWB",
"output": "50\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 "
},
{
"input": "1\nW",
"output": "0"
},
{
"input": "2\nWW",
"output": "0"
},
{
"input": "2\nWB",
"output": "1\n1 "
},
{
"input": "2\nBW",
"output": "1\n1 "
},
{
"input": "3\nBBB",
"output": "1\n3 "
},
{
"input": "3\nBWB",
"output": "2\n1 1 "
},
{
"input": "3\nWBB",
"output": "1\n2 "
},
{
"input": "3\nWWB",
"output": "1\n1 "
},
{
"input": "3\nWBW",
"output": "1\n1 "
},
{
"input": "3\nBWW",
"output": "1\n1 "
},
{
"input": "3\nWWW",
"output": "0"
},
{
"input": "100\nBBBWWWWWWBBWWBBWWWBBWBBBBBBBBBBBWBBBWBBWWWBBWWBBBWBWWBBBWWBBBWBBBBBWWWBWWBBWWWWWWBWBBWWBWWWBWBWWWWWB",
"output": "21\n3 2 2 2 11 3 2 2 3 1 3 3 5 1 2 1 2 1 1 1 1 "
},
{
"input": "5\nBBBWB",
"output": "2\n3 1 "
},
{
"input": "5\nBWWWB",
"output": "2\n1 1 "
},
{
"input": "5\nWWWWB",
"output": "1\n1 "
},
{
"input": "5\nBWWWW",
"output": "1\n1 "
},
{
"input": "5\nBBBWW",
"output": "1\n3 "
},
{
"input": "5\nWWBBB",
"output": "1\n3 "
},
{
"input": "10\nBBBBBWWBBB",
"output": "2\n5 3 "
},
{
"input": "10\nBBBBWBBWBB",
"output": "3\n4 2 2 "
},
{
"input": "20\nBBBBBWWBWBBWBWWBWBBB",
"output": "6\n5 1 2 1 1 3 "
},
{
"input": "20\nBBBWWWWBBWWWBWBWWBBB",
"output": "5\n3 2 1 1 3 "
},
{
"input": "20\nBBBBBBBBWBBBWBWBWBBB",
"output": "5\n8 3 1 1 3 "
},
{
"input": "20\nBBBWBWBWWWBBWWWWBWBB",
"output": "6\n3 1 1 2 1 2 "
},
{
"input": "40\nBBBBBBWWWWBWBWWWBWWWWWWWWWWWBBBBBBBBBBBB",
"output": "5\n6 1 1 1 12 "
},
{
"input": "40\nBBBBBWBWWWBBWWWBWBWWBBBBWWWWBWBWBBBBBBBB",
"output": "9\n5 1 2 1 1 4 1 1 8 "
},
{
"input": "50\nBBBBBBBBBBBWWWWBWBWWWWBBBBBBBBWWWWWWWBWWWWBWBBBBBB",
"output": "7\n11 1 1 8 1 1 6 "
},
{
"input": "50\nWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW",
"output": "0"
},
{
"input": "50\nBBBBBWWWWWBWWWBWWWWWBWWWBWWWWWWBBWBBWWWWBWWWWWWWBW",
"output": "9\n5 1 1 1 1 2 2 1 1 "
},
{
"input": "50\nWWWWBWWBWWWWWWWWWWWWWWWWWWWWWWWWWBWBWBWWWWWWWBBBBB",
"output": "6\n1 1 1 1 1 5 "
},
{
"input": "50\nBBBBBWBWBWWBWBWWWWWWBWBWBWWWWWWWWWWWWWBWBWWWWBWWWB",
"output": "12\n5 1 1 1 1 1 1 1 1 1 1 1 "
},
{
"input": "50\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "1\n50 "
},
{
"input": "100\nBBBBBBBBBBBWBWWWWBWWBBWBBWWWWWWWWWWBWBWWBWWWWWWWWWWWBBBWWBBWWWWWBWBWWWWBWWWWWWWWWWWBWWWWWBBBBBBBBBBB",
"output": "15\n11 1 1 2 2 1 1 1 3 2 1 1 1 1 11 "
},
{
"input": "100\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "1\n100 "
},
{
"input": "100\nBBBBBBBBBBBBBBBBBBBBWBWBWWWWWBWWWWWWWWWWWWWWBBWWWBWWWWBWWBWWWWWWBWWWWWWWWWWWWWBWBBBBBBBBBBBBBBBBBBBB",
"output": "11\n20 1 1 1 2 1 1 1 1 1 20 "
},
{
"input": "100\nBBBBWWWWWWWWWWWWWWWWWWWWWWWWWBWBWWWWWBWBWWWWWWBBWWWWWWWWWWWWBWWWWBWWWWWWWWWWWWBWWWWWWWBWWWWWWWBBBBBB",
"output": "11\n4 1 1 1 1 2 1 1 1 1 6 "
},
{
"input": "5\nBWBWB",
"output": "3\n1 1 1 "
},
{
"input": "10\nWWBWWWBWBB",
"output": "3\n1 1 2 "
},
{
"input": "50\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "1\n50 "
},
{
"input": "50\nBBBBBBBBBBBBBBBBBWWBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "2\n17 31 "
},
{
"input": "100\nBBBBBBBBBBBBBBBBBBBBBBBBWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB",
"output": "2\n24 42 "
},
{
"input": "90\nWWBWWBWBBWBBWWBWBWBBBWBWBBBWBWBWBWBWBWBWBWBBBBBWBBWWWWBWBBWBWWBBBWBWBWWBWBWBWBWWWWWWBWBBBB",
"output": "30\n1 1 2 2 1 1 3 1 3 1 1 1 1 1 1 1 5 2 1 2 1 3 1 1 1 1 1 1 1 4 "
},
{
"input": "100\nBWWWBWBWBBBBBWBWWBWBWWWBWBWBWWBBWWBBBWBBBWWBWBWWBBBBWBWBBBWBWBBWWWWWWBWWBBBBWBWBWWBWBWWWBWBWWBWBWWWB",
"output": "31\n1 1 1 5 1 1 1 1 1 1 2 3 3 1 1 4 1 3 1 2 1 4 1 1 1 1 1 1 1 1 1 "
},
{
"input": "90\nWBWBBBBBBWWWBBWWBWWWBBWWBWWWBWBBWBWBBWWWWBWBWBBWBBWBWWWBBWBBWWWWBWBBWWWBBBWBBWBWBBBBWWBWWB",
"output": "25\n1 6 2 1 2 1 1 2 1 2 1 1 2 2 1 2 2 1 2 3 2 1 4 1 1 "
},
{
"input": "80\nBBWWBBBWBBWWWWBBWBWBBWWWWWBWBBWWBWBWBWBWBWWBWWBWWWBWBBWBBWBBWBBBWWBBBBBBBWBBBWBB",
"output": "23\n2 3 2 2 1 2 1 2 1 1 1 1 1 1 1 1 2 2 2 3 7 3 2 "
},
{
"input": "65\nWWWWBWWWBBBBBWWWWWWBBBWWBBBBWWWWWWWWBBBWWWWBWBWWBBWWWWBWWWBBWBBBB",
"output": "11\n1 5 3 4 3 1 1 2 1 2 4 "
}
] | 1,618,637,795
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 46
| 0
|
n = int(input())
s = input()
li = []
count = 0
for i in s:
if i=='B':
count+=1
else:
if count!=0:
li.append(count)
count =0
if len(li)!=0:
print(len(li))
print(*li)
else:
print(len(li))
|
Title: One-dimensional Japanese Crossword
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
Output Specification:
The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Demo Input:
['3\nBBW\n', '5\nBWBWB\n', '4\nWWWW\n', '4\nBBBB\n', '13\nWBBBBWWBWBBBW\n']
Demo Output:
['1\n2 ', '3\n1 1 1 ', '0\n', '1\n4 ', '3\n4 1 3 ']
Note:
The last sample case correspond to the picture in the statement.
|
```python
n = int(input())
s = input()
li = []
count = 0
for i in s:
if i=='B':
count+=1
else:
if count!=0:
li.append(count)
count =0
if len(li)!=0:
print(len(li))
print(*li)
else:
print(len(li))
```
| 0
|
|
489
|
C
|
Given Length and Sum of Digits...
|
PROGRAMMING
| 1,400
|
[
"dp",
"greedy",
"implementation"
] | null | null |
You have a positive integer *m* and a non-negative integer *s*. Your task is to find the smallest and the largest of the numbers that have length *m* and sum of digits *s*. The required numbers should be non-negative integers written in the decimal base without leading zeroes.
|
The single line of the input contains a pair of integers *m*, *s* (1<=≤<=*m*<=≤<=100,<=0<=≤<=*s*<=≤<=900) — the length and the sum of the digits of the required numbers.
|
In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes).
|
[
"2 15\n",
"3 0\n"
] |
[
"69 96\n",
"-1 -1\n"
] |
none
| 1,500
|
[
{
"input": "2 15",
"output": "69 96"
},
{
"input": "3 0",
"output": "-1 -1"
},
{
"input": "2 1",
"output": "10 10"
},
{
"input": "3 10",
"output": "109 910"
},
{
"input": "100 100",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000099999999999 9999999999910000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "1 900",
"output": "-1 -1"
},
{
"input": "1 9",
"output": "9 9"
},
{
"input": "1 0",
"output": "0 0"
},
{
"input": "1 1",
"output": "1 1"
},
{
"input": "1 2",
"output": "2 2"
},
{
"input": "1 8",
"output": "8 8"
},
{
"input": "1 10",
"output": "-1 -1"
},
{
"input": "1 11",
"output": "-1 -1"
},
{
"input": "2 0",
"output": "-1 -1"
},
{
"input": "2 1",
"output": "10 10"
},
{
"input": "2 2",
"output": "11 20"
},
{
"input": "2 8",
"output": "17 80"
},
{
"input": "2 10",
"output": "19 91"
},
{
"input": "2 11",
"output": "29 92"
},
{
"input": "2 16",
"output": "79 97"
},
{
"input": "2 17",
"output": "89 98"
},
{
"input": "2 18",
"output": "99 99"
},
{
"input": "2 19",
"output": "-1 -1"
},
{
"input": "2 20",
"output": "-1 -1"
},
{
"input": "2 900",
"output": "-1 -1"
},
{
"input": "3 1",
"output": "100 100"
},
{
"input": "3 2",
"output": "101 200"
},
{
"input": "3 3",
"output": "102 300"
},
{
"input": "3 9",
"output": "108 900"
},
{
"input": "3 10",
"output": "109 910"
},
{
"input": "3 20",
"output": "299 992"
},
{
"input": "3 21",
"output": "399 993"
},
{
"input": "3 26",
"output": "899 998"
},
{
"input": "3 27",
"output": "999 999"
},
{
"input": "3 28",
"output": "-1 -1"
},
{
"input": "3 100",
"output": "-1 -1"
},
{
"input": "100 0",
"output": "-1 -1"
},
{
"input": "100 1",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 2",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 9",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000008 9000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 10",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000009 9100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 11",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000019 9200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 296",
"output": "1000000000000000000000000000000000000000000000000000000000000000000799999999999999999999999999999999 9999999999999999999999999999999980000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 297",
"output": "1000000000000000000000000000000000000000000000000000000000000000000899999999999999999999999999999999 9999999999999999999999999999999990000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 298",
"output": "1000000000000000000000000000000000000000000000000000000000000000000999999999999999999999999999999999 9999999999999999999999999999999991000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 299",
"output": "1000000000000000000000000000000000000000000000000000000000000000001999999999999999999999999999999999 9999999999999999999999999999999992000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 300",
"output": "1000000000000000000000000000000000000000000000000000000000000000002999999999999999999999999999999999 9999999999999999999999999999999993000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 301",
"output": "1000000000000000000000000000000000000000000000000000000000000000003999999999999999999999999999999999 9999999999999999999999999999999994000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 895",
"output": "4999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999994"
},
{
"input": "100 896",
"output": "5999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999995"
},
{
"input": "100 897",
"output": "6999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999996"
},
{
"input": "100 898",
"output": "7999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999997"
},
{
"input": "100 899",
"output": "8999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999998"
},
{
"input": "100 900",
"output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"
},
{
"input": "99 900",
"output": "-1 -1"
},
{
"input": "99 891",
"output": "999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"
},
{
"input": "99 892",
"output": "-1 -1"
},
{
"input": "96 364",
"output": "100000000000000000000000000000000000000000000000000000039999999999999999999999999999999999999999 999999999999999999999999999999999999999940000000000000000000000000000000000000000000000000000000"
},
{
"input": "97 206",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000079999999999999999999999 9999999999999999999999800000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "98 250",
"output": "10000000000000000000000000000000000000000000000000000000000000000000006999999999999999999999999999 99999999999999999999999999970000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "99 96",
"output": "100000000000000000000000000000000000000000000000000000000000000000000000000000000000000059999999999 999999999960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 215",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000799999999999999999999999 9999999999999999999999980000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "5 18",
"output": "10089 99000"
},
{
"input": "2 19",
"output": "-1 -1"
},
{
"input": "3 3",
"output": "102 300"
}
] | 1,687,861,171
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 31
| 0
|
m,s = map(int,input().split())
mn = 10**(m-1)
mx = str((10**m )-1)
if s>=1 and s<=len(mx)*9:
num = ''
if s<10:
mxim = int(str(s) + '0'*(m-1))
minm = int(str(1) + '0'*(m-2) + str(s-1))
else:
n = s//9
l = s%9
if l != 0:
mxim = int('9'*n + str(l))
minm = int(str(l)+'9'*n)
else:
mxim = int('9'*n )
minm = int('9'*n )
print(minm,mxim)
else:
print(-1,-1)
|
Title: Given Length and Sum of Digits...
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a positive integer *m* and a non-negative integer *s*. Your task is to find the smallest and the largest of the numbers that have length *m* and sum of digits *s*. The required numbers should be non-negative integers written in the decimal base without leading zeroes.
Input Specification:
The single line of the input contains a pair of integers *m*, *s* (1<=≤<=*m*<=≤<=100,<=0<=≤<=*s*<=≤<=900) — the length and the sum of the digits of the required numbers.
Output Specification:
In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes).
Demo Input:
['2 15\n', '3 0\n']
Demo Output:
['69 96\n', '-1 -1\n']
Note:
none
|
```python
m,s = map(int,input().split())
mn = 10**(m-1)
mx = str((10**m )-1)
if s>=1 and s<=len(mx)*9:
num = ''
if s<10:
mxim = int(str(s) + '0'*(m-1))
minm = int(str(1) + '0'*(m-2) + str(s-1))
else:
n = s//9
l = s%9
if l != 0:
mxim = int('9'*n + str(l))
minm = int(str(l)+'9'*n)
else:
mxim = int('9'*n )
minm = int('9'*n )
print(minm,mxim)
else:
print(-1,-1)
```
| 0
|
|
443
|
A
|
Anton and Letters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
|
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
|
Print a single number — the number of distinct letters in Anton's set.
|
[
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] |
[
"3\n",
"2\n",
"0\n"
] |
none
| 500
|
[
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{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, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,689,069,194
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
a=map(int,input().split())
b=list(set(a))
c=len(b)
print(c)
|
Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none
|
```python
a=map(int,input().split())
b=list(set(a))
c=len(b)
print(c)
```
| -1
|
|
177
|
A1
|
Good Matrix Elements
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good:
- Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it.
Help the Smart Beaver count the sum of good elements of the given matrix.
|
The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix.
The input limitations for getting 30 points are:
- 1<=≤<=*n*<=≤<=5
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=101
|
Print a single integer — the sum of good matrix elements.
|
[
"3\n1 2 3\n4 5 6\n7 8 9\n",
"5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n"
] |
[
"45\n",
"17\n"
] |
In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure.
| 30
|
[
{
"input": "3\n1 2 3\n4 5 6\n7 8 9",
"output": "45"
},
{
"input": "5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1",
"output": "17"
},
{
"input": "1\n3",
"output": "3"
},
{
"input": "5\n27 7 3 11 72\n19 49 68 19 59\n41 25 37 64 65\n8 39 96 62 90\n13 37 43 26 33",
"output": "756"
},
{
"input": "3\n19 7 16\n12 15 5\n15 15 5",
"output": "109"
},
{
"input": "3\n36 4 33\n11 46 32\n20 49 34",
"output": "265"
},
{
"input": "3\n79 91 74\n33 82 22\n18 28 54",
"output": "481"
},
{
"input": "5\n7 0 8 1 7\n5 1 1 0 4\n4 2 8 1 6\n1 2 3 2 7\n6 0 1 9 6",
"output": "65"
},
{
"input": "5\n27 20 28 11 17\n25 21 1 20 14\n14 22 28 1 6\n1 2 23 2 7\n6 0 1 29 6",
"output": "225"
},
{
"input": "5\n57 50 58 41 17\n25 21 1 50 44\n44 22 28 31 36\n31 32 23 32 37\n6 0 31 59 6",
"output": "495"
},
{
"input": "5\n57 80 28 41 47\n85 51 61 50 74\n44 82 28 31 36\n31 32 23 32 37\n66 60 31 59 6",
"output": "705"
},
{
"input": "5\n13 58 10 17 43\n61 73 100 0 9\n52 38 16 22 96\n11 4 14 67 62\n70 89 7 98 83",
"output": "708"
},
{
"input": "5\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "5\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "5\n0 1 0 1 0\n0 0 0 0 1\n0 0 5 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "5"
},
{
"input": "5\n0 0 0 0 1\n0 1 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "3\n0 0 0\n0 0 0\n0 0 0",
"output": "0"
}
] | 1,549,573,414
| 2,147,483,647
|
Python 3
|
OK
|
TESTS1
| 17
| 248
| 0
|
x=int(input())
z=0;s=0;l=(x-1)
for i in range(x):
y=input().split()
if i<((x//2)-1) :
z+=int(y[s])+int(y[l])+int(y[x//2])
s+=1;l-=1
elif i==((x//2)-1) :
z += int(y[s]) + int(y[l]) + int(y[x // 2])
elif i ==(x//2):
for r in range(x):
z+=int(y[r])
else:
z+=int(y[x//2])+int(y[l])+int(y[s])
s-=1;l+=1
print(z)
|
Title: Good Matrix Elements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Smart Beaver from ABBYY got hooked on square matrices. Now he is busy studying an *n*<=×<=*n* size matrix, where *n* is odd. The Smart Beaver considers the following matrix elements good:
- Elements of the main diagonal. - Elements of the secondary diagonal. - Elements of the "middle" row — the row which has exactly rows above it and the same number of rows below it. - Elements of the "middle" column — the column that has exactly columns to the left of it and the same number of columns to the right of it.
Help the Smart Beaver count the sum of good elements of the given matrix.
Input Specification:
The first line of input data contains a single odd integer *n*. Each of the next *n* lines contains *n* integers *a**ij* (0<=≤<=*a**ij*<=≤<=100) separated by single spaces — the elements of the given matrix.
The input limitations for getting 30 points are:
- 1<=≤<=*n*<=≤<=5
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=101
Output Specification:
Print a single integer — the sum of good matrix elements.
Demo Input:
['3\n1 2 3\n4 5 6\n7 8 9\n', '5\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n1 1 1 1 1\n']
Demo Output:
['45\n', '17\n']
Note:
In the first sample all matrix elements will be good. Good elements in the second sample are shown on the figure.
|
```python
x=int(input())
z=0;s=0;l=(x-1)
for i in range(x):
y=input().split()
if i<((x//2)-1) :
z+=int(y[s])+int(y[l])+int(y[x//2])
s+=1;l-=1
elif i==((x//2)-1) :
z += int(y[s]) + int(y[l]) + int(y[x // 2])
elif i ==(x//2):
for r in range(x):
z+=int(y[r])
else:
z+=int(y[x//2])+int(y[l])+int(y[s])
s-=1;l+=1
print(z)
```
| 3
|
|
754
|
B
|
Ilya and tic-tac-toe game
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
Ilya is an experienced player in tic-tac-toe on the 4<=×<=4 field. He always starts and plays with Xs. He played a lot of games today with his friend Arseny. The friends became tired and didn't finish the last game. It was Ilya's turn in the game when they left it. Determine whether Ilya could have won the game by making single turn or not.
The rules of tic-tac-toe on the 4<=×<=4 field are as follows. Before the first turn all the field cells are empty. The two players take turns placing their signs into empty cells (the first player places Xs, the second player places Os). The player who places Xs goes first, the another one goes second. The winner is the player who first gets three of his signs in a row next to each other (horizontal, vertical or diagonal).
|
The tic-tac-toe position is given in four lines.
Each of these lines contains four characters. Each character is '.' (empty cell), 'x' (lowercase English letter x), or 'o' (lowercase English letter o). It is guaranteed that the position is reachable playing tic-tac-toe, and it is Ilya's turn now (in particular, it means that the game is not finished). It is possible that all the cells are empty, it means that the friends left without making single turn.
|
Print single line: "YES" in case Ilya could have won by making single turn, and "NO" otherwise.
|
[
"xx..\n.oo.\nx...\noox.\n",
"x.ox\nox..\nx.o.\noo.x\n",
"x..x\n..oo\no...\nx.xo\n",
"o.x.\no...\n.x..\nooxx\n"
] |
[
"YES\n",
"NO\n",
"YES\n",
"NO\n"
] |
In the first example Ilya had two winning moves: to the empty cell in the left column and to the leftmost empty cell in the first row.
In the second example it wasn't possible to win by making single turn.
In the third example Ilya could have won by placing X in the last row between two existing Xs.
In the fourth example it wasn't possible to win by making single turn.
| 1,000
|
[
{
"input": "xx..\n.oo.\nx...\noox.",
"output": "YES"
},
{
"input": "x.ox\nox..\nx.o.\noo.x",
"output": "NO"
},
{
"input": "x..x\n..oo\no...\nx.xo",
"output": "YES"
},
{
"input": "o.x.\no...\n.x..\nooxx",
"output": "NO"
},
{
"input": ".xox\no.x.\nx.o.\n..o.",
"output": "YES"
},
{
"input": "o.oo\n.x.o\nx.x.\n.x..",
"output": "YES"
},
{
"input": "xxox\no.x.\nx.oo\nxo.o",
"output": "YES"
},
{
"input": ".xox\n.x..\nxoo.\noox.",
"output": "NO"
},
{
"input": "...x\n.x.o\n.o..\n.x.o",
"output": "NO"
},
{
"input": "oo.x\nxo.o\no.xx\n.oxx",
"output": "YES"
},
{
"input": ".x.o\n..o.\n..ox\nxox.",
"output": "NO"
},
{
"input": "....\n.x..\nx...\n..oo",
"output": "YES"
},
{
"input": "....\n....\n.x.o\n..xo",
"output": "YES"
},
{
"input": "o..o\nx..x\n.o.x\nxo..",
"output": "YES"
},
{
"input": "ox.o\nx..x\nx..o\noo.x",
"output": "NO"
},
{
"input": ".xox\n.x.o\nooxo\n..x.",
"output": "YES"
},
{
"input": "x..o\no..o\n..x.\nx.xo",
"output": "YES"
},
{
"input": "xxoo\no.oo\n...x\nx..x",
"output": "NO"
},
{
"input": "xoox\n.xx.\no..o\n..xo",
"output": "YES"
},
{
"input": "..o.\nxxox\n....\n.oxo",
"output": "YES"
},
{
"input": "xoox\nxxox\noo..\n.ox.",
"output": "YES"
},
{
"input": "..ox\n.o..\nx..o\n.oxx",
"output": "NO"
},
{
"input": ".oo.\n.x..\nx...\nox..",
"output": "YES"
},
{
"input": "o.xx\nxo.o\n...o\n..x.",
"output": "YES"
},
{
"input": "x...\n.ox.\n.oo.\n.xox",
"output": "NO"
},
{
"input": "xoxx\n..x.\no.oo\nx.o.",
"output": "YES"
},
{
"input": ".x.x\n.o.o\no.xx\nx.oo",
"output": "YES"
},
{
"input": "...o\nxo.x\n.x..\nxoo.",
"output": "YES"
},
{
"input": "o...\n...o\noxx.\n.xxo",
"output": "YES"
},
{
"input": "xxox\no..o\nx..o\noxox",
"output": "NO"
},
{
"input": "x.x.\nox.o\n.o.o\nxox.",
"output": "YES"
},
{
"input": "xxo.\n...x\nooxx\n.o.o",
"output": "YES"
},
{
"input": "xoxo\no..x\n.xo.\nox..",
"output": "YES"
},
{
"input": ".o..\nox..\n.o.x\n.x..",
"output": "NO"
},
{
"input": ".oxo\nx...\n.o..\n.xox",
"output": "NO"
},
{
"input": ".oxx\n..o.\n.o.x\n.ox.",
"output": "YES"
},
{
"input": ".xxo\n...o\n..ox\nox..",
"output": "YES"
},
{
"input": "x...\nxo..\noxo.\n..ox",
"output": "NO"
},
{
"input": "xoxo\nx.ox\n....\noxo.",
"output": "YES"
},
{
"input": "x..o\nxo.x\no.xo\nxoox",
"output": "NO"
},
{
"input": ".x..\no..x\n.oo.\nxox.",
"output": "NO"
},
{
"input": "xxox\no.x.\nxo.o\nxo.o",
"output": "NO"
},
{
"input": ".xo.\nx.oo\n...x\n.o.x",
"output": "NO"
},
{
"input": "ox.o\n...x\n..oo\nxxox",
"output": "NO"
},
{
"input": "oox.\nxoo.\no.x.\nx..x",
"output": "NO"
},
{
"input": "oxox\nx.oo\nooxx\nxxo.",
"output": "NO"
},
{
"input": "....\nxo.x\n..x.\noo..",
"output": "NO"
},
{
"input": ".ox.\nx..o\nxo.x\noxo.",
"output": "YES"
},
{
"input": ".xox\nxo..\n..oo\n.x..",
"output": "NO"
},
{
"input": "xxo.\n.oo.\n..x.\n..xo",
"output": "NO"
},
{
"input": "ox..\n..oo\n..x.\nxxo.",
"output": "NO"
},
{
"input": "xxo.\nx..x\noo.o\noxox",
"output": "YES"
},
{
"input": "xx..\noxxo\nxo.o\noox.",
"output": "YES"
},
{
"input": "x..o\no..o\no..x\nxxox",
"output": "NO"
},
{
"input": "oxo.\nxx.x\nooxx\n.o.o",
"output": "YES"
},
{
"input": ".o.x\no..o\nx..x\n..xo",
"output": "NO"
},
{
"input": "xo..\n....\nx...\n..o.",
"output": "YES"
},
{
"input": ".x..\no...\n...x\n.o..",
"output": "YES"
},
{
"input": "...x\n....\n.x.o\n..o.",
"output": "YES"
},
{
"input": "o..x\n....\n...x\n..o.",
"output": "YES"
},
{
"input": ".oo.\nx...\n....\n..x.",
"output": "YES"
},
{
"input": ".o..\n.x..\n..o.\n.x..",
"output": "YES"
},
{
"input": "..o.\n.x..\n....\no..x",
"output": "YES"
},
{
"input": "..o.\n..x.\n....\n.ox.",
"output": "YES"
},
{
"input": ".o..\no..x\n....\n.x..",
"output": "YES"
},
{
"input": "....\n..ox\n....\n.o.x",
"output": "YES"
},
{
"input": ".o..\n....\no...\nx.x.",
"output": "YES"
},
{
"input": "....\n.o..\n....\nox.x",
"output": "YES"
},
{
"input": "oxo.\nxxox\noo.o\nxoxx",
"output": "YES"
},
{
"input": ".xx.\n...x\noo.o\no..x",
"output": "YES"
},
{
"input": "x...\n.x..\n....\noo..",
"output": "YES"
},
{
"input": "oxox\n..ox\nxoxo\nxoxo",
"output": "YES"
},
{
"input": "....\n...x\n...x\noo..",
"output": "YES"
}
] | 1,607,637,443
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 95
| 155
| 0
|
s=[]
for i in range(4):
s1=input()
s.append(s1)
a=set()
for i in range(4):
s1=s[i][0:3]
s2=s[i][1:4]
s3=s[0][i]+s[1][i]+s[2][i]
s4=s[1][i]+s[2][i]+s[3][i]
a.add(s1)
a.add(s2)
a.add(s3)
a.add(s4)
for i in range(0,2):
for j in range(2):
a.add(s[i][j]+s[i+1][j+1]+s[i+2][j+2])
a.add(s[i][3-j]+s[i+1][2-j]+s[i+2][1-j])
s1='xx.'
s2='x.x'
s3='.xx'
if(s1 in a or s2 in a or s3 in a):
print("YES")
else:
print("NO")
|
Title: Ilya and tic-tac-toe game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ilya is an experienced player in tic-tac-toe on the 4<=×<=4 field. He always starts and plays with Xs. He played a lot of games today with his friend Arseny. The friends became tired and didn't finish the last game. It was Ilya's turn in the game when they left it. Determine whether Ilya could have won the game by making single turn or not.
The rules of tic-tac-toe on the 4<=×<=4 field are as follows. Before the first turn all the field cells are empty. The two players take turns placing their signs into empty cells (the first player places Xs, the second player places Os). The player who places Xs goes first, the another one goes second. The winner is the player who first gets three of his signs in a row next to each other (horizontal, vertical or diagonal).
Input Specification:
The tic-tac-toe position is given in four lines.
