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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
670
|
A
|
Holidays
|
PROGRAMMING
| 900
|
[
"brute force",
"constructive algorithms",
"greedy",
"math"
] | null | null |
On the planet Mars a year lasts exactly *n* days (there are no leap years on Mars). But Martians have the same weeks as earthlings — 5 work days and then 2 days off. Your task is to determine the minimum possible and the maximum possible number of days off per year on Mars.
|
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=1<=000<=000) — the number of days in a year on Mars.
|
Print two integers — the minimum possible and the maximum possible number of days off per year on Mars.
|
[
"14\n",
"2\n"
] |
[
"4 4\n",
"0 2\n"
] |
In the first sample there are 14 days in a year on Mars, and therefore independently of the day a year starts with there will be exactly 4 days off .
In the second sample there are only 2 days in a year on Mars, and they can both be either work days or days off.
| 500
|
[
{
"input": "14",
"output": "4 4"
},
{
"input": "2",
"output": "0 2"
},
{
"input": "1",
"output": "0 1"
},
{
"input": "3",
"output": "0 2"
},
{
"input": "4",
"output": "0 2"
},
{
"input": "5",
"output": "0 2"
},
{
"input": "6",
"output": "1 2"
},
{
"input": "7",
"output": "2 2"
},
{
"input": "8",
"output": "2 3"
},
{
"input": "9",
"output": "2 4"
},
{
"input": "10",
"output": "2 4"
},
{
"input": "11",
"output": "2 4"
},
{
"input": "12",
"output": "2 4"
},
{
"input": "13",
"output": "3 4"
},
{
"input": "1000000",
"output": "285714 285715"
},
{
"input": "16",
"output": "4 6"
},
{
"input": "17",
"output": "4 6"
},
{
"input": "18",
"output": "4 6"
},
{
"input": "19",
"output": "4 6"
},
{
"input": "20",
"output": "5 6"
},
{
"input": "21",
"output": "6 6"
},
{
"input": "22",
"output": "6 7"
},
{
"input": "23",
"output": "6 8"
},
{
"input": "24",
"output": "6 8"
},
{
"input": "25",
"output": "6 8"
},
{
"input": "26",
"output": "6 8"
},
{
"input": "27",
"output": "7 8"
},
{
"input": "28",
"output": "8 8"
},
{
"input": "29",
"output": "8 9"
},
{
"input": "30",
"output": "8 10"
},
{
"input": "100",
"output": "28 30"
},
{
"input": "99",
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},
{
"input": "98",
"output": "28 28"
},
{
"input": "97",
"output": "27 28"
},
{
"input": "96",
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{
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},
{
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},
{
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"output": "26 28"
},
{
"input": "92",
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},
{
"input": "91",
"output": "26 26"
},
{
"input": "90",
"output": "25 26"
},
{
"input": "89",
"output": "24 26"
},
{
"input": "88",
"output": "24 26"
},
{
"input": "87",
"output": "24 26"
},
{
"input": "86",
"output": "24 26"
},
{
"input": "85",
"output": "24 25"
},
{
"input": "84",
"output": "24 24"
},
{
"input": "83",
"output": "23 24"
},
{
"input": "82",
"output": "22 24"
},
{
"input": "81",
"output": "22 24"
},
{
"input": "80",
"output": "22 24"
},
{
"input": "1000",
"output": "285 286"
},
{
"input": "999",
"output": "284 286"
},
{
"input": "998",
"output": "284 286"
},
{
"input": "997",
"output": "284 286"
},
{
"input": "996",
"output": "284 286"
},
{
"input": "995",
"output": "284 285"
},
{
"input": "994",
"output": "284 284"
},
{
"input": "993",
"output": "283 284"
},
{
"input": "992",
"output": "282 284"
},
{
"input": "991",
"output": "282 284"
},
{
"input": "990",
"output": "282 284"
},
{
"input": "989",
"output": "282 284"
},
{
"input": "988",
"output": "282 283"
},
{
"input": "987",
"output": "282 282"
},
{
"input": "986",
"output": "281 282"
},
{
"input": "985",
"output": "280 282"
},
{
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"output": "280 282"
},
{
"input": "983",
"output": "280 282"
},
{
"input": "982",
"output": "280 282"
},
{
"input": "981",
"output": "280 281"
},
{
"input": "980",
"output": "280 280"
},
{
"input": "10000",
"output": "2856 2858"
},
{
"input": "9999",
"output": "2856 2858"
},
{
"input": "9998",
"output": "2856 2858"
},
{
"input": "9997",
"output": "2856 2857"
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{
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"output": "2856 2856"
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{
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"output": "2853 2854"
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{
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"output": "2852 2854"
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"output": "2852 2852"
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{
"input": "9981",
"output": "2851 2852"
},
{
"input": "9980",
"output": "2850 2852"
},
{
"input": "100000",
"output": "28570 28572"
},
{
"input": "99999",
"output": "28570 28572"
},
{
"input": "99998",
"output": "28570 28572"
},
{
"input": "99997",
"output": "28570 28572"
},
{
"input": "99996",
"output": "28570 28571"
},
{
"input": "99995",
"output": "28570 28570"
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{
"input": "99994",
"output": "28569 28570"
},
{
"input": "99993",
"output": "28568 28570"
},
{
"input": "99992",
"output": "28568 28570"
},
{
"input": "99991",
"output": "28568 28570"
},
{
"input": "99990",
"output": "28568 28570"
},
{
"input": "99989",
"output": "28568 28569"
},
{
"input": "99988",
"output": "28568 28568"
},
{
"input": "99987",
"output": "28567 28568"
},
{
"input": "99986",
"output": "28566 28568"
},
{
"input": "99985",
"output": "28566 28568"
},
{
"input": "99984",
"output": "28566 28568"
},
{
"input": "99983",
"output": "28566 28568"
},
{
"input": "99982",
"output": "28566 28567"
},
{
"input": "99981",
"output": "28566 28566"
},
{
"input": "99980",
"output": "28565 28566"
},
{
"input": "999999",
"output": "285714 285714"
},
{
"input": "999998",
"output": "285713 285714"
},
{
"input": "999997",
"output": "285712 285714"
},
{
"input": "999996",
"output": "285712 285714"
},
{
"input": "999995",
"output": "285712 285714"
},
{
"input": "999994",
"output": "285712 285714"
},
{
"input": "999993",
"output": "285712 285713"
},
{
"input": "999992",
"output": "285712 285712"
},
{
"input": "999991",
"output": "285711 285712"
},
{
"input": "999990",
"output": "285710 285712"
},
{
"input": "999989",
"output": "285710 285712"
},
{
"input": "999988",
"output": "285710 285712"
},
{
"input": "999987",
"output": "285710 285712"
},
{
"input": "999986",
"output": "285710 285711"
},
{
"input": "999985",
"output": "285710 285710"
},
{
"input": "999984",
"output": "285709 285710"
},
{
"input": "999983",
"output": "285708 285710"
},
{
"input": "999982",
"output": "285708 285710"
},
{
"input": "999981",
"output": "285708 285710"
},
{
"input": "999980",
"output": "285708 285710"
},
{
"input": "234123",
"output": "66892 66893"
},
{
"input": "234122",
"output": "66892 66892"
},
{
"input": "234121",
"output": "66891 66892"
},
{
"input": "234120",
"output": "66890 66892"
},
{
"input": "234119",
"output": "66890 66892"
},
{
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"output": "66890 66892"
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{
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"output": "66890 66892"
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{
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{
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},
{
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},
{
"input": "234112",
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{
"input": "234111",
"output": "66888 66890"
},
{
"input": "234110",
"output": "66888 66890"
},
{
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},
{
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{
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},
{
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{
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{
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{
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{
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{
"input": "868523",
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{
"input": "868522",
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{
"input": "868521",
"output": "248148 248150"
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{
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"output": "248148 248150"
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{
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{
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},
{
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},
{
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{
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},
{
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},
{
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"output": "35259 35260"
},
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"output": "35258 35260"
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},
{
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"output": "35256 35258"
},
{
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"output": "35256 35258"
},
{
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"output": "35256 35258"
},
{
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"output": "35256 35258"
},
{
"input": "123397",
"output": "35256 35257"
},
{
"input": "123396",
"output": "35256 35256"
},
{
"input": "123395",
"output": "35255 35256"
},
{
"input": "123394",
"output": "35254 35256"
},
{
"input": "123393",
"output": "35254 35256"
},
{
"input": "15",
"output": "4 5"
}
] | 1,693,657,313
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 31
| 0
|
n = int(input())
min_days_off = (n // 7) * 2
max_days_off = min_days_off + min(2, n % 7)
print(min_days_off, max_days_off)
|
Title: Holidays
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On the planet Mars a year lasts exactly *n* days (there are no leap years on Mars). But Martians have the same weeks as earthlings — 5 work days and then 2 days off. Your task is to determine the minimum possible and the maximum possible number of days off per year on Mars.
Input Specification:
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=1<=000<=000) — the number of days in a year on Mars.
Output Specification:
Print two integers — the minimum possible and the maximum possible number of days off per year on Mars.
Demo Input:
['14\n', '2\n']
Demo Output:
['4 4\n', '0 2\n']
Note:
In the first sample there are 14 days in a year on Mars, and therefore independently of the day a year starts with there will be exactly 4 days off .
In the second sample there are only 2 days in a year on Mars, and they can both be either work days or days off.
|
```python
n = int(input())
min_days_off = (n // 7) * 2
max_days_off = min_days_off + min(2, n % 7)
print(min_days_off, max_days_off)
```
| 0
|
|
462
|
B
|
Appleman and Card Game
|
PROGRAMMING
| 1,300
|
[
"greedy"
] | null | null |
Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
|
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman.
|
Print a single integer – the answer to the problem.
|
[
"15 10\nDZFDFZDFDDDDDDF\n",
"6 4\nYJSNPI\n"
] |
[
"82\n",
"4\n"
] |
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
| 1,000
|
[
{
"input": "15 10\nDZFDFZDFDDDDDDF",
"output": "82"
},
{
"input": "6 4\nYJSNPI",
"output": "4"
},
{
"input": "5 3\nAOWBY",
"output": "3"
},
{
"input": "1 1\nV",
"output": "1"
},
{
"input": "2 1\nWT",
"output": "1"
},
{
"input": "2 2\nBL",
"output": "2"
},
{
"input": "5 1\nFACJT",
"output": "1"
},
{
"input": "5 5\nMJDIJ",
"output": "7"
},
{
"input": "15 5\nAZBIPTOFTJCJJIK",
"output": "13"
},
{
"input": "100 1\nEVEEVEEEGGECFEHEFVFVFHVHEEEEEFCVEEEEEEVFVEEVEEHEEVEFEVVEFEEEFEVECEHGHEEFGEEVCEECCECEFHEVEEEEEEGEEHVH",
"output": "1"
},
{
"input": "100 15\nKKTFFUTFCKUIKKKKFIFFKTUKUUKUKKIKKKTIFKTKUCFFKKKIIKKKKKKTFKFKKIRKKKFKUUKIKUUUFFKKKKTUZKITUIKKIKUKKTIK",
"output": "225"
},
{
"input": "100 50\nYYIYYAAAIEAAYAYAEAIIIAAEAAYEAEYYYIAEYAYAYYAAAIAYAEAAYAYYIYAAYYAAAAAAIYYYAAYAAEAAYAIEIYIYAYAYAYIIAAEY",
"output": "1972"
},
{
"input": "100 90\nFAFAOOAOOAFAOTFAFAFFATAAAOFAAOAFBAAAFBOAOFFFOAOAFAPFOFAOFAAFOAAAAFAAFOFAAOFPPAAOOAAOOFFOFFFOFAOTOFAF",
"output": "2828"
},
{
"input": "100 99\nBFFBBFBFBQFFFFFQBFFBFFBQFBFQFBBFQFFFBFFFBFQFQFBFFBBFYQFBFFFFFFFBQQFQBFBQBQFFFBQQFFFBQFYFBFBFFFBBBQQY",
"output": "3713"
},
{
"input": "100 100\nMQSBDAJABILIBCUEOWGWCEXMUTEYQKAIWGINXVQEOFDUBSVULROQHQRZZAALVQFEFRAAAYUIMGCAFQGIAEFBETRECGSFQJNXHHDN",
"output": "514"
},
{
"input": "100 50\nBMYIXQSJNHGFVFPJBIOBXIKSFNUFPVODCUBQYSIIQNVNXXCWXWRHKFEUPPIIDDGRDBJLZDCBMNJMYRMWFIHOSTDJJHXHPNRKWNFD",
"output": "328"
},
{
"input": "100 50\nENFNEMLJEMDMFMNNGNIMNINALGLLLAEMENEMNLMMIEIJNAINBJEJMFJLLIMINELGFLAIAMJMHMGNLIEFJIEEFEFGLLLDLMEAEIMM",
"output": "748"
}
] | 1,599,641,630
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 41
| 155
| 4,198,400
|
n,k=map(int,input().split())
s=list(input())
d={}
for i in s:
try:
d[i]+=1
except:
d[i]=1
new=[]
for keys in d:
new.append(d[keys])
new.sort(reverse=True)
i=0
total=0
while(k>0):
if new[i]<=k:
total+=(new[i]**2)
k-=new[i]
i+=1
else:
total+=k**2
break
print(total)
|
Title: Appleman and Card Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman.
Output Specification:
Print a single integer – the answer to the problem.
Demo Input:
['15 10\nDZFDFZDFDDDDDDF\n', '6 4\nYJSNPI\n']
Demo Output:
['82\n', '4\n']
Note:
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
|
```python
n,k=map(int,input().split())
s=list(input())
d={}
for i in s:
try:
d[i]+=1
except:
d[i]=1
new=[]
for keys in d:
new.append(d[keys])
new.sort(reverse=True)
i=0
total=0
while(k>0):
if new[i]<=k:
total+=(new[i]**2)
k-=new[i]
i+=1
else:
total+=k**2
break
print(total)
```
| 3
|
|
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"
}
] | 1,680,093,596
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
a="abcdefghijklmnopqrstuvwxyz"
word=input()
if sum([c in a for c in word]) < (len(word)+1)/2:
print(word.upper())
else:
print(word.lower())
|
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
a="abcdefghijklmnopqrstuvwxyz"
word=input()
if sum([c in a for c in word]) < (len(word)+1)/2:
print(word.upper())
else:
print(word.lower())
```
| 0
|
160
|
A
|
Twins
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] | null | null |
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
|
In the single line print the single number — the minimum needed number of coins.
|
[
"2\n3 3\n",
"3\n2 1 2\n"
] |
[
"2\n",
"2\n"
] |
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.
| 500
|
[
{
"input": "2\n3 3",
"output": "2"
},
{
"input": "3\n2 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "5\n4 2 2 2 2",
"output": "3"
},
{
"input": "7\n1 10 1 2 1 1 1",
"output": "1"
},
{
"input": "5\n3 2 3 3 1",
"output": "3"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "4"
},
{
"input": "7\n10 10 5 5 5 5 1",
"output": "3"
},
{
"input": "20\n2 1 2 2 2 1 1 2 1 2 2 1 1 1 1 2 1 1 1 1",
"output": "8"
},
{
"input": "20\n4 2 4 4 3 4 2 2 4 2 3 1 1 2 2 3 3 3 1 4",
"output": "8"
},
{
"input": "20\n35 26 41 40 45 46 22 26 39 23 11 15 47 42 18 15 27 10 45 40",
"output": "8"
},
{
"input": "20\n7 84 100 10 31 35 41 2 63 44 57 4 63 11 23 49 98 71 16 90",
"output": "6"
},
{
"input": "50\n19 2 12 26 17 27 10 26 17 17 5 24 11 15 3 9 16 18 19 1 25 23 18 6 2 7 25 7 21 25 13 29 16 9 25 3 14 30 18 4 10 28 6 10 8 2 2 4 8 28",
"output": "14"
},
{
"input": "70\n2 18 18 47 25 5 14 9 19 46 36 49 33 32 38 23 32 39 8 29 31 17 24 21 10 15 33 37 46 21 22 11 20 35 39 13 11 30 28 40 39 47 1 17 24 24 21 46 12 2 20 43 8 16 44 11 45 10 13 44 31 45 45 46 11 10 33 35 23 42",
"output": "22"
},
{
"input": "100\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",
"output": "51"
},
{
"input": "100\n1 2 2 1 2 1 1 2 1 1 1 2 2 1 1 1 2 2 2 1 2 1 1 1 1 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 1 1 1 1 2 2 1 2 1 2 1 2 2 2 1 2 1 2 2 1 1 2 2 1 1 2 2 2 1 1 2 1 1 2 2 1 2 1 1 2 2 1 2 1 1 2 2 1 1 1 1 2 1 1 1 1 2 2 2 2",
"output": "37"
},
{
"input": "100\n1 2 3 2 1 2 2 3 1 3 3 2 2 1 1 2 2 1 1 1 1 2 3 3 2 1 1 2 2 2 3 3 3 2 1 3 1 3 3 2 3 1 2 2 2 3 2 1 1 3 3 3 3 2 1 1 2 3 2 2 3 2 3 2 2 3 2 2 2 2 3 3 3 1 3 3 1 1 2 3 2 2 2 2 3 3 3 2 1 2 3 1 1 2 3 3 1 3 3 2",
"output": "36"
},
{
"input": "100\n5 5 4 3 5 1 2 5 1 1 3 5 4 4 1 1 1 1 5 4 4 5 1 5 5 1 2 1 3 1 5 1 3 3 3 2 2 2 1 1 5 1 3 4 1 1 3 2 5 2 2 5 5 4 4 1 3 4 3 3 4 5 3 3 3 1 2 1 4 2 4 4 1 5 1 3 5 5 5 5 3 4 4 3 1 2 5 2 3 5 4 2 4 5 3 2 4 2 4 3",
"output": "33"
},
{
"input": "100\n3 4 8 10 8 6 4 3 7 7 6 2 3 1 3 10 1 7 9 3 5 5 2 6 2 9 1 7 4 2 4 1 6 1 7 10 2 5 3 7 6 4 6 2 8 8 8 6 6 10 3 7 4 3 4 1 7 9 3 6 3 6 1 4 9 3 8 1 10 1 4 10 7 7 9 5 3 8 10 2 1 10 8 7 10 8 5 3 1 2 1 10 6 1 5 3 3 5 7 2",
"output": "30"
},
{
"input": "100\n16 9 11 8 11 4 9 17 4 8 4 10 9 10 6 3 3 15 1 6 1 15 12 18 6 14 13 18 1 7 18 4 10 7 10 12 3 16 14 4 10 8 10 7 19 13 15 1 4 8 16 10 6 4 3 16 11 10 7 3 4 16 1 20 1 11 4 16 10 7 7 12 18 19 3 17 19 3 4 19 2 12 11 3 18 20 2 2 14 4 20 13 13 11 16 20 19 14 7 2",
"output": "29"
},
{
"input": "100\n2 46 4 6 38 19 15 34 10 35 37 30 3 25 5 45 40 45 33 31 6 20 10 44 11 9 2 14 35 5 9 23 20 2 48 22 25 35 38 31 24 33 35 16 4 30 27 10 12 22 6 24 12 30 23 21 14 12 32 21 7 12 25 43 18 34 34 28 47 13 28 43 18 39 44 42 35 26 35 14 8 29 32 20 29 3 20 6 20 9 9 27 8 42 10 37 42 27 8 1",
"output": "30"
},
{
"input": "100\n85 50 17 89 65 89 5 20 86 26 16 21 85 14 44 31 87 31 6 2 48 67 8 80 79 1 48 36 97 1 5 30 79 50 78 12 2 55 76 100 54 40 26 81 97 96 68 56 87 14 51 17 54 37 52 33 69 62 38 63 74 15 62 78 9 19 67 2 60 58 93 60 18 96 55 48 34 7 79 82 32 58 90 67 20 50 27 15 7 89 98 10 11 15 99 49 4 51 77 52",
"output": "29"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "30"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "4\n4 1 2 3",
"output": "2"
},
{
"input": "5\n2 2 2 4 4",
"output": "2"
},
{
"input": "4\n100 99 99 100",
"output": "2"
}
] | 1,692,360,192
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 4
| 92
| 0
|
h=int(input())
g=list(map(int,input().split()))
y=0
for i in range(h):
y+=g[i]
if y>sum(g[i+1:]):break
print(i+1)
|
Title: Twins
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Imagine that you have a twin brother or sister. Having another person that looks exactly like you seems very unusual. It's hard to say if having something of an alter ego is good or bad. And if you do have a twin, then you very well know what it's like.
Now let's imagine a typical morning in your family. You haven't woken up yet, and Mom is already going to work. She has been so hasty that she has nearly forgotten to leave the two of her darling children some money to buy lunches in the school cafeteria. She fished in the purse and found some number of coins, or to be exact, *n* coins of arbitrary values *a*1,<=*a*2,<=...,<=*a**n*. But as Mom was running out of time, she didn't split the coins for you two. So she scribbled a note asking you to split the money equally.
As you woke up, you found Mom's coins and read her note. "But why split the money equally?" — you thought. After all, your twin is sleeping and he won't know anything. So you decided to act like that: pick for yourself some subset of coins so that the sum of values of your coins is strictly larger than the sum of values of the remaining coins that your twin will have. However, you correctly thought that if you take too many coins, the twin will suspect the deception. So, you've decided to stick to the following strategy to avoid suspicions: you take the minimum number of coins, whose sum of values is strictly more than the sum of values of the remaining coins. On this basis, determine what minimum number of coins you need to take to divide them in the described manner.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of coins. The second line contains a sequence of *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100) — the coins' values. All numbers are separated with spaces.
Output Specification:
In the single line print the single number — the minimum needed number of coins.
Demo Input:
['2\n3 3\n', '3\n2 1 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample you will have to take 2 coins (you and your twin have sums equal to 6, 0 correspondingly). If you take 1 coin, you get sums 3, 3. If you take 0 coins, you get sums 0, 6. Those variants do not satisfy you as your sum should be strictly more that your twins' sum.
In the second sample one coin isn't enough for us, too. You can pick coins with values 1, 2 or 2, 2. In any case, the minimum number of coins equals 2.
|
```python
h=int(input())
g=list(map(int,input().split()))
y=0
for i in range(h):
y+=g[i]
if y>sum(g[i+1:]):break
print(i+1)
```
| 0
|
|
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"
}
] | 1,696,616,657
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 92
| 0
|
x = input()
total = 0
for i in range(len(x)):
if x[i] == '4' or x[i] == '7':
total += 1
if total > 7:
break
if total == 4 or total == 7:
print('YES')
else:
print('NO')
|
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 = input()
total = 0
for i in range(len(x)):
if x[i] == '4' or x[i] == '7':
total += 1
if total > 7:
break
if total == 4 or total == 7:
print('YES')
else:
print('NO')
```
| 3.977
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,679,330,082
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 46
| 0
|
a=input()
b=input()
d=len(a)
c=""
for i in range(d):
if((int(a[i])+int(b[i]))==2):
c=c+"0"
else:
c=c+str(int(a[i])+int(b[i]))
print(c)
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
a=input()
b=input()
d=len(a)
c=""
for i in range(d):
if((int(a[i])+int(b[i]))==2):
c=c+"0"
else:
c=c+str(int(a[i])+int(b[i]))
print(c)
```
| 3.9885
|
774
|
C
|
Maximum Number
|
PROGRAMMING
| 1,200
|
[
"*special",
"constructive algorithms",
"greedy",
"implementation"
] | null | null |
Stepan has the newest electronic device with a display. Different digits can be shown on it. Each digit is shown on a seven-section indicator like it is shown on the picture below.
So, for example, to show the digit 3 on the display, 5 sections must be highlighted; and for the digit 6, 6 sections must be highlighted.
The battery of the newest device allows to highlight at most *n* sections on the display.
Stepan wants to know the maximum possible integer number which can be shown on the display of his newest device. Your task is to determine this number. Note that this number must not contain leading zeros. Assume that the size of the display is enough to show any integer.
|
The first line contains the integer *n* (2<=≤<=*n*<=≤<=100<=000) — the maximum number of sections which can be highlighted on the display.
|
Print the maximum integer which can be shown on the display of Stepan's newest device.
|
[
"2\n",
"3\n"
] |
[
"1\n",
"7\n"
] |
none
| 0
|
[
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "7"
},
{
"input": "4",
"output": "11"
},
{
"input": "5",
"output": "71"
},
{
"input": "6",
"output": "111"
},
{
"input": "85651",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "85666",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "99999",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "100000",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "99998",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "9",
"output": "7111"
},
{
"input": "99997",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "99996",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "99995",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "156",
"output": "111111111111111111111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "255",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111"
},
{
"input": "4568",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "5431",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "6782",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "8343",
"output": "7111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
},
{
"input": "9514",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111..."
}
] | 1,546,792,107
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 78
| 0
|
t = int(input())
for z in range(t):
n = int(input())
d = {0:6, 1:2,2:5,3:5,4:4,5:5,6:6,7:3,8:7,9:6}
s = 0
for i in str(n):
s+=i
|
Title: Maximum Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Stepan has the newest electronic device with a display. Different digits can be shown on it. Each digit is shown on a seven-section indicator like it is shown on the picture below.
So, for example, to show the digit 3 on the display, 5 sections must be highlighted; and for the digit 6, 6 sections must be highlighted.
The battery of the newest device allows to highlight at most *n* sections on the display.
Stepan wants to know the maximum possible integer number which can be shown on the display of his newest device. Your task is to determine this number. Note that this number must not contain leading zeros. Assume that the size of the display is enough to show any integer.
Input Specification:
The first line contains the integer *n* (2<=≤<=*n*<=≤<=100<=000) — the maximum number of sections which can be highlighted on the display.
Output Specification:
Print the maximum integer which can be shown on the display of Stepan's newest device.
Demo Input:
['2\n', '3\n']
Demo Output:
['1\n', '7\n']
Note:
none
|
```python
t = int(input())
for z in range(t):
n = int(input())
d = {0:6, 1:2,2:5,3:5,4:4,5:5,6:6,7:3,8:7,9:6}
s = 0
for i in str(n):
s+=i
```
| -1
|
|
358
|
A
|
Dima and Continuous Line
|
PROGRAMMING
| 1,400
|
[
"brute force",
"implementation"
] | null | null |
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of *n* distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the *n*-th point. Two points with coordinates (*x*1,<=0) and (*x*2,<=0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=103). The second line contains *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=106<=≤<=*x**i*<=≤<=106) — the *i*-th point has coordinates (*x**i*,<=0). The points are not necessarily sorted by their *x* coordinate.
|
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
|
[
"4\n0 10 5 15\n",
"4\n0 15 5 10\n"
] |
[
"yes\n",
"no\n"
] |
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
| 500
|
[
{
"input": "4\n0 10 5 15",
"output": "yes"
},
{
"input": "4\n0 15 5 10",
"output": "no"
},
{
"input": "5\n0 1000 2000 3000 1500",
"output": "yes"
},
{
"input": "5\n-724093 710736 -383722 -359011 439613",
"output": "no"
},
{
"input": "50\n384672 661179 -775591 -989608 611120 442691 601796 502406 384323 -315945 -934146 873993 -156910 -94123 -930137 208544 816236 466922 473696 463604 794454 -872433 -149791 -858684 -467655 -555239 623978 -217138 -408658 493342 -733576 -350871 711210 884148 -426172 519986 -356885 527171 661680 977247 141654 906254 -961045 -759474 -48634 891473 -606365 -513781 -966166 27696",
"output": "yes"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "no"
},
{
"input": "11\n1 11 10 2 3 9 8 4 5 7 6",
"output": "no"
},
{
"input": "10\n3 2 4 5 1 6 9 7 8 10",
"output": "yes"
},
{
"input": "11\n3 4 2 5 1 6 11 7 10 8 9",
"output": "no"
},
{
"input": "15\n0 -1 1 2 3 13 12 4 11 10 5 6 7 9 8",
"output": "no"
},
{
"input": "16\n6 7 8 9 5 10 11 12 13 14 15 4 16 2 1 3",
"output": "yes"
},
{
"input": "1\n0",
"output": "no"
},
{
"input": "4\n3 1 4 2",
"output": "yes"
},
{
"input": "5\n0 2 4 -2 5",
"output": "no"
},
{
"input": "5\n1 9 8 7 0",
"output": "yes"
},
{
"input": "3\n5 10 0",
"output": "no"
},
{
"input": "6\n1 3 -1 5 2 4",
"output": "yes"
},
{
"input": "4\n3 2 4 1",
"output": "no"
},
{
"input": "4\n10 5 15 0",
"output": "no"
},
{
"input": "2\n-5 -10",
"output": "no"
},
{
"input": "3\n1 0 3",
"output": "no"
},
{
"input": "4\n-2 -4 1 -3",
"output": "yes"
},
{
"input": "4\n3 6 0 2",
"output": "no"
},
{
"input": "4\n-9 10 -10 0",
"output": "yes"
},
{
"input": "4\n5 10 1 15",
"output": "no"
},
{
"input": "3\n1 0 2",
"output": "no"
},
{
"input": "4\n2 3 4 1",
"output": "no"
},
{
"input": "4\n7 5 9 12",
"output": "no"
}
] | 1,588,000,581
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 93
| 0
|
n=int(input())
x = list(map(int, input().split()))
if(n<=3):
print("no")
else:
for i in range(n-3):
if(x[i]<x[i+2]<x[i+1]<x[i+3] or x[i+2]<x[i]<x[i+3]<x[i+1]):
print("yes")
else:
print("no")
|
Title: Dima and Continuous Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of *n* distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the *n*-th point. Two points with coordinates (*x*1,<=0) and (*x*2,<=0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=103). The second line contains *n* distinct integers *x*1,<=*x*2,<=...,<=*x**n* (<=-<=106<=≤<=*x**i*<=≤<=106) — the *i*-th point has coordinates (*x**i*,<=0). The points are not necessarily sorted by their *x* coordinate.
Output Specification:
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Demo Input:
['4\n0 10 5 15\n', '4\n0 15 5 10\n']
Demo Output:
['yes\n', 'no\n']
Note:
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
|
```python
n=int(input())
x = list(map(int, input().split()))
if(n<=3):
print("no")
else:
for i in range(n-3):
if(x[i]<x[i+2]<x[i+1]<x[i+3] or x[i+2]<x[i]<x[i+3]<x[i+1]):
print("yes")
else:
print("no")
```
| 0
|
|
73
|
A
|
The Elder Trolls IV: Oblivon
|
PROGRAMMING
| 1,600
|
[
"greedy",
"math"
] |
A. The Elder Trolls IV: Oblivon
|
2
|
256
|
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size *x*<=×<=*y*<=×<=*z*, consisting of undestructable cells 1<=×<=1<=×<=1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most *k* times.
Vasya's character uses absolutely thin sword with infinite length.
|
The first line of input contains four integer numbers *x*,<=*y*,<=*z*,<=*k* (1<=≤<=*x*,<=*y*,<=*z*<=≤<=106,<=0<=≤<=*k*<=≤<=109).
|
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
|
[
"2 2 2 3\n",
"2 2 2 1\n"
] |
[
"8",
"2"
] |
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
| 500
|
[
{
"input": "2 2 2 3",
"output": "8"
},
{
"input": "2 2 2 1",
"output": "2"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "1 2 3 3",
"output": "6"
},
{
"input": "20 4 5 12",
"output": "120"
},
{
"input": "100 500 100500 1000000000",
"output": "5025000000"
},
{
"input": "2 5 5 9",
"output": "50"
},
{
"input": "11 1 11 11",
"output": "42"
},
{
"input": "100500 5000 500 100000000",
"output": "251250000000"
},
{
"input": "2 2 2 0",
"output": "1"
},
{
"input": "1000000 1000000 1000000 2444441",
"output": "540974149875309150"
},
{
"input": "1000000 1000000 1000000 1000000000",
"output": "1000000000000000000"
},
{
"input": "1000000 1000000 1000000 2999996",
"output": "999999000000000000"
},
{
"input": "1000000 1000000 1000000 2999997",
"output": "1000000000000000000"
},
{
"input": "999999 1000000 999997 999999999",
"output": "999996000003000000"
},
{
"input": "500000 1000000 750000 100000",
"output": "37040370459260"
},
{
"input": "999999 1 999998 1333333",
"output": "444445555556"
},
{
"input": "500000 10000 1000000 29998",
"output": "1000100000000"
},
{
"input": "10000 500000 1000000 29999",
"output": "1000200010000"
},
{
"input": "10000 1000000 500000 29996",
"output": "999900000000"
},
{
"input": "999999 123456 987654 0",
"output": "1"
},
{
"input": "1 1 1 0",
"output": "1"
},
{
"input": "219482 801483 941695 280976",
"output": "821595067700400"
},
{
"input": "808994 288453 204353 580644",
"output": "7250580779648149"
},
{
"input": "428676 64403 677407 626161",
"output": "5081000961597840"
},
{
"input": "559002 326875 150818 157621",
"output": "145045169133102"
},
{
"input": "178008 590076 624581 201286",
"output": "302062187173952"
},
{
"input": "797745 854005 98703 735186",
"output": "9996502351557447"
},
{
"input": "418223 118667 573175 776998",
"output": "12857677898465963"
},
{
"input": "39436 384053 48008 313346",
"output": "427693170156640"
},
{
"input": "661377 149342 523189 353305",
"output": "1633415415004970"
},
{
"input": "781081 414037 495753 892089",
"output": "26294515330164544"
},
{
"input": "402353 679460 969495 930195",
"output": "29810031851367496"
},
{
"input": "593408 709898 624186 915570",
"output": "28425961712082871"
},
{
"input": "1000 988 1000000 3000",
"output": "1002820000"
},
{
"input": "91839 2 3 50",
"output": "288"
},
{
"input": "999900 999990 4 129",
"output": "16384"
},
{
"input": "1024 100000 4 13",
"output": "144"
},
{
"input": "1000000 1000000 1000000 1000000000",
"output": "1000000000000000000"
},
{
"input": "1 1000000 1 1000000000",
"output": "1000000"
},
{
"input": "2 1000 1000000 1000000000",
"output": "2000000000"
},
{
"input": "999999 2 1000000 1000000000",
"output": "1999998000000"
},
{
"input": "1000 1 1 1",
"output": "2"
},
{
"input": "1000 1 1 998",
"output": "999"
},
{
"input": "1000 1 1 1000",
"output": "1000"
},
{
"input": "33334 66667 1000000 100000",
"output": "37040370459260"
},
{
"input": "1 1000000 2 23123",
"output": "46246"
}
] | 1,699,805,619
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 60
| 0
|
y=int(input())
s=input()
A=0
D=0
for i in range(s):
if i=="A":
A+=1
else:
D+=1
if A>D:
print("Anton")
elif D>A:
print("Danik")
else:
print("Friendship")
|
Title: The Elder Trolls IV: Oblivon
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya plays The Elder Trolls IV: Oblivon. Oh, those creators of computer games! What they do not come up with! Absolutely unique monsters have been added to the The Elder Trolls IV: Oblivon. One of these monsters is Unkillable Slug. Why it is "Unkillable"? Firstly, because it can be killed with cutting weapon only, so lovers of two-handed amber hammers should find suitable knife themselves. Secondly, it is necessary to make so many cutting strokes to Unkillable Slug. Extremely many. Too many!
Vasya has already promoted his character to 80-th level and in order to gain level 81 he was asked to kill Unkillable Slug. The monster has a very interesting shape. It looks like a rectangular parallelepiped with size *x*<=×<=*y*<=×<=*z*, consisting of undestructable cells 1<=×<=1<=×<=1. At one stroke Vasya can cut the Slug along an imaginary grid, i.e. cut with a plane parallel to one of the parallelepiped side. Monster dies when amount of parts it is divided reaches some critical value.
All parts of monster do not fall after each cut, they remains exactly on its places. I. e. Vasya can cut several parts with one cut.
Vasya wants to know what the maximum number of pieces he can cut the Unkillable Slug into striking him at most *k* times.
Vasya's character uses absolutely thin sword with infinite length.
Input Specification:
The first line of input contains four integer numbers *x*,<=*y*,<=*z*,<=*k* (1<=≤<=*x*,<=*y*,<=*z*<=≤<=106,<=0<=≤<=*k*<=≤<=109).
Output Specification:
Output the only number — the answer for the problem.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
Demo Input:
['2 2 2 3\n', '2 2 2 1\n']
Demo Output:
['8', '2']
Note:
In the first sample Vasya make 3 pairwise perpendicular cuts. He cuts monster on two parts with the first cut, then he divides each part on two with the second cut, and finally he divides each of the 4 parts on two.
|
```python
y=int(input())
s=input()
A=0
D=0
for i in range(s):
if i=="A":
A+=1
else:
D+=1
if A>D:
print("Anton")
elif D>A:
print("Danik")
else:
print("Friendship")
```
| -1
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly.
Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of actions Valentin did.
The next *n* lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types:
1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and *w* is the word that Valentin said. 1. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and *w* is the word that Valentin said. 1. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and *s* is the guess — a lowercase English letter.
All words consist only of lowercase English letters. The total length of all words does not exceed 105.
It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter.
|
Output a single integer — the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined.
|
[
"5\n! abc\n. ad\n. b\n! cd\n? c\n",
"8\n! hello\n! codeforces\n? c\n. o\n? d\n? h\n. l\n? e\n",
"7\n! ababahalamaha\n? a\n? b\n? a\n? b\n? a\n? h\n"
] |
[
"1\n",
"2\n",
"0\n"
] |
In the first test case after the first action it becomes clear that the selected letter is one of the following: *a*, *b*, *c*. After the second action we can note that the selected letter is not *a*. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is *c*, but Valentin pronounces the word *cd* and gets an excessive electric shock.
In the second test case after the first two electric shocks we understand that the selected letter is *e* or *o*. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is *e*, but Valentin makes 3 more actions before he makes a correct hypothesis.
In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks.
| 0
|
[
{
"input": "5\n! abc\n. ad\n. b\n! cd\n? c",
"output": "1"
},
{
"input": "8\n! hello\n! codeforces\n? c\n. o\n? d\n? h\n. l\n? e",
"output": "2"
},
{
"input": "7\n! ababahalamaha\n? a\n? b\n? a\n? b\n? a\n? h",
"output": "0"
},
{
"input": "4\n! abcd\n! cdef\n? d\n? c",
"output": "0"
},
{
"input": "1\n? q",
"output": "0"
},
{
"input": "15\n. r\n? e\n. s\n. rw\n? y\n. fj\n. zftyd\n? r\n! wq\n? w\n? p\n. ours\n. dto\n. lbyfru\n? q",
"output": "2"
},
{
"input": "3\n. abcdefghijklmnopqrstuvwxy\n? a\n? z",
"output": "1"
},
{
"input": "3\n. abcdefghijklmnopqrstuvwxy\n! z\n? z",
"output": "1"
}
] | 1,514,214,166
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 62
| 5,529,600
|
s=set('abcdefghijklmnopqrstuvwxyz')
d=0
for _ in range(int(input())):
a,b=input().split()
if a=='!':s&=set(b)
else:s-=set(b)
if len(s) == 1 and a in '?!': d += 1
print(d)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valentin participates in a show called "Shockers". The rules are quite easy: jury selects one letter which Valentin doesn't know. He should make a small speech, but every time he pronounces a word that contains the selected letter, he receives an electric shock. He can make guesses which letter is selected, but for each incorrect guess he receives an electric shock too. The show ends when Valentin guesses the selected letter correctly.
Valentin can't keep in mind everything, so he could guess the selected letter much later than it can be uniquely determined and get excessive electric shocks. Excessive electric shocks are those which Valentin got after the moment the selected letter can be uniquely determined. You should find out the number of excessive electric shocks.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of actions Valentin did.
The next *n* lines contain descriptions of his actions, each line contains description of one action. Each action can be of one of three types:
1. Valentin pronounced some word and didn't get an electric shock. This action is described by the string ". w" (without quotes), in which "." is a dot (ASCII-code 46), and *w* is the word that Valentin said. 1. Valentin pronounced some word and got an electric shock. This action is described by the string "! w" (without quotes), in which "!" is an exclamation mark (ASCII-code 33), and *w* is the word that Valentin said. 1. Valentin made a guess about the selected letter. This action is described by the string "? s" (without quotes), in which "?" is a question mark (ASCII-code 63), and *s* is the guess — a lowercase English letter.
All words consist only of lowercase English letters. The total length of all words does not exceed 105.
It is guaranteed that last action is a guess about the selected letter. Also, it is guaranteed that Valentin didn't make correct guesses about the selected letter before the last action. Moreover, it's guaranteed that if Valentin got an electric shock after pronouncing some word, then it contains the selected letter; and also if Valentin didn't get an electric shock after pronouncing some word, then it does not contain the selected letter.
Output Specification:
Output a single integer — the number of electric shocks that Valentin could have avoided if he had told the selected letter just after it became uniquely determined.
Demo Input:
['5\n! abc\n. ad\n. b\n! cd\n? c\n', '8\n! hello\n! codeforces\n? c\n. o\n? d\n? h\n. l\n? e\n', '7\n! ababahalamaha\n? a\n? b\n? a\n? b\n? a\n? h\n']
Demo Output:
['1\n', '2\n', '0\n']
Note:
In the first test case after the first action it becomes clear that the selected letter is one of the following: *a*, *b*, *c*. After the second action we can note that the selected letter is not *a*. Valentin tells word "b" and doesn't get a shock. After that it is clear that the selected letter is *c*, but Valentin pronounces the word *cd* and gets an excessive electric shock.
In the second test case after the first two electric shocks we understand that the selected letter is *e* or *o*. Valentin tries some words consisting of these letters and after the second word it's clear that the selected letter is *e*, but Valentin makes 3 more actions before he makes a correct hypothesis.
In the third example the selected letter can be uniquely determined only when Valentin guesses it, so he didn't get excessive electric shocks.
|
```python
s=set('abcdefghijklmnopqrstuvwxyz')
d=0
for _ in range(int(input())):
a,b=input().split()
if a=='!':s&=set(b)
else:s-=set(b)
if len(s) == 1 and a in '?!': d += 1
print(d)
```
| 0
|
|
401
|
A
|
Vanya and Cards
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Vanya loves playing. He even has a special set of cards to play with. Each card has a single integer. The number on the card can be positive, negative and can even be equal to zero. The only limit is, the number on each card doesn't exceed *x* in the absolute value.
Natasha doesn't like when Vanya spends a long time playing, so she hid all of his cards. Vanya became sad and started looking for the cards but he only found *n* of them. Vanya loves the balance, so he wants the sum of all numbers on found cards equal to zero. On the other hand, he got very tired of looking for cards. Help the boy and say what is the minimum number of cards does he need to find to make the sum equal to zero?
You can assume that initially Vanya had infinitely many cards with each integer number from <=-<=*x* to *x*.
|
The first line contains two integers: *n* (1<=≤<=*n*<=≤<=1000) — the number of found cards and *x* (1<=≤<=*x*<=≤<=1000) — the maximum absolute value of the number on a card. The second line contains *n* space-separated integers — the numbers on found cards. It is guaranteed that the numbers do not exceed *x* in their absolute value.
|
Print a single number — the answer to the problem.
|
[
"3 2\n-1 1 2\n",
"2 3\n-2 -2\n"
] |
[
"1\n",
"2\n"
] |
In the first sample, Vanya needs to find a single card with number -2.
In the second sample, Vanya needs to find two cards with number 2. He can't find a single card with the required number as the numbers on the lost cards do not exceed 3 in their absolute value.
| 500
|
[
{
"input": "3 2\n-1 1 2",
"output": "1"
},
{
"input": "2 3\n-2 -2",
"output": "2"
},
{
"input": "4 4\n1 2 3 4",
"output": "3"
},
{
"input": "2 2\n-1 -1",
"output": "1"
},
{
"input": "15 5\n-2 -1 2 -4 -3 4 -4 -2 -2 2 -2 -1 1 -4 -2",
"output": "4"
},
{
"input": "15 16\n-15 -5 -15 -14 -8 15 -15 -12 -5 -3 5 -7 3 8 -15",
"output": "6"
},
{
"input": "1 4\n-3",
"output": "1"
},
{
"input": "10 7\n6 4 6 6 -3 4 -1 2 3 3",
"output": "5"
},
{
"input": "2 1\n1 -1",
"output": "0"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "8 13\n-11 -1 -11 12 -2 -2 -10 -11",
"output": "3"
},
{
"input": "16 11\n3 -7 7 -9 -2 -3 -4 -2 -6 8 10 7 1 4 6 7",
"output": "2"
},
{
"input": "67 15\n-2 -2 6 -4 -7 4 3 13 -9 -4 11 -7 -6 -11 1 11 -1 11 14 10 -8 7 5 11 -13 1 -1 7 -14 9 -11 -11 13 -4 12 -11 -8 -5 -11 6 10 -2 6 9 9 6 -11 -2 7 -10 -1 9 -8 -5 1 -7 -2 3 -1 -13 -6 -9 -8 10 13 -3 9",
"output": "1"
},
{
"input": "123 222\n44 -190 -188 -185 -55 17 190 176 157 176 -24 -113 -54 -61 -53 53 -77 68 -12 -114 -217 163 -122 37 -37 20 -108 17 -140 -210 218 19 -89 54 18 197 111 -150 -36 -131 -172 36 67 16 -202 72 169 -137 -34 -122 137 -72 196 -17 -104 180 -102 96 -69 -184 21 -15 217 -61 175 -221 62 173 -93 -106 122 -135 58 7 -110 -108 156 -141 -102 -50 29 -204 -46 -76 101 -33 -190 99 52 -197 175 -71 161 -140 155 10 189 -217 -97 -170 183 -88 83 -149 157 -208 154 -3 77 90 74 165 198 -181 -166 -4 -200 -89 -200 131 100 -61 -149",
"output": "8"
},
{
"input": "130 142\n58 -50 43 -126 84 -92 -108 -92 57 127 12 -135 -49 89 141 -112 -31 47 75 -19 80 81 -5 17 10 4 -26 68 -102 -10 7 -62 -135 -123 -16 55 -72 -97 -34 21 21 137 130 97 40 -18 110 -52 73 52 85 103 -134 -107 88 30 66 97 126 82 13 125 127 -87 81 22 45 102 13 95 4 10 -35 39 -43 -112 -5 14 -46 19 61 -44 -116 137 -116 -80 -39 92 -75 29 -65 -15 5 -108 -114 -129 -5 52 -21 118 -41 35 -62 -125 130 -95 -11 -75 19 108 108 127 141 2 -130 54 96 -81 -102 140 -58 -102 132 50 -126 82 6 45 -114 -42",
"output": "5"
},
{
"input": "7 12\n2 5 -1 -4 -7 4 3",
"output": "1"
},
{
"input": "57 53\n-49 7 -41 7 38 -51 -23 8 45 1 -24 26 37 28 -31 -40 38 25 -32 -47 -3 20 -40 -32 -44 -36 5 33 -16 -5 28 10 -22 3 -10 -51 -32 -51 27 -50 -22 -12 41 3 15 24 30 -12 -34 -15 -29 38 -10 -35 -9 6 -51",
"output": "8"
},
{
"input": "93 273\n-268 -170 -163 19 -69 18 -244 35 -34 125 -224 -48 179 -247 127 -150 271 -49 -102 201 84 -151 -70 -46 -16 216 240 127 3 218 -209 223 -227 -201 228 -8 203 46 -100 -207 126 255 40 -58 -217 93 172 -97 23 183 102 -92 -157 -117 173 47 144 -235 -227 -62 -128 13 -151 158 110 -116 68 -2 -148 -206 -52 79 -152 -223 74 -149 -69 232 38 -70 -256 -213 -236 132 -189 -200 199 -57 -108 -53 269 -101 -134",
"output": "8"
},
{
"input": "1 1000\n997",
"output": "1"
},
{
"input": "4 3\n2 -1 -2 -1",
"output": "1"
},
{
"input": "1 1\n-1",
"output": "1"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "2 2\n1 -1",
"output": "0"
},
{
"input": "2 2\n-1 1",
"output": "0"
},
{
"input": "2 3\n-1 1",
"output": "0"
},
{
"input": "2 2\n-2 2",
"output": "0"
},
{
"input": "2 2\n2 2",
"output": "2"
},
{
"input": "4 2\n-1 -1 -1 -1",
"output": "2"
},
{
"input": "4 1\n-1 -1 -1 1",
"output": "2"
},
{
"input": "3 2\n2 2 2",
"output": "3"
},
{
"input": "10 300\n300 300 300 300 300 300 300 300 300 300",
"output": "10"
}
] | 1,614,233,387
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 47
| 108
| 0
|
n,m = map(int,input().split())
t = list(map(int,input().split()))
u= sum(t)
if u==0:print(0)
else:
if abs(u)<=m:print(1)
else:
if abs(u)%m==0:print(abs(u)//m)
else:print((abs(u)//m)+1)
|
Title: Vanya and Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya loves playing. He even has a special set of cards to play with. Each card has a single integer. The number on the card can be positive, negative and can even be equal to zero. The only limit is, the number on each card doesn't exceed *x* in the absolute value.
Natasha doesn't like when Vanya spends a long time playing, so she hid all of his cards. Vanya became sad and started looking for the cards but he only found *n* of them. Vanya loves the balance, so he wants the sum of all numbers on found cards equal to zero. On the other hand, he got very tired of looking for cards. Help the boy and say what is the minimum number of cards does he need to find to make the sum equal to zero?
You can assume that initially Vanya had infinitely many cards with each integer number from <=-<=*x* to *x*.
Input Specification:
The first line contains two integers: *n* (1<=≤<=*n*<=≤<=1000) — the number of found cards and *x* (1<=≤<=*x*<=≤<=1000) — the maximum absolute value of the number on a card. The second line contains *n* space-separated integers — the numbers on found cards. It is guaranteed that the numbers do not exceed *x* in their absolute value.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['3 2\n-1 1 2\n', '2 3\n-2 -2\n']
Demo Output:
['1\n', '2\n']
Note:
In the first sample, Vanya needs to find a single card with number -2.
In the second sample, Vanya needs to find two cards with number 2. He can't find a single card with the required number as the numbers on the lost cards do not exceed 3 in their absolute value.
|
```python
n,m = map(int,input().split())
t = list(map(int,input().split()))
u= sum(t)
if u==0:print(0)
else:
if abs(u)<=m:print(1)
else:
if abs(u)%m==0:print(abs(u)//m)
else:print((abs(u)//m)+1)
```
| 3
|
|
552
|
A
|
Vanya and Table
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Vanya has a table consisting of 100 rows, each row contains 100 cells. The rows are numbered by integers from 1 to 100 from bottom to top, the columns are numbered from 1 to 100 from left to right.
In this table, Vanya chose *n* rectangles with sides that go along borders of squares (some rectangles probably occur multiple times). After that for each cell of the table he counted the number of rectangles it belongs to and wrote this number into it. Now he wants to find the sum of values in all cells of the table and as the table is too large, he asks you to help him find the result.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of rectangles.
Each of the following *n* lines contains four integers *x*1,<=*y*1,<=*x*2,<=*y*2 (1<=≤<=*x*1<=≤<=*x*2<=≤<=100, 1<=≤<=*y*1<=≤<=*y*2<=≤<=100), where *x*1 and *y*1 are the number of the column and row of the lower left cell and *x*2 and *y*2 are the number of the column and row of the upper right cell of a rectangle.
|
In a single line print the sum of all values in the cells of the table.
|
[
"2\n1 1 2 3\n2 2 3 3\n",
"2\n1 1 3 3\n1 1 3 3\n"
] |
[
"10\n",
"18\n"
] |
Note to the first sample test:
Values of the table in the first three rows and columns will be as follows:
121
121
110
So, the sum of values will be equal to 10.
Note to the second sample test:
Values of the table in the first three rows and columns will be as follows:
222
222
222
So, the sum of values will be equal to 18.
| 500
|
[
{
"input": "2\n1 1 2 3\n2 2 3 3",
"output": "10"
},
{
"input": "2\n1 1 3 3\n1 1 3 3",
"output": "18"
},
{
"input": "5\n4 11 20 15\n7 5 12 20\n10 8 16 12\n7 5 12 15\n2 2 20 13",
"output": "510"
},
{
"input": "5\n4 11 20 20\n6 11 20 16\n5 2 19 15\n11 3 18 15\n3 2 14 11",
"output": "694"
},
{
"input": "5\n1 1 1 100\n1 1 1 100\n1 1 1 100\n1 1 1 100\n1 1 1 100",
"output": "500"
},
{
"input": "1\n1 1 1 1",
"output": "1"
},
{
"input": "1\n100 100 100 100",
"output": "1"
},
{
"input": "1\n1 1 1 100",
"output": "100"
},
{
"input": "3\n1 1 1 1\n1 2 1 2\n1 3 1 3",
"output": "3"
},
{
"input": "1\n1 1 100 100",
"output": "10000"
}
] | 1,618,458,453
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 124
| 2,150,400
|
def proA(arr):
for i in arr:
x,y,a,b=i
ans+= (a-x)*(b-y)
return ans
t=int(input())
for i in range(t):
#n=int(input())
arr.append(list(map(int,input().split())))
#proD(arr)
print(proA(arr))
|
Title: Vanya and Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya has a table consisting of 100 rows, each row contains 100 cells. The rows are numbered by integers from 1 to 100 from bottom to top, the columns are numbered from 1 to 100 from left to right.
In this table, Vanya chose *n* rectangles with sides that go along borders of squares (some rectangles probably occur multiple times). After that for each cell of the table he counted the number of rectangles it belongs to and wrote this number into it. Now he wants to find the sum of values in all cells of the table and as the table is too large, he asks you to help him find the result.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of rectangles.
Each of the following *n* lines contains four integers *x*1,<=*y*1,<=*x*2,<=*y*2 (1<=≤<=*x*1<=≤<=*x*2<=≤<=100, 1<=≤<=*y*1<=≤<=*y*2<=≤<=100), where *x*1 and *y*1 are the number of the column and row of the lower left cell and *x*2 and *y*2 are the number of the column and row of the upper right cell of a rectangle.
Output Specification:
In a single line print the sum of all values in the cells of the table.
Demo Input:
['2\n1 1 2 3\n2 2 3 3\n', '2\n1 1 3 3\n1 1 3 3\n']
Demo Output:
['10\n', '18\n']
Note:
Note to the first sample test:
Values of the table in the first three rows and columns will be as follows:
121
121
110
So, the sum of values will be equal to 10.
Note to the second sample test:
Values of the table in the first three rows and columns will be as follows:
222
222
222
So, the sum of values will be equal to 18.
|
```python
def proA(arr):
for i in arr:
x,y,a,b=i
ans+= (a-x)*(b-y)
return ans
t=int(input())
for i in range(t):
#n=int(input())
arr.append(list(map(int,input().split())))
#proD(arr)
print(proA(arr))
```
| -1
|
|
599
|
A
|
Patrick and Shopping
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
|
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
|
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
|
[
"10 20 30\n",
"1 1 5\n"
] |
[
"60\n",
"4\n"
] |
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
| 500
|
[
{
"input": "10 20 30",
"output": "60"
},
{
"input": "1 1 5",
"output": "4"
},
{
"input": "100 33 34",
"output": "134"
},
{
"input": "777 777 777",
"output": "2331"
},
{
"input": "2 2 8",
"output": "8"
},
{
"input": "12 34 56",
"output": "92"
},
{
"input": "789 101112 131415",
"output": "203802"
},
{
"input": "27485716 99999999 35182",
"output": "55041796"
},
{
"input": "1 293548 5",
"output": "12"
},
{
"input": "12059 259855 5874875",
"output": "543828"
},
{
"input": "46981 105809 585858",
"output": "305580"
},
{
"input": "9889 1221 2442",
"output": "7326"
},
{
"input": "100500 200600 300700",
"output": "601800"
},
{
"input": "318476 318476 318476",
"output": "955428"
},
{
"input": "23985 3353 75633",
"output": "54676"
},
{
"input": "120 1298 2222",
"output": "2836"
},
{
"input": "98437 23487 666672",
"output": "243848"
},
{
"input": "100000000 100000000 100000000",
"output": "300000000"
},
{
"input": "2 5 2",
"output": "8"
},
{
"input": "1 1000 1",
"output": "4"
},
{
"input": "1 100000000 1",
"output": "4"
}
] | 1,628,346,755
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 46
| 6,656,000
|
x,y,z=map(int, input().split())
print([(x+y)*2,x+y+z][max(x,y)*2>=z])
|
Title: Patrick and Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
Input Specification:
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
Output Specification:
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
Demo Input:
['10 20 30\n', '1 1 5\n']
Demo Output:
['60\n', '4\n']
Note:
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
|
```python
x,y,z=map(int, input().split())
print([(x+y)*2,x+y+z][max(x,y)*2>=z])
```
| 0
|
|
478
|
B
|
Random Teams
|
PROGRAMMING
| 1,300
|
[
"combinatorics",
"constructive algorithms",
"greedy",
"math"
] | null | null |
*n* participants of the competition were split into *m* teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
|
The only line of input contains two integers *n* and *m*, separated by a single space (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the number of participants and the number of teams respectively.
|
The only line of the output should contain two integers *k**min* and *k**max* — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
|
[
"5 1\n",
"3 2\n",
"6 3\n"
] |
[
"10 10\n",
"1 1\n",
"3 6\n"
] |
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
| 1,000
|
[
{
"input": "5 1",
"output": "10 10"
},
{
"input": "3 2",
"output": "1 1"
},
{
"input": "6 3",
"output": "3 6"
},
{
"input": "5 3",
"output": "2 3"
},
{
"input": "10 2",
"output": "20 36"
},
{
"input": "10 6",
"output": "4 10"
},
{
"input": "1000000000 1",
"output": "499999999500000000 499999999500000000"
},
{
"input": "5000000 12",
"output": "1041664166668 12499942500066"
},
{
"input": "1833 195",
"output": "7722 1342341"
},
{
"input": "1000000000 1000000000",
"output": "0 0"
},
{
"input": "1000000000 1000000",
"output": "499500000000 499000500499500000"
},
{
"input": "1000000000 32170",
"output": "15541930838100 499967831017438365"
},
{
"input": "1000000 1000",
"output": "499500000 499000999500"
},
{
"input": "1234 1123",
"output": "111 6216"
},
{
"input": "599222887 298488",
"output": "601178656545 179355218158217800"
},
{
"input": "999999999 500000000",
"output": "499999999 124999999750000000"
},
{
"input": "1000000000 384842119",
"output": "845473643 189209609585784021"
},
{
"input": "1000000000 384842119",
"output": "845473643 189209609585784021"
},
{
"input": "1000000000 2",
"output": "249999999500000000 499999998500000001"
},
{
"input": "1000000000 999999999",
"output": "1 1"
},
{
"input": "38447 383",
"output": "1910550 724453080"
},
{
"input": "100000000 99999799",
"output": "201 20301"
},
{
"input": "1 1",
"output": "0 0"
},
{
"input": "2 1",
"output": "1 1"
},
{
"input": "2 2",
"output": "0 0"
},
{
"input": "10 10",
"output": "0 0"
}
] | 1,677,739,633
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 26
| 93
| 0
|
a,b = ([int(x) for x in input().split()])
c=a-(b-1)
d=(c*(c-1))//2
e=a//b
f=a%b
g=b-f
h=g*(e*(e-1))//2 + f*(e*(e+1))//2
print(h,d)
|
Title: Random Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
*n* participants of the competition were split into *m* teams in some manner so that each team has at least one participant. After the competition each pair of participants from the same team became friends.
Your task is to write a program that will find the minimum and the maximum number of pairs of friends that could have formed by the end of the competition.
Input Specification:
The only line of input contains two integers *n* and *m*, separated by a single space (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the number of participants and the number of teams respectively.
Output Specification:
The only line of the output should contain two integers *k**min* and *k**max* — the minimum possible number of pairs of friends and the maximum possible number of pairs of friends respectively.
Demo Input:
['5 1\n', '3 2\n', '6 3\n']
Demo Output:
['10 10\n', '1 1\n', '3 6\n']
Note:
In the first sample all the participants get into one team, so there will be exactly ten pairs of friends.
In the second sample at any possible arrangement one team will always have two participants and the other team will always have one participant. Thus, the number of pairs of friends will always be equal to one.
In the third sample minimum number of newly formed friendships can be achieved if participants were split on teams consisting of 2 people, maximum number can be achieved if participants were split on teams of 1, 1 and 4 people.
|
```python
a,b = ([int(x) for x in input().split()])
c=a-(b-1)
d=(c*(c-1))//2
e=a//b
f=a%b
g=b-f
h=g*(e*(e-1))//2 + f*(e*(e+1))//2
print(h,d)
```
| 3
|
|
190
|
A
|
Vasya and the Bus
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
One day Vasya heard a story: "In the city of High Bertown a bus number 62 left from the bus station. It had *n* grown-ups and *m* kids..."
The latter events happen to be of no importance to us. Vasya is an accountant and he loves counting money. So he wondered what maximum and minimum sum of money these passengers could have paid for the ride.
The bus fare equals one berland ruble in High Bertown. However, not everything is that easy — no more than one child can ride for free with each grown-up passenger. That means that a grown-up passenger who rides with his *k* (*k*<=><=0) children, pays overall *k* rubles: a ticket for himself and (*k*<=-<=1) tickets for his children. Also, a grown-up can ride without children, in this case he only pays one ruble.
We know that in High Bertown children can't ride in a bus unaccompanied by grown-ups.
Help Vasya count the minimum and the maximum sum in Berland rubles, that all passengers of this bus could have paid in total.
|
The input file consists of a single line containing two space-separated numbers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=105) — the number of the grown-ups and the number of the children in the bus, correspondingly.
|
If *n* grown-ups and *m* children could have ridden in the bus, then print on a single line two space-separated integers — the minimum and the maximum possible total bus fare, correspondingly.
Otherwise, print "Impossible" (without the quotes).
|
[
"1 2\n",
"0 5\n",
"2 2\n"
] |
[
"2 2",
"Impossible",
"2 3"
] |
In the first sample a grown-up rides with two children and pays two rubles.
In the second sample there are only children in the bus, so the situation is impossible.
In the third sample there are two cases: - Each of the two grown-ups rides with one children and pays one ruble for the tickets. In this case the passengers pay two rubles in total. - One of the grown-ups ride with two children's and pays two rubles, the another one rides alone and pays one ruble for himself. So, they pay three rubles in total.
| 500
|
[
{
"input": "1 2",
"output": "2 2"
},
{
"input": "0 5",
"output": "Impossible"
},
{
"input": "2 2",
"output": "2 3"
},
{
"input": "2 7",
"output": "7 8"
},
{
"input": "4 10",
"output": "10 13"
},
{
"input": "6 0",
"output": "6 6"
},
{
"input": "7 1",
"output": "7 7"
},
{
"input": "0 0",
"output": "0 0"
},
{
"input": "71 24",
"output": "71 94"
},
{
"input": "16 70",
"output": "70 85"
},
{
"input": "0 1",
"output": "Impossible"
},
{
"input": "1 0",
"output": "1 1"
},
{
"input": "1 1",
"output": "1 1"
},
{
"input": "63 82",
"output": "82 144"
},
{
"input": "8 26",
"output": "26 33"
},
{
"input": "21 27",
"output": "27 47"
},
{
"input": "0 38",
"output": "Impossible"
},
{
"input": "46 84",
"output": "84 129"
},
{
"input": "59 96",
"output": "96 154"
},
{
"input": "63028 0",
"output": "63028 63028"
},
{
"input": "9458 0",
"output": "9458 9458"
},
{
"input": "80236 0",
"output": "80236 80236"
},
{
"input": "26666 0",
"output": "26666 26666"
},
{
"input": "59617 0",
"output": "59617 59617"
},
{
"input": "0 6048",
"output": "Impossible"
},
{
"input": "63028 28217",
"output": "63028 91244"
},
{
"input": "9458 39163",
"output": "39163 48620"
},
{
"input": "80236 14868",
"output": "80236 95103"
},
{
"input": "26666 52747",
"output": "52747 79412"
},
{
"input": "59617 28452",
"output": "59617 88068"
},
{
"input": "6048 4158",
"output": "6048 10205"
},
{
"input": "76826 4210",
"output": "76826 81035"
},
{
"input": "23256 15156",
"output": "23256 38411"
},
{
"input": "56207 53035",
"output": "56207 109241"
},
{
"input": "2637 28740",
"output": "28740 31376"
},
{
"input": "73415 4445",
"output": "73415 77859"
},
{
"input": "82019 4498",
"output": "82019 86516"
},
{
"input": "28449 80204",
"output": "80204 108652"
},
{
"input": "99227 53323",
"output": "99227 152549"
},
{
"input": "45657 29028",
"output": "45657 74684"
},
{
"input": "78608 4733",
"output": "78608 83340"
},
{
"input": "25038 4786",
"output": "25038 29823"
},
{
"input": "95816 80492",
"output": "95816 176307"
},
{
"input": "42246 94024",
"output": "94024 136269"
},
{
"input": "0 100000",
"output": "Impossible"
},
{
"input": "100000 0",
"output": "100000 100000"
},
{
"input": "1 100000",
"output": "100000 100000"
},
{
"input": "100000 1",
"output": "100000 100000"
},
{
"input": "63028 63028",
"output": "63028 126055"
},
{
"input": "9458 9458",
"output": "9458 18915"
},
{
"input": "80236 80236",
"output": "80236 160471"
},
{
"input": "26666 26666",
"output": "26666 53331"
},
{
"input": "59617 59617",
"output": "59617 119233"
},
{
"input": "100000 100000",
"output": "100000 199999"
}
] | 1,600,714,288
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 248
| 0
|
n,m = map(int,input().split())
if n==0 :
print("IMPOSSIBLE")
else:
if m==0 :
print(n,n)
elif n==m :
print(n,n+m-1)
elif n>m :
print(n,n+m-1)
elif n<m :
print((m//n)*n + m%n ,n+m-1)
|
Title: Vasya and the Bus
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya heard a story: "In the city of High Bertown a bus number 62 left from the bus station. It had *n* grown-ups and *m* kids..."
The latter events happen to be of no importance to us. Vasya is an accountant and he loves counting money. So he wondered what maximum and minimum sum of money these passengers could have paid for the ride.
The bus fare equals one berland ruble in High Bertown. However, not everything is that easy — no more than one child can ride for free with each grown-up passenger. That means that a grown-up passenger who rides with his *k* (*k*<=><=0) children, pays overall *k* rubles: a ticket for himself and (*k*<=-<=1) tickets for his children. Also, a grown-up can ride without children, in this case he only pays one ruble.
We know that in High Bertown children can't ride in a bus unaccompanied by grown-ups.
Help Vasya count the minimum and the maximum sum in Berland rubles, that all passengers of this bus could have paid in total.
Input Specification:
The input file consists of a single line containing two space-separated numbers *n* and *m* (0<=≤<=*n*,<=*m*<=≤<=105) — the number of the grown-ups and the number of the children in the bus, correspondingly.
Output Specification:
If *n* grown-ups and *m* children could have ridden in the bus, then print on a single line two space-separated integers — the minimum and the maximum possible total bus fare, correspondingly.
Otherwise, print "Impossible" (without the quotes).
Demo Input:
['1 2\n', '0 5\n', '2 2\n']
Demo Output:
['2 2', 'Impossible', '2 3']
Note:
In the first sample a grown-up rides with two children and pays two rubles.
In the second sample there are only children in the bus, so the situation is impossible.
In the third sample there are two cases: - Each of the two grown-ups rides with one children and pays one ruble for the tickets. In this case the passengers pay two rubles in total. - One of the grown-ups ride with two children's and pays two rubles, the another one rides alone and pays one ruble for himself. So, they pay three rubles in total.
|
```python
n,m = map(int,input().split())
if n==0 :
print("IMPOSSIBLE")
else:
if m==0 :
print(n,n)
elif n==m :
print(n,n+m-1)
elif n>m :
print(n,n+m-1)
elif n<m :
print((m//n)*n + m%n ,n+m-1)
```
| 0
|
|
227
|
B
|
Effective Approach
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
|
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
|
[
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] |
[
"1 2\n",
"2 1\n",
"6 6\n"
] |
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).
| 1,000
|
[
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4",
"output": "9 2"
},
{
"input": "10\n5 2 10 8 3 1 9 7 6 4\n9\n2 5 9 2 3 2 5 9 8",
"output": "31 68"
},
{
"input": "9\n3 8 4 7 1 2 5 6 9\n3\n2 7 1",
"output": "15 15"
},
{
"input": "9\n5 3 8 4 2 6 1 7 9\n4\n6 1 9 2",
"output": "27 13"
},
{
"input": "4\n1 3 2 4\n4\n3 1 2 3",
"output": "8 12"
},
{
"input": "3\n1 2 3\n8\n3 2 1 1 2 3 1 2",
"output": "15 17"
}
] | 1,639,776,660
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 6
| 2,000
| 10,240,000
|
import sys
def solve(L,B):
S=0
S2=0
for e in B:
S=S+L.index(e)+1
print(S,len(B)*len(L)-(S-len(B)))
_ = sys.stdin.readline()
L = list(map(int,sys.stdin.readline().split()))
_ = sys.stdin.readline()
B = list(map(int,sys.stdin.readline().split()))
solve(L,B)
|
Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element).
|
```python
import sys
def solve(L,B):
S=0
S2=0
for e in B:
S=S+L.index(e)+1
print(S,len(B)*len(L)-(S-len(B)))
_ = sys.stdin.readline()
L = list(map(int,sys.stdin.readline().split()))
_ = sys.stdin.readline()
B = list(map(int,sys.stdin.readline().split()))
solve(L,B)
```
| 0
|
|
937
|
A
|
Olympiad
|
PROGRAMMING
| 800
|
[
"implementation",
"sortings"
] | null | null |
The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score.
|
Print a single integer — the desired number of ways.
|
[
"4\n1 3 3 2\n",
"3\n1 1 1\n",
"4\n42 0 0 42\n"
] |
[
"3\n",
"1\n",
"1\n"
] |
There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything.
| 500
|
[
{
"input": "4\n1 3 3 2",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "4\n42 0 0 42",
"output": "1"
},
{
"input": "10\n1 0 1 0 1 0 0 0 0 1",
"output": "1"
},
{
"input": "10\n572 471 540 163 50 30 561 510 43 200",
"output": "10"
},
{
"input": "100\n122 575 426 445 172 81 247 429 97 202 175 325 382 384 417 356 132 502 328 537 57 339 518 211 479 306 140 168 268 16 140 263 593 249 391 310 555 468 231 180 157 18 334 328 276 155 21 280 322 545 111 267 467 274 291 304 235 34 365 180 21 95 501 552 325 331 302 353 296 22 289 399 7 466 32 302 568 333 75 192 284 10 94 128 154 512 9 480 243 521 551 492 420 197 207 125 367 117 438 600",
"output": "94"
},
{
"input": "100\n600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600",
"output": "1"
},
{
"input": "78\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12",
"output": "13"
},
{
"input": "34\n220 387 408 343 184 447 197 307 337 414 251 319 426 322 347 242 208 412 188 185 241 235 216 259 331 372 322 284 444 384 214 297 389 391",
"output": "33"
},
{
"input": "100\n1 2 1 0 3 0 2 0 0 1 2 0 1 3 0 3 3 1 3 0 0 2 1 2 2 1 3 3 3 3 3 2 0 0 2 1 2 3 2 3 0 1 1 3 3 2 0 3 1 0 2 2 2 1 2 3 2 1 0 3 0 2 0 3 0 2 1 0 3 1 0 2 2 1 3 1 3 0 2 3 3 1 1 3 1 3 0 3 2 0 2 3 3 0 2 0 2 0 1 3",
"output": "3"
},
{
"input": "100\n572 471 540 163 50 30 561 510 43 200 213 387 500 424 113 487 357 333 294 337 435 202 447 494 485 465 161 344 470 559 104 356 393 207 224 213 511 514 60 386 149 216 392 229 429 173 165 401 395 150 127 579 344 390 529 296 225 425 318 79 465 447 177 110 367 212 459 270 41 500 277 567 125 436 178 9 214 342 203 112 144 24 79 155 495 556 40 549 463 281 241 316 2 246 1 396 510 293 332 55",
"output": "93"
},
{
"input": "99\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12 2 3 9 3 7 13 7 13 0 11 8 12 2 5 9 4 0 6 6 2 13",
"output": "13"
},
{
"input": "99\n1 0 1 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 0 0 0 1 1 0 0 0 0 0 1",
"output": "1"
},
{
"input": "99\n21 74 25 44 71 80 46 28 96 1 74 24 81 83 16 55 31 1 27 36 56 38 17 10 78 5 39 67 67 15 39 62 92 48 90 9 54 67 30 79 56 17 33 27 75 54 20 79 21 44 10 66 66 73 90 3 34 33 64 79 20 94 0 51 24 30 1 52 95 21 88 98 6 65 31 1 67 32 74 91 83 9 93 27 53 11 8 79 42 20 50 91 19 96 6 24 66 16 37",
"output": "61"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "2\n0 600",
"output": "1"
},
{
"input": "4\n1 1 1 2",
"output": "2"
},
{
"input": "4\n0 0 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "2\n0 5",
"output": "1"
},
{
"input": "5\n1 0 0 1 2",
"output": "2"
}
] | 1,664,013,427
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 46
| 0
|
no3=int(input())
no4=list(map(int,input().split()))
while(0 in no4):
no4.remove(0)
no4=set(no4)
print(len(no4))
|
Title: Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score.
Output Specification:
Print a single integer — the desired number of ways.
Demo Input:
['4\n1 3 3 2\n', '3\n1 1 1\n', '4\n42 0 0 42\n']
Demo Output:
['3\n', '1\n', '1\n']
Note:
There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything.
|
```python
no3=int(input())
no4=list(map(int,input().split()))
while(0 in no4):
no4.remove(0)
no4=set(no4)
print(len(no4))
```
| 3
|
|
902
|
B
|
Coloring a Tree
|
PROGRAMMING
| 1,200
|
[
"dfs and similar",
"dsu",
"greedy"
] | null | null |
You are given a rooted tree with *n* vertices. The vertices are numbered from 1 to *n*, the root is the vertex number 1.
Each vertex has a color, let's denote the color of vertex *v* by *c**v*. Initially *c**v*<==<=0.
You have to color the tree into the given colors using the smallest possible number of steps. On each step you can choose a vertex *v* and a color *x*, and then color all vectices in the subtree of *v* (including *v* itself) in color *x*. In other words, for every vertex *u*, such that the path from root to *u* passes through *v*, set *c**u*<==<=*x*.
It is guaranteed that you have to color each vertex in a color different from 0.
You can learn what a rooted tree is using the link: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory)).
|
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=104) — the number of vertices in the tree.
The second line contains *n*<=-<=1 integers *p*2,<=*p*3,<=...,<=*p**n* (1<=≤<=*p**i*<=<<=*i*), where *p**i* means that there is an edge between vertices *i* and *p**i*.
The third line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=*n*), where *c**i* is the color you should color the *i*-th vertex into.
It is guaranteed that the given graph is a tree.
|
Print a single integer — the minimum number of steps you have to perform to color the tree into given colors.
|
[
"6\n1 2 2 1 5\n2 1 1 1 1 1\n",
"7\n1 1 2 3 1 4\n3 3 1 1 1 2 3\n"
] |
[
"3\n",
"5\n"
] |
The tree from the first sample is shown on the picture (numbers are vetices' indices):
<img class="tex-graphics" src="https://espresso.codeforces.com/10324ccdc37f95343acc4f3c6050d8c334334ffa.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On first step we color all vertices in the subtree of vertex 1 into color 2 (numbers are colors):
<img class="tex-graphics" src="https://espresso.codeforces.com/1c7bb267e2c1a006132248a43121400189309e2f.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On seond step we color all vertices in the subtree of vertex 5 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/2201a6d49b89ba850ff0d0bdcbb3f8e9dd3871a8.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On third step we color all vertices in the subtree of vertex 2 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/6fa977fcdebdde94c47695151e0427b33d0102c5.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The tree from the second sample is shown on the picture (numbers are vetices' indices):
<img class="tex-graphics" src="https://espresso.codeforces.com/d70f9ae72a2ed429dd6531cac757e375dd3c953d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On first step we color all vertices in the subtree of vertex 1 into color 3 (numbers are colors):
<img class="tex-graphics" src="https://espresso.codeforces.com/7289e8895d0dd56c47b6b17969b9cf77b36786b5.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On second step we color all vertices in the subtree of vertex 3 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/819001df7229138db3a407713744d1e3be88b64e.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On third step we color all vertices in the subtree of vertex 6 into color 2:
<img class="tex-graphics" src="https://espresso.codeforces.com/80ebbd870a0a339636a21b9acdaf9de046458b43.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On fourth step we color all vertices in the subtree of vertex 4 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/ed836aa723ac0176abde4e32988e3ac205014e93.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On fith step we color all vertices in the subtree of vertex 7 into color 3:
<img class="tex-graphics" src="https://espresso.codeforces.com/8132909e11b41c27b8df2f0b0c10bc841f35e58a.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 1,000
|
[
{
"input": "6\n1 2 2 1 5\n2 1 1 1 1 1",
"output": "3"
},
{
"input": "7\n1 1 2 3 1 4\n3 3 1 1 1 2 3",
"output": "5"
},
{
"input": "2\n1\n2 2",
"output": "1"
},
{
"input": "3\n1 1\n2 2 2",
"output": "1"
},
{
"input": "4\n1 2 1\n1 2 3 4",
"output": "4"
},
{
"input": "4\n1 2 3\n4 1 2 4",
"output": "4"
},
{
"input": "5\n1 2 1 4\n1 1 1 2 2",
"output": "2"
},
{
"input": "3\n1 2\n2 1 1",
"output": "2"
},
{
"input": "4\n1 1 1\n3 1 3 1",
"output": "3"
},
{
"input": "4\n1 1 2\n4 1 4 1",
"output": "2"
},
{
"input": "4\n1 2 2\n3 1 2 3",
"output": "4"
},
{
"input": "3\n1 1\n1 2 2",
"output": "3"
}
] | 1,644,863,382
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 50
| 124
| 5,734,400
|
from collections import defaultdict as di
from collections import deque as dq
n=int(input())
coupl = di(list)
nodess = [int(x)-1 for x in input().split()]
for i in range(n-1):
node = nodess[i]
coupl[i+1].append(node)
coupl[node].append(i+1)
colors = [int(x)-1 for x in input().split()]
Q=dq()
found=set()
Q.append((0,0))
parent = [-1]*n
order = []
while Q:
node,p = Q.popleft()
if node in found:
continue
found.add(node)
parent[node]=p
order.append(node)
for nei in coupl[node]:
Q.append((nei,node))
color = [-1]*n
count = 0
for node in order:
if colors[node]==color[parent[node]]:
pass
else:
count+=1
color[node]=colors[node]
print(count)
|
Title: Coloring a Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a rooted tree with *n* vertices. The vertices are numbered from 1 to *n*, the root is the vertex number 1.
Each vertex has a color, let's denote the color of vertex *v* by *c**v*. Initially *c**v*<==<=0.
You have to color the tree into the given colors using the smallest possible number of steps. On each step you can choose a vertex *v* and a color *x*, and then color all vectices in the subtree of *v* (including *v* itself) in color *x*. In other words, for every vertex *u*, such that the path from root to *u* passes through *v*, set *c**u*<==<=*x*.
It is guaranteed that you have to color each vertex in a color different from 0.
You can learn what a rooted tree is using the link: [https://en.wikipedia.org/wiki/Tree_(graph_theory)](https://en.wikipedia.org/wiki/Tree_(graph_theory)).
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=104) — the number of vertices in the tree.
The second line contains *n*<=-<=1 integers *p*2,<=*p*3,<=...,<=*p**n* (1<=≤<=*p**i*<=<<=*i*), where *p**i* means that there is an edge between vertices *i* and *p**i*.
The third line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=*n*), where *c**i* is the color you should color the *i*-th vertex into.
It is guaranteed that the given graph is a tree.
Output Specification:
Print a single integer — the minimum number of steps you have to perform to color the tree into given colors.
Demo Input:
['6\n1 2 2 1 5\n2 1 1 1 1 1\n', '7\n1 1 2 3 1 4\n3 3 1 1 1 2 3\n']
Demo Output:
['3\n', '5\n']
Note:
The tree from the first sample is shown on the picture (numbers are vetices' indices):
<img class="tex-graphics" src="https://espresso.codeforces.com/10324ccdc37f95343acc4f3c6050d8c334334ffa.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On first step we color all vertices in the subtree of vertex 1 into color 2 (numbers are colors):
<img class="tex-graphics" src="https://espresso.codeforces.com/1c7bb267e2c1a006132248a43121400189309e2f.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On seond step we color all vertices in the subtree of vertex 5 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/2201a6d49b89ba850ff0d0bdcbb3f8e9dd3871a8.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On third step we color all vertices in the subtree of vertex 2 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/6fa977fcdebdde94c47695151e0427b33d0102c5.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The tree from the second sample is shown on the picture (numbers are vetices' indices):
<img class="tex-graphics" src="https://espresso.codeforces.com/d70f9ae72a2ed429dd6531cac757e375dd3c953d.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On first step we color all vertices in the subtree of vertex 1 into color 3 (numbers are colors):
<img class="tex-graphics" src="https://espresso.codeforces.com/7289e8895d0dd56c47b6b17969b9cf77b36786b5.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On second step we color all vertices in the subtree of vertex 3 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/819001df7229138db3a407713744d1e3be88b64e.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On third step we color all vertices in the subtree of vertex 6 into color 2:
<img class="tex-graphics" src="https://espresso.codeforces.com/80ebbd870a0a339636a21b9acdaf9de046458b43.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On fourth step we color all vertices in the subtree of vertex 4 into color 1:
<img class="tex-graphics" src="https://espresso.codeforces.com/ed836aa723ac0176abde4e32988e3ac205014e93.png" style="max-width: 100.0%;max-height: 100.0%;"/>
On fith step we color all vertices in the subtree of vertex 7 into color 3:
<img class="tex-graphics" src="https://espresso.codeforces.com/8132909e11b41c27b8df2f0b0c10bc841f35e58a.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
from collections import defaultdict as di
from collections import deque as dq
n=int(input())
coupl = di(list)
nodess = [int(x)-1 for x in input().split()]
for i in range(n-1):
node = nodess[i]
coupl[i+1].append(node)
coupl[node].append(i+1)
colors = [int(x)-1 for x in input().split()]
Q=dq()
found=set()
Q.append((0,0))
parent = [-1]*n
order = []
while Q:
node,p = Q.popleft()
if node in found:
continue
found.add(node)
parent[node]=p
order.append(node)
for nei in coupl[node]:
Q.append((nei,node))
color = [-1]*n
count = 0
for node in order:
if colors[node]==color[parent[node]]:
pass
else:
count+=1
color[node]=colors[node]
print(count)
```
| 3
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,653,576,405
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 15
| 0
|
s=int(input())
s1=int(input())
a=str(s)
y=str(s1)
x=""
for i in range(len(a)):
if(a[i]==y[i]):
x=x+"0"
else:
x=x+"1"
print("".join(list(x)))
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
s=int(input())
s1=int(input())
a=str(s)
y=str(s1)
x=""
for i in range(len(a)):
if(a[i]==y[i]):
x=x+"0"
else:
x=x+"1"
print("".join(list(x)))
```
| -1
|
365
|
A
|
Good Number
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
|
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
|
Print a single integer — the number of *k*-good numbers in *a*.
|
[
"10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n",
"2 1\n1\n10\n"
] |
[
"10\n",
"1\n"
] |
none
| 500
|
[
{
"input": "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560",
"output": "10"
},
{
"input": "2 1\n1\n10",
"output": "1"
},
{
"input": "1 0\n1000000000",
"output": "1"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "6 0\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "1 3\n1000000000",
"output": "0"
},
{
"input": "1 9\n1000000000",
"output": "0"
},
{
"input": "1 0\n1",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 3\n1",
"output": "0"
},
{
"input": "1 9\n1",
"output": "0"
},
{
"input": "2 8\n123456780\n123",
"output": "1"
},
{
"input": "2 8\n12345678\n1230",
"output": "0"
},
{
"input": "6 1\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "6 2\n10\n102\n120\n1032\n1212103\n1999999",
"output": "4"
},
{
"input": "6 3\n10\n102\n120\n1032\n1212103\n1999999",
"output": "2"
},
{
"input": "6 9\n10\n102\n120\n1032\n1212103\n1999999",
"output": "0"
},
{
"input": "1 2\n2220",
"output": "0"
}
] | 1,672,005,283
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 46
| 2,867,200
|
n, k = input().split()
# n, k = 10, 6
k = int(k)
numbers = []
counter = 0
if k > 4:
for _ in range(int(n)):
i = input()
for j in range(k, 10):
if j not in i:
break
else:
counter += 1
else:
for _ in range(int(n)):
i = input()
for j in range(9, k - 1, -1):
if j not in i:
break
else:
counter += 1
print(counter)
# counter = 0
# for i in numbers:
# if k > 0:
# flag = True
# if i[0] > i[1] * k:
# flag = False
# if flag:
# counter += 1
# else:
# flag = True
# if i[0] > 0:
# flag = False
# if flag:
# counter += 1
# print(counter)
|
Title: Good Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
Input Specification:
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single integer — the number of *k*-good numbers in *a*.
Demo Input:
['10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n', '2 1\n1\n10\n']
Demo Output:
['10\n', '1\n']
Note:
none
|
```python
n, k = input().split()
# n, k = 10, 6
k = int(k)
numbers = []
counter = 0
if k > 4:
for _ in range(int(n)):
i = input()
for j in range(k, 10):
if j not in i:
break
else:
counter += 1
else:
for _ in range(int(n)):
i = input()
for j in range(9, k - 1, -1):
if j not in i:
break
else:
counter += 1
print(counter)
# counter = 0
# for i in numbers:
# if k > 0:
# flag = True
# if i[0] > i[1] * k:
# flag = False
# if flag:
# counter += 1
# else:
# flag = True
# if i[0] > 0:
# flag = False
# if flag:
# counter += 1
# print(counter)
```
| -1
|
|
433
|
A
|
Kitahara Haruki's Gift
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends?
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple.
|
In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes).
|
[
"3\n100 200 100\n",
"4\n100 100 100 200\n"
] |
[
"YES\n",
"NO\n"
] |
In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa.
| 500
|
[
{
"input": "3\n100 200 100",
"output": "YES"
},
{
"input": "4\n100 100 100 200",
"output": "NO"
},
{
"input": "1\n100",
"output": "NO"
},
{
"input": "1\n200",
"output": "NO"
},
{
"input": "2\n100 100",
"output": "YES"
},
{
"input": "2\n200 200",
"output": "YES"
},
{
"input": "100\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "YES"
},
{
"input": "100\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "NO"
},
{
"input": "52\n200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 100 200 100 200 200 200 100 200 200",
"output": "YES"
},
{
"input": "2\n100 200",
"output": "NO"
},
{
"input": "2\n200 100",
"output": "NO"
},
{
"input": "3\n100 100 100",
"output": "NO"
},
{
"input": "3\n200 200 200",
"output": "NO"
},
{
"input": "3\n200 100 200",
"output": "NO"
},
{
"input": "4\n100 100 100 100",
"output": "YES"
},
{
"input": "4\n200 200 200 200",
"output": "YES"
},
{
"input": "100\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "YES"
},
{
"input": "100\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 100 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "NO"
},
{
"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",
"output": "YES"
},
{
"input": "100\n100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "NO"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "YES"
},
{
"input": "100\n100 100 100 100 100 100 100 100 200 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "NO"
},
{
"input": "99\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "NO"
},
{
"input": "99\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "NO"
},
{
"input": "99\n200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "YES"
},
{
"input": "99\n200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200",
"output": "NO"
},
{
"input": "99\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",
"output": "NO"
},
{
"input": "99\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "YES"
},
{
"input": "99\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "NO"
},
{
"input": "100\n100 100 200 100 100 200 200 200 200 100 200 100 100 100 200 100 100 100 100 200 100 100 100 100 100 100 200 100 100 200 200 100 100 100 200 200 200 100 200 200 100 200 100 100 200 100 200 200 100 200 200 100 100 200 200 100 200 200 100 100 200 100 200 100 200 200 200 200 200 100 200 200 200 200 200 200 100 100 200 200 200 100 100 100 200 100 100 200 100 100 100 200 200 100 100 200 200 200 200 100",
"output": "YES"
},
{
"input": "100\n100 100 200 200 100 200 100 100 100 100 100 100 200 100 200 200 200 100 100 200 200 200 200 200 100 200 100 200 100 100 100 200 100 100 200 100 200 100 100 100 200 200 100 100 100 200 200 200 200 200 100 200 200 100 100 100 100 200 100 100 200 100 100 100 100 200 200 200 100 200 100 200 200 200 100 100 200 200 200 200 100 200 100 200 200 100 200 100 200 200 200 200 200 200 100 100 100 200 200 100",
"output": "NO"
},
{
"input": "100\n100 200 100 100 200 200 200 200 100 200 200 200 200 200 200 200 200 200 100 100 100 200 200 200 200 200 100 200 200 200 200 100 200 200 100 100 200 100 100 100 200 100 100 100 200 100 200 100 200 200 200 100 100 200 100 200 100 200 100 100 100 200 100 200 100 100 100 100 200 200 200 200 100 200 200 100 200 100 100 100 200 100 100 100 100 100 200 100 100 100 200 200 200 100 200 100 100 100 200 200",
"output": "YES"
},
{
"input": "99\n100 200 200 200 100 200 100 200 200 100 100 100 100 200 100 100 200 100 200 100 100 200 100 100 200 200 100 100 100 100 200 200 200 200 200 100 100 200 200 100 100 100 100 200 200 100 100 100 100 100 200 200 200 100 100 100 200 200 200 100 200 100 100 100 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 100 200 100 200 200 200 200 100 200 100 100 100 100 100 100 100 100 100",
"output": "YES"
},
{
"input": "99\n100 200 100 100 100 100 200 200 100 200 100 100 200 100 100 100 100 100 100 200 100 100 100 100 100 100 100 200 100 200 100 100 100 100 100 100 100 200 200 200 200 200 200 200 100 200 100 200 100 200 100 200 100 100 200 200 200 100 200 200 200 200 100 200 100 200 200 200 200 100 200 100 200 200 100 200 200 200 200 200 100 100 200 100 100 100 100 200 200 200 100 100 200 200 200 200 200 200 200",
"output": "NO"
},
{
"input": "99\n200 100 100 100 200 200 200 100 100 100 100 100 100 100 100 100 200 200 100 200 200 100 200 100 100 200 200 200 100 200 100 200 200 100 200 100 200 200 200 100 100 200 200 200 200 100 100 100 100 200 200 200 200 100 200 200 200 100 100 100 200 200 200 100 200 100 200 100 100 100 200 100 200 200 100 200 200 200 100 100 100 200 200 200 100 200 200 200 100 100 100 200 100 200 100 100 100 200 200",
"output": "YES"
},
{
"input": "56\n100 200 200 200 200 200 100 200 100 100 200 100 100 100 100 100 200 200 200 100 200 100 100 200 200 200 100 200 100 200 200 100 100 100 100 100 200 100 200 100 200 200 200 100 100 200 200 200 200 200 200 200 200 200 200 100",
"output": "YES"
},
{
"input": "72\n200 100 200 200 200 100 100 200 200 100 100 100 100 200 100 200 100 100 100 100 200 100 200 100 100 200 100 100 200 100 200 100 100 200 100 200 100 100 200 200 200 200 200 100 100 200 200 200 200 100 100 100 200 200 100 100 100 100 100 200 100 100 200 100 100 200 200 100 100 200 100 200",
"output": "YES"
},
{
"input": "32\n200 200 200 100 100 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 200 200 200 200 200",
"output": "YES"
},
{
"input": "48\n200 200 200 200 200 200 100 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 200 100 200 100 200 200 200 200 200 200",
"output": "NO"
},
{
"input": "60\n100 100 200 200 100 200 100 200 100 100 100 100 100 100 200 100 100 100 200 100 200 100 100 100 100 100 200 100 200 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 200 100 100 100",
"output": "YES"
},
{
"input": "24\n200 200 100 100 200 100 200 200 100 200 200 200 200 200 100 200 200 200 200 200 200 200 200 100",
"output": "YES"
},
{
"input": "40\n100 100 200 200 200 200 100 100 100 200 100 100 200 200 100 100 100 100 100 200 100 200 200 100 200 200 200 100 100 100 100 100 200 200 100 200 100 100 200 100",
"output": "NO"
},
{
"input": "5\n200 200 200 200 200",
"output": "NO"
},
{
"input": "9\n100 100 100 200 100 100 200 100 200",
"output": "YES"
},
{
"input": "1\n200",
"output": "NO"
},
{
"input": "7\n200 200 200 100 200 200 200",
"output": "NO"
},
{
"input": "4\n100 100 200 200",
"output": "YES"
},
{
"input": "6\n100 100 100 200 200 200",
"output": "NO"
},
{
"input": "4\n200 100 100 200",
"output": "YES"
},
{
"input": "5\n100 100 100 100 200",
"output": "YES"
}
] | 1,636,710,121
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include<stdio.h>
int main()
{
int i,a[101],b,c,d,e,f,g,m,n;
scanf("%d",&i);
c=0;
d=0;
e=0;
g=0;
for(b=1;b<=i;b++)
{
scanf("%d",&a[b]);
getchar();
if(a[b]==100)
d++;
if(a[b]==200)
e++;
c=c+a[b];
}
for(m=0;m<=d;m++)
{
for(n=0;n<=e;n++)
{
f=100*m+200*n;
if(f==c/2)
{
g++;
}
}
}
if(g==0)
printf("No");
if(g!=0)
printf("Yes");
return 0;
}
|
Title: Kitahara Haruki's Gift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends?
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple.
Output Specification:
In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes).
Demo Input:
['3\n100 200 100\n', '4\n100 100 100 200\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa.
|
```python
#include<stdio.h>
int main()
{
int i,a[101],b,c,d,e,f,g,m,n;
scanf("%d",&i);
c=0;
d=0;
e=0;
g=0;
for(b=1;b<=i;b++)
{
scanf("%d",&a[b]);
getchar();
if(a[b]==100)
d++;
if(a[b]==200)
e++;
c=c+a[b];
}
for(m=0;m<=d;m++)
{
for(n=0;n<=e;n++)
{
f=100*m+200*n;
if(f==c/2)
{
g++;
}
}
}
if(g==0)
printf("No");
if(g!=0)
printf("Yes");
return 0;
}
```
| -1
|
|
804
|
A
|
Find Amir
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"greedy",
"math"
] | null | null |
A few years ago Sajjad left his school and register to another one due to security reasons. Now he wishes to find Amir, one of his schoolmates and good friends.
There are *n* schools numerated from 1 to *n*. One can travel between each pair of them, to do so, he needs to buy a ticket. The ticker between schools *i* and *j* costs and can be used multiple times. Help Sajjad to find the minimum cost he needs to pay for tickets to visit all schools. He can start and finish in any school.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of schools.
|
Print single integer: the minimum cost of tickets needed to visit all schools.
|
[
"2\n",
"10\n"
] |
[
"0\n",
"4\n"
] |
In the first example we can buy a ticket between the schools that costs <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c67d72010e0498bfd065a6a38fdeaec90358507b.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
| 500
|
[
{
"input": "2",
"output": "0"
},
{
"input": "10",
"output": "4"
},
{
"input": "43670",
"output": "21834"
},
{
"input": "4217",
"output": "2108"
},
{
"input": "17879",
"output": "8939"
},
{
"input": "31809",
"output": "15904"
},
{
"input": "40873",
"output": "20436"
},
{
"input": "77859",
"output": "38929"
},
{
"input": "53022",
"output": "26510"
},
{
"input": "79227",
"output": "39613"
},
{
"input": "100000",
"output": "49999"
},
{
"input": "82801",
"output": "41400"
},
{
"input": "5188",
"output": "2593"
},
{
"input": "86539",
"output": "43269"
},
{
"input": "12802",
"output": "6400"
},
{
"input": "20289",
"output": "10144"
},
{
"input": "32866",
"output": "16432"
},
{
"input": "33377",
"output": "16688"
},
{
"input": "31775",
"output": "15887"
},
{
"input": "60397",
"output": "30198"
},
{
"input": "100000",
"output": "49999"
},
{
"input": "99999",
"output": "49999"
},
{
"input": "99998",
"output": "49998"
},
{
"input": "99997",
"output": "49998"
},
{
"input": "99996",
"output": "49997"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "1"
},
{
"input": "1",
"output": "0"
},
{
"input": "3",
"output": "1"
}
] | 1,655,516,842
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
x = int(input())
print((x-1)/2)
|
Title: Find Amir
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A few years ago Sajjad left his school and register to another one due to security reasons. Now he wishes to find Amir, one of his schoolmates and good friends.
There are *n* schools numerated from 1 to *n*. One can travel between each pair of them, to do so, he needs to buy a ticket. The ticker between schools *i* and *j* costs and can be used multiple times. Help Sajjad to find the minimum cost he needs to pay for tickets to visit all schools. He can start and finish in any school.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of schools.
Output Specification:
Print single integer: the minimum cost of tickets needed to visit all schools.