Each of these lines contains four characters. Each character is '.' (empty cell), 'x' (lowercase English letter x), or 'o' (lowercase English letter o). It is guaranteed that the position is reachable playing tic-tac-toe, and it is Ilya's turn now (in particular, it means that the game is not finished). It is possible that all the cells are empty, it means that the friends left without making single turn.
Output Specification:
Print single line: "YES" in case Ilya could have won by making single turn, and "NO" otherwise.
Demo Input:
['xx..\n.oo.\nx...\noox.\n', 'x.ox\nox..\nx.o.\noo.x\n', 'x..x\n..oo\no...\nx.xo\n', 'o.x.\no...\n.x..\nooxx\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'NO\n']
Note:
In the first example Ilya had two winning moves: to the empty cell in the left column and to the leftmost empty cell in the first row.
In the second example it wasn't possible to win by making single turn.
In the third example Ilya could have won by placing X in the last row between two existing Xs.
In the fourth example it wasn't possible to win by making single turn.
|
```python
s=[]
for i in range(4):
s1=input()
s.append(s1)
a=set()
for i in range(4):
s1=s[i][0:3]
s2=s[i][1:4]
s3=s[0][i]+s[1][i]+s[2][i]
s4=s[1][i]+s[2][i]+s[3][i]
a.add(s1)
a.add(s2)
a.add(s3)
a.add(s4)
for i in range(0,2):
for j in range(2):
a.add(s[i][j]+s[i+1][j+1]+s[i+2][j+2])
a.add(s[i][3-j]+s[i+1][2-j]+s[i+2][1-j])
s1='xx.'
s2='x.x'
s3='.xx'
if(s1 in a or s2 in a or s3 in a):
print("YES")
else:
print("NO")
```
| 3
|
|
381
|
A
|
Sereja and Dima
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"two pointers"
] | null | null |
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
|
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
|
[
"4\n4 1 2 10\n",
"7\n1 2 3 4 5 6 7\n"
] |
[
"12 5\n",
"16 12\n"
] |
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
| 500
|
[
{
"input": "4\n4 1 2 10",
"output": "12 5"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "16 12"
},
{
"input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13",
"output": "613 418"
},
{
"input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24",
"output": "644 500"
},
{
"input": "1\n3",
"output": "3 0"
},
{
"input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646",
"output": "6848 6568"
},
{
"input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727",
"output": "9562 9561"
},
{
"input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12",
"output": "315 315"
},
{
"input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304",
"output": "3238 2222"
},
{
"input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325",
"output": "5246 4864"
},
{
"input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71",
"output": "8147 7807"
},
{
"input": "1\n1",
"output": "1 0"
}
] | 1,669,581,455
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 46
| 0
|
card = int(input(''))
arr = list(map(int,input().split(' ')))
khaled = 0
yahya = 0
start = 0
end = card -1
winner = 1
while card != 0 :
result = 0
card -=1
if arr[start] > arr[end]:
result = arr[start]
start = start +1
else :
result = arr[end]
end = end - 1
if winner == 1 :
khaled = khaled +result
winner = 2
else :
yahya = yahya + result
winner = 1
print (khaled , yahya)
|
Title: Sereja and Dima
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output Specification:
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Demo Input:
['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n']
Demo Output:
['12 5\n', '16 12\n']
Note:
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
|
```python
card = int(input(''))
arr = list(map(int,input().split(' ')))
khaled = 0
yahya = 0
start = 0
end = card -1
winner = 1
while card != 0 :
result = 0
card -=1
if arr[start] > arr[end]:
result = arr[start]
start = start +1
else :
result = arr[end]
end = end - 1
if winner == 1 :
khaled = khaled +result
winner = 2
else :
yahya = yahya + result
winner = 1
print (khaled , yahya)
```
| 3
|
|
821
|
C
|
Okabe and Boxes
|
PROGRAMMING
| 1,500
|
[
"data structures",
"greedy",
"trees"
] | null | null |
Okabe and Super Hacker Daru are stacking and removing boxes. There are *n* boxes numbered from 1 to *n*. Initially there are no boxes on the stack.
Okabe, being a control freak, gives Daru 2*n* commands: *n* of which are to add a box to the top of the stack, and *n* of which are to remove a box from the top of the stack and throw it in the trash. Okabe wants Daru to throw away the boxes in the order from 1 to *n*. Of course, this means that it might be impossible for Daru to perform some of Okabe's remove commands, because the required box is not on the top of the stack.
That's why Daru can decide to wait until Okabe looks away and then reorder the boxes in the stack in any way he wants. He can do it at any point of time between Okabe's commands, but he can't add or remove boxes while he does it.
Tell Daru the minimum number of times he needs to reorder the boxes so that he can successfully complete all of Okabe's commands. It is guaranteed that every box is added before it is required to be removed.
|
The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of boxes.
Each of the next 2*n* lines of input starts with a string "add" or "remove". If the line starts with the "add", an integer *x* (1<=≤<=*x*<=≤<=*n*) follows, indicating that Daru should add the box with number *x* to the top of the stack.
It is guaranteed that exactly *n* lines contain "add" operations, all the boxes added are distinct, and *n* lines contain "remove" operations. It is also guaranteed that a box is always added before it is required to be removed.
|
Print the minimum number of times Daru needs to reorder the boxes to successfully complete all of Okabe's commands.
|
[
"3\nadd 1\nremove\nadd 2\nadd 3\nremove\nremove\n",
"7\nadd 3\nadd 2\nadd 1\nremove\nadd 4\nremove\nremove\nremove\nadd 6\nadd 7\nadd 5\nremove\nremove\nremove\n"
] |
[
"1\n",
"2\n"
] |
In the first sample, Daru should reorder the boxes after adding box 3 to the stack.
In the second sample, Daru should reorder the boxes after adding box 4 and box 7 to the stack.
| 1,500
|
[
{
"input": "3\nadd 1\nremove\nadd 2\nadd 3\nremove\nremove",
"output": "1"
},
{
"input": "7\nadd 3\nadd 2\nadd 1\nremove\nadd 4\nremove\nremove\nremove\nadd 6\nadd 7\nadd 5\nremove\nremove\nremove",
"output": "2"
},
{
"input": "4\nadd 1\nadd 3\nremove\nadd 4\nadd 2\nremove\nremove\nremove",
"output": "2"
},
{
"input": "2\nadd 1\nremove\nadd 2\nremove",
"output": "0"
},
{
"input": "1\nadd 1\nremove",
"output": "0"
},
{
"input": "15\nadd 12\nadd 7\nadd 10\nadd 11\nadd 5\nadd 2\nadd 1\nadd 6\nadd 8\nremove\nremove\nadd 15\nadd 4\nadd 13\nadd 9\nadd 3\nadd 14\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "2"
},
{
"input": "14\nadd 7\nadd 2\nadd 13\nadd 5\nadd 12\nadd 6\nadd 4\nadd 1\nadd 14\nremove\nadd 10\nremove\nadd 9\nadd 8\nadd 11\nadd 3\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "3"
},
{
"input": "11\nadd 10\nadd 9\nadd 11\nadd 1\nadd 5\nadd 6\nremove\nadd 3\nadd 8\nadd 2\nadd 4\nremove\nremove\nremove\nremove\nremove\nadd 7\nremove\nremove\nremove\nremove\nremove",
"output": "2"
},
{
"input": "3\nadd 3\nadd 2\nadd 1\nremove\nremove\nremove",
"output": "0"
},
{
"input": "4\nadd 1\nadd 3\nadd 4\nremove\nadd 2\nremove\nremove\nremove",
"output": "1"
},
{
"input": "6\nadd 3\nadd 4\nadd 5\nadd 1\nadd 6\nremove\nadd 2\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "16\nadd 1\nadd 2\nadd 3\nadd 4\nadd 5\nadd 6\nadd 7\nadd 8\nadd 9\nadd 10\nadd 11\nadd 12\nadd 13\nadd 14\nadd 15\nadd 16\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "2\nadd 2\nadd 1\nremove\nremove",
"output": "0"
},
{
"input": "17\nadd 1\nadd 2\nadd 3\nadd 4\nadd 5\nadd 6\nadd 7\nadd 8\nadd 9\nadd 10\nadd 11\nadd 12\nadd 13\nadd 14\nadd 15\nadd 16\nadd 17\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "18\nadd 1\nadd 2\nadd 3\nadd 4\nadd 5\nadd 6\nadd 7\nadd 8\nadd 9\nadd 10\nadd 11\nadd 12\nadd 13\nadd 14\nadd 15\nadd 16\nadd 17\nadd 18\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "4\nadd 1\nadd 2\nremove\nremove\nadd 4\nadd 3\nremove\nremove",
"output": "1"
},
{
"input": "19\nadd 1\nadd 2\nadd 3\nadd 4\nadd 5\nadd 6\nadd 7\nadd 8\nadd 9\nadd 10\nadd 11\nadd 12\nadd 13\nadd 14\nadd 15\nadd 16\nadd 17\nadd 18\nadd 19\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "5\nadd 4\nadd 3\nadd 1\nremove\nadd 2\nremove\nremove\nadd 5\nremove\nremove",
"output": "1"
},
{
"input": "7\nadd 4\nadd 6\nadd 1\nadd 5\nadd 7\nremove\nadd 2\nremove\nadd 3\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "8\nadd 1\nadd 2\nadd 3\nadd 7\nadd 8\nremove\nremove\nremove\nadd 6\nadd 5\nadd 4\nremove\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "4\nadd 1\nadd 4\nremove\nadd 3\nadd 2\nremove\nremove\nremove",
"output": "1"
},
{
"input": "7\nadd 1\nadd 2\nadd 3\nadd 5\nadd 7\nremove\nremove\nremove\nadd 4\nremove\nremove\nadd 6\nremove\nremove",
"output": "1"
},
{
"input": "4\nadd 4\nadd 1\nadd 2\nremove\nremove\nadd 3\nremove\nremove",
"output": "1"
},
{
"input": "5\nadd 1\nadd 3\nadd 4\nadd 5\nremove\nadd 2\nremove\nremove\nremove\nremove",
"output": "1"
},
{
"input": "5\nadd 2\nadd 1\nremove\nremove\nadd 5\nadd 3\nremove\nadd 4\nremove\nremove",
"output": "0"
},
{
"input": "9\nadd 3\nadd 2\nadd 1\nadd 4\nadd 6\nadd 9\nremove\nremove\nremove\nremove\nadd 5\nremove\nremove\nadd 8\nadd 7\nremove\nremove\nremove",
"output": "1"
},
{
"input": "10\nadd 9\nadd 10\nadd 4\nadd 3\nadd 2\nadd 1\nremove\nremove\nremove\nremove\nadd 8\nadd 7\nadd 5\nadd 6\nremove\nremove\nremove\nremove\nremove\nremove",
"output": "1"
}
] | 1,596,410,483
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 8
| 3,000
| 30,924,800
|
n = int(input())
stack = []
nxt = 1
ans = 0
l = 0
for _ in range(2*n):
cmd = input()
if cmd[:3] == "add":
_,b = cmd.split()
b = int(b)
stack.append(b)
else:
if not stack: nxt += 1;continue
if stack.pop() == nxt:
nxt += 1
else:
stack = []
nxt += 1
ans += 1
print(ans)
|
Title: Okabe and Boxes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Okabe and Super Hacker Daru are stacking and removing boxes. There are *n* boxes numbered from 1 to *n*. Initially there are no boxes on the stack.
Okabe, being a control freak, gives Daru 2*n* commands: *n* of which are to add a box to the top of the stack, and *n* of which are to remove a box from the top of the stack and throw it in the trash. Okabe wants Daru to throw away the boxes in the order from 1 to *n*. Of course, this means that it might be impossible for Daru to perform some of Okabe's remove commands, because the required box is not on the top of the stack.
That's why Daru can decide to wait until Okabe looks away and then reorder the boxes in the stack in any way he wants. He can do it at any point of time between Okabe's commands, but he can't add or remove boxes while he does it.
Tell Daru the minimum number of times he needs to reorder the boxes so that he can successfully complete all of Okabe's commands. It is guaranteed that every box is added before it is required to be removed.
Input Specification:
The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of boxes.
Each of the next 2*n* lines of input starts with a string "add" or "remove". If the line starts with the "add", an integer *x* (1<=≤<=*x*<=≤<=*n*) follows, indicating that Daru should add the box with number *x* to the top of the stack.
It is guaranteed that exactly *n* lines contain "add" operations, all the boxes added are distinct, and *n* lines contain "remove" operations. It is also guaranteed that a box is always added before it is required to be removed.
Output Specification:
Print the minimum number of times Daru needs to reorder the boxes to successfully complete all of Okabe's commands.
Demo Input:
['3\nadd 1\nremove\nadd 2\nadd 3\nremove\nremove\n', '7\nadd 3\nadd 2\nadd 1\nremove\nadd 4\nremove\nremove\nremove\nadd 6\nadd 7\nadd 5\nremove\nremove\nremove\n']
Demo Output:
['1\n', '2\n']
Note:
In the first sample, Daru should reorder the boxes after adding box 3 to the stack.
In the second sample, Daru should reorder the boxes after adding box 4 and box 7 to the stack.
|
```python
n = int(input())
stack = []
nxt = 1
ans = 0
l = 0
for _ in range(2*n):
cmd = input()
if cmd[:3] == "add":
_,b = cmd.split()
b = int(b)
stack.append(b)
else:
if not stack: nxt += 1;continue
if stack.pop() == nxt:
nxt += 1
else:
stack = []
nxt += 1
ans += 1
print(ans)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word — some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction — it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word *s*. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word *abacabaca* is given. This word can be obtained in the following ways: , where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {*aca*,<=*ba*,<=*ca*}.
|
The only line contains a string *s* (5<=≤<=|*s*|<=≤<=104) consisting of lowercase English letters.
|
On the first line print integer *k* — a number of distinct possible suffixes. On the next *k* lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
|
[
"abacabaca\n",
"abaca\n"
] |
[
"3\naca\nba\nca\n",
"0\n"
] |
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
| 0
|
[
{
"input": "abacabaca",
"output": "3\naca\nba\nca"
},
{
"input": "abaca",
"output": "0"
},
{
"input": "gzqgchv",
"output": "1\nhv"
},
{
"input": "iosdwvzerqfi",
"output": "9\ner\nerq\nfi\nqfi\nrq\nvz\nvze\nze\nzer"
},
{
"input": "oawtxikrpvfuzugjweki",
"output": "25\neki\nfu\nfuz\ngj\ngjw\nik\nikr\njw\njwe\nki\nkr\nkrp\npv\npvf\nrp\nrpv\nug\nugj\nuz\nuzu\nvf\nvfu\nwe\nzu\nzug"
},
{
"input": "abcdexyzzzz",
"output": "5\nxyz\nyz\nyzz\nzz\nzzz"
},
{
"input": "affviytdmexpwfqplpyrlniprbdphrcwlboacoqec",
"output": "67\nac\naco\nbd\nbdp\nbo\nboa\nco\ncoq\ncw\ncwl\ndm\ndme\ndp\ndph\nec\nex\nexp\nfq\nfqp\nhr\nhrc\nip\nipr\nlb\nlbo\nln\nlni\nlp\nlpy\nme\nmex\nni\nnip\noa\noac\noq\nph\nphr\npl\nplp\npr\nprb\npw\npwf\npy\npyr\nqec\nqp\nqpl\nrb\nrbd\nrc\nrcw\nrl\nrln\ntd\ntdm\nwf\nwfq\nwl\nwlb\nxp\nxpw\nyr\nyrl\nyt\nytd"
},
{
"input": "lmnxtobrknqjvnzwadpccrlvisxyqbxxmghvl",
"output": "59\nad\nadp\nbr\nbrk\nbx\nbxx\ncc\nccr\ncr\ncrl\ndp\ndpc\ngh\nhvl\nis\nisx\njv\njvn\nkn\nknq\nlv\nlvi\nmg\nmgh\nnq\nnqj\nnz\nnzw\nob\nobr\npc\npcc\nqb\nqbx\nqj\nqjv\nrk\nrkn\nrl\nrlv\nsx\nsxy\nvi\nvis\nvl\nvn\nvnz\nwa\nwad\nxm\nxmg\nxx\nxxm\nxy\nxyq\nyq\nyqb\nzw\nzwa"
},
{
"input": "tbdbdpkluawodlrwldjgplbiylrhuywkhafbkiuoppzsjxwbaqqiwagprqtoauowtaexrhbmctcxwpmplkyjnpwukzwqrqpv",
"output": "170\nae\naex\naf\nafb\nag\nagp\naq\naqq\nau\nauo\naw\nawo\nba\nbaq\nbi\nbiy\nbk\nbki\nbm\nbmc\nct\nctc\ncx\ncxw\ndj\ndjg\ndl\ndlr\nex\nexr\nfb\nfbk\ngp\ngpl\ngpr\nha\nhaf\nhb\nhbm\nhu\nhuy\niu\niuo\niw\niwa\niy\niyl\njg\njgp\njn\njnp\njx\njxw\nkh\nkha\nki\nkiu\nkl\nklu\nky\nkyj\nkz\nkzw\nlb\nlbi\nld\nldj\nlk\nlky\nlr\nlrh\nlrw\nlu\nlua\nmc\nmct\nmp\nmpl\nnp\nnpw\noa\noau\nod\nodl\nop\nopp\now\nowt\npk\npkl\npl\nplb\nplk\npm\npmp\npp\nppz\npr\nprq\npv\npw\npwu\npz\npzs\nqi\nqiw\nqpv\nqq\nqqi\nqr\nqrq\nqt\nq..."
},
{
"input": "caqmjjtwmqxytcsawfufvlofqcqdwnyvywvbbhmpzqwqqxieptiaguwvqdrdftccsglgfezrzhstjcxdknftpyslyqdmkwdolwbusyrgyndqllgesktvgarpfkiglxgtcfepclqhgfbfmkymsszrtynlxbosmrvntsqwccdtahkpnelwiqn",
"output": "323\nag\nagu\nah\nahk\nar\narp\naw\nawf\nbb\nbbh\nbf\nbfm\nbh\nbhm\nbo\nbos\nbu\nbus\ncc\nccd\nccs\ncd\ncdt\ncf\ncfe\ncl\nclq\ncq\ncqd\ncs\ncsa\ncsg\ncx\ncxd\ndf\ndft\ndk\ndkn\ndm\ndmk\ndo\ndol\ndq\ndql\ndr\ndrd\ndt\ndta\ndw\ndwn\nel\nelw\nep\nepc\nept\nes\nesk\nez\nezr\nfb\nfbf\nfe\nfep\nfez\nfk\nfki\nfm\nfmk\nfq\nfqc\nft\nftc\nftp\nfu\nfuf\nfv\nfvl\nga\ngar\nge\nges\ngf\ngfb\ngfe\ngl\nglg\nglx\ngt\ngtc\ngu\nguw\ngy\ngyn\nhg\nhgf\nhk\nhkp\nhm\nhmp\nhs\nhst\nia\niag\nie\niep\nig\nigl\niqn\njc\njcx\njt\njtw..."
},
{
"input": "prntaxhysjfcfmrjngdsitlguahtpnwgbaxptubgpwcfxqehrulbxfcjssgocqncscduvyvarvwxzvmjoatnqfsvsilubexmwugedtzavyamqjqtkxzuslielibjnvkpvyrbndehsqcaqzcrmomqqwskwcypgqoawxdutnxmeivnfpzwvxiyscbfnloqjhjacsfnkfmbhgzpujrqdbaemjsqphokkiplblbflvadcyykcqrdohfasstobwrobslaofbasylwiizrpozvhtwyxtzl",
"output": "505\nac\nacs\nad\nadc\nae\naem\nah\naht\nam\namq\nao\naof\naq\naqz\nar\narv\nas\nass\nasy\nat\natn\nav\navy\naw\nawx\nax\naxp\nba\nbae\nbas\nbax\nbe\nbex\nbf\nbfl\nbfn\nbg\nbgp\nbh\nbhg\nbj\nbjn\nbl\nblb\nbn\nbnd\nbs\nbsl\nbw\nbwr\nbx\nbxf\nca\ncaq\ncb\ncbf\ncd\ncdu\ncf\ncfm\ncfx\ncj\ncjs\ncq\ncqn\ncqr\ncr\ncrm\ncs\ncsc\ncsf\ncy\ncyp\ncyy\ndb\ndba\ndc\ndcy\nde\ndeh\ndo\ndoh\nds\ndsi\ndt\ndtz\ndu\ndut\nduv\ned\nedt\neh\nehr\nehs\nei\neiv\nel\neli\nem\nemj\nex\nexm\nfa\nfas\nfb\nfba\nfc\nfcf\nfcj\nfl\nflv\nf..."
},
{
"input": "gvtgnjyfvnuhagulgmjlqzpvxsygmikofsnvkuplnkxeibnicygpvfvtebppadpdnrxjodxdhxqceaulbfxogwrigstsjudhkgwkhseuwngbppisuzvhzzxxbaggfngmevksbrntpprxvcczlalutdzhwmzbalkqmykmodacjrmwhwugyhwlrbnqxsznldmaxpndwmovcolowxhj",
"output": "375\nac\nacj\nad\nadp\nag\nagg\nagu\nal\nalk\nalu\nau\naul\nax\naxp\nba\nbag\nbal\nbf\nbfx\nbn\nbni\nbnq\nbp\nbpp\nbr\nbrn\ncc\nccz\nce\ncea\ncj\ncjr\nco\ncol\ncy\ncyg\ncz\nczl\nda\ndac\ndh\ndhk\ndhx\ndm\ndma\ndn\ndnr\ndp\ndpd\ndw\ndwm\ndx\ndxd\ndz\ndzh\nea\neau\neb\nebp\nei\neib\neu\neuw\nev\nevk\nfn\nfng\nfs\nfsn\nfv\nfvn\nfvt\nfx\nfxo\ngb\ngbp\ngf\ngfn\ngg\nggf\ngm\ngme\ngmi\ngmj\ngp\ngpv\ngs\ngst\ngu\ngul\ngw\ngwk\ngwr\ngy\ngyh\nha\nhag\nhj\nhk\nhkg\nhs\nhse\nhw\nhwl\nhwm\nhwu\nhx\nhxq\nhz\nhzz\nib\nib..."
},
{
"input": "topqexoicgzjmssuxnswdhpwbsqwfhhziwqibjgeepcvouhjezlomobgireaxaceppoxfxvkwlvgwtjoiplihbpsdhczddwfvcbxqqmqtveaunshmobdlkmmfyajjlkhxnvfmibtbbqswrhcfwytrccgtnlztkddrevkfovunuxtzhhhnorecyfgmlqcwjfjtqegxagfiuqtpjpqlwiefofpatxuqxvikyynncsueynmigieototnbcwxavlbgeqao",
"output": "462\nac\nace\nag\nagf\naj\najj\nao\nat\natx\nau\naun\nav\navl\nax\naxa\nbb\nbbq\nbc\nbcw\nbd\nbdl\nbg\nbge\nbgi\nbj\nbjg\nbp\nbps\nbq\nbqs\nbs\nbsq\nbt\nbtb\nbx\nbxq\ncb\ncbx\ncc\nccg\nce\ncep\ncf\ncfw\ncg\ncgt\ncgz\ncs\ncsu\ncv\ncvo\ncw\ncwj\ncwx\ncy\ncyf\ncz\nczd\ndd\nddr\nddw\ndh\ndhc\ndhp\ndl\ndlk\ndr\ndre\ndw\ndwf\nea\neau\neax\nec\necy\nee\neep\nef\nefo\neg\negx\neo\neot\nep\nepc\nepp\neq\nev\nevk\ney\neyn\nez\nezl\nfg\nfgm\nfh\nfhh\nfi\nfiu\nfj\nfjt\nfm\nfmi\nfo\nfof\nfov\nfp\nfpa\nfv\nfvc\nfw\nfwy\n..."
},
{
"input": "lcrjhbybgamwetyrppxmvvxiyufdkcotwhmptefkqxjhrknjdponulsynpkgszhbkeinpnjdonjfwzbsaweqwlsvuijauwezfydktfljxgclpxpknhygdqyiapvzudyyqomgnsrdhhxhsrdfrwnxdolkmwmw",
"output": "276\nam\namw\nap\napv\nau\nauw\naw\nawe\nbg\nbga\nbk\nbke\nbs\nbsa\nby\nbyb\ncl\nclp\nco\ncot\ndf\ndfr\ndh\ndhh\ndk\ndkc\ndkt\ndo\ndol\ndon\ndp\ndpo\ndq\ndqy\ndy\ndyy\nef\nefk\nei\nein\neq\neqw\net\nety\nez\nezf\nfd\nfdk\nfk\nfkq\nfl\nflj\nfr\nfrw\nfw\nfwz\nfy\nfyd\nga\ngam\ngc\ngcl\ngd\ngdq\ngn\ngns\ngs\ngsz\nhb\nhbk\nhh\nhhx\nhm\nhmp\nhr\nhrk\nhs\nhsr\nhx\nhxh\nhy\nhyg\nia\niap\nij\nija\nin\ninp\niy\niyu\nja\njau\njd\njdo\njdp\njf\njfw\njh\njhr\njx\njxg\nkc\nkco\nke\nkei\nkg\nkgs\nkm\nkmw\nkn\nknh\nknj\n..."
},
{
"input": "hzobjysjhbebobkoror",
"output": "20\nbe\nbeb\nbko\nbo\nbob\neb\nebo\nhb\nhbe\njh\njhb\nko\nkor\nob\nor\nror\nsj\nsjh\nys\nysj"
},
{
"input": "safgmgpzljarfswowdxqhuhypxcmiddyvehjtnlflzknznrukdsbatxoytzxkqngopeipbythhbhfkvlcdxwqrxumbtbgiosjnbeorkzsrfarqofsrcwsfpyheaszjpkjysrcxbzebkxzovdchhososo",
"output": "274\nar\narf\narq\nas\nasz\nat\natx\nba\nbat\nbe\nbeo\nbg\nbgi\nbh\nbhf\nbk\nbkx\nbt\nbtb\nby\nbyt\nbz\nbze\ncd\ncdx\nch\nchh\ncm\ncmi\ncw\ncws\ncx\ncxb\ndc\ndch\ndd\nddy\nds\ndsb\ndx\ndxq\ndxw\ndy\ndyv\nea\neas\neb\nebk\neh\nehj\nei\neip\neo\neor\nfa\nfar\nfk\nfkv\nfl\nflz\nfp\nfpy\nfs\nfsr\nfsw\ngi\ngio\ngo\ngop\ngp\ngpz\nhb\nhbh\nhe\nhea\nhf\nhfk\nhh\nhhb\nhj\nhjt\nhos\nhu\nhuh\nhy\nhyp\nid\nidd\nio\nios\nip\nipb\nja\njar\njn\njnb\njp\njpk\njt\njtn\njy\njys\nkd\nkds\nkj\nkjy\nkn\nknz\nkq\nkqn\nkv\nkvl\n..."
},
{
"input": "glaoyryxrgsysy",
"output": "10\ngs\ngsy\nrgs\nry\nryx\nsy\nxr\nysy\nyx\nyxr"
},
{
"input": "aaaaaxyxxxx",
"output": "5\nxx\nxxx\nxyx\nyx\nyxx"
},
{
"input": "aaaaax",
"output": "0"
},
{
"input": "aaaaaxx",
"output": "1\nxx"
},
{
"input": "aaaaaaa",
"output": "1\naa"
},
{
"input": "aaaaaxxx",
"output": "2\nxx\nxxx"
},
{
"input": "aaaaayxx",
"output": "2\nxx\nyxx"
},
{
"input": "aaaaaxyz",
"output": "2\nxyz\nyz"
},
{
"input": "aaaaaxyxy",
"output": "2\nxy\nyxy"
},
{
"input": "aaaxyyxyy",
"output": "3\nxyy\nyx\nyy"
},
{
"input": "aaaaaxxxxxx",
"output": "2\nxx\nxxx"
},
{
"input": "aaaaaxxxxx",
"output": "2\nxx\nxxx"
},
{
"input": "aaaaaxyzxyxy",
"output": "5\nxy\nyxy\nyzx\nzx\nzxy"
},
{
"input": "aaaaadddgggg",
"output": "6\ndd\nddg\ndg\ndgg\ngg\nggg"
},
{
"input": "abcdeabzzzzzzzz",
"output": "5\nab\nabz\nbz\nzz\nzzz"
},
{
"input": "bbbbbccaaaaaa",
"output": "4\naa\naaa\nca\ncca"
},
{
"input": "xxxxxababc",
"output": "5\nab\naba\nabc\nba\nbc"
},
{
"input": "dddddaabbbbbb",
"output": "4\naab\nab\nbb\nbbb"
},
{
"input": "xxxxxababe",
"output": "5\nab\naba\nabe\nba\nbe"
},
{
"input": "aaaaababaaaaaaaaaaaa",
"output": "6\naa\naaa\nab\nba\nbaa\nbab"
}
] | 1,462,480,198
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 1
| 62
| 4,608,000
|
def solve(data):
n=len(data)
res=[]
if n<=6:
return 0
else:
for i in range(5,n-2,1):
if (i != n-3):
res.append(data[i:i+2])
if (i != n-4):
res.append(data[i:i+3])
res.append(data[-2:])
res = sorted(list(set(res)))
return res
data=input()
sol=solve(data)
print(len(sol))
for x in sol:
print(x)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
First-rate specialists graduate from Berland State Institute of Peace and Friendship. You are one of the most talented students in this university. The education is not easy because you need to have fundamental knowledge in different areas, which sometimes are not related to each other.