Demo Input:
['2\n', '10\n']
Demo Output:
['0\n', '4\n']
Note:
In the first example we can buy a ticket between the schools that costs <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c67d72010e0498bfd065a6a38fdeaec90358507b.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
|
```python
x = int(input())
print((x-1)/2)
```
| 0
|
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,609,510,637
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 4
| 248
| 0
|
n=int(input())
l=list(map(int,input().split()))
a,b,c,d=0,0,0,0
for i in range(n):
if l[i]%2==0:
a+=1
b=i
else:
c+=1
d=i
if b==1:
print(b+1)
elif c==1:
print(d+1)
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
n=int(input())
l=list(map(int,input().split()))
a,b,c,d=0,0,0,0
for i in range(n):
if l[i]%2==0:
a+=1
b=i
else:
c+=1
d=i
if b==1:
print(b+1)
elif c==1:
print(d+1)
```
| 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"
}
] | 1,665,558,208
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 46
| 530
| 16,793,600
|
from collections import defaultdict
n,m = list(map(int,input().split()))
arr = list(map(int,input().split()))
parent = [i for i in range(n+1)]
rank = [0]*(n+1)
def find(p):
while p != parent[p]:
p = parent[p]
return p
def union(x, y):
x = find(x)
y = find(y)
if x!=y:
if rank[x] > rank[y]:
parent[y] = x
elif rank[y] > rank[x]:
parent[x] = y
else:
parent[y] = x
rank[x] += 1
for _ in range(m):
a,b = list(map(int,input().split()))
union(a,b)
for i in range(1,len(parent)):
parent[i] = find(parent[i])
gold = 0
collect = defaultdict(list)
for i in range(1,len(parent)):
collect[parent[i]].append(i)
for v in collect.values():
mn = 10**9
for c in v:
mn = min(mn,arr[c-1])
gold += mn
print(gold)
|
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
from collections import defaultdict
n,m = list(map(int,input().split()))
arr = list(map(int,input().split()))
parent = [i for i in range(n+1)]
rank = [0]*(n+1)
def find(p):
while p != parent[p]:
p = parent[p]
return p
def union(x, y):
x = find(x)
y = find(y)
if x!=y:
if rank[x] > rank[y]:
parent[y] = x
elif rank[y] > rank[x]:
parent[x] = y
else:
parent[y] = x
rank[x] += 1
for _ in range(m):
a,b = list(map(int,input().split()))
union(a,b)
for i in range(1,len(parent)):
parent[i] = find(parent[i])
gold = 0
collect = defaultdict(list)
for i in range(1,len(parent)):
collect[parent[i]].append(i)
for v in collect.values():
mn = 10**9
for c in v:
mn = min(mn,arr[c-1])
gold += mn
print(gold)
```
| 3
|
|
633
|
A
|
Ebony and Ivory
|
PROGRAMMING
| 1,100
|
[
"brute force",
"math",
"number theory"
] | null | null |
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals *a* units of damage while Ivory deals *b* units of damage. In order to break the shield Dante has to deal exactly *c* units of damage. Find out if this is possible.
|
The first line of the input contains three integers *a*, *b*, *c* (1<=≤<=*a*,<=*b*<=≤<=100,<=1<=≤<=*c*<=≤<=10<=000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
|
Print "Yes" (without quotes) if Dante can deal exactly *c* damage to the shield and "No" (without quotes) otherwise.
|
[
"4 6 15\n",
"3 2 7\n",
"6 11 6\n"
] |
[
"No\n",
"Yes\n",
"Yes\n"
] |
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
| 250
|
[
{
"input": "4 6 15",
"output": "No"
},
{
"input": "3 2 7",
"output": "Yes"
},
{
"input": "6 11 6",
"output": "Yes"
},
{
"input": "3 12 15",
"output": "Yes"
},
{
"input": "5 5 10",
"output": "Yes"
},
{
"input": "6 6 7",
"output": "No"
},
{
"input": "1 1 20",
"output": "Yes"
},
{
"input": "12 14 19",
"output": "No"
},
{
"input": "15 12 26",
"output": "No"
},
{
"input": "2 4 8",
"output": "Yes"
},
{
"input": "4 5 30",
"output": "Yes"
},
{
"input": "4 5 48",
"output": "Yes"
},
{
"input": "2 17 105",
"output": "Yes"
},
{
"input": "10 25 282",
"output": "No"
},
{
"input": "6 34 323",
"output": "No"
},
{
"input": "2 47 464",
"output": "Yes"
},
{
"input": "4 53 113",
"output": "Yes"
},
{
"input": "6 64 546",
"output": "Yes"
},
{
"input": "1 78 725",
"output": "Yes"
},
{
"input": "1 84 811",
"output": "Yes"
},
{
"input": "3 100 441",
"output": "Yes"
},
{
"input": "20 5 57",
"output": "No"
},
{
"input": "14 19 143",
"output": "No"
},
{
"input": "17 23 248",
"output": "No"
},
{
"input": "11 34 383",
"output": "Yes"
},
{
"input": "20 47 568",
"output": "Yes"
},
{
"input": "16 58 410",
"output": "Yes"
},
{
"input": "11 70 1199",
"output": "Yes"
},
{
"input": "16 78 712",
"output": "Yes"
},
{
"input": "20 84 562",
"output": "No"
},
{
"input": "19 100 836",
"output": "Yes"
},
{
"input": "23 10 58",
"output": "No"
},
{
"input": "25 17 448",
"output": "Yes"
},
{
"input": "22 24 866",
"output": "Yes"
},
{
"input": "24 35 67",
"output": "No"
},
{
"input": "29 47 264",
"output": "Yes"
},
{
"input": "23 56 45",
"output": "No"
},
{
"input": "25 66 1183",
"output": "Yes"
},
{
"input": "21 71 657",
"output": "Yes"
},
{
"input": "29 81 629",
"output": "No"
},
{
"input": "23 95 2226",
"output": "Yes"
},
{
"input": "32 4 62",
"output": "No"
},
{
"input": "37 15 789",
"output": "Yes"
},
{
"input": "39 24 999",
"output": "Yes"
},
{
"input": "38 32 865",
"output": "No"
},
{
"input": "32 50 205",
"output": "No"
},
{
"input": "31 57 1362",
"output": "Yes"
},
{
"input": "38 68 1870",
"output": "Yes"
},
{
"input": "36 76 549",
"output": "No"
},
{
"input": "35 84 1257",
"output": "No"
},
{
"input": "39 92 2753",
"output": "Yes"
},
{
"input": "44 1 287",
"output": "Yes"
},
{
"input": "42 12 830",
"output": "No"
},
{
"input": "42 27 9",
"output": "No"
},
{
"input": "49 40 1422",
"output": "No"
},
{
"input": "44 42 2005",
"output": "No"
},
{
"input": "50 55 2479",
"output": "No"
},
{
"input": "48 65 917",
"output": "No"
},
{
"input": "45 78 152",
"output": "No"
},
{
"input": "43 90 4096",
"output": "Yes"
},
{
"input": "43 94 4316",
"output": "Yes"
},
{
"input": "60 7 526",
"output": "Yes"
},
{
"input": "53 11 735",
"output": "Yes"
},
{
"input": "52 27 609",
"output": "Yes"
},
{
"input": "57 32 992",
"output": "Yes"
},
{
"input": "52 49 421",
"output": "No"
},
{
"input": "57 52 2634",
"output": "Yes"
},
{
"input": "54 67 3181",
"output": "Yes"
},
{
"input": "52 73 638",
"output": "No"
},
{
"input": "57 84 3470",
"output": "No"
},
{
"input": "52 100 5582",
"output": "No"
},
{
"input": "62 1 501",
"output": "Yes"
},
{
"input": "63 17 858",
"output": "Yes"
},
{
"input": "70 24 1784",
"output": "Yes"
},
{
"input": "65 32 1391",
"output": "Yes"
},
{
"input": "62 50 2775",
"output": "No"
},
{
"input": "62 58 88",
"output": "No"
},
{
"input": "66 68 3112",
"output": "Yes"
},
{
"input": "61 71 1643",
"output": "No"
},
{
"input": "69 81 3880",
"output": "No"
},
{
"input": "63 100 1960",
"output": "Yes"
},
{
"input": "73 6 431",
"output": "Yes"
},
{
"input": "75 19 736",
"output": "Yes"
},
{
"input": "78 25 247",
"output": "No"
},
{
"input": "79 36 2854",
"output": "Yes"
},
{
"input": "80 43 1864",
"output": "Yes"
},
{
"input": "76 55 2196",
"output": "Yes"
},
{
"input": "76 69 4122",
"output": "Yes"
},
{
"input": "76 76 4905",
"output": "No"
},
{
"input": "75 89 3056",
"output": "Yes"
},
{
"input": "73 100 3111",
"output": "Yes"
},
{
"input": "84 9 530",
"output": "No"
},
{
"input": "82 18 633",
"output": "No"
},
{
"input": "85 29 2533",
"output": "Yes"
},
{
"input": "89 38 2879",
"output": "Yes"
},
{
"input": "89 49 2200",
"output": "Yes"
},
{
"input": "88 60 4140",
"output": "Yes"
},
{
"input": "82 68 1299",
"output": "No"
},
{
"input": "90 76 2207",
"output": "No"
},
{
"input": "83 84 4923",
"output": "Yes"
},
{
"input": "89 99 7969",
"output": "Yes"
},
{
"input": "94 9 168",
"output": "No"
},
{
"input": "91 20 1009",
"output": "No"
},
{
"input": "93 23 2872",
"output": "Yes"
},
{
"input": "97 31 3761",
"output": "Yes"
},
{
"input": "99 46 1341",
"output": "Yes"
},
{
"input": "98 51 2845",
"output": "No"
},
{
"input": "93 66 3412",
"output": "No"
},
{
"input": "95 76 3724",
"output": "Yes"
},
{
"input": "91 87 6237",
"output": "Yes"
},
{
"input": "98 97 7886",
"output": "Yes"
},
{
"input": "12 17 15",
"output": "No"
},
{
"input": "93 94 95",
"output": "No"
},
{
"input": "27 43 27",
"output": "Yes"
},
{
"input": "17 43 68",
"output": "Yes"
},
{
"input": "44 12 12",
"output": "Yes"
},
{
"input": "44 50 150",
"output": "Yes"
},
{
"input": "1 1 10000",
"output": "Yes"
},
{
"input": "2 3 10000",
"output": "Yes"
},
{
"input": "100 1 10",
"output": "Yes"
},
{
"input": "3 2 1",
"output": "No"
},
{
"input": "1 1 1",
"output": "Yes"
},
{
"input": "9 9 10000",
"output": "No"
},
{
"input": "2 3 9995",
"output": "Yes"
},
{
"input": "3 5 4",
"output": "No"
},
{
"input": "99 98 100",
"output": "No"
},
{
"input": "6 10 2",
"output": "No"
},
{
"input": "1 6 5",
"output": "Yes"
},
{
"input": "1 4 3",
"output": "Yes"
},
{
"input": "3 2 3",
"output": "Yes"
},
{
"input": "1 7 6",
"output": "Yes"
},
{
"input": "2 3 9871",
"output": "Yes"
},
{
"input": "10 5 5",
"output": "Yes"
},
{
"input": "10 8 2",
"output": "No"
}
] | 1,681,464,856
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 134
| 93
| 1,536,000
|
a, b, c = map(int, input().split())
ans = "NO"
for i in range(c+1):
tmp = c - a*i
if tmp >= 0 and tmp%b == 0:
#print(i, (c-a*i)//b)
ans = "YES"
break
print(ans)
|
Title: Ebony and Ivory
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.
For every bullet that hits the shield, Ebony deals *a* units of damage while Ivory deals *b* units of damage. In order to break the shield Dante has to deal exactly *c* units of damage. Find out if this is possible.
Input Specification:
The first line of the input contains three integers *a*, *b*, *c* (1<=≤<=*a*,<=*b*<=≤<=100,<=1<=≤<=*c*<=≤<=10<=000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.
Output Specification:
Print "Yes" (without quotes) if Dante can deal exactly *c* damage to the shield and "No" (without quotes) otherwise.
Demo Input:
['4 6 15\n', '3 2 7\n', '6 11 6\n']
Demo Output:
['No\n', 'Yes\n', 'Yes\n']
Note:
In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.
|
```python
a, b, c = map(int, input().split())
ans = "NO"
for i in range(c+1):
tmp = c - a*i
if tmp >= 0 and tmp%b == 0:
#print(i, (c-a*i)//b)
ans = "YES"
break
print(ans)
```
| 3
|
|
400
|
B
|
Inna and New Matrix of Candies
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation",
"schedules"
] | null | null |
Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Candy Matrix 2: Reload".
The field for the new game is a rectangle table of size *n*<=×<=*m*. Each line of the table contains one cell with a dwarf figurine, one cell with a candy, the other cells of the line are empty. The game lasts for several moves. During each move the player should choose all lines of the matrix where dwarf is not on the cell with candy and shout "Let's go!". After that, all the dwarves from the chosen lines start to simultaneously move to the right. During each second, each dwarf goes to the adjacent cell that is located to the right of its current cell. The movement continues until one of the following events occurs:
- some dwarf in one of the chosen lines is located in the rightmost cell of his row; - some dwarf in the chosen lines is located in the cell with the candy.
The point of the game is to transport all the dwarves to the candy cells.
Inna is fabulous, as she came up with such an interesting game. But what about you? Your task is to play this game optimally well. Specifically, you should say by the given game field what minimum number of moves the player needs to reach the goal of the game.
|
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*<=≤<=1000; 2<=≤<=*m*<=≤<=1000).
Next *n* lines each contain *m* characters — the game field for the "Candy Martix 2: Reload". Character "*" represents an empty cell of the field, character "G" represents a dwarf and character "S" represents a candy. The matrix doesn't contain other characters. It is guaranteed that each line contains exactly one character "G" and one character "S".
|
In a single line print a single integer — either the minimum number of moves needed to achieve the aim of the game, or -1, if the aim cannot be achieved on the given game field.
|
[
"3 4\n*G*S\nG**S\n*G*S\n",
"1 3\nS*G\n"
] |
[
"2\n",
"-1\n"
] |
none
| 1,000
|
[
{
"input": "3 4\n*G*S\nG**S\n*G*S",
"output": "2"
},
{
"input": "1 3\nS*G",
"output": "-1"
},
{
"input": "10 10\nG********S\n*G*******S\n**G******S\n***G*****S\n****G****S\n*****G***S\n******G**S\n*******G*S\n********GS\nG********S",
"output": "9"
},
{
"input": "5 10\nG***S*****\nG****S****\n***GS*****\nG*S*******\nG***S*****",
"output": "4"
},
{
"input": "4 8\nG*S*****\n****G*S*\nG*****S*\n**G***S*",
"output": "3"
},
{
"input": "4 10\n***G****S*\n*****GS***\nG****S****\nG*******S*",
"output": "3"
},
{
"input": "1 2\nSG",
"output": "-1"
},
{
"input": "1 2\nGS",
"output": "1"
},
{
"input": "1 4\nSG**",
"output": "-1"
}
] | 1,461,713,289
| 4,089
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 4,608,000
|
'''
Created on 27.4.2016
@author: Ivan
'''
import sys
class Pair:
def __init__(self, x, y):
self.x = x
self.y = y
def __str__(self):
return str(self.x) + " " + str(self.y)
def __eq__(self, other):
return self.x == other.x and self.y == other.y
def __hash__(self):
return hash(str(self))
def main():
line = input()
parts = line.split(" ")
a = int(parts[0])
b = int(parts[1])
list = []
for i in range(0, a):
list.append(input())
pairs = []
res = 0
for l in list:
x = l.find("G")
y = l.find("S")
print()
if y < x:
print("-1")
return
pairs.append(Pair(x, y))
pairs = set(pairs)
print(str(len(pairs)))
if __name__ == '__main__':
main()
|
Title: Inna and New Matrix of Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Candy Matrix 2: Reload".
The field for the new game is a rectangle table of size *n*<=×<=*m*. Each line of the table contains one cell with a dwarf figurine, one cell with a candy, the other cells of the line are empty. The game lasts for several moves. During each move the player should choose all lines of the matrix where dwarf is not on the cell with candy and shout "Let's go!". After that, all the dwarves from the chosen lines start to simultaneously move to the right. During each second, each dwarf goes to the adjacent cell that is located to the right of its current cell. The movement continues until one of the following events occurs:
- some dwarf in one of the chosen lines is located in the rightmost cell of his row; - some dwarf in the chosen lines is located in the cell with the candy.
The point of the game is to transport all the dwarves to the candy cells.
Inna is fabulous, as she came up with such an interesting game. But what about you? Your task is to play this game optimally well. Specifically, you should say by the given game field what minimum number of moves the player needs to reach the goal of the game.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*<=≤<=1000; 2<=≤<=*m*<=≤<=1000).
Next *n* lines each contain *m* characters — the game field for the "Candy Martix 2: Reload". Character "*" represents an empty cell of the field, character "G" represents a dwarf and character "S" represents a candy. The matrix doesn't contain other characters. It is guaranteed that each line contains exactly one character "G" and one character "S".
Output Specification:
In a single line print a single integer — either the minimum number of moves needed to achieve the aim of the game, or -1, if the aim cannot be achieved on the given game field.
Demo Input:
['3 4\n*G*S\nG**S\n*G*S\n', '1 3\nS*G\n']
Demo Output:
['2\n', '-1\n']
Note:
none
|
```python
'''
Created on 27.4.2016
@author: Ivan
'''
import sys
class Pair:
def __init__(self, x, y):
self.x = x
self.y = y
def __str__(self):
return str(self.x) + " " + str(self.y)
def __eq__(self, other):
return self.x == other.x and self.y == other.y
def __hash__(self):
return hash(str(self))
def main():
line = input()
parts = line.split(" ")
a = int(parts[0])
b = int(parts[1])
list = []
for i in range(0, a):
list.append(input())
pairs = []
res = 0
for l in list:
x = l.find("G")
y = l.find("S")
print()
if y < x:
print("-1")
return
pairs.append(Pair(x, y))
pairs = set(pairs)
print(str(len(pairs)))
if __name__ == '__main__':
main()
```
| 0
|
|
254
|
A
|
Cards with Numbers
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"sortings"
] | null | null |
Petya has got 2*n* cards, each card contains some integer. The numbers on the cards can be the same. Let's index all cards by consecutive integers from 1 to 2*n*. We'll denote the number that is written on a card with number *i*, as *a**i*. In order to play one entertaining game with his friends, Petya needs to split the cards into pairs so that each pair had equal numbers on the cards. Help Petya do that.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains the sequence of 2*n* positive integers *a*1,<=*a*2,<=...,<=*a*2*n* (1<=≤<=*a**i*<=≤<=5000) — the numbers that are written on the cards. The numbers on the line are separated by single spaces.
|
If it is impossible to divide the cards into pairs so that cards in each pair had the same numbers, print on a single line integer -1. But if the required partition exists, then print *n* pairs of integers, a pair per line — the indices of the cards that form the pairs.
Separate the numbers on the lines by spaces. You can print the pairs and the numbers in the pairs in any order. If there are multiple solutions, print any of them.
|
[
"3\n20 30 10 30 20 10\n",
"1\n1 2\n"
] |
[
"4 2\n1 5\n6 3\n",
"-1"
] |
none
| 500
|
[
{
"input": "3\n20 30 10 30 20 10",
"output": "4 2\n1 5\n6 3"
},
{
"input": "1\n1 2",
"output": "-1"
},
{
"input": "5\n2 2 2 2 2 1 2 2 1 2",
"output": "2 1\n3 4\n7 5\n6 9\n10 8"
},
{
"input": "5\n2 1 2 2 1 1 1 1 1 2",
"output": "3 1\n2 5\n7 6\n8 9\n10 4"
},
{
"input": "5\n1 2 2 2 1 2 2 1 2 1",
"output": "3 2\n1 5\n6 4\n7 9\n10 8"
},
{
"input": "5\n3 3 1 1 1 3 2 3 1 2",
"output": "2 1\n3 4\n8 6\n5 9\n10 7"
},
{
"input": "5\n1 1 3 1 3 3 3 1 1 1",
"output": "2 1\n3 5\n7 6\n4 8\n10 9"
},
{
"input": "5\n3 1 1 1 2 3 3 3 2 1",
"output": "3 2\n1 6\n8 7\n5 9\n10 4"
},
{
"input": "5\n3 3 2 2 3 3 1 3 1 3",
"output": "2 1\n3 4\n6 5\n7 9\n10 8"
},
{
"input": "5\n4 1 3 1 4 1 2 2 3 1",
"output": "4 2\n1 5\n8 7\n3 9\n10 6"
},
{
"input": "100\n8 6 7 8 7 9 1 7 3 3 5 8 7 8 5 4 8 4 8 1 2 8 3 7 8 7 6 5 7 9 6 10 7 6 7 8 6 8 9 5 1 5 6 1 4 8 4 8 7 2 6 2 6 6 2 8 2 8 7 1 5 4 4 6 4 9 7 5 1 8 1 3 9 2 3 2 4 7 6 10 5 3 4 10 8 9 6 7 2 7 10 1 8 10 4 1 1 1 2 7 5 4 9 10 6 8 3 1 10 9 9 6 1 5 8 6 6 3 3 4 10 10 8 9 7 10 9 3 7 6 3 2 10 8 5 8 5 5 5 10 8 5 7 6 10 7 7 9 10 10 9 9 3 6 5 6 8 1 9 8 2 4 8 8 6 8 10 2 3 5 2 6 8 4 8 6 4 5 10 8 1 10 5 2 5 6 8 2 6 8 1 3 4 5 7 5 6 9 2 8",
"output": "4 1\n3 5\n10 9\n8 13\n14 12\n11 15\n18 16\n17 19\n20 7\n22 25\n26 24\n2 27\n30 6\n29 33\n34 31\n36 38\n40 28\n37 43\n44 41\n45 47\n48 46\n35 49\n50 21\n51 53\n55 52\n56 58\n61 42\n62 63\n64 54\n39 66\n67 59\n60 69\n72 23\n57 74\n77 65\n32 80\n81 68\n75 82\n85 70\n73 86\n87 79\n78 88\n89 76\n84 91\n92 71\n83 95\n97 96\n90 100\n104 94\n93 106\n108 98\n103 110\n112 105\n101 114\n117 116\n107 118\n120 102\n109 121\n123 115\n111 124\n126 122\n119 128\n129 125\n99 132\n136 134\n135 137\n139 138\n133 140\n144 130..."
},
{
"input": "100\n7 3 8 8 1 9 6 6 3 3 8 2 7 9 9 10 2 10 4 4 9 3 6 5 2 6 3 6 3 5 2 3 8 2 5 10 3 9 7 2 1 6 7 4 8 3 9 10 9 4 3 3 7 1 4 2 2 5 6 6 1 7 9 1 8 1 2 2 5 9 7 7 6 4 6 10 1 1 8 1 5 6 4 9 5 4 4 10 6 4 5 1 9 1 7 8 6 10 3 2 4 7 10 4 8 10 6 7 8 4 1 3 8 3 2 1 9 4 2 4 3 1 6 8 6 2 2 5 6 8 6 10 1 6 4 2 7 3 6 10 6 5 6 6 3 9 4 6 4 1 5 4 4 2 8 4 10 3 7 6 6 10 2 5 5 6 1 6 1 9 9 1 10 5 10 1 1 5 7 5 2 1 4 2 3 3 3 5 1 8 10 3 3 5 9 6 3 6 8 1",
"output": "4 3\n7 8\n9 2\n1 13\n14 6\n12 17\n18 16\n19 20\n21 15\n10 22\n26 23\n27 29\n30 24\n25 31\n33 11\n32 37\n40 34\n5 41\n42 28\n39 43\n47 38\n36 48\n50 44\n46 51\n57 56\n35 58\n60 59\n54 61\n62 53\n49 63\n65 45\n64 66\n68 67\n71 72\n74 55\n73 75\n78 77\n69 81\n84 70\n83 86\n88 76\n82 89\n90 87\n85 91\n92 80\n79 96\n99 52\n95 102\n103 98\n101 104\n107 97\n105 109\n111 94\n112 114\n115 100\n93 117\n118 110\n116 122\n124 113\n123 125\n126 119\n129 131\n132 106\n120 135\n136 127\n108 137\n138 121\n134 139\n142 128..."
},
{
"input": "100\n6 3 6 8 8 4 3 7 10 3 1 3 9 5 10 10 6 7 6 6 2 3 8 8 7 6 4 9 6 7 4 4 10 4 7 3 2 7 10 8 6 7 9 1 3 5 3 7 9 1 1 7 1 1 7 7 8 3 2 7 4 8 7 8 10 3 1 7 2 7 9 8 8 8 5 2 8 1 2 7 8 7 8 8 5 10 10 4 9 10 8 7 8 8 7 7 3 6 4 3 4 8 10 8 6 3 7 1 8 6 3 3 7 10 3 9 3 5 10 9 9 2 8 7 2 3 2 1 10 9 6 2 8 7 2 2 5 3 10 6 7 2 1 1 5 10 7 5 4 9 7 7 8 1 1 3 3 7 10 5 9 8 6 8 2 2 1 7 8 9 6 2 2 6 2 9 10 2 10 9 6 3 3 10 6 5 3 6 6 3 6 10 8 7 4 8 6 3 4 7",
"output": "-1"
},
{
"input": "1\n2 2",
"output": "2 1"
},
{
"input": "2\n1 2 4 7",
"output": "-1"
}
] | 1,686,132,668
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 35
| 498
| 75,468,800
|
import string
from collections import deque
alph = string.ascii_lowercase
def main():
with open('input.txt', 'r') as fin:
f = fin.readlines()
n = int(f[0])
a = list(map(int, f[1].split()))
d = {i:0 for i in a}
for i in a:
d[i] += 1
flag = 1
for i in list(d.keys()):
if d[i] % 2:
flag = 0
break
with open('output.txt', 'w') as fout:
if not flag:
fout.write(str(-1))
else:
dick = {i:[] for i in a}
for i in range(2*n):
dick[a[i]].append(i)
ans = []
for i in list(dick.keys()):
for j in range(0, len(dick[i]), 2):
ans.append([dick[i][j], dick[i][j+1]])
for i in range(n):
fout.write(str(ans[i][0]+1) + ' ' + str(ans[i][1]+1) + '\n')
if __name__ == "__main__":
main()
|
Title: Cards with Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has got 2*n* cards, each card contains some integer. The numbers on the cards can be the same. Let's index all cards by consecutive integers from 1 to 2*n*. We'll denote the number that is written on a card with number *i*, as *a**i*. In order to play one entertaining game with his friends, Petya needs to split the cards into pairs so that each pair had equal numbers on the cards. Help Petya do that.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains the sequence of 2*n* positive integers *a*1,<=*a*2,<=...,<=*a*2*n* (1<=≤<=*a**i*<=≤<=5000) — the numbers that are written on the cards. The numbers on the line are separated by single spaces.
Output Specification:
If it is impossible to divide the cards into pairs so that cards in each pair had the same numbers, print on a single line integer -1. But if the required partition exists, then print *n* pairs of integers, a pair per line — the indices of the cards that form the pairs.
Separate the numbers on the lines by spaces. You can print the pairs and the numbers in the pairs in any order. If there are multiple solutions, print any of them.
Demo Input:
['3\n20 30 10 30 20 10\n', '1\n1 2\n']
Demo Output:
['4 2\n1 5\n6 3\n', '-1']
Note:
none
|
```python
import string
from collections import deque
alph = string.ascii_lowercase
def main():
with open('input.txt', 'r') as fin:
f = fin.readlines()
n = int(f[0])
a = list(map(int, f[1].split()))
d = {i:0 for i in a}
for i in a:
d[i] += 1
flag = 1
for i in list(d.keys()):
if d[i] % 2:
flag = 0
break
with open('output.txt', 'w') as fout:
if not flag:
fout.write(str(-1))
else:
dick = {i:[] for i in a}
for i in range(2*n):
dick[a[i]].append(i)
ans = []
for i in list(dick.keys()):
for j in range(0, len(dick[i]), 2):
ans.append([dick[i][j], dick[i][j+1]])
for i in range(n):
fout.write(str(ans[i][0]+1) + ' ' + str(ans[i][1]+1) + '\n')
if __name__ == "__main__":
main()
```
| 3
|
|
599
|
A
|
Patrick and Shopping
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
|
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
|
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
|
[
"10 20 30\n",
"1 1 5\n"
] |
[
"60\n",
"4\n"
] |
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
| 500
|
[
{
"input": "10 20 30",
"output": "60"
},
{
"input": "1 1 5",
"output": "4"
},
{
"input": "100 33 34",
"output": "134"
},
{
"input": "777 777 777",
"output": "2331"
},
{
"input": "2 2 8",
"output": "8"
},
{
"input": "12 34 56",
"output": "92"
},
{
"input": "789 101112 131415",
"output": "203802"
},
{
"input": "27485716 99999999 35182",
"output": "55041796"
},
{
"input": "1 293548 5",
"output": "12"
},
{
"input": "12059 259855 5874875",
"output": "543828"
},
{
"input": "46981 105809 585858",
"output": "305580"
},
{
"input": "9889 1221 2442",
"output": "7326"
},
{
"input": "100500 200600 300700",
"output": "601800"
},
{
"input": "318476 318476 318476",
"output": "955428"
},
{
"input": "23985 3353 75633",
"output": "54676"
},
{
"input": "120 1298 2222",
"output": "2836"
},
{
"input": "98437 23487 666672",
"output": "243848"
},
{
"input": "100000000 100000000 100000000",
"output": "300000000"
},
{
"input": "2 5 2",
"output": "8"
},
{
"input": "1 1000 1",
"output": "4"
},
{
"input": "1 100000000 1",
"output": "4"
}
] | 1,587,792,836
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 93
| 0
|
x, y, z = map(int, input().split())
tracking = (x + y) * 2
if tracking < z:
print(tracking)
else:
print(x+y+z)
|
Title: Patrick and Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
Input Specification:
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
Output Specification:
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
Demo Input:
['10 20 30\n', '1 1 5\n']
Demo Output:
['60\n', '4\n']
Note:
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
|
```python
x, y, z = map(int, input().split())
tracking = (x + y) * 2
if tracking < z:
print(tracking)
else:
print(x+y+z)
```
| 0
|
|
999
|
C
|
Alphabetic Removals
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
You are given a string $s$ consisting of $n$ lowercase Latin letters. Polycarp wants to remove exactly $k$ characters ($k \le n$) from the string $s$. Polycarp uses the following algorithm $k$ times:
- if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; - if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; - ... - remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly $k$ times, thus removing exactly $k$ characters.
Help Polycarp find the resulting string.
|
The first line of input contains two integers $n$ and $k$ ($1 \le k \le n \le 4 \cdot 10^5$) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string $s$ consisting of $n$ lowercase Latin letters.
|
Print the string that will be obtained from $s$ after Polycarp removes exactly $k$ letters using the above algorithm $k$ times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
|
[
"15 3\ncccaabababaccbc\n",
"15 9\ncccaabababaccbc\n",
"1 1\nu\n"
] |
[
"cccbbabaccbc\n",
"cccccc\n",
""
] |
none
| 0
|
[
{
"input": "15 3\ncccaabababaccbc",
"output": "cccbbabaccbc"
},
{
"input": "15 9\ncccaabababaccbc",
"output": "cccccc"
},
{
"input": "5 2\nzyzyx",
"output": "zzy"
},
{
"input": "4 3\nhack",
"output": "k"
},
{
"input": "4 3\nzzzz",
"output": "z"
},
{
"input": "6 5\naaccdd",
"output": "d"
},
{
"input": "2 1\nzz",
"output": "z"
},
{
"input": "14 5\nhxehmvkybeklnj",
"output": "xmvkyklnj"
}
] | 1,682,240,564
| 2,264
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 4
| 2,000
| 56,934,400
|
n,k=map(int,input().split())
a=[i for i in input()]
d={}
for i in a:
d[i]=d.get(i,0)+1
p=list(d.keys())
p.sort()
z="a"
flag=True
for i in p:
if k>=d[i]:
k-=d[i]
else:
z=i
flag=False
break
t=""
if flag:
z=chr(ord(p[-1])+1)
for i in a:
if i>z:
t=t+i
elif i==z:
if k>0:
k-=1
else:
t=t+i
print(t)
|
Title: Alphabetic Removals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string $s$ consisting of $n$ lowercase Latin letters. Polycarp wants to remove exactly $k$ characters ($k \le n$) from the string $s$. Polycarp uses the following algorithm $k$ times:
- if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; - if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item; - ... - remove the leftmost occurrence of the letter 'z' and stop the algorithm.
This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly $k$ times, thus removing exactly $k$ characters.
Help Polycarp find the resulting string.
Input Specification:
The first line of input contains two integers $n$ and $k$ ($1 \le k \le n \le 4 \cdot 10^5$) — the length of the string and the number of letters Polycarp will remove.
The second line contains the string $s$ consisting of $n$ lowercase Latin letters.
Output Specification:
Print the string that will be obtained from $s$ after Polycarp removes exactly $k$ letters using the above algorithm $k$ times.
If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).
Demo Input:
['15 3\ncccaabababaccbc\n', '15 9\ncccaabababaccbc\n', '1 1\nu\n']
Demo Output:
['cccbbabaccbc\n', 'cccccc\n', '']
Note:
none
|
```python
n,k=map(int,input().split())
a=[i for i in input()]
d={}
for i in a:
d[i]=d.get(i,0)+1
p=list(d.keys())
p.sort()
z="a"
flag=True
for i in p:
if k>=d[i]:
k-=d[i]
else:
z=i
flag=False
break
t=""
if flag:
z=chr(ord(p[-1])+1)
for i in a:
if i>z:
t=t+i
elif i==z:
if k>0:
k-=1
else:
t=t+i
print(t)
```
| 0
|
|
621
|
A
|
Wet Shark and Odd and Even
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.
|
The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.
|
Print the maximum possible even sum that can be obtained if we use some of the given integers.
|
[
"3\n1 2 3\n",
"5\n999999999 999999999 999999999 999999999 999999999\n"
] |
[
"6",
"3999999996"
] |
In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999.
| 500
|
[
{
"input": "3\n1 2 3",
"output": "6"
},
{
"input": "5\n999999999 999999999 999999999 999999999 999999999",
"output": "3999999996"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "15\n39 52 88 78 46 95 84 98 55 3 68 42 6 18 98",
"output": "870"
},
{
"input": "15\n59 96 34 48 8 72 67 90 15 85 7 90 97 47 25",
"output": "840"
},
{
"input": "15\n87 37 91 29 58 45 51 74 70 71 47 38 91 89 44",
"output": "922"
},
{
"input": "15\n11 81 49 7 11 14 30 67 29 50 90 81 77 18 59",
"output": "674"
},
{
"input": "15\n39 21 95 89 73 90 9 55 85 32 30 21 68 59 82",
"output": "848"
},
{
"input": "15\n59 70 48 54 26 67 84 39 40 18 77 69 70 88 93",
"output": "902"
},
{
"input": "15\n87 22 98 32 88 36 72 31 100 97 17 16 60 22 20",
"output": "798"
},
{
"input": "15\n15 63 51 13 37 9 43 19 55 79 57 60 50 59 31",
"output": "632"
},
{
"input": "1\n4",
"output": "4"
},
{
"input": "2\n1 4",
"output": "4"
},
{
"input": "3\n1 2 4",
"output": "6"
},
{
"input": "2\n9 3",
"output": "12"
},
{
"input": "2\n1000000000 1001",
"output": "1000000000"
},
{
"input": "3\n1 8 4",
"output": "12"
},
{
"input": "3\n7 4 4",
"output": "8"
},
{
"input": "5\n2 3 4 5 3",
"output": "14"
},
{
"input": "2\n4 5",
"output": "4"
},
{
"input": "3\n2 4 5",
"output": "6"
},
{
"input": "3\n2 2 3",
"output": "4"
},
{
"input": "2\n2 3",
"output": "2"
},
{
"input": "4\n2 3 7 7",
"output": "16"
},
{
"input": "2\n999999999 2",
"output": "2"
},
{
"input": "2\n2 5",
"output": "2"
},
{
"input": "3\n5 3 1",
"output": "8"
},
{
"input": "4\n3 2 5 7",
"output": "14"
}
] | 1,617,174,720
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 8
| 139
| 7,168,000
|
n=int(input())
m=input().split()
sum1=0
weak=[]
strong=[]
for i in range(n):
m[i]=int(m[i])
for i in m:
if i%2==0:
weak.append(i)
if i%2==1:
strong.append(i)
sum1+=sum(weak)
strong.sort()
if len(strong)%2==0:
sum1+=sum(strong)
if len(strong)%2==1:
sum1+=sum(strong[0:len(strong)-1])
print(sum1)
|
Title: Wet Shark and Odd and Even
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.
Input Specification:
The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.
Output Specification:
Print the maximum possible even sum that can be obtained if we use some of the given integers.
Demo Input:
['3\n1 2 3\n', '5\n999999999 999999999 999999999 999999999 999999999\n']
Demo Output:
['6', '3999999996']
Note:
In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999.
|
```python
n=int(input())
m=input().split()
sum1=0
weak=[]
strong=[]
for i in range(n):
m[i]=int(m[i])
for i in m:
if i%2==0:
weak.append(i)
if i%2==1:
strong.append(i)
sum1+=sum(weak)
strong.sort()
if len(strong)%2==0:
sum1+=sum(strong)
if len(strong)%2==1:
sum1+=sum(strong[0:len(strong)-1])
print(sum1)
```
| 0
|
|
233
|
A
|
Perfect Permutation
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*.
|
A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size.
|
If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces.
|
[
"1\n",
"2\n",
"4\n"
] |
[
"-1\n",
"2 1 \n",
"2 1 4 3 \n"
] |
none
| 500
|
[
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "2 1 "
},
{
"input": "4",
"output": "2 1 4 3 "
},
{
"input": "3",
"output": "-1"
},
{
"input": "5",
"output": "-1"
},
{
"input": "6",
"output": "2 1 4 3 6 5 "
},
{
"input": "7",
"output": "-1"
},
{
"input": "20",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 "
},
{
"input": "8",
"output": "2 1 4 3 6 5 8 7 "
},
{
"input": "9",
"output": "-1"
},
{
"input": "10",
"output": "2 1 4 3 6 5 8 7 10 9 "
},
{
"input": "11",
"output": "-1"
},
{
"input": "21",
"output": "-1"
},
{
"input": "50",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 "
},
{
"input": "51",
"output": "-1"
},
{
"input": "52",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 "
},
{
"input": "84",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 "
},
{
"input": "86",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 "
},
{
"input": "100",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 100 99 "
},
{
"input": "98",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 98 97 "
},
{
"input": "96",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 50 49 52 51 54 53 56 55 58 57 60 59 62 61 64 63 66 65 68 67 70 69 72 71 74 73 76 75 78 77 80 79 82 81 84 83 86 85 88 87 90 89 92 91 94 93 96 95 "
},
{
"input": "33",
"output": "-1"
},
{
"input": "34",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 "
},
{
"input": "36",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 "
},
{
"input": "38",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 "
},
{
"input": "40",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 "
},
{
"input": "42",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 "
},
{
"input": "44",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 "
},
{
"input": "46",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 "
},
{
"input": "48",
"output": "2 1 4 3 6 5 8 7 10 9 12 11 14 13 16 15 18 17 20 19 22 21 24 23 26 25 28 27 30 29 32 31 34 33 36 35 38 37 40 39 42 41 44 43 46 45 48 47 "
}
] | 1,634,081,550
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 154
| 6,758,400
|
# import inbuilt standard input output
from sys import stdin, stdout
def main():
len_ = int(stdin.readline())
if len_%2 == 0:
str_ = []
for j in range(0, len_, 2):
str_.append(str(j + 2))
str_.append(str(j + 1))
string_ = " ".join(str_)
stdout.write(string_)
else:
stdout.write("-1")
if __name__ == "__main__":
main()
|
Title: Perfect Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*.
Input Specification:
A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size.
Output Specification:
If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces.
Demo Input:
['1\n', '2\n', '4\n']
Demo Output:
['-1\n', '2 1 \n', '2 1 4 3 \n']
Note:
none
|
```python
# import inbuilt standard input output
from sys import stdin, stdout
def main():
len_ = int(stdin.readline())
if len_%2 == 0:
str_ = []
for j in range(0, len_, 2):
str_.append(str(j + 2))
str_.append(str(j + 1))
string_ = " ".join(str_)
stdout.write(string_)
else:
stdout.write("-1")
if __name__ == "__main__":
main()
```
| 3
|
|
92
|
B
|
Binary Number
|
PROGRAMMING
| 1,300
|
[
"greedy"
] |
B. Binary Number
|
1
|
256
|
Little walrus Fangy loves math very much. That's why when he is bored he plays with a number performing some operations.
Fangy takes some positive integer *x* and wants to get a number one from it. While *x* is not equal to 1, Fangy repeats the following action: if *x* is odd, then he adds 1 to it, otherwise he divides *x* by 2. Fangy knows that for any positive integer number the process ends in finite time.
How many actions should Fangy perform to get a number one from number *x*?
|
The first line contains a positive integer *x* in a binary system. It is guaranteed that the first digit of *x* is different from a zero and the number of its digits does not exceed 106.
|
Print the required number of actions.
|
[
"1\n",
"1001001\n",
"101110\n"
] |
[
"0\n",
"12\n",
"8\n"
] |
Let's consider the third sample. Number 101110 is even, which means that we should divide it by 2. After the dividing Fangy gets an odd number 10111 and adds one to it. Number 11000 can be divided by 2 three times in a row and get number 11. All that's left is to increase the number by one (we get 100), and then divide it by 2 two times in a row. As a result, we get 1.
| 1,000
|
[
{
"input": "1",
"output": "0"
},
{
"input": "1001001",
"output": "12"
},
{
"input": "101110",
"output": "8"
},
{
"input": "11",
"output": "3"
},
{
"input": "11110001101",
"output": "16"
},
{
"input": "1010101001001111000111110011111000010101011111101010",
"output": "74"
},
{
"input": "1100000010010100111011100011110101111",
"output": "55"
},
{
"input": "11000111111110110110100110110101111100010100110110010",
"output": "74"
},
{
"input": "11100000110100011110101001101111100000011001111000011110000000111110111",
"output": "106"
},
{
"input": "1000101100110000000001111010110000000010001001111110011011000011101011001001010010111",
"output": "133"
},
{
"input": "1000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "72"
},
{
"input": "10000100000",
"output": "16"
}
] | 1,605,937,841
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 10
| 186
| 8,601,600
|
s = input()[::-1]
pr = 0
ans = 0
if s == '1':
print(0)
exit()
for i in s:
if i == '0' and pr == 0:
ans += 1
elif i == '0' and pr == 1:
ans += 2
elif i == '1' and pr == 1:
ans += 1
elif i == '1' and pr == 0:
ans += 2
pr = 1
print(ans)
|
Title: Binary Number
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Little walrus Fangy loves math very much. That's why when he is bored he plays with a number performing some operations.
Fangy takes some positive integer *x* and wants to get a number one from it. While *x* is not equal to 1, Fangy repeats the following action: if *x* is odd, then he adds 1 to it, otherwise he divides *x* by 2. Fangy knows that for any positive integer number the process ends in finite time.
How many actions should Fangy perform to get a number one from number *x*?
Input Specification:
The first line contains a positive integer *x* in a binary system. It is guaranteed that the first digit of *x* is different from a zero and the number of its digits does not exceed 106.
Output Specification:
Print the required number of actions.
Demo Input:
['1\n', '1001001\n', '101110\n']
Demo Output:
['0\n', '12\n', '8\n']
Note:
Let's consider the third sample. Number 101110 is even, which means that we should divide it by 2. After the dividing Fangy gets an odd number 10111 and adds one to it. Number 11000 can be divided by 2 three times in a row and get number 11. All that's left is to increase the number by one (we get 100), and then divide it by 2 two times in a row. As a result, we get 1.
|
```python
s = input()[::-1]
pr = 0
ans = 0
if s == '1':
print(0)
exit()
for i in s:
if i == '0' and pr == 0:
ans += 1
elif i == '0' and pr == 1:
ans += 2
elif i == '1' and pr == 1:
ans += 1
elif i == '1' and pr == 0:
ans += 2
pr = 1
print(ans)
```
| 0
|
757
|
B
|
Bash's Big Day
|
PROGRAMMING
| 1,400
|
[
"greedy",
"math",
"number theory"
] | null | null |
Bash has set out on a journey to become the greatest Pokemon master. To get his first Pokemon, he went to Professor Zulu's Lab. Since Bash is Professor Zulu's favourite student, Zulu allows him to take as many Pokemon from his lab as he pleases.
But Zulu warns him that a group of *k*<=><=1 Pokemon with strengths {*s*1,<=*s*2,<=*s*3,<=...,<=*s**k*} tend to fight among each other if *gcd*(*s*1,<=*s*2,<=*s*3,<=...,<=*s**k*)<==<=1 (see notes for *gcd* definition).
Bash, being smart, does not want his Pokemon to fight among each other. However, he also wants to maximize the number of Pokemon he takes from the lab. Can you help Bash find out the maximum number of Pokemon he can take?
Note: A Pokemon cannot fight with itself.
|
The input consists of two lines.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105), the number of Pokemon in the lab.
The next line contains *n* space separated integers, where the *i*-th of them denotes *s**i* (1<=≤<=*s**i*<=≤<=105), the strength of the *i*-th Pokemon.
|
Print single integer — the maximum number of Pokemons Bash can take.
|
[
"3\n2 3 4\n",
"5\n2 3 4 6 7\n"
] |
[
"2\n",
"3\n"
] |
*gcd* (greatest common divisor) of positive integers set {*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>} is the maximum positive integer that divides all the integers {*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>}.
In the first sample, we can take Pokemons with strengths {2, 4} since *gcd*(2, 4) = 2.
In the second sample, we can take Pokemons with strengths {2, 4, 6}, and there is no larger group with *gcd* ≠ 1.
| 1,000
|
[
{
"input": "3\n2 3 4",
"output": "2"
},
{
"input": "5\n2 3 4 6 7",
"output": "3"
},
{
"input": "3\n5 6 4",
"output": "2"
},
{
"input": "8\n41 74 4 27 85 39 100 36",
"output": "4"
},
{
"input": "6\n89 20 86 81 62 23",
"output": "3"
},
{
"input": "71\n23 84 98 8 14 4 42 56 83 87 28 22 32 50 5 96 90 1 59 74 77 88 71 38 62 36 85 97 99 6 81 20 49 57 66 9 45 41 29 68 35 19 27 76 78 72 55 25 46 48 26 53 39 31 94 34 63 37 64 16 79 24 82 17 12 3 89 61 80 30 10",
"output": "38"
},
{
"input": "95\n72 38 75 62 87 30 11 65 35 16 73 23 18 48 19 4 22 42 14 60 49 83 59 15 51 27 80 97 37 100 64 81 54 71 52 20 5 98 78 86 26 55 25 57 36 3 8 74 82 21 29 1 76 2 79 61 39 9 89 77 70 63 56 28 92 53 31 45 93 47 67 99 58 12 84 44 32 34 69 40 13 7 66 68 17 85 6 90 33 91 94 24 46 10 50",
"output": "48"
},
{
"input": "44\n39706 21317 26213 55086 10799 31825 29024 6565 96535 11412 14642 91901 41932 24538 81351 53861 63403 34199 82286 32594 29684 42753 16857 73821 71085 36306 70080 11233 21023 8551 85406 95390 92375 52675 77938 46265 74855 5229 5856 66713 65730 24525 84078 20684",
"output": "19"
},
{
"input": "35\n45633 86983 46174 48399 33926 51395 76300 6387 48852 82808 28694 79864 4482 35982 21956 76522 19656 74518 28480 71481 25700 46815 14170 95705 8535 96993 29029 8898 97637 62710 14615 22864 69849 27068 68557",
"output": "20"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n10 7 9 8 3 3 10 7 3 3",
"output": "5"
},
{
"input": "9\n10 10 6 10 9 1 8 3 5",
"output": "5"
},
{
"input": "7\n9 4 2 3 3 9 8",
"output": "4"
},
{
"input": "1\n4",
"output": "1"
},
{
"input": "6\n1623 45906 37856 34727 27156 12598",
"output": "4"
},
{
"input": "30\n83172 59163 67334 83980 5932 8773 77649 41428 62789 28159 17183 10199 41496 59500 14614 10468 54886 64679 42382 57021 50499 95643 77239 61434 16181 30505 59152 55972 18265 70566",
"output": "15"
},
{
"input": "23\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 22 16 2 13 16",
"output": "22"
},
{
"input": "46\n12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 12553 15 1 18 28 20 6 31 16 5 23 21 38 3 11 18 11 3 25 33",
"output": "27"
},
{
"input": "43\n8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8831 8 23 40 33 11 5 21 16 19 15 41 30 28 31 5 32 16 5 38 11 21 34",
"output": "21"
},
{
"input": "25\n58427 26687 48857 46477 7039 25423 58757 48119 38113 40637 22391 48337 4157 10597 8167 19031 64613 70913 69313 18047 17159 77491 13499 70949 24107",
"output": "1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "2\n3 6",
"output": "2"
},
{
"input": "5\n1 1 1 1 1",
"output": "1"
},
{
"input": "5\n3 3 3 3 3",
"output": "5"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "2\n541 541",
"output": "2"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n99989 99989",
"output": "2"
},
{
"input": "3\n3 9 27",
"output": "3"
},
{
"input": "2\n1009 1009",
"output": "2"
},
{
"input": "4\n1 1 1 1",
"output": "1"
},
{
"input": "6\n2 10 20 5 15 25",
"output": "5"
},
{
"input": "3\n3 3 6",
"output": "3"
},
{
"input": "3\n457 457 457",
"output": "3"
},
{
"input": "2\n34 17",
"output": "2"
},
{
"input": "3\n12 24 3",
"output": "3"
},
{
"input": "10\n99991 99991 99991 99991 99991 99991 99991 99991 99991 99991",
"output": "10"
},
{
"input": "2\n1009 2018",
"output": "2"
},
{
"input": "3\n3 3 3",
"output": "3"
},
{
"input": "7\n6 9 12 15 21 27 33",
"output": "7"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "2\n557 557",
"output": "2"
},
{
"input": "3\n1 1 2",
"output": "1"
},
{
"input": "5\n2 2 101 101 101",
"output": "3"
},
{
"input": "2\n122 3721",
"output": "2"
},
{
"input": "2\n49201 98402",
"output": "2"
},
{
"input": "2\n88258 44129",
"output": "2"
},
{
"input": "2\n7919 47514",
"output": "2"
},
{
"input": "5\n1 2 1 1 1",
"output": "1"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "5\n1 1 1 1 2",
"output": "1"
},
{
"input": "3\n10007 20014 30021",
"output": "3"
},
{
"input": "4\n10 10 5 5",
"output": "4"
},
{
"input": "2\n6 9",
"output": "2"
},
{
"input": "3\n6 12 9",
"output": "3"
},
{
"input": "2\n14 7",
"output": "2"
},
{
"input": "2\n199 199",
"output": "2"
},
{
"input": "3\n49999 49999 99998",
"output": "3"
},
{
"input": "5\n1009 1009 1009 1009 1009",
"output": "5"
},
{
"input": "3\n17 17 17",
"output": "3"
},
{
"input": "2\n503 1509",
"output": "2"
},
{
"input": "3\n49999 99998 99998",
"output": "3"
},
{
"input": "2\n28657 28657",
"output": "2"
},
{
"input": "3\n121 22 33",
"output": "3"
},
{
"input": "2\n10079 20158",
"output": "2"
},
{
"input": "2\n6 15",
"output": "2"
},
{
"input": "4\n6 6 6 3",
"output": "4"
},
{
"input": "2\n991 1982",
"output": "2"
},
{
"input": "3\n20362 30543 50905",
"output": "3"
},
{
"input": "2\n443 886",
"output": "2"
},
{
"input": "4\n3 7 7 21",
"output": "3"
},
{
"input": "3\n3 6 9",
"output": "3"
},
{
"input": "2\n217 31",
"output": "2"
},
{
"input": "6\n10 10 10 5 5 5",
"output": "6"
},
{
"input": "4\n49441 25225 9081 4036",
"output": "4"
},
{
"input": "2\n4 9",
"output": "1"
},
{
"input": "2\n18 27",
"output": "2"
},
{
"input": "2\n13 26",
"output": "2"
},
{
"input": "4\n6 6 6 15",
"output": "4"
}
] | 1,615,566,308
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 134
| 467
| 10,854,400
|
from collections import defaultdict
n = int(input())
a = list(map(int,input().split()))
factors = defaultdict(int)
for i in range(n):
root = int(a[i]**0.5)
for j in range(2,root+1):
if a[i]%j==0:
factors[j]+=1
while a[i]%j==0:
a[i]//=j
if a[i]>1:
factors[a[i]]+=1
ans = 1
for i in factors:
ans = max(ans,factors[i])
print(ans)
|
Title: Bash's Big Day
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bash has set out on a journey to become the greatest Pokemon master. To get his first Pokemon, he went to Professor Zulu's Lab. Since Bash is Professor Zulu's favourite student, Zulu allows him to take as many Pokemon from his lab as he pleases.
But Zulu warns him that a group of *k*<=><=1 Pokemon with strengths {*s*1,<=*s*2,<=*s*3,<=...,<=*s**k*} tend to fight among each other if *gcd*(*s*1,<=*s*2,<=*s*3,<=...,<=*s**k*)<==<=1 (see notes for *gcd* definition).
Bash, being smart, does not want his Pokemon to fight among each other. However, he also wants to maximize the number of Pokemon he takes from the lab. Can you help Bash find out the maximum number of Pokemon he can take?
Note: A Pokemon cannot fight with itself.
Input Specification:
The input consists of two lines.
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105), the number of Pokemon in the lab.
The next line contains *n* space separated integers, where the *i*-th of them denotes *s**i* (1<=≤<=*s**i*<=≤<=105), the strength of the *i*-th Pokemon.
Output Specification:
Print single integer — the maximum number of Pokemons Bash can take.
Demo Input:
['3\n2 3 4\n', '5\n2 3 4 6 7\n']
Demo Output:
['2\n', '3\n']
Note:
*gcd* (greatest common divisor) of positive integers set {*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>} is the maximum positive integer that divides all the integers {*a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>}.
In the first sample, we can take Pokemons with strengths {2, 4} since *gcd*(2, 4) = 2.
In the second sample, we can take Pokemons with strengths {2, 4, 6}, and there is no larger group with *gcd* ≠ 1.
|
```python
from collections import defaultdict
n = int(input())
a = list(map(int,input().split()))
factors = defaultdict(int)
for i in range(n):
root = int(a[i]**0.5)
for j in range(2,root+1):
if a[i]%j==0:
factors[j]+=1
while a[i]%j==0:
a[i]//=j
if a[i]>1:
factors[a[i]]+=1
ans = 1
for i in factors:
ans = max(ans,factors[i])
print(ans)
```
| 3
|
|
366
|
B
|
Dima and To-do List
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
You helped Dima to have a great weekend, but it's time to work. Naturally, Dima, as all other men who have girlfriends, does everything wrong.
Inna and Dima are now in one room. Inna tells Dima off for everything he does in her presence. After Inna tells him off for something, she goes to another room, walks there in circles muttering about how useless her sweetheart is. During that time Dima has time to peacefully complete *k*<=-<=1 tasks. Then Inna returns and tells Dima off for the next task he does in her presence and goes to another room again. It continues until Dima is through with his tasks.
Overall, Dima has *n* tasks to do, each task has a unique number from 1 to *n*. Dima loves order, so he does tasks consecutively, starting from some task. For example, if Dima has 6 tasks to do in total, then, if he starts from the 5-th task, the order is like that: first Dima does the 5-th task, then the 6-th one, then the 1-st one, then the 2-nd one, then the 3-rd one, then the 4-th one.
Inna tells Dima off (only lovingly and appropriately!) so often and systematically that he's very well learned the power with which she tells him off for each task. Help Dima choose the first task so that in total he gets told off with as little power as possible.
|
The first line of the input contains two integers *n*,<=*k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=103), where *a**i* is the power Inna tells Dima off with if she is present in the room while he is doing the *i*-th task.
It is guaranteed that *n* is divisible by *k*.
|
In a single line print the number of the task Dima should start with to get told off with as little power as possible. If there are multiple solutions, print the one with the minimum number of the first task to do.
|
[
"6 2\n3 2 1 6 5 4\n",
"10 5\n1 3 5 7 9 9 4 1 8 5\n"
] |
[
"1\n",
"3\n"
] |
Explanation of the first example.
If Dima starts from the first task, Inna tells him off with power 3, then Dima can do one more task (as *k* = 2), then Inna tells him off for the third task with power 1, then she tells him off for the fifth task with power 5. Thus, Dima gets told off with total power 3 + 1 + 5 = 9. If Dima started from the second task, for example, then Inna would tell him off for tasks 2, 4 and 6 with power 2 + 6 + 4 = 12.
Explanation of the second example.
In the second example *k* = 5, thus, Dima manages to complete 4 tasks in-between the telling off sessions. Thus, Inna tells Dima off for tasks number 1 and 6 (if he starts from 1 or 6), 2 and 7 (if he starts from 2 or 7) and so on. The optimal answer is to start from task 3 or 8, 3 has a smaller number, so the answer is 3.
| 1,000
|
[
{
"input": "6 2\n3 2 1 6 5 4",
"output": "1"
},
{
"input": "10 5\n1 3 5 7 9 9 4 1 8 5",
"output": "3"
},
{
"input": "20 4\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "10 10\n8 4 5 7 6 9 2 2 3 5",
"output": "7"
},
{
"input": "50 10\n1 2 3 4 5 6 7 8 9 10 10 1 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "4 2\n2 1 1 3",
"output": "1"
},
{
"input": "15 5\n5 5 5 5 5 1 2 3 4 5 1 2 3 4 5",
"output": "1"
},
{
"input": "20 10\n3 3 3 3 3 3 3 3 2 3 3 3 3 3 3 3 3 3 6 4",
"output": "1"
}
] | 1,633,281,724
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 5
| 1,000
| 6,963,200
|
n, k = map(int, input().split())
tasks = list(map(int, input().split()))
min_sum = 10 ** 7
min_start = 1
for i in range(n):
temp = 0
for j in range(i, i+n, k):
temp += tasks[j % n]
if temp < min_sum:
min_sum = temp
min_start = i + 1
print(min_start)
|
Title: Dima and To-do List
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You helped Dima to have a great weekend, but it's time to work. Naturally, Dima, as all other men who have girlfriends, does everything wrong.
Inna and Dima are now in one room. Inna tells Dima off for everything he does in her presence. After Inna tells him off for something, she goes to another room, walks there in circles muttering about how useless her sweetheart is. During that time Dima has time to peacefully complete *k*<=-<=1 tasks. Then Inna returns and tells Dima off for the next task he does in her presence and goes to another room again. It continues until Dima is through with his tasks.
Overall, Dima has *n* tasks to do, each task has a unique number from 1 to *n*. Dima loves order, so he does tasks consecutively, starting from some task. For example, if Dima has 6 tasks to do in total, then, if he starts from the 5-th task, the order is like that: first Dima does the 5-th task, then the 6-th one, then the 1-st one, then the 2-nd one, then the 3-rd one, then the 4-th one.
Inna tells Dima off (only lovingly and appropriately!) so often and systematically that he's very well learned the power with which she tells him off for each task. Help Dima choose the first task so that in total he gets told off with as little power as possible.
Input Specification:
The first line of the input contains two integers *n*,<=*k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=103), where *a**i* is the power Inna tells Dima off with if she is present in the room while he is doing the *i*-th task.
It is guaranteed that *n* is divisible by *k*.
Output Specification:
In a single line print the number of the task Dima should start with to get told off with as little power as possible. If there are multiple solutions, print the one with the minimum number of the first task to do.
Demo Input:
['6 2\n3 2 1 6 5 4\n', '10 5\n1 3 5 7 9 9 4 1 8 5\n']
Demo Output:
['1\n', '3\n']
Note:
Explanation of the first example.
If Dima starts from the first task, Inna tells him off with power 3, then Dima can do one more task (as *k* = 2), then Inna tells him off for the third task with power 1, then she tells him off for the fifth task with power 5. Thus, Dima gets told off with total power 3 + 1 + 5 = 9. If Dima started from the second task, for example, then Inna would tell him off for tasks 2, 4 and 6 with power 2 + 6 + 4 = 12.
Explanation of the second example.
In the second example *k* = 5, thus, Dima manages to complete 4 tasks in-between the telling off sessions. Thus, Inna tells Dima off for tasks number 1 and 6 (if he starts from 1 or 6), 2 and 7 (if he starts from 2 or 7) and so on. The optimal answer is to start from task 3 or 8, 3 has a smaller number, so the answer is 3.
|
```python
n, k = map(int, input().split())
tasks = list(map(int, input().split()))
min_sum = 10 ** 7
min_start = 1
for i in range(n):
temp = 0
for j in range(i, i+n, k):
temp += tasks[j % n]
if temp < min_sum:
min_sum = temp
min_start = i + 1
print(min_start)
```
| 0
|
|
465
|
A
|
inc ARG
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Sergey is testing a next-generation processor. Instead of bytes the processor works with memory cells consisting of *n* bits. These bits are numbered from 1 to *n*. An integer is stored in the cell in the following way: the least significant bit is stored in the first bit of the cell, the next significant bit is stored in the second bit, and so on; the most significant bit is stored in the *n*-th bit.
Now Sergey wants to test the following instruction: "add 1 to the value of the cell". As a result of the instruction, the integer that is written in the cell must be increased by one; if some of the most significant bits of the resulting number do not fit into the cell, they must be discarded.
Sergey wrote certain values of the bits in the cell and is going to add one to its value. How many bits of the cell will change after the operation?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of bits in the cell.
The second line contains a string consisting of *n* characters — the initial state of the cell. The first character denotes the state of the first bit of the cell. The second character denotes the second least significant bit and so on. The last character denotes the state of the most significant bit.
|
Print a single integer — the number of bits in the cell which change their state after we add 1 to the cell.
|
[
"4\n1100\n",
"4\n1111\n"
] |
[
"3\n",
"4\n"
] |
In the first sample the cell ends up with value 0010, in the second sample — with 0000.
| 500
|
[
{
"input": "4\n1100",
"output": "3"
},
{
"input": "4\n1111",
"output": "4"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n00",
"output": "1"
},
{
"input": "2\n01",
"output": "1"
},
{
"input": "2\n10",
"output": "2"
},
{
"input": "2\n11",
"output": "2"
},
{
"input": "10\n0000000000",
"output": "1"
},
{
"input": "20\n11111111110110001100",
"output": "11"
},
{
"input": "50\n01011110100010000001010000100001001101001101101011",
"output": "1"
},
{
"input": "60\n111111111101111111111111111111111111111111111111111111111111",
"output": "11"
},
{
"input": "60\n111111111111111111111111111111111111111111111111111111111111",
"output": "60"
},
{
"input": "66\n111111010010011001110011000111000100011110011001111110011111111101",
"output": "7"
},
{
"input": "90\n000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "91\n1011111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "2"
},
{
"input": "92\n11111111101011111011111111111111111111111011101111111111011111111111111101111111101011111101",
"output": "10"
},
{
"input": "100\n0001011110100011001100100010111001000001111101101001001001001011110100101101010000000110100101110010",
"output": "1"
},
{
"input": "100\n0111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "1"
},
{
"input": "100\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "100\n0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "100"
},
{
"input": "100\n1011001110001000011111110011000100001110010110111101110110011011011000010100110001111100000010110010",
"output": "2"
},
{
"input": "100\n1101111011001111111111110011110111101110111111111111111111111111111111011111111111110111111111111111",
"output": "3"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "100"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110",
"output": "100"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111101",
"output": "99"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111100000",
"output": "96"
},
{
"input": "100\n1111111111111111111111111111111110011000110110001010101100100100001000010010010010000100110000010101",
"output": "34"
}
] | 1,633,950,364
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 61
| 6,963,200
|
n = int(input())
s = input()
change = 0
in_hand = 1
for i in range(n):
if (int(s[i]) + in_hand) == 0:
pass
elif (int(s[i]) + in_hand) == 1 and s[i] == "1":
pass
elif (int(s[i]) + in_hand) == 1 and in_hand == 1:
in_hand = 0
change += 1
else:
change += 1
print(change)
|
Title: inc ARG
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sergey is testing a next-generation processor. Instead of bytes the processor works with memory cells consisting of *n* bits. These bits are numbered from 1 to *n*. An integer is stored in the cell in the following way: the least significant bit is stored in the first bit of the cell, the next significant bit is stored in the second bit, and so on; the most significant bit is stored in the *n*-th bit.
Now Sergey wants to test the following instruction: "add 1 to the value of the cell". As a result of the instruction, the integer that is written in the cell must be increased by one; if some of the most significant bits of the resulting number do not fit into the cell, they must be discarded.
Sergey wrote certain values of the bits in the cell and is going to add one to its value. How many bits of the cell will change after the operation?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of bits in the cell.
The second line contains a string consisting of *n* characters — the initial state of the cell. The first character denotes the state of the first bit of the cell. The second character denotes the second least significant bit and so on. The last character denotes the state of the most significant bit.
Output Specification:
Print a single integer — the number of bits in the cell which change their state after we add 1 to the cell.
Demo Input:
['4\n1100\n', '4\n1111\n']
Demo Output:
['3\n', '4\n']
Note:
In the first sample the cell ends up with value 0010, in the second sample — with 0000.
|
```python
n = int(input())
s = input()
change = 0
in_hand = 1
for i in range(n):
if (int(s[i]) + in_hand) == 0:
pass
elif (int(s[i]) + in_hand) == 1 and s[i] == "1":
pass
elif (int(s[i]) + in_hand) == 1 and in_hand == 1:
in_hand = 0
change += 1
else:
change += 1
print(change)
```
| 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"
}
] | 1,686,974,575
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 92
| 0
|
def can_divide_watermelon(w):
# Check if the weight is even and greater than 2
if w % 2 == 0 and w > 2:
return "YES"
else:
return "NO"
# Get the input weight of the watermelon
w = int(input())
# Check if the watermelon can be divided as desired
result = can_divide_watermelon(w)
# Print the result
print(result)
|
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
def can_divide_watermelon(w):
# Check if the weight is even and greater than 2
if w % 2 == 0 and w > 2:
return "YES"
else:
return "NO"
# Get the input weight of the watermelon
w = int(input())
# Check if the watermelon can be divided as desired
result = can_divide_watermelon(w)
# Print the result
print(result)
```
| 3.954
|
776
|
A
|
A Serial Killer
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"strings"
] | null | null |
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
|
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=≤<=*n*<=≤<=1000), the number of days.
Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
|
Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
|
[
"ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n",
"icm codeforces\n1\ncodeforces technex\n"
] |
[
"ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n",
"icm codeforces\nicm technex\n"
] |
In first example, the killer starts with ross and rachel.
- After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears.
| 500
|
[
{
"input": "ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler",
"output": "ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler"
},
{
"input": "icm codeforces\n1\ncodeforces technex",
"output": "icm codeforces\nicm technex"
},
{
"input": "a b\n3\na c\nb d\nd e",
"output": "a b\nc b\nc d\nc e"
},
{
"input": "ze udggmyop\n4\nze szhrbmft\nudggmyop mjorab\nszhrbmft ojdtfnzxj\nojdtfnzxj yjlkg",
"output": "ze udggmyop\nszhrbmft udggmyop\nszhrbmft mjorab\nojdtfnzxj mjorab\nyjlkg mjorab"
},
{
"input": "q s\n10\nq b\nb j\ns g\nj f\nf m\ng c\nc a\nm d\nd z\nz o",
"output": "q s\nb s\nj s\nj g\nf g\nm g\nm c\nm a\nd a\nz a\no a"
},
{
"input": "iii iiiiii\n7\niii iiiiiiiiii\niiiiiiiiii iiii\niiii i\niiiiii iiiiiiii\niiiiiiii iiiiiiiii\ni iiiii\niiiii ii",
"output": "iii iiiiii\niiiiiiiiii iiiiii\niiii iiiiii\ni iiiiii\ni iiiiiiii\ni iiiiiiiii\niiiii iiiiiiiii\nii iiiiiiiii"
},
{
"input": "bwyplnjn zkms\n26\nzkms nzmcsytxh\nnzmcsytxh yujsb\nbwyplnjn gtbzhudpb\ngtbzhudpb hpk\nyujsb xvy\nhpk wrwnfokml\nwrwnfokml ndouuikw\nndouuikw ucgrja\nucgrja tgfmpldz\nxvy nycrfphn\nnycrfphn quvs\nquvs htdy\nhtdy k\ntgfmpldz xtdpkxm\nxtdpkxm suwqxs\nk fv\nsuwqxs qckllwy\nqckllwy diun\nfv lefa\nlefa gdoqjysx\ndiun dhpz\ngdoqjysx bdmqdyt\ndhpz dgz\ndgz v\nbdmqdyt aswy\naswy ydkayhlrnm",
"output": "bwyplnjn zkms\nbwyplnjn nzmcsytxh\nbwyplnjn yujsb\ngtbzhudpb yujsb\nhpk yujsb\nhpk xvy\nwrwnfokml xvy\nndouuikw xvy\nucgrja xvy\ntgfmpldz xvy\ntgfmpldz nycrfphn\ntgfmpldz quvs\ntgfmpldz htdy\ntgfmpldz k\nxtdpkxm k\nsuwqxs k\nsuwqxs fv\nqckllwy fv\ndiun fv\ndiun lefa\ndiun gdoqjysx\ndhpz gdoqjysx\ndhpz bdmqdyt\ndgz bdmqdyt\nv bdmqdyt\nv aswy\nv ydkayhlrnm"
},
{
"input": "wxz hbeqwqp\n7\nhbeqwqp cpieghnszh\ncpieghnszh tlqrpd\ntlqrpd ttwrtio\nttwrtio xapvds\nxapvds zk\nwxz yryk\nzk b",
"output": "wxz hbeqwqp\nwxz cpieghnszh\nwxz tlqrpd\nwxz ttwrtio\nwxz xapvds\nwxz zk\nyryk zk\nyryk b"
},
{
"input": "wced gnsgv\n23\ngnsgv japawpaf\njapawpaf nnvpeu\nnnvpeu a\na ddupputljq\nddupputljq qyhnvbh\nqyhnvbh pqwijl\nwced khuvs\nkhuvs bjkh\npqwijl ysacmboc\nbjkh srf\nsrf jknoz\njknoz hodf\nysacmboc xqtkoyh\nhodf rfp\nxqtkoyh bivgnwqvoe\nbivgnwqvoe nknf\nnknf wuig\nrfp e\ne bqqknq\nwuig sznhhhu\nbqqknq dhrtdld\ndhrtdld n\nsznhhhu bguylf",
"output": "wced gnsgv\nwced japawpaf\nwced nnvpeu\nwced a\nwced ddupputljq\nwced qyhnvbh\nwced pqwijl\nkhuvs pqwijl\nbjkh pqwijl\nbjkh ysacmboc\nsrf ysacmboc\njknoz ysacmboc\nhodf ysacmboc\nhodf xqtkoyh\nrfp xqtkoyh\nrfp bivgnwqvoe\nrfp nknf\nrfp wuig\ne wuig\nbqqknq wuig\nbqqknq sznhhhu\ndhrtdld sznhhhu\nn sznhhhu\nn bguylf"
},
{
"input": "qqqqqqqqqq qqqqqqqq\n3\nqqqqqqqq qqqqqqqqq\nqqqqqqqqq qqqqq\nqqqqq q",
"output": "qqqqqqqqqq qqqqqqqq\nqqqqqqqqqq qqqqqqqqq\nqqqqqqqqqq qqqqq\nqqqqqqqqqq q"
},
{
"input": "wwwww w\n8\nwwwww wwwwwwww\nwwwwwwww wwwwwwwww\nwwwwwwwww wwwwwwwwww\nw www\nwwwwwwwwww wwww\nwwww ww\nwww wwwwww\nwwwwww wwwwwww",
"output": "wwwww w\nwwwwwwww w\nwwwwwwwww w\nwwwwwwwwww w\nwwwwwwwwww www\nwwww www\nww www\nww wwwwww\nww wwwwwww"
},
{
"input": "k d\n17\nk l\nd v\nv z\nl r\nz i\nr s\ns p\np w\nw j\nj h\ni c\nh m\nm q\nc o\no g\nq x\nx n",
"output": "k d\nl d\nl v\nl z\nr z\nr i\ns i\np i\nw i\nj i\nh i\nh c\nm c\nq c\nq o\nq g\nx g\nn g"
}
] | 1,530,190,368
| 468
|
Python 3
|
OK
|
TESTS
| 57
| 108
| 0
|
a, b = input().split(' ')
n = int(input())
print(a, b)
for i in range(n):
c, d = input().split(' ')
if c == a:
a = d
else:
b = d
print(a, b)
|
Title: A Serial Killer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input Specification:
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=≤<=*n*<=≤<=1000), the number of days.
Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output Specification:
Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Demo Input:
['ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n', 'icm codeforces\n1\ncodeforces technex\n']
Demo Output:
['ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n', 'icm codeforces\nicm technex\n']
Note:
In first example, the killer starts with ross and rachel.
- After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears.
|
```python
a, b = input().split(' ')
n = int(input())
print(a, b)
for i in range(n):
c, d = input().split(' ')
if c == a:
a = d
else:
b = d
print(a, b)
```
| 3
|
|
551
|
A
|
GukiZ and Contest
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"sortings"
] | null | null |
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
|
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
|
[
"3\n1 3 3\n",
"1\n1\n",
"5\n3 5 3 4 5\n"
] |
[
"3 1 1\n",
"1\n",
"4 1 4 3 1\n"
] |
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
| 500
|
[
{
"input": "3\n1 3 3",
"output": "3 1 1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "5\n3 5 3 4 5",
"output": "4 1 4 3 1"
},
{
"input": "7\n1 3 5 4 2 2 1",
"output": "6 3 1 2 4 4 6"
},
{
"input": "11\n5 6 4 2 9 7 6 6 6 6 7",
"output": "9 4 10 11 1 2 4 4 4 4 2"
},
{
"input": "1\n2000",
"output": "1"
},
{
"input": "2\n2000 2000",
"output": "1 1"
},
{
"input": "3\n500 501 502",
"output": "3 2 1"
},
{
"input": "10\n105 106 1 1 1 11 1000 999 1000 999",
"output": "6 5 8 8 8 7 1 3 1 3"
},
{
"input": "6\n1 2 3 4 5 6",
"output": "6 5 4 3 2 1"
},
{
"input": "7\n6 5 4 3 2 1 1",
"output": "1 2 3 4 5 6 6"
},
{
"input": "8\n153 100 87 14 10 8 6 5",
"output": "1 2 3 4 5 6 7 8"
},
{
"input": "70\n11 54 37 62 1 46 13 17 38 47 28 15 63 5 61 34 49 66 32 59 3 41 58 28 23 62 41 64 20 5 14 41 10 37 51 32 65 46 61 8 15 19 16 44 31 42 19 46 66 25 26 58 60 5 19 18 69 53 20 40 45 27 24 41 32 23 57 56 62 10",
"output": "62 18 35 7 70 23 61 56 34 22 42 58 6 66 10 37 21 2 38 13 69 29 14 42 48 7 29 5 50 66 60 29 63 35 20 38 4 23 10 65 58 52 57 27 41 28 52 23 2 46 45 14 12 66 52 55 1 19 50 33 26 44 47 29 38 48 16 17 7 63"
},
{
"input": "5\n1 2000 1 1 2000",
"output": "3 1 3 3 1"
}
] | 1,619,871,237
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 36
| 1,123
| 102,400
|
from sys import stdin
n=int(stdin.readline())
a=[int(x) for x in stdin.readline().split()]
for i in range(n):
c=0
for j in range(n):
if a[i]<a[j]:
c+=1
print(c+1,end=" ")
|
Title: GukiZ and Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.
In total, *n* students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to *n*. Let's denote the rating of *i*-th student as *a**i*. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.
He thinks that each student will take place equal to . In particular, if student *A* has rating strictly lower then student *B*, *A* will get the strictly better position than *B*, and if two students have equal ratings, they will share the same position.
GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000), number of GukiZ's students.
The second line contains *n* numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=2000) where *a**i* is the rating of *i*-th student (1<=≤<=*i*<=≤<=*n*).
Output Specification:
In a single line, print the position after the end of the contest for each of *n* students in the same order as they appear in the input.
Demo Input:
['3\n1 3 3\n', '1\n1\n', '5\n3 5 3 4 5\n']
Demo Output:
['3 1 1\n', '1\n', '4 1 4 3 1\n']
Note:
In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.
In the second sample, first student is the only one on the contest.
In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3 are the last sharing fourth position.
|
```python
from sys import stdin
n=int(stdin.readline())
a=[int(x) for x in stdin.readline().split()]
for i in range(n):
c=0
for j in range(n):
if a[i]<a[j]:
c+=1
print(c+1,end=" ")
```
| 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"
}
] | 1,674,982,934
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 296
| 0
|
n = int(input().strip())
cnt = 0
for i in range(n):
s = input().strip()
if s == "Icosahedron":
cnt += 20
if s == "Cube":
cnt += 6
if s == "Tetrahedron":
cnt += 4
if s == "Dodecahedron":
cnt += 12
if s == "Octahedron":
cnt += 8
print(cnt)
|
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
n = int(input().strip())
cnt = 0
for i in range(n):
s = input().strip()
if s == "Icosahedron":
cnt += 20
if s == "Cube":
cnt += 6
if s == "Tetrahedron":
cnt += 4
if s == "Dodecahedron":
cnt += 12
if s == "Octahedron":
cnt += 8
print(cnt)
```
| 3
|
|
912
|
B
|
New Year's Eve
|
PROGRAMMING
| 1,300
|
[
"bitmasks",
"constructive algorithms",
"number theory"
] | null | null |
Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains *n* sweet candies from the good ol' bakery, each labeled from 1 to *n* corresponding to its tastiness. No two candies have the same tastiness.
The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!
A xor-sum of a sequence of integers *a*1,<=*a*2,<=...,<=*a**m* is defined as the bitwise XOR of all its elements: , here denotes the bitwise XOR operation; more about bitwise XOR can be found [here.](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than *k* candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.
|
The sole string contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1018).
|
Output one number — the largest possible xor-sum.
|
[
"4 3\n",
"6 6\n"
] |
[
"7\n",
"7\n"
] |
In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.
In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7.
| 1,000
|
[
{
"input": "4 3",
"output": "7"
},
{
"input": "6 6",
"output": "7"
},
{
"input": "2 2",
"output": "3"
},
{
"input": "1022 10",
"output": "1023"
},
{
"input": "415853337373441 52",
"output": "562949953421311"
},
{
"input": "75 12",
"output": "127"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1152921504606846975"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000000000000 2",
"output": "1152921504606846975"
},
{
"input": "49194939 22",
"output": "67108863"
},
{
"input": "228104606 17",
"output": "268435455"
},
{
"input": "817034381 7",
"output": "1073741823"
},
{
"input": "700976748 4",
"output": "1073741823"
},
{
"input": "879886415 9",
"output": "1073741823"
},
{
"input": "18007336 10353515",
"output": "33554431"
},
{
"input": "196917003 154783328",
"output": "268435455"
},
{
"input": "785846777 496205300",
"output": "1073741823"
},
{
"input": "964756444 503568330",
"output": "1073741823"
},
{
"input": "848698811 317703059",
"output": "1073741823"
},
{
"input": "676400020444788 1",
"output": "676400020444788"
},
{
"input": "502643198528213 1",
"output": "502643198528213"
},
{
"input": "815936580997298686 684083143940282566",
"output": "1152921504606846975"
},
{
"input": "816762824175382110 752185261508428780",
"output": "1152921504606846975"
},
{
"input": "327942415253132295 222598158321260499",
"output": "576460752303423487"
},
{
"input": "328768654136248423 284493129147496637",
"output": "576460752303423487"
},
{
"input": "329594893019364551 25055600080496801",
"output": "576460752303423487"
},
{
"input": "921874985256864012 297786684518764536",
"output": "1152921504606846975"
},
{
"input": "922701224139980141 573634416190460758",
"output": "1152921504606846975"
},
{
"input": "433880815217730325 45629641110945892",
"output": "576460752303423487"
},
{
"input": "434707058395813749 215729375494216481",
"output": "576460752303423487"
},
{
"input": "435533301573897173 34078453236225189",
"output": "576460752303423487"
},
{
"input": "436359544751980597 199220719961060641",
"output": "576460752303423487"
},
{
"input": "437185783635096725 370972992240105630",
"output": "576460752303423487"
},
{
"input": "438012026813180149 111323110116193830",
"output": "576460752303423487"
},
{
"input": "438838269991263573 295468957052046146",
"output": "576460752303423487"
},
{
"input": "439664513169346997 46560240538186155",
"output": "576460752303423487"
},
{
"input": "440490752052463125 216165966013438147",
"output": "576460752303423487"
},
{
"input": "441316995230546549 401964286420555423",
"output": "576460752303423487"
},
{
"input": "952496582013329437 673506882352402278",
"output": "1152921504606846975"
},
{
"input": "1000000000000000000 1",
"output": "1000000000000000000"
},
{
"input": "2147483647 1",
"output": "2147483647"
},
{
"input": "2147483647 2",
"output": "2147483647"
},
{
"input": "2147483647 31",
"output": "2147483647"
},
{
"input": "8 2",
"output": "15"
},
{
"input": "3 3",
"output": "3"
},
{
"input": "4 1",
"output": "4"
},
{
"input": "10 2",
"output": "15"
},
{
"input": "288230376151711743 2",
"output": "288230376151711743"
},
{
"input": "5 2",
"output": "7"
},
{
"input": "576460752303423487 2",
"output": "576460752303423487"
},
{
"input": "36028797018963967 123",
"output": "36028797018963967"
},
{
"input": "1125899906842623 2",
"output": "1125899906842623"
},
{
"input": "576460752303423489 5",
"output": "1152921504606846975"
},
{
"input": "288230376151711743 3",
"output": "288230376151711743"
},
{
"input": "36028797018963967 345",
"output": "36028797018963967"
},
{
"input": "18014398509481984 30",
"output": "36028797018963967"
},
{
"input": "8 8",
"output": "15"
},
{
"input": "8 1",
"output": "8"
}
] | 1,656,389,029
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 58
| 109
| 512,000
|
import sys
input = sys.stdin.readline
n, k = map(int, input().split())
if(k == 1):
print(n)
exit()
for i in range(61, -1, -1):
c = 1 << i
if(n & c):
break
print(2 * c - 1)
|
Title: New Year's Eve
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains *n* sweet candies from the good ol' bakery, each labeled from 1 to *n* corresponding to its tastiness. No two candies have the same tastiness.
The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!
A xor-sum of a sequence of integers *a*1,<=*a*2,<=...,<=*a**m* is defined as the bitwise XOR of all its elements: , here denotes the bitwise XOR operation; more about bitwise XOR can be found [here.](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than *k* candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.
Input Specification:
The sole string contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1018).
Output Specification:
Output one number — the largest possible xor-sum.
Demo Input:
['4 3\n', '6 6\n']
Demo Output:
['7\n', '7\n']
Note:
In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.
In the second sample case, one can, for example, take all six candies and obtain the xor-sum of 7.
|
```python
import sys
input = sys.stdin.readline
n, k = map(int, input().split())
if(k == 1):
print(n)
exit()
for i in range(61, -1, -1):
c = 1 << i
if(n & c):
break
print(2 * c - 1)
```
| 3
|
|
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"
}
] | 1,688,464,057
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 46
| 0
|
s = input()
l = []
for i in s:
if i != '{' and i != '}' and i != ' ' and i != ',':
l.append(i)
c = 0
print(len(set(l)))
|
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
s = input()
l = []
for i in s:
if i != '{' and i != '}' and i != ' ' and i != ',':
l.append(i)
c = 0
print(len(set(l)))
```
| 3
|
|
29
|
A
|
Spit Problem
|
PROGRAMMING
| 1,000
|
[
"brute force"
] |
A. Spit Problem
|
2
|
256
|
In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to know if in the zoo there are two camels, which spitted at each other. Help him to solve this task.
The trajectory of a camel's spit is an arc, i.e. if the camel in position *x* spits *d* meters right, he can hit only the camel in position *x*<=+<=*d*, if such a camel exists.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the amount of camels in the zoo. Each of the following *n* lines contains two integers *x**i* and *d**i* (<=-<=104<=≤<=*x**i*<=≤<=104,<=1<=≤<=|*d**i*|<=≤<=2·104) — records in Bob's notepad. *x**i* is a position of the *i*-th camel, and *d**i* is a distance at which the *i*-th camel spitted. Positive values of *d**i* correspond to the spits right, negative values correspond to the spits left. No two camels may stand in the same position.
|
If there are two camels, which spitted at each other, output YES. Otherwise, output NO.
|
[
"2\n0 1\n1 -1\n",
"3\n0 1\n1 1\n2 -2\n",
"5\n2 -10\n3 10\n0 5\n5 -5\n10 1\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "2\n0 1\n1 -1",
"output": "YES"
},
{
"input": "3\n0 1\n1 1\n2 -2",
"output": "NO"
},
{
"input": "5\n2 -10\n3 10\n0 5\n5 -5\n10 1",
"output": "YES"
},
{
"input": "10\n-9897 -1144\n-4230 -6350\n2116 -3551\n-3635 4993\n3907 -9071\n-2362 4120\n-6542 984\n5807 3745\n7594 7675\n-5412 -6872",
"output": "NO"
},
{
"input": "11\n-1536 3809\n-2406 -8438\n-1866 395\n5636 -490\n-6867 -7030\n7525 3575\n-6796 2908\n3884 4629\n-2862 -6122\n-8984 6122\n7137 -326",
"output": "YES"
},
{
"input": "12\n-9765 1132\n-1382 -215\n-9405 7284\n-2040 3947\n-9360 3150\n6425 9386\n806 -2278\n-2121 -7284\n5663 -1608\n-8377 9297\n6245 708\n8470 6024",
"output": "YES"
},
{
"input": "15\n8122 -9991\n-4068 -3386\n8971 3731\n3458 5161\n-8700 7562\n2691 8735\n-1510 -3892\n5183 -3753\n-7018 6637\n-7454 3386\n-818 -6377\n6771 -8647\n-7357 -1246\n-6186 1922\n9889 -3627",
"output": "YES"
},
{
"input": "20\n-5264 6424\n-3664 -7459\n-2780 -9859\n-3317 6842\n5681 -8092\n1555 1904\n-6684 1414\n6593 -1253\n-5708 -1202\n335 1733\n-926 7579\n3459 -1904\n-4486 4006\n6201 3616\n2847 -5255\n8438 7057\n8171 6042\n-9102 3545\n7731 -233\n6264 6563",
"output": "YES"
},
{
"input": "30\n-398 -1774\n313 -6974\n2346 -4657\n8552 -9647\n-5265 1538\n8195 4864\n-5641 -5219\n-1394 8563\n-1190 1992\n-4669 -1156\n7574 256\n9206 -2414\n4140 -549\n-294 2169\n7029 -2871\n3808 -9799\n3141 5690\n4648 -2680\n-5990 9800\n-2299 1697\n6077 -7177\n-400 -9724\n-4644 -2392\n-2198 -9531\n-2105 9386\n-8165 -4201\n-1589 -7916\n2518 -7840\n4173 -6949\n-3368 -9943",
"output": "NO"
}
] | 1,595,796,375
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 218
| 6,656,000
|
# -*- coding: utf-8 -*-
lista,dist,cont = [],[],0
n = int(input())
for i in range(n):
x,y = map(int,input().split())
lista.append(x)
dist.append(x+y)
for i in range(n):
if dist[i] in lista:
if lista[i] == dist[lista.index(dist[i])]: cont = 1; break
if cont == 0: print('NO')
else: print('YES')
|
Title: Spit Problem
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In a Berland's zoo there is an enclosure with camels. It is known that camels like to spit. Bob watched these interesting animals for the whole day and registered in his notepad where each animal spitted. Now he wants to know if in the zoo there are two camels, which spitted at each other. Help him to solve this task.
The trajectory of a camel's spit is an arc, i.e. if the camel in position *x* spits *d* meters right, he can hit only the camel in position *x*<=+<=*d*, if such a camel exists.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the amount of camels in the zoo. Each of the following *n* lines contains two integers *x**i* and *d**i* (<=-<=104<=≤<=*x**i*<=≤<=104,<=1<=≤<=|*d**i*|<=≤<=2·104) — records in Bob's notepad. *x**i* is a position of the *i*-th camel, and *d**i* is a distance at which the *i*-th camel spitted. Positive values of *d**i* correspond to the spits right, negative values correspond to the spits left. No two camels may stand in the same position.
Output Specification:
If there are two camels, which spitted at each other, output YES. Otherwise, output NO.
Demo Input:
['2\n0 1\n1 -1\n', '3\n0 1\n1 1\n2 -2\n', '5\n2 -10\n3 10\n0 5\n5 -5\n10 1\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none
|
```python
# -*- coding: utf-8 -*-
lista,dist,cont = [],[],0
n = int(input())
for i in range(n):
x,y = map(int,input().split())
lista.append(x)
dist.append(x+y)
for i in range(n):
if dist[i] in lista:
if lista[i] == dist[lista.index(dist[i])]: cont = 1; break
if cont == 0: print('NO')
else: print('YES')
```
| 3.933102
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,577,463,825
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 186
| 0
|
t=int(input())
l1=list(map(int,input().split(" ")[1:]))
only_odd = [num for num in l1 if num % 2 == 1]
only_even = [num for num in l1 if num % 2 == 0]
if(len(only_odd)>len(only_even)):
for elem in only_even:
print(elem)
else:
for elem in only_odd:
print(elem)
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
t=int(input())
l1=list(map(int,input().split(" ")[1:]))
only_odd = [num for num in l1 if num % 2 == 1]
only_even = [num for num in l1 if num % 2 == 0]
if(len(only_odd)>len(only_even)):
for elem in only_even:
print(elem)
else:
for elem in only_odd:
print(elem)
```
| 0
|
557
|
A
|
Ilya and Diplomas
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation",
"math"
] | null | null |
Soon a school Olympiad in Informatics will be held in Berland, *n* schoolchildren will participate there.
At a meeting of the jury of the Olympiad it was decided that each of the *n* participants, depending on the results, will get a diploma of the first, second or third degree. Thus, each student will receive exactly one diploma.
They also decided that there must be given at least *min*1 and at most *max*1 diplomas of the first degree, at least *min*2 and at most *max*2 diplomas of the second degree, and at least *min*3 and at most *max*3 diplomas of the third degree.
After some discussion it was decided to choose from all the options of distributing diplomas satisfying these limitations the one that maximizes the number of participants who receive diplomas of the first degree. Of all these options they select the one which maximizes the number of the participants who receive diplomas of the second degree. If there are multiple of these options, they select the option that maximizes the number of diplomas of the third degree.
Choosing the best option of distributing certificates was entrusted to Ilya, one of the best programmers of Berland. However, he found more important things to do, so it is your task now to choose the best option of distributing of diplomas, based on the described limitations.
It is guaranteed that the described limitations are such that there is a way to choose such an option of distributing diplomas that all *n* participants of the Olympiad will receive a diploma of some degree.
|
The first line of the input contains a single integer *n* (3<=≤<=*n*<=≤<=3·106) — the number of schoolchildren who will participate in the Olympiad.
The next line of the input contains two integers *min*1 and *max*1 (1<=≤<=*min*1<=≤<=*max*1<=≤<=106) — the minimum and maximum limits on the number of diplomas of the first degree that can be distributed.
The third line of the input contains two integers *min*2 and *max*2 (1<=≤<=*min*2<=≤<=*max*2<=≤<=106) — the minimum and maximum limits on the number of diplomas of the second degree that can be distributed.
The next line of the input contains two integers *min*3 and *max*3 (1<=≤<=*min*3<=≤<=*max*3<=≤<=106) — the minimum and maximum limits on the number of diplomas of the third degree that can be distributed.
It is guaranteed that *min*1<=+<=*min*2<=+<=*min*3<=≤<=*n*<=≤<=*max*1<=+<=*max*2<=+<=*max*3.
|
In the first line of the output print three numbers, showing how many diplomas of the first, second and third degree will be given to students in the optimal variant of distributing diplomas.
The optimal variant of distributing diplomas is the one that maximizes the number of students who receive diplomas of the first degree. Of all the suitable options, the best one is the one which maximizes the number of participants who receive diplomas of the second degree. If there are several of these options, the best one is the one that maximizes the number of diplomas of the third degree.
|
[
"6\n1 5\n2 6\n3 7\n",
"10\n1 2\n1 3\n1 5\n",
"6\n1 3\n2 2\n2 2\n"
] |
[
"1 2 3 \n",
"2 3 5 \n",
"2 2 2 \n"
] |
none
| 500
|
[
{
"input": "6\n1 5\n2 6\n3 7",
"output": "1 2 3 "
},
{
"input": "10\n1 2\n1 3\n1 5",
"output": "2 3 5 "
},
{
"input": "6\n1 3\n2 2\n2 2",
"output": "2 2 2 "
},
{
"input": "55\n1 1000000\n40 50\n10 200",
"output": "5 40 10 "
},
{
"input": "3\n1 1\n1 1\n1 1",
"output": "1 1 1 "
},
{
"input": "3\n1 1000000\n1 1000000\n1 1000000",
"output": "1 1 1 "
},
{
"input": "1000\n100 400\n300 500\n400 1200",
"output": "300 300 400 "
},
{
"input": "3000000\n1 1000000\n1 1000000\n1 1000000",
"output": "1000000 1000000 1000000 "
},
{
"input": "11\n3 5\n3 5\n3 5",
"output": "5 3 3 "
},
{
"input": "12\n3 5\n3 5\n3 5",
"output": "5 4 3 "
},
{
"input": "13\n3 5\n3 5\n3 5",
"output": "5 5 3 "
},
{
"input": "3000000\n1000000 1000000\n1000000 1000000\n1000000 1000000",
"output": "1000000 1000000 1000000 "
},
{
"input": "50\n1 100\n1 100\n1 100",
"output": "48 1 1 "
},
{
"input": "1279\n123 670\n237 614\n846 923",
"output": "196 237 846 "
},
{
"input": "1589\n213 861\n5 96\n506 634",
"output": "861 96 632 "
},
{
"input": "2115\n987 987\n112 483\n437 959",
"output": "987 483 645 "
},
{
"input": "641\n251 960\n34 370\n149 149",
"output": "458 34 149 "
},
{
"input": "1655\n539 539\n10 425\n605 895",
"output": "539 425 691 "
},
{
"input": "1477\n210 336\n410 837\n448 878",
"output": "336 693 448 "
},
{
"input": "1707\n149 914\n190 422\n898 899",
"output": "619 190 898 "
},
{
"input": "1529\n515 515\n563 869\n169 451",
"output": "515 845 169 "
},
{
"input": "1543\n361 994\n305 407\n102 197",
"output": "994 407 142 "
},
{
"input": "1107\n471 849\n360 741\n71 473",
"output": "676 360 71 "
},
{
"input": "1629279\n267360 999930\n183077 674527\n202618 786988",
"output": "999930 426731 202618 "
},
{
"input": "1233589\n2850 555444\n500608 921442\n208610 607343",
"output": "524371 500608 208610 "
},
{
"input": "679115\n112687 183628\n101770 982823\n81226 781340",
"output": "183628 414261 81226 "
},
{
"input": "1124641\n117999 854291\n770798 868290\n76651 831405",
"output": "277192 770798 76651 "
},
{
"input": "761655\n88152 620061\n60403 688549\n79370 125321",
"output": "620061 62224 79370 "
},
{
"input": "2174477\n276494 476134\n555283 954809\n319941 935631",
"output": "476134 954809 743534 "
},
{
"input": "1652707\n201202 990776\n34796 883866\n162979 983308",
"output": "990776 498952 162979 "
},
{
"input": "2065529\n43217 891429\n434379 952871\n650231 855105",
"output": "891429 523869 650231 "
},
{
"input": "1702543\n405042 832833\n50931 747750\n381818 796831",
"output": "832833 487892 381818 "
},
{
"input": "501107\n19061 859924\n126478 724552\n224611 489718",
"output": "150018 126478 224611 "
},
{
"input": "1629279\n850831 967352\n78593 463906\n452094 885430",
"output": "967352 209833 452094 "
},
{
"input": "1233589\n2850 157021\n535109 748096\n392212 475634",
"output": "157021 684356 392212 "
},
{
"input": "679115\n125987 786267\n70261 688983\n178133 976789",
"output": "430721 70261 178133 "
},
{
"input": "1124641\n119407 734250\n213706 860770\n102149 102149",
"output": "734250 288242 102149 "
},
{
"input": "761655\n325539 325539\n280794 792505\n18540 106895",
"output": "325539 417576 18540 "
},
{
"input": "2174477\n352351 791072\n365110 969163\n887448 955610",
"output": "791072 495957 887448 "
},
{
"input": "1652707\n266774 638522\n65688 235422\n924898 992826",
"output": "638522 89287 924898 "
},
{
"input": "2065529\n608515 608515\n751563 864337\n614898 705451",
"output": "608515 842116 614898 "
},
{
"input": "1702543\n5784 996578\n47395 300407\n151614 710197",
"output": "996578 300407 405558 "
},
{
"input": "501107\n8073 390048\n190494 647328\n274071 376923",
"output": "36542 190494 274071 "
},
{
"input": "200\n50 50\n100 100\n50 50",
"output": "50 100 50 "
},
{
"input": "14\n1 100\n1 100\n8 9",
"output": "5 1 8 "
},
{
"input": "300\n200 400\n50 100\n40 80",
"output": "210 50 40 "
},
{
"input": "10\n3 6\n3 6\n3 6",
"output": "4 3 3 "
},
{
"input": "14\n3 6\n3 6\n3 6",
"output": "6 5 3 "
},
{
"input": "17\n3 6\n3 6\n3 6",
"output": "6 6 5 "
},
{
"input": "1000000\n300000 600000\n300000 600000\n300000 600000",
"output": "400000 300000 300000 "
},
{
"input": "1400000\n300000 600000\n300000 600000\n300000 600000",
"output": "600000 500000 300000 "
},
{
"input": "1700000\n300000 600000\n300000 600000\n300000 600000",
"output": "600000 600000 500000 "
},
{
"input": "561\n400 400\n80 80\n81 81",
"output": "400 80 81 "
},
{
"input": "2000\n100 1000\n1 1\n1 2000",
"output": "1000 1 999 "
},
{
"input": "1000002\n1 1000000\n1 1000000\n999999 1000000",
"output": "2 1 999999 "
},
{
"input": "1000002\n1 1000000\n1 1000000\n1000000 1000000",
"output": "1 1 1000000 "
}
] | 1,435,676,980
| 580
|
Python 3
|
OK
|
TESTS
| 56
| 155
| 6,348,800
|
import math,string,itertools,fractions,heapq,collections,re,array,bisect,copy
from itertools import chain, dropwhile, permutations, combinations
from collections import defaultdict, deque
# Guide:
# 1. construct complex data types while reading (e.g. graph adj list)
# 2. avoid any non-necessary time/memory usage
# 3. avoid templates and write more from scratch
# 4. switch to "flat" implementations
def VI(): return list(map(int,input().split()))
def I(): return int(input())
def LIST(n,m=None): return [0]*n if m is None else [[0]*m for i in range(n)]
def ELIST(n): return [[] for i in range(n)]
def MI(n=None,m=None): # input matrix of integers
if n is None: n,m = VI()
arr = LIST(n)
for i in range(n): arr[i] = VI()
return arr
def MS(n=None,m=None): # input matrix of strings
if n is None: n,m = VI()
arr = LIST(n)
for i in range(n): arr[i] = input()
return arr
def MIT(n=None,m=None): # input transposed matrix/array of integers
if n is None: n,m = VI()
a = MI(n,m)
arr = LIST(m,n)
for i,l in enumerate(a):
for j,x in enumerate(l):
arr[j][i] = x
return arr
def run(n,x,l,r):
s = 0
curr = 1
for i in range(n):
skip = (l[i]-curr) // x
s += r[i]-curr-skip*x+1
curr = r[i]+1
print(s)
def main(info=0):
n = I()
an, ax = VI()
bn, bx = VI()
cn, cx = VI()
m = [an, bn, cn]
if sum(m) < n and ax>an:
m[0] += min(ax-an, n-sum(m))
if sum(m) < n and bx>bn:
m[1] += min(bx-bn, n-sum(m))
if sum(m) < n and cx>cn:
m[2] += min(cx-cn, n-sum(m))
print(m[0], m[1], m[2])
if __name__ == "__main__":
main()
|
Title: Ilya and Diplomas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Soon a school Olympiad in Informatics will be held in Berland, *n* schoolchildren will participate there.
At a meeting of the jury of the Olympiad it was decided that each of the *n* participants, depending on the results, will get a diploma of the first, second or third degree. Thus, each student will receive exactly one diploma.
They also decided that there must be given at least *min*1 and at most *max*1 diplomas of the first degree, at least *min*2 and at most *max*2 diplomas of the second degree, and at least *min*3 and at most *max*3 diplomas of the third degree.
After some discussion it was decided to choose from all the options of distributing diplomas satisfying these limitations the one that maximizes the number of participants who receive diplomas of the first degree. Of all these options they select the one which maximizes the number of the participants who receive diplomas of the second degree. If there are multiple of these options, they select the option that maximizes the number of diplomas of the third degree.
Choosing the best option of distributing certificates was entrusted to Ilya, one of the best programmers of Berland. However, he found more important things to do, so it is your task now to choose the best option of distributing of diplomas, based on the described limitations.
It is guaranteed that the described limitations are such that there is a way to choose such an option of distributing diplomas that all *n* participants of the Olympiad will receive a diploma of some degree.
Input Specification:
The first line of the input contains a single integer *n* (3<=≤<=*n*<=≤<=3·106) — the number of schoolchildren who will participate in the Olympiad.
The next line of the input contains two integers *min*1 and *max*1 (1<=≤<=*min*1<=≤<=*max*1<=≤<=106) — the minimum and maximum limits on the number of diplomas of the first degree that can be distributed.
The third line of the input contains two integers *min*2 and *max*2 (1<=≤<=*min*2<=≤<=*max*2<=≤<=106) — the minimum and maximum limits on the number of diplomas of the second degree that can be distributed.
The next line of the input contains two integers *min*3 and *max*3 (1<=≤<=*min*3<=≤<=*max*3<=≤<=106) — the minimum and maximum limits on the number of diplomas of the third degree that can be distributed.
It is guaranteed that *min*1<=+<=*min*2<=+<=*min*3<=≤<=*n*<=≤<=*max*1<=+<=*max*2<=+<=*max*3.