For example, you should know linguistics very well. You learn a structure of Reberland language as foreign language. In this language words are constructed according to the following rules. First you need to choose the "root" of the word — some string which has more than 4 letters. Then several strings with the length 2 or 3 symbols are appended to this word. The only restriction — it is not allowed to append the same string twice in a row. All these strings are considered to be suffixes of the word (this time we use word "suffix" to describe a morpheme but not the few last characters of the string as you may used to).
Here is one exercise that you have found in your task list. You are given the word *s*. Find all distinct strings with the length 2 or 3, which can be suffixes of this word according to the word constructing rules in Reberland language.
Two strings are considered distinct if they have different length or there is a position in which corresponding characters do not match.
Let's look at the example: the word *abacabaca* is given. This word can be obtained in the following ways: , where the root of the word is overlined, and suffixes are marked by "corners". Thus, the set of possible suffixes for this word is {*aca*,<=*ba*,<=*ca*}.
Input Specification:
The only line contains a string *s* (5<=≤<=|*s*|<=≤<=104) consisting of lowercase English letters.
Output Specification:
On the first line print integer *k* — a number of distinct possible suffixes. On the next *k* lines print suffixes.
Print suffixes in lexicographical (alphabetical) order.
Demo Input:
['abacabaca\n', 'abaca\n']
Demo Output:
['3\naca\nba\nca\n', '0\n']
Note:
The first test was analysed in the problem statement.
In the second example the length of the string equals 5. The length of the root equals 5, so no string can be used as a suffix.
|
```python
def solve(data):
n=len(data)
res=[]
if n<=6:
return 0
else:
for i in range(5,n-2,1):
if (i != n-3):
res.append(data[i:i+2])
if (i != n-4):
res.append(data[i:i+3])
res.append(data[-2:])
res = sorted(list(set(res)))
return res
data=input()
sol=solve(data)
print(len(sol))
for x in sol:
print(x)
```
| -1
|
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,609,182,175
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 108
| 0
|
n = input()
h1 = n.find("h")
e1 = n.find("e")
l1 = n.find("ll")
o1 = n.find("o")
if h1 != -1 and e1 != -1 and l1 !=0 and o1 != -1:
if h1 < e1 and e1 < l1 and l1 < o1:
print("YES")
exit(0)
print("NO")
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
n = input()
h1 = n.find("h")
e1 = n.find("e")
l1 = n.find("ll")
o1 = n.find("o")
if h1 != -1 and e1 != -1 and l1 !=0 and o1 != -1:
if h1 < e1 and e1 < l1 and l1 < o1:
print("YES")
exit(0)
print("NO")
```
| 0
|
92
|
A
|
Chips
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] |
A. Chips
|
2
|
256
|
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
|
Print the number of chips the presenter ended up with.
|
[
"4 11\n",
"17 107\n",
"3 8\n"
] |
[
"0\n",
"2\n",
"1\n"
] |
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.
| 500
|
[
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "25 9712",
"output": "11"
},
{
"input": "9 7601",
"output": "5"
},
{
"input": "1 9058",
"output": "0"
},
{
"input": "29 7772",
"output": "26"
},
{
"input": "45 9465",
"output": "14"
},
{
"input": "46 866",
"output": "5"
},
{
"input": "29 1241",
"output": "20"
},
{
"input": "17 4248",
"output": "12"
},
{
"input": "20 8082",
"output": "11"
},
{
"input": "50 9555",
"output": "0"
},
{
"input": "4 7455",
"output": "2"
},
{
"input": "36 880",
"output": "4"
},
{
"input": "24 7440",
"output": "9"
},
{
"input": "44 7888",
"output": "12"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "50 10000",
"output": "40"
},
{
"input": "1 10000",
"output": "0"
},
{
"input": "50 1",
"output": "0"
},
{
"input": "50 50",
"output": "5"
}
] | 1,598,358,206
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 216
| 0
|
n,m=map(int,input().split())
x=0
flag=0
while x<m:
for i in range(1,n+1):
if x<m:
x+=i
if x>m:
x-=i
flag=1
break
if flag==1:
break
print(m-x)
|
Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip.
|
```python
n,m=map(int,input().split())
x=0
flag=0
while x<m:
for i in range(1,n+1):
if x<m:
x+=i
if x>m:
x-=i
flag=1
break
if flag==1:
break
print(m-x)
```
| 3.946
|
1,011
|
A
|
Stages
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once.
|
The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once.
|
Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all.
|
[
"5 3\nxyabd\n",
"7 4\nproblem\n",
"2 2\nab\n",
"12 1\nabaabbaaabbb\n"
] |
[
"29",
"34",
"-1",
"1"
] |
In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1.
| 500
|
[
{
"input": "5 3\nxyabd",
"output": "29"
},
{
"input": "7 4\nproblem",
"output": "34"
},
{
"input": "2 2\nab",
"output": "-1"
},
{
"input": "12 1\nabaabbaaabbb",
"output": "1"
},
{
"input": "50 13\nqwertyuiopasdfghjklzxcvbnmaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "169"
},
{
"input": "50 14\nqwertyuiopasdfghjklzxcvbnmaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "1 1\na",
"output": "1"
},
{
"input": "50 1\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "1"
},
{
"input": "50 2\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "13 13\nuwgmkyqeiaocs",
"output": "169"
},
{
"input": "13 13\nhzdxpbfvrltnj",
"output": "182"
},
{
"input": "1 1\nn",
"output": "14"
},
{
"input": "10 8\nsmzeblyjqw",
"output": "113"
},
{
"input": "20 20\ntzmvhskkyugkuuxpvtbh",
"output": "-1"
},
{
"input": "30 15\nwjzolzzkfulwgioksfxmcxmnnjtoav",
"output": "-1"
},
{
"input": "40 30\nxumfrflllrrgswehqtsskefixhcxjrxbjmrpsshv",
"output": "-1"
},
{
"input": "50 31\nahbyyoxltryqdmvenemaqnbakglgqolxnaifnqtoclnnqiabpz",
"output": "-1"
},
{
"input": "10 7\niuiukrxcml",
"output": "99"
},
{
"input": "38 2\nvjzarfykmrsrvwbwfwldsulhxtykmjbnwmdufa",
"output": "5"
},
{
"input": "12 6\nfwseyrarkwcd",
"output": "61"
},
{
"input": "2 2\nac",
"output": "4"
},
{
"input": "1 1\nc",
"output": "3"
},
{
"input": "2 2\nad",
"output": "5"
},
{
"input": "2 1\nac",
"output": "1"
},
{
"input": "4 3\nadjz",
"output": "15"
},
{
"input": "3 3\naoz",
"output": "42"
},
{
"input": "3 1\nzzz",
"output": "26"
},
{
"input": "2 1\nxz",
"output": "24"
},
{
"input": "5 1\naaddd",
"output": "1"
}
] | 1,532,824,383
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 109
| 0
|
n = input("n:")
k = input("k:")
if k==0 :
print("0")
else :
n=int(n)
k=int(k)
a = []
for i in range(0,n) :
a.append(ord(input("->"))-96)
a = sorted(a)
j = 1
s=a[0]
i=1
while j<k and i<len(a) :
if a[i]>a[i-1]+1 :
s +=a[i]
j+=1
i+=1
if j<k :
print(-1)
else :
print(s)
|
Title: Stages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once.
Input Specification:
The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once.
Output Specification:
Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all.
Demo Input:
['5 3\nxyabd\n', '7 4\nproblem\n', '2 2\nab\n', '12 1\nabaabbaaabbb\n']
Demo Output:
['29', '34', '-1', '1']
Note:
In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1.
|
```python
n = input("n:")
k = input("k:")
if k==0 :
print("0")
else :
n=int(n)
k=int(k)
a = []
for i in range(0,n) :
a.append(ord(input("->"))-96)
a = sorted(a)
j = 1
s=a[0]
i=1
while j<k and i<len(a) :
if a[i]>a[i-1]+1 :
s +=a[i]
j+=1
i+=1
if j<k :
print(-1)
else :
print(s)
```
| -1
|
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,574,775,530
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 93
| 0
|
word = input()
count = 0
for a in 'holle':
if a in word:
b = word.find(a)
word = word[b:]
count += 1
if count >= 5:
print('YES')
else:
print('NO')
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
word = input()
count = 0
for a in 'holle':
if a in word:
b = word.find(a)
word = word[b:]
count += 1
if count >= 5:
print('YES')
else:
print('NO')
```
| 0
|
108
|
A
|
Palindromic Times
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] |
A. Palindromic Times
|
2
|
256
|
Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him.
|
The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits.
|
Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time.
|
[
"12:21\n",
"23:59\n"
] |
[
"13:31\n",
"00:00\n"
] |
none
| 500
|
[
{
"input": "12:21",
"output": "13:31"
},
{
"input": "23:59",
"output": "00:00"
},
{
"input": "15:51",
"output": "20:02"
},
{
"input": "10:44",
"output": "11:11"
},
{
"input": "04:02",
"output": "04:40"
},
{
"input": "02:11",
"output": "02:20"
},
{
"input": "12:15",
"output": "12:21"
},
{
"input": "07:07",
"output": "10:01"
},
{
"input": "00:17",
"output": "01:10"
},
{
"input": "04:55",
"output": "05:50"
},
{
"input": "02:17",
"output": "02:20"
},
{
"input": "07:56",
"output": "10:01"
},
{
"input": "00:29",
"output": "01:10"
},
{
"input": "23:31",
"output": "23:32"
},
{
"input": "19:30",
"output": "20:02"
},
{
"input": "12:14",
"output": "12:21"
},
{
"input": "17:32",
"output": "20:02"
},
{
"input": "03:44",
"output": "04:40"
},
{
"input": "07:15",
"output": "10:01"
},
{
"input": "18:42",
"output": "20:02"
},
{
"input": "08:56",
"output": "10:01"
},
{
"input": "04:50",
"output": "05:50"
},
{
"input": "14:32",
"output": "14:41"
},
{
"input": "23:23",
"output": "23:32"
},
{
"input": "08:35",
"output": "10:01"
},
{
"input": "03:32",
"output": "04:40"
},
{
"input": "07:59",
"output": "10:01"
},
{
"input": "14:12",
"output": "14:41"
},
{
"input": "23:52",
"output": "00:00"
},
{
"input": "16:36",
"output": "20:02"
},
{
"input": "17:50",
"output": "20:02"
},
{
"input": "06:59",
"output": "10:01"
},
{
"input": "16:50",
"output": "20:02"
},
{
"input": "00:00",
"output": "01:10"
},
{
"input": "23:59",
"output": "00:00"
},
{
"input": "23:33",
"output": "00:00"
}
] | 1,513,303,590
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 5,529,600
|
s = input()
l = s.split(':')
if l[1] > l[0][::-1]:
l[0] = str(int(l[0]) + 1)
if l[0] == '24':
l[0] = '00'
if len(l[0])==1:
l[0] = '0' + l[0]
if l[0][1] >= '6':
if l[0][0] == '0':
print('10:01')
else:
print('20:02')
else:
print(l[0]+':'+l[0][::-1])
|
Title: Palindromic Times
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Tattah is asleep if and only if Tattah is attending a lecture. This is a well-known formula among Tattah's colleagues.
On a Wednesday afternoon, Tattah was attending Professor HH's lecture. At 12:21, right before falling asleep, he was staring at the digital watch around Saher's wrist. He noticed that the digits on the clock were the same when read from both directions i.e. a palindrome.
In his sleep, he started dreaming about such rare moments of the day when the time displayed on a digital clock is a palindrome. As soon as he woke up, he felt destined to write a program that finds the next such moment.
However, he still hasn't mastered the skill of programming while sleeping, so your task is to help him.
Input Specification:
The first and only line of the input starts with a string with the format "HH:MM" where "HH" is from "00" to "23" and "MM" is from "00" to "59". Both "HH" and "MM" have exactly two digits.
Output Specification:
Print the palindromic time of day that comes soonest after the time given in the input. If the input time is palindromic, output the soonest palindromic time after the input time.
Demo Input:
['12:21\n', '23:59\n']
Demo Output:
['13:31\n', '00:00\n']
Note:
none
|
```python
s = input()
l = s.split(':')
if l[1] > l[0][::-1]:
l[0] = str(int(l[0]) + 1)
if l[0] == '24':
l[0] = '00'
if len(l[0])==1:
l[0] = '0' + l[0]
if l[0][1] >= '6':
if l[0][0] == '0':
print('10:01')
else:
print('20:02')
else:
print(l[0]+':'+l[0][::-1])
```
| 0
|
260
|
A
|
Adding Digits
|
PROGRAMMING
| 1,400
|
[
"implementation",
"math"
] | null | null |
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
|
The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105).
|
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
|
[
"5 4 5\n",
"12 11 1\n",
"260 150 10\n"
] |
[
"524848\n",
"121\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "5 4 5",
"output": "524848"
},
{
"input": "12 11 1",
"output": "121"
},
{
"input": "260 150 10",
"output": "-1"
},
{
"input": "78843 5684 42717",
"output": "-1"
},
{
"input": "93248 91435 1133",
"output": "-1"
},
{
"input": "100000 10 64479",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 21 73839",
"output": "9999990000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99991 623 36438",
"output": "9999150000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99999 334 94854",
"output": "9999960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99252 9827 84849",
"output": "9925270000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99313 9833 10561",
"output": "9931330000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "94885 55815 11417",
"output": "9488550000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99492 58525 53481",
"output": "9949250000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99858 28531 79193",
"output": "9985850000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "99136 47208 42607",
"output": "9913680000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "63270 19953 5555",
"output": "-1"
},
{
"input": "10240 128 100000",
"output": "1024000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "12 11 3",
"output": "12100"
},
{
"input": "14 12 99998",
"output": "1440000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000..."
},
{
"input": "1 11 3",
"output": "1100"
},
{
"input": "3 40 1",
"output": "-1"
},
{
"input": "150 100 10",
"output": "1500000000000"
},
{
"input": "5 10 1",
"output": "50"
},
{
"input": "1 15 10",
"output": "15000000000"
},
{
"input": "3 13 2",
"output": "390"
}
] | 1,636,198,397
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 25
| 93
| 18,227,200
|
n,a,b = map(int,input().split())
m=0
for i in range(10):
if((n*10+i)%a==0):
n = n*10+i
m=1
break
if m:
print(str(n)+'0'*(b-1))
else:
print(-1)
|
Title: Adding Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
Input Specification:
The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105).
Output Specification:
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
Demo Input:
['5 4 5\n', '12 11 1\n', '260 150 10\n']
Demo Output:
['524848\n', '121\n', '-1\n']
Note:
none
|
```python
n,a,b = map(int,input().split())
m=0
for i in range(10):
if((n*10+i)%a==0):
n = n*10+i
m=1
break
if m:
print(str(n)+'0'*(b-1))
else:
print(-1)
```
| 3
|
|
110
|
A
|
Nearly Lucky Number
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Nearly Lucky Number
|
2
|
256
|
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number *n* is a nearly lucky number.
|
The only line contains an integer *n* (1<=≤<=*n*<=≤<=1018).
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
|
Print on the single line "YES" if *n* is a nearly lucky number. Otherwise, print "NO" (without the quotes).
|
[
"40047\n",
"7747774\n",
"1000000000000000000\n"
] |
[
"NO\n",
"YES\n",
"NO\n"
] |
In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".
In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".
In the third sample there are no lucky digits, so the answer is "NO".
| 500
|
[
{
"input": "40047",
"output": "NO"
},
{
"input": "7747774",
"output": "YES"
},
{
"input": "1000000000000000000",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "4",
"output": "NO"
},
{
"input": "474404774",
"output": "NO"
},
{
"input": "4744000695826",
"output": "YES"
},
{
"input": "10000000004744744",
"output": "YES"
},
{
"input": "446486416781684178",
"output": "YES"
},
{
"input": "999999999",
"output": "NO"
},
{
"input": "7777",
"output": "YES"
},
{
"input": "87414417444",
"output": "NO"
},
{
"input": "111222333444555667",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "4700",
"output": "NO"
},
{
"input": "3794555488744477",
"output": "NO"
},
{
"input": "444444444444444444",
"output": "NO"
},
{
"input": "474447447774444774",
"output": "NO"
},
{
"input": "777777777777777",
"output": "NO"
},
{
"input": "34777745021000000",
"output": "NO"
},
{
"input": "963",
"output": "NO"
},
{
"input": "855474448854788540",
"output": "NO"
},
{
"input": "999999999999994744",
"output": "YES"
},
{
"input": "400000000474",
"output": "YES"
},
{
"input": "123456789123456789",
"output": "YES"
},
{
"input": "740577777584945874",
"output": "NO"
},
{
"input": "7777777",
"output": "YES"
},
{
"input": "4444000111222333",
"output": "YES"
},
{
"input": "9847745885202111",
"output": "YES"
},
{
"input": "123456000000",
"output": "NO"
},
{
"input": "4744447444444",
"output": "NO"
},
{
"input": "7477",
"output": "YES"
},
{
"input": "4747477",
"output": "YES"
},
{
"input": "777777777444444444",
"output": "NO"
}
] | 1,695,987,934
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 62
| 0
|
x="YES"
c=input()
for i in range(len(c)):
if(c[i]=='4' or c[i]=='7'):
continue
else:
x="NO"
break
print(x)
|
Title: Nearly Lucky Number
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number *n* is a nearly lucky number.
Input Specification:
The only line contains an integer *n* (1<=≤<=*n*<=≤<=1018).
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output Specification:
Print on the single line "YES" if *n* is a nearly lucky number. Otherwise, print "NO" (without the quotes).
Demo Input:
['40047\n', '7747774\n', '1000000000000000000\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".
In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".
In the third sample there are no lucky digits, so the answer is "NO".
|
```python
x="YES"
c=input()
for i in range(len(c)):
if(c[i]=='4' or c[i]=='7'):
continue
else:
x="NO"
break
print(x)
```
| 0
|
1
|
A
|
Theatre Square
|
PROGRAMMING
| 1,000
|
[
"math"
] |
A. Theatre Square
|
1
|
256
|
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
|
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
|
Write the needed number of flagstones.
|
[
"6 6 4\n"
] |
[
"4\n"
] |
none
| 0
|
[
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3",
"output": "1"
},
{
"input": "2 3 4",
"output": "1"
},
{
"input": "1000000000 1000000000 1",
"output": "1000000000000000000"
},
{
"input": "12 13 4",
"output": "12"
},
{
"input": "222 332 5",
"output": "3015"
},
{
"input": "1000 1000 10",
"output": "10000"
},
{
"input": "1001 1000 10",
"output": "10100"
},
{
"input": "100 10001 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "1000000000 1000000000 999999999",
"output": "4"
},
{
"input": "1000000000 1000000000 192",
"output": "27126743055556"
},
{
"input": "1000000000 987654321 1",
"output": "987654321000000000"
},
{
"input": "456784567 1000000000 51",
"output": "175618850864484"
},
{
"input": "39916800 134217728 40320",
"output": "3295710"
}
] | 1,664,886,730
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
n, m, a = map(int, input().split())
if m%a > 1 and n%a > 1:
cnt = ((m//a)+1%(m%a))*((n//a)+1%(n%a))
if m%a == 1 and n%a==1:
cnt = ((m//a)+1)*((n//a)+1)
else:
cnt = (m//a)*(n//a)
print(cnt)
|
Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none
|
```python
n, m, a = map(int, input().split())
if m%a > 1 and n%a > 1:
cnt = ((m//a)+1%(m%a))*((n//a)+1%(n%a))
if m%a == 1 and n%a==1:
cnt = ((m//a)+1)*((n//a)+1)
else:
cnt = (m//a)*(n//a)
print(cnt)
```
| 0
|
272
|
A
|
Dima and Friends
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
|
In a single line print the answer to the problem.
|
[
"1\n1\n",
"1\n2\n",
"2\n3 5\n"
] |
[
"3\n",
"2\n",
"3\n"
] |
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers.
| 500
|
[
{
"input": "1\n1",
"output": "3"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "2\n3 5",
"output": "3"
},
{
"input": "1\n5",
"output": "3"
},
{
"input": "5\n4 4 3 5 1",
"output": "4"
},
{
"input": "6\n2 3 2 2 1 3",
"output": "4"
},
{
"input": "8\n2 2 5 3 4 3 3 2",
"output": "4"
},
{
"input": "7\n4 1 3 2 2 4 5",
"output": "4"
},
{
"input": "3\n3 5 1",
"output": "4"
},
{
"input": "95\n4 2 3 4 4 5 2 2 4 4 3 5 3 3 3 5 4 2 5 4 2 1 1 3 4 2 1 3 5 4 2 1 1 5 1 1 2 2 4 4 5 4 5 5 2 1 2 2 2 4 5 5 2 4 3 4 4 3 5 2 4 1 5 4 5 1 3 2 4 2 2 1 5 3 1 5 3 4 3 3 2 1 2 2 1 3 1 5 2 3 1 1 2 5 2",
"output": "5"
},
{
"input": "31\n3 2 3 3 3 3 4 4 1 5 5 4 2 4 3 2 2 1 4 4 1 2 3 1 1 5 5 3 4 4 1",
"output": "4"
},
{
"input": "42\n3 1 2 2 5 1 2 2 4 5 4 5 2 5 4 5 4 4 1 4 3 3 4 4 4 4 3 2 1 3 4 5 5 2 1 2 1 5 5 2 4 4",
"output": "5"
},
{
"input": "25\n4 5 5 5 3 1 1 4 4 4 3 5 4 4 1 4 4 1 2 4 2 5 4 5 3",
"output": "5"
},
{
"input": "73\n3 4 3 4 5 1 3 4 2 1 4 2 2 3 5 3 1 4 2 3 2 1 4 5 3 5 2 2 4 3 2 2 5 3 2 3 5 1 3 1 1 4 5 2 4 2 5 1 4 3 1 3 1 4 2 3 3 3 3 5 5 2 5 2 5 4 3 1 1 5 5 2 3",
"output": "4"
},
{
"input": "46\n1 4 4 5 4 5 2 3 5 5 3 2 5 4 1 3 2 2 1 4 3 1 5 5 2 2 2 2 4 4 1 1 4 3 4 3 1 4 2 2 4 2 3 2 5 2",
"output": "4"
},
{
"input": "23\n5 2 1 1 4 2 5 5 3 5 4 5 5 1 1 5 2 4 5 3 4 4 3",
"output": "5"
},
{
"input": "6\n4 2 3 1 3 5",
"output": "4"
},
{
"input": "15\n5 5 5 3 5 4 1 3 3 4 3 4 1 4 4",
"output": "5"
},
{
"input": "93\n1 3 1 4 3 3 5 3 1 4 5 4 3 2 2 4 3 1 4 1 2 3 3 3 2 5 1 3 1 4 5 1 1 1 4 2 1 2 3 1 1 1 5 1 5 5 1 2 5 4 3 2 2 4 4 2 5 4 5 5 3 1 3 1 2 1 3 1 1 2 3 4 4 5 5 3 2 1 3 3 5 1 3 5 4 4 1 3 3 4 2 3 2",
"output": "5"
},
{
"input": "96\n1 5 1 3 2 1 2 2 2 2 3 4 1 1 5 4 4 1 2 3 5 1 4 4 4 1 3 3 1 4 5 4 1 3 5 3 4 4 3 2 1 1 4 4 5 1 1 2 5 1 2 3 1 4 1 2 2 2 3 2 3 3 2 5 2 2 3 3 3 3 2 1 2 4 5 5 1 5 3 2 1 4 3 5 5 5 3 3 5 3 4 3 4 2 1 3",
"output": "5"
},
{
"input": "49\n1 4 4 3 5 2 2 1 5 1 2 1 2 5 1 4 1 4 5 2 4 5 3 5 2 4 2 1 3 4 2 1 4 2 1 1 3 3 2 3 5 4 3 4 2 4 1 4 1",
"output": "5"
},
{
"input": "73\n4 1 3 3 3 1 5 2 1 4 1 1 3 5 1 1 4 5 2 1 5 4 1 5 3 1 5 2 4 5 1 4 3 3 5 2 2 3 3 2 5 1 4 5 2 3 1 4 4 3 5 2 3 5 1 4 3 5 1 2 4 1 3 3 5 4 2 4 2 4 1 2 5",
"output": "5"
},
{
"input": "41\n5 3 5 4 2 5 4 3 1 1 1 5 4 3 4 3 5 4 2 5 4 1 1 3 2 4 5 3 5 1 5 5 1 1 1 4 4 1 2 4 3",
"output": "5"
},
{
"input": "100\n3 3 1 4 2 4 4 3 1 5 1 1 4 4 3 4 4 3 5 4 5 2 4 3 4 1 2 4 5 4 2 1 5 4 1 1 4 3 2 4 1 2 1 4 4 5 5 4 4 5 3 2 5 1 4 2 2 1 1 2 5 2 5 1 5 3 1 4 3 2 4 3 2 2 4 5 5 1 2 3 1 4 1 2 2 2 5 5 2 3 2 4 3 1 1 2 1 2 1 2",
"output": "5"
},
{
"input": "100\n2 1 1 3 5 4 4 2 3 4 3 4 5 4 5 4 2 4 5 3 4 5 4 1 1 4 4 1 1 2 5 4 2 4 5 3 2 5 4 3 4 5 1 3 4 2 5 4 5 4 5 2 4 1 2 5 3 1 4 4 5 3 4 3 1 2 5 4 2 5 4 1 5 3 5 4 1 2 5 3 1 1 1 1 5 3 4 3 5 1 1 5 5 1 1 2 2 1 5 1",
"output": "5"
},
{
"input": "100\n4 4 3 3 2 5 4 4 2 1 4 4 4 5 4 1 2 1 5 2 4 3 4 1 4 1 2 5 1 4 5 4 2 1 2 5 3 4 5 5 2 1 2 2 2 2 2 3 2 5 1 2 2 3 2 5 5 1 3 4 5 2 1 3 4 2 2 4 4 3 3 3 2 3 2 1 5 5 5 2 1 4 2 3 5 1 4 4 2 3 2 5 5 4 3 5 1 3 5 5",
"output": "5"
},
{
"input": "100\n4 4 2 5 4 2 2 3 4 4 3 2 3 3 1 3 4 3 3 4 1 3 1 4 5 3 4 3 1 1 1 3 3 2 3 4 3 4 2 2 1 5 1 4 5 1 1 1 3 3 1 1 3 2 5 4 2 5 2 4 5 4 4 1 1 2 1 1 4 5 1 1 5 3 3 2 5 5 5 1 4 1 4 1 1 3 2 3 4 4 2 5 5 2 5 1 1 3 5 3",
"output": "5"
},
{
"input": "100\n4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "4"
},
{
"input": "100\n5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5",
"output": "5"
}
] | 1,587,222,581
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 5
| 310
| 0
|
'''
INPUT SHORTCUTS
N, K = map(int,input().split())
N ,A,B = map(int,input().split())
string = str(input())
arr = list(map(int,input().split()))
N = int(input())
'''
N = int(input())
arr = list(map(int,input().split()))
add = sum(arr)
if add%(N+1)==0:
print(2)
else:
print(3)
|
Title: Dima and Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima and his friends have been playing hide and seek at Dima's place all night. As a result, Dima's place got messy. In the morning they decided that they need to clean the place.
To decide who exactly would clean the apartment, the friends want to play a counting-out game. First, all the guys stand in a circle, and then each of them shows some number of fingers on one hand (one to five), and then the boys count in a circle, starting from Dima, the number of people, respective to the total number of fingers shown. The person on who the countdown stops will clean the apartment.
For example, if Dima and one of his friends played hide and seek, and 7 fingers were shown during the counting-out, then Dima would clean the place. If there were 2 or say, 8 fingers shown, then his friend would clean the place.
Dima knows how many fingers each of his friends will show during the counting-out. Now he is interested in the number of ways to show some number of fingers on one hand (one to five), so that he did not have to clean the place. Help Dima.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of Dima's friends. Dima himself isn't considered to be his own friend. The second line contains *n* positive integers, not exceeding 5, representing, how many fingers the Dima's friends will show.
The numbers in the lines are separated by a single space.
Output Specification:
In a single line print the answer to the problem.
Demo Input:
['1\n1\n', '1\n2\n', '2\n3 5\n']
Demo Output:
['3\n', '2\n', '3\n']
Note:
In the first sample Dima can show 1, 3 or 5 fingers. If Dima shows 3 fingers, then the counting-out will go like that: Dima, his friend, Dima, his friend.
In the second sample Dima can show 2 or 4 fingers.
|
```python
'''
INPUT SHORTCUTS
N, K = map(int,input().split())
N ,A,B = map(int,input().split())
string = str(input())
arr = list(map(int,input().split()))
N = int(input())
'''
N = int(input())
arr = list(map(int,input().split()))
add = sum(arr)
if add%(N+1)==0:
print(2)
else:
print(3)
```
| 0
|
|
459
|
B
|
Pashmak and Flowers
|
PROGRAMMING
| 1,300
|
[
"combinatorics",
"implementation",
"sortings"
] | null | null |
Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!