Output Specification:
In the first line of the output print three numbers, showing how many diplomas of the first, second and third degree will be given to students in the optimal variant of distributing diplomas.
The optimal variant of distributing diplomas is the one that maximizes the number of students who receive diplomas of the first degree. Of all the suitable options, the best one is the one which maximizes the number of participants who receive diplomas of the second degree. If there are several of these options, the best one is the one that maximizes the number of diplomas of the third degree.
Demo Input:
['6\n1 5\n2 6\n3 7\n', '10\n1 2\n1 3\n1 5\n', '6\n1 3\n2 2\n2 2\n']
Demo Output:
['1 2 3 \n', '2 3 5 \n', '2 2 2 \n']
Note:
none
|
```python
import math,string,itertools,fractions,heapq,collections,re,array,bisect,copy
from itertools import chain, dropwhile, permutations, combinations
from collections import defaultdict, deque
# Guide:
# 1. construct complex data types while reading (e.g. graph adj list)
# 2. avoid any non-necessary time/memory usage
# 3. avoid templates and write more from scratch
# 4. switch to "flat" implementations
def VI(): return list(map(int,input().split()))
def I(): return int(input())
def LIST(n,m=None): return [0]*n if m is None else [[0]*m for i in range(n)]
def ELIST(n): return [[] for i in range(n)]
def MI(n=None,m=None): # input matrix of integers
if n is None: n,m = VI()
arr = LIST(n)
for i in range(n): arr[i] = VI()
return arr
def MS(n=None,m=None): # input matrix of strings
if n is None: n,m = VI()
arr = LIST(n)
for i in range(n): arr[i] = input()
return arr
def MIT(n=None,m=None): # input transposed matrix/array of integers
if n is None: n,m = VI()
a = MI(n,m)
arr = LIST(m,n)
for i,l in enumerate(a):
for j,x in enumerate(l):
arr[j][i] = x
return arr
def run(n,x,l,r):
s = 0
curr = 1
for i in range(n):
skip = (l[i]-curr) // x
s += r[i]-curr-skip*x+1
curr = r[i]+1
print(s)
def main(info=0):
n = I()
an, ax = VI()
bn, bx = VI()
cn, cx = VI()
m = [an, bn, cn]
if sum(m) < n and ax>an:
m[0] += min(ax-an, n-sum(m))
if sum(m) < n and bx>bn:
m[1] += min(bx-bn, n-sum(m))
if sum(m) < n and cx>cn:
m[2] += min(cx-cn, n-sum(m))
print(m[0], m[1], m[2])
if __name__ == "__main__":
main()
```
| 3
|
|
122
|
A
|
Lucky Division
|
PROGRAMMING
| 1,000
|
[
"brute force",
"number theory"
] | null | null |
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
|
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
|
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
|
[
"47\n",
"16\n",
"78\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
| 500
|
[
{
"input": "47",
"output": "YES"
},
{
"input": "16",
"output": "YES"
},
{
"input": "78",
"output": "NO"
},
{
"input": "48",
"output": "YES"
},
{
"input": "100",
"output": "YES"
},
{
"input": "107",
"output": "NO"
},
{
"input": "77",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "480",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "49",
"output": "YES"
},
{
"input": "56",
"output": "YES"
},
{
"input": "124",
"output": "YES"
},
{
"input": "1000",
"output": "YES"
},
{
"input": "999",
"output": "NO"
},
{
"input": "298",
"output": "NO"
},
{
"input": "274",
"output": "NO"
},
{
"input": "998",
"output": "NO"
},
{
"input": "42",
"output": "YES"
},
{
"input": "788",
"output": "YES"
},
{
"input": "70",
"output": "YES"
},
{
"input": "444",
"output": "YES"
},
{
"input": "777",
"output": "YES"
},
{
"input": "799",
"output": "YES"
},
{
"input": "25",
"output": "NO"
},
{
"input": "882",
"output": "YES"
},
{
"input": "88",
"output": "YES"
},
{
"input": "11",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "7",
"output": "YES"
},
{
"input": "8",
"output": "YES"
},
{
"input": "94",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "141",
"output": "YES"
}
] | 1,696,858,089
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 92
| 0
|
divisors = [4,7,44,47,74,77,444,447,474,744,774,777]
n = int(input())
for number in divisors:
if n%number == 0:
print('YES')
exit()
print('NO')
|
Title: Lucky Division
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
Output Specification:
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['47\n', '16\n', '78\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
|
```python
divisors = [4,7,44,47,74,77,444,447,474,744,774,777]
n = int(input())
for number in divisors:
if n%number == 0:
print('YES')
exit()
print('NO')
```
| 0
|
|
119
|
A
|
Epic Game
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
|
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
|
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
|
[
"3 5 9\n",
"1 1 100\n"
] |
[
"0",
"1"
] |
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
| 500
|
[
{
"input": "3 5 9",
"output": "0"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "23 12 16",
"output": "1"
},
{
"input": "95 26 29",
"output": "1"
},
{
"input": "73 32 99",
"output": "1"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "41 12 65",
"output": "1"
},
{
"input": "13 61 100",
"output": "1"
},
{
"input": "100 100 10",
"output": "0"
},
{
"input": "12 24 26",
"output": "1"
},
{
"input": "73 21 96",
"output": "1"
},
{
"input": "17 22 81",
"output": "1"
},
{
"input": "14 88 97",
"output": "1"
},
{
"input": "42 81 17",
"output": "0"
},
{
"input": "34 36 19",
"output": "1"
},
{
"input": "21 87 12",
"output": "1"
},
{
"input": "97 92 98",
"output": "1"
},
{
"input": "49 25 62",
"output": "1"
},
{
"input": "29 17 44",
"output": "1"
},
{
"input": "10 1 10",
"output": "0"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "100 100 1",
"output": "0"
},
{
"input": "11 99 55",
"output": "0"
},
{
"input": "8 13 21",
"output": "0"
},
{
"input": "34 55 89",
"output": "0"
},
{
"input": "21 34 55",
"output": "1"
},
{
"input": "13 21 34",
"output": "1"
},
{
"input": "34 21 55",
"output": "0"
},
{
"input": "10 20 84",
"output": "1"
},
{
"input": "4 2 5",
"output": "0"
},
{
"input": "3 2 3",
"output": "0"
},
{
"input": "31 23 1",
"output": "0"
},
{
"input": "100 100 100",
"output": "0"
},
{
"input": "100 1 100",
"output": "0"
},
{
"input": "1 100 100",
"output": "1"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "17 23 100",
"output": "1"
},
{
"input": "99 98 100",
"output": "1"
},
{
"input": "1 1 2",
"output": "1"
},
{
"input": "2 1 1",
"output": "0"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "2 1 2",
"output": "0"
},
{
"input": "10 10 3",
"output": "1"
},
{
"input": "2 2 1",
"output": "0"
},
{
"input": "9 5 1",
"output": "0"
}
] | 1,600,965,284
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 45
| 218
| 0
|
"""
Solution to Codeforces Problem 119A : https://codeforces.com/problemset/problem/119/A
"""
import math
def solve(a, b, n):
while 0 < n:
for ix, value in enumerate([a, b]):
n -= math.gcd(value, n)
if n == 0:
print(ix)
if __name__ == "__main__":
solve(*list(map(int, input().split(' '))))
|
Title: Epic Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
Input Specification:
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
Output Specification:
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
Demo Input:
['3 5 9\n', '1 1 100\n']
Demo Output:
['0', '1']
Note:
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that.
|
```python
"""
Solution to Codeforces Problem 119A : https://codeforces.com/problemset/problem/119/A
"""
import math
def solve(a, b, n):
while 0 < n:
for ix, value in enumerate([a, b]):
n -= math.gcd(value, n)
if n == 0:
print(ix)
if __name__ == "__main__":
solve(*list(map(int, input().split(' '))))
```
| 3
|
|
701
|
A
|
Cards
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation"
] | null | null |
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
|
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card.
|
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
|
[
"6\n1 5 7 4 4 3\n",
"4\n10 10 10 10\n"
] |
[
"1 3\n6 2\n4 5\n",
"1 2\n3 4\n"
] |
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable.
| 500
|
[
{
"input": "6\n1 5 7 4 4 3",
"output": "1 3\n6 2\n4 5"
},
{
"input": "4\n10 10 10 10",
"output": "1 4\n2 3"
},
{
"input": "100\n2 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",
"output": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"input": "4\n82 46 8 44",
"output": "3 1\n4 2"
},
{
"input": "2\n35 50",
"output": "1 2"
},
{
"input": "8\n24 39 49 38 44 64 44 50",
"output": "1 6\n4 8\n2 3\n5 7"
},
{
"input": "100\n23 44 35 88 10 78 8 84 46 19 69 36 81 60 46 12 53 22 83 73 6 18 80 14 54 39 74 42 34 20 91 70 32 11 80 53 70 21 24 12 87 68 35 39 8 84 81 70 8 54 73 2 60 71 4 33 65 48 69 58 55 57 78 61 45 50 55 72 86 37 5 11 12 81 32 19 22 11 22 82 23 56 61 84 47 59 31 38 31 90 57 1 24 38 68 27 80 9 37 14",
"output": "92 31\n52 90\n55 4\n71 41\n21 69\n7 84\n45 46\n49 8\n98 19\n5 80\n34 74\n72 47\n78 13\n16 97\n40 35\n73 23\n24 63\n100 6\n22 27\n10 51\n76 20\n30 68\n38 54\n18 48\n77 37\n79 32\n1 59\n81 11\n39 95\n93 42\n96 57\n87 83\n89 64\n33 53\n75 14\n56 86\n29 60\n3 91\n43 62\n12 82\n70 67\n99 61\n88 50\n94 25\n26 36\n44 17\n28 66\n2 58\n65 85\n9 15"
},
{
"input": "12\n22 83 2 67 55 12 40 93 83 73 12 28",
"output": "3 8\n6 9\n11 2\n1 10\n12 4\n7 5"
},
{
"input": "16\n10 33 36 32 48 25 31 27 45 13 37 26 22 21 15 43",
"output": "1 5\n10 9\n15 16\n14 11\n13 3\n6 2\n12 4\n8 7"
},
{
"input": "20\n18 13 71 60 28 10 20 65 65 12 13 14 64 68 6 50 72 7 66 58",
"output": "15 17\n18 3\n6 14\n10 19\n2 9\n11 8\n12 13\n1 4\n7 20\n5 16"
},
{
"input": "24\n59 39 25 22 46 21 24 70 60 11 46 42 44 37 13 37 41 58 72 23 25 61 58 62",
"output": "10 19\n15 8\n6 24\n4 22\n20 9\n7 1\n3 23\n21 18\n14 11\n16 5\n2 13\n17 12"
},
{
"input": "28\n22 1 51 31 83 35 3 64 59 10 61 25 19 53 55 80 78 8 82 22 67 4 27 64 33 6 85 76",
"output": "2 27\n7 5\n22 19\n26 16\n18 17\n10 28\n13 21\n1 24\n20 8\n12 11\n23 9\n4 15\n25 14\n6 3"
},
{
"input": "32\n41 42 22 68 40 52 66 16 73 25 41 21 36 60 46 30 24 55 35 10 54 52 70 24 20 56 3 34 35 6 51 8",
"output": "27 9\n30 23\n32 4\n20 7\n8 14\n25 26\n12 18\n3 21\n17 22\n24 6\n10 31\n16 15\n28 2\n19 11\n29 1\n13 5"
},
{
"input": "36\n1 10 61 43 27 49 55 33 7 30 45 78 69 34 38 19 36 49 55 11 30 63 46 24 16 68 71 18 11 52 72 24 60 68 8 41",
"output": "1 12\n9 31\n35 27\n2 13\n20 34\n29 26\n25 22\n28 3\n16 33\n24 19\n32 7\n5 30\n10 18\n21 6\n8 23\n14 11\n17 4\n15 36"
},
{
"input": "40\n7 30 13 37 37 56 45 28 61 28 23 33 44 63 58 52 21 2 42 19 10 32 9 7 61 15 58 20 45 4 46 24 35 17 50 4 20 48 41 55",
"output": "18 14\n30 25\n36 9\n1 27\n24 15\n23 6\n21 40\n3 16\n26 35\n34 38\n20 31\n28 29\n37 7\n17 13\n11 19\n32 39\n8 5\n10 4\n2 33\n22 12"
},
{
"input": "44\n7 12 46 78 24 68 86 22 71 79 85 14 58 72 26 46 54 39 35 13 31 45 81 21 15 8 47 64 69 87 57 6 18 80 47 29 36 62 34 67 59 48 75 25",
"output": "32 30\n1 7\n26 11\n2 23\n20 34\n12 10\n25 4\n33 43\n24 14\n8 9\n5 29\n44 6\n15 40\n36 28\n21 38\n39 41\n19 13\n37 31\n18 17\n22 42\n3 35\n16 27"
},
{
"input": "48\n57 38 16 25 34 57 29 38 60 51 72 78 22 39 10 33 20 16 12 3 51 74 9 88 4 70 56 65 86 18 33 12 77 78 52 87 68 85 81 5 61 2 52 39 80 13 74 30",
"output": "42 24\n20 36\n25 29\n40 38\n23 39\n15 45\n19 34\n32 12\n46 33\n3 47\n18 22\n30 11\n17 26\n13 37\n4 28\n7 41\n48 9\n16 6\n31 1\n5 27\n2 43\n8 35\n14 21\n44 10"
},
{
"input": "52\n57 12 13 40 68 31 18 4 31 18 65 3 62 32 6 3 49 48 51 33 53 40 9 32 47 53 58 19 14 23 32 38 39 69 19 20 62 52 68 17 39 22 54 59 3 2 52 9 67 68 24 39",
"output": "46 34\n12 50\n16 39\n45 5\n8 49\n15 11\n23 37\n48 13\n2 44\n3 27\n29 1\n40 43\n7 26\n10 21\n28 47\n35 38\n36 19\n42 17\n30 18\n51 25\n6 22\n9 4\n14 52\n24 41\n31 33\n20 32"
},
{
"input": "56\n53 59 66 68 71 25 48 32 12 61 72 69 30 6 56 55 25 49 60 47 46 46 66 19 31 9 23 15 10 12 71 53 51 32 39 31 66 66 17 52 12 7 7 22 49 12 71 29 63 7 47 29 18 39 27 26",
"output": "14 11\n42 47\n43 31\n50 5\n26 12\n29 4\n9 38\n30 37\n41 23\n46 3\n28 49\n39 10\n53 19\n24 2\n44 15\n27 16\n6 32\n17 1\n56 40\n55 33\n48 45\n52 18\n13 7\n25 51\n36 20\n8 22\n34 21\n35 54"
},
{
"input": "60\n47 63 20 68 46 12 45 44 14 38 28 73 60 5 20 18 70 64 37 47 26 47 37 61 29 61 23 28 30 68 55 22 25 60 38 7 63 12 38 15 14 30 11 5 70 15 53 52 7 57 49 45 55 37 45 28 50 2 31 30",
"output": "58 12\n14 45\n44 17\n36 30\n49 4\n43 18\n6 37\n38 2\n9 26\n41 24\n40 34\n46 13\n16 50\n3 53\n15 31\n32 47\n27 48\n33 57\n21 51\n11 22\n28 20\n56 1\n25 5\n29 55\n42 52\n60 7\n59 8\n19 39\n23 35\n54 10"
},
{
"input": "64\n63 39 19 5 48 56 49 45 29 68 25 59 37 69 62 26 60 44 60 6 67 68 2 40 56 6 19 12 17 70 23 11 59 37 41 55 30 68 72 14 38 34 3 71 2 4 55 15 31 66 15 51 36 72 18 7 6 14 43 33 8 35 57 18",
"output": "23 54\n45 39\n43 44\n46 30\n4 14\n20 38\n26 22\n57 10\n56 21\n61 50\n32 1\n28 15\n40 19\n58 17\n48 33\n51 12\n29 63\n55 25\n64 6\n3 47\n27 36\n31 52\n11 7\n16 5\n9 8\n37 18\n49 59\n60 35\n42 24\n62 2\n53 41\n13 34"
},
{
"input": "68\n58 68 40 55 62 15 10 54 19 18 69 27 15 53 8 18 8 33 15 49 20 9 70 8 18 64 14 59 9 64 3 35 46 11 5 65 58 55 28 58 4 55 64 5 68 24 4 58 23 45 58 50 38 68 5 15 20 9 5 53 20 63 69 68 15 53 65 65",
"output": "31 23\n41 63\n47 11\n35 64\n44 54\n55 45\n59 2\n15 68\n17 67\n24 36\n22 43\n29 30\n58 26\n7 62\n34 5\n27 28\n6 51\n13 48\n19 40\n56 37\n65 1\n10 42\n16 38\n25 4\n9 8\n21 66\n57 60\n61 14\n49 52\n46 20\n12 33\n39 50\n18 3\n32 53"
},
{
"input": "72\n61 13 55 23 24 55 44 33 59 19 14 17 66 40 27 33 29 37 28 74 50 56 59 65 64 17 42 56 73 51 64 23 22 26 38 22 36 47 60 14 52 28 14 12 6 41 73 5 64 67 61 74 54 34 45 34 44 4 34 49 18 72 44 47 31 19 11 31 5 4 45 50",
"output": "58 52\n70 20\n48 47\n69 29\n45 62\n67 50\n44 13\n2 24\n11 49\n40 31\n43 25\n12 51\n26 1\n61 39\n10 23\n66 9\n33 28\n36 22\n4 6\n32 3\n5 53\n34 41\n15 30\n19 72\n42 21\n17 60\n65 64\n68 38\n8 71\n16 55\n54 63\n56 57\n59 7\n37 27\n18 46\n35 14"
},
{
"input": "76\n73 37 73 67 26 45 43 74 47 31 43 81 4 3 39 79 48 81 67 39 67 66 43 67 80 51 34 79 5 58 45 10 39 50 9 78 6 18 75 17 45 17 51 71 34 53 33 11 17 15 11 69 50 41 13 74 10 33 77 41 11 64 36 74 17 32 3 10 27 20 5 73 52 41 7 57",
"output": "14 18\n67 12\n13 25\n29 28\n71 16\n37 36\n75 59\n35 39\n32 64\n57 56\n68 8\n48 72\n51 3\n61 1\n55 44\n50 52\n40 24\n42 21\n49 19\n65 4\n38 22\n70 62\n5 30\n69 76\n10 46\n66 73\n47 43\n58 26\n27 53\n45 34\n63 17\n2 9\n15 41\n20 31\n33 6\n54 23\n60 11\n74 7"
},
{
"input": "80\n18 38 65 1 20 9 57 2 36 26 15 17 33 61 65 27 10 35 49 42 40 32 19 33 12 36 56 31 10 41 8 54 56 60 5 47 61 43 23 19 20 30 7 6 38 60 29 58 35 64 30 51 6 17 30 24 47 1 37 47 34 36 48 28 5 25 47 19 30 39 36 23 31 28 46 46 59 43 19 49",
"output": "4 15\n58 3\n8 50\n35 37\n65 14\n44 46\n53 34\n43 77\n31 48\n6 7\n17 33\n29 27\n25 32\n11 52\n12 80\n54 19\n1 63\n23 67\n40 60\n68 57\n79 36\n5 76\n41 75\n39 78\n72 38\n56 20\n66 30\n10 21\n16 70\n64 45\n74 2\n47 59\n42 71\n51 62\n55 26\n69 9\n28 49\n73 18\n22 61\n13 24"
},
{
"input": "84\n59 41 54 14 42 55 29 28 41 73 40 15 1 1 66 49 76 59 68 60 42 81 19 23 33 12 80 81 42 22 54 54 2 22 22 28 27 60 36 57 17 76 38 20 40 65 23 9 81 50 25 13 46 36 59 53 6 35 47 40 59 19 67 46 63 49 12 33 23 49 33 23 32 62 60 70 44 1 6 63 28 16 70 69",
"output": "13 49\n14 28\n78 22\n33 27\n57 42\n79 17\n48 10\n26 83\n67 76\n52 84\n4 19\n12 63\n82 15\n41 46\n23 80\n62 65\n44 74\n30 75\n34 38\n35 20\n24 61\n47 55\n69 18\n72 1\n51 40\n37 6\n8 32\n36 31\n81 3\n7 56\n73 50\n25 70\n68 66\n71 16\n58 59\n39 64\n54 53\n43 77\n11 29\n45 21\n60 5\n2 9"
},
{
"input": "88\n10 28 71 6 58 66 45 52 13 71 39 1 10 29 30 70 14 17 15 38 4 60 5 46 66 41 40 58 2 57 32 44 21 26 13 40 64 63 56 33 46 8 30 43 67 55 44 28 32 62 14 58 42 67 45 59 32 68 10 31 51 6 42 34 9 12 51 27 20 14 62 42 16 5 1 14 30 62 40 59 58 26 25 15 27 47 21 57",
"output": "12 10\n75 3\n29 16\n21 58\n23 54\n74 45\n4 25\n62 6\n42 37\n65 38\n1 78\n13 71\n59 50\n66 22\n9 80\n35 56\n17 81\n51 52\n70 28\n76 5\n19 88\n84 30\n73 39\n18 46\n69 8\n33 67\n87 61\n83 86\n34 41\n82 24\n68 55\n85 7\n2 47\n48 32\n14 44\n15 72\n43 63\n77 53\n60 26\n31 79\n49 36\n57 27\n40 11\n64 20"
},
{
"input": "92\n17 37 81 15 29 70 73 42 49 23 44 77 27 44 74 11 43 66 15 41 60 36 33 11 2 76 16 51 45 21 46 16 85 29 76 79 16 6 60 13 25 44 62 28 43 35 63 24 76 71 62 15 57 72 45 10 71 59 74 14 53 13 58 72 14 72 73 11 25 1 57 42 86 63 50 30 64 38 10 77 75 24 58 8 54 12 43 30 27 71 52 34",
"output": "70 73\n25 33\n38 3\n84 36\n56 80\n79 12\n16 49\n24 35\n68 26\n86 81\n40 59\n62 15\n60 67\n65 7\n4 66\n19 64\n52 54\n27 90\n32 57\n37 50\n1 6\n30 18\n10 77\n48 74\n82 47\n41 51\n69 43\n13 39\n89 21\n44 58\n5 83\n34 63\n76 71\n88 53\n23 85\n92 61\n46 91\n22 28\n2 75\n78 9\n20 31\n8 55\n72 29\n17 42\n45 14\n87 11"
},
{
"input": "96\n77 7 47 19 73 31 46 13 89 69 52 9 26 77 6 87 55 45 71 2 79 1 80 20 4 82 64 20 75 86 84 24 77 56 16 54 53 35 74 73 40 29 63 20 83 39 58 16 31 41 40 16 11 90 30 48 62 39 55 8 50 3 77 73 75 66 14 90 18 54 38 10 53 22 67 38 27 91 62 37 85 13 92 7 18 83 10 3 86 54 80 59 34 16 39 43",
"output": "22 83\n20 78\n62 68\n88 54\n25 9\n15 16\n2 89\n84 30\n60 81\n12 31\n72 86\n87 45\n53 26\n8 91\n82 23\n67 21\n35 63\n48 33\n52 14\n94 1\n69 65\n85 29\n4 39\n24 64\n28 40\n44 5\n74 19\n32 10\n13 75\n77 66\n42 27\n55 43\n6 79\n49 57\n93 92\n38 47\n80 34\n71 59\n76 17\n46 90\n58 70\n95 36\n41 73\n51 37\n50 11\n96 61\n18 56\n7 3"
},
{
"input": "4\n100 100 1 1",
"output": "3 2\n4 1"
},
{
"input": "100\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",
"output": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"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",
"output": "1 100\n2 99\n3 98\n4 97\n5 96\n6 95\n7 94\n8 93\n9 92\n10 91\n11 90\n12 89\n13 88\n14 87\n15 86\n16 85\n17 84\n18 83\n19 82\n20 81\n21 80\n22 79\n23 78\n24 77\n25 76\n26 75\n27 74\n28 73\n29 72\n30 71\n31 70\n32 69\n33 68\n34 67\n35 66\n36 65\n37 64\n38 63\n39 62\n40 61\n41 60\n42 59\n43 58\n44 57\n45 56\n46 55\n47 54\n48 53\n49 52\n50 51"
},
{
"input": "4\n3 4 4 5",
"output": "1 4\n2 3"
},
{
"input": "4\n1 1 2 2",
"output": "1 4\n2 3"
},
{
"input": "4\n1 2 3 4",
"output": "1 4\n2 3"
}
] | 1,688,707,456
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 31
| 0
|
def cards(n, card_values):
target_sum = sum(card_values) // (n // 2)
player_1 = []
player_2 = []
for i in range(n-1):
for j in range(i+1, n):
card_sum = card_values[i] + card_values[j]
if card_sum == target_sum:
if (i+1 not in player_2) and (j+1 not in player_1):
player_1.append(i+1)
player_2.append(j+1)
break
return player_1, player_2
n = int(input())
card_values = list(map(int, input().split()))
p1, p2 = cards(n, card_values)
# zip_player = zip(p1, p2)
# for z in zip_player:
# print(z)
for i in range(len(p1)):
print(f"{p1[i]} {p2[i]}")
|
Title: Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cards (*n* is even) in the deck. Each card has a positive integer written on it. *n*<=/<=2 people will play new card game. At the beginning of the game each player gets two cards, each card is given to exactly one player.
Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100) — the number of cards in the deck. It is guaranteed that *n* is even.
The second line contains the sequence of *n* positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is equal to the number written on the *i*-th card.
Output Specification:
Print *n*<=/<=2 pairs of integers, the *i*-th pair denote the cards that should be given to the *i*-th player. Each card should be given to exactly one player. Cards are numbered in the order they appear in the input.
It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.
Demo Input:
['6\n1 5 7 4 4 3\n', '4\n10 10 10 10\n']
Demo Output:
['1 3\n6 2\n4 5\n', '1 2\n3 4\n']
Note:
In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.
In the second sample, all values *a*<sub class="lower-index">*i*</sub> are equal. Thus, any distribution is acceptable.
|
```python
def cards(n, card_values):
target_sum = sum(card_values) // (n // 2)
player_1 = []
player_2 = []
for i in range(n-1):
for j in range(i+1, n):
card_sum = card_values[i] + card_values[j]
if card_sum == target_sum:
if (i+1 not in player_2) and (j+1 not in player_1):
player_1.append(i+1)
player_2.append(j+1)
break
return player_1, player_2
n = int(input())
card_values = list(map(int, input().split()))
p1, p2 = cards(n, card_values)
# zip_player = zip(p1, p2)
# for z in zip_player:
# print(z)
for i in range(len(p1)):
print(f"{p1[i]} {p2[i]}")
```
| 0
|
|
363
|
B
|
Fence
|
PROGRAMMING
| 1,100
|
[
"brute force",
"dp"
] | null | null |
There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).
|
The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence.
|
Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them.
|
[
"7 3\n1 2 6 1 1 7 1\n"
] |
[
"3\n"
] |
In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.
| 1,000
|
[
{
"input": "7 3\n1 2 6 1 1 7 1",
"output": "3"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "2 1\n10 20",
"output": "1"
},
{
"input": "10 5\n1 2 3 1 2 2 3 1 4 5",
"output": "1"
},
{
"input": "10 2\n3 1 4 1 4 6 2 1 4 6",
"output": "7"
},
{
"input": "2 2\n20 10",
"output": "1"
},
{
"input": "2 1\n20 1",
"output": "2"
},
{
"input": "3 1\n1 2 3",
"output": "1"
},
{
"input": "3 1\n2 1 3",
"output": "2"
},
{
"input": "3 1\n3 2 1",
"output": "3"
},
{
"input": "3 2\n1 2 3",
"output": "1"
},
{
"input": "3 2\n3 2 1",
"output": "2"
},
{
"input": "3 3\n1 2 3",
"output": "1"
},
{
"input": "4 2\n9 8 11 7",
"output": "1"
},
{
"input": "4 2\n10 1 2 3",
"output": "2"
},
{
"input": "6 3\n56 56 56 2 1 2",
"output": "4"
},
{
"input": "8 3\n1 1 1 1 2 60 90 1",
"output": "1"
},
{
"input": "4 1\n1 5 2 2",
"output": "1"
},
{
"input": "4 2\n4 6 7 4",
"output": "1"
},
{
"input": "10 4\n1 1 1 4 4 4 4 4 4 3",
"output": "1"
},
{
"input": "6 3\n1 2 1 3 1 1",
"output": "1"
},
{
"input": "5 2\n100 100 100 1 1",
"output": "4"
}
] | 1,687,377,345
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 51
| 108
| 19,046,400
|
#author: sushmanth
from sys import stdin
input = stdin.readline
inp = lambda : list(map(int,input().split()))
def answer():
s = 0
for i in range(k):
s += a[i]
ans , index = s , 1
for i in range(k , n):
s -= a[i - k]
s += a[i]
if(ans > s):
ans = s
index = i - k + 2
return index
for T in range(1):
n , k = inp()
a = inp()
print(answer())
|
Title: Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).
Input Specification:
The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence.
Output Specification:
Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them.
Demo Input:
['7 3\n1 2 6 1 1 7 1\n']
Demo Output:
['3\n']
Note:
In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8.
|
```python
#author: sushmanth
from sys import stdin
input = stdin.readline
inp = lambda : list(map(int,input().split()))
def answer():
s = 0
for i in range(k):
s += a[i]
ans , index = s , 1
for i in range(k , n):
s -= a[i - k]
s += a[i]
if(ans > s):
ans = s
index = i - k + 2
return index
for T in range(1):
n , k = inp()
a = inp()
print(answer())
```
| 3
|
|
660
|
C
|
Hard Process
|
PROGRAMMING
| 1,600
|
[
"binary search",
"dp",
"two pointers"
] | null | null |
You are given an array *a* with *n* elements. Each element of *a* is either 0 or 1.
Let's denote the length of the longest subsegment of consecutive elements in *a*, consisting of only numbers one, as *f*(*a*). You can change no more than *k* zeroes to ones to maximize *f*(*a*).
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=3·105,<=0<=≤<=*k*<=≤<=*n*) — the number of elements in *a* and the parameter *k*.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1) — the elements of *a*.
|
On the first line print a non-negative integer *z* — the maximal value of *f*(*a*) after no more than *k* changes of zeroes to ones.
On the second line print *n* integers *a**j* — the elements of the array *a* after the changes.
If there are multiple answers, you can print any one of them.
|
[
"7 1\n1 0 0 1 1 0 1\n",
"10 2\n1 0 0 1 0 1 0 1 0 1\n"
] |
[
"4\n1 0 0 1 1 1 1\n",
"5\n1 0 0 1 1 1 1 1 0 1\n"
] |
none
| 0
|
[
{
"input": "7 1\n1 0 0 1 1 0 1",
"output": "4\n1 0 0 1 1 1 1"
},
{
"input": "10 2\n1 0 0 1 0 1 0 1 0 1",
"output": "5\n1 0 0 1 1 1 1 1 0 1"
},
{
"input": "1 0\n0",
"output": "0\n0"
},
{
"input": "1 0\n0",
"output": "0\n0"
},
{
"input": "7 0\n0 1 0 0 0 1 0",
"output": "1\n0 1 0 0 0 1 0"
},
{
"input": "7 2\n1 0 1 1 0 0 0",
"output": "5\n1 1 1 1 1 0 0"
},
{
"input": "7 3\n1 0 0 0 0 1 0",
"output": "4\n1 1 1 1 0 1 0"
},
{
"input": "3 0\n1 1 1",
"output": "3\n1 1 1"
},
{
"input": "1 0\n1",
"output": "1\n1"
},
{
"input": "2 0\n0 1",
"output": "1\n0 1"
},
{
"input": "1 1\n0",
"output": "1\n1"
},
{
"input": "3 1\n0 0 1",
"output": "2\n0 1 1"
},
{
"input": "2 0\n0 0",
"output": "0\n0 0"
},
{
"input": "3 1\n0 0 0",
"output": "1\n1 0 0"
},
{
"input": "7 7\n1 0 1 0 0 0 0",
"output": "7\n1 1 1 1 1 1 1"
},
{
"input": "1 1\n1",
"output": "1\n1"
},
{
"input": "3 0\n0 0 0",
"output": "0\n0 0 0"
},
{
"input": "7 3\n1 1 1 1 1 1 1",
"output": "7\n1 1 1 1 1 1 1"
},
{
"input": "10 3\n0 0 1 0 0 1 0 0 1 1",
"output": "6\n0 0 1 0 1 1 1 1 1 1"
},
{
"input": "2 1\n0 0",
"output": "1\n1 0"
},
{
"input": "2 2\n1 0",
"output": "2\n1 1"
}
] | 1,691,773,140
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
n = int(input())
m = {}
def fun(i, le, d, i1, a):
if (i, le, d, i1) in m:
return m[(i, le, d, i1)]
if i == n:
m[(i, le, d, i1)] = le
return le
if le == 0:
m[(i, le, d, i1)] = max(fun(i + 1, 1, 0, i, a), fun(i + 1, 0, 0, i + 1, a))
return m[(i, le, d, i1)]
if a[i] - a[i1] >= d:
m[(i, le, d, i1)] = max(fun(i + 1, le + 1, a[i] - a[i1], i, a), fun(i + 1, le, d, i1, a))
return m[(i, le, d, i1)]
m[(i, le, d, i1)] = fun(i + 1, le, d, i1, a)
return m[(i, le, d, i1)]
a = list(map(int, input().split()))
print(fun(0, 0, 0, 0, a))
|
Title: Hard Process
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a* with *n* elements. Each element of *a* is either 0 or 1.
Let's denote the length of the longest subsegment of consecutive elements in *a*, consisting of only numbers one, as *f*(*a*). You can change no more than *k* zeroes to ones to maximize *f*(*a*).
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=3·105,<=0<=≤<=*k*<=≤<=*n*) — the number of elements in *a* and the parameter *k*.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1) — the elements of *a*.
Output Specification:
On the first line print a non-negative integer *z* — the maximal value of *f*(*a*) after no more than *k* changes of zeroes to ones.
On the second line print *n* integers *a**j* — the elements of the array *a* after the changes.
If there are multiple answers, you can print any one of them.
Demo Input:
['7 1\n1 0 0 1 1 0 1\n', '10 2\n1 0 0 1 0 1 0 1 0 1\n']
Demo Output:
['4\n1 0 0 1 1 1 1\n', '5\n1 0 0 1 1 1 1 1 0 1\n']
Note:
none
|
```python
n = int(input())
m = {}
def fun(i, le, d, i1, a):
if (i, le, d, i1) in m:
return m[(i, le, d, i1)]
if i == n:
m[(i, le, d, i1)] = le
return le
if le == 0:
m[(i, le, d, i1)] = max(fun(i + 1, 1, 0, i, a), fun(i + 1, 0, 0, i + 1, a))
return m[(i, le, d, i1)]
if a[i] - a[i1] >= d:
m[(i, le, d, i1)] = max(fun(i + 1, le + 1, a[i] - a[i1], i, a), fun(i + 1, le, d, i1, a))
return m[(i, le, d, i1)]
m[(i, le, d, i1)] = fun(i + 1, le, d, i1, a)
return m[(i, le, d, i1)]
a = list(map(int, input().split()))
print(fun(0, 0, 0, 0, a))
```
| -1
|
|
474
|
B
|
Worms
|
PROGRAMMING
| 1,200
|
[
"binary search",
"implementation"
] | null | null |
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole *n* ordered piles of worms such that *i*-th pile contains *a**i* worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to *a*1, worms in second pile are labeled with numbers *a*1<=+<=1 to *a*1<=+<=*a*2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), the number of piles.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=103, *a*1<=+<=*a*2<=+<=...<=+<=*a**n*<=≤<=106), where *a**i* is the number of worms in the *i*-th pile.
The third line contains single integer *m* (1<=≤<=*m*<=≤<=105), the number of juicy worms said by Marmot.
The fourth line contains *m* integers *q*1,<=*q*2,<=...,<=*q**m* (1<=≤<=*q**i*<=≤<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*), the labels of the juicy worms.
|
Print *m* lines to the standard output. The *i*-th line should contain an integer, representing the number of the pile where the worm labeled with the number *q**i* is.
|
[
"5\n2 7 3 4 9\n3\n1 25 11\n"
] |
[
"1\n5\n3\n"
] |
For the sample input:
- The worms with labels from [1, 2] are in the first pile. - The worms with labels from [3, 9] are in the second pile. - The worms with labels from [10, 12] are in the third pile. - The worms with labels from [13, 16] are in the fourth pile. - The worms with labels from [17, 25] are in the fifth pile.
| 1,000
|
[
{
"input": "5\n2 7 3 4 9\n3\n1 25 11",
"output": "1\n5\n3"
}
] | 1,689,325,429
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 77
| 2,867,200
|
import numpy as np
n = int(input())
nlist = list(map(int,input().split()[:n]))
n_array = np.array(nlist)
# nlist = [[32,4,3],[32,43,4],[4,78,9],[4,3,4]]
m = int(input())
mlist = list(map(int,input().split()[:m]))
m_array = np.array(mlist)
for i in range(m):
sum = 0
tar = m_array[i]
for j in range(n):
sum = sum + n_array[j]
if sum >= tar :
print(j+1)
break
|
Title: Worms
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole *n* ordered piles of worms such that *i*-th pile contains *a**i* worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to *a*1, worms in second pile are labeled with numbers *a*1<=+<=1 to *a*1<=+<=*a*2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), the number of piles.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=103, *a*1<=+<=*a*2<=+<=...<=+<=*a**n*<=≤<=106), where *a**i* is the number of worms in the *i*-th pile.
The third line contains single integer *m* (1<=≤<=*m*<=≤<=105), the number of juicy worms said by Marmot.
The fourth line contains *m* integers *q*1,<=*q*2,<=...,<=*q**m* (1<=≤<=*q**i*<=≤<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*), the labels of the juicy worms.
Output Specification:
Print *m* lines to the standard output. The *i*-th line should contain an integer, representing the number of the pile where the worm labeled with the number *q**i* is.
Demo Input:
['5\n2 7 3 4 9\n3\n1 25 11\n']
Demo Output:
['1\n5\n3\n']
Note:
For the sample input:
- The worms with labels from [1, 2] are in the first pile. - The worms with labels from [3, 9] are in the second pile. - The worms with labels from [10, 12] are in the third pile. - The worms with labels from [13, 16] are in the fourth pile. - The worms with labels from [17, 25] are in the fifth pile.
|
```python
import numpy as np
n = int(input())
nlist = list(map(int,input().split()[:n]))
n_array = np.array(nlist)
# nlist = [[32,4,3],[32,43,4],[4,78,9],[4,3,4]]
m = int(input())
mlist = list(map(int,input().split()[:m]))
m_array = np.array(mlist)
for i in range(m):
sum = 0
tar = m_array[i]
for j in range(n):
sum = sum + n_array[j]
if sum >= tar :
print(j+1)
break
```
| -1
|
|
777
|
B
|
Game of Credit Cards
|
PROGRAMMING
| 1,300
|
[
"data structures",
"dp",
"greedy",
"sortings"
] | null | null |
After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards.
Rules of this game are simple: each player bring his favourite *n*-digit credit card. Then both players name the digits written on their cards one by one. If two digits are not equal, then the player, whose digit is smaller gets a flick (knock in the forehead usually made with a forefinger) from the other player. For example, if *n*<==<=3, Sherlock's card is 123 and Moriarty's card has number 321, first Sherlock names 1 and Moriarty names 3 so Sherlock gets a flick. Then they both digit 2 so no one gets a flick. Finally, Sherlock names 3, while Moriarty names 1 and gets a flick.
Of course, Sherlock will play honestly naming digits one by one in the order they are given, while Moriary, as a true villain, plans to cheat. He is going to name his digits in some other order (however, he is not going to change the overall number of occurences of each digit). For example, in case above Moriarty could name 1, 2, 3 and get no flicks at all, or he can name 2, 3 and 1 to give Sherlock two flicks.
Your goal is to find out the minimum possible number of flicks Moriarty will get (no one likes flicks) and the maximum possible number of flicks Sherlock can get from Moriarty. Note, that these two goals are different and the optimal result may be obtained by using different strategies.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of digits in the cards Sherlock and Moriarty are going to use.
The second line contains *n* digits — Sherlock's credit card number.
The third line contains *n* digits — Moriarty's credit card number.
|
First print the minimum possible number of flicks Moriarty will get. Then print the maximum possible number of flicks that Sherlock can get from Moriarty.
|
[
"3\n123\n321\n",
"2\n88\n00\n"
] |
[
"0\n2\n",
"2\n0\n"
] |
First sample is elaborated in the problem statement. In the second sample, there is no way Moriarty can avoid getting two flicks.
| 1,000
|
[
{
"input": "3\n123\n321",
"output": "0\n2"
},
{
"input": "2\n88\n00",
"output": "2\n0"
},
{
"input": "1\n4\n5",
"output": "0\n1"
},
{
"input": "1\n8\n7",
"output": "1\n0"
},
{
"input": "2\n55\n55",
"output": "0\n0"
},
{
"input": "3\n534\n432",
"output": "1\n1"
},
{
"input": "3\n486\n024",
"output": "2\n0"
},
{
"input": "5\n22222\n22222",
"output": "0\n0"
},
{
"input": "5\n72471\n05604",
"output": "2\n3"
},
{
"input": "5\n72471\n72471",
"output": "0\n3"
},
{
"input": "5\n72471\n41772",
"output": "0\n3"
},
{
"input": "8\n99999999\n99999999",
"output": "0\n0"
},
{
"input": "8\n01234567\n01234567",
"output": "0\n7"
},
{
"input": "8\n07070707\n76543210",
"output": "3\n4"
},
{
"input": "8\n88888888\n98769876",
"output": "4\n2"
},
{
"input": "8\n23456789\n01234567",
"output": "2\n5"
},
{
"input": "5\n11222\n22111",
"output": "1\n2"
},
{
"input": "9\n777777777\n777777777",
"output": "0\n0"
},
{
"input": "9\n353589343\n280419388",
"output": "3\n5"
},
{
"input": "10\n8104381743\n8104381743",
"output": "0\n8"
},
{
"input": "10\n8104381743\n8418134730",
"output": "0\n8"
},
{
"input": "10\n1111122222\n2222211111",
"output": "0\n5"
},
{
"input": "100\n6317494220822818719411404030346382869796138932712461187067886456209071515048745855973784223939110171\n6017563370120161528504797580620647099370814387367549926215651181421345104088127581963662589996747937",
"output": "2\n86"
},
{
"input": "200\n89017372169770060638462517044634884577600285180365791227828841983192945639410766634962149214165390392679430585962408483864914959904870801002680423351144765728677610509881245693518626747873607530273392\n34804385048475325130258121398275821439066233953856051421626677848150660724595847484466293487857039579239864150199164135152374201294909986469345076423004279983866383035554588630496127880705497919788390",
"output": "6\n175"
},
{
"input": "3\n112\n111",
"output": "1\n0"
},
{
"input": "5\n66666\n11119",
"output": "4\n1"
},
{
"input": "3\n232\n123",
"output": "1\n1"
}
] | 1,589,601,472
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 14
| 529
| 6,963,200
|
n=int(input())
s=input()
m=input()
mt=[]
st=[]
for i in range(0,n):
countm=0
for j in range(0,n):
if(s[i]>m[j]):
countm+=1
mt.append(countm)
for i in range(0,n):
counts=0
for j in range(0,n):
if(m[i]>s[j]):
counts+=1
st.append(counts)
print(min(mt))
print(max(st))
|
Title: Game of Credit Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards.
Rules of this game are simple: each player bring his favourite *n*-digit credit card. Then both players name the digits written on their cards one by one. If two digits are not equal, then the player, whose digit is smaller gets a flick (knock in the forehead usually made with a forefinger) from the other player. For example, if *n*<==<=3, Sherlock's card is 123 and Moriarty's card has number 321, first Sherlock names 1 and Moriarty names 3 so Sherlock gets a flick. Then they both digit 2 so no one gets a flick. Finally, Sherlock names 3, while Moriarty names 1 and gets a flick.
Of course, Sherlock will play honestly naming digits one by one in the order they are given, while Moriary, as a true villain, plans to cheat. He is going to name his digits in some other order (however, he is not going to change the overall number of occurences of each digit). For example, in case above Moriarty could name 1, 2, 3 and get no flicks at all, or he can name 2, 3 and 1 to give Sherlock two flicks.
Your goal is to find out the minimum possible number of flicks Moriarty will get (no one likes flicks) and the maximum possible number of flicks Sherlock can get from Moriarty. Note, that these two goals are different and the optimal result may be obtained by using different strategies.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of digits in the cards Sherlock and Moriarty are going to use.
The second line contains *n* digits — Sherlock's credit card number.
The third line contains *n* digits — Moriarty's credit card number.
Output Specification:
First print the minimum possible number of flicks Moriarty will get. Then print the maximum possible number of flicks that Sherlock can get from Moriarty.
Demo Input:
['3\n123\n321\n', '2\n88\n00\n']
Demo Output:
['0\n2\n', '2\n0\n']
Note:
First sample is elaborated in the problem statement. In the second sample, there is no way Moriarty can avoid getting two flicks.
|
```python
n=int(input())
s=input()
m=input()
mt=[]
st=[]
for i in range(0,n):
countm=0
for j in range(0,n):
if(s[i]>m[j]):
countm+=1
mt.append(countm)
for i in range(0,n):
counts=0
for j in range(0,n):
if(m[i]>s[j]):
counts+=1
st.append(counts)
print(min(mt))
print(max(st))
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,666,654,234
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 2
| 124
| 0
|
m, n = map(int, input().split())
a = max(m, n)
b = min(m, n)
ans = 0
while a > 0 and b > 0:
num = a // 2
ans += num * 2
a -= 2 * num
b -= num
m = a
n = b
a = max(m, n)
b = min(m, n)
# print(f"ans={ans}, a={a}, b={b} ")
print(ans)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
m, n = map(int, input().split())
a = max(m, n)
b = min(m, n)
ans = 0
while a > 0 and b > 0:
num = a // 2
ans += num * 2
a -= 2 * num
b -= num
m = a
n = b
a = max(m, n)
b = min(m, n)
# print(f"ans={ans}, a={a}, b={b} ")
print(ans)
```
| 0
|
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"
}
] | 1,632,933,109
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 37
| 108
| 20,172,800
|
n,k=map(int,input().split())
print(((n+1)//k+((n+1)%k>0))*k)
|
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
n,k=map(int,input().split())
print(((n+1)//k+((n+1)%k>0))*k)
```
| 3
|
|
595
|
A
|
Vitaly and Night
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment.
Vitaly sees a building of *n* floors and 2·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on.
Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping.
|
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of floors in the house and the number of flats on each floor respectively.
Next *n* lines describe the floors from top to bottom and contain 2·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'.
|
Print a single integer — the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping.
|
[
"2 2\n0 0 0 1\n1 0 1 1\n",
"1 3\n1 1 0 1 0 0\n"
] |
[
"3\n",
"2\n"
] |
In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off.
In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
| 500
|
[
{
"input": "2 2\n0 0 0 1\n1 0 1 1",
"output": "3"
},
{
"input": "1 3\n1 1 0 1 0 0",
"output": "2"
},
{
"input": "3 3\n1 1 1 1 1 1\n1 1 0 1 1 0\n1 0 0 0 1 1",
"output": "8"
},
{
"input": "1 5\n1 0 1 1 1 0 1 1 1 1",
"output": "5"
},
{
"input": "1 100\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 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 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": "99"
},
{
"input": "1 100\n0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 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 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 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 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 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "6"
},
{
"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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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": "0"
},
{
"input": "100 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n0 1\n1 1\n1 1\n1 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "100"
},
{
"input": "100 1\n0 0\n0 0\n0 1\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\n0 0\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 1\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\n0 0\n0 0\n0 0\n0 1\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\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\n0 0\n0 0\n1 0",
"output": "8"
},
{
"input": "100 1\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\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\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\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\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\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\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "1 1\n0 0",
"output": "0"
},
{
"input": "1 1\n0 1",
"output": "1"
},
{
"input": "1 1\n1 0",
"output": "1"
},
{
"input": "1 1\n1 1",
"output": "1"
}
] | 1,589,809,169
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 62
| 6,656,000
|
houselevers, houses = map(int, input().split())
counter = 0
list1 = []
for i in range(houselevers):
list1.extend([int(h) for h in input().split()])
for b in range(0, houses*2, 2):
if list1[b] + list1[b+1] > 0:
counter += 1
print(counter)
|
Title: Vitaly and Night
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly decided to look which windows are lit in the house he was passing by at that moment.
Vitaly sees a building of *n* floors and 2·*m* windows on each floor. On each floor there are *m* flats numbered from 1 to *m*, and two consecutive windows correspond to each flat. If we number the windows from 1 to 2·*m* from left to right, then the *j*-th flat of the *i*-th floor has windows 2·*j*<=-<=1 and 2·*j* in the corresponding row of windows (as usual, floors are enumerated from the bottom). Vitaly thinks that people in the flat aren't sleeping at that moment if at least one of the windows corresponding to this flat has lights on.
Given the information about the windows of the given house, your task is to calculate the number of flats where, according to Vitaly, people aren't sleeping.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of floors in the house and the number of flats on each floor respectively.
Next *n* lines describe the floors from top to bottom and contain 2·*m* characters each. If the *i*-th window of the given floor has lights on, then the *i*-th character of this line is '1', otherwise it is '0'.
Output Specification:
Print a single integer — the number of flats that have lights on in at least one window, that is, the flats where, according to Vitaly, people aren't sleeping.
Demo Input:
['2 2\n0 0 0 1\n1 0 1 1\n', '1 3\n1 1 0 1 0 0\n']
Demo Output:
['3\n', '2\n']
Note:
In the first test case the house has two floors, two flats on each floor. That is, in total there are 4 flats. The light isn't on only on the second floor in the left flat. That is, in both rooms of the flat the light is off.