Your task is to write a program which calculates two things:
1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way.
|
The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109).
|
The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively.
|
[
"2\n1 2\n",
"3\n1 4 5\n",
"5\n3 1 2 3 1\n"
] |
[
"1 1",
"4 1",
"2 4"
] |
In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:
1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers.
| 500
|
[
{
"input": "2\n1 2",
"output": "1 1"
},
{
"input": "3\n1 4 5",
"output": "4 1"
},
{
"input": "5\n3 1 2 3 1",
"output": "2 4"
},
{
"input": "2\n1 1",
"output": "0 1"
},
{
"input": "3\n1 1 1",
"output": "0 3"
},
{
"input": "4\n1 1 1 1",
"output": "0 6"
},
{
"input": "5\n1 1 1 1 1",
"output": "0 10"
},
{
"input": "5\n2 2 2 2 2",
"output": "0 10"
},
{
"input": "10\n2 2 2 2 2 2 2 2 2 2",
"output": "0 45"
},
{
"input": "3\n2 2 2",
"output": "0 3"
},
{
"input": "3\n3 3 3",
"output": "0 3"
},
{
"input": "2\n10000000 100000000",
"output": "90000000 1"
},
{
"input": "5\n5 5 5 5 5",
"output": "0 10"
},
{
"input": "5\n3 3 3 3 3",
"output": "0 10"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "0 15"
},
{
"input": "2\n5 6",
"output": "1 1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "0 45"
},
{
"input": "10\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "0 45"
},
{
"input": "4\n4 4 4 4",
"output": "0 6"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "0 21"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "0 55"
},
{
"input": "8\n8 8 8 8 8 8 8 8",
"output": "0 28"
},
{
"input": "2\n3 2",
"output": "1 1"
}
] | 1,697,645,709
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 124
| 19,660,800
|
n = int(input())
a = list(map(int, input().split()))
count = 0
x = max(a)
y = min(a)
if a.count(x) == a.count(y) and a.count(x) > 1:
count= a.count(x) + a.count(y)
dif = x - y
else:
count+=1
dif = x - y
print(dif, count)
|
Title: Pashmak and Flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!
Your task is to write a program which calculates two things:
1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109).
Output Specification:
The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively.
Demo Input:
['2\n1 2\n', '3\n1 4 5\n', '5\n3 1 2 3 1\n']
Demo Output:
['1 1', '4 1', '2 4']
Note:
In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:
1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers.
|
```python
n = int(input())
a = list(map(int, input().split()))
count = 0
x = max(a)
y = min(a)
if a.count(x) == a.count(y) and a.count(x) > 1:
count= a.count(x) + a.count(y)
dif = x - y
else:
count+=1
dif = x - y
print(dif, count)
```
| 0
|
|
394
|
A
|
Counting Sticks
|
PROGRAMMING
| 0
|
[
"brute force",
"implementation"
] | null | null |
When new students come to the Specialized Educational and Scientific Centre (SESC) they need to start many things from the beginning. Sometimes the teachers say (not always unfairly) that we cannot even count. So our teachers decided to teach us arithmetics from the start. And what is the best way to teach students add and subtract? — That's right, using counting sticks! An here's our new task:
An expression of counting sticks is an expression of type:
Sign + consists of two crossed sticks: one vertical and one horizontal. Sign = consists of two horizontal sticks. The expression is arithmetically correct if *A*<=+<=*B*<==<=*C*.
We've got an expression that looks like *A*<=+<=*B*<==<=*C* given by counting sticks. Our task is to shift at most one stick (or we can shift nothing) so that the expression became arithmetically correct. Note that we cannot remove the sticks from the expression, also we cannot shift the sticks from the signs + and =.
We really aren't fabulous at arithmetics. Can you help us?
|
The single line contains the initial expression. It is guaranteed that the expression looks like *A*<=+<=*B*<==<=*C*, where 1<=≤<=*A*,<=*B*,<=*C*<=≤<=100.
|
If there isn't a way to shift the stick so the expression becomes correct, print on a single line "Impossible" (without the quotes). If there is a way, print the resulting expression. Follow the format of the output from the test samples. Don't print extra space characters.
If there are multiple correct answers, print any of them. For clarifications, you are recommended to see the test samples.
|
[
"||+|=|||||\n",
"|||||+||=||\n",
"|+|=||||||\n",
"||||+||=||||||\n"
] |
[
"|||+|=||||\n",
"Impossible\n",
"Impossible\n",
"||||+||=||||||\n"
] |
In the first sample we can shift stick from the third group of sticks to the first one.
In the second sample we cannot shift vertical stick from + sign to the second group of sticks. So we cannot make a - sign.
There is no answer in the third sample because we cannot remove sticks from the expression.
In the forth sample the initial expression is already arithmetically correct and that is why we don't have to shift sticks.
| 500
|
[
{
"input": "||+|=|||||",
"output": "|||+|=||||"
},
{
"input": "|||||+||=||",
"output": "Impossible"
},
{
"input": "|+|=||||||",
"output": "Impossible"
},
{
"input": "||||+||=||||||",
"output": "||||+||=||||||"
},
{
"input": "||||||||||||+|||||||||||=||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "||||||||||||||||||+||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||=|||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+|=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+|=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "|+|=|",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||||=|",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=|",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||=|",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||||||||||=|",
"output": "Impossible"
},
{
"input": "||+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||+|=|",
"output": "|+|=||"
},
{
"input": "|+||=|",
"output": "|+|=||"
},
{
"input": "|+|=||",
"output": "|+|=||"
},
{
"input": "|||+|=|",
"output": "Impossible"
},
{
"input": "|||+|=|",
"output": "Impossible"
},
{
"input": "|||||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "||||||||||||||||||||||||||||||||||||||||||||||||||+|||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
},
{
"input": "|+|=|||",
"output": "Impossible"
},
{
"input": "|+|=||||",
"output": "||+|=|||"
},
{
"input": "|+||=|",
"output": "|+|=||"
},
{
"input": "|+||||||=|||||",
"output": "|+|||||=||||||"
},
{
"input": "|+|||=||",
"output": "|+||=|||"
},
{
"input": "|+||||=|||",
"output": "|+|||=||||"
},
{
"input": "|+|||||=||||",
"output": "|+||||=|||||"
},
{
"input": "||+||=||",
"output": "|+||=|||"
},
{
"input": "||+|||=|||",
"output": "|+|||=||||"
},
{
"input": "|||||+||=||||||||||",
"output": "Impossible"
},
{
"input": "|||+|||=||||",
"output": "||+|||=|||||"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "|||||+|||||=||||||||||||",
"output": "||||||+|||||=|||||||||||"
},
{
"input": "||+|||||||=|||||||",
"output": "|+|||||||=||||||||"
},
{
"input": "||||+||||=||||||",
"output": "|||+||||=|||||||"
},
{
"input": "||||+|||=|||||",
"output": "|||+|||=||||||"
},
{
"input": "||+|||=|||||||",
"output": "|||+|||=||||||"
},
{
"input": "||+|=|",
"output": "|+|=||"
},
{
"input": "|||+|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "||+|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||"
},
{
"input": "||+|=||||||",
"output": "Impossible"
},
{
"input": "||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||+||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||=||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||",
"output": "Impossible"
}
] | 1,580,495,625
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 140
| 819,200
|
import re
def get_actual(input_string):
a, b, c = map(str, re.split('[+ =]', input_string))
a_count = len(a)
b_count = len(b)
c_count = len(c)
if c_count - b_count == a_count:
return input_string
elif (c_count - (b_count + a_count)) >= 1:
#diff = c_count - (b_count + a_count)
c_count = c_count - 1
a_count = a_count + 1
if a_count + b_count == c_count:
a = "".join(str(x) for x in ['|' * a_count])
b = "".join(str(x) for x in ['|' * b_count])
c = "".join(str(x) for x in ['|' * c_count])
return a + '+' + b + '=' + c
else:
return "Impossible"
else:
return "Impossible"
input_string = input()
print(get_actual(input_string))
|
Title: Counting Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When new students come to the Specialized Educational and Scientific Centre (SESC) they need to start many things from the beginning. Sometimes the teachers say (not always unfairly) that we cannot even count. So our teachers decided to teach us arithmetics from the start. And what is the best way to teach students add and subtract? — That's right, using counting sticks! An here's our new task:
An expression of counting sticks is an expression of type:
Sign + consists of two crossed sticks: one vertical and one horizontal. Sign = consists of two horizontal sticks. The expression is arithmetically correct if *A*<=+<=*B*<==<=*C*.
We've got an expression that looks like *A*<=+<=*B*<==<=*C* given by counting sticks. Our task is to shift at most one stick (or we can shift nothing) so that the expression became arithmetically correct. Note that we cannot remove the sticks from the expression, also we cannot shift the sticks from the signs + and =.
We really aren't fabulous at arithmetics. Can you help us?
Input Specification:
The single line contains the initial expression. It is guaranteed that the expression looks like *A*<=+<=*B*<==<=*C*, where 1<=≤<=*A*,<=*B*,<=*C*<=≤<=100.
Output Specification:
If there isn't a way to shift the stick so the expression becomes correct, print on a single line "Impossible" (without the quotes). If there is a way, print the resulting expression. Follow the format of the output from the test samples. Don't print extra space characters.
If there are multiple correct answers, print any of them. For clarifications, you are recommended to see the test samples.
Demo Input:
['||+|=|||||\n', '|||||+||=||\n', '|+|=||||||\n', '||||+||=||||||\n']
Demo Output:
['|||+|=||||\n', 'Impossible\n', 'Impossible\n', '||||+||=||||||\n']
Note:
In the first sample we can shift stick from the third group of sticks to the first one.
In the second sample we cannot shift vertical stick from + sign to the second group of sticks. So we cannot make a - sign.
There is no answer in the third sample because we cannot remove sticks from the expression.
In the forth sample the initial expression is already arithmetically correct and that is why we don't have to shift sticks.
|
```python
import re
def get_actual(input_string):
a, b, c = map(str, re.split('[+ =]', input_string))
a_count = len(a)
b_count = len(b)
c_count = len(c)
if c_count - b_count == a_count:
return input_string
elif (c_count - (b_count + a_count)) >= 1:
#diff = c_count - (b_count + a_count)
c_count = c_count - 1
a_count = a_count + 1
if a_count + b_count == c_count:
a = "".join(str(x) for x in ['|' * a_count])
b = "".join(str(x) for x in ['|' * b_count])
c = "".join(str(x) for x in ['|' * c_count])
return a + '+' + b + '=' + c
else:
return "Impossible"
else:
return "Impossible"
input_string = input()
print(get_actual(input_string))
```
| 0
|
|
893
|
C
|
Rumor
|
PROGRAMMING
| 1,300
|
[
"dfs and similar",
"graphs",
"greedy"
] | null | null |
Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it.
Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it.
Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on.
The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest?
Take a look at the notes if you think you haven't understood the problem completely.
|
The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends.
The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor.
Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once.
|
Print one number — the minimum amount of gold Vova has to spend in order to finish the quest.
|
[
"5 2\n2 5 3 4 8\n1 4\n4 5\n",
"10 0\n1 2 3 4 5 6 7 8 9 10\n",
"10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n"
] |
[
"10\n",
"55\n",
"15\n"
] |
In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor.
In the second example Vova has to bribe everyone.
In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters.
| 0
|
[
{
"input": "5 2\n2 5 3 4 8\n1 4\n4 5",
"output": "10"
},
{
"input": "10 0\n1 2 3 4 5 6 7 8 9 10",
"output": "55"
},
{
"input": "10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10",
"output": "15"
},
{
"input": "1 0\n0",
"output": "0"
},
{
"input": "1 0\n1000000000",
"output": "1000000000"
},
{
"input": "2 0\n0 0",
"output": "0"
},
{
"input": "2 0\n1000000000 0",
"output": "1000000000"
},
{
"input": "2 0\n0 1000000000",
"output": "1000000000"
},
{
"input": "2 0\n1000000000 1000000000",
"output": "2000000000"
},
{
"input": "2 1\n0 0\n1 2",
"output": "0"
}
] | 1,625,973,881
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 46
| 951
| 39,526,400
|
# _
#####################################################################################################################
nCharacters, nPairs = map(int, input().split(' '))
characters, golds = set(), {}
for iCharacter, gold in enumerate(input().split(' ')):
character = str(iCharacter+1)
characters.add(character)
golds[character] = int(gold)
friendsMap = {}
for i in range(nPairs):
pair = input().split(' ')
if friendsMap.get(pair[0], False):
friendsMap[pair[0]].add(pair[1])
else:
friendsMap[pair[0]] = {pair[1]}
if friendsMap.get(pair[1], False):
friendsMap[pair[1]].add(pair[0])
else:
friendsMap[pair[1]] = {pair[0]}
def findMinGold(character, minGold, charactersToConsider):
for friend in friendsMap[character]:
mutualFriendsGroup.add(friend)
if golds[friend] < minGold:
minGold = golds[friend]
if friendsMap[friend] - mutualFriendsGroup:
charactersToConsider.append(friend)
return minGold
minTotalGold, charactersInAGroup = 0, set()
for character in friendsMap:
if character not in charactersInAGroup:
mutualFriendsGroup, charactersToConsider = {character}, []
minGold = findMinGold(character, golds[character], charactersToConsider)
while charactersToConsider:
nextCharactersToConsider = []
for nextCharacter in charactersToConsider:
if golds[nextCharacter] < minGold:
minGold = golds[nextCharacter]
minGold = findMinGold(nextCharacter, minGold, nextCharactersToConsider)
charactersToConsider = nextCharactersToConsider
charactersInAGroup |= mutualFriendsGroup
minTotalGold += minGold
print(minTotalGold + sum(golds[character] for character in characters-charactersInAGroup))
|
Title: Rumor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it.
Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it.
Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on.
The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest?
Take a look at the notes if you think you haven't understood the problem completely.
Input Specification:
The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends.
The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor.
Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once.
Output Specification:
Print one number — the minimum amount of gold Vova has to spend in order to finish the quest.
Demo Input:
['5 2\n2 5 3 4 8\n1 4\n4 5\n', '10 0\n1 2 3 4 5 6 7 8 9 10\n', '10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n']
Demo Output:
['10\n', '55\n', '15\n']
Note:
In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor.
In the second example Vova has to bribe everyone.
In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters.
|
```python
# _
#####################################################################################################################
nCharacters, nPairs = map(int, input().split(' '))
characters, golds = set(), {}
for iCharacter, gold in enumerate(input().split(' ')):
character = str(iCharacter+1)
characters.add(character)
golds[character] = int(gold)
friendsMap = {}
for i in range(nPairs):
pair = input().split(' ')
if friendsMap.get(pair[0], False):
friendsMap[pair[0]].add(pair[1])
else:
friendsMap[pair[0]] = {pair[1]}
if friendsMap.get(pair[1], False):
friendsMap[pair[1]].add(pair[0])
else:
friendsMap[pair[1]] = {pair[0]}
def findMinGold(character, minGold, charactersToConsider):
for friend in friendsMap[character]:
mutualFriendsGroup.add(friend)
if golds[friend] < minGold:
minGold = golds[friend]
if friendsMap[friend] - mutualFriendsGroup:
charactersToConsider.append(friend)
return minGold
minTotalGold, charactersInAGroup = 0, set()
for character in friendsMap:
if character not in charactersInAGroup:
mutualFriendsGroup, charactersToConsider = {character}, []
minGold = findMinGold(character, golds[character], charactersToConsider)
while charactersToConsider:
nextCharactersToConsider = []
for nextCharacter in charactersToConsider:
if golds[nextCharacter] < minGold:
minGold = golds[nextCharacter]
minGold = findMinGold(nextCharacter, minGold, nextCharactersToConsider)
charactersToConsider = nextCharactersToConsider
charactersInAGroup |= mutualFriendsGroup
minTotalGold += minGold
print(minTotalGold + sum(golds[character] for character in characters-charactersInAGroup))
```
| 3
|
|
4
|
A
|
Watermelon
|
PROGRAMMING
| 800
|
[
"brute force",
"math"
] |
A. Watermelon
|
1
|
64
|
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
|
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
|
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
|
[
"8\n"
] |
[
"YES\n"
] |
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
| 0
|
[
{
"input": "8",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "6",
"output": "YES"
},
{
"input": "10",
"output": "YES"
},
{
"input": "9",
"output": "NO"
},
{
"input": "53",
"output": "NO"
},
{
"input": "77",
"output": "NO"
},
{
"input": "32",
"output": "YES"
},
{
"input": "44",
"output": "YES"
},
{
"input": "98",
"output": "YES"
},
{
"input": "99",
"output": "NO"
},
{
"input": "90",
"output": "YES"
},
{
"input": "67",
"output": "NO"
},
{
"input": "100",
"output": "YES"
},
{
"input": "88",
"output": "YES"
}
] | 1,691,089,851
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 92
| 0
|
w = int(input())
if ((w%2)!=0):
print("NO")
elif(w<=2):
print("NO")
else:
print("YES")
|
Title: Watermelon
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
Input Specification:
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Output Specification:
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
Demo Input:
['8\n']
Demo Output:
['YES\n']
Note:
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos).
|
```python
w = int(input())
if ((w%2)!=0):
print("NO")
elif(w<=2):
print("NO")
else:
print("YES")
```
| 3.954
|
248
|
A
|
Cupboards
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
|
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
|
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
|
[
"5\n0 1\n1 0\n0 1\n1 1\n0 1\n"
] |
[
"3\n"
] |
none
| 500
|
[
{
"input": "5\n0 1\n1 0\n0 1\n1 1\n0 1",
"output": "3"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1\n1 1\n1 1",
"output": "1"
},
{
"input": "8\n0 1\n1 0\n0 1\n1 1\n0 1\n1 0\n0 1\n1 0",
"output": "7"
},
{
"input": "8\n1 0\n1 0\n1 0\n0 1\n0 1\n1 1\n1 1\n0 1",
"output": "6"
},
{
"input": "15\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "5\n1 0\n1 0\n1 0\n0 1\n0 1",
"output": "4"
}
] | 1,632,839,537
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 57
| 186
| 6,963,200
|
n = int(input())
countl0,countl1 = 0,0
countr0,countr1 = 0,0
for i in range(n):
l,r = map(int, input().split())
if l==0:
countl0 += 1
else:
countl1 += 1
if r==0:
countr0 += 1
else:
countr1 += 1
print(min(countr1,countr0) + min(countl0,countl1))
|
Title: Cupboards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
Input Specification:
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
Output Specification:
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
Demo Input:
['5\n0 1\n1 0\n0 1\n1 1\n0 1\n']
Demo Output:
['3\n']
Note:
none
|
```python
n = int(input())
countl0,countl1 = 0,0
countr0,countr1 = 0,0
for i in range(n):
l,r = map(int, input().split())
if l==0:
countl0 += 1
else:
countl1 += 1
if r==0:
countr0 += 1
else:
countr1 += 1
print(min(countr1,countr0) + min(countl0,countl1))
```
| 3
|
|
779
|
C
|
Dishonest Sellers
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"greedy",
"sortings"
] | null | null |
Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*.
Not all of sellers are honest, so now some products could be more expensive than after a week of discounts.
Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items.
|
In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now.
The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now).
The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week).
|
Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now.
|
[
"3 1\n5 4 6\n3 1 5\n",
"5 3\n3 4 7 10 3\n4 5 5 12 5\n"
] |
[
"10\n",
"25\n"
] |
In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10.
In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25.
| 1,000
|
[
{
"input": "3 1\n5 4 6\n3 1 5",
"output": "10"
},
{
"input": "5 3\n3 4 7 10 3\n4 5 5 12 5",
"output": "25"
},
{
"input": "1 0\n9\n8",
"output": "8"
},
{
"input": "2 0\n4 10\n1 2",
"output": "3"
},
{
"input": "4 2\n19 5 17 13\n3 18 8 10",
"output": "29"
},
{
"input": "5 3\n28 17 20 45 45\n39 12 41 27 9",
"output": "101"
},
{
"input": "10 5\n87 96 19 81 10 88 7 49 36 21\n11 75 28 28 74 17 64 19 81 31",
"output": "243"
},
{
"input": "50 45\n80 125 152 122 85 62 123 195 157 73 8 127 47 193 70 132 131 38 98 97 185 98 68 166 114 53 144 106 141 106 191 123 44 199 35 18 84 169 46 52 129 102 155 130 63 169 116 145 154 41\n155 187 79 106 136 173 17 146 9 47 173 21 37 183 110 74 185 14 82 156 147 25 162 195 26 25 22 80 171 180 7 92 53 18 100 17 159 184 37 127 116 192 163 49 94 95 118 102 8 30",
"output": "4680"
},
{
"input": "5 2\n1 1 1 1 1\n2 2 2 2 2",
"output": "5"
},
{
"input": "5 2\n2 4 4 4 4\n3 1 1 1 1",
"output": "9"
},
{
"input": "6 3\n9 9 9 2 2 2\n6 6 6 1 1 1",
"output": "24"
},
{
"input": "5 3\n10 17 23 26 35\n7 18 18 29 32",
"output": "103"
},
{
"input": "2 1\n10 12\n5 6",
"output": "16"
}
] | 1,551,559,263
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 92
| 0
|
N, K = input().strip().split()
N, K = int(N), int(K) #wants to buy at least k now(with discounts), wants to buy n in total
discounts = input().strip().split()
normal = input().strip().split()
items = []
for i in range(N):
discounts[i] = int(discounts[i])
normal[i] = int(normal[i])
items.append([discounts[i], normal[i], min(discounts[i], normal[i])])
money = 0
items = sorted(items)
for i in range(K):
item = items.pop(0)
#print("buy for " + str(item[0]))
money += item[0]
items = sorted(items, key= lambda k: k[2])
#print(items)
for i in range(N-K):
item = items.pop(0)
#print("buy for " + str(item[2]))
money += item[2]
print(money)
|
Title: Dishonest Sellers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*.
Not all of sellers are honest, so now some products could be more expensive than after a week of discounts.
Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items.
Input Specification:
In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now.
The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now).
The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week).
Output Specification:
Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now.
Demo Input:
['3 1\n5 4 6\n3 1 5\n', '5 3\n3 4 7 10 3\n4 5 5 12 5\n']
Demo Output:
['10\n', '25\n']
Note:
In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10.
In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25.
|
```python
N, K = input().strip().split()
N, K = int(N), int(K) #wants to buy at least k now(with discounts), wants to buy n in total
discounts = input().strip().split()
normal = input().strip().split()
items = []
for i in range(N):
discounts[i] = int(discounts[i])
normal[i] = int(normal[i])
items.append([discounts[i], normal[i], min(discounts[i], normal[i])])
money = 0
items = sorted(items)
for i in range(K):
item = items.pop(0)
#print("buy for " + str(item[0]))
money += item[0]
items = sorted(items, key= lambda k: k[2])
#print(items)
for i in range(N-K):
item = items.pop(0)
#print("buy for " + str(item[2]))
money += item[2]
print(money)
```
| 0
|
|
766
|
B
|
Mahmoud and a Triangle
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"geometry",
"greedy",
"math",
"number theory",
"sortings"
] | null | null |
Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.
|
The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has.
|
In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.
|
[
"5\n1 5 3 2 4\n",
"3\n4 1 2\n"
] |
[
"YES\n",
"NO\n"
] |
For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.
| 1,000
|
[
{
"input": "5\n1 5 3 2 4",
"output": "YES"
},
{
"input": "3\n4 1 2",
"output": "NO"
},
{
"input": "30\n197 75 517 39724 7906061 1153471 3 15166 168284 3019844 272293 316 16 24548 42 118 5792 5 9373 1866366 4886214 24 2206 712886 104005 1363 836 64273 440585 3576",
"output": "NO"
},
{
"input": "30\n229017064 335281886 247217656 670601882 743442492 615491486 544941439 911270108 474843964 803323771 177115397 62179276 390270885 754889875 881720571 902691435 154083299 328505383 761264351 182674686 94104683 357622370 573909964 320060691 33548810 247029007 812823597 946798893 813659359 710111761",
"output": "YES"
},
{
"input": "40\n740553458 532562042 138583675 75471987 487348843 476240280 972115023 103690894 546736371 915774563 35356828 819948191 138721993 24257926 761587264 767176616 608310208 78275645 386063134 227581756 672567198 177797611 87579917 941781518 274774331 843623616 981221615 630282032 118843963 749160513 354134861 132333165 405839062 522698334 29698277 541005920 856214146 167344951 398332403 68622974",
"output": "YES"
},
{
"input": "40\n155 1470176 7384 765965701 1075 4 561554 6227772 93 16304522 1744 662 3 292572860 19335 908613 42685804 347058 20 132560 3848974 69067081 58 2819 111752888 408 81925 30 11951 4564 251 26381275 473392832 50628 180819969 2378797 10076746 9 214492 31291",
"output": "NO"
},
{
"input": "3\n1 1000000000 1000000000",
"output": "YES"
},
{
"input": "4\n1 1000000000 1000000000 1000000000",
"output": "YES"
},
{
"input": "3\n1 1000000000 1",
"output": "NO"
},
{
"input": "5\n1 2 3 5 2",
"output": "YES"
},
{
"input": "41\n19 161 4090221 118757367 2 45361275 1562319 596751 140871 97 1844 310910829 10708344 6618115 698 1 87059 33 2527892 12703 73396090 17326460 3 368811 20550 813975131 10 53804 28034805 7847 2992 33254 1139 227930 965568 261 4846 503064297 192153458 57 431",
"output": "NO"
},
{
"input": "42\n4317083 530966905 202811311 104 389267 35 1203 18287479 125344279 21690 859122498 65 859122508 56790 1951 148683 457 1 22 2668100 8283 2 77467028 13405 11302280 47877251 328155592 35095 29589769 240574 4 10 1019123 6985189 629846 5118 169 1648973 91891 741 282 3159",
"output": "YES"
},
{
"input": "43\n729551585 11379 5931704 330557 1653 15529406 729551578 278663905 1 729551584 2683 40656510 29802 147 1400284 2 126260 865419 51 17 172223763 86 1 534861 450887671 32 234 25127103 9597697 48226 7034 389 204294 2265706 65783617 4343 3665990 626 78034 106440137 5 18421 1023",
"output": "YES"
},
{
"input": "44\n719528276 2 235 444692918 24781885 169857576 18164 47558 15316043 9465834 64879816 2234575 1631 853530 8 1001 621 719528259 84 6933 31 1 3615623 719528266 40097928 274835337 1381044 11225 2642 5850203 6 527506 18 104977753 76959 29393 49 4283 141 201482 380 1 124523 326015",
"output": "YES"
},
{
"input": "45\n28237 82 62327732 506757 691225170 5 970 4118 264024506 313192 367 14713577 73933 691225154 6660 599 691225145 3473403 51 427200630 1326718 2146678 100848386 1569 27 163176119 193562 10784 45687 819951 38520653 225 119620 1 3 691225169 691225164 17445 23807072 1 9093493 5620082 2542 139 14",
"output": "YES"
},
{
"input": "44\n165580141 21 34 55 1 89 144 17711 2 377 610 987 2584 13 5 4181 6765 10946 1597 8 28657 3 233 75025 121393 196418 317811 9227465 832040 1346269 2178309 3524578 5702887 1 14930352 102334155 24157817 39088169 63245986 701408733 267914296 433494437 514229 46368",
"output": "NO"
},
{
"input": "3\n1 1000000000 999999999",
"output": "NO"
},
{
"input": "5\n1 1 1 1 1",
"output": "YES"
},
{
"input": "10\n1 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000",
"output": "NO"
},
{
"input": "5\n2 3 4 10 20",
"output": "YES"
},
{
"input": "6\n18 23 40 80 160 161",
"output": "YES"
},
{
"input": "4\n5 6 7 888",
"output": "YES"
},
{
"input": "9\n1 1 2 2 4 5 10 10 20",
"output": "YES"
},
{
"input": "7\n3 150 900 4 500 1500 5",
"output": "YES"
},
{
"input": "3\n2 2 3",
"output": "YES"
},
{
"input": "7\n1 2 100 200 250 1000000 2000000",
"output": "YES"
},
{
"input": "8\n2 3 5 5 5 6 6 13",
"output": "YES"
},
{
"input": "3\n2 3 4",
"output": "YES"
},
{
"input": "6\n1 1 1 4 5 100",
"output": "YES"
},
{
"input": "13\n1 2 3 5 8 13 22 34 55 89 144 233 377",
"output": "YES"
},
{
"input": "4\n2 3 4 8",
"output": "YES"
},
{
"input": "3\n5 6 7",
"output": "YES"
},
{
"input": "5\n1 4 5 6 1000000",
"output": "YES"
},
{
"input": "4\n5 6 7 20",
"output": "YES"
},
{
"input": "6\n1 1 1 1 1 65",
"output": "YES"
},
{
"input": "4\n3 4 5 100",
"output": "YES"
},
{
"input": "3\n2 4 5",
"output": "YES"
},
{
"input": "7\n1 1 1 1 1 10 1000",
"output": "YES"
},
{
"input": "4\n1 1 2 3",
"output": "NO"
},
{
"input": "11\n1 2 5 6 7 8 9 17 18 19 100",
"output": "YES"
},
{
"input": "4\n5 16 20 200",
"output": "YES"
},
{
"input": "5\n17 6 3 3 1",
"output": "YES"
},
{
"input": "3\n1 1 1",
"output": "YES"
},
{
"input": "6\n1 1 1 2 3 5",
"output": "YES"
},
{
"input": "4\n2 4 6 6",
"output": "YES"
},
{
"input": "9\n1 2 4 4 4 4 7 8 20",
"output": "YES"
},
{
"input": "9\n1 1 2 5 5 5 10 10 20",
"output": "YES"
},
{
"input": "7\n3 150 600 4 1700 6000 5",
"output": "YES"
},
{
"input": "5\n5761 20966 27841 28800 29399",
"output": "YES"
},
{
"input": "9\n1 2 3 6 7 10 11 12 24",
"output": "YES"
},
{
"input": "4\n1 2 1 1",
"output": "YES"
},
{
"input": "5\n1 1 2 3 4",
"output": "YES"
}
] | 1,648,130,370
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 87
| 124
| 8,089,600
|
n = int(input())
arr = []
aux = input().split()
for i in range(n):
arr.append(int(aux[i]))
def isNonDegTr(arr, n):
if(n < 3):
return False
arr.sort()
for i in range(n-2):
if(arr[i]+arr[i+1]>arr[i+2]):
return True
if(isNonDegTr(arr, n)):
print("YES")
else:
print("NO")
|
Title: Mahmoud and a Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.