In the second test case the house has one floor and the first floor has three flats. The light is on in the leftmost flat (in both windows) and in the middle flat (in one window). In the right flat the light is off.
|
```python
houselevers, houses = map(int, input().split())
counter = 0
list1 = []
for i in range(houselevers):
list1.extend([int(h) for h in input().split()])
for b in range(0, houses*2, 2):
if list1[b] + list1[b+1] > 0:
counter += 1
print(counter)
```
| 0
|
|
293
|
A
|
Weird Game
|
PROGRAMMING
| 1,500
|
[
"games",
"greedy"
] | null | null |
Yaroslav, Andrey and Roman can play cubes for hours and hours. But the game is for three, so when Roman doesn't show up, Yaroslav and Andrey play another game.
Roman leaves a word for each of them. Each word consists of 2·*n* binary characters "0" or "1". After that the players start moving in turns. Yaroslav moves first. During a move, a player must choose an integer from 1 to 2·*n*, which hasn't been chosen by anybody up to that moment. Then the player takes a piece of paper and writes out the corresponding character from his string.
Let's represent Yaroslav's word as *s*<==<=*s*1*s*2... *s*2*n*. Similarly, let's represent Andrey's word as *t*<==<=*t*1*t*2... *t*2*n*. Then, if Yaroslav choose number *k* during his move, then he is going to write out character *s**k* on the piece of paper. Similarly, if Andrey choose number *r* during his move, then he is going to write out character *t**r* on the piece of paper.
The game finishes when no player can make a move. After the game is over, Yaroslav makes some integer from the characters written on his piece of paper (Yaroslav can arrange these characters as he wants). Andrey does the same. The resulting numbers can contain leading zeroes. The person with the largest number wins. If the numbers are equal, the game ends with a draw.
You are given two strings *s* and *t*. Determine the outcome of the game provided that Yaroslav and Andrey play optimally well.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=106). The second line contains string *s* — Yaroslav's word. The third line contains string *t* — Andrey's word.
It is guaranteed that both words consist of 2·*n* characters "0" and "1".
|
Print "First", if both players play optimally well and Yaroslav wins. If Andrey wins, print "Second" and if the game ends with a draw, print "Draw". Print the words without the quotes.
|
[
"2\n0111\n0001\n",
"3\n110110\n001001\n",
"3\n111000\n000111\n",
"4\n01010110\n00101101\n",
"4\n01100000\n10010011\n"
] |
[
"First\n",
"First\n",
"Draw\n",
"First\n",
"Second\n"
] |
none
| 500
|
[
{
"input": "2\n0111\n0001",
"output": "First"
},
{
"input": "3\n110110\n001001",
"output": "First"
},
{
"input": "3\n111000\n000111",
"output": "Draw"
},
{
"input": "4\n01010110\n00101101",
"output": "First"
},
{
"input": "4\n01100000\n10010011",
"output": "Second"
},
{
"input": "4\n10001001\n10101101",
"output": "Draw"
},
{
"input": "3\n000000\n000100",
"output": "Draw"
},
{
"input": "2\n0000\n1110",
"output": "Second"
},
{
"input": "4\n11111111\n10100110",
"output": "First"
},
{
"input": "4\n10100111\n11011000",
"output": "First"
},
{
"input": "4\n00101011\n11110100",
"output": "Draw"
},
{
"input": "4\n11000011\n00111100",
"output": "Draw"
},
{
"input": "4\n11101111\n01000110",
"output": "First"
},
{
"input": "4\n01110111\n00101110",
"output": "First"
},
{
"input": "4\n11011111\n10110110",
"output": "First"
},
{
"input": "4\n01101010\n11111110",
"output": "Second"
},
{
"input": "4\n01111111\n10011001",
"output": "First"
},
{
"input": "4\n01010100\n10011111",
"output": "Second"
},
{
"input": "4\n01111011\n01001011",
"output": "First"
},
{
"input": "4\n11011010\n11011001",
"output": "Draw"
},
{
"input": "4\n11001101\n10001010",
"output": "First"
},
{
"input": "4\n01101111\n10111101",
"output": "Draw"
},
{
"input": "4\n10111100\n00000101",
"output": "First"
},
{
"input": "4\n01111000\n11111010",
"output": "Second"
},
{
"input": "4\n11001100\n00000111",
"output": "First"
},
{
"input": "4\n01110111\n10101101",
"output": "First"
},
{
"input": "4\n01001000\n11111100",
"output": "Second"
},
{
"input": "4\n01011011\n01010010",
"output": "First"
},
{
"input": "4\n00101101\n01001001",
"output": "First"
},
{
"input": "4\n00110110\n10000100",
"output": "First"
},
{
"input": "4\n10010000\n01000110",
"output": "Draw"
},
{
"input": "4\n00000100\n10001111",
"output": "Second"
},
{
"input": "4\n01110100\n01110100",
"output": "Draw"
},
{
"input": "4\n11000001\n11010001",
"output": "Draw"
},
{
"input": "4\n11001000\n00011000",
"output": "First"
},
{
"input": "4\n10110011\n01011111",
"output": "Draw"
},
{
"input": "4\n10000100\n11010100",
"output": "Second"
},
{
"input": "4\n01011011\n10101110",
"output": "Draw"
},
{
"input": "10\n00000000000111111111\n00000000011111111111",
"output": "Draw"
},
{
"input": "1\n11\n11",
"output": "Draw"
},
{
"input": "1\n11\n00",
"output": "First"
},
{
"input": "1\n00\n01",
"output": "Draw"
},
{
"input": "2\n0111\n1001",
"output": "First"
},
{
"input": "1\n01\n11",
"output": "Draw"
}
] | 1,603,271,980
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 8
| 2,000
| 5,427,200
|
n = int(input())
s=input()
t=input()
s_1 = 0
t_1 = 0
for i in range(2*n):
if s[i]=='1':
s_1+=1
if t[i]=='1':
t_1+=1
first = 0
second = 0
r = 0
for i in range(2*n):
if s[i]=='1' and t[i]=='0':
first+=1
elif s[i]=='0' and t[i]=='1':
second+=1
elif s[i]=='1' and t[i]=='1':
r+=1
if r%2==1:
first+=1
if s_1>t_1:
print('First')
elif first==second or first==second-1:
print('Draw')
elif first>second:
print('First')
elif first < second:
print('Second')
|
Title: Weird Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yaroslav, Andrey and Roman can play cubes for hours and hours. But the game is for three, so when Roman doesn't show up, Yaroslav and Andrey play another game.
Roman leaves a word for each of them. Each word consists of 2·*n* binary characters "0" or "1". After that the players start moving in turns. Yaroslav moves first. During a move, a player must choose an integer from 1 to 2·*n*, which hasn't been chosen by anybody up to that moment. Then the player takes a piece of paper and writes out the corresponding character from his string.
Let's represent Yaroslav's word as *s*<==<=*s*1*s*2... *s*2*n*. Similarly, let's represent Andrey's word as *t*<==<=*t*1*t*2... *t*2*n*. Then, if Yaroslav choose number *k* during his move, then he is going to write out character *s**k* on the piece of paper. Similarly, if Andrey choose number *r* during his move, then he is going to write out character *t**r* on the piece of paper.
The game finishes when no player can make a move. After the game is over, Yaroslav makes some integer from the characters written on his piece of paper (Yaroslav can arrange these characters as he wants). Andrey does the same. The resulting numbers can contain leading zeroes. The person with the largest number wins. If the numbers are equal, the game ends with a draw.
You are given two strings *s* and *t*. Determine the outcome of the game provided that Yaroslav and Andrey play optimally well.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=106). The second line contains string *s* — Yaroslav's word. The third line contains string *t* — Andrey's word.
It is guaranteed that both words consist of 2·*n* characters "0" and "1".
Output Specification:
Print "First", if both players play optimally well and Yaroslav wins. If Andrey wins, print "Second" and if the game ends with a draw, print "Draw". Print the words without the quotes.
Demo Input:
['2\n0111\n0001\n', '3\n110110\n001001\n', '3\n111000\n000111\n', '4\n01010110\n00101101\n', '4\n01100000\n10010011\n']
Demo Output:
['First\n', 'First\n', 'Draw\n', 'First\n', 'Second\n']
Note:
none
|
```python
n = int(input())
s=input()
t=input()
s_1 = 0
t_1 = 0
for i in range(2*n):
if s[i]=='1':
s_1+=1
if t[i]=='1':
t_1+=1
first = 0
second = 0
r = 0
for i in range(2*n):
if s[i]=='1' and t[i]=='0':
first+=1
elif s[i]=='0' and t[i]=='1':
second+=1
elif s[i]=='1' and t[i]=='1':
r+=1
if r%2==1:
first+=1
if s_1>t_1:
print('First')
elif first==second or first==second-1:
print('Draw')
elif first>second:
print('First')
elif first < second:
print('Second')
```
| 0
|
|
545
|
C
|
Woodcutters
|
PROGRAMMING
| 1,500
|
[
"dp",
"greedy"
] | null | null |
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate.
|
Print a single number — the maximum number of trees that you can cut down by the given rules.
|
[
"5\n1 2\n2 1\n5 10\n10 9\n19 1\n",
"5\n1 2\n2 1\n5 10\n10 9\n20 1\n"
] |
[
"3\n",
"4\n"
] |
In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
| 1,750
|
[
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n19 1",
"output": "3"
},
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n20 1",
"output": "4"
},
{
"input": "4\n10 4\n15 1\n19 3\n20 1",
"output": "4"
},
{
"input": "35\n1 7\n3 11\n6 12\n7 6\n8 5\n9 11\n15 3\n16 10\n22 2\n23 3\n25 7\n27 3\n34 5\n35 10\n37 3\n39 4\n40 5\n41 1\n44 1\n47 7\n48 11\n50 6\n52 5\n57 2\n58 7\n60 4\n62 1\n67 3\n68 12\n69 8\n70 1\n71 5\n72 5\n73 6\n74 4",
"output": "10"
},
{
"input": "40\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1",
"output": "2"
},
{
"input": "67\n1 1\n3 8\n4 10\n7 8\n9 2\n10 1\n11 5\n12 8\n13 4\n16 6\n18 3\n19 3\n22 5\n24 6\n27 5\n28 3\n29 3\n30 5\n32 5\n33 10\n34 7\n35 8\n36 5\n41 3\n42 2\n43 5\n46 4\n48 4\n49 9\n52 4\n53 9\n55 1\n56 4\n59 7\n68 7\n69 4\n71 9\n72 10\n74 5\n76 4\n77 9\n80 7\n81 9\n82 5\n83 5\n84 9\n85 7\n86 9\n87 4\n88 7\n89 10\n90 3\n91 5\n92 10\n93 5\n94 8\n95 4\n96 2\n97 10\n98 1\n99 3\n100 1\n101 5\n102 4\n103 8\n104 8\n105 8",
"output": "5"
},
{
"input": "1\n1000000000 1000000000",
"output": "1"
},
{
"input": "10\n7 12\n10 2\n12 2\n15 1\n19 2\n20 1\n53 25\n63 10\n75 12\n87 1",
"output": "9"
},
{
"input": "3\n1 1\n1000 1000\n1000000000 1000000000",
"output": "3"
},
{
"input": "2\n1 999999999\n1000000000 1000000000",
"output": "2"
},
{
"input": "10\n999999900 1000000000\n999999901 1000000000\n999999902 1000000000\n999999903 1000000000\n999999904 1000000000\n999999905 1000000000\n999999906 1000000000\n999999907 1000000000\n999999908 1000000000\n999999909 1000000000",
"output": "2"
},
{
"input": "2\n100000000 1000000000\n1000000000 1000000000",
"output": "2"
}
] | 1,679,820,327
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 67
| 108
| 3,993,600
|
from sys import stdin,stdout
input,print = stdin.readline,stdout.write
n=int(input())
back=-10**9
prev=-1
count=0
for i in range(n):
pos,curr=map(int,input().split())
#print(str(back),str(prev),str(pos),str(curr),str(count))
dist=pos-back
if prev==-1:
if curr<dist:
count+=1
else:
prev=curr
else:
if prev+curr<dist:
count+=2
prev=-1
elif prev<dist:
count+=1
prev=curr
elif curr<dist:
count+=1
prev=-1
else:
prev=curr
back=pos
if prev!=-1:
count+=1
print(str(count))
|
Title: Woodcutters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate.
Output Specification:
Print a single number — the maximum number of trees that you can cut down by the given rules.
Demo Input:
['5\n1 2\n2 1\n5 10\n10 9\n19 1\n', '5\n1 2\n2 1\n5 10\n10 9\n20 1\n']
Demo Output:
['3\n', '4\n']
Note:
In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
|
```python
from sys import stdin,stdout
input,print = stdin.readline,stdout.write
n=int(input())
back=-10**9
prev=-1
count=0
for i in range(n):
pos,curr=map(int,input().split())
#print(str(back),str(prev),str(pos),str(curr),str(count))
dist=pos-back
if prev==-1:
if curr<dist:
count+=1
else:
prev=curr
else:
if prev+curr<dist:
count+=2
prev=-1
elif prev<dist:
count+=1
prev=curr
elif curr<dist:
count+=1
prev=-1
else:
prev=curr
back=pos
if prev!=-1:
count+=1
print(str(count))
```
| 3
|
|
371
|
A
|
K-Periodic Array
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation",
"math"
] | null | null |
This task will exclusively concentrate only on the arrays where all elements equal 1 and/or 2.
Array *a* is *k*-period if its length is divisible by *k* and there is such array *b* of length *k*, that *a* is represented by array *b* written exactly times consecutively. In other words, array *a* is *k*-periodic, if it has period of length *k*.
For example, any array is *n*-periodic, where *n* is the array length. Array [2,<=1,<=2,<=1,<=2,<=1] is at the same time 2-periodic and 6-periodic and array [1,<=2,<=1,<=1,<=2,<=1,<=1,<=2,<=1] is at the same time 3-periodic and 9-periodic.
For the given array *a*, consisting only of numbers one and two, find the minimum number of elements to change to make the array *k*-periodic. If the array already is *k*-periodic, then the required value equals 0.
|
The first line of the input contains a pair of integers *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100), where *n* is the length of the array and the value *n* is divisible by *k*. The second line contains the sequence of elements of the given array *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=2), *a**i* is the *i*-th element of the array.
|
Print the minimum number of array elements we need to change to make the array *k*-periodic. If the array already is *k*-periodic, then print 0.
|
[
"6 2\n2 1 2 2 2 1\n",
"8 4\n1 1 2 1 1 1 2 1\n",
"9 3\n2 1 1 1 2 1 1 1 2\n"
] |
[
"1\n",
"0\n",
"3\n"
] |
In the first sample it is enough to change the fourth element from 2 to 1, then the array changes to [2, 1, 2, 1, 2, 1].
In the second sample, the given array already is 4-periodic.
In the third sample it is enough to replace each occurrence of number two by number one. In this case the array will look as [1, 1, 1, 1, 1, 1, 1, 1, 1] — this array is simultaneously 1-, 3- and 9-periodic.
| 500
|
[
{
"input": "6 2\n2 1 2 2 2 1",
"output": "1"
},
{
"input": "8 4\n1 1 2 1 1 1 2 1",
"output": "0"
},
{
"input": "9 3\n2 1 1 1 2 1 1 1 2",
"output": "3"
},
{
"input": "1 1\n2",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "0"
},
{
"input": "2 2\n2 2",
"output": "0"
},
{
"input": "100 1\n1 2 1 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 1 2 1 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 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "8"
},
{
"input": "2 1\n1 2",
"output": "1"
},
{
"input": "2 2\n2 1",
"output": "0"
},
{
"input": "3 1\n2 1 2",
"output": "1"
},
{
"input": "3 3\n1 2 1",
"output": "0"
},
{
"input": "4 2\n2 1 2 2",
"output": "1"
},
{
"input": "10 2\n2 2 2 1 1 2 2 2 2 1",
"output": "3"
},
{
"input": "10 5\n2 2 1 2 1 1 2 1 1 1",
"output": "2"
},
{
"input": "20 4\n2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 1 2",
"output": "0"
},
{
"input": "20 5\n2 2 1 1 1 2 1 1 1 1 2 2 1 1 2 2 2 1 1 2",
"output": "3"
},
{
"input": "20 10\n1 2 2 2 2 1 1 1 2 1 1 2 2 2 2 1 2 2 2 1",
"output": "2"
},
{
"input": "100 2\n2 2 1 2 1 2 1 2 1 2 1 2 1 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 1 2 1 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2",
"output": "5"
},
{
"input": "100 4\n1 2 1 1 1 2 1 1 1 2 1 1 1 2 1 1 1 2 2 2 1 2 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 2 1 1 1 2 2 1 1 1 1 1 1 2 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 2 1 1 1 2 1 1 1 2 1 1 1 2 2 1 1 2 1 1 1 2 1 2 1 2 1 1 1 2 1 1 1 2 1 1",
"output": "8"
},
{
"input": "100 5\n2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 1 2 2 2 1 1 2 1 2 2 2 2 2 2 2 1 2 2 2",
"output": "16"
},
{
"input": "100 10\n2 1 1 1 1 2 2 2 1 1 2 1 1 2 1 2 1 2 1 1 2 1 1 1 1 2 1 2 1 1 2 1 1 1 1 2 2 2 1 1 2 1 1 1 1 2 1 2 1 1 2 1 1 1 1 2 1 2 2 1 2 1 1 1 1 2 1 2 1 1 2 1 2 1 1 2 1 2 1 1 2 1 1 1 1 2 1 2 1 1 2 1 1 1 2 2 1 2 1 1",
"output": "6"
},
{
"input": "100 20\n2 2 2 1 1 2 1 2 1 2 1 1 2 2 2 2 2 1 2 1 2 2 2 2 1 2 1 2 1 1 1 1 2 2 2 2 1 2 1 1 2 2 2 2 1 2 1 2 1 2 1 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 1 2 1 1 1 1 2 2 2 2 2 1 1 2 2 1 2 2 1 2 1 2 1 2 1 1 2 2 1 2 2 1 1 1",
"output": "13"
},
{
"input": "100 25\n2 2 1 2 2 2 2 2 1 2 2 1 2 1 1 2 1 2 1 2 2 2 1 2 2 2 1 1 2 1 2 1 2 1 2 2 1 2 1 1 2 2 2 1 2 2 1 2 2 2 2 1 1 2 1 2 2 1 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 1 2 1 1 2 2 2 2 2 1 2 2 1 1 2 1 2 2 2 1 2 2 2 2 2 2",
"output": "15"
},
{
"input": "100 10\n2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 1 2 1 2 1 2 2 2 2 2 1 2 1 2 1",
"output": "0"
}
] | 1,661,538,055
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 24
| 46
| 0
|
def contarMudancas(n, k):
vetor = list(map(int, input().split(' ')))
cont1 = []
cont2 = []
for i in range(k):
cont1.append(0)
cont2.append(0)
for i in range(n):
if vetor[i] == 1:
cont1[i % k] += 1
else:
cont2[i % k] += 1
minMudancas = 0
for i in range(k):
if cont1[i] >= cont2[i]:
minMudancas += cont2[i]
else:
minMudancas += cont1[i]
return minMudancas
if __name__ == "__main__":
nk = list(map(int, input().split(' ')))
print(contarMudancas(nk[0], nk[1]))
|
Title: K-Periodic Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This task will exclusively concentrate only on the arrays where all elements equal 1 and/or 2.
Array *a* is *k*-period if its length is divisible by *k* and there is such array *b* of length *k*, that *a* is represented by array *b* written exactly times consecutively. In other words, array *a* is *k*-periodic, if it has period of length *k*.
For example, any array is *n*-periodic, where *n* is the array length. Array [2,<=1,<=2,<=1,<=2,<=1] is at the same time 2-periodic and 6-periodic and array [1,<=2,<=1,<=1,<=2,<=1,<=1,<=2,<=1] is at the same time 3-periodic and 9-periodic.
For the given array *a*, consisting only of numbers one and two, find the minimum number of elements to change to make the array *k*-periodic. If the array already is *k*-periodic, then the required value equals 0.
Input Specification:
The first line of the input contains a pair of integers *n*, *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100), where *n* is the length of the array and the value *n* is divisible by *k*. The second line contains the sequence of elements of the given array *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=2), *a**i* is the *i*-th element of the array.
Output Specification:
Print the minimum number of array elements we need to change to make the array *k*-periodic. If the array already is *k*-periodic, then print 0.
Demo Input:
['6 2\n2 1 2 2 2 1\n', '8 4\n1 1 2 1 1 1 2 1\n', '9 3\n2 1 1 1 2 1 1 1 2\n']
Demo Output:
['1\n', '0\n', '3\n']
Note:
In the first sample it is enough to change the fourth element from 2 to 1, then the array changes to [2, 1, 2, 1, 2, 1].
In the second sample, the given array already is 4-periodic.
In the third sample it is enough to replace each occurrence of number two by number one. In this case the array will look as [1, 1, 1, 1, 1, 1, 1, 1, 1] — this array is simultaneously 1-, 3- and 9-periodic.
|
```python
def contarMudancas(n, k):
vetor = list(map(int, input().split(' ')))
cont1 = []
cont2 = []
for i in range(k):
cont1.append(0)
cont2.append(0)
for i in range(n):
if vetor[i] == 1:
cont1[i % k] += 1
else:
cont2[i % k] += 1
minMudancas = 0
for i in range(k):
if cont1[i] >= cont2[i]:
minMudancas += cont2[i]
else:
minMudancas += cont1[i]
return minMudancas
if __name__ == "__main__":
nk = list(map(int, input().split(' ')))
print(contarMudancas(nk[0], nk[1]))
```
| 3
|
|
320
|
A
|
Magic Numbers
|
PROGRAMMING
| 900
|
[
"brute force",
"greedy"
] | null | null |
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a magic number or not.
|
The first line of input contains an integer *n*, (1<=≤<=*n*<=≤<=109). This number doesn't contain leading zeros.
|
Print "YES" if *n* is a magic number or print "NO" if it's not.
|
[
"114114\n",
"1111\n",
"441231\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "114114",
"output": "YES"
},
{
"input": "1111",
"output": "YES"
},
{
"input": "441231",
"output": "NO"
},
{
"input": "1",
"output": "YES"
},
{
"input": "14",
"output": "YES"
},
{
"input": "114",
"output": "YES"
},
{
"input": "9",
"output": "NO"
},
{
"input": "414",
"output": "NO"
},
{
"input": "1000000000",
"output": "NO"
},
{
"input": "144144144",
"output": "YES"
},
{
"input": "1444",
"output": "NO"
},
{
"input": "11",
"output": "YES"
},
{
"input": "141414141",
"output": "YES"
},
{
"input": "11110111",
"output": "NO"
},
{
"input": "114114144",
"output": "YES"
},
{
"input": "444",
"output": "NO"
},
{
"input": "9999",
"output": "NO"
},
{
"input": "111444",
"output": "NO"
},
{
"input": "11114",
"output": "YES"
},
{
"input": "41111",
"output": "NO"
},
{
"input": "114414441",
"output": "NO"
},
{
"input": "144414441",
"output": "NO"
},
{
"input": "144244144",
"output": "NO"
},
{
"input": "111111111",
"output": "YES"
},
{
"input": "144444444",
"output": "NO"
},
{
"input": "444444444",
"output": "NO"
},
{
"input": "141441441",
"output": "YES"
},
{
"input": "441",
"output": "NO"
},
{
"input": "15",
"output": "NO"
},
{
"input": "14444",
"output": "NO"
},
{
"input": "11444",
"output": "NO"
},
{
"input": "144",
"output": "YES"
},
{
"input": "1414414",
"output": "YES"
},
{
"input": "141444",
"output": "NO"
},
{
"input": "14144",
"output": "YES"
},
{
"input": "4",
"output": "NO"
},
{
"input": "1144",
"output": "YES"
},
{
"input": "141111444",
"output": "NO"
},
{
"input": "14414414",
"output": "YES"
},
{
"input": "141414144",
"output": "YES"
},
{
"input": "1414",
"output": "YES"
},
{
"input": "1441",
"output": "YES"
},
{
"input": "12",
"output": "NO"
},
{
"input": "144144141",
"output": "YES"
},
{
"input": "144144",
"output": "YES"
},
{
"input": "14414411",
"output": "YES"
},
{
"input": "14414",
"output": "YES"
}
] | 1,696,489,048
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 60
| 0
|
def check3(i) :
if(n[i] not in ch): return 0
if(n[i] != '1'): return 0
if((i+1)<ln) :
if(n[i+1] not in ch): return 0
if(n[i+1] != '4') :
return 1
else :
if((i+2)<ln) :
if(n[i+2] not in ch): return 0
if(n[i+2] != '4'): return 2
else: return 3
return 1
n = input()
ln = len(n)
ch = '14';
i = 0
while(i<ln) :
c3 = check3(i)
if(c3 == 0) :
print('NO')
break
else: i+=c3
else: print('YES')
|
Title: Magic Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A magic number is a number formed by concatenation of numbers 1, 14 and 144. We can use each of these numbers any number of times. Therefore 14144, 141414 and 1411 are magic numbers but 1444, 514 and 414 are not.
You're given a number. Determine if it is a magic number or not.
Input Specification:
The first line of input contains an integer *n*, (1<=≤<=*n*<=≤<=109). This number doesn't contain leading zeros.
Output Specification:
Print "YES" if *n* is a magic number or print "NO" if it's not.
Demo Input:
['114114\n', '1111\n', '441231\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none
|
```python
def check3(i) :
if(n[i] not in ch): return 0
if(n[i] != '1'): return 0
if((i+1)<ln) :
if(n[i+1] not in ch): return 0
if(n[i+1] != '4') :
return 1
else :
if((i+2)<ln) :
if(n[i+2] not in ch): return 0
if(n[i+2] != '4'): return 2
else: return 3
return 1
n = input()
ln = len(n)
ch = '14';
i = 0
while(i<ln) :
c3 = check3(i)
if(c3 == 0) :
print('NO')
break
else: i+=c3
else: print('YES')
```
| 0
|
|
765
|
E
|
Tree Folding
|
PROGRAMMING
| 2,200
|
[
"dfs and similar",
"dp",
"greedy",
"implementation",
"trees"
] | null | null |
Vanya wants to minimize a tree. He can perform the following operation multiple times: choose a vertex *v*, and two disjoint (except for *v*) paths of equal length *a*0<==<=*v*, *a*1, ..., *a**k*, and *b*0<==<=*v*, *b*1, ..., *b**k*. Additionally, vertices *a*1, ..., *a**k*, *b*1, ..., *b**k* must not have any neighbours in the tree other than adjacent vertices of corresponding paths. After that, one of the paths may be merged into the other, that is, the vertices *b*1, ..., *b**k* can be effectively erased:
Help Vanya determine if it possible to make the tree into a path via a sequence of described operations, and if the answer is positive, also determine the shortest length of such path.
|
The first line of input contains the number of vertices *n* (2<=≤<=*n*<=≤<=2·105).
Next *n*<=-<=1 lines describe edges of the tree. Each of these lines contains two space-separated integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — indices of endpoints of the corresponding edge. It is guaranteed that the given graph is a tree.
|
If it is impossible to obtain a path, print -1. Otherwise, print the minimum number of edges in a possible path.
|
[
"6\n1 2\n2 3\n2 4\n4 5\n1 6\n",
"7\n1 2\n1 3\n3 4\n1 5\n5 6\n6 7\n"
] |
[
"3\n",
"-1\n"
] |
In the first sample case, a path of three edges is obtained after merging paths 2 - 1 - 6 and 2 - 4 - 5.
It is impossible to perform any operation in the second sample case. For example, it is impossible to merge paths 1 - 3 - 4 and 1 - 5 - 6, since vertex 6 additionally has a neighbour 7 that is not present in the corresponding path.
| 2,500
|
[
{
"input": "6\n1 2\n2 3\n2 4\n4 5\n1 6",
"output": "3"
},
{
"input": "7\n1 2\n1 3\n3 4\n1 5\n5 6\n6 7",
"output": "-1"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "3\n3 1\n1 2",
"output": "1"
},
{
"input": "10\n5 10\n7 8\n8 3\n2 6\n3 2\n9 7\n4 5\n10 1\n6 4",
"output": "9"
},
{
"input": "11\n11 9\n6 7\n7 1\n8 11\n5 6\n3 5\n9 3\n10 8\n2 4\n4 10",
"output": "5"
},
{
"input": "10\n4 2\n7 4\n2 6\n2 5\n4 8\n10 3\n2 9\n9 1\n5 10",
"output": "-1"
},
{
"input": "11\n8 9\n2 7\n1 11\n3 2\n9 1\n8 5\n8 6\n5 4\n4 10\n8 3",
"output": "1"
},
{
"input": "12\n12 6\n6 7\n8 11\n4 8\n10 4\n12 3\n2 10\n6 2\n12 9\n4 1\n9 5",
"output": "-1"
},
{
"input": "4\n4 1\n4 3\n4 2",
"output": "1"
},
{
"input": "5\n1 5\n2 3\n2 4\n1 2",
"output": "3"
},
{
"input": "6\n1 6\n3 1\n6 4\n5 3\n2 5",
"output": "5"
},
{
"input": "7\n5 6\n5 7\n5 1\n7 4\n6 3\n3 2",
"output": "-1"
},
{
"input": "8\n6 1\n4 7\n4 8\n8 5\n7 6\n4 3\n4 2",
"output": "-1"
},
{
"input": "3\n1 3\n3 2",
"output": "1"
},
{
"input": "5\n5 4\n4 3\n3 1\n5 2",
"output": "1"
},
{
"input": "9\n1 2\n1 3\n1 4\n1 5\n1 6\n6 7\n6 8\n8 9",
"output": "3"
}
] | 1,681,926,453
| 5,553
|
PyPy 3
|
OK
|
TESTS
| 54
| 904
| 82,534,400
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
G = [set() for _ in range(n + 1)]
for _ in range(n - 1):
u, v = map(int, input().split())
G[u].add(v)
G[v].add(u)
q = []
x = [set() for _ in range(n + 1)]
cnt = [len(g) for g in G]
for i in range(1, n + 1):
if cnt[i] == 1:
q.append(i)
x[i].add(0)
k = 0
ans = n - 1
while len(q) ^ k and k < n - 1:
i = q[k]
u = x[i].pop()
j = G[i].pop()
G[j].remove(i)
if u + 1 in x[j] or not len(x[j]):
cnt[j] -= 1
if u + 1 in x[j]:
ans -= u + 1
x[j].add(u + 1)
if len(x[j]) == 1 and cnt[j] == 1:
q.append(j)
k += 1
c = 0
for y in x:
if len(y) >= 3:
ans = -1
elif len(y):
c += 1
if c ^ 1:
ans = -1
while not ans % 2:
ans //= 2
print(ans)
|
Title: Tree Folding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya wants to minimize a tree. He can perform the following operation multiple times: choose a vertex *v*, and two disjoint (except for *v*) paths of equal length *a*0<==<=*v*, *a*1, ..., *a**k*, and *b*0<==<=*v*, *b*1, ..., *b**k*. Additionally, vertices *a*1, ..., *a**k*, *b*1, ..., *b**k* must not have any neighbours in the tree other than adjacent vertices of corresponding paths. After that, one of the paths may be merged into the other, that is, the vertices *b*1, ..., *b**k* can be effectively erased:
Help Vanya determine if it possible to make the tree into a path via a sequence of described operations, and if the answer is positive, also determine the shortest length of such path.
Input Specification:
The first line of input contains the number of vertices *n* (2<=≤<=*n*<=≤<=2·105).
Next *n*<=-<=1 lines describe edges of the tree. Each of these lines contains two space-separated integers *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*) — indices of endpoints of the corresponding edge. It is guaranteed that the given graph is a tree.
Output Specification:
If it is impossible to obtain a path, print -1. Otherwise, print the minimum number of edges in a possible path.
Demo Input:
['6\n1 2\n2 3\n2 4\n4 5\n1 6\n', '7\n1 2\n1 3\n3 4\n1 5\n5 6\n6 7\n']
Demo Output:
['3\n', '-1\n']
Note:
In the first sample case, a path of three edges is obtained after merging paths 2 - 1 - 6 and 2 - 4 - 5.
It is impossible to perform any operation in the second sample case. For example, it is impossible to merge paths 1 - 3 - 4 and 1 - 5 - 6, since vertex 6 additionally has a neighbour 7 that is not present in the corresponding path.
|
```python
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
G = [set() for _ in range(n + 1)]
for _ in range(n - 1):
u, v = map(int, input().split())
G[u].add(v)
G[v].add(u)
q = []
x = [set() for _ in range(n + 1)]
cnt = [len(g) for g in G]
for i in range(1, n + 1):
if cnt[i] == 1:
q.append(i)
x[i].add(0)
k = 0
ans = n - 1
while len(q) ^ k and k < n - 1:
i = q[k]
u = x[i].pop()
j = G[i].pop()
G[j].remove(i)
if u + 1 in x[j] or not len(x[j]):
cnt[j] -= 1
if u + 1 in x[j]:
ans -= u + 1
x[j].add(u + 1)
if len(x[j]) == 1 and cnt[j] == 1:
q.append(j)
k += 1
c = 0
for y in x:
if len(y) >= 3:
ans = -1
elif len(y):
c += 1
if c ^ 1:
ans = -1
while not ans % 2:
ans //= 2
print(ans)
```
| 3
|
|
581
|
A
|
Vasya the Hipster
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
|
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
|
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
|
[
"3 1\n",
"2 3\n",
"7 3\n"
] |
[
"1 1\n",
"2 0\n",
"3 2\n"
] |
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
| 500
|
[
{
"input": "3 1",
"output": "1 1"
},
{
"input": "2 3",
"output": "2 0"
},
{
"input": "7 3",
"output": "3 2"
},
{
"input": "100 100",
"output": "100 0"
},
{
"input": "4 10",
"output": "4 3"
},
{
"input": "6 10",
"output": "6 2"
},
{
"input": "6 11",
"output": "6 2"
},
{
"input": "10 40",
"output": "10 15"
},
{
"input": "11 56",
"output": "11 22"
},
{
"input": "34 30",
"output": "30 2"
},
{
"input": "33 33",
"output": "33 0"
},
{
"input": "100 45",
"output": "45 27"
},
{
"input": "100 23",
"output": "23 38"
},
{
"input": "45 12",
"output": "12 16"
},
{
"input": "1 1",
"output": "1 0"
},
{
"input": "1 100",
"output": "1 49"
},
{
"input": "100 1",
"output": "1 49"
},
{
"input": "68 59",
"output": "59 4"
},
{
"input": "45 99",
"output": "45 27"
},
{
"input": "99 100",
"output": "99 0"
},
{
"input": "100 98",
"output": "98 1"
},
{
"input": "59 12",
"output": "12 23"
},
{
"input": "86 4",
"output": "4 41"
},
{
"input": "68 21",
"output": "21 23"
},
{
"input": "100 11",
"output": "11 44"
},
{
"input": "100 10",
"output": "10 45"
},
{
"input": "15 45",
"output": "15 15"
},
{
"input": "11 32",
"output": "11 10"
},
{
"input": "34 96",
"output": "34 31"
},
{
"input": "89 89",
"output": "89 0"
}
] | 1,662,631,552
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 46
| 0
|
m,n=sorted(map(int,input().split()));print(m,(n-m)//2)
|
Title: Vasya the Hipster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
Input Specification:
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
Output Specification:
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
Demo Input:
['3 1\n', '2 3\n', '7 3\n']
Demo Output:
['1 1\n', '2 0\n', '3 2\n']
Note:
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
|
```python
m,n=sorted(map(int,input().split()));print(m,(n-m)//2)
```
| 3
|
|
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"
}
] | 1,687,011,853
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 31
| 0
|
m = int(input("m= "))
n = int(input("n= "))
a = int(input("a= "))
if n%a == 0:
k = n/a
else:
k = (n//a)+1
if m%a == 0:
p = m/a
else:
p = (m//a)+1
print(k*p)
|
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
m = int(input("m= "))
n = int(input("n= "))
a = int(input("a= "))
if n%a == 0:
k = n/a
else:
k = (n//a)+1
if m%a == 0:
p = m/a
else:
p = (m//a)+1
print(k*p)
```
| -1
|
131
|
A
|
cAPS lOCK
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
|
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
|
Print the result of the given word's processing.
|
[
"cAPS\n",
"Lock\n"
] |
[
"Caps",
"Lock\n"
] |
none
| 500
|
[
{
"input": "cAPS",
"output": "Caps"
},
{
"input": "Lock",
"output": "Lock"
},
{
"input": "cAPSlOCK",
"output": "cAPSlOCK"
},
{
"input": "CAPs",
"output": "CAPs"
},
{
"input": "LoCK",
"output": "LoCK"
},
{
"input": "OOPS",
"output": "oops"
},
{
"input": "oops",
"output": "oops"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "a"
},
{
"input": "aA",
"output": "Aa"
},
{
"input": "Zz",
"output": "Zz"
},
{
"input": "Az",
"output": "Az"
},
{
"input": "zA",
"output": "Za"
},
{
"input": "AAA",
"output": "aaa"
},
{
"input": "AAa",
"output": "AAa"
},
{
"input": "AaR",
"output": "AaR"
},
{
"input": "Tdr",
"output": "Tdr"
},
{
"input": "aTF",
"output": "Atf"
},
{
"input": "fYd",
"output": "fYd"
},
{
"input": "dsA",
"output": "dsA"
},
{
"input": "fru",
"output": "fru"
},
{
"input": "hYBKF",
"output": "Hybkf"
},
{
"input": "XweAR",
"output": "XweAR"
},
{
"input": "mogqx",
"output": "mogqx"
},
{
"input": "eOhEi",
"output": "eOhEi"
},
{
"input": "nkdku",
"output": "nkdku"
},
{
"input": "zcnko",
"output": "zcnko"
},
{
"input": "lcccd",
"output": "lcccd"
},
{
"input": "vwmvg",
"output": "vwmvg"
},
{
"input": "lvchf",
"output": "lvchf"
},
{
"input": "IUNVZCCHEWENCHQQXQYPUJCRDZLUXCLJHXPHBXEUUGNXOOOPBMOBRIBHHMIRILYJGYYGFMTMFSVURGYHUWDRLQVIBRLPEVAMJQYO",
"output": "iunvzcchewenchqqxqypujcrdzluxcljhxphbxeuugnxooopbmobribhhmirilyjgyygfmtmfsvurgyhuwdrlqvibrlpevamjqyo"
},
{
"input": "OBHSZCAMDXEJWOZLKXQKIVXUUQJKJLMMFNBPXAEFXGVNSKQLJGXHUXHGCOTESIVKSFMVVXFVMTEKACRIWALAGGMCGFEXQKNYMRTG",
"output": "obhszcamdxejwozlkxqkivxuuqjkjlmmfnbpxaefxgvnskqljgxhuxhgcotesivksfmvvxfvmtekacriwalaggmcgfexqknymrtg"
},
{
"input": "IKJYZIKROIYUUCTHSVSKZTETNNOCMAUBLFJCEVANCADASMZRCNLBZPQRXESHEEMOMEPCHROSRTNBIDXYMEPJSIXSZQEBTEKKUHFS",
"output": "ikjyzikroiyuucthsvskztetnnocmaublfjcevancadasmzrcnlbzpqrxesheemomepchrosrtnbidxymepjsixszqebtekkuhfs"
},
{
"input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE",
"output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype"
},
{
"input": "uCKJZRGZJCPPLEEYJTUNKOQSWGBMTBQEVPYFPIPEKRVYQNTDPANOIXKMPINNFUSZWCURGBDPYTEKBEKCPMVZPMWAOSHJYMGKOMBQ",
"output": "Uckjzrgzjcppleeyjtunkoqswgbmtbqevpyfpipekrvyqntdpanoixkmpinnfuszwcurgbdpytekbekcpmvzpmwaoshjymgkombq"
},
{
"input": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR",
"output": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR"
},
{
"input": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE",
"output": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE"
},
{
"input": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ",
"output": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ"
},
{
"input": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm",
"output": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm"
},
{
"input": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm",
"output": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm"
},
{
"input": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg",
"output": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg"
},
{
"input": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc",
"output": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc"
},
{
"input": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv",
"output": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv"
},
{
"input": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect",
"output": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect"
},
{
"input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE",
"output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype"
},
{
"input": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu",
"output": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu"
},
{
"input": "aBACABa",
"output": "aBACABa"
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAaa",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAaa"
},
{
"input": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA"
},
{
"input": "dDDDDDDDDDDDDD",
"output": "Dddddddddddddd"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "z",
"output": "Z"
},
{
"input": "AZ",
"output": "az"
},
{
"input": "Z",
"output": "z"
},
{
"input": "aAAAA",
"output": "Aaaaa"
},
{
"input": "F",
"output": "f"
}
] | 1,693,068,402
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 46
| 0
|
def word_change(word):
if word.isupper() or word[0].islower() and word[1:].isupper():
return word.swapcase()
else:
return word
word = input()
res = word_change(word)
print(res)
|
Title: cAPS lOCK
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input Specification:
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output Specification:
Print the result of the given word's processing.
Demo Input:
['cAPS\n', 'Lock\n']
Demo Output:
['Caps', 'Lock\n']
Note:
none
|
```python
def word_change(word):
if word.isupper() or word[0].islower() and word[1:].isupper():
return word.swapcase()
else:
return word
word = input()
res = word_change(word)
print(res)
```
| 0
|
|
312
|
B
|
Archer
|
PROGRAMMING
| 1,300
|
[
"math",
"probabilities"
] | null | null |
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
|
A single line contains four integers .
|
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
|
[
"1 2 1 2\n"
] |
[
"0.666666666667"
] |
none
| 1,000
|
[
{
"input": "1 2 1 2",
"output": "0.666666666667"
},
{
"input": "1 3 1 3",
"output": "0.600000000000"
},
{
"input": "1 3 2 3",
"output": "0.428571428571"
},
{
"input": "3 4 3 4",
"output": "0.800000000000"
},
{
"input": "1 2 10 11",
"output": "0.523809523810"
},
{
"input": "4 5 4 5",
"output": "0.833333333333"
},
{
"input": "466 701 95 721",
"output": "0.937693791148"
},
{
"input": "268 470 444 885",
"output": "0.725614009325"
},
{
"input": "632 916 713 821",
"output": "0.719292895126"
},
{
"input": "269 656 918 992",
"output": "0.428937461623"
},
{
"input": "71 657 187 695",
"output": "0.310488463257"
},
{
"input": "435 852 973 978",
"output": "0.511844133157"
},
{
"input": "518 816 243 359",
"output": "0.719734031025"
},
{
"input": "882 962 311 811",
"output": "0.966386645447"
},
{
"input": "684 774 580 736",
"output": "0.906051574446"
},
{
"input": "486 868 929 999",
"output": "0.577723252958"
},
{
"input": "132 359 996 998",
"output": "0.368154532345"
},
{
"input": "933 977 266 450",
"output": "0.972879407907"
},
{
"input": "298 833 615 872",
"output": "0.441270817024"
},
{
"input": "34 554 14 958",
"output": "0.817324099167"
},
{
"input": "836 934 800 905",
"output": "0.906105535462"
},
{
"input": "482 815 69 509",
"output": "0.914365577772"
},
{
"input": "284 423 137 521",
"output": "0.885974839378"
},
{
"input": "648 881 486 703",
"output": "0.800911421248"
},
{
"input": "450 885 755 836",
"output": "0.533901011176"
},
{
"input": "533 773 823 998",
"output": "0.729222130525"
},
{
"input": "897 957 92 898",
"output": "0.993193806364"
},
{
"input": "699 925 441 928",
"output": "0.866816866175"
},
{
"input": "64 704 148 603",
"output": "0.289486317811"
},
{
"input": "719 735 626 990",
"output": "0.986124079764"
},
{
"input": "1 1000 1 1000",
"output": "0.500250125063"
}
] | 1,621,718,969
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 4
| 93
| 0
|
a,b,c,d = map(int, input().split())
g = a*b
others = b*c - a*c + a*d
print(g/others)
|
Title: Archer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
SmallR is an archer. SmallR is taking a match of archer with Zanoes. They try to shoot in the target in turns, and SmallR shoots first. The probability of shooting the target each time is for SmallR while for Zanoes. The one who shoots in the target first should be the winner.
Output the probability that SmallR will win the match.
Input Specification:
A single line contains four integers .
Output Specification:
Print a single real number, the probability that SmallR will win the match.
The answer will be considered correct if the absolute or relative error doesn't exceed 10<=-<=6.
Demo Input:
['1 2 1 2\n']
Demo Output:
['0.666666666667']
Note:
none
|
```python
a,b,c,d = map(int, input().split())
g = a*b
others = b*c - a*c + a*d
print(g/others)
```
| 0
|
|
16
|
B
|
Burglar and Matches
|
PROGRAMMING
| 900
|
[
"greedy",
"implementation",
"sortings"
] |
B. Burglar and Matches
|
0
|
64
|
A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal.
|
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer.
|
Output the only number — answer to the problem.
|
[
"7 3\n5 10\n2 5\n3 6\n",
"3 3\n1 3\n2 2\n3 1\n"
] |
[
"62\n",
"7\n"
] |
none
| 0
|
[
{
"input": "7 3\n5 10\n2 5\n3 6",
"output": "62"
},
{
"input": "3 3\n1 3\n2 2\n3 1",
"output": "7"
},
{
"input": "1 1\n1 2",
"output": "2"
},
{
"input": "1 2\n1 9\n1 6",
"output": "9"
},
{
"input": "1 10\n1 1\n1 9\n1 3\n1 9\n1 7\n1 10\n1 4\n1 7\n1 3\n1 1",
"output": "10"
},
{
"input": "2 1\n2 1",
"output": "2"
},
{
"input": "2 2\n2 4\n1 4",
"output": "8"
},
{
"input": "2 3\n1 7\n1 2\n1 5",
"output": "12"
},
{
"input": "4 1\n2 2",
"output": "4"
},
{
"input": "4 2\n1 10\n4 4",
"output": "22"
},
{
"input": "4 3\n1 4\n6 4\n1 7",
"output": "19"
},
{
"input": "5 1\n10 5",
"output": "25"
},
{
"input": "5 2\n3 9\n2 2",
"output": "31"
},
{
"input": "5 5\n2 9\n3 1\n2 1\n1 8\n2 8",
"output": "42"
},
{
"input": "5 10\n1 3\n1 2\n1 9\n1 10\n1 1\n1 5\n1 10\n1 2\n1 3\n1 7",
"output": "41"
},
{
"input": "10 1\n9 4",
"output": "36"
},
{
"input": "10 2\n14 3\n1 3",
"output": "30"
},
{
"input": "10 7\n4 8\n1 10\n1 10\n1 2\n3 3\n1 3\n1 10",
"output": "71"
},
{
"input": "10 10\n1 8\n2 10\n1 9\n1 1\n1 9\n1 6\n1 4\n2 5\n1 2\n1 4",
"output": "70"
},
{
"input": "10 4\n1 5\n5 2\n1 9\n3 3",
"output": "33"
},
{
"input": "100 5\n78 6\n29 10\n3 6\n7 3\n2 4",
"output": "716"
},
{
"input": "1000 7\n102 10\n23 6\n79 4\n48 1\n34 10\n839 8\n38 4",
"output": "8218"
},
{
"input": "10000 10\n336 2\n2782 5\n430 10\n1893 7\n3989 10\n2593 8\n165 6\n1029 2\n2097 4\n178 10",
"output": "84715"
},
{
"input": "100000 3\n2975 2\n35046 4\n61979 9",
"output": "703945"
},
{
"input": "1000000 4\n314183 9\n304213 4\n16864 5\n641358 9",
"output": "8794569"
},
{
"input": "10000000 10\n360313 10\n416076 1\n435445 9\n940322 7\n1647581 7\n4356968 10\n3589256 2\n2967933 5\n2747504 7\n1151633 3",
"output": "85022733"
},
{
"input": "100000000 7\n32844337 7\n11210848 7\n47655987 1\n33900472 4\n9174763 2\n32228738 10\n29947408 5",
"output": "749254060"
},
{
"input": "200000000 10\n27953106 7\n43325979 4\n4709522 1\n10975786 4\n67786538 8\n48901838 7\n15606185 6\n2747583 1\n100000000 1\n633331 3",
"output": "1332923354"
},
{
"input": "200000000 9\n17463897 9\n79520463 1\n162407 4\n41017993 8\n71054118 4\n9447587 2\n5298038 9\n3674560 7\n20539314 5",
"output": "996523209"
},
{
"input": "200000000 8\n6312706 6\n2920548 2\n16843192 3\n1501141 2\n13394704 6\n10047725 10\n4547663 6\n54268518 6",
"output": "630991750"
},
{
"input": "200000000 7\n25621043 2\n21865270 1\n28833034 1\n22185073 5\n100000000 2\n13891017 9\n61298710 8",
"output": "931584598"
},
{
"input": "200000000 6\n7465600 6\n8453505 10\n4572014 8\n8899499 3\n86805622 10\n64439238 6",
"output": "1447294907"
},
{
"input": "200000000 5\n44608415 6\n100000000 9\n51483223 9\n44136047 1\n52718517 1",
"output": "1634907859"
},
{
"input": "200000000 4\n37758556 10\n100000000 6\n48268521 3\n20148178 10",
"output": "1305347138"
},
{
"input": "200000000 3\n65170000 7\n20790088 1\n74616133 4",
"output": "775444620"
},
{
"input": "200000000 2\n11823018 6\n100000000 9",
"output": "970938108"
},
{
"input": "200000000 1\n100000000 6",
"output": "600000000"
},
{
"input": "200000000 10\n12097724 9\n41745972 5\n26982098 9\n14916995 7\n21549986 7\n3786630 9\n8050858 7\n27994924 4\n18345001 5\n8435339 5",
"output": "1152034197"
},
{
"input": "200000000 10\n55649 8\n10980981 9\n3192542 8\n94994808 4\n3626106 1\n100000000 6\n5260110 9\n4121453 2\n15125061 4\n669569 6",
"output": "1095537357"
},
{
"input": "10 20\n1 7\n1 7\n1 8\n1 3\n1 10\n1 7\n1 7\n1 9\n1 3\n1 1\n1 2\n1 1\n1 3\n1 10\n1 9\n1 8\n1 8\n1 6\n1 7\n1 5",
"output": "83"
},
{
"input": "10000000 20\n4594 7\n520836 8\n294766 6\n298672 4\n142253 6\n450626 1\n1920034 9\n58282 4\n1043204 1\n683045 1\n1491746 5\n58420 4\n451217 2\n129423 4\n246113 5\n190612 8\n912923 6\n473153 6\n783733 6\n282411 10",
"output": "54980855"
},
{
"input": "200000000 20\n15450824 5\n839717 10\n260084 8\n1140850 8\n28744 6\n675318 3\n25161 2\n5487 3\n6537698 9\n100000000 5\n7646970 9\n16489 6\n24627 3\n1009409 5\n22455 1\n25488456 4\n484528 9\n32663641 3\n750968 4\n5152 6",
"output": "939368573"
},
{
"input": "200000000 20\n16896 2\n113 3\n277 2\n299 7\n69383562 2\n3929 8\n499366 4\n771846 5\n9 4\n1278173 7\n90 2\n54 7\n72199858 10\n17214 5\n3 10\n1981618 3\n3728 2\n141 8\n2013578 9\n51829246 5",
"output": "1158946383"
},
{
"input": "200000000 20\n983125 2\n7453215 9\n9193588 2\n11558049 7\n28666199 1\n34362244 1\n5241493 5\n15451270 4\n19945845 8\n6208681 3\n38300385 7\n6441209 8\n21046742 7\n577198 10\n3826434 8\n9764276 8\n6264675 7\n8567063 3\n3610303 4\n2908232 3",
"output": "1131379312"
},
{
"input": "10 15\n1 6\n2 6\n3 4\n1 3\n1 2\n1 5\n1 6\n1 2\n2 9\n1 10\n1 3\n1 7\n1 8\n1 2\n2 9",
"output": "79"
},
{
"input": "10000000 15\n111 5\n914124 3\n3 9\n177790 1\n2352 3\n32138 9\n104477 1\n1223 4\n18 6\n6655580 4\n57643 10\n94309 2\n37 1\n227002 10\n1733193 7",
"output": "45116295"
},
{
"input": "200000000 15\n7069868 1\n5567826 8\n2310059 10\n13539782 7\n38420939 4\n29911411 8\n52256316 1\n12265839 9\n2074265 1\n24896428 9\n72470695 5\n3236301 1\n3890243 2\n65168965 8\n65724 6",
"output": "1489289257"
},
{
"input": "200000000 15\n12044094 7\n2475138 10\n944451 7\n4854766 2\n3809145 10\n7727571 2\n43908937 6\n2745883 1\n427511 2\n100000000 5\n190914 6\n554889 3\n288798 4\n1848572 5\n893874 3",
"output": "961871671"
},
{
"input": "200000000 15\n6334191 7\n1927941 4\n5175933 10\n468389 1\n433043 10\n6863198 5\n7480646 4\n14774279 10\n2921129 8\n18325627 7\n6973152 9\n8277324 9\n21522856 2\n2058070 1\n2444742 4",
"output": "664376069"
}
] | 1,584,631,281
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 218
| 307,200
|
n,m=map(int,input().split())
l=[]
for i in range(m):
x,y=map(int,input().split())
l.append((x,y))
l.sort(key=lambda x:x[1],reverse=True)
ans=0
for i in range(m):
if l[i][0]>=n:
x=n
else:
x=l[i][0]
n-=x
ans+=x*l[i][1]
print(ans)
|
Title: Burglar and Matches
Time Limit: 0 seconds
Memory Limit: 64 megabytes
Problem Description:
A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer.
Output Specification:
Output the only number — answer to the problem.
Demo Input:
['7 3\n5 10\n2 5\n3 6\n', '3 3\n1 3\n2 2\n3 1\n']
Demo Output:
['62\n', '7\n']
Note:
none
|
```python
n,m=map(int,input().split())
l=[]
for i in range(m):
x,y=map(int,input().split())
l.append((x,y))
l.sort(key=lambda x:x[1],reverse=True)
ans=0
for i in range(m):
if l[i][0]>=n:
x=n
else:
x=l[i][0]
n-=x
ans+=x*l[i][1]
print(ans)
```
| 3
|
63
|
A
|
Sinking Ship
|
PROGRAMMING
| 900
|
[
"implementation",
"sortings",
"strings"
] |
A. Sinking Ship
|
2
|
256
|
The ship crashed into a reef and is sinking. Now the entire crew must be evacuated. All *n* crew members have already lined up in a row (for convenience let's label them all from left to right with positive integers from 1 to *n*) and await further instructions. However, one should evacuate the crew properly, in a strict order. Specifically:
The first crew members to leave the ship are rats. Then women and children (both groups have the same priority) leave the ship. After that all men are evacuated from the ship. The captain leaves the sinking ship last.
If we cannot determine exactly who should leave the ship first for any two members of the crew by the rules from the previous paragraph, then the one who stands to the left in the line leaves the ship first (or in other words, the one whose number in the line is less).
For each crew member we know his status as a crew member, and also his name. All crew members have different names. Determine the order in which to evacuate the crew.
|
The first line contains an integer *n*, which is the number of people in the crew (1<=≤<=*n*<=≤<=100). Then follow *n* lines. The *i*-th of those lines contains two words — the name of the crew member who is *i*-th in line, and his status on the ship. The words are separated by exactly one space. There are no other spaces in the line. The names consist of Latin letters, the first letter is uppercase, the rest are lowercase. The length of any name is from 1 to 10 characters. The status can have the following values: rat for a rat, woman for a woman, child for a child, man for a man, captain for the captain. The crew contains exactly one captain.
|
Print *n* lines. The *i*-th of them should contain the name of the crew member who must be the *i*-th one to leave the ship.
|
[
"6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman\n"
] |
[
"Teddy\nAlice\nBob\nJulia\nCharlie\nJack\n"
] |
none
| 500
|
[
{
"input": "6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman",
"output": "Teddy\nAlice\nBob\nJulia\nCharlie\nJack"
},
{
"input": "1\nA captain",
"output": "A"
},
{
"input": "1\nAbcdefjhij captain",
"output": "Abcdefjhij"
},
{
"input": "5\nA captain\nB man\nD woman\nC child\nE rat",
"output": "E\nD\nC\nB\nA"
},
{
"input": "10\nCap captain\nD child\nC woman\nA woman\nE child\nMan man\nB child\nF woman\nRat rat\nRatt rat",
"output": "Rat\nRatt\nD\nC\nA\nE\nB\nF\nMan\nCap"
},
{
"input": "5\nJoyxnkypf captain\nDxssgr woman\nKeojmnpd rat\nGdv man\nHnw man",
"output": "Keojmnpd\nDxssgr\nGdv\nHnw\nJoyxnkypf"
},
{
"input": "11\nJue rat\nWyglbyphk rat\nGjlgu child\nGi man\nAttx rat\nTheorpkgx man\nYm rat\nX child\nB captain\nEnualf rat\nKktsgyuyv woman",
"output": "Jue\nWyglbyphk\nAttx\nYm\nEnualf\nGjlgu\nX\nKktsgyuyv\nGi\nTheorpkgx\nB"
},
{
"input": "22\nWswwcvvm woman\nBtmfats rat\nI rat\nOcmtsnwx man\nUrcqv rat\nYghnogt woman\nWtyfc man\nWqle child\nUjfrelpu rat\nDstixj man\nAhksnio woman\nKhkvaap woman\nSjppvwm rat\nEgdmsv rat\nDank rat\nNquicjnw rat\nLh captain\nTdyaqaqln rat\nQtj rat\nTfgwijvq rat\nNbiso child\nNqthvbf woman",
"output": "Btmfats\nI\nUrcqv\nUjfrelpu\nSjppvwm\nEgdmsv\nDank\nNquicjnw\nTdyaqaqln\nQtj\nTfgwijvq\nWswwcvvm\nYghnogt\nWqle\nAhksnio\nKhkvaap\nNbiso\nNqthvbf\nOcmtsnwx\nWtyfc\nDstixj\nLh"
},
{
"input": "36\nKqxmtwmsf child\nIze woman\nDlpr child\nK woman\nF captain\nRjwfeuhba rat\nBbv rat\nS rat\nMnmg woman\nSmzyx woman\nSr man\nQmhroracn rat\nSoqpuqock rat\nPibdq man\nIlrkrptx rat\nZaecfyqka man\nMmersfs child\nVvvocqi man\nHjeqxvq rat\nMpmb woman\nWmgu woman\nCerelmhoxi child\nA man\nDylv man\nXrdgmmtcpq woman\nXj woman\nCeh child\nOfccguo man\nNevr man\nAknkwdkq child\nBmo man\nLqo woman\nVej child\nT child\nRlijc man\nNsdtpdg woman",
"output": "Rjwfeuhba\nBbv\nS\nQmhroracn\nSoqpuqock\nIlrkrptx\nHjeqxvq\nKqxmtwmsf\nIze\nDlpr\nK\nMnmg\nSmzyx\nMmersfs\nMpmb\nWmgu\nCerelmhoxi\nXrdgmmtcpq\nXj\nCeh\nAknkwdkq\nLqo\nVej\nT\nNsdtpdg\nSr\nPibdq\nZaecfyqka\nVvvocqi\nA\nDylv\nOfccguo\nNevr\nBmo\nRlijc\nF"
},
{
"input": "2\nA captain\nB man",
"output": "B\nA"
},
{
"input": "2\nB woman\nA captain",
"output": "B\nA"
},
{
"input": "2\nX child\nY captain",
"output": "X\nY"
},
{
"input": "2\nX captain\nY rat",
"output": "Y\nX"
},
{
"input": "6\nAbdelrhman captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman",
"output": "Teddy\nAlice\nBob\nJulia\nCharlie\nAbdelrhman"
},
{
"input": "4\nAratos captain\nMind child\nAman woman\nCarrot man",
"output": "Mind\nAman\nCarrot\nAratos"
}
] | 1,579,368,616
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 216
| 307,200
|
n=int(input())
r=[]
w=[]
m=[]
c=[]
for p in range(n):
a,b=[i for i in input().split()]
if b=="rat":
r.insert(0,a)
elif b=="woman" or b=="child":
w.insert(0,a)
elif b=="man":
m.insert(0,a)
elif b=="captain":
c.insert(0,a)
r.sort()
w.sort()
m.sort()
for i in range(len(r)):
print(r[i])
for j in range(len(w)):
print(w[j])
for k in range(len(m)):
print(m[k])
print(c[0])
|
Title: Sinking Ship
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The ship crashed into a reef and is sinking. Now the entire crew must be evacuated. All *n* crew members have already lined up in a row (for convenience let's label them all from left to right with positive integers from 1 to *n*) and await further instructions. However, one should evacuate the crew properly, in a strict order. Specifically:
The first crew members to leave the ship are rats. Then women and children (both groups have the same priority) leave the ship. After that all men are evacuated from the ship. The captain leaves the sinking ship last.
If we cannot determine exactly who should leave the ship first for any two members of the crew by the rules from the previous paragraph, then the one who stands to the left in the line leaves the ship first (or in other words, the one whose number in the line is less).
For each crew member we know his status as a crew member, and also his name. All crew members have different names. Determine the order in which to evacuate the crew.
Input Specification:
The first line contains an integer *n*, which is the number of people in the crew (1<=≤<=*n*<=≤<=100). Then follow *n* lines. The *i*-th of those lines contains two words — the name of the crew member who is *i*-th in line, and his status on the ship. The words are separated by exactly one space. There are no other spaces in the line. The names consist of Latin letters, the first letter is uppercase, the rest are lowercase. The length of any name is from 1 to 10 characters. The status can have the following values: rat for a rat, woman for a woman, child for a child, man for a man, captain for the captain. The crew contains exactly one captain.
Output Specification:
Print *n* lines. The *i*-th of them should contain the name of the crew member who must be the *i*-th one to leave the ship.
Demo Input:
['6\nJack captain\nAlice woman\nCharlie man\nTeddy rat\nBob child\nJulia woman\n']
Demo Output:
['Teddy\nAlice\nBob\nJulia\nCharlie\nJack\n']
Note:
none
|
```python
n=int(input())
r=[]
w=[]
m=[]
c=[]
for p in range(n):
a,b=[i for i in input().split()]
if b=="rat":
r.insert(0,a)
elif b=="woman" or b=="child":
w.insert(0,a)
elif b=="man":
m.insert(0,a)
elif b=="captain":
c.insert(0,a)
r.sort()
w.sort()
m.sort()
for i in range(len(r)):
print(r[i])
for j in range(len(w)):
print(w[j])
for k in range(len(m)):
print(m[k])
print(c[0])
```
| 0
|
849
|
A
|
Odds and Ends
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Where do odds begin, and where do they end? Where does hope emerge, and will they ever break?
Given an integer sequence *a*1,<=*a*2,<=...,<=*a**n* of length *n*. Decide whether it is possible to divide it into an odd number of non-empty subsegments, the each of which has an odd length and begins and ends with odd numbers.
A subsegment is a contiguous slice of the whole sequence. For example, {3,<=4,<=5} and {1} are subsegments of sequence {1,<=2,<=3,<=4,<=5,<=6}, while {1,<=2,<=4} and {7} are not.
|
The first line of input contains a non-negative integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence.
The second line contains *n* space-separated non-negative integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — the elements of the sequence.
|
Output "Yes" if it's possible to fulfill the requirements, and "No" otherwise.
You can output each letter in any case (upper or lower).
|
[
"3\n1 3 5\n",
"5\n1 0 1 5 1\n",
"3\n4 3 1\n",
"4\n3 9 9 3\n"
] |
[
"Yes\n",
"Yes\n",
"No\n",
"No\n"
] |
In the first example, divide the sequence into 1 subsegment: {1, 3, 5} and the requirements will be met.
In the second example, divide the sequence into 3 subsegments: {1, 0, 1}, {5}, {1}.
In the third example, one of the subsegments must start with 4 which is an even number, thus the requirements cannot be met.
In the fourth example, the sequence can be divided into 2 subsegments: {3, 9, 9}, {3}, but this is not a valid solution because 2 is an even number.
| 500
|
[
{
"input": "3\n1 3 5",
"output": "Yes"
},
{
"input": "5\n1 0 1 5 1",
"output": "Yes"
},
{
"input": "3\n4 3 1",
"output": "No"
},
{
"input": "4\n3 9 9 3",
"output": "No"
},
{
"input": "1\n1",
"output": "Yes"
},
{
"input": "5\n100 99 100 99 99",
"output": "No"
},
{
"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",
"output": "No"
},
{
"input": "1\n0",
"output": "No"
},
{
"input": "2\n1 1",
"output": "No"
},
{
"input": "2\n10 10",
"output": "No"
},
{
"input": "2\n54 21",
"output": "No"
},
{
"input": "5\n0 0 0 0 0",
"output": "No"
},
{
"input": "5\n67 92 0 26 43",
"output": "Yes"
},
{
"input": "15\n45 52 35 80 68 80 93 57 47 32 69 23 63 90 43",
"output": "Yes"
},
{
"input": "15\n81 28 0 82 71 64 63 89 87 92 38 30 76 72 36",
"output": "No"
},
{
"input": "50\n49 32 17 59 77 98 65 50 85 10 40 84 65 34 52 25 1 31 61 45 48 24 41 14 76 12 33 76 44 86 53 33 92 58 63 93 50 24 31 79 67 50 72 93 2 38 32 14 87 99",
"output": "No"
},
{
"input": "55\n65 69 53 66 11 100 68 44 43 17 6 66 24 2 6 6 61 72 91 53 93 61 52 96 56 42 6 8 79 49 76 36 83 58 8 43 2 90 71 49 80 21 75 13 76 54 95 61 58 82 40 33 73 61 46",
"output": "No"
},
{
"input": "99\n73 89 51 85 42 67 22 80 75 3 90 0 52 100 90 48 7 15 41 1 54 2 23 62 86 68 2 87 57 12 45 34 68 54 36 49 27 46 22 70 95 90 57 91 90 79 48 89 67 92 28 27 25 37 73 66 13 89 7 99 62 53 48 24 73 82 62 88 26 39 21 86 50 95 26 27 60 6 56 14 27 90 55 80 97 18 37 36 70 2 28 53 36 77 39 79 82 42 69",
"output": "Yes"
},
{
"input": "99\n99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99",
"output": "Yes"
},
{
"input": "100\n61 63 34 45 20 91 31 28 40 27 94 1 73 5 69 10 56 94 80 23 79 99 59 58 13 56 91 59 77 78 88 72 80 72 70 71 63 60 41 41 41 27 83 10 43 14 35 48 0 78 69 29 63 33 42 67 1 74 51 46 79 41 37 61 16 29 82 28 22 14 64 49 86 92 82 55 54 24 75 58 95 31 3 34 26 23 78 91 49 6 30 57 27 69 29 54 42 0 61 83",
"output": "No"
},
{
"input": "6\n1 2 2 2 2 1",
"output": "No"
},
{
"input": "3\n1 2 1",
"output": "Yes"
},
{
"input": "4\n1 3 2 3",
"output": "No"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "No"
},
{
"input": "6\n1 1 0 0 1 1",
"output": "No"
},
{
"input": "4\n1 4 9 3",
"output": "No"
},
{
"input": "4\n1 0 1 1",
"output": "No"
},
{
"input": "10\n1 0 0 1 1 1 1 1 1 1",
"output": "No"
},
{
"input": "10\n9 2 5 7 8 3 1 9 4 9",
"output": "No"
},
{
"input": "99\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 2",
"output": "No"
},
{
"input": "6\n1 2 1 2 2 1",
"output": "No"
},
{
"input": "6\n1 0 1 0 0 1",
"output": "No"
},
{
"input": "4\n1 3 4 7",
"output": "No"
},
{
"input": "8\n1 1 1 2 1 1 1 1",
"output": "No"
},
{
"input": "3\n1 1 2",
"output": "No"
},
{
"input": "5\n1 2 1 2 1",
"output": "Yes"
},
{
"input": "5\n5 4 4 2 1",
"output": "Yes"
},
{
"input": "6\n1 3 3 3 3 1",
"output": "No"
},
{
"input": "7\n1 2 1 2 2 2 1",
"output": "Yes"
},
{
"input": "4\n1 2 2 1",
"output": "No"
},
{
"input": "6\n1 2 3 4 6 5",
"output": "No"
},
{
"input": "5\n1 1 2 2 2",
"output": "No"
},
{
"input": "5\n1 0 0 1 1",
"output": "Yes"
},
{
"input": "3\n1 2 4",
"output": "No"
},
{
"input": "3\n1 0 2",
"output": "No"
},
{
"input": "5\n1 1 1 0 1",
"output": "Yes"
},
{
"input": "4\n3 9 2 3",
"output": "No"
},
{
"input": "6\n1 1 1 4 4 1",
"output": "No"
},
{
"input": "6\n1 2 3 5 6 7",
"output": "No"
},
{
"input": "6\n1 1 1 2 2 1",
"output": "No"
},
{
"input": "6\n1 1 1 0 0 1",
"output": "No"
},
{
"input": "5\n1 2 2 5 5",
"output": "Yes"
},
{
"input": "5\n1 3 2 4 5",
"output": "Yes"
},
{
"input": "8\n1 2 3 5 7 8 8 5",
"output": "No"
},
{
"input": "10\n1 1 1 2 1 1 1 1 1 1",
"output": "No"
},
{
"input": "4\n1 0 0 1",
"output": "No"
},
{
"input": "7\n1 0 1 1 0 0 1",
"output": "Yes"
},
{
"input": "7\n1 4 5 7 6 6 3",
"output": "Yes"
},
{
"input": "4\n2 2 2 2",
"output": "No"
},
{
"input": "5\n2 3 4 5 6",
"output": "No"
},
{
"input": "4\n1 1 2 1",
"output": "No"
},
{
"input": "3\n1 2 3",
"output": "Yes"
},
{
"input": "6\n1 3 3 2 2 3",
"output": "No"
},
{
"input": "4\n1 1 2 3",
"output": "No"
},
{
"input": "4\n1 2 3 5",
"output": "No"
},
{
"input": "5\n3 4 4 3 3",
"output": "Yes"
},
{
"input": "4\n3 2 2 3",
"output": "No"
},
{
"input": "6\n1 1 1 1 2 1",
"output": "No"
},
{
"input": "6\n1 1 2 2 1 1",
"output": "No"
},
{
"input": "10\n3 4 2 4 3 2 2 4 4 3",
"output": "No"
},
{
"input": "7\n1 2 4 3 2 4 5",
"output": "Yes"
},
{
"input": "28\n75 51 25 52 13 7 34 29 5 59 68 56 13 2 9 37 59 83 18 32 36 30 20 43 92 76 78 67",
"output": "No"
},
{
"input": "79\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 18",
"output": "No"
},
{
"input": "100\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",
"output": "No"
}
] | 1,623,912,710
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 74
| 93
| 0
|
n = int(input())
daf = list(map(int, input().split()))
if n%2 == 0 or daf[0]%2 == 0 or daf[-1]%2 == 0:
print("NO")
else:
print("YES")
|
Title: Odds and Ends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Where do odds begin, and where do they end? Where does hope emerge, and will they ever break?
Given an integer sequence *a*1,<=*a*2,<=...,<=*a**n* of length *n*. Decide whether it is possible to divide it into an odd number of non-empty subsegments, the each of which has an odd length and begins and ends with odd numbers.
A subsegment is a contiguous slice of the whole sequence. For example, {3,<=4,<=5} and {1} are subsegments of sequence {1,<=2,<=3,<=4,<=5,<=6}, while {1,<=2,<=4} and {7} are not.
Input Specification:
The first line of input contains a non-negative integer *n* (1<=≤<=*n*<=≤<=100) — the length of the sequence.
The second line contains *n* space-separated non-negative integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — the elements of the sequence.
Output Specification:
Output "Yes" if it's possible to fulfill the requirements, and "No" otherwise.
You can output each letter in any case (upper or lower).
Demo Input:
['3\n1 3 5\n', '5\n1 0 1 5 1\n', '3\n4 3 1\n', '4\n3 9 9 3\n']
Demo Output:
['Yes\n', 'Yes\n', 'No\n', 'No\n']
Note:
In the first example, divide the sequence into 1 subsegment: {1, 3, 5} and the requirements will be met.
In the second example, divide the sequence into 3 subsegments: {1, 0, 1}, {5}, {1}.
In the third example, one of the subsegments must start with 4 which is an even number, thus the requirements cannot be met.
In the fourth example, the sequence can be divided into 2 subsegments: {3, 9, 9}, {3}, but this is not a valid solution because 2 is an even number.
|
```python
n = int(input())
daf = list(map(int, input().split()))
if n%2 == 0 or daf[0]%2 == 0 or daf[-1]%2 == 0:
print("NO")
else:
print("YES")
```
| 3
|
|
808
|
D
|
Array Division
|
PROGRAMMING
| 1,900
|
[
"binary search",
"data structures",
"implementation"
] | null | null |
Vasya has an array *a* consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the second part. It is not always possible, so Vasya will move some element before dividing the array (Vasya will erase some element and insert it into an arbitrary position).
Inserting an element in the same position he was erased from is also considered moving.
Can Vasya divide the array after choosing the right element to move and its new position?
|
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2... *a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.
|
Print YES if Vasya can divide the array after moving one element. Otherwise print NO.
|
[
"3\n1 3 2\n",
"5\n1 2 3 4 5\n",
"5\n2 2 3 4 5\n"
] |
[
"YES\n",
"NO\n",
"YES\n"
] |
In the first example Vasya can move the second element to the end of the array.
In the second example no move can make the division possible.
In the third example Vasya can move the fourth element by one position to the left.
| 0
|
[
{
"input": "3\n1 3 2",
"output": "YES"
},
{
"input": "5\n1 2 3 4 5",
"output": "NO"
},
{
"input": "5\n2 2 3 4 5",
"output": "YES"
},
{
"input": "5\n72 32 17 46 82",
"output": "NO"
},
{
"input": "6\n26 10 70 11 69 57",
"output": "NO"
},
{
"input": "7\n4 7 10 7 5 5 1",
"output": "NO"
},
{
"input": "8\n9 5 5 10 4 9 5 8",
"output": "NO"
},
{
"input": "10\n9 6 8 5 5 2 8 9 2 2",
"output": "YES"
},
{
"input": "15\n4 8 10 3 1 4 5 9 3 2 1 7 7 3 8",
"output": "NO"
},
{
"input": "20\n71 83 54 6 10 64 91 98 94 49 65 68 14 39 91 60 74 100 17 13",
"output": "NO"
},
{
"input": "20\n2 8 10 4 6 6 4 1 2 2 6 9 5 1 9 1 9 8 10 6",
"output": "NO"
},
{
"input": "100\n9 9 72 55 14 8 55 58 35 67 3 18 73 92 41 49 15 60 18 66 9 26 97 47 43 88 71 97 19 34 48 96 79 53 8 24 69 49 12 23 77 12 21 88 66 9 29 13 61 69 54 77 41 13 4 68 37 74 7 6 29 76 55 72 89 4 78 27 29 82 18 83 12 4 32 69 89 85 66 13 92 54 38 5 26 56 17 55 29 4 17 39 29 94 3 67 85 98 21 14",
"output": "YES"
},
{
"input": "100\n89 38 63 73 77 4 99 74 30 5 69 57 97 37 88 71 36 59 19 63 46 20 33 58 61 98 100 31 33 53 99 96 34 17 44 95 54 52 22 77 67 88 20 88 26 43 12 23 96 94 14 7 57 86 56 54 32 8 3 43 97 56 74 22 5 100 12 60 93 12 44 68 31 63 7 71 21 29 19 38 50 47 97 43 50 59 88 40 51 61 20 68 32 66 70 48 19 55 91 53",
"output": "NO"
},
{
"input": "100\n80 100 88 52 25 87 85 8 92 62 35 66 74 39 58 41 55 53 23 73 90 72 36 44 97 67 16 54 3 8 25 34 84 47 77 39 93 19 49 20 29 44 21 48 21 56 82 59 8 31 94 95 84 54 72 20 95 91 85 1 67 19 76 28 31 63 87 98 55 28 16 20 36 91 93 39 94 69 80 97 100 96 68 26 91 45 22 84 20 36 20 92 53 75 58 51 60 26 76 25",
"output": "NO"
},
{
"input": "100\n27 95 57 29 91 85 83 36 72 86 39 5 79 61 78 93 100 97 73 23 82 66 41 92 38 92 100 96 48 56 66 47 5 32 69 13 95 23 46 62 99 83 57 66 98 82 81 57 37 37 81 64 45 76 72 43 99 76 86 22 37 39 93 80 99 36 53 83 3 32 52 9 78 34 47 100 33 72 19 40 29 56 77 32 79 72 15 88 100 98 56 50 22 81 88 92 58 70 21 19",
"output": "NO"
},
{
"input": "100\n35 31 83 11 7 94 57 58 30 26 2 99 33 58 98 6 3 52 13 66 21 53 26 94 100 5 1 3 91 13 97 49 86 25 63 90 88 98 57 57 34 81 32 16 65 94 59 83 44 14 46 18 28 89 75 95 87 57 52 18 46 80 31 43 38 54 69 75 82 9 64 96 75 40 96 52 67 85 86 38 95 55 16 57 17 20 22 7 63 3 12 16 42 87 46 12 51 95 67 80",
"output": "NO"
},
{
"input": "6\n1 4 3 100 100 6",
"output": "YES"
},
{
"input": "6\n6 100 100 3 4 1",
"output": "YES"
},
{
"input": "6\n4 2 3 7 1 1",
"output": "YES"
},
{
"input": "4\n6 1 4 5",
"output": "NO"
},
{
"input": "3\n228 114 114",
"output": "YES"
},
{
"input": "3\n229 232 444",
"output": "NO"
},
{
"input": "3\n322 324 555",
"output": "NO"
},
{
"input": "3\n69 34 5",
"output": "NO"
},
{
"input": "6\n5 4 1 2 2 2",
"output": "YES"
},
{
"input": "3\n545 237 546",
"output": "NO"
},
{
"input": "5\n2 3 1 1 1",
"output": "YES"
},
{
"input": "6\n2 2 10 2 2 2",
"output": "YES"
},
{
"input": "5\n5 4 6 5 6",
"output": "NO"
},
{
"input": "5\n6 1 1 1 1",
"output": "NO"
},
{
"input": "2\n1 3",
"output": "NO"
},
{
"input": "5\n5 2 2 3 4",
"output": "YES"
},
{
"input": "2\n2 2",
"output": "YES"
},
{
"input": "5\n1 2 6 1 2",
"output": "YES"
},
{
"input": "5\n1 1 8 5 1",
"output": "YES"
},
{
"input": "10\n73 67 16 51 56 71 37 49 90 6",
"output": "NO"
},
{
"input": "1\n10",
"output": "NO"
},
{
"input": "1\n1",
"output": "NO"
},
{
"input": "2\n1 1",
"output": "YES"
},
{
"input": "5\n8 2 7 5 4",
"output": "YES"
},
{
"input": "1\n2",
"output": "NO"
},
{
"input": "16\n9 10 2 1 6 7 6 5 8 3 2 10 8 4 9 2",
"output": "YES"
},
{
"input": "4\n8 2 2 4",
"output": "YES"
},
{
"input": "19\n9 9 3 2 4 5 5 7 8 10 8 10 1 2 2 6 5 3 3",
"output": "NO"
},
{
"input": "11\n7 2 1 8 8 2 4 10 8 7 1",
"output": "YES"
},
{
"input": "6\n10 20 30 40 99 1",
"output": "YES"
},
{
"input": "10\n3 7 9 2 10 1 9 6 4 1",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "YES"
},
{
"input": "2\n9 3",
"output": "NO"
},
{
"input": "7\n1 2 3 12 1 2 3",
"output": "YES"
},
{
"input": "6\n2 4 4 5 8 5",
"output": "YES"
},
{
"input": "18\n2 10 3 6 6 6 10 8 8 1 10 9 9 3 1 9 7 4",
"output": "YES"
},
{
"input": "20\n9 6 6 10 4 4 8 7 4 10 10 2 10 5 9 5 3 10 1 9",
"output": "NO"
},
{
"input": "12\n3 8 10 2 4 4 6 9 5 10 10 3",
"output": "YES"
},
{
"input": "11\n9 2 7 7 7 3 7 5 4 10 7",
"output": "NO"
},
{
"input": "5\n1 1 4 1 1",
"output": "YES"
},
{
"input": "2\n4 4",
"output": "YES"
},
{
"input": "2\n7 1",
"output": "NO"
},
{
"input": "5\n10 5 6 7 6",
"output": "YES"
},
{
"input": "11\n4 3 10 3 7 8 4 9 2 1 1",
"output": "YES"
},
{
"input": "6\n705032704 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "NO"
},
{
"input": "8\n1 5 6 8 3 1 7 3",
"output": "YES"
},
{
"input": "20\n8 6 3 6 3 5 10 2 6 1 7 6 9 10 8 3 5 9 3 8",
"output": "YES"
},
{
"input": "11\n2 4 8 3 4 7 9 10 5 3 3",
"output": "YES"
},
{
"input": "7\n6 4 2 24 6 4 2",
"output": "YES"
},
{
"input": "17\n7 1 1 1 8 9 1 10 8 8 7 9 7 9 1 6 5",
"output": "NO"
},
{
"input": "7\n7 10 1 2 6 2 2",
"output": "NO"
},
{
"input": "5\n10 10 40 10 10",
"output": "YES"
},
{
"input": "3\n4 3 13",
"output": "NO"
},
{
"input": "5\n5 2 10 2 1",
"output": "YES"
},
{
"input": "7\n7 4 5 62 20 20 6",
"output": "YES"
},
{
"input": "6\n1 5 2 20 10 2",
"output": "YES"
},
{
"input": "2\n5 6",
"output": "NO"
},
{
"input": "14\n5 2 9 7 5 8 3 2 2 4 9 1 3 10",
"output": "YES"
},
{
"input": "5\n1 2 3 4 2",
"output": "YES"
},
{
"input": "5\n2 2 2 5 5",
"output": "NO"
},
{
"input": "11\n1 1 1 1 1 10 1 1 1 1 1",
"output": "YES"
},
{
"input": "9\n8 4 13 19 11 1 8 2 8",
"output": "YES"
},
{
"input": "6\n14 16 14 14 15 11",
"output": "YES"
},
{
"input": "9\n14 19 1 13 11 3 1 1 7",
"output": "YES"
},
{
"input": "6\n16 13 3 7 4 15",
"output": "YES"
},
{
"input": "4\n11 7 12 14",
"output": "NO"
},
{
"input": "3\n3 2 1",
"output": "YES"
},
{
"input": "5\n2 1 3 6 4",
"output": "YES"
},
{
"input": "5\n3 4 8 11 2",
"output": "YES"
},
{
"input": "5\n1 2 10 3 4",
"output": "YES"
},
{
"input": "6\n8 15 12 14 15 4",
"output": "YES"
},
{
"input": "5\n1 2 4 4 5",
"output": "YES"
},
{
"input": "3\n2 4 2",
"output": "YES"
},
{
"input": "5\n2 3 1 6 4",
"output": "YES"
},
{
"input": "7\n1 2 3 12 3 2 1",
"output": "YES"
},
{
"input": "3\n3 4 13",
"output": "NO"
},
{
"input": "6\n1 1 1 1 1000000000 1000000000",
"output": "YES"
},
{
"input": "6\n19 6 5 13 6 13",
"output": "YES"
},
{
"input": "8\n2 2 2 5 1 2 3 3",
"output": "YES"
}
] | 1,694,010,253
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 32
| 171
| 18,124,800
|
n = int(input())
arr = list(map(int, input().split(' ')))
pref_sum = [0] * (n + 1)
for ind in range(1, n + 1):
pref_sum[ind] = pref_sum[ind - 1] + arr[ind - 1]
prev_els = set()
prev_els.add(arr[0])
ind = 1
half_sum = sum(arr) / 2
while ind < n:
prev_els.add(arr[ind])
if abs(half_sum - pref_sum[ind]) in prev_els:
print("YES")
break
else:
ind += 1
else:
print("NO")
|
Title: Array Division
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has an array *a* consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the second part. It is not always possible, so Vasya will move some element before dividing the array (Vasya will erase some element and insert it into an arbitrary position).
Inserting an element in the same position he was erased from is also considered moving.
Can Vasya divide the array after choosing the right element to move and its new position?
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2... *a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.
Output Specification:
Print YES if Vasya can divide the array after moving one element. Otherwise print NO.
Demo Input:
['3\n1 3 2\n', '5\n1 2 3 4 5\n', '5\n2 2 3 4 5\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first example Vasya can move the second element to the end of the array.
In the second example no move can make the division possible.
In the third example Vasya can move the fourth element by one position to the left.
|
```python
n = int(input())
arr = list(map(int, input().split(' ')))
pref_sum = [0] * (n + 1)
for ind in range(1, n + 1):
pref_sum[ind] = pref_sum[ind - 1] + arr[ind - 1]
prev_els = set()
prev_els.add(arr[0])
ind = 1
half_sum = sum(arr) / 2
while ind < n:
prev_els.add(arr[ind])
if abs(half_sum - pref_sum[ind]) in prev_els:
print("YES")
break
else:
ind += 1
else:
print("NO")
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,681,145,342
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
a, b = map(int, input().split())
if a % 2 == 0 and b % 2 == 0:
print(a*b//2)
else:
if a % 2 == 1 and b % 2 == 0:
print((a-1)*b//2 + (a-1)//2)
else:
print((b-1)*a//2 + (b-1)//2)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
a, b = map(int, input().split())
if a % 2 == 0 and b % 2 == 0:
print(a*b//2)
else:
if a % 2 == 1 and b % 2 == 0:
print((a-1)*b//2 + (a-1)//2)
else:
print((b-1)*a//2 + (b-1)//2)
```
| 0
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,616,539,848
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 31
| 124
| 0
|
M,N=map(int,input().split())
if M==1 and N==1:
print("No es posible")
else:
print(int(M*N) // 2)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
M,N=map(int,input().split())
if M==1 and N==1:
print("No es posible")
else:
print(int(M*N) // 2)
```
| 0
|
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"
}
] | 1,516,373,136
| 636
|
PyPy 3
|
OK
|
TESTS
| 254
| 109
| 19,968,000
|
n = int(input())
h, m = [int(z) for z in input().split()]
if "7" in str(h) or "7" in str(m):
print(0)
exit(0)
def time(x):
t = x // 60
m = x % 60
return t, m
cur = h * 60 + m
cnt = 0
while True:
cur -= n
cnt += 1
if cur < 0:
cur += 24 * 60
a, b = time(cur)
if "7" in str(a) or "7" in str(b):
print(cnt)
exit(0)
|
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())
h, m = [int(z) for z in input().split()]
if "7" in str(h) or "7" in str(m):
print(0)
exit(0)
def time(x):
t = x // 60
m = x % 60
return t, m
cur = h * 60 + m
cnt = 0
while True:
cur -= n
cnt += 1
if cur < 0:
cur += 24 * 60
a, b = time(cur)
if "7" in str(a) or "7" in str(b):
print(cnt)
exit(0)
```
| 3
|
|
507
|
C
|
Guess Your Way Out!
|
PROGRAMMING
| 1,700
|
[
"implementation",
"math",
"trees"
] | null | null |
Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height *h*. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node.
Let's index all the leaf nodes from the left to the right from 1 to 2*h*. The exit is located at some node *n* where 1<=≤<=*n*<=≤<=2*h*, the player doesn't know where the exit is so he has to guess his way out!
Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules:
- Character 'L' means "go to the left child of the current node"; - Character 'R' means "go to the right child of the current node"; - If the destination node is already visited, Amr skips current command, otherwise he moves to the destination node; - If Amr skipped two consecutive commands, he goes back to the parent of the current node before executing next command; - If he reached a leaf node that is not the exit, he returns to the parent of the current node; - If he reaches an exit, the game is finished.
Now Amr wonders, if he follows this algorithm, how many nodes he is going to visit before reaching the exit?
|
Input consists of two integers *h*,<=*n* (1<=≤<=*h*<=≤<=50, 1<=≤<=*n*<=≤<=2*h*).
|
Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm.
|
[
"1 2\n",
"2 3\n",
"3 6\n",
"10 1024\n"
] |
[
"2",
"5",
"10",
"2046"
] |
A perfect binary tree of height *h* is a binary tree consisting of *h* + 1 levels. Level 0 consists of a single node called root, level *h* consists of 2<sup class="upper-index">*h*</sup> nodes called leaves. Each node that is not a leaf has exactly two children, left and right one.
Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit.
<img class="tex-graphics" src="https://espresso.codeforces.com/e9d0715dc8cd9b4f6ac7a0fb137563f857660adc.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 1,500
|
[
{
"input": "1 2",
"output": "2"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "3 6",
"output": "10"
},
{
"input": "10 1024",
"output": "2046"
},
{
"input": "10 577",
"output": "1345"
},
{
"input": "11 550",
"output": "408"
},
{
"input": "19 12783",
"output": "503251"
},
{
"input": "28 72803174",
"output": "50649698"
},
{
"input": "39 457181784666",
"output": "830699159852"
},
{
"input": "12 955",
"output": "2871"
},
{
"input": "13 154",
"output": "7770"
},
{
"input": "14 2334",
"output": "9440"
},
{
"input": "15 15512",
"output": "14926"
},
{
"input": "16 21395",
"output": "2899"
},
{
"input": "17 80239",
"output": "177237"
},
{
"input": "18 153276",
"output": "328766"
},
{
"input": "20 589266",
"output": "1505684"
},
{
"input": "21 1687606",
"output": "3522472"
},
{
"input": "24 14428281",
"output": "26969983"
},
{
"input": "29 113463931",
"output": "347736449"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "3 8",
"output": "14"
},
{
"input": "31 1819651953",
"output": "3412135549"
},
{
"input": "33 2599588275",
"output": "1357401405"
},
{
"input": "38 262402936512",
"output": "519008349260"
},
{
"input": "4 13",
"output": "27"
},
{
"input": "40 615535158153",
"output": "1572205271927"
},
{
"input": "42 1042128038474",
"output": "3195908899134"
},
{
"input": "45 17519319833295",
"output": "17381304930499"
},
{
"input": "46 34999315964173",
"output": "34646522010881"
},
{
"input": "49 295606900104348",
"output": "820858833984106"
},
{
"input": "50 905353992267944",
"output": "1871650493613618"
},
{
"input": "3 5",
"output": "11"
},
{
"input": "4 14",
"output": "26"
},
{
"input": "6 40",
"output": "88"
},
{
"input": "7 31",
"output": "95"
},
{
"input": "8 19",
"output": "205"
},
{
"input": "10 359",
"output": "91"
},
{
"input": "11 349",
"output": "1057"
},
{
"input": "13 4796",
"output": "10298"
},
{
"input": "20 742273",
"output": "1182599"
},
{
"input": "22 3343393",
"output": "7009189"
},
{
"input": "24 3543583",
"output": "10865127"
},
{
"input": "25 678676",
"output": "31527640"
},
{
"input": "27 109473899",
"output": "209022797"
},
{
"input": "29 19827102",
"output": "478963048"
},
{
"input": "8 204",
"output": "422"
},
{
"input": "30 414940886",
"output": "372407442"
},
{
"input": "32 3786259360",
"output": "8003335020"
},
{
"input": "35 31233562499",
"output": "60951693197"
},
{
"input": "38 99361414961",
"output": "28342263489"
},
{
"input": "9 43",
"output": "391"
},
{
"input": "40 874338951117",
"output": "1800799608767"
},
{
"input": "33 2696188969",
"output": "505562011"
},
{
"input": "35 12080044014",
"output": "2415167450"
},
{
"input": "12 2715",
"output": "4185"
},
{
"input": "29 524109003",
"output": "1052258991"
},
{
"input": "50 1",
"output": "1125899906842623"
},
{
"input": "50 1125899906842624",
"output": "2251799813685246"
},
{
"input": "50 562949953421312",
"output": "562949953421312"
},
{
"input": "50 844424930131968",
"output": "1407374883553280"
},
{
"input": "50 375299968947542",
"output": "50"
}
] | 1,614,601,977
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 62
| 93
| 0
|
from sys import stdin,stdout
nmbr = lambda: int(input())
lst = lambda: list(map(int, input().split()))
def fn(s,e,left,dpth):
global ans
if s==e:return
mid=(s+e)>>1
left^=1
nodes=(1<<(h-dpth+1))-1
if s<=n<=mid:
if left==0:
ans+=nodes//2
fn(s,mid,1,dpth+1)
else:
if left==1:
ans+=nodes//2
fn(mid+1,e,0,dpth+1)
for _ in range(1):#nmbr()):
h,n=lst()
ans=0
end=(1<<h)
mid=(1+end)//2
if 1<=n<=mid:fn(1,mid,1,1)
else:
ans+=(1<<(h))-1
fn(mid+1,end,0,1)
# fn(1,end,1,0)
print(ans+h)
|
Title: Guess Your Way Out!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height *h*. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node.
Let's index all the leaf nodes from the left to the right from 1 to 2*h*. The exit is located at some node *n* where 1<=≤<=*n*<=≤<=2*h*, the player doesn't know where the exit is so he has to guess his way out!
Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules:
- Character 'L' means "go to the left child of the current node"; - Character 'R' means "go to the right child of the current node"; - If the destination node is already visited, Amr skips current command, otherwise he moves to the destination node; - If Amr skipped two consecutive commands, he goes back to the parent of the current node before executing next command; - If he reached a leaf node that is not the exit, he returns to the parent of the current node; - If he reaches an exit, the game is finished.
Now Amr wonders, if he follows this algorithm, how many nodes he is going to visit before reaching the exit?
Input Specification:
Input consists of two integers *h*,<=*n* (1<=≤<=*h*<=≤<=50, 1<=≤<=*n*<=≤<=2*h*).
Output Specification:
Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm.
Demo Input:
['1 2\n', '2 3\n', '3 6\n', '10 1024\n']
Demo Output:
['2', '5', '10', '2046']
Note:
A perfect binary tree of height *h* is a binary tree consisting of *h* + 1 levels. Level 0 consists of a single node called root, level *h* consists of 2<sup class="upper-index">*h*</sup> nodes called leaves. Each node that is not a leaf has exactly two children, left and right one.
Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit.
<img class="tex-graphics" src="https://espresso.codeforces.com/e9d0715dc8cd9b4f6ac7a0fb137563f857660adc.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
from sys import stdin,stdout
nmbr = lambda: int(input())
lst = lambda: list(map(int, input().split()))
def fn(s,e,left,dpth):
global ans
if s==e:return
mid=(s+e)>>1
left^=1
nodes=(1<<(h-dpth+1))-1
if s<=n<=mid:
if left==0:
ans+=nodes//2
fn(s,mid,1,dpth+1)
else:
if left==1:
ans+=nodes//2
fn(mid+1,e,0,dpth+1)
for _ in range(1):#nmbr()):
h,n=lst()
ans=0
end=(1<<h)
mid=(1+end)//2
if 1<=n<=mid:fn(1,mid,1,1)
else:
ans+=(1<<(h))-1
fn(mid+1,end,0,1)
# fn(1,end,1,0)
print(ans+h)
```
| 3
|
|
177
|
G1
|
Fibonacci Strings
|
PROGRAMMING
| 2,400
|
[
"strings"
] | null | null |
Fibonacci strings are defined as follows:
- *f*1 = «a» - *f*2 = «b» - *f**n* = *f**n*<=-<=1 *f**n*<=-<=2, *n*<=><=2
Thus, the first five Fibonacci strings are: "a", "b", "ba", "bab", "babba".
You are given a Fibonacci string and *m* strings *s**i*. For each string *s**i*, find the number of times it occurs in the given Fibonacci string as a substring.
|
The first line contains two space-separated integers *k* and *m* — the number of a Fibonacci string and the number of queries, correspondingly.
Next *m* lines contain strings *s**i* that correspond to the queries. It is guaranteed that strings *s**i* aren't empty and consist only of characters "a" and "b".
The input limitations for getting 30 points are:
- 1<=≤<=*k*<=≤<=3000 - 1<=≤<=*m*<=≤<=3000 - The total length of strings *s**i* doesn't exceed 3000
The input limitations for getting 100 points are:
- 1<=≤<=*k*<=≤<=1018 - 1<=≤<=*m*<=≤<=104 - The total length of strings *s**i* doesn't exceed 105
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
|
For each string *s**i* print the number of times it occurs in the given Fibonacci string as a substring. Since the numbers can be large enough, print them modulo 1000000007 (109<=+<=7). Print the answers for the strings in the order in which they are given in the input.
|
[
"6 5\na\nb\nab\nba\naba\n"
] |
[
"3\n5\n3\n3\n1\n"
] |
none
| 30
|
[
{
"input": "6 5\na\nb\nab\nba\naba",
"output": "3\n5\n3\n3\n1"
},
{
"input": "10 10\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab",
"output": "12\n21\n21\n0\n12\n21\n21\n0\n12\n21"
},
{
"input": "10 10\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb",
"output": "0\n12\n21\n0\n12\n8\n0\n0\n0\n12"
},
{
"input": "1 10\nb\na\nb\na\nb\na\nb\na\nb\na",
"output": "0\n1\n0\n1\n0\n1\n0\n1\n0\n1"
},
{
"input": "2 10\nb\na\nb\na\nb\na\nb\na\nb\na",
"output": "1\n0\n1\n0\n1\n0\n1\n0\n1\n0"
},
{
"input": "3 10\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb",
"output": "0\n0\n0\n0\n0\n0\n0\n0\n0\n0"
},
{
"input": "15 100\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb\nbab\naab\nbba\naba\nbaa\naaa\nbbb\nabb\nbab\naab",
"output": "0\n144\n232\n0\n144\n88\n0\n0\n0\n144\n232\n0\n144\n88\n0\n0\n0\n144\n232\n0\n144\n88\n0\n0\n0\n144\n232\n0\n144\n88\n0\n0\n0\n144\n232\n0\n144\n88\n0\n0\n0\n144\n232\n0\n144\n88\n0\n0\n0\n144\n232\n0\n144\n88\n0\n0\n0\n144\n232\n0\n144\n88\n0\n0\n0\n144\n232\n0\n144\n88\n0\n0\n0\n144\n232\n0\n144\n88\n0\n0\n0\n144\n232\n0\n144\n88\n0\n0\n0\n144\n232\n0\n144\n88\n0\n0\n0\n144\n232\n0"
},
{
"input": "15 100\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na\nb\na",
"output": "377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233\n377\n233"
},
{
"input": "15 100\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa\nbb\nab\nba\naa",
"output": "144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0\n144\n232\n233\n0"
},
{
"input": "50 100\na\na\nb\na\na\nb\nb\nb\na\na\nb\nb\na\na\na\na\nb\nb\na\nb\nb\nb\na\na\na\na\na\na\nb\nb\na\nb\na\nb\na\nb\na\nb\nb\nb\na\na\nb\na\na\na\nb\na\nb\na\na\na\nb\na\nb\na\na\nb\na\na\nb\na\na\na\na\nb\na\nb\na\nb\nb\na\nb\nb\nb\na\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\na\na\na\nb\nb\nb\nb\na\na\na\nb\nb\nb\na",
"output": "807526948\n807526948\n778742000\n807526948\n807526948\n778742000\n778742000\n778742000\n807526948\n807526948\n778742000\n778742000\n807526948\n807526948\n807526948\n807526948\n778742000\n778742000\n807526948\n778742000\n778742000\n778742000\n807526948\n807526948\n807526948\n807526948\n807526948\n807526948\n778742000\n778742000\n807526948\n778742000\n807526948\n778742000\n807526948\n778742000\n807526948\n778742000\n778742000\n778742000\n807526948\n807526948\n778742000\n807526948\n807526948\n807526948\n77874..."