Input Specification:
The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has.
Output Specification:
In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.
Demo Input:
['5\n1 5 3 2 4\n', '3\n4 1 2\n']
Demo Output:
['YES\n', 'NO\n']
Note:
For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle.
|
```python
n = int(input())
arr = []
aux = input().split()
for i in range(n):
arr.append(int(aux[i]))
def isNonDegTr(arr, n):
if(n < 3):
return False
arr.sort()
for i in range(n-2):
if(arr[i]+arr[i+1]>arr[i+2]):
return True
if(isNonDegTr(arr, n)):
print("YES")
else:
print("NO")
```
| 3
|
|
455
|
A
|
Boredom
|
PROGRAMMING
| 1,500
|
[
"dp"
] | null | null |
Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence *a* consisting of *n* integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it *a**k*) and delete it, at that all elements equal to *a**k*<=+<=1 and *a**k*<=-<=1 also must be deleted from the sequence. That step brings *a**k* points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) that shows how many numbers are in Alex's sequence.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105).
|
Print a single integer — the maximum number of points that Alex can earn.
|
[
"2\n1 2\n",
"3\n1 2 3\n",
"9\n1 2 1 3 2 2 2 2 3\n"
] |
[
"2\n",
"4\n",
"10\n"
] |
Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.
| 500
|
[
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n1 2 3",
"output": "4"
},
{
"input": "9\n1 2 1 3 2 2 2 2 3",
"output": "10"
},
{
"input": "5\n3 3 4 5 4",
"output": "11"
},
{
"input": "5\n5 3 5 3 4",
"output": "16"
},
{
"input": "5\n4 2 3 2 5",
"output": "9"
},
{
"input": "10\n10 5 8 9 5 6 8 7 2 8",
"output": "46"
},
{
"input": "10\n1 1 1 1 1 1 2 3 4 4",
"output": "14"
},
{
"input": "100\n6 6 8 9 7 9 6 9 5 7 7 4 5 3 9 1 10 3 4 5 8 9 6 5 6 4 10 9 1 4 1 7 1 4 9 10 8 2 9 9 10 5 8 9 5 6 8 7 2 8 7 6 2 6 10 8 6 2 5 5 3 2 8 8 5 3 6 2 1 4 7 2 7 3 7 4 10 10 7 5 4 7 5 10 7 1 1 10 7 7 7 2 3 4 2 8 4 7 4 4",
"output": "296"
},
{
"input": "100\n6 1 5 7 10 10 2 7 3 7 2 10 7 6 3 5 5 5 3 7 2 4 2 7 7 4 2 8 2 10 4 7 9 1 1 7 9 7 1 10 10 9 5 6 10 1 7 5 8 1 1 5 3 10 2 4 3 5 2 7 4 9 5 10 1 3 7 6 6 9 3 6 6 10 1 10 6 1 10 3 4 1 7 9 2 7 8 9 3 3 2 4 6 6 1 2 9 4 1 2",
"output": "313"
},
{
"input": "100\n7 6 3 8 8 3 10 5 3 8 6 4 6 9 6 7 3 9 10 7 5 5 9 10 7 2 3 8 9 5 4 7 9 3 6 4 9 10 7 6 8 7 6 6 10 3 7 4 5 7 7 5 1 5 4 8 7 3 3 4 7 8 5 9 2 2 3 1 6 4 6 6 6 1 7 10 7 4 5 3 9 2 4 1 5 10 9 3 9 6 8 5 2 1 10 4 8 5 10 9",
"output": "298"
},
{
"input": "100\n2 10 9 1 2 6 7 2 2 8 9 9 9 5 6 2 5 1 1 10 7 4 5 5 8 1 9 4 10 1 9 3 1 8 4 10 8 8 2 4 6 5 1 4 2 2 1 2 8 5 3 9 4 10 10 7 8 6 1 8 2 6 7 1 6 7 3 10 10 3 7 7 6 9 6 8 8 10 4 6 4 3 3 3 2 3 10 6 8 5 5 10 3 7 3 1 1 1 5 5",
"output": "312"
},
{
"input": "100\n4 9 7 10 4 7 2 6 1 9 1 8 7 5 5 7 6 7 9 8 10 5 3 5 7 10 3 2 1 3 8 9 4 10 4 7 6 4 9 6 7 1 9 4 3 5 8 9 2 7 10 5 7 5 3 8 10 3 8 9 3 4 3 10 6 5 1 8 3 2 5 8 4 7 5 3 3 2 6 9 9 8 2 7 6 3 2 2 8 8 4 5 6 9 2 3 2 2 5 2",
"output": "287"
},
{
"input": "100\n4 8 10 1 8 8 8 1 10 3 1 8 6 8 6 1 10 3 3 3 3 7 2 1 1 6 10 1 7 9 8 10 3 8 6 2 1 6 5 6 10 8 9 7 4 3 10 5 3 9 10 5 10 8 8 5 7 8 9 5 3 9 9 2 7 8 1 10 4 9 2 8 10 10 5 8 5 1 7 3 4 5 2 5 9 3 2 5 6 2 3 10 1 5 9 6 10 4 10 8",
"output": "380"
},
{
"input": "100\n4 8 10 1 8 8 8 1 10 3 1 8 6 8 6 1 10 3 3 3 3 7 2 1 1 6 10 1 7 9 8 10 3 8 6 2 1 6 5 6 10 8 9 7 4 3 10 5 3 9 10 5 10 8 8 5 7 8 9 5 3 9 9 2 7 8 1 10 4 9 2 8 10 10 5 8 5 1 7 3 4 5 2 5 9 3 2 5 6 2 3 10 1 5 9 6 10 4 10 8",
"output": "380"
},
{
"input": "100\n10 5 8 4 4 4 1 4 5 8 3 10 2 4 1 10 8 1 1 6 8 4 2 9 1 3 1 7 7 9 3 5 5 8 6 9 9 4 8 1 3 3 2 6 1 5 4 5 3 5 5 6 7 5 7 9 3 5 4 9 2 6 8 1 1 7 7 3 8 9 8 7 3 2 4 1 6 1 3 9 4 2 2 8 5 10 1 8 8 5 1 5 6 9 4 5 6 5 10 2",
"output": "265"
},
{
"input": "100\n7 5 1 8 5 6 6 2 6 2 7 7 3 6 2 4 4 2 10 2 2 2 10 6 6 1 5 10 9 1 5 9 8 9 4 1 10 5 7 5 7 6 4 8 8 1 7 8 3 8 2 1 8 4 10 3 5 6 6 10 9 6 5 1 10 7 6 9 9 2 10 10 9 1 2 1 7 7 4 10 1 10 5 5 3 8 9 8 1 4 10 2 4 5 4 4 1 6 2 9",
"output": "328"
},
{
"input": "100\n5 6 10 7 1 7 10 1 9 1 5 1 4 1 3 3 7 9 1 6 1 6 5 7 1 6 3 1 3 6 3 8 2 4 1 5 2 10 7 3 10 4 10 1 5 4 2 9 7 9 5 7 10 4 1 4 8 9 3 1 3 7 7 4 3 7 7 10 6 9 5 5 6 5 3 9 8 8 5 5 4 10 9 4 10 4 1 8 3 5 4 10 9 3 10 4 10 7 10 9",
"output": "324"
},
{
"input": "10\n7 4 5 3 9 1 10 3 4 5",
"output": "34"
},
{
"input": "10\n8 9 6 5 6 4 10 9 1 4",
"output": "39"
},
{
"input": "10\n1 7 1 4 9 10 8 2 9 9",
"output": "40"
},
{
"input": "1\n100000",
"output": "100000"
}
] | 1,668,917,616
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 47
| 233
| 7,270,400
|
from collections import Counter
input()
s = Counter(map(int,input().split()))
m = max(s)
dp = [0]*(m+1)
dp[1] = s[1]
dp[2] = max(dp[1],2*s[2])
for i in range(3,m+1):
dp[i] = max(dp[i-2] + i*s[i], dp[i-1])
print(dp[-1])
|
Title: Boredom
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence *a* consisting of *n* integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it *a**k*) and delete it, at that all elements equal to *a**k*<=+<=1 and *a**k*<=-<=1 also must be deleted from the sequence. That step brings *a**k* points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) that shows how many numbers are in Alex's sequence.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
Print a single integer — the maximum number of points that Alex can earn.
Demo Input:
['2\n1 2\n', '3\n1 2 3\n', '9\n1 2 1 3 2 2 2 2 3\n']
Demo Output:
['2\n', '4\n', '10\n']
Note:
Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.
|
```python
from collections import Counter
input()
s = Counter(map(int,input().split()))
m = max(s)
dp = [0]*(m+1)
dp[1] = s[1]
dp[2] = max(dp[1],2*s[2])
for i in range(3,m+1):
dp[i] = max(dp[i-2] + i*s[i], dp[i-1])
print(dp[-1])
```
| 3
|
|
133
|
A
|
HQ9+
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
HQ9+ is a joke programming language which has only four one-character instructions:
- "H" prints "Hello, World!",- "Q" prints the source code of the program itself,- "9" prints the lyrics of "99 Bottles of Beer" song, - "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
|
The input will consist of a single line *p* which will give a program in HQ9+. String *p* will contain between 1 and 100 characters, inclusive. ASCII-code of each character of *p* will be between 33 (exclamation mark) and 126 (tilde), inclusive.
|
Output "YES", if executing the program will produce any output, and "NO" otherwise.
|
[
"Hi!\n",
"Codeforces\n"
] |
[
"YES\n",
"NO\n"
] |
In the first case the program contains only one instruction — "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
| 500
|
[
{
"input": "Hi!",
"output": "YES"
},
{
"input": "Codeforces",
"output": "NO"
},
{
"input": "a+b=c",
"output": "NO"
},
{
"input": "hq-lowercase",
"output": "NO"
},
{
"input": "Q",
"output": "YES"
},
{
"input": "9",
"output": "YES"
},
{
"input": "H",
"output": "YES"
},
{
"input": "+",
"output": "NO"
},
{
"input": "~",
"output": "NO"
},
{
"input": "dEHsbM'gS[\\brZ_dpjXw8f?L[4E\"s4Zc9*(,j:>p$}m7HD[_9nOWQ\\uvq2mHWR",
"output": "YES"
},
{
"input": "tt6l=RHOfStm.;Qd$-}zDes*E,.F7qn5-b%HC",
"output": "YES"
},
{
"input": "@F%K2=%RyL/",
"output": "NO"
},
{
"input": "juq)k(FT.^G=G\\zcqnO\"uJIE1_]KFH9S=1c\"mJ;F9F)%>&.WOdp09+k`Yc6}\"6xw,Aos:M\\_^^:xBb[CcsHm?J",
"output": "YES"
},
{
"input": "6G_\"Fq#<AWyHG=Rci1t%#Jc#x<Fpg'N@t%F=``YO7\\Zd;6PkMe<#91YgzTC)",
"output": "YES"
},
{
"input": "Fvg_~wC>SO4lF}*c`Q;mII9E{4.QodbqN]C",
"output": "YES"
},
{
"input": "p-UXsbd&f",
"output": "NO"
},
{
"input": "<]D7NMA)yZe=`?RbP5lsa.l_Mg^V:\"-0x+$3c,q&L%18Ku<HcA\\s!^OQblk^x{35S'>yz8cKgVHWZ]kV0>_",
"output": "YES"
},
{
"input": "f.20)8b+.R}Gy!DbHU3v(.(=Q^`z[_BaQ}eO=C1IK;b2GkD\\{\\Bf\"!#qh]",
"output": "YES"
},
{
"input": "}do5RU<(w<q[\"-NR)IAH_HyiD{",
"output": "YES"
},
{
"input": "Iy^.,Aw*,5+f;l@Q;jLK'G5H-r1Pfmx?ei~`CjMmUe{K:lS9cu4ay8rqRh-W?Gqv!e-j*U)!Mzn{E8B6%~aSZ~iQ_QwlC9_cX(o8",
"output": "YES"
},
{
"input": "sKLje,:q>-D,;NvQ3,qN3-N&tPx0nL/,>Ca|z\"k2S{NF7btLa3_TyXG4XZ:`(t&\"'^M|@qObZxv",
"output": "YES"
},
{
"input": "%z:c@1ZsQ@\\6U/NQ+M9R>,$bwG`U1+C\\18^:S},;kw!&4r|z`",
"output": "YES"
},
{
"input": "OKBB5z7ud81[Tn@P\"nDUd,>@",
"output": "NO"
},
{
"input": "y{0;neX]w0IenPvPx0iXp+X|IzLZZaRzBJ>q~LhMhD$x-^GDwl;,a'<bAqH8QrFwbK@oi?I'W.bZ]MlIQ/x(0YzbTH^l.)]0Bv",
"output": "YES"
},
{
"input": "EL|xIP5_+Caon1hPpQ0[8+r@LX4;b?gMy>;/WH)pf@Ur*TiXu*e}b-*%acUA~A?>MDz#!\\Uh",
"output": "YES"
},
{
"input": "UbkW=UVb>;z6)p@Phr;^Dn.|5O{_i||:Rv|KJ_ay~V(S&Jp",
"output": "NO"
},
{
"input": "!3YPv@2JQ44@)R2O_4`GO",
"output": "YES"
},
{
"input": "Kba/Q,SL~FMd)3hOWU'Jum{9\"$Ld4:GW}D]%tr@G{hpG:PV5-c'VIZ~m/6|3I?_4*1luKnOp`%p|0H{[|Y1A~4-ZdX,Rw2[\\",
"output": "YES"
},
{
"input": "NRN*=v>;oU7[acMIJn*n^bWm!cm3#E7Efr>{g-8bl\"DN4~_=f?[T;~Fq#&)aXq%</GcTJD^e$@Extm[e\"C)q_L",
"output": "NO"
},
{
"input": "y#<fv{_=$MP!{D%I\\1OqjaqKh[pqE$KvYL<9@*V'j8uH0/gQdA'G;&y4Cv6&",
"output": "YES"
},
{
"input": "+SE_Pg<?7Fh,z&uITQut2a-mk8X8La`c2A}",
"output": "YES"
},
{
"input": "Uh3>ER](J",
"output": "NO"
},
{
"input": "!:!{~=9*\\P;Z6F?HC5GadFz)>k*=u|+\"Cm]ICTmB!`L{&oS/z6b~#Snbp/^\\Q>XWU-vY+/dP.7S=-#&whS@,",
"output": "YES"
},
{
"input": "KimtYBZp+ISeO(uH;UldoE6eAcp|9u?SzGZd6j-e}[}u#e[Cx8.qgY]$2!",
"output": "YES"
},
{
"input": "[:[SN-{r>[l+OggH3v3g{EPC*@YBATT@",
"output": "YES"
},
{
"input": "'jdL(vX",
"output": "NO"
},
{
"input": "Q;R+aay]cL?Zh*uG\"YcmO*@Dts*Gjp}D~M7Z96+<4?9I3aH~0qNdO(RmyRy=ci,s8qD_kwj;QHFzD|5,5",
"output": "YES"
},
{
"input": "{Q@#<LU_v^qdh%gGxz*pu)Y\"]k-l-N30WAxvp2IE3:jD0Wi4H/xWPH&s",
"output": "YES"
},
{
"input": "~@Gb(S&N$mBuBUMAky-z^{5VwLNTzYg|ZUZncL@ahS?K*As<$iNUARM3r43J'jJB)$ujfPAq\"G<S9flGyakZg!2Z.-NJ|2{F>]",
"output": "YES"
},
{
"input": "Jp5Aa>aP6fZ!\\6%A}<S}j{O4`C6y$8|i3IW,WHy&\"ioE&7zP\"'xHAY;:x%@SnS]Mr{R|})gU",
"output": "YES"
},
{
"input": "ZA#:U)$RI^sE\\vuAt]x\"2zipI!}YEu2<j$:H0_9/~eB?#->",
"output": "YES"
},
{
"input": "&ppw0._:\\p-PuWM@l}%%=",
"output": "NO"
},
{
"input": "P(^pix\"=oiEZu8?@d@J(I`Xp5TN^T3\\Z7P5\"ZrvZ{2Fwz3g-8`U!)(1$a<g+9Q|COhDoH;HwFY02Pa|ZGp$/WZBR=>6Jg!yr",
"output": "YES"
},
{
"input": "`WfODc\\?#ax~1xu@[ao+o_rN|L7%v,p,nDv>3+6cy.]q3)+A6b!q*Hc+#.t4f~vhUa~$^q",
"output": "YES"
},
{
"input": ",)TH9N}'6t2+0Yg?S#6/{_.,!)9d}h'wG|sY&'Ul4D0l0",
"output": "YES"
},
{
"input": "VXB&r9Z)IlKOJ:??KDA",
"output": "YES"
},
{
"input": "\")1cL>{o\\dcYJzu?CefyN^bGRviOH&P7rJS3PT4:0V3F)%\\}L=AJouYsj_>j2|7^1NWu*%NbOP>ngv-ls<;b-4Sd3Na0R",
"output": "YES"
},
{
"input": "2Y}\\A)>row{~c[g>:'.|ZC8%UTQ/jcdhK%6O)QRC.kd@%y}LJYk=V{G5pQK/yKJ%{G3C",
"output": "YES"
},
{
"input": "O.&=qt(`z(",
"output": "NO"
},
{
"input": "_^r6fyIc/~~;>l%9?aVEi7-{=,[<aMiB'-scSg$$|\"jAzY0N>QkHHGBZj2c\"=fhRlWd5;5K|GgU?7h]!;wl@",
"output": "YES"
},
{
"input": "+/`sAd&eB29E=Nu87${.u6GY@$^a$,}s^!p!F}B-z8<<wORb<S7;HM1a,gp",
"output": "YES"
},
{
"input": "U_ilyOGMT+QiW/M8/D(1=6a7)_FA,h4`8",
"output": "YES"
},
{
"input": "!0WKT:$O",
"output": "NO"
},
{
"input": "1EE*I%EQz6$~pPu7|(r7nyPQt4uGU@]~H'4uII?b1_Wn)K?ZRHrr0z&Kr;}aO3<mN=3:{}QgPxI|Ncm4#)",
"output": "YES"
},
{
"input": "[u3\"$+!:/.<Dp1M7tH}:zxjt],^kv}qP;y12\"`^'/u*h%AFmPJ>e1#Yly",
"output": "YES"
},
{
"input": "'F!_]tB<A&UO+p?7liE>(x&RFgG2~\\(",
"output": "NO"
},
{
"input": "Qv)X8",
"output": "YES"
},
{
"input": "aGv7,J@&g1(}E3g6[LuDZwZl2<v7IwQA%\"R(?ouBD>_=y\"3Kf%^>vON<a^T\\G^ootgE@whWmZo=[ex|F",
"output": "YES"
},
{
"input": "e{}2vQ+/r@p0}cLKNe4MCk",
"output": "YES"
},
{
"input": "mzbmweyydiadtlcouegmdbyfwurpwbpuvhifnuapwyndmhtqvkgkbhtytszotwflegsjzzszfwtzfpnscguemwrczqxycivdqnkH",
"output": "YES"
},
{
"input": "Qzbmweyydiadtlcouegmdbyfwurpwbpuvhifnuapwyndmhtqvkgkbhtytszotwflegsjzzszfwtzfpnscguemwrczqxycivdqnky",
"output": "YES"
},
{
"input": "mzbmweyydiadtlcouegmdbyfwurpwb9uvhifnuapwyndmhtqvkgkbhtytszotwflegsjzzszfwtzfpnscguemwrczqxycivdqnky",
"output": "YES"
},
{
"input": "1H1",
"output": "YES"
},
{
"input": "+Q",
"output": "YES"
},
{
"input": "1ab",
"output": "NO"
},
{
"input": "!",
"output": "NO"
},
{
"input": "0+",
"output": "NO"
},
{
"input": "+H",
"output": "YES"
},
{
"input": "cH",
"output": "YES"
},
{
"input": "+8",
"output": "NO"
},
{
"input": "8+",
"output": "NO"
},
{
"input": "++++++++++++++++++++++++++",
"output": "NO"
},
{
"input": "(+)",
"output": "NO"
},
{
"input": "H+",
"output": "YES"
},
{
"input": "a!",
"output": "NO"
},
{
"input": "++++++++++++++",
"output": "NO"
},
{
"input": "+++++++++++++++++++++++++++++++++++++++++++++++++++++++++",
"output": "NO"
},
{
"input": "8",
"output": "NO"
},
{
"input": "3",
"output": "NO"
},
{
"input": "HQ9",
"output": "YES"
},
{
"input": "+++H",
"output": "YES"
},
{
"input": "++++++++++",
"output": "NO"
},
{
"input": "HHHHHHH",
"output": "YES"
},
{
"input": "abacabaH",
"output": "YES"
},
{
"input": "+G",
"output": "NO"
}
] | 1,690,708,890
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
#from collections import Counter
#import math
count_time=1
if count_time:
import time
start_time = time.time()
#------------------------------------------
p=input()
j=None
ok=False
for i in p:
if i=='+':
j=chr(ord(j)+1)
if j in 'HQ9':
ok=True
break
elif i in 'HQ9':
ok=True
break
if ok:print('YES')
else:print('NO')
#----------------------------------------------------
if count_time:
end_time = time.time()
print('----------------\nRunning time: {} s'
.format(end_time - start_time))
|
Title: HQ9+
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
HQ9+ is a joke programming language which has only four one-character instructions:
- "H" prints "Hello, World!",- "Q" prints the source code of the program itself,- "9" prints the lyrics of "99 Bottles of Beer" song, - "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input Specification:
The input will consist of a single line *p* which will give a program in HQ9+. String *p* will contain between 1 and 100 characters, inclusive. ASCII-code of each character of *p* will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output Specification:
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Demo Input:
['Hi!\n', 'Codeforces\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first case the program contains only one instruction — "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions.
|
```python
#from collections import Counter
#import math
count_time=1
if count_time:
import time
start_time = time.time()
#------------------------------------------
p=input()
j=None
ok=False
for i in p:
if i=='+':
j=chr(ord(j)+1)
if j in 'HQ9':
ok=True
break
elif i in 'HQ9':
ok=True
break
if ok:print('YES')
else:print('NO')
#----------------------------------------------------
if count_time:
end_time = time.time()
print('----------------\nRunning time: {} s'
.format(end_time - start_time))
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
|
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
|
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
|
[
"aaba\nabaa\n",
"aabb\nabab\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
| 0
|
[
{
"input": "aaba\nabaa",
"output": "YES"
},
{
"input": "aabb\nabab",
"output": "NO"
},
{
"input": "a\na",
"output": "YES"
},
{
"input": "a\nb",
"output": "NO"
},
{
"input": "ab\nab",
"output": "YES"
},
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "ab\nbb",
"output": "NO"
},
{
"input": "zzaa\naazz",
"output": "YES"
},
{
"input": "azza\nzaaz",
"output": "YES"
},
{
"input": "abc\nabc",
"output": "YES"
},
{
"input": "abc\nacb",
"output": "NO"
},
{
"input": "azzz\nzzaz",
"output": "YES"
},
{
"input": "abcd\ndcab",
"output": "YES"
},
{
"input": "abcd\ncdab",
"output": "YES"
},
{
"input": "abcd\ndcba",
"output": "YES"
},
{
"input": "abcd\nacbd",
"output": "NO"
},
{
"input": "oloaxgddgujq\noloaxgujqddg",
"output": "YES"
},
{
"input": "uwzwdxfmosmqatyv\ndxfmzwwusomqvyta",
"output": "YES"
},
{
"input": "hagnzomowtledfdotnll\nledfdotnllomowthagnz",
"output": "YES"
},
{
"input": "snyaydaeobufdg\nsnyaydaeobufdg",
"output": "YES"
},
{
"input": "baaaaa\nabaaaa",
"output": "NO"
},
{
"input": "hhiisug\nmzdjwju",
"output": "NO"
},
{
"input": "bbbabbabaaab\naaaabbabbbbb",
"output": "NO"
},
{
"input": "bbaaab\naababb",
"output": "NO"
},
{
"input": "aabbaaaa\naaaaabab",
"output": "NO"
},
{
"input": "aab\naba",
"output": "NO"
},
{
"input": "abcddd\nbacddd",
"output": "NO"
},
{
"input": "qgiufelsfhanx\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "aabaababaaba\naababaaababa",
"output": "NO"
},
{
"input": "nocdqzdriyyil\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "zdmctxl\nkojqhgw",
"output": "NO"
},
{
"input": "yhwepqwyhwepqwyhwepqweahnqtueahnqtueahnqtuyhwepqwyhwepqwyhwepqwyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtueahnqtueahnqtueahnqtueahnqtu\neahnqtueahnqtueahnqtuyhwepqweahnqtuyhwepqwyhwepqweahnqtuyhwepqweahnqtuyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtuyhwepqwyhwepqwyhwepqw",
"output": "NO"
},
{
"input": "abc\nbac",
"output": "NO"
},
{
"input": "ottceez\npcstdvz",
"output": "NO"
}
] | 1,619,302,706
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 31
| 409,600
|
def comparassion(palavra_1, palavra_2):
if(palavra_1 == palavra_2):
return True
if(len(palavra_1)!=len(palavra_2)):
return False
center = len(palavra_1)//2
return (comparassion(palavra_1[center:], palavra_2[center:]) and comparassion(palavra_1[:center], palavra_2[:center])) or (comparassion(palavra_1[center:], palavra_2[:center]) and comparassion(palavra_1[:center], palavra_2[center:]))
palavra_1 = input()
palavra_2 = input()
print("YES" if comparassion(palavra_1, palavra_2) else "NO")
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input Specification:
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
Output Specification:
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Demo Input:
['aaba\nabaa\n', 'aabb\nabab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa".
|
```python
def comparassion(palavra_1, palavra_2):
if(palavra_1 == palavra_2):
return True
if(len(palavra_1)!=len(palavra_2)):
return False
center = len(palavra_1)//2
return (comparassion(palavra_1[center:], palavra_2[center:]) and comparassion(palavra_1[:center], palavra_2[:center])) or (comparassion(palavra_1[center:], palavra_2[:center]) and comparassion(palavra_1[:center], palavra_2[center:]))
palavra_1 = input()
palavra_2 = input()
print("YES" if comparassion(palavra_1, palavra_2) else "NO")
```
| -1
|
|
989
|
B
|
A Tide of Riverscape
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"strings"
] | null | null |
"Time," Mino thinks aloud.
"What?"
"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."
"And what are you recording?"
"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.
Doubtfully, Kanno peeks at Mino's records.
The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).
You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.
In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$.
|
The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively.
The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.
|
Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).
|
[
"10 7\n1.0.1.0.1.\n",
"10 6\n1.0.1.1000\n",
"10 9\n1........1\n"
] |
[
"1000100010\n",
"1001101000\n",
"No\n"
] |
In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.
In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.
In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.
Note that there are multiple acceptable answers for the first two examples, you can print any of them.
| 1,000
|
[
{
"input": "10 7\n1.0.1.0.1.",
"output": "1000100010"
},
{
"input": "10 6\n1.0.1.1000",
"output": "1001101000"
},
{
"input": "10 9\n1........1",
"output": "No"
},
{
"input": "1 1\n.",
"output": "No"
},
{
"input": "5 1\n0...1",
"output": "00001"
},
{
"input": "17 10\n..1.100..1..0.100",
"output": "00101000010000100"
},
{
"input": "2 1\n0.",
"output": "01"
},
{
"input": "2 1\n..",
"output": "01"
},
{
"input": "3 1\n.0.",
"output": "001"
},
{
"input": "3 1\n00.",
"output": "001"
},
{
"input": "3 2\n0..",
"output": "001"
},
{
"input": "3 2\n0.0",
"output": "No"
},
{
"input": "3 2\n1..",
"output": "100"
},
{
"input": "3 2\n.1.",
"output": "011"
},
{
"input": "3 2\n1.0",
"output": "100"
},
{
"input": "3 3\n1..",
"output": "No"
},
{
"input": "3 3\n.00",
"output": "No"
},
{
"input": "5 3\n0.000",
"output": "01000"
},
{
"input": "10 6\n10010.1001",
"output": "No"
},
{
"input": "75 38\n00.0.1.0.0110.1.00010..100.1110..110..00.0.1.0.0110.1.00010..100.1110..110.",
"output": "000001000011001000010001000111000110000000010000110010000100010001110001101"
},
{
"input": "128 108\n01100.110...000.0001.1.11.11.010010.01100.0.1.01.0.0011.11001.000101...1.0.0..100.0110.0110.0.0101.0.0.0001.01100.110...100.0001",
"output": "01100011000000000001010110110010010001100000100100000110110010000101000100000010000110001100000101000000001001100011000010000001"
},
{
"input": "5 4\n.101.",
"output": "01011"
},
{
"input": "4 2\n101.",
"output": "1011"
},
{
"input": "5 4\n.1011",
"output": "01011"
},
{
"input": "2 1\n..",
"output": "01"
},
{
"input": "5 3\n00.11",
"output": "00011"
},
{
"input": "10 8\n1111.00000",
"output": "1111000000"
},
{
"input": "10 3\n11111111.1",
"output": "1111111101"
},
{
"input": "3 2\n1.0",
"output": "100"
},
{
"input": "6 4\n11..10",
"output": "110010"
},
{
"input": "4 2\n.111",
"output": "0111"
},
{
"input": "3 2\n01.",
"output": "011"
},
{
"input": "5 4\n10.00",
"output": "10000"
},
{
"input": "10 9\n1........0",
"output": "1000000000"
},
{
"input": "2 1\n0.",
"output": "01"
},
{
"input": "8 4\n111111..",
"output": "11111100"
},
{
"input": "3 2\n0.1",
"output": "001"
},
{
"input": "4 1\n111.",
"output": "1110"
},
{
"input": "3 1\n01.",
"output": "010"
},
{
"input": "10 7\n000....111",
"output": "0000000111"
}
] | 1,528,725,459
| 1,359
|
Python 3
|
CHALLENGED
|
CHALLENGES
| 8
| 140
| 716,800
|
import re
import math
import decimal
import bisect
def read():
return input().strip()
def iread():
return int(input().strip())
def viread():
return [int(x) for x in input().strip().split()]
# code goes here
n, p = viread()
line = read()
for i, c in enumerate(line):
try:
# print("at i = {}, c = {}, line[i + p] = {}".format(i, c, line[i + p]))
# print("c != '.': ", (c != '.'))
# print("line[i + p] == '.': ", (line[i + p] == '.'))
# print("line[i + p] != c: ", (line[i + p] != c))
# print("--------------")
# print("line[i + p] != '.': ", (line[i + p] != '.'))
# print("c == '.': ", (c == '.'))
# print("c != line[i + p]: ", (c != line[i + p]))
if c != '.' and (line[i + p] == '.' or line[i + p] != c):
# we can
# print("first condition is true")
notc = ('0' if c == '1' else '1').encode()[0]
newline = bytearray(line.encode())
if line[i + p] == '.':
newline[i + p] = notc
toprint = re.sub(r'\.', '0', newline.decode())
print(toprint)
exit(0)
elif line[i + p] != '.' and (c == '.' or c != line[i + p]):
# we can
# print("second condition is true")
notc = ('0' if line[i + p] == '1' else '1').encode()[0]
newline = bytearray(line.encode())
if c == '.':
newline[i] = notc
toprint = re.sub(r'\.', '0', newline.decode())
print(toprint)
exit(0)
# print("neither true, go to next...\n=================================")
except IndexError:
break
print("No")
|
Title: A Tide of Riverscape
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Time," Mino thinks aloud.
"What?"
"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."
"And what are you recording?"
"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.
Doubtfully, Kanno peeks at Mino's records.
The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).
You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.
In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1 \leq i \leq \lvert s \rvert - p$, the $i$-th and $(i + p)$-th characters of $s$ are the same. Here $\lvert s \rvert$ is the length of $s$.
Input Specification:
The first line contains two space-separated integers $n$ and $p$ ($1 \leq p \leq n \leq 2000$) — the length of the given string and the supposed period, respectively.
The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.
Output Specification:
Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).
Demo Input:
['10 7\n1.0.1.0.1.\n', '10 6\n1.0.1.1000\n', '10 9\n1........1\n']
Demo Output:
['1000100010\n', '1001101000\n', 'No\n']
Note:
In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.
In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.
In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.
Note that there are multiple acceptable answers for the first two examples, you can print any of them.
|
```python
import re
import math
import decimal
import bisect
def read():
return input().strip()
def iread():
return int(input().strip())
def viread():
return [int(x) for x in input().strip().split()]
# code goes here
n, p = viread()
line = read()
for i, c in enumerate(line):
try:
# print("at i = {}, c = {}, line[i + p] = {}".format(i, c, line[i + p]))
# print("c != '.': ", (c != '.'))
# print("line[i + p] == '.': ", (line[i + p] == '.'))
# print("line[i + p] != c: ", (line[i + p] != c))
# print("--------------")
# print("line[i + p] != '.': ", (line[i + p] != '.'))
# print("c == '.': ", (c == '.'))
# print("c != line[i + p]: ", (c != line[i + p]))
if c != '.' and (line[i + p] == '.' or line[i + p] != c):
# we can
# print("first condition is true")
notc = ('0' if c == '1' else '1').encode()[0]
newline = bytearray(line.encode())
if line[i + p] == '.':
newline[i + p] = notc
toprint = re.sub(r'\.', '0', newline.decode())
print(toprint)
exit(0)
elif line[i + p] != '.' and (c == '.' or c != line[i + p]):
# we can
# print("second condition is true")
notc = ('0' if line[i + p] == '1' else '1').encode()[0]
newline = bytearray(line.encode())
if c == '.':
newline[i] = notc
toprint = re.sub(r'\.', '0', newline.decode())
print(toprint)
exit(0)
# print("neither true, go to next...\n=================================")
except IndexError:
break
print("No")
```
| -1
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Students went into a class to write a test and sat in some way. The teacher thought: "Probably they sat in this order to copy works of each other. I need to rearrange them in such a way that students that were neighbors are not neighbors in a new seating."
The class can be represented as a matrix with *n* rows and *m* columns with a student in each cell. Two students are neighbors if cells in which they sit have a common side.
Let's enumerate students from 1 to *n*·*m* in order of rows. So a student who initially sits in the cell in row *i* and column *j* has a number (*i*<=-<=1)·*m*<=+<=*j*. You have to find a matrix with *n* rows and *m* columns in which all numbers from 1 to *n*·*m* appear exactly once and adjacent numbers in the original matrix are not adjacent in it, or determine that there is no such matrix.
|
The only line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105; *n*·*m*<=≤<=105) — the number of rows and the number of columns in the required matrix.
|
If there is no such matrix, output "NO" (without quotes).
Otherwise in the first line output "YES" (without quotes), and in the next *n* lines output *m* integers which form the required matrix.
|
[
"2 4\n",
"2 1\n"
] |
[
"YES\n5 4 7 2 \n3 6 1 8 \n",
"NO\n"
] |
In the first test case the matrix initially looks like this:
It's easy to see that there are no two students that are adjacent in both matrices.
In the second test case there are only two possible seatings and in both of them students with numbers 1 and 2 are neighbors.
| 0
|
[
{
"input": "2 4",
"output": "YES\n5 4 7 2 \n3 6 1 8 "
},
{
"input": "2 1",
"output": "NO"
},
{
"input": "1 1",
"output": "YES\n1"
},
{
"input": "1 2",
"output": "NO"
},
{
"input": "1 3",
"output": "NO"
},
{
"input": "2 2",
"output": "NO"
},
{
"input": "2 3",
"output": "NO"
},
{
"input": "3 1",
"output": "NO"
},
{
"input": "3 2",
"output": "NO"
},
{
"input": "3 3",
"output": "YES\n6 1 8\n7 5 3\n2 9 4"
},
{
"input": "1 4",
"output": "YES\n2 4 1 3"
},
{
"input": "4 1",
"output": "YES\n2\n4\n1\n3"
},
{
"input": "4 2",
"output": "YES\n2 5 \n7 4 \n6 1 \n3 8 "
},
{
"input": "1 100000",
"output": "YES\n1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 101 103 105 107 109 111 113 115 117 119 121 123 125 127 129 131 133 135 137 139 141 143 145 147 149 151 153 155 157 159 161 163 165 167 169 171 173 175 177 179 181 183 185 187 189 191 193 195 197 199 201 203 205 207 209 211 213 215 217 219 221 223 225 227 229 231 233 235 237 239 241 243 245 247 249 251 253 255 257 259 261 263 265 267 269 271 273 275 277 279 2..."
},
{
"input": "100000 1",
"output": "YES\n1\n3\n5\n7\n9\n11\n13\n15\n17\n19\n21\n23\n25\n27\n29\n31\n33\n35\n37\n39\n41\n43\n45\n47\n49\n51\n53\n55\n57\n59\n61\n63\n65\n67\n69\n71\n73\n75\n77\n79\n81\n83\n85\n87\n89\n91\n93\n95\n97\n99\n101\n103\n105\n107\n109\n111\n113\n115\n117\n119\n121\n123\n125\n127\n129\n131\n133\n135\n137\n139\n141\n143\n145\n147\n149\n151\n153\n155\n157\n159\n161\n163\n165\n167\n169\n171\n173\n175\n177\n179\n181\n183\n185\n187\n189\n191\n193\n195\n197\n199\n201\n203\n205\n207\n209\n211\n213\n215\n217\n219\n221\n223\n2..."
},
{
"input": "316 316",
"output": "YES\n317 4 319 6 321 8 323 10 325 12 327 14 329 16 331 18 333 20 335 22 337 24 339 26 341 28 343 30 345 32 347 34 349 36 351 38 353 40 355 42 357 44 359 46 361 48 363 50 365 52 367 54 369 56 371 58 373 60 375 62 377 64 379 66 381 68 383 70 385 72 387 74 389 76 391 78 393 80 395 82 397 84 399 86 401 88 403 90 405 92 407 94 409 96 411 98 413 100 415 102 417 104 419 106 421 108 423 110 425 112 427 114 429 116 431 118 433 120 435 122 437 124 439 126 441 128 443 130 445 132 447 134 449 136 451 138 453 140 455 1..."
},
{
"input": "315 316",
"output": "YES\n317 4 319 6 321 8 323 10 325 12 327 14 329 16 331 18 333 20 335 22 337 24 339 26 341 28 343 30 345 32 347 34 349 36 351 38 353 40 355 42 357 44 359 46 361 48 363 50 365 52 367 54 369 56 371 58 373 60 375 62 377 64 379 66 381 68 383 70 385 72 387 74 389 76 391 78 393 80 395 82 397 84 399 86 401 88 403 90 405 92 407 94 409 96 411 98 413 100 415 102 417 104 419 106 421 108 423 110 425 112 427 114 429 116 431 118 433 120 435 122 437 124 439 126 441 128 443 130 445 132 447 134 449 136 451 138 453 140 455 1..."
},
{
"input": "316 315",
"output": "YES\n2 633 4 635 6 637 8 639 10 641 12 643 14 645 16 647 18 649 20 651 22 653 24 655 26 657 28 659 30 661 32 663 34 665 36 667 38 669 40 671 42 673 44 675 46 677 48 679 50 681 52 683 54 685 56 687 58 689 60 691 62 693 64 695 66 697 68 699 70 701 72 703 74 705 76 707 78 709 80 711 82 713 84 715 86 717 88 719 90 721 92 723 94 725 96 727 98 729 100 731 102 733 104 735 106 737 108 739 110 741 112 743 114 745 116 747 118 749 120 751 122 753 124 755 126 757 128 759 130 761 132 763 134 765 136 767 138 769 140 771..."
},
{
"input": "315 315",
"output": "YES\n316 4 318 6 320 8 322 10 324 12 326 14 328 16 330 18 332 20 334 22 336 24 338 26 340 28 342 30 344 32 346 34 348 36 350 38 352 40 354 42 356 44 358 46 360 48 362 50 364 52 366 54 368 56 370 58 372 60 374 62 376 64 378 66 380 68 382 70 384 72 386 74 388 76 390 78 392 80 394 82 396 84 398 86 400 88 402 90 404 92 406 94 408 96 410 98 412 100 414 102 416 104 418 106 420 108 422 110 424 112 426 114 428 116 430 118 432 120 434 122 436 124 438 126 440 128 442 130 444 132 446 134 448 136 450 138 452 140 454 1..."
},
{
"input": "100 1000",
"output": "YES\n1001 4 1003 6 1005 8 1007 10 1009 12 1011 14 1013 16 1015 18 1017 20 1019 22 1021 24 1023 26 1025 28 1027 30 1029 32 1031 34 1033 36 1035 38 1037 40 1039 42 1041 44 1043 46 1045 48 1047 50 1049 52 1051 54 1053 56 1055 58 1057 60 1059 62 1061 64 1063 66 1065 68 1067 70 1069 72 1071 74 1073 76 1075 78 1077 80 1079 82 1081 84 1083 86 1085 88 1087 90 1089 92 1091 94 1093 96 1095 98 1097 100 1099 102 1101 104 1103 106 1105 108 1107 110 1109 112 1111 114 1113 116 1115 118 1117 120 1119 122 1121 124 1123 126..."
},
{
"input": "1000 100",
"output": "YES\n2 203 4 205 6 207 8 209 10 211 12 213 14 215 16 217 18 219 20 221 22 223 24 225 26 227 28 229 30 231 32 233 34 235 36 237 38 239 40 241 42 243 44 245 46 247 48 249 50 251 52 253 54 255 56 257 58 259 60 261 62 263 64 265 66 267 68 269 70 271 72 273 74 275 76 277 78 279 80 281 82 283 84 285 86 287 88 289 90 291 92 293 94 295 96 297 98 299 100 201 \n301 102 303 104 305 106 307 108 309 110 311 112 313 114 315 116 317 118 319 120 321 122 323 124 325 126 327 128 329 130 331 132 333 134 335 136 337 138 339 1..."
},
{
"input": "10 10000",
"output": "YES\n10001 4 10003 6 10005 8 10007 10 10009 12 10011 14 10013 16 10015 18 10017 20 10019 22 10021 24 10023 26 10025 28 10027 30 10029 32 10031 34 10033 36 10035 38 10037 40 10039 42 10041 44 10043 46 10045 48 10047 50 10049 52 10051 54 10053 56 10055 58 10057 60 10059 62 10061 64 10063 66 10065 68 10067 70 10069 72 10071 74 10073 76 10075 78 10077 80 10079 82 10081 84 10083 86 10085 88 10087 90 10089 92 10091 94 10093 96 10095 98 10097 100 10099 102 10101 104 10103 106 10105 108 10107 110 10109 112 10111 1..."
},
{
"input": "10000 10",
"output": "YES\n2 23 4 25 6 27 8 29 10 21 \n31 12 33 14 35 16 37 18 39 20 \n22 43 24 45 26 47 28 49 30 41 \n51 32 53 34 55 36 57 38 59 40 \n42 63 44 65 46 67 48 69 50 61 \n71 52 73 54 75 56 77 58 79 60 \n62 83 64 85 66 87 68 89 70 81 \n91 72 93 74 95 76 97 78 99 80 \n82 103 84 105 86 107 88 109 90 101 \n111 92 113 94 115 96 117 98 119 100 \n102 123 104 125 106 127 108 129 110 121 \n131 112 133 114 135 116 137 118 139 120 \n122 143 124 145 126 147 128 149 130 141 \n151 132 153 134 155 136 157 138 159 140 \n142 163 144..."
},
{
"input": "100 1",
"output": "YES\n1\n3\n5\n7\n9\n11\n13\n15\n17\n19\n21\n23\n25\n27\n29\n31\n33\n35\n37\n39\n41\n43\n45\n47\n49\n51\n53\n55\n57\n59\n61\n63\n65\n67\n69\n71\n73\n75\n77\n79\n81\n83\n85\n87\n89\n91\n93\n95\n97\n99\n2\n4\n6\n8\n10\n12\n14\n16\n18\n20\n22\n24\n26\n28\n30\n32\n34\n36\n38\n40\n42\n44\n46\n48\n50\n52\n54\n56\n58\n60\n62\n64\n66\n68\n70\n72\n74\n76\n78\n80\n82\n84\n86\n88\n90\n92\n94\n96\n98\n100"
},
{
"input": "1 100",
"output": "YES\n1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100 "
},
{
"input": "100 2",
"output": "YES\n2 5 \n7 4 \n6 9 \n11 8 \n10 13 \n15 12 \n14 17 \n19 16 \n18 21 \n23 20 \n22 25 \n27 24 \n26 29 \n31 28 \n30 33 \n35 32 \n34 37 \n39 36 \n38 41 \n43 40 \n42 45 \n47 44 \n46 49 \n51 48 \n50 53 \n55 52 \n54 57 \n59 56 \n58 61 \n63 60 \n62 65 \n67 64 \n66 69 \n71 68 \n70 73 \n75 72 \n74 77 \n79 76 \n78 81 \n83 80 \n82 85 \n87 84 \n86 89 \n91 88 \n90 93 \n95 92 \n94 97 \n99 96 \n98 101 \n103 100 \n102 105 \n107 104 \n106 109 \n111 108 \n110 113 \n115 112 \n114 117 \n119 116 \n118 121 \n123 120 \n122 125 \n..."
},
{
"input": "2 100",
"output": "YES\n101 4 103 6 105 8 107 10 109 12 111 14 113 16 115 18 117 20 119 22 121 24 123 26 125 28 127 30 129 32 131 34 133 36 135 38 137 40 139 42 141 44 143 46 145 48 147 50 149 52 151 54 153 56 155 58 157 60 159 62 161 64 163 66 165 68 167 70 169 72 171 74 173 76 175 78 177 80 179 82 181 84 183 86 185 88 187 90 189 92 191 94 193 96 195 98 197 100 199 2 \n3 102 5 104 7 106 9 108 11 110 13 112 15 114 17 116 19 118 21 120 23 122 25 124 27 126 29 128 31 130 33 132 35 134 37 136 39 138 41 140 43 142 45 144 47 146 ..."
},
{
"input": "100 3",
"output": "YES\n2 9 7 \n10 5 12 \n8 15 13 \n16 11 18 \n14 21 19 \n22 17 24 \n20 27 25 \n28 23 30 \n26 33 31 \n34 29 36 \n32 39 37 \n40 35 42 \n38 45 43 \n46 41 48 \n44 51 49 \n52 47 54 \n50 57 55 \n58 53 60 \n56 63 61 \n64 59 66 \n62 69 67 \n70 65 72 \n68 75 73 \n76 71 78 \n74 81 79 \n82 77 84 \n80 87 85 \n88 83 90 \n86 93 91 \n94 89 96 \n92 99 97 \n100 95 102 \n98 105 103 \n106 101 108 \n104 111 109 \n112 107 114 \n110 117 115 \n118 113 120 \n116 123 121 \n124 119 126 \n122 129 127 \n130 125 132 \n128 135 133 \n136 ..."
},
{
"input": "3 100",
"output": "YES\n101 4 103 6 105 8 107 10 109 12 111 14 113 16 115 18 117 20 119 22 121 24 123 26 125 28 127 30 129 32 131 34 133 36 135 38 137 40 139 42 141 44 143 46 145 48 147 50 149 52 151 54 153 56 155 58 157 60 159 62 161 64 163 66 165 68 167 70 169 72 171 74 173 76 175 78 177 80 179 82 181 84 183 86 185 88 187 90 189 92 191 94 193 96 195 98 197 100 199 2 \n203 102 205 104 207 106 209 108 211 110 213 112 215 114 217 116 219 118 221 120 223 122 225 124 227 126 229 128 231 130 233 132 235 134 237 136 239 138 241 1..."
},
{
"input": "100 4",
"output": "YES\n2 11 4 9 \n13 6 15 8 \n10 19 12 17 \n21 14 23 16 \n18 27 20 25 \n29 22 31 24 \n26 35 28 33 \n37 30 39 32 \n34 43 36 41 \n45 38 47 40 \n42 51 44 49 \n53 46 55 48 \n50 59 52 57 \n61 54 63 56 \n58 67 60 65 \n69 62 71 64 \n66 75 68 73 \n77 70 79 72 \n74 83 76 81 \n85 78 87 80 \n82 91 84 89 \n93 86 95 88 \n90 99 92 97 \n101 94 103 96 \n98 107 100 105 \n109 102 111 104 \n106 115 108 113 \n117 110 119 112 \n114 123 116 121 \n125 118 127 120 \n122 131 124 129 \n133 126 135 128 \n130 139 132 137 \n141 134 143 ..."
},
{
"input": "4 100",
"output": "YES\n101 4 103 6 105 8 107 10 109 12 111 14 113 16 115 18 117 20 119 22 121 24 123 26 125 28 127 30 129 32 131 34 133 36 135 38 137 40 139 42 141 44 143 46 145 48 147 50 149 52 151 54 153 56 155 58 157 60 159 62 161 64 163 66 165 68 167 70 169 72 171 74 173 76 175 78 177 80 179 82 181 84 183 86 185 88 187 90 189 92 191 94 193 96 195 98 197 100 199 2 \n203 102 205 104 207 106 209 108 211 110 213 112 215 114 217 116 219 118 221 120 223 122 225 124 227 126 229 128 231 130 233 132 235 134 237 136 239 138 241 1..."
},
{
"input": "101 1",
"output": "YES\n1\n3\n5\n7\n9\n11\n13\n15\n17\n19\n21\n23\n25\n27\n29\n31\n33\n35\n37\n39\n41\n43\n45\n47\n49\n51\n53\n55\n57\n59\n61\n63\n65\n67\n69\n71\n73\n75\n77\n79\n81\n83\n85\n87\n89\n91\n93\n95\n97\n99\n101\n2\n4\n6\n8\n10\n12\n14\n16\n18\n20\n22\n24\n26\n28\n30\n32\n34\n36\n38\n40\n42\n44\n46\n48\n50\n52\n54\n56\n58\n60\n62\n64\n66\n68\n70\n72\n74\n76\n78\n80\n82\n84\n86\n88\n90\n92\n94\n96\n98\n100"
},
{
"input": "1 101",
"output": "YES\n1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 101 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100 "
},
{
"input": "101 2",
"output": "YES\n2 5 \n7 4 \n6 9 \n11 8 \n10 13 \n15 12 \n14 17 \n19 16 \n18 21 \n23 20 \n22 25 \n27 24 \n26 29 \n31 28 \n30 33 \n35 32 \n34 37 \n39 36 \n38 41 \n43 40 \n42 45 \n47 44 \n46 49 \n51 48 \n50 53 \n55 52 \n54 57 \n59 56 \n58 61 \n63 60 \n62 65 \n67 64 \n66 69 \n71 68 \n70 73 \n75 72 \n74 77 \n79 76 \n78 81 \n83 80 \n82 85 \n87 84 \n86 89 \n91 88 \n90 93 \n95 92 \n94 97 \n99 96 \n98 101 \n103 100 \n102 105 \n107 104 \n106 109 \n111 108 \n110 113 \n115 112 \n114 117 \n119 116 \n118 121 \n123 120 \n122 125 \n..."
},
{
"input": "2 101",
"output": "YES\n102 4 104 6 106 8 108 10 110 12 112 14 114 16 116 18 118 20 120 22 122 24 124 26 126 28 128 30 130 32 132 34 134 36 136 38 138 40 140 42 142 44 144 46 146 48 148 50 150 52 152 54 154 56 156 58 158 60 160 62 162 64 164 66 166 68 168 70 170 72 172 74 174 76 176 78 178 80 180 82 182 84 184 86 186 88 188 90 190 92 192 94 194 96 196 98 198 100 200 1 202 \n3 103 5 105 7 107 9 109 11 111 13 113 15 115 17 117 19 119 21 121 23 123 25 125 27 127 29 129 31 131 33 133 35 135 37 137 39 139 41 141 43 143 45 145 47 ..."
},
{
"input": "101 3",
"output": "YES\n2 9 7 \n10 5 12 \n8 15 13 \n16 11 18 \n14 21 19 \n22 17 24 \n20 27 25 \n28 23 30 \n26 33 31 \n34 29 36 \n32 39 37 \n40 35 42 \n38 45 43 \n46 41 48 \n44 51 49 \n52 47 54 \n50 57 55 \n58 53 60 \n56 63 61 \n64 59 66 \n62 69 67 \n70 65 72 \n68 75 73 \n76 71 78 \n74 81 79 \n82 77 84 \n80 87 85 \n88 83 90 \n86 93 91 \n94 89 96 \n92 99 97 \n100 95 102 \n98 105 103 \n106 101 108 \n104 111 109 \n112 107 114 \n110 117 115 \n118 113 120 \n116 123 121 \n124 119 126 \n122 129 127 \n130 125 132 \n128 135 133 \n136 ..."
},
{
"input": "3 101",
"output": "YES\n102 4 104 6 106 8 108 10 110 12 112 14 114 16 116 18 118 20 120 22 122 24 124 26 126 28 128 30 130 32 132 34 134 36 136 38 138 40 140 42 142 44 144 46 146 48 148 50 150 52 152 54 154 56 156 58 158 60 160 62 162 64 164 66 166 68 168 70 170 72 172 74 174 76 176 78 178 80 180 82 182 84 184 86 186 88 188 90 190 92 192 94 194 96 196 98 198 100 200 1 202 \n205 103 207 105 209 107 211 109 213 111 215 113 217 115 219 117 221 119 223 121 225 123 227 125 229 127 231 129 233 131 235 133 237 135 239 137 241 139 2..."
},
{
"input": "101 4",
"output": "YES\n2 11 4 9 \n13 6 15 8 \n10 19 12 17 \n21 14 23 16 \n18 27 20 25 \n29 22 31 24 \n26 35 28 33 \n37 30 39 32 \n34 43 36 41 \n45 38 47 40 \n42 51 44 49 \n53 46 55 48 \n50 59 52 57 \n61 54 63 56 \n58 67 60 65 \n69 62 71 64 \n66 75 68 73 \n77 70 79 72 \n74 83 76 81 \n85 78 87 80 \n82 91 84 89 \n93 86 95 88 \n90 99 92 97 \n101 94 103 96 \n98 107 100 105 \n109 102 111 104 \n106 115 108 113 \n117 110 119 112 \n114 123 116 121 \n125 118 127 120 \n122 131 124 129 \n133 126 135 128 \n130 139 132 137 \n141 134 143 ..."
},
{
"input": "4 101",
"output": "YES\n102 4 104 6 106 8 108 10 110 12 112 14 114 16 116 18 118 20 120 22 122 24 124 26 126 28 128 30 130 32 132 34 134 36 136 38 138 40 140 42 142 44 144 46 146 48 148 50 150 52 152 54 154 56 156 58 158 60 160 62 162 64 164 66 166 68 168 70 170 72 172 74 174 76 176 78 178 80 180 82 182 84 184 86 186 88 188 90 190 92 192 94 194 96 196 98 198 100 200 1 202 \n205 103 207 105 209 107 211 109 213 111 215 113 217 115 219 117 221 119 223 121 225 123 227 125 229 127 231 129 233 131 235 133 237 135 239 137 241 139 2..."
},
{
"input": "6 16666",
"output": "YES\n16667 4 16669 6 16671 8 16673 10 16675 12 16677 14 16679 16 16681 18 16683 20 16685 22 16687 24 16689 26 16691 28 16693 30 16695 32 16697 34 16699 36 16701 38 16703 40 16705 42 16707 44 16709 46 16711 48 16713 50 16715 52 16717 54 16719 56 16721 58 16723 60 16725 62 16727 64 16729 66 16731 68 16733 70 16735 72 16737 74 16739 76 16741 78 16743 80 16745 82 16747 84 16749 86 16751 88 16753 90 16755 92 16757 94 16759 96 16761 98 16763 100 16765 102 16767 104 16769 106 16771 108 16773 110 16775 112 16777 1..."
},
{
"input": "314 315",
"output": "YES\n316 4 318 6 320 8 322 10 324 12 326 14 328 16 330 18 332 20 334 22 336 24 338 26 340 28 342 30 344 32 346 34 348 36 350 38 352 40 354 42 356 44 358 46 360 48 362 50 364 52 366 54 368 56 370 58 372 60 374 62 376 64 378 66 380 68 382 70 384 72 386 74 388 76 390 78 392 80 394 82 396 84 398 86 400 88 402 90 404 92 406 94 408 96 410 98 412 100 414 102 416 104 418 106 420 108 422 110 424 112 426 114 428 116 430 118 432 120 434 122 436 124 438 126 440 128 442 130 444 132 446 134 448 136 450 138 452 140 454 1..."
},
{
"input": "2 20",
"output": "YES\n21 4 23 6 25 8 27 10 29 12 31 14 33 16 35 18 37 20 39 2 \n3 22 5 24 7 26 9 28 11 30 13 32 15 34 17 36 19 38 1 40 "
}
] | 1,514,045,032
| 7,132
|
PyPy 3
|
WRONG_ANSWER
|
PRETESTS
| 0
| 93
| 25,497,600
|
#!/usr/bin/env python3
from random import shuffle
def solve():
n, m = get([int])
if n+m < 4:
return 'NO'
L = [(i-1)*m + j for i in range(1,n+1) for j in range(1,m+1)]
shuffle(L)
L = list(zip(*([iter(L)]*m)))
L = [' '.join(map(str, row)) for row in L]
return '\n'.join(L)
_testcases = """
2 1
a
""".strip()
# ======================= B O I L E R P L A T E ======================= #
# Practicality beats purity
from bisect import bisect_left, bisect_right
from collections import defaultdict
from functools import lru_cache
from heapq import heapify, heappop, heappush
from operator import itemgetter, attrgetter
import bisect
import collections
import functools
import heapq
import itertools
import math
import operator
import re
import string
import sys
inf = float('inf')
cache = lru_cache(None)
sys.setrecursionlimit(10000)
def tree():
return collections.defaultdict(tree)
def equal(x, y, epsilon=1e-6):
# https://code.google.com/codejam/kickstart/resources/faq#real-number-behavior
if -epsilon <= x - y <= epsilon:
return True
if -epsilon <= x <= epsilon or -epsilon <= y <= epsilon:
return False
return (-epsilon <= (x - y) / x <= epsilon or -epsilon <= (x - y) / y <= epsilon)
def get(_type): # For easy input
if type(_type) == list:
if len(_type) == 1:
_type = _type[0]
return list(map(_type, input().strip().split()))
else:
return [_type[i](inp) for i, inp in enumerate(input().strip().split())]
else:
return _type(input().strip())
if __name__ == '__main__':
printRecursionTree = timeit = lambda x: x
_p, print = print, lambda *a, **b: None
_p(solve())
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Students went into a class to write a test and sat in some way. The teacher thought: "Probably they sat in this order to copy works of each other. I need to rearrange them in such a way that students that were neighbors are not neighbors in a new seating."
The class can be represented as a matrix with *n* rows and *m* columns with a student in each cell. Two students are neighbors if cells in which they sit have a common side.
Let's enumerate students from 1 to *n*·*m* in order of rows. So a student who initially sits in the cell in row *i* and column *j* has a number (*i*<=-<=1)·*m*<=+<=*j*. You have to find a matrix with *n* rows and *m* columns in which all numbers from 1 to *n*·*m* appear exactly once and adjacent numbers in the original matrix are not adjacent in it, or determine that there is no such matrix.
Input Specification:
The only line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105; *n*·*m*<=≤<=105) — the number of rows and the number of columns in the required matrix.
Output Specification:
If there is no such matrix, output "NO" (without quotes).
Otherwise in the first line output "YES" (without quotes), and in the next *n* lines output *m* integers which form the required matrix.
Demo Input:
['2 4\n', '2 1\n']
Demo Output:
['YES\n5 4 7 2 \n3 6 1 8 \n', 'NO\n']
Note:
In the first test case the matrix initially looks like this:
It's easy to see that there are no two students that are adjacent in both matrices.
In the second test case there are only two possible seatings and in both of them students with numbers 1 and 2 are neighbors.
|
```python
#!/usr/bin/env python3
from random import shuffle
def solve():
n, m = get([int])
if n+m < 4:
return 'NO'
L = [(i-1)*m + j for i in range(1,n+1) for j in range(1,m+1)]
shuffle(L)
L = list(zip(*([iter(L)]*m)))
L = [' '.join(map(str, row)) for row in L]
return '\n'.join(L)
_testcases = """
2 1
a
""".strip()
# ======================= B O I L E R P L A T E ======================= #
# Practicality beats purity
from bisect import bisect_left, bisect_right
from collections import defaultdict
from functools import lru_cache
from heapq import heapify, heappop, heappush
from operator import itemgetter, attrgetter
import bisect
import collections
import functools
import heapq
import itertools
import math
import operator
import re
import string
import sys
inf = float('inf')
cache = lru_cache(None)
sys.setrecursionlimit(10000)
def tree():
return collections.defaultdict(tree)
def equal(x, y, epsilon=1e-6):
# https://code.google.com/codejam/kickstart/resources/faq#real-number-behavior
if -epsilon <= x - y <= epsilon:
return True
if -epsilon <= x <= epsilon or -epsilon <= y <= epsilon:
return False
return (-epsilon <= (x - y) / x <= epsilon or -epsilon <= (x - y) / y <= epsilon)
def get(_type): # For easy input
if type(_type) == list:
if len(_type) == 1:
_type = _type[0]
return list(map(_type, input().strip().split()))
else:
return [_type[i](inp) for i, inp in enumerate(input().strip().split())]
else:
return _type(input().strip())
if __name__ == '__main__':
printRecursionTree = timeit = lambda x: x
_p, print = print, lambda *a, **b: None
_p(solve())
```
| 0
|
|
612
|
C
|
Replace To Make Regular Bracket Sequence
|
PROGRAMMING
| 1,400
|
[
"data structures",
"expression parsing",
"math"
] | null | null |
You are given string *s* consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let *s*1 and *s*2 be a RBS then the strings <*s*1>*s*2, {*s*1}*s*2, [*s*1]*s*2, (*s*1)*s*2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string *s* RBS.
|
The only line contains a non empty string *s*, consisting of only opening and closing brackets of four kinds. The length of *s* does not exceed 106.
|
If it's impossible to get RBS from *s* print Impossible.
Otherwise print the least number of replaces needed to get RBS from *s*.
|
[
"[<}){}\n",
"{()}[]\n",
"]]\n"
] |
[
"2",
"0",
"Impossible"
] |
none
| 0
|
[
{
"input": "[<}){}",
"output": "2"
},
{
"input": "{()}[]",
"output": "0"
},
{
"input": "]]",
"output": "Impossible"
},
{
"input": ">",
"output": "Impossible"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{]",
"output": "1"
},
{
"input": "{]",
"output": "1"
},
{
"input": "{]",
"output": "1"
},
{
"input": "[]{[]({)([",
"output": "Impossible"
},
{
"input": "(([{>}{[{[)]]>>]",
"output": "7"
},
{
"input": "((<>)[]<]><]",
"output": "3"
},
{
"input": "[[([[(>]>)))[<)>",
"output": "6"
},
{
"input": "({)[}<)](}",
"output": "5"
},
{
"input": "(}{)[<][)(]}",
"output": "6"
},
{
"input": ">}({>]{[}<{<{{)[]]{)]>]]]<(][{)<<<{<<)>)()[>{<]]{}<>}}}}(>}<})(][>{((<{<)]}>)))][>[}[])<]){]]][<[)([",
"output": "Impossible"
},
{
"input": "<<[<{{<([({<<[)<>(]]){})>[](])[)))[[}>]<)>[[>{>>>[<]}<>>)[>]<{)<[><(<]][>(>]>][(<][{]}(()<[()[>><<])<]})]<]}{)",
"output": "45"
},
{
"input": "[<<{{((}[}<<)<)>})(][{>}})((>)<[)[>}[})[)>()[()[((}<<(>)<>](<>(}[>})[[[{)<}<<(}{>>}[<([[])<><)]<{>}[>>>{({>)}]})>)",
"output": "42"
},
{
"input": "(<[([(<({>(}{]>[(})])}])()<<}{]{[>]>(>>[(>>}[){(}<[{(()]{{<(<{][[{<><{<{)<>>]}}}{)(}{})}[<))>>}((({>){({}{{]}]>>}})>))",
"output": "43"
},
{
"input": "{(]}<([]<]{>]<{<({{{<>))}[({(}{)[}({>]}}<<)}<]))(<>(>{>{{{)<}({<]<>{)(>[)>{({}<([<[[)]><>{]}}(>]{}<)[})]}]]}]}>}",
"output": "40"
},
{
"input": "<(([)]<[}>{)>][[(<()({{{>>((]<}<{{<>}><<[)}[[)([[)[)}<]<{(<>){([)><{[([[][({])})<][(}]}>>[){)[({>){(}[}})>}>]>)]}))})>}]",
"output": "45"
},
{
"input": ")(",
"output": "Impossible"
},
{
"input": "[[[[[[[[",
"output": "Impossible"
}
] | 1,532,831,523
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 93
| 0
|
st = input()
a1 = st.count('(')-st.count(')')
a2 = st.count('<')-st.count('>')
a3 = st.count('{')-st.count('}')
a4 = st.count('[')-st.count(']')
if a1+a2+a3+a4 ==0:
print(abs(a1)+abs(a2)+abs(a3)+abs(4))
else:
print("Impossible")
|
Title: Replace To Make Regular Bracket Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given string *s* consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let *s*1 and *s*2 be a RBS then the strings <*s*1>*s*2, {*s*1}*s*2, [*s*1]*s*2, (*s*1)*s*2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string *s* RBS.
Input Specification:
The only line contains a non empty string *s*, consisting of only opening and closing brackets of four kinds. The length of *s* does not exceed 106.
Output Specification:
If it's impossible to get RBS from *s* print Impossible.
Otherwise print the least number of replaces needed to get RBS from *s*.
Demo Input:
['[<}){}\n', '{()}[]\n', ']]\n']
Demo Output:
['2', '0', 'Impossible']
Note:
none
|
```python
st = input()
a1 = st.count('(')-st.count(')')
a2 = st.count('<')-st.count('>')
a3 = st.count('{')-st.count('}')
a4 = st.count('[')-st.count(']')
if a1+a2+a3+a4 ==0:
print(abs(a1)+abs(a2)+abs(a3)+abs(4))
else:
print("Impossible")
```
| 0
|
|
34
|
A
|
Reconnaissance 2
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Reconnaissance 2
|
2
|
256
|
*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
|
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.
|
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
|
[
"5\n10 12 13 15 10\n",
"4\n10 20 30 40\n"
] |
[
"5 1\n",
"1 2\n"
] |
none
| 500
|
[
{
"input": "5\n10 12 13 15 10",
"output": "5 1"
},
{
"input": "4\n10 20 30 40",
"output": "1 2"
},
{
"input": "6\n744 359 230 586 944 442",
"output": "2 3"
},
{
"input": "5\n826 747 849 687 437",
"output": "1 2"
},
{
"input": "5\n999 999 993 969 999",
"output": "1 2"
},
{
"input": "5\n4 24 6 1 15",
"output": "3 4"
},
{
"input": "2\n511 32",
"output": "1 2"
},
{
"input": "3\n907 452 355",
"output": "2 3"
},
{
"input": "4\n303 872 764 401",
"output": "4 1"
},
{
"input": "10\n684 698 429 694 956 812 594 170 937 764",
"output": "1 2"
},
{
"input": "20\n646 840 437 946 640 564 936 917 487 752 844 734 468 969 674 646 728 642 514 695",
"output": "7 8"
},
{
"input": "30\n996 999 998 984 989 1000 996 993 1000 983 992 999 999 1000 979 992 987 1000 996 1000 1000 989 981 996 995 999 999 989 999 1000",
"output": "12 13"
},
{
"input": "50\n93 27 28 4 5 78 59 24 19 134 31 128 118 36 90 32 32 1 44 32 33 13 31 10 12 25 38 50 25 12 4 22 28 53 48 83 4 25 57 31 71 24 8 7 28 86 23 80 101 58",
"output": "16 17"
},
{
"input": "88\n1000 1000 1000 1000 1000 998 998 1000 1000 1000 1000 999 999 1000 1000 1000 999 1000 997 999 997 1000 999 998 1000 999 1000 1000 1000 999 1000 999 999 1000 1000 999 1000 999 1000 1000 998 1000 1000 1000 998 998 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 999 1000 1000 999 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 998 1000 1000 1000 998 1000 1000 998 1000 999 1000 1000 1000 1000",
"output": "1 2"
},
{
"input": "99\n4 4 21 6 5 3 13 2 6 1 3 4 1 3 1 9 11 1 6 17 4 5 20 4 1 9 5 11 3 4 14 1 3 3 1 4 3 5 27 1 1 2 10 7 11 4 19 7 11 6 11 13 3 1 10 7 2 1 16 1 9 4 29 13 2 12 14 2 21 1 9 8 26 12 12 5 2 14 7 8 8 8 9 4 12 2 6 6 7 16 8 14 2 10 20 15 3 7 4",
"output": "1 2"
},
{
"input": "100\n713 572 318 890 577 657 646 146 373 783 392 229 455 871 20 593 573 336 26 381 280 916 907 732 820 713 111 840 570 446 184 711 481 399 788 647 492 15 40 530 549 506 719 782 126 20 778 996 712 761 9 74 812 418 488 175 103 585 900 3 604 521 109 513 145 708 990 361 682 827 791 22 596 780 596 385 450 643 158 496 876 975 319 783 654 895 891 361 397 81 682 899 347 623 809 557 435 279 513 438",
"output": "86 87"
},
{
"input": "100\n31 75 86 68 111 27 22 22 26 30 54 163 107 75 160 122 14 23 17 26 27 20 43 58 59 71 21 148 9 32 43 91 133 286 132 70 90 156 84 14 77 93 23 18 13 72 18 131 33 28 72 175 30 86 249 20 14 208 28 57 63 199 6 10 24 30 62 267 43 479 60 28 138 1 45 3 19 47 7 166 116 117 50 140 28 14 95 85 93 43 61 15 2 70 10 51 7 95 9 25",
"output": "7 8"
},
{
"input": "100\n896 898 967 979 973 709 961 968 806 967 896 967 826 975 936 903 986 856 851 931 852 971 786 837 949 978 686 936 952 909 965 749 908 916 943 973 983 975 939 886 964 928 960 976 907 788 994 773 949 871 947 980 945 985 726 981 887 943 907 990 931 874 840 867 948 951 961 904 888 901 976 967 994 921 828 970 972 722 755 970 860 855 914 869 714 899 969 978 898 862 642 939 904 936 819 934 884 983 955 964",
"output": "1 2"
},
{
"input": "100\n994 927 872 970 815 986 952 996 965 1000 877 986 978 999 950 990 936 997 993 960 921 860 895 869 943 998 983 968 973 953 999 990 995 871 853 979 973 963 953 938 997 989 993 964 960 973 946 975 1000 962 920 746 989 957 904 965 920 979 966 961 1000 993 975 952 846 971 991 979 985 969 984 973 956 1000 952 778 983 974 956 927 995 997 980 997 1000 970 960 970 988 983 947 904 935 972 1000 863 992 996 932 967",
"output": "81 82"
},
{
"input": "100\n48 108 63 21 27 8 49 21 75 8 24 42 149 18 8 28 21 18 25 35 59 70 59 33 40 1 67 34 120 82 4 115 72 87 3 15 15 63 37 12 40 27 83 14 38 20 14 58 93 10 31 3 39 6 197 77 54 16 31 146 9 49 14 8 77 82 5 11 80 116 8 61 50 24 7 103 29 11 3 3 1 12 46 24 21 131 39 29 36 2 107 40 16 99 31 41 29 48 17 17",
"output": "36 37"
}
] | 1,532,040,691
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 248
| 0
|
n = int(input())
values = [int(k) for k in input().split()]
minimo = 1000000000
j = 0
for i in range(-1,n-1):
if abs(values[i] - values[i+1])< minimo:
minimo = abs(values[i] - values[i+1])
j = i
if j<0:
print(n, 1)
else:
print(j+1, j+2)
|
Title: Reconnaissance 2
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
*n* soldiers stand in a circle. For each soldier his height *a**i* is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |*a**i*<=-<=*a**j*| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000). The soldier heights are given in clockwise or counterclockwise direction.
Output Specification:
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Demo Input:
['5\n10 12 13 15 10\n', '4\n10 20 30 40\n']
Demo Output:
['5 1\n', '1 2\n']
Note:
none
|
```python
n = int(input())
values = [int(k) for k in input().split()]
minimo = 1000000000
j = 0
for i in range(-1,n-1):
if abs(values[i] - values[i+1])< minimo:
minimo = abs(values[i] - values[i+1])
j = i
if j<0:
print(n, 1)
else:
print(j+1, j+2)
```
| 3.938
|
710
|
B
|
Optimal Point on a Line
|
PROGRAMMING
| 1,400
|
[
"brute force",
"sortings"
] | null | null |
You are given *n* points on a line with their coordinates *x**i*. Find the point *x* so the sum of distances to the given points is minimal.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of points on the line.
The second line contains *n* integers *x**i* (<=-<=109<=≤<=*x**i*<=≤<=109) — the coordinates of the given *n* points.
|
Print the only integer *x* — the position of the optimal point on the line. If there are several optimal points print the position of the leftmost one. It is guaranteed that the answer is always the integer.
|
[
"4\n1 2 3 4\n"
] |
[
"2\n"
] |
none
| 0
|
[
{
"input": "4\n1 2 3 4",
"output": "2"
},
{
"input": "5\n-1 -10 2 6 7",
"output": "2"
},
{
"input": "10\n-68 10 87 22 30 89 82 -97 -52 25",
"output": "22"
},
{
"input": "100\n457 827 807 17 871 935 907 -415 536 170 551 -988 865 758 -457 -892 -875 -488 684 19 0 555 -807 -624 -239 826 318 811 20 -732 -91 460 551 -610 555 -493 -154 442 -141 946 -913 -104 704 -380 699 32 106 -455 -518 214 -464 -861 243 -798 -472 559 529 -844 -32 871 -459 236 387 626 -318 -580 -611 -842 790 486 64 951 81 78 -693 403 -731 309 678 696 891 846 -106 918 212 -44 994 606 -829 -454 243 -477 -402 -818 -819 -310 -837 -209 736 424",
"output": "64"
},
{
"input": "2\n-1 0",
"output": "-1"
},
{
"input": "48\n-777 -767 -764 -713 -688 -682 -606 -586 -585 -483 -465 -440 -433 -397 -390 -377 -299 -252 -159 -147 -96 -29 -15 15 52 109 124 129 142 218 231 314 320 339 442 496 505 548 575 576 594 624 694 827 891 979 981 981",
"output": "15"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 1 1 1 1 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "4\n-1 -1 0 1",
"output": "-1"
},
{
"input": "10\n0 0 0 0 0 0 0 0 0 1000000000",
"output": "0"
},
{
"input": "2\n1 -1",
"output": "-1"
},
{
"input": "2\n100 50",
"output": "50"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "1\n10",
"output": "10"
},
{
"input": "3\n606194955 -856471310 117647402",
"output": "117647402"
},
{
"input": "2\n615002717 -843553590",
"output": "-843553590"
},
{
"input": "2\n-1 2",
"output": "-1"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n2",
"output": "2"
},
{
"input": "5\n-638512131 348325781 -550537933 -618161835 -567935532",
"output": "-567935532"
},
{
"input": "1\n120",
"output": "120"
},
{
"input": "2\n-1000000000 1000000000",
"output": "-1000000000"
},
{
"input": "1\n618309368",
"output": "618309368"
}
] | 1,471,985,078
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 296
| 27,443,200
|
n = int(input())
A = list(map(int, input().split()))
A.sort()
print(A[n//2 - (1 if n % 2 == 0 else 0)])
|
Title: Optimal Point on a Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* points on a line with their coordinates *x**i*. Find the point *x* so the sum of distances to the given points is minimal.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of points on the line.
The second line contains *n* integers *x**i* (<=-<=109<=≤<=*x**i*<=≤<=109) — the coordinates of the given *n* points.
Output Specification:
Print the only integer *x* — the position of the optimal point on the line. If there are several optimal points print the position of the leftmost one. It is guaranteed that the answer is always the integer.
Demo Input:
['4\n1 2 3 4\n']
Demo Output:
['2\n']
Note:
none
|
```python
n = int(input())
A = list(map(int, input().split()))
A.sort()
print(A[n//2 - (1 if n % 2 == 0 else 0)])
```
| 3
|
|
676
|
C
|
Vasya and String
|
PROGRAMMING
| 1,500
|
[
"binary search",
"dp",
"strings",
"two pointers"
] | null | null |
High school student Vasya got a string of length *n* as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than *k* characters of the original string. What is the maximum beauty of the string he can achieve?
|
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000,<=0<=≤<=*k*<=≤<=*n*) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only.
|
Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than *k* characters.
|
[
"4 2\nabba\n",
"8 1\naabaabaa\n"
] |
[
"4\n",
"5\n"
] |
In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
| 1,500
|
[
{
"input": "4 2\nabba",
"output": "4"
},
{
"input": "8 1\naabaabaa",
"output": "5"
},
{
"input": "1 0\na",
"output": "1"
},
{
"input": "1 1\nb",
"output": "1"
},
{
"input": "1 0\nb",
"output": "1"
},
{
"input": "1 1\na",
"output": "1"
},
{
"input": "10 10\nbbbbbbbbbb",
"output": "10"
},
{
"input": "10 2\nbbbbbbbbbb",
"output": "10"
},
{
"input": "10 1\nbbabbabbba",
"output": "6"
},
{
"input": "10 10\nbbabbbaabb",
"output": "10"
},
{
"input": "10 9\nbabababbba",
"output": "10"
},
{
"input": "10 4\nbababbaaab",
"output": "9"
},
{
"input": "10 10\naabaaabaaa",
"output": "10"
},
{
"input": "10 10\naaaabbbaaa",
"output": "10"
},
{
"input": "10 1\nbaaaaaaaab",
"output": "9"
},
{
"input": "10 5\naaaaabaaaa",
"output": "10"
},
{
"input": "10 4\naaaaaaaaaa",
"output": "10"
},
{
"input": "100 10\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "100"
},
{
"input": "100 7\nbbbbabbbbbaabbbabbbbbbbbbbbabbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbbbabbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbab",
"output": "93"
},
{
"input": "100 30\nbbaabaaabbbbbbbbbbaababababbbbbbaabaabbbbbbbbabbbbbabbbbabbbbbbbbaabbbbbbbbbabbbbbabbbbbbbbbaaaaabba",
"output": "100"
},
{
"input": "100 6\nbaababbbaabbabbaaabbabbaabbbbbbbbaabbbabbbbaabbabbbbbabababbbbabbbbbbabbbbbbbbbaaaabbabbbbaabbabaabb",
"output": "34"
},
{
"input": "100 45\naabababbabbbaaabbbbbbaabbbabbaabbbbbabbbbbbbbabbbbbbabbaababbaabbababbbbbbababbbbbaabbbbbbbaaaababab",
"output": "100"
},
{
"input": "100 2\nababaabababaaababbaaaabbaabbbababbbaaabbbbabababbbabababaababaaabaabbbbaaabbbabbbbbabbbbbbbaabbabbba",
"output": "17"
},
{
"input": "100 25\nbabbbaaababaaabbbaabaabaabbbabbabbbbaaaaaaabaaabaaaaaaaaaabaaaabaaabbbaaabaaababaaabaabbbbaaaaaaaaaa",
"output": "80"
},
{
"input": "100 14\naabaaaaabababbabbabaaaabbaaaabaaabbbaaabaaaaaaaabaaaaabbaaaaaaaaabaaaaaaabbaababaaaababbbbbabaaaabaa",
"output": "61"
},
{
"input": "100 8\naaaaabaaaabaabaaaaaaaabaaaabaaaaaaaaaaaaaabaaaaabaaaaaaaaaaaaaaaaabaaaababaabaaaaaaaaaaaaabbabaaaaaa",
"output": "76"
},
{
"input": "100 12\naaaaaaaaaaaaaaaabaaabaaaaaaaaaabbaaaabbabaaaaaaaaaaaaaaaaaaaaabbaaabaaaaaaaaaaaabaaaaaaaabaaaaaaaaaa",
"output": "100"
},
{
"input": "100 65\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "100"
},
{
"input": "10 0\nbbbbbbbbbb",
"output": "10"
},
{
"input": "10 0\nbbbbabbbbb",
"output": "5"
},
{
"input": "10 0\nbbabbbabba",
"output": "3"
},
{
"input": "10 0\nbaabbbbaba",
"output": "4"
},
{
"input": "10 0\naababbbbaa",
"output": "4"
},
{
"input": "10 2\nabbbbbaaba",
"output": "8"
},
{
"input": "10 0\nabbaaabaaa",
"output": "3"
},
{
"input": "10 0\naabbaaabaa",
"output": "3"
},
{
"input": "10 1\naaaaaababa",
"output": "8"
},
{
"input": "10 0\nbaaaaaaaaa",
"output": "9"
},
{
"input": "10 0\naaaaaaaaaa",
"output": "10"
},
{
"input": "100 0\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "100"
},
{
"input": "100 0\nbbbbbbbbbbabbbbaaabbbbbbbbbbbabbbabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbabbbbbbbbbabbbbbbbbbbbbbab",
"output": "40"
},
{
"input": "100 11\nbaabbbbbababbbbabbbbbbbabbbbbbbbbbbbbbabbbbbbababbbbababbbbaaabbbbabbbbbabbbbbbbbabababbbabbbbbbbabb",
"output": "65"
},
{
"input": "100 8\nbbababbbbbaabbbaaababbbbababababbbbababbabbbabbbbbaabbbabbbababbabbbbabbbabbbbaabbbbabbbaabbbbaaaabb",
"output": "33"
},
{
"input": "100 21\nabbaaaabbbababaabbbababbbbbbbbabbaababababbbabbbaaabbaaabbbbabbabbbabbbabaababbbabbbbbabbbbbbabbbbab",
"output": "65"
},
{
"input": "100 9\nabbbaabaabaaaaaaabbabbbababbaaabbbaaabbaabaaaaabbbbbabbaabaabbbbbaaaaababbaaabbabaabaaabababbaababbb",
"output": "26"
},
{
"input": "100 5\naababababbaaaaaaaabbbabaaaabbabaaaabbaabaaaaabababbabaabaaabaaaaaaaabaababbabbaaabaabbabbaaaaabbabba",
"output": "22"
},
{
"input": "100 9\naababaabaaaaaaaaabbbaabaaaaaaabaaaaaaaaaaaaabaaabaabaabbbbabbaababbabbaaaabbababaabaababaabaaaaaaaaa",
"output": "49"
},
{
"input": "100 6\naaaaabbaaaaaaaaaaabaaaabaaaaaaaaabaaabaaaaaabaaaaaaaaaaabaabaaaabaaaaaaaaaaaaaaabaabbaaaaaaaaaaaaaaa",
"output": "56"
},
{
"input": "100 7\nabaaabaabaabaaaaaabaaaaaaaaaaaaaaaaaaaaaaaaaaaaabaaabaaaaaaabbabaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaba",
"output": "86"
},
{
"input": "100 0\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "100"
}
] | 1,691,815,733
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 117
| 109
| 102,400
|
def max_beauty(n, k, s):
max_len = 0
for c in ['a', 'b']:
start = 0
changes = 0
for end in range(n):
if s[end] != c:
changes += 1
while changes > k:
if s[start] != c:
changes -= 1
start += 1
max_len = max(max_len, end - start + 1)
return max_len
# Read input
n, k = map(int, input().split())
s = input()
# Calculate and print the result
result = max_beauty(n, k, s)
print(result)
|
Title: Vasya and String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
High school student Vasya got a string of length *n* as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than *k* characters of the original string. What is the maximum beauty of the string he can achieve?
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000,<=0<=≤<=*k*<=≤<=*n*) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only.
Output Specification:
Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than *k* characters.
Demo Input:
['4 2\nabba\n', '8 1\naabaabaa\n']
Demo Output:
['4\n', '5\n']
Note:
In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
|
```python
def max_beauty(n, k, s):
max_len = 0
for c in ['a', 'b']:
start = 0
changes = 0
for end in range(n):
if s[end] != c:
changes += 1
while changes > k:
if s[start] != c:
changes -= 1
start += 1
max_len = max(max_len, end - start + 1)
return max_len
# Read input
n, k = map(int, input().split())
s = input()
# Calculate and print the result
result = max_beauty(n, k, s)
print(result)
```
| 3
|
|
45
|
A
|
Codecraft III
|
PROGRAMMING
| 900
|
[
"implementation"
] |
A. Codecraft III
|
2
|
256
|
Today Vasya visited a widely known site and learned that the continuation of his favourite game Codecraft II will appear after exactly *k* months. He looked at the calendar and learned that at the moment is the month number *s*. Vasya immediately got interested in what month Codecraft III will appear. Help him understand that.
All the twelve months in Vasya's calendar are named using their usual English names: January, February, March, April, May, June, July, August, September, October, November, December.
|
The first input line contains the name of the current month. It is guaranteed that it is a proper English name of one of twelve months. The first letter is uppercase, the rest are lowercase. The second line contains integer *k* (0<=≤<=*k*<=≤<=100) — the number of months left till the appearance of Codecraft III.
|
Print starting from an uppercase letter the name of the month in which the continuation of Codeforces II will appear. The printed name must be contained in the list January, February, March, April, May, June, July, August, September, October, November, December.
|
[
"November\n3\n",
"May\n24\n"
] |
[
"February\n",
"May\n"
] |
none
| 0
|
[
{
"input": "November\n3",
"output": "February"
},
{
"input": "May\n24",
"output": "May"
},
{
"input": "April\n0",
"output": "April"
},
{
"input": "September\n0",
"output": "September"
},
{
"input": "August\n0",
"output": "August"
},
{
"input": "June\n1",
"output": "July"
},
{
"input": "July\n2",
"output": "September"
},
{
"input": "September\n3",
"output": "December"
},
{
"input": "July\n4",
"output": "November"
},
{
"input": "August\n24",
"output": "August"
},
{
"input": "May\n48",
"output": "May"
},
{
"input": "November\n47",
"output": "October"
},
{
"input": "December\n49",
"output": "January"
},
{
"input": "June\n99",
"output": "September"
},
{
"input": "March\n100",
"output": "July"
},
{
"input": "December\n1",
"output": "January"
},
{
"input": "January\n11",
"output": "December"
},
{
"input": "December\n0",
"output": "December"
},
{
"input": "January\n0",
"output": "January"
},
{
"input": "July\n77",
"output": "December"
},
{
"input": "February\n11",
"output": "January"
},
{
"input": "February\n22",
"output": "December"
},
{
"input": "July\n33",
"output": "April"
},
{
"input": "May\n44",
"output": "January"
},
{
"input": "June\n97",
"output": "July"
}
] | 1,390,484,477
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
months = ["January", "February", "March", "April", "May",
"June", "July", "August", "September",
"October", "November", "December"]
current_month = raw_input()
months_left = int(raw_input())
i = months.index(current_month)
print months[(i+months_left)%12]
|
Title: Codecraft III
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Today Vasya visited a widely known site and learned that the continuation of his favourite game Codecraft II will appear after exactly *k* months. He looked at the calendar and learned that at the moment is the month number *s*. Vasya immediately got interested in what month Codecraft III will appear. Help him understand that.
All the twelve months in Vasya's calendar are named using their usual English names: January, February, March, April, May, June, July, August, September, October, November, December.
Input Specification:
The first input line contains the name of the current month. It is guaranteed that it is a proper English name of one of twelve months. The first letter is uppercase, the rest are lowercase. The second line contains integer *k* (0<=≤<=*k*<=≤<=100) — the number of months left till the appearance of Codecraft III.
Output Specification:
Print starting from an uppercase letter the name of the month in which the continuation of Codeforces II will appear. The printed name must be contained in the list January, February, March, April, May, June, July, August, September, October, November, December.
Demo Input:
['November\n3\n', 'May\n24\n']
Demo Output:
['February\n', 'May\n']
Note:
none
|
```python
months = ["January", "February", "March", "April", "May",
"June", "July", "August", "September",
"October", "November", "December"]
current_month = raw_input()
months_left = int(raw_input())
i = months.index(current_month)
print months[(i+months_left)%12]
```
| -1
|
854
|
A
|
Fraction
|
PROGRAMMING
| 800
|
[
"brute force",
"constructive algorithms",
"math"
] | null | null |
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
|
In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.
|
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
|
[
"3\n",
"4\n",
"12\n"
] |
[
"1 2\n",
"1 3\n",
"5 7\n"
] |
none
| 500
|
[
{
"input": "3",
"output": "1 2"
},
{
"input": "4",
"output": "1 3"
},
{
"input": "12",
"output": "5 7"
},
{
"input": "34",
"output": "15 19"
},
{
"input": "13",
"output": "6 7"
},
{
"input": "11",
"output": "5 6"
},
{
"input": "24",
"output": "11 13"
},
{
"input": "17",
"output": "8 9"
},
{
"input": "10",
"output": "3 7"
},
{
"input": "69",
"output": "34 35"
},
{
"input": "100",
"output": "49 51"
},
{
"input": "57",
"output": "28 29"
},
{
"input": "1000",
"output": "499 501"
},
{
"input": "999",
"output": "499 500"
},
{
"input": "998",
"output": "497 501"
},
{
"input": "997",
"output": "498 499"
},
{
"input": "996",
"output": "497 499"
},
{
"input": "995",
"output": "497 498"
},
{
"input": "994",
"output": "495 499"
},
{
"input": "5",
"output": "2 3"
},
{
"input": "6",
"output": "1 5"
},
{
"input": "8",
"output": "3 5"
},
{
"input": "9",
"output": "4 5"
},
{
"input": "423",
"output": "211 212"
},
{
"input": "876",
"output": "437 439"
},
{
"input": "29",
"output": "14 15"
}
] | 1,531,915,544
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 109
| 0
|
from math import gcd
n = int(input())
for a, b in zip(range((n - 1) // 2, 0, -1), range((n + 2) // 2, n)):
if gcd(a, b) == 1:
print(a, b)
break
|
Title: Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
Input Specification:
In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.
Output Specification:
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
Demo Input:
['3\n', '4\n', '12\n']
Demo Output:
['1 2\n', '1 3\n', '5 7\n']
Note:
none
|
```python
from math import gcd
n = int(input())
for a, b in zip(range((n - 1) // 2, 0, -1), range((n + 2) // 2, n)):
if gcd(a, b) == 1:
print(a, b)
break
```
| 3
|
|
678
|
A
|
Johny Likes Numbers
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
|
The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109).
|
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
|
[
"5 3\n",
"25 13\n",
"26 13\n"
] |
[
"6\n",
"26\n",
"39\n"
] |
none
| 0
|
[
{
"input": "5 3",
"output": "6"
},
{
"input": "25 13",
"output": "26"
},
{
"input": "26 13",
"output": "39"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "8 8",
"output": "16"
},
{
"input": "14 15",
"output": "15"
},
{
"input": "197 894",
"output": "894"
},
{
"input": "6058 8581",
"output": "8581"
},
{
"input": "97259 41764",
"output": "125292"
},
{
"input": "453145 333625",
"output": "667250"
},
{
"input": "2233224 4394826",
"output": "4394826"
},
{
"input": "76770926 13350712",
"output": "80104272"
},
{
"input": "687355301 142098087",
"output": "710490435"
},
{
"input": "1000000000 999999999",
"output": "1999999998"
},
{
"input": "1000000000 1000000000",
"output": "2000000000"
},
{
"input": "999999999 1000000000",
"output": "1000000000"
},
{
"input": "1000000000 1",
"output": "1000000001"
},
{
"input": "1000000000 2",
"output": "1000000002"
},
{
"input": "999999999 1",
"output": "1000000000"
},
{
"input": "100000000 1",
"output": "100000001"
},
{
"input": "999999999 500000000",
"output": "1000000000"
},
{
"input": "999999990 10",
"output": "1000000000"
},
{
"input": "1000000000 999999997",
"output": "1999999994"
},
{
"input": "999999999 2",
"output": "1000000000"
},
{
"input": "999999984 1",
"output": "999999985"
},
{
"input": "999999983 1",
"output": "999999984"
},
{
"input": "666666666 1",
"output": "666666667"
},
{
"input": "1000000000 990000000",
"output": "1980000000"
},
{
"input": "41 48",
"output": "48"
},
{
"input": "123456 2",
"output": "123458"
},
{
"input": "111 111",
"output": "222"
},
{
"input": "878787 1",
"output": "878788"
},
{
"input": "121 1",
"output": "122"
},
{
"input": "114514 114514",
"output": "229028"
},
{
"input": "500000001 1000000000",
"output": "1000000000"
},
{
"input": "999999997 1",
"output": "999999998"
},
{
"input": "100000000 10",
"output": "100000010"
}
] | 1,624,679,818
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 77
| 0
|
a, b = map(int, input().split())
c = a % b
if c == 0:
print(a+b)
else:
print(a + (b - c))
|
Title: Johny Likes Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109).
Output Specification:
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
Demo Input:
['5 3\n', '25 13\n', '26 13\n']
Demo Output:
['6\n', '26\n', '39\n']
Note:
none
|
```python
a, b = map(int, input().split())
c = a % b
if c == 0:
print(a+b)
else:
print(a + (b - c))
```
| 3
|
|
468
|
A
|
24 Game
|
PROGRAMMING
| 1,500
|
[
"constructive algorithms",
"greedy",
"math"
] | null | null |
Little X used to play a card game called "24 Game", but recently he has found it too easy. So he invented a new game.
Initially you have a sequence of *n* integers: 1,<=2,<=...,<=*n*. In a single step, you can pick two of them, let's denote them *a* and *b*, erase them from the sequence, and append to the sequence either *a*<=+<=*b*, or *a*<=-<=*b*, or *a*<=×<=*b*.
After *n*<=-<=1 steps there is only one number left. Can you make this number equal to 24?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105).
|
If it's possible, print "YES" in the first line. Otherwise, print "NO" (without the quotes).
If there is a way to obtain 24 as the result number, in the following *n*<=-<=1 lines print the required operations an operation per line. Each operation should be in form: "*a* *op* *b* = *c*". Where *a* and *b* are the numbers you've picked at this operation; *op* is either "+", or "-", or "*"; *c* is the result of corresponding operation. Note, that the absolute value of *c* mustn't be greater than 1018. The result of the last operation must be equal to 24. Separate operator sign and equality sign from numbers with spaces.
If there are multiple valid answers, you may print any of them.
|
[
"1\n",
"8\n"
] |
[
"NO\n",
"YES\n8 * 7 = 56\n6 * 5 = 30\n3 - 4 = -1\n1 - 2 = -1\n30 - -1 = 31\n56 - 31 = 25\n25 + -1 = 24\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "NO"
},
{
"input": "8",
"output": "YES\n8 * 7 = 56\n6 * 5 = 30\n3 - 4 = -1\n1 - 2 = -1\n30 - -1 = 31\n56 - 31 = 25\n25 + -1 = 24"
},
{
"input": "12",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24"
},
{
"input": "100",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "1000",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "987",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24\n9 - 8 = 1\n24 * 1 = 24\n11 - 10 = 1\n24 * 1 = 24\n13 - 12 = 1\n24 * 1 = 24\n15 - 14 = 1\n24 * 1 = 24\n17 - 16 = 1\n24 * 1 = 24\n19 - 18 = 1\n24 * 1 = 24\n21 - 20 = 1\n24 * 1 = 24\n23 - 22 = 1\n24 * 1 = 24\n25 - 24 = 1\n24 * 1 = 24\n27 - 26 = 1\n24 * 1 = 24\n29 - 28 = 1\n24 * 1 = 24\n31 - 30 = 1\n24 * 1 = 24\n33 - 32 = 1\n24 * 1 = 24\n35 - 34 = 1\n24 * 1 = 24\n37 - 36 = 1\n24 * 1 = 24\n39 - 38 = 1\n24 * 1 = 24\n41 - 40 = 1\n24 * 1 = 2..."
},
{
"input": "2",
"output": "NO"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24"
},
{
"input": "5",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24"
},
{
"input": "6",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24"
},
{
"input": "7",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24"
},
{
"input": "100000",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "99999",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24\n9 - 8 = 1\n24 * 1 = 24\n11 - 10 = 1\n24 * 1 = 24\n13 - 12 = 1\n24 * 1 = 24\n15 - 14 = 1\n24 * 1 = 24\n17 - 16 = 1\n24 * 1 = 24\n19 - 18 = 1\n24 * 1 = 24\n21 - 20 = 1\n24 * 1 = 24\n23 - 22 = 1\n24 * 1 = 24\n25 - 24 = 1\n24 * 1 = 24\n27 - 26 = 1\n24 * 1 = 24\n29 - 28 = 1\n24 * 1 = 24\n31 - 30 = 1\n24 * 1 = 24\n33 - 32 = 1\n24 * 1 = 24\n35 - 34 = 1\n24 * 1 = 24\n37 - 36 = 1\n24 * 1 = 24\n39 - 38 = 1\n24 * 1 = 24\n41 - 40 = 1\n24 * 1 = 2..."
},
{
"input": "99998",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "99997",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24\n9 - 8 = 1\n24 * 1 = 24\n11 - 10 = 1\n24 * 1 = 24\n13 - 12 = 1\n24 * 1 = 24\n15 - 14 = 1\n24 * 1 = 24\n17 - 16 = 1\n24 * 1 = 24\n19 - 18 = 1\n24 * 1 = 24\n21 - 20 = 1\n24 * 1 = 24\n23 - 22 = 1\n24 * 1 = 24\n25 - 24 = 1\n24 * 1 = 24\n27 - 26 = 1\n24 * 1 = 24\n29 - 28 = 1\n24 * 1 = 24\n31 - 30 = 1\n24 * 1 = 24\n33 - 32 = 1\n24 * 1 = 24\n35 - 34 = 1\n24 * 1 = 24\n37 - 36 = 1\n24 * 1 = 24\n39 - 38 = 1\n24 * 1 = 24\n41 - 40 = 1\n24 * 1 = 2..."
},
{
"input": "580",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "422",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "116",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "447",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24\n9 - 8 = 1\n24 * 1 = 24\n11 - 10 = 1\n24 * 1 = 24\n13 - 12 = 1\n24 * 1 = 24\n15 - 14 = 1\n24 * 1 = 24\n17 - 16 = 1\n24 * 1 = 24\n19 - 18 = 1\n24 * 1 = 24\n21 - 20 = 1\n24 * 1 = 24\n23 - 22 = 1\n24 * 1 = 24\n25 - 24 = 1\n24 * 1 = 24\n27 - 26 = 1\n24 * 1 = 24\n29 - 28 = 1\n24 * 1 = 24\n31 - 30 = 1\n24 * 1 = 24\n33 - 32 = 1\n24 * 1 = 24\n35 - 34 = 1\n24 * 1 = 24\n37 - 36 = 1\n24 * 1 = 24\n39 - 38 = 1\n24 * 1 = 24\n41 - 40 = 1\n24 * 1 = 2..."
},
{
"input": "62052",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "25770",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "56118",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "86351",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24\n9 - 8 = 1\n24 * 1 = 24\n11 - 10 = 1\n24 * 1 = 24\n13 - 12 = 1\n24 * 1 = 24\n15 - 14 = 1\n24 * 1 = 24\n17 - 16 = 1\n24 * 1 = 24\n19 - 18 = 1\n24 * 1 = 24\n21 - 20 = 1\n24 * 1 = 24\n23 - 22 = 1\n24 * 1 = 24\n25 - 24 = 1\n24 * 1 = 24\n27 - 26 = 1\n24 * 1 = 24\n29 - 28 = 1\n24 * 1 = 24\n31 - 30 = 1\n24 * 1 = 24\n33 - 32 = 1\n24 * 1 = 24\n35 - 34 = 1\n24 * 1 = 24\n37 - 36 = 1\n24 * 1 = 24\n39 - 38 = 1\n24 * 1 = 24\n41 - 40 = 1\n24 * 1 = 2..."
},
{
"input": "48108",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "33373",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24\n9 - 8 = 1\n24 * 1 = 24\n11 - 10 = 1\n24 * 1 = 24\n13 - 12 = 1\n24 * 1 = 24\n15 - 14 = 1\n24 * 1 = 24\n17 - 16 = 1\n24 * 1 = 24\n19 - 18 = 1\n24 * 1 = 24\n21 - 20 = 1\n24 * 1 = 24\n23 - 22 = 1\n24 * 1 = 24\n25 - 24 = 1\n24 * 1 = 24\n27 - 26 = 1\n24 * 1 = 24\n29 - 28 = 1\n24 * 1 = 24\n31 - 30 = 1\n24 * 1 = 24\n33 - 32 = 1\n24 * 1 = 24\n35 - 34 = 1\n24 * 1 = 24\n37 - 36 = 1\n24 * 1 = 24\n39 - 38 = 1\n24 * 1 = 24\n41 - 40 = 1\n24 * 1 = 2..."
},
{
"input": "9782",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "19082",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24\n6 - 5 = 1\n24 * 1 = 24\n8 - 7 = 1\n24 * 1 = 24\n10 - 9 = 1\n24 * 1 = 24\n12 - 11 = 1\n24 * 1 = 24\n14 - 13 = 1\n24 * 1 = 24\n16 - 15 = 1\n24 * 1 = 24\n18 - 17 = 1\n24 * 1 = 24\n20 - 19 = 1\n24 * 1 = 24\n22 - 21 = 1\n24 * 1 = 24\n24 - 23 = 1\n24 * 1 = 24\n26 - 25 = 1\n24 * 1 = 24\n28 - 27 = 1\n24 * 1 = 24\n30 - 29 = 1\n24 * 1 = 24\n32 - 31 = 1\n24 * 1 = 24\n34 - 33 = 1\n24 * 1 = 24\n36 - 35 = 1\n24 * 1 = 24\n38 - 37 = 1\n24 * 1 = 24\n40 - 39 = 1\n24 * 1 = 24\n42 - 41..."
},
{
"input": "4",
"output": "YES\n3 * 4 = 12\n2 * 1 = 2\n12 * 2 = 24"
},
{
"input": "7",
"output": "YES\n5 - 3 = 2\n2 * 4 = 8\n1 + 2 = 3\n8 * 3 = 24\n7 - 6 = 1\n24 * 1 = 24"
},
{
"input": "3",
"output": "NO"
}
] | 1,642,655,359
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 31
| 264
| 9,932,800
|
n = int(input())
if n in [1, 2, 3]:
print("NO")
else:
print("YES")
if n % 2 == 0:
for i in range((n - 4) // 2):
print(f"{n - 2 * i} - {n - 2 * i - 1} = 1")
print("4 * 3 = 12", "12 * 2 = 24", "24 * 1 = 24", sep = '\n')
for j in range((n - 4) // 2):
print("24 * 1 = 24")
else:
for i in range((n - 5) // 2):
print(f"{n - 2 * i} - {n - 2 * i - 1} = 1")
print("2 + 1 = 3", "5 - 3 = 2", "2 * 3 = 6", "6 * 4 = 24", sep = '\n')
for j in range((n - 5) // 2):
print("24 * 1 = 24")
|
Title: 24 Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little X used to play a card game called "24 Game", but recently he has found it too easy. So he invented a new game.
Initially you have a sequence of *n* integers: 1,<=2,<=...,<=*n*. In a single step, you can pick two of them, let's denote them *a* and *b*, erase them from the sequence, and append to the sequence either *a*<=+<=*b*, or *a*<=-<=*b*, or *a*<=×<=*b*.
After *n*<=-<=1 steps there is only one number left. Can you make this number equal to 24?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105).
Output Specification:
If it's possible, print "YES" in the first line. Otherwise, print "NO" (without the quotes).
If there is a way to obtain 24 as the result number, in the following *n*<=-<=1 lines print the required operations an operation per line. Each operation should be in form: "*a* *op* *b* = *c*". Where *a* and *b* are the numbers you've picked at this operation; *op* is either "+", or "-", or "*"; *c* is the result of corresponding operation. Note, that the absolute value of *c* mustn't be greater than 1018. The result of the last operation must be equal to 24. Separate operator sign and equality sign from numbers with spaces.
If there are multiple valid answers, you may print any of them.
Demo Input:
['1\n', '8\n']
Demo Output:
['NO\n', 'YES\n8 * 7 = 56\n6 * 5 = 30\n3 - 4 = -1\n1 - 2 = -1\n30 - -1 = 31\n56 - 31 = 25\n25 + -1 = 24\n']
Note:
none
|
```python
n = int(input())
if n in [1, 2, 3]:
print("NO")
else:
print("YES")
if n % 2 == 0:
for i in range((n - 4) // 2):
print(f"{n - 2 * i} - {n - 2 * i - 1} = 1")
print("4 * 3 = 12", "12 * 2 = 24", "24 * 1 = 24", sep = '\n')
for j in range((n - 4) // 2):
print("24 * 1 = 24")
else:
for i in range((n - 5) // 2):
print(f"{n - 2 * i} - {n - 2 * i - 1} = 1")
print("2 + 1 = 3", "5 - 3 = 2", "2 * 3 = 6", "6 * 4 = 24", sep = '\n')
for j in range((n - 5) // 2):
print("24 * 1 = 24")
```
| 3
|
|
118
|
A
|
String Task
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
|
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
|
Print the resulting string. It is guaranteed that this string is not empty.
|
[
"tour\n",
"Codeforces\n",
"aBAcAba\n"
] |
[
".t.r\n",
".c.d.f.r.c.s\n",
".b.c.b\n"
] |
none
| 500
|
[
{
"input": "tour",
"output": ".t.r"
},
{
"input": "Codeforces",
"output": ".c.d.f.r.c.s"
},
{
"input": "aBAcAba",
"output": ".b.c.b"
},
{
"input": "obn",
"output": ".b.n"
},
{
"input": "wpwl",
"output": ".w.p.w.l"
},
{
"input": "ggdvq",
"output": ".g.g.d.v.q"
},
{
"input": "pumesz",
"output": ".p.m.s.z"
},
{
"input": "g",
"output": ".g"
},
{
"input": "zjuotps",
"output": ".z.j.t.p.s"
},
{
"input": "jzbwuehe",
"output": ".j.z.b.w.h"
},
{
"input": "tnkgwuugu",
"output": ".t.n.k.g.w.g"
},
{
"input": "kincenvizh",
"output": ".k.n.c.n.v.z.h"
},
{
"input": "xattxjenual",
"output": ".x.t.t.x.j.n.l"
},
{
"input": "ktajqhpqsvhw",
"output": ".k.t.j.q.h.p.q.s.v.h.w"
},
{
"input": "xnhcigytnqcmy",
"output": ".x.n.h.c.g.t.n.q.c.m"
},
{
"input": "jfmtbejyilxcec",
"output": ".j.f.m.t.b.j.l.x.c.c"
},
{
"input": "D",
"output": ".d"
},
{
"input": "ab",
"output": ".b"
},
{
"input": "Ab",
"output": ".b"
},
{
"input": "aB",
"output": ".b"
},
{
"input": "AB",
"output": ".b"
},
{
"input": "ba",
"output": ".b"
},
{
"input": "bA",
"output": ".b"
},
{
"input": "Ba",
"output": ".b"
},
{
"input": "BA",
"output": ".b"
},
{
"input": "aab",
"output": ".b"
},
{
"input": "baa",
"output": ".b"
},
{
"input": "femOZeCArKCpUiHYnbBPTIOFmsHmcpObtPYcLCdjFrUMIyqYzAokKUiiKZRouZiNMoiOuGVoQzaaCAOkquRjmmKKElLNqCnhGdQM",
"output": ".f.m.z.c.r.k.c.p.h.n.b.b.p.t.f.m.s.h.m.c.p.b.t.p.c.l.c.d.j.f.r.m.q.z.k.k.k.z.r.z.n.m.g.v.q.z.c.k.q.r.j.m.m.k.k.l.l.n.q.c.n.h.g.d.q.m"
},
{
"input": "VMBPMCmMDCLFELLIISUJDWQRXYRDGKMXJXJHXVZADRZWVWJRKFRRNSAWKKDPZZLFLNSGUNIVJFBEQsMDHSBJVDTOCSCgZWWKvZZN",
"output": ".v.m.b.p.m.c.m.m.d.c.l.f.l.l.s.j.d.w.q.r.x.r.d.g.k.m.x.j.x.j.h.x.v.z.d.r.z.w.v.w.j.r.k.f.r.r.n.s.w.k.k.d.p.z.z.l.f.l.n.s.g.n.v.j.f.b.q.s.m.d.h.s.b.j.v.d.t.c.s.c.g.z.w.w.k.v.z.z.n"
},
{
"input": "MCGFQQJNUKuAEXrLXibVjClSHjSxmlkQGTKZrRaDNDomIPOmtSgjJAjNVIVLeUGUAOHNkCBwNObVCHOWvNkLFQQbFnugYVMkJruJ",
"output": ".m.c.g.f.q.q.j.n.k.x.r.l.x.b.v.j.c.l.s.h.j.s.x.m.l.k.q.g.t.k.z.r.r.d.n.d.m.p.m.t.s.g.j.j.j.n.v.v.l.g.h.n.k.c.b.w.n.b.v.c.h.w.v.n.k.l.f.q.q.b.f.n.g.v.m.k.j.r.j"
},
{
"input": "iyaiuiwioOyzUaOtAeuEYcevvUyveuyioeeueoeiaoeiavizeeoeyYYaaAOuouueaUioueauayoiuuyiuovyOyiyoyioaoyuoyea",
"output": ".w.z.t.c.v.v.v.v.z.v"
},
{
"input": "yjnckpfyLtzwjsgpcrgCfpljnjwqzgVcufnOvhxplvflxJzqxnhrwgfJmPzifgubvspffmqrwbzivatlmdiBaddiaktdsfPwsevl",
"output": ".j.n.c.k.p.f.l.t.z.w.j.s.g.p.c.r.g.c.f.p.l.j.n.j.w.q.z.g.v.c.f.n.v.h.x.p.l.v.f.l.x.j.z.q.x.n.h.r.w.g.f.j.m.p.z.f.g.b.v.s.p.f.f.m.q.r.w.b.z.v.t.l.m.d.b.d.d.k.t.d.s.f.p.w.s.v.l"
},
{
"input": "RIIIUaAIYJOiuYIUWFPOOAIuaUEZeIooyUEUEAoIyIHYOEAlVAAIiLUAUAeiUIEiUMuuOiAgEUOIAoOUYYEYFEoOIIVeOOAOIIEg",
"output": ".r.j.w.f.p.z.h.l.v.l.m.g.f.v.g"
},
{
"input": "VBKQCFBMQHDMGNSGBQVJTGQCNHHRJMNKGKDPPSQRRVQTZNKBZGSXBPBRXPMVFTXCHZMSJVBRNFNTHBHGJLMDZJSVPZZBCCZNVLMQ",
"output": ".v.b.k.q.c.f.b.m.q.h.d.m.g.n.s.g.b.q.v.j.t.g.q.c.n.h.h.r.j.m.n.k.g.k.d.p.p.s.q.r.r.v.q.t.z.n.k.b.z.g.s.x.b.p.b.r.x.p.m.v.f.t.x.c.h.z.m.s.j.v.b.r.n.f.n.t.h.b.h.g.j.l.m.d.z.j.s.v.p.z.z.b.c.c.z.n.v.l.m.q"
},
{
"input": "iioyoaayeuyoolyiyoeuouiayiiuyTueyiaoiueyioiouyuauouayyiaeoeiiigmioiououeieeeyuyyaYyioiiooaiuouyoeoeg",
"output": ".l.t.g.m.g"
},
{
"input": "ueyiuiauuyyeueykeioouiiauzoyoeyeuyiaoaiiaaoaueyaeydaoauexuueafouiyioueeaaeyoeuaueiyiuiaeeayaioeouiuy",
"output": ".k.z.d.x.f"
},
{
"input": "FSNRBXLFQHZXGVMKLQDVHWLDSLKGKFMDRQWMWSSKPKKQBNDZRSCBLRSKCKKFFKRDMZFZGCNSMXNPMZVDLKXGNXGZQCLRTTDXLMXQ",
"output": ".f.s.n.r.b.x.l.f.q.h.z.x.g.v.m.k.l.q.d.v.h.w.l.d.s.l.k.g.k.f.m.d.r.q.w.m.w.s.s.k.p.k.k.q.b.n.d.z.r.s.c.b.l.r.s.k.c.k.k.f.f.k.r.d.m.z.f.z.g.c.n.s.m.x.n.p.m.z.v.d.l.k.x.g.n.x.g.z.q.c.l.r.t.t.d.x.l.m.x.q"
},
{
"input": "EYAYAYIOIOYOOAUOEUEUOUUYIYUUMOEOIIIAOIUOAAOIYOIOEUIERCEYYAOIOIGYUIAOYUEOEUAEAYPOYEYUUAUOAOEIYIEYUEEY",
"output": ".m.r.c.g.p"
},
{
"input": "jvvzcdcxjstbbksmqjsngxkgtttdxrljjxtwptgwwqzpvqchvgrkqlzxmptzblxhhsmrkmzzmgdfskhtmmnqzzflpmqdctvrfgtx",
"output": ".j.v.v.z.c.d.c.x.j.s.t.b.b.k.s.m.q.j.s.n.g.x.k.g.t.t.t.d.x.r.l.j.j.x.t.w.p.t.g.w.w.q.z.p.v.q.c.h.v.g.r.k.q.l.z.x.m.p.t.z.b.l.x.h.h.s.m.r.k.m.z.z.m.g.d.f.s.k.h.t.m.m.n.q.z.z.f.l.p.m.q.d.c.t.v.r.f.g.t.x"
},
{
"input": "YB",
"output": ".b"
},
{
"input": "fly",
"output": ".f.l"
},
{
"input": "YyyYYYyyYxdwdawdDAWDdaddYYYY",
"output": ".x.d.w.d.w.d.d.w.d.d.d.d"
}
] | 1,695,631,130
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 62
| 0
|
s=input()
vowels=["a","o","y","e","u","i"]
output=""
for letter in s:
letter=letter.lower()
if letter in vowels :
continue
output+="."+letter
print(output)
|
Title: String Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya started to attend programming lessons. On the first lesson his task was to write a simple program. The program was supposed to do the following: in the given string, consisting if uppercase and lowercase Latin letters, it:
- deletes all the vowels, - inserts a character "." before each consonant, - replaces all uppercase consonants with corresponding lowercase ones.
Vowels are letters "A", "O", "Y", "E", "U", "I", and the rest are consonants. The program's input is exactly one string, it should return the output as a single string, resulting after the program's processing the initial string.
Help Petya cope with this easy task.
Input Specification:
The first line represents input string of Petya's program. This string only consists of uppercase and lowercase Latin letters and its length is from 1 to 100, inclusive.
Output Specification:
Print the resulting string. It is guaranteed that this string is not empty.
Demo Input:
['tour\n', 'Codeforces\n', 'aBAcAba\n']
Demo Output:
['.t.r\n', '.c.d.f.r.c.s\n', '.b.c.b\n']
Note:
none
|
```python
s=input()
vowels=["a","o","y","e","u","i"]
output=""
for letter in s:
letter=letter.lower()
if letter in vowels :
continue
output+="."+letter
print(output)
```
| 3
|
Subsets and Splits
Successful Python Submissions
Retrieves all records from the train dataset where the verdict is 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Retrieves records of users with a rating of 1600 or higher and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a rating above 2000 and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a 'OK' verdict, providing a basic overview of a specific category within the dataset.