},
{
"input": "50 100\nb\naa\na\na\na\nb\na\nb\nbb\na\naa\naa\naa\nba\naa\na\na\naa\na\na\naa\nb\nbb\naa\naa\nbb\nb\nbb\na\naa\naa\naa\na\na\nb\nb\na\naa\nb\naa\nab\nb\nb\na\naa\nbb\nb\nb\na\nb\na\na\nb\na\nbb\nb\nab\nbb\na\naa\naa\nb\nb\na\nbb\nba\naa\nba\nbb\nba\nbb\na\nb\na\nba\na\nab\na\nbb\nab\na\nab\nbb\na\na\nb\na\nba\na\nbb\nb\nab\nab\naa\na\nab\nab\nbb\nab\nab",
"output": "778742000\n0\n807526948\n807526948\n807526948\n778742000\n807526948\n778742000\n971215058\n807526948\n0\n0\n0\n807526948\n0\n807526948\n807526948\n0\n807526948\n807526948\n0\n778742000\n971215058\n0\n0\n971215058\n778742000\n971215058\n807526948\n0\n0\n0\n807526948\n807526948\n778742000\n778742000\n807526948\n0\n778742000\n0\n807526948\n778742000\n778742000\n807526948\n0\n971215058\n778742000\n778742000\n807526948\n778742000\n807526948\n807526948\n778742000\n807526948\n971215058\n778742000\n807526948\n9712..."
},
{
"input": "50 100\nb\naab\nab\naaa\na\naab\naaa\nbb\na\na\nb\naab\nbbb\naa\nbbb\nb\nab\nab\nbbb\nb\nbaa\na\nbab\nbbb\na\naba\nab\na\nba\nb\nbba\naba\nba\nbba\nb\nb\nb\nb\nab\na\nabb\nab\nbb\nbba\nb\nbbb\nbb\nb\naba\naab\nba\nbb\na\na\nbb\nba\nbaa\nba\nba\na\nb\nbb\nb\nbaa\nbab\nbba\nb\nbb\nbbb\nb\naba\nbba\nbba\naaa\nab\na\nbbb\nab\nb\nba\nbab\nb\naab\nbb\naba\nb\na\naa\nbaa\nbbb\naa\naba\nb\nbb\nba\nb\nb\naaa\nab\nba",
"output": "778742000\n0\n807526948\n0\n807526948\n0\n0\n971215058\n807526948\n807526948\n778742000\n0\n0\n0\n0\n778742000\n807526948\n807526948\n0\n778742000\n0\n807526948\n807526948\n0\n807526948\n836311896\n807526948\n807526948\n807526948\n778742000\n971215058\n836311896\n807526948\n971215058\n778742000\n778742000\n778742000\n778742000\n807526948\n807526948\n971215058\n807526948\n971215058\n971215058\n778742000\n0\n971215058\n778742000\n836311896\n0\n807526948\n971215058\n807526948\n807526948\n971215058\n807526948\n..."
},
{
"input": "50 100\nbb\naa\nb\nbaa\nbbba\naa\nba\na\nabba\nbaa\naa\naab\nab\nbabb\naabb\nbaa\nbaaa\nbaa\naab\nbba\nbb\naba\naaba\nbab\naaba\naa\naaaa\nbabb\nbbb\naaba\naaa\nab\nbab\nb\nb\naa\naaab\naa\nbba\nbaa\nbabb\nbaba\nba\naaba\nbba\nba\nab\nabb\nb\nba\nbbb\nba\naaa\nbbb\nbaa\nbb\na\naaa\naaba\nab\nbba\nba\nb\nbbb\naaa\na\na\nb\nb\naba\nbb\nba\na\nb\nbaa\nb\naaaa\na\naaab\nbaba\nba\nbb\nbaba\nab\nbaaa\nbbbb\na\naabb\nab\nb\nb\naaa\nb\nb\nabab\nabb\nb\nb\nbb\nb",
"output": "971215058\n0\n778742000\n0\n0\n0\n807526948\n807526948\n971215058\n0\n0\n0\n807526948\n971215058\n0\n0\n0\n0\n0\n971215058\n971215058\n836311896\n0\n807526948\n0\n0\n0\n971215058\n0\n0\n0\n807526948\n807526948\n778742000\n778742000\n0\n0\n0\n971215058\n0\n971215058\n836311896\n807526948\n0\n971215058\n807526948\n807526948\n971215058\n778742000\n807526948\n0\n807526948\n0\n0\n0\n971215058\n807526948\n0\n0\n807526948\n971215058\n807526948\n778742000\n0\n0\n807526948\n807526948\n778742000\n778742000\n83631189..."
},
{
"input": "1 14\na\nb\naa\nab\nba\nbb\naaa\naab\naba\nabb\nbaa\nbab\nbba\nbbb",
"output": "1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0"
},
{
"input": "2 14\na\nb\naa\nab\nba\nbb\naaa\naab\naba\nabb\nbaa\nbab\nbba\nbbb",
"output": "0\n1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0"
},
{
"input": "3 14\na\nb\naa\nab\nba\nbb\naaa\naab\naba\nabb\nbaa\nbab\nbba\nbbb",
"output": "1\n1\n0\n0\n1\n0\n0\n0\n0\n0\n0\n0\n0\n0"
},
{
"input": "4 14\na\nb\naa\nab\nba\nbb\naaa\naab\naba\nabb\nbaa\nbab\nbba\nbbb",
"output": "1\n2\n0\n1\n1\n0\n0\n0\n0\n0\n0\n1\n0\n0"
},
{
"input": "5 14\na\nb\naa\nab\nba\nbb\naaa\naab\naba\nabb\nbaa\nbab\nbba\nbbb",
"output": "2\n3\n0\n1\n2\n1\n0\n0\n0\n1\n0\n1\n1\n0"
}
] | 1,685,408,480
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS1
| 2
| 92
| 0
|
MOD = 10**9 + 7
def count_queries(k, queries):
fib1 = "a"
fib2 = "b"
for _ in range(2, k):
fib1, fib2 = fib2, fib2 + fib1
fib_k = fib2 if k > 1 else fib1
counts = []
for query in queries:
count = fib_k.count(query)
counts.append(count % MOD)
return counts
def main():
k, m = map(int, input().split())
queries = [input() for _ in range(m)]
counts = count_queries(k, queries)
for count in counts:
print(count)
if __name__ == "__main__":
main()
|
Title: Fibonacci Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fibonacci strings are defined as follows:
- *f*1 = «a» - *f*2 = «b» - *f**n* = *f**n*<=-<=1 *f**n*<=-<=2, *n*<=><=2
Thus, the first five Fibonacci strings are: "a", "b", "ba", "bab", "babba".
You are given a Fibonacci string and *m* strings *s**i*. For each string *s**i*, find the number of times it occurs in the given Fibonacci string as a substring.
Input Specification:
The first line contains two space-separated integers *k* and *m* — the number of a Fibonacci string and the number of queries, correspondingly.
Next *m* lines contain strings *s**i* that correspond to the queries. It is guaranteed that strings *s**i* aren't empty and consist only of characters "a" and "b".
The input limitations for getting 30 points are:
- 1<=≤<=*k*<=≤<=3000 - 1<=≤<=*m*<=≤<=3000 - The total length of strings *s**i* doesn't exceed 3000
The input limitations for getting 100 points are:
- 1<=≤<=*k*<=≤<=1018 - 1<=≤<=*m*<=≤<=104 - The total length of strings *s**i* doesn't exceed 105
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Output Specification:
For each string *s**i* print the number of times it occurs in the given Fibonacci string as a substring. Since the numbers can be large enough, print them modulo 1000000007 (109<=+<=7). Print the answers for the strings in the order in which they are given in the input.
Demo Input:
['6 5\na\nb\nab\nba\naba\n']
Demo Output:
['3\n5\n3\n3\n1\n']
Note:
none
|
```python
MOD = 10**9 + 7
def count_queries(k, queries):
fib1 = "a"
fib2 = "b"
for _ in range(2, k):
fib1, fib2 = fib2, fib2 + fib1
fib_k = fib2 if k > 1 else fib1
counts = []
for query in queries:
count = fib_k.count(query)
counts.append(count % MOD)
return counts
def main():
k, m = map(int, input().split())
queries = [input() for _ in range(m)]
counts = count_queries(k, queries)
for count in counts:
print(count)
if __name__ == "__main__":
main()
```
| 0
|
|
845
|
A
|
Chess Tourney
|
PROGRAMMING
| 1,100
|
[
"implementation",
"sortings"
] | null | null |
Berland annual chess tournament is coming!
Organizers have gathered 2·*n* chess players who should be divided into two teams with *n* people each. The first team is sponsored by BerOil and the second team is sponsored by BerMobile. Obviously, organizers should guarantee the win for the team of BerOil.
Thus, organizers should divide all 2·*n* players into two teams with *n* people each in such a way that the first team always wins.
Every chess player has its rating *r**i*. It is known that chess player with the greater rating always wins the player with the lower rating. If their ratings are equal then any of the players can win.
After teams assignment there will come a drawing to form *n* pairs of opponents: in each pair there is a player from the first team and a player from the second team. Every chess player should be in exactly one pair. Every pair plays once. The drawing is totally random.
Is it possible to divide all 2·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing?
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100).
The second line contains 2·*n* integers *a*1,<=*a*2,<=... *a*2*n* (1<=≤<=*a**i*<=≤<=1000).
|
If it's possible to divide all 2·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing, then print "YES". Otherwise print "NO".
|
[
"2\n1 3 2 4\n",
"1\n3 3\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 0
|
[
{
"input": "2\n1 3 2 4",
"output": "YES"
},
{
"input": "1\n3 3",
"output": "NO"
},
{
"input": "5\n1 1 1 1 2 2 3 3 3 3",
"output": "NO"
},
{
"input": "5\n1 1 1 1 1 2 2 2 2 2",
"output": "YES"
},
{
"input": "10\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "1\n2 3",
"output": "YES"
},
{
"input": "100\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",
"output": "NO"
},
{
"input": "35\n919 240 231 858 456 891 959 965 758 30 431 73 505 694 874 543 975 445 16 147 904 690 940 278 562 127 724 314 30 233 389 442 353 652 581 383 340 445 487 283 85 845 578 946 228 557 906 572 919 388 686 181 958 955 736 438 991 170 632 593 475 264 178 344 159 414 739 590 348 884",
"output": "YES"
},
{
"input": "5\n1 2 3 4 10 10 6 7 8 9",
"output": "YES"
},
{
"input": "2\n1 1 1 2",
"output": "NO"
},
{
"input": "2\n10 4 4 4",
"output": "NO"
},
{
"input": "2\n2 3 3 3",
"output": "NO"
},
{
"input": "4\n1 2 3 4 5 4 6 7",
"output": "NO"
},
{
"input": "4\n2 5 4 5 8 3 1 5",
"output": "YES"
},
{
"input": "4\n8 2 2 4 1 4 10 9",
"output": "NO"
},
{
"input": "2\n3 8 10 2",
"output": "YES"
},
{
"input": "3\n1 3 4 4 5 6",
"output": "NO"
},
{
"input": "2\n3 3 3 4",
"output": "NO"
},
{
"input": "2\n1 1 2 2",
"output": "YES"
},
{
"input": "2\n1 1 3 3",
"output": "YES"
},
{
"input": "2\n1 2 3 2",
"output": "NO"
},
{
"input": "10\n1 2 7 3 9 4 1 5 10 3 6 1 10 7 8 5 7 6 1 4",
"output": "NO"
},
{
"input": "3\n1 2 3 3 4 5",
"output": "NO"
},
{
"input": "2\n2 2 1 1",
"output": "YES"
},
{
"input": "7\n1 2 3 4 5 6 7 7 8 9 10 11 12 19",
"output": "NO"
},
{
"input": "5\n1 2 3 4 5 3 3 5 6 7",
"output": "YES"
},
{
"input": "4\n1 1 2 2 3 3 3 3",
"output": "YES"
},
{
"input": "51\n576 377 63 938 667 992 959 997 476 94 652 272 108 410 543 456 942 800 917 163 931 584 357 890 895 318 544 179 268 130 649 916 581 350 573 223 495 26 377 695 114 587 380 424 744 434 332 249 318 522 908 815 313 384 981 773 585 747 376 812 538 525 997 896 859 599 437 163 878 14 224 733 369 741 473 178 153 678 12 894 630 921 505 635 128 404 64 499 208 325 343 996 970 39 380 80 12 756 580 57 934 224",
"output": "YES"
},
{
"input": "3\n3 3 3 2 3 2",
"output": "NO"
},
{
"input": "2\n5 3 3 6",
"output": "YES"
},
{
"input": "2\n1 2 2 3",
"output": "NO"
},
{
"input": "2\n1 3 2 2",
"output": "NO"
},
{
"input": "2\n1 3 3 4",
"output": "NO"
},
{
"input": "2\n1 2 2 2",
"output": "NO"
},
{
"input": "3\n1 2 7 19 19 7",
"output": "NO"
},
{
"input": "3\n1 2 3 3 5 6",
"output": "NO"
},
{
"input": "2\n1 2 2 4",
"output": "NO"
},
{
"input": "2\n6 6 5 5",
"output": "YES"
},
{
"input": "2\n3 1 3 1",
"output": "YES"
},
{
"input": "3\n1 2 3 3 1 1",
"output": "YES"
},
{
"input": "3\n3 2 1 3 4 5",
"output": "NO"
},
{
"input": "3\n4 5 6 4 2 1",
"output": "NO"
},
{
"input": "3\n1 1 2 3 2 4",
"output": "NO"
},
{
"input": "3\n100 99 1 1 1 1",
"output": "NO"
},
{
"input": "3\n1 2 3 6 5 3",
"output": "NO"
},
{
"input": "2\n2 2 1 2",
"output": "NO"
},
{
"input": "4\n1 2 3 4 5 6 7 4",
"output": "NO"
},
{
"input": "3\n1 2 3 1 1 1",
"output": "NO"
},
{
"input": "3\n6 5 3 3 1 3",
"output": "NO"
},
{
"input": "2\n1 2 1 2",
"output": "YES"
},
{
"input": "3\n1 2 5 6 8 6",
"output": "YES"
},
{
"input": "5\n1 2 3 4 5 3 3 3 3 3",
"output": "NO"
},
{
"input": "2\n1 2 4 2",
"output": "NO"
},
{
"input": "3\n7 7 4 5 319 19",
"output": "NO"
},
{
"input": "3\n1 2 4 4 3 5",
"output": "YES"
},
{
"input": "3\n3 2 3 4 5 2",
"output": "NO"
},
{
"input": "5\n1 2 3 4 4 5 3 6 7 8",
"output": "NO"
},
{
"input": "3\n3 3 4 4 5 1",
"output": "YES"
},
{
"input": "2\n3 4 3 3",
"output": "NO"
},
{
"input": "2\n2 5 4 4",
"output": "NO"
},
{
"input": "5\n1 2 3 3 4 5 6 7 8 4",
"output": "NO"
},
{
"input": "3\n1 2 3 3 5 5",
"output": "NO"
},
{
"input": "2\n3 4 4 4",
"output": "NO"
},
{
"input": "2\n1 4 5 4",
"output": "NO"
},
{
"input": "2\n1 2 3 3",
"output": "YES"
},
{
"input": "2\n1 1 2 1",
"output": "NO"
},
{
"input": "4\n1 1 1 1 2 2 2 2",
"output": "YES"
},
{
"input": "4\n1 2 3 5 6 7 8 5",
"output": "NO"
},
{
"input": "2\n4 3 3 1",
"output": "NO"
},
{
"input": "3\n3 1 2 4 3 5",
"output": "NO"
},
{
"input": "3\n1 2 3 3 4 6",
"output": "NO"
},
{
"input": "4\n2 2 2 4 5 5 5 5",
"output": "YES"
},
{
"input": "2\n1 3 4 3",
"output": "NO"
},
{
"input": "2\n3 3 2 3",
"output": "NO"
},
{
"input": "2\n1 2 1 1",
"output": "NO"
},
{
"input": "3\n1 3 4 4 2 5",
"output": "YES"
},
{
"input": "4\n4 7 1 2 3 5 6 4",
"output": "NO"
},
{
"input": "2\n3 2 2 2",
"output": "NO"
},
{
"input": "1\n2 1",
"output": "YES"
},
{
"input": "2\n3 3 1 2",
"output": "YES"
},
{
"input": "1\n8 6",
"output": "YES"
},
{
"input": "7\n6 7 6 7 3 1 9 4 6 10 8 2 5 7",
"output": "NO"
},
{
"input": "2\n3 9 2 1",
"output": "YES"
},
{
"input": "2\n3 3 3 3",
"output": "NO"
}
] | 1,503,331,932
| 4,032
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 0
|
n=int(input())
x=input().split()
i=0
q=0
while(i<2*n):
if(x[i]>=x[i+1]):
q=1
print("NO")
break
i=i+2
if(q==0):
print("YES")
|
Title: Chess Tourney
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Berland annual chess tournament is coming!
Organizers have gathered 2·*n* chess players who should be divided into two teams with *n* people each. The first team is sponsored by BerOil and the second team is sponsored by BerMobile. Obviously, organizers should guarantee the win for the team of BerOil.
Thus, organizers should divide all 2·*n* players into two teams with *n* people each in such a way that the first team always wins.
Every chess player has its rating *r**i*. It is known that chess player with the greater rating always wins the player with the lower rating. If their ratings are equal then any of the players can win.
After teams assignment there will come a drawing to form *n* pairs of opponents: in each pair there is a player from the first team and a player from the second team. Every chess player should be in exactly one pair. Every pair plays once. The drawing is totally random.
Is it possible to divide all 2·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing?
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100).
The second line contains 2·*n* integers *a*1,<=*a*2,<=... *a*2*n* (1<=≤<=*a**i*<=≤<=1000).
Output Specification:
If it's possible to divide all 2·*n* players into two teams with *n* people each so that the player from the first team in every pair wins regardless of the results of the drawing, then print "YES". Otherwise print "NO".
Demo Input:
['2\n1 3 2 4\n', '1\n3 3\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
n=int(input())
x=input().split()
i=0
q=0
while(i<2*n):
if(x[i]>=x[i+1]):
q=1
print("NO")
break
i=i+2
if(q==0):
print("YES")
```
| 0
|
|
214
|
A
|
System of Equations
|
PROGRAMMING
| 800
|
[
"brute force"
] | null | null |
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system.
|
A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space.
|
On a single line print the answer to the problem.
|
[
"9 3\n",
"14 28\n",
"4 20\n"
] |
[
"1\n",
"1\n",
"0\n"
] |
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair.
| 500
|
[
{
"input": "9 3",
"output": "1"
},
{
"input": "14 28",
"output": "1"
},
{
"input": "4 20",
"output": "0"
},
{
"input": "18 198",
"output": "1"
},
{
"input": "22 326",
"output": "1"
},
{
"input": "26 104",
"output": "1"
},
{
"input": "14 10",
"output": "0"
},
{
"input": "8 20",
"output": "0"
},
{
"input": "2 8",
"output": "0"
},
{
"input": "20 11",
"output": "0"
},
{
"input": "57 447",
"output": "1"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "66 296",
"output": "1"
},
{
"input": "75 683",
"output": "1"
},
{
"input": "227 975",
"output": "1"
},
{
"input": "247 499",
"output": "1"
},
{
"input": "266 116",
"output": "1"
},
{
"input": "286 916",
"output": "1"
},
{
"input": "307 341",
"output": "1"
},
{
"input": "451 121",
"output": "1"
},
{
"input": "471 921",
"output": "1"
},
{
"input": "502 346",
"output": "1"
},
{
"input": "535 59",
"output": "1"
},
{
"input": "555 699",
"output": "1"
},
{
"input": "747 351",
"output": "1"
},
{
"input": "790 64",
"output": "1"
},
{
"input": "810 704",
"output": "1"
},
{
"input": "855 225",
"output": "1"
},
{
"input": "902 34",
"output": "1"
},
{
"input": "922 514",
"output": "1"
},
{
"input": "971 131",
"output": "1"
},
{
"input": "991 931",
"output": "1"
},
{
"input": "840 780",
"output": "0"
},
{
"input": "102 595",
"output": "0"
},
{
"input": "139 433",
"output": "0"
},
{
"input": "968 288",
"output": "0"
},
{
"input": "563 354",
"output": "0"
},
{
"input": "994 975",
"output": "0"
},
{
"input": "456 221",
"output": "0"
},
{
"input": "205 210",
"output": "0"
},
{
"input": "1 11",
"output": "0"
},
{
"input": "1000 1000",
"output": "0"
},
{
"input": "3 3",
"output": "0"
},
{
"input": "11 99",
"output": "0"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "11 1",
"output": "0"
},
{
"input": "6 6",
"output": "1"
},
{
"input": "100 452",
"output": "0"
},
{
"input": "420 380",
"output": "0"
},
{
"input": "31 31",
"output": "0"
},
{
"input": "2 6",
"output": "0"
},
{
"input": "112 134",
"output": "0"
},
{
"input": "13 13",
"output": "0"
},
{
"input": "1 571",
"output": "0"
}
] | 1,696,081,131
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 42
| 122
| 0
|
n, m = map(int, input().split())
c = 0
for b in range(100):
a = m - b ** 2
if a ** 2 + b == n:
c += 1
print(c)
|
Title: System of Equations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system.
Input Specification:
A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space.
Output Specification:
On a single line print the answer to the problem.
Demo Input:
['9 3\n', '14 28\n', '4 20\n']
Demo Output:
['1\n', '1\n', '0\n']
Note:
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair.
|
```python
n, m = map(int, input().split())
c = 0
for b in range(100):
a = m - b ** 2
if a ** 2 + b == n:
c += 1
print(c)
```
| 0
|
|
776
|
A
|
A Serial Killer
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation",
"strings"
] | null | null |
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
|
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=≤<=*n*<=≤<=1000), the number of days.
Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
|
Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
|
[
"ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n",
"icm codeforces\n1\ncodeforces technex\n"
] |
[
"ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n",
"icm codeforces\nicm technex\n"
] |
In first example, the killer starts with ross and rachel.
- After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears.
| 500
|
[
{
"input": "ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler",
"output": "ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler"
},
{
"input": "icm codeforces\n1\ncodeforces technex",
"output": "icm codeforces\nicm technex"
},
{
"input": "a b\n3\na c\nb d\nd e",
"output": "a b\nc b\nc d\nc e"
},
{
"input": "ze udggmyop\n4\nze szhrbmft\nudggmyop mjorab\nszhrbmft ojdtfnzxj\nojdtfnzxj yjlkg",
"output": "ze udggmyop\nszhrbmft udggmyop\nszhrbmft mjorab\nojdtfnzxj mjorab\nyjlkg mjorab"
},
{
"input": "q s\n10\nq b\nb j\ns g\nj f\nf m\ng c\nc a\nm d\nd z\nz o",
"output": "q s\nb s\nj s\nj g\nf g\nm g\nm c\nm a\nd a\nz a\no a"
},
{
"input": "iii iiiiii\n7\niii iiiiiiiiii\niiiiiiiiii iiii\niiii i\niiiiii iiiiiiii\niiiiiiii iiiiiiiii\ni iiiii\niiiii ii",
"output": "iii iiiiii\niiiiiiiiii iiiiii\niiii iiiiii\ni iiiiii\ni iiiiiiii\ni iiiiiiiii\niiiii iiiiiiiii\nii iiiiiiiii"
},
{
"input": "bwyplnjn zkms\n26\nzkms nzmcsytxh\nnzmcsytxh yujsb\nbwyplnjn gtbzhudpb\ngtbzhudpb hpk\nyujsb xvy\nhpk wrwnfokml\nwrwnfokml ndouuikw\nndouuikw ucgrja\nucgrja tgfmpldz\nxvy nycrfphn\nnycrfphn quvs\nquvs htdy\nhtdy k\ntgfmpldz xtdpkxm\nxtdpkxm suwqxs\nk fv\nsuwqxs qckllwy\nqckllwy diun\nfv lefa\nlefa gdoqjysx\ndiun dhpz\ngdoqjysx bdmqdyt\ndhpz dgz\ndgz v\nbdmqdyt aswy\naswy ydkayhlrnm",
"output": "bwyplnjn zkms\nbwyplnjn nzmcsytxh\nbwyplnjn yujsb\ngtbzhudpb yujsb\nhpk yujsb\nhpk xvy\nwrwnfokml xvy\nndouuikw xvy\nucgrja xvy\ntgfmpldz xvy\ntgfmpldz nycrfphn\ntgfmpldz quvs\ntgfmpldz htdy\ntgfmpldz k\nxtdpkxm k\nsuwqxs k\nsuwqxs fv\nqckllwy fv\ndiun fv\ndiun lefa\ndiun gdoqjysx\ndhpz gdoqjysx\ndhpz bdmqdyt\ndgz bdmqdyt\nv bdmqdyt\nv aswy\nv ydkayhlrnm"
},
{
"input": "wxz hbeqwqp\n7\nhbeqwqp cpieghnszh\ncpieghnszh tlqrpd\ntlqrpd ttwrtio\nttwrtio xapvds\nxapvds zk\nwxz yryk\nzk b",
"output": "wxz hbeqwqp\nwxz cpieghnszh\nwxz tlqrpd\nwxz ttwrtio\nwxz xapvds\nwxz zk\nyryk zk\nyryk b"
},
{
"input": "wced gnsgv\n23\ngnsgv japawpaf\njapawpaf nnvpeu\nnnvpeu a\na ddupputljq\nddupputljq qyhnvbh\nqyhnvbh pqwijl\nwced khuvs\nkhuvs bjkh\npqwijl ysacmboc\nbjkh srf\nsrf jknoz\njknoz hodf\nysacmboc xqtkoyh\nhodf rfp\nxqtkoyh bivgnwqvoe\nbivgnwqvoe nknf\nnknf wuig\nrfp e\ne bqqknq\nwuig sznhhhu\nbqqknq dhrtdld\ndhrtdld n\nsznhhhu bguylf",
"output": "wced gnsgv\nwced japawpaf\nwced nnvpeu\nwced a\nwced ddupputljq\nwced qyhnvbh\nwced pqwijl\nkhuvs pqwijl\nbjkh pqwijl\nbjkh ysacmboc\nsrf ysacmboc\njknoz ysacmboc\nhodf ysacmboc\nhodf xqtkoyh\nrfp xqtkoyh\nrfp bivgnwqvoe\nrfp nknf\nrfp wuig\ne wuig\nbqqknq wuig\nbqqknq sznhhhu\ndhrtdld sznhhhu\nn sznhhhu\nn bguylf"
},
{
"input": "qqqqqqqqqq qqqqqqqq\n3\nqqqqqqqq qqqqqqqqq\nqqqqqqqqq qqqqq\nqqqqq q",
"output": "qqqqqqqqqq qqqqqqqq\nqqqqqqqqqq qqqqqqqqq\nqqqqqqqqqq qqqqq\nqqqqqqqqqq q"
},
{
"input": "wwwww w\n8\nwwwww wwwwwwww\nwwwwwwww wwwwwwwww\nwwwwwwwww wwwwwwwwww\nw www\nwwwwwwwwww wwww\nwwww ww\nwww wwwwww\nwwwwww wwwwwww",
"output": "wwwww w\nwwwwwwww w\nwwwwwwwww w\nwwwwwwwwww w\nwwwwwwwwww www\nwwww www\nww www\nww wwwwww\nww wwwwwww"
},
{
"input": "k d\n17\nk l\nd v\nv z\nl r\nz i\nr s\ns p\np w\nw j\nj h\ni c\nh m\nm q\nc o\no g\nq x\nx n",
"output": "k d\nl d\nl v\nl z\nr z\nr i\ns i\np i\nw i\nj i\nh i\nh c\nm c\nq c\nq o\nq g\nx g\nn g"
}
] | 1,640,866,945
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 57
| 108
| 0
|
a, b = map(str, input().split())
n = int(input())
lst = [a, b]
lst_output = [a, b]
for i in range(n):
c, d = map(str, input().split())
lst.append(c)
lst.append(d)
def all_index(lst1, word):
all_index1 = []
for k in range(len(lst1)):
if lst1[k] == word:
all_index1.append(k)
return all_index1
for i in range(2, len(lst), 2):
lst_idx = all_index(lst_output, f"{lst[i]}")
for w in range(len(lst_idx)):
if lst_idx[w] == i-2:
lst_output += [lst[i+1], lst_output[i-1]]
elif lst_idx[w] == i-1:
lst_output += [lst_output[i-2], lst[i+1]]
for z in range(0, len(lst_output), 2):
print(lst_output[z], end=" "), print(lst_output[z+1])
|
Title: A Serial Killer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input Specification:
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer *n* (1<=≤<=*n*<=≤<=1000), the number of days.
Next *n* lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output Specification:
Output *n*<=+<=1 lines, the *i*-th line should contain the two persons from which the killer selects for the *i*-th murder. The (*n*<=+<=1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Demo Input:
['ross rachel\n4\nross joey\nrachel phoebe\nphoebe monica\nmonica chandler\n', 'icm codeforces\n1\ncodeforces technex\n']
Demo Output:
['ross rachel\njoey rachel\njoey phoebe\njoey monica\njoey chandler\n', 'icm codeforces\nicm technex\n']
Note:
In first example, the killer starts with ross and rachel.
- After day 1, ross is killed and joey appears. - After day 2, rachel is killed and phoebe appears. - After day 3, phoebe is killed and monica appears. - After day 4, monica is killed and chandler appears.
|
```python
a, b = map(str, input().split())
n = int(input())
lst = [a, b]
lst_output = [a, b]
for i in range(n):
c, d = map(str, input().split())
lst.append(c)
lst.append(d)
def all_index(lst1, word):
all_index1 = []
for k in range(len(lst1)):
if lst1[k] == word:
all_index1.append(k)
return all_index1
for i in range(2, len(lst), 2):
lst_idx = all_index(lst_output, f"{lst[i]}")
for w in range(len(lst_idx)):
if lst_idx[w] == i-2:
lst_output += [lst[i+1], lst_output[i-1]]
elif lst_idx[w] == i-1:
lst_output += [lst_output[i-2], lst[i+1]]
for z in range(0, len(lst_output), 2):
print(lst_output[z], end=" "), print(lst_output[z+1])
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
|
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
|
Print the *k*-rounding of *n*.
|
[
"375 4\n",
"10000 1\n",
"38101 0\n",
"123456789 8\n"
] |
[
"30000\n",
"10000\n",
"38101\n",
"12345678900000000\n"
] |
none
| 0
|
[
{
"input": "375 4",
"output": "30000"
},
{
"input": "10000 1",
"output": "10000"
},
{
"input": "38101 0",
"output": "38101"
},
{
"input": "123456789 8",
"output": "12345678900000000"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "2 0",
"output": "2"
},
{
"input": "100 0",
"output": "100"
},
{
"input": "1000000000 0",
"output": "1000000000"
},
{
"input": "160 2",
"output": "800"
},
{
"input": "3 0",
"output": "3"
},
{
"input": "10 0",
"output": "10"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "3 1",
"output": "30"
},
{
"input": "4 1",
"output": "20"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "6 1",
"output": "30"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "8 1",
"output": "40"
},
{
"input": "9 1",
"output": "90"
},
{
"input": "10 1",
"output": "10"
},
{
"input": "11 1",
"output": "110"
},
{
"input": "12 1",
"output": "60"
},
{
"input": "16 2",
"output": "400"
},
{
"input": "2 2",
"output": "100"
},
{
"input": "1 2",
"output": "100"
},
{
"input": "5 2",
"output": "100"
},
{
"input": "15 2",
"output": "300"
},
{
"input": "36 2",
"output": "900"
},
{
"input": "1 8",
"output": "100000000"
},
{
"input": "8 8",
"output": "100000000"
},
{
"input": "96 8",
"output": "300000000"
},
{
"input": "175 8",
"output": "700000000"
},
{
"input": "9999995 8",
"output": "199999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "12345678 8",
"output": "617283900000000"
},
{
"input": "78125 8",
"output": "100000000"
},
{
"input": "390625 8",
"output": "100000000"
},
{
"input": "1953125 8",
"output": "500000000"
},
{
"input": "9765625 8",
"output": "2500000000"
},
{
"input": "68359375 8",
"output": "17500000000"
},
{
"input": "268435456 8",
"output": "104857600000000"
},
{
"input": "125829120 8",
"output": "9830400000000"
},
{
"input": "128000 8",
"output": "400000000"
},
{
"input": "300000 8",
"output": "300000000"
},
{
"input": "3711871 8",
"output": "371187100000000"
},
{
"input": "55555 8",
"output": "1111100000000"
},
{
"input": "222222222 8",
"output": "11111111100000000"
},
{
"input": "479001600 8",
"output": "7484400000000"
},
{
"input": "655360001 7",
"output": "6553600010000000"
},
{
"input": "655360001 8",
"output": "65536000100000000"
},
{
"input": "1000000000 1",
"output": "1000000000"
},
{
"input": "1000000000 7",
"output": "1000000000"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "10000000 8",
"output": "100000000"
},
{
"input": "1000000 8",
"output": "100000000"
},
{
"input": "10000009 8",
"output": "1000000900000000"
},
{
"input": "10000005 8",
"output": "200000100000000"
},
{
"input": "10000002 8",
"output": "500000100000000"
},
{
"input": "999999997 8",
"output": "99999999700000000"
},
{
"input": "999999997 7",
"output": "9999999970000000"
},
{
"input": "999999995 8",
"output": "19999999900000000"
},
{
"input": "123 8",
"output": "12300000000"
},
{
"input": "24 2",
"output": "600"
},
{
"input": "16 4",
"output": "10000"
},
{
"input": "123456787 8",
"output": "12345678700000000"
},
{
"input": "100000000 8",
"output": "100000000"
},
{
"input": "7 1",
"output": "70"
},
{
"input": "101 1",
"output": "1010"
},
{
"input": "50 2",
"output": "100"
},
{
"input": "999999818 1",
"output": "4999999090"
},
{
"input": "2 1",
"output": "10"
},
{
"input": "123 1",
"output": "1230"
},
{
"input": "16 1",
"output": "80"
},
{
"input": "1 1",
"output": "10"
},
{
"input": "1000000000 8",
"output": "1000000000"
},
{
"input": "15304 6",
"output": "1913000000"
},
{
"input": "3 8",
"output": "300000000"
},
{
"input": "4 2",
"output": "100"
},
{
"input": "100000 7",
"output": "10000000"
},
{
"input": "5 8",
"output": "100000000"
},
{
"input": "16724 6",
"output": "4181000000"
},
{
"input": "999999999 1",
"output": "9999999990"
},
{
"input": "999999990 8",
"output": "9999999900000000"
},
{
"input": "999999999 8",
"output": "99999999900000000"
},
{
"input": "100000 1",
"output": "100000"
},
{
"input": "8 3",
"output": "1000"
},
{
"input": "16768 6",
"output": "262000000"
},
{
"input": "123456789 1",
"output": "1234567890"
},
{
"input": "2 8",
"output": "100000000"
},
{
"input": "999999937 8",
"output": "99999993700000000"
},
{
"input": "5 1",
"output": "10"
},
{
"input": "2000000 7",
"output": "10000000"
},
{
"input": "1999998 2",
"output": "99999900"
},
{
"input": "125 3",
"output": "1000"
}
] | 1,554,167,201
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 96
| 109
| 0
|
entrada = input()
a = int(entrada.split()[0])
k = int(entrada.split()[1])
"""
if(a%s == 0):
print(a)
else:
i = 2
mmc = a*s
p = sqrt(a)
while i <= p:
if s%i == 0 and a%i == 0:
mmc = int(mmc/i)
a = int(a/i)
s = int(s/i)
else:
i+=1
print(mmc)
"""
ult = int(a%10)
a2 = a
while(a2%10**k != 0):
a3 = a2
ult = int(a2%10)
while(ult == 0 and a3 > 1):
a3 = a3/10
ult = int(a3%10)
if(ult % 2 == 0):
a2 = int(a2*5)
elif(ult == 5):
a2 = int(a2*2)
else:
a2 = int(a2*10)
print(a2)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a given positive integer *n* denote its *k*-rounding as the minimum positive integer *x*, such that *x* ends with *k* or more zeros in base 10 and is divisible by *n*.
For example, 4-rounding of 375 is 375·80<==<=30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.
Write a program that will perform the *k*-rounding of *n*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=109, 0<=≤<=*k*<=≤<=8).
Output Specification:
Print the *k*-rounding of *n*.
Demo Input:
['375 4\n', '10000 1\n', '38101 0\n', '123456789 8\n']
Demo Output:
['30000\n', '10000\n', '38101\n', '12345678900000000\n']
Note:
none
|
```python
entrada = input()
a = int(entrada.split()[0])
k = int(entrada.split()[1])
"""
if(a%s == 0):
print(a)
else:
i = 2
mmc = a*s
p = sqrt(a)
while i <= p:
if s%i == 0 and a%i == 0:
mmc = int(mmc/i)
a = int(a/i)
s = int(s/i)
else:
i+=1
print(mmc)
"""
ult = int(a%10)
a2 = a
while(a2%10**k != 0):
a3 = a2
ult = int(a2%10)
while(ult == 0 and a3 > 1):
a3 = a3/10
ult = int(a3%10)
if(ult % 2 == 0):
a2 = int(a2*5)
elif(ult == 5):
a2 = int(a2*2)
else:
a2 = int(a2*10)
print(a2)
```
| 3
|
|
997
|
A
|
Convert to Ones
|
PROGRAMMING
| 1,500
|
[
"brute force",
"greedy",
"implementation",
"math"
] | null | null |
You've got a string $a_1, a_2, \dots, a_n$, consisting of zeros and ones.
Let's call a sequence of consecutive elements $a_i, a_{i<=+<=1}, \ldots,<=a_j$ ($1\leq<=i\leq<=j\leq<=n$) a substring of string $a$.
You can apply the following operations any number of times:
- Choose some substring of string $a$ (for example, you can choose entire string) and reverse it, paying $x$ coins for it (for example, «0101101» $\to$ «0111001»); - Choose some substring of string $a$ (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying $y$ coins for it (for example, «0101101» $\to$ «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
|
The first line of input contains integers $n$, $x$ and $y$ ($1<=\leq<=n<=\leq<=300\,000, 0 \leq x, y \leq 10^9$) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string $a$ of length $n$, consisting of zeros and ones.
|
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print $0$, if you do not need to perform any operations.
|
[
"5 1 10\n01000\n",
"5 10 1\n01000\n",
"7 2 3\n1111111\n"
] |
[
"11\n",
"2\n",
"0\n"
] |
In the first sample, at first you need to reverse substring $[1 \dots 2]$, and then you need to invert substring $[2 \dots 5]$.
Then the string was changed as follows:
«01000» $\to$ «10000» $\to$ «11111».
The total cost of operations is $1 + 10 = 11$.
In the second sample, at first you need to invert substring $[1 \dots 1]$, and then you need to invert substring $[3 \dots 5]$.
Then the string was changed as follows:
«01000» $\to$ «11000» $\to$ «11111».
The overall cost is $1 + 1 = 2$.
In the third example, string already consists only of ones, so the answer is $0$.
| 500
|
[
{
"input": "5 1 10\n01000",
"output": "11"
},
{
"input": "5 10 1\n01000",
"output": "2"
},
{
"input": "7 2 3\n1111111",
"output": "0"
},
{
"input": "1 60754033 959739508\n0",
"output": "959739508"
},
{
"input": "1 431963980 493041212\n1",
"output": "0"
},
{
"input": "1 314253869 261764879\n0",
"output": "261764879"
},
{
"input": "1 491511050 399084767\n1",
"output": "0"
},
{
"input": "2 163093925 214567542\n00",
"output": "214567542"
},
{
"input": "2 340351106 646854722\n10",
"output": "646854722"
},
{
"input": "2 222640995 489207317\n01",
"output": "489207317"
},
{
"input": "2 399898176 552898277\n11",
"output": "0"
},
{
"input": "2 690218164 577155357\n00",
"output": "577155357"
},
{
"input": "2 827538051 754412538\n10",
"output": "754412538"
},
{
"input": "2 636702427 259825230\n01",
"output": "259825230"
},
{
"input": "2 108926899 102177825\n11",
"output": "0"
},
{
"input": "3 368381052 440077270\n000",
"output": "440077270"
},
{
"input": "3 505700940 617334451\n100",
"output": "617334451"
},
{
"input": "3 499624340 643020827\n010",
"output": "1142645167"
},
{
"input": "3 75308005 971848814\n110",
"output": "971848814"
},
{
"input": "3 212627893 854138703\n001",
"output": "854138703"
},
{
"input": "3 31395883 981351561\n101",
"output": "981351561"
},
{
"input": "3 118671447 913685773\n011",
"output": "913685773"
},
{
"input": "3 255991335 385910245\n111",
"output": "0"
},
{
"input": "3 688278514 268200134\n000",
"output": "268200134"
},
{
"input": "3 825598402 445457315\n100",
"output": "445457315"
},
{
"input": "3 300751942 45676507\n010",
"output": "91353014"
},
{
"input": "3 517900980 438071829\n110",
"output": "438071829"
},
{
"input": "3 400190869 280424424\n001",
"output": "280424424"
},
{
"input": "3 577448050 344115384\n101",
"output": "344115384"
},
{
"input": "3 481435271 459737939\n011",
"output": "459737939"
},
{
"input": "3 931962412 913722450\n111",
"output": "0"
},
{
"input": "4 522194562 717060616\n0000",
"output": "717060616"
},
{
"input": "4 659514449 894317797\n1000",
"output": "894317797"
},
{
"input": "4 71574977 796834337\n0100",
"output": "868409314"
},
{
"input": "4 248832158 934154224\n1100",
"output": "934154224"
},
{
"input": "4 71474110 131122047\n0010",
"output": "202596157"
},
{
"input": "4 308379228 503761290\n1010",
"output": "812140518"
},
{
"input": "4 272484957 485636409\n0110",
"output": "758121366"
},
{
"input": "4 662893590 704772137\n1110",
"output": "704772137"
},
{
"input": "4 545183479 547124732\n0001",
"output": "547124732"
},
{
"input": "4 684444619 722440661\n1001",
"output": "722440661"
},
{
"input": "4 477963686 636258459\n0101",
"output": "1114222145"
},
{
"input": "4 360253575 773578347\n1101",
"output": "773578347"
},
{
"input": "4 832478048 910898234\n0011",
"output": "910898234"
},
{
"input": "4 343185412 714767937\n1011",
"output": "714767937"
},
{
"input": "4 480505300 892025118\n0111",
"output": "892025118"
},
{
"input": "4 322857895 774315007\n1111",
"output": "0"
},
{
"input": "4 386548854 246539479\n0000",
"output": "246539479"
},
{
"input": "4 523868742 128829368\n1000",
"output": "128829368"
},
{
"input": "4 956155921 11119257\n0100",
"output": "22238514"
},
{
"input": "4 188376438 93475808\n1100",
"output": "93475808"
},
{
"input": "4 754947032 158668188\n0010",
"output": "317336376"
},
{
"input": "4 927391856 637236921\n1010",
"output": "1274473842"
},
{
"input": "4 359679035 109461393\n0110",
"output": "218922786"
},
{
"input": "4 991751283 202031630\n1110",
"output": "202031630"
},
{
"input": "4 339351517 169008463\n0001",
"output": "169008463"
},
{
"input": "4 771638697 346265644\n1001",
"output": "346265644"
},
{
"input": "4 908958584 523522825\n0101",
"output": "1047045650"
},
{
"input": "4 677682252 405812714\n1101",
"output": "405812714"
},
{
"input": "4 815002139 288102603\n0011",
"output": "288102603"
},
{
"input": "4 952322026 760327076\n1011",
"output": "760327076"
},
{
"input": "4 663334158 312481698\n0111",
"output": "312481698"
},
{
"input": "4 840591339 154834293\n1111",
"output": "0"
},
{
"input": "14 3 11\n10110100011001",
"output": "20"
},
{
"input": "19 1 1\n1010101010101010101",
"output": "9"
},
{
"input": "1 10 1\n1",
"output": "0"
},
{
"input": "1 100 1\n1",
"output": "0"
},
{
"input": "5 1000 1\n11111",
"output": "0"
},
{
"input": "5 10 1\n11111",
"output": "0"
},
{
"input": "7 3 2\n1111111",
"output": "0"
},
{
"input": "5 1 10\n10101",
"output": "11"
},
{
"input": "1 3 2\n1",
"output": "0"
},
{
"input": "2 10 1\n11",
"output": "0"
},
{
"input": "4 148823922 302792601\n1010",
"output": "451616523"
},
{
"input": "1 2 1\n1",
"output": "0"
},
{
"input": "5 2 3\n00011",
"output": "3"
},
{
"input": "1 5 0\n1",
"output": "0"
},
{
"input": "7 2 3\n1001001",
"output": "5"
},
{
"input": "10 1 1000000000\n1111010111",
"output": "1000000001"
},
{
"input": "25 999999998 999999999\n1011001110101010100111001",
"output": "7999999985"
},
{
"input": "2 0 1\n00",
"output": "1"
},
{
"input": "2 1 100\n10",
"output": "100"
},
{
"input": "7 20 3\n1111111",
"output": "0"
},
{
"input": "1 1 0\n1",
"output": "0"
},
{
"input": "3 1 10\n010",
"output": "11"
},
{
"input": "2 1 0\n11",
"output": "0"
},
{
"input": "7 100 3\n1111111",
"output": "0"
},
{
"input": "5 1 1000\n10101",
"output": "1001"
},
{
"input": "5 2 1\n11111",
"output": "0"
},
{
"input": "1 1000 1\n1",
"output": "0"
},
{
"input": "1 799543940 488239239\n1",
"output": "0"
},
{
"input": "6 1 1000\n010101",
"output": "1002"
},
{
"input": "5 11 1\n11111",
"output": "0"
},
{
"input": "5 2 3\n10101",
"output": "5"
},
{
"input": "3 10 1\n111",
"output": "0"
},
{
"input": "7 9 10\n1001011",
"output": "19"
},
{
"input": "5 5 6\n10101",
"output": "11"
},
{
"input": "1 1000000000 0\n1",
"output": "0"
},
{
"input": "4 0 1\n0101",
"output": "1"
},
{
"input": "8 2 3\n10101010",
"output": "9"
},
{
"input": "6 3 100\n010101",
"output": "106"
},
{
"input": "3 3 2\n111",
"output": "0"
},
{
"input": "1 20 1\n1",
"output": "0"
},
{
"input": "2 1 2\n01",
"output": "2"
}
] | 1,585,382,284
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 115
| 171
| 3,072,000
|
import sys
n,x,y=map(int,sys.stdin.readline().split())
s=input()
a=0
for i in range(n-1):
if s[i]=='0':
if s[i]!=s[i+1]:
a+=1
if s[n-1]=='0':
a+=1
if a==0:
print(0)
else:
print((a-1)*(min(x,y))+y)
|
Title: Convert to Ones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a string $a_1, a_2, \dots, a_n$, consisting of zeros and ones.
Let's call a sequence of consecutive elements $a_i, a_{i<=+<=1}, \ldots,<=a_j$ ($1\leq<=i\leq<=j\leq<=n$) a substring of string $a$.
You can apply the following operations any number of times:
- Choose some substring of string $a$ (for example, you can choose entire string) and reverse it, paying $x$ coins for it (for example, «0101101» $\to$ «0111001»); - Choose some substring of string $a$ (for example, you can choose entire string or just one symbol) and replace each symbol to the opposite one (zeros are replaced by ones, and ones — by zeros), paying $y$ coins for it (for example, «0101101» $\to$ «0110001»).
You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.
What is the minimum number of coins you need to spend to get a string consisting only of ones?
Input Specification:
The first line of input contains integers $n$, $x$ and $y$ ($1<=\leq<=n<=\leq<=300\,000, 0 \leq x, y \leq 10^9$) — length of the string, cost of the first operation (substring reverse) and cost of the second operation (inverting all elements of substring).
The second line contains the string $a$ of length $n$, consisting of zeros and ones.
Output Specification:
Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print $0$, if you do not need to perform any operations.
Demo Input:
['5 1 10\n01000\n', '5 10 1\n01000\n', '7 2 3\n1111111\n']
Demo Output:
['11\n', '2\n', '0\n']
Note:
In the first sample, at first you need to reverse substring $[1 \dots 2]$, and then you need to invert substring $[2 \dots 5]$.
Then the string was changed as follows:
«01000» $\to$ «10000» $\to$ «11111».
The total cost of operations is $1 + 10 = 11$.
In the second sample, at first you need to invert substring $[1 \dots 1]$, and then you need to invert substring $[3 \dots 5]$.
Then the string was changed as follows:
«01000» $\to$ «11000» $\to$ «11111».
The overall cost is $1 + 1 = 2$.
In the third example, string already consists only of ones, so the answer is $0$.
|
```python
import sys
n,x,y=map(int,sys.stdin.readline().split())
s=input()
a=0
for i in range(n-1):
if s[i]=='0':
if s[i]!=s[i+1]:
a+=1
if s[n-1]=='0':
a+=1
if a==0:
print(0)
else:
print((a-1)*(min(x,y))+y)
```
| 3
|
|
915
|
A
|
Garden
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding.
|
The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.
|
Print one integer number — the minimum number of hours required to water the garden.
|
[
"3 6\n2 3 5\n",
"6 7\n1 2 3 4 5 6\n"
] |
[
"2\n",
"7\n"
] |
In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1.
| 0
|
[
{
"input": "3 6\n2 3 5",
"output": "2"
},
{
"input": "6 7\n1 2 3 4 5 6",
"output": "7"
},
{
"input": "5 97\n1 10 50 97 2",
"output": "1"
},
{
"input": "5 97\n1 10 50 100 2",
"output": "97"
},
{
"input": "100 100\n2 46 24 18 86 90 31 38 84 49 58 28 15 80 14 24 87 56 62 87 41 87 55 71 87 32 41 56 91 32 24 75 43 42 35 30 72 53 31 26 54 61 87 85 36 75 44 31 7 38 77 57 61 54 70 77 45 96 39 57 11 8 91 42 52 15 42 30 92 41 27 26 34 27 3 80 32 86 26 97 63 91 30 75 14 7 19 23 45 11 8 43 44 73 11 56 3 55 63 16",
"output": "50"
},
{
"input": "100 91\n13 13 62 96 74 47 81 46 78 21 20 42 4 73 25 30 76 74 58 28 25 52 42 48 74 40 82 9 25 29 17 22 46 64 57 95 81 39 47 86 40 95 97 35 31 98 45 98 47 78 52 63 58 14 89 97 17 95 28 22 20 36 68 38 95 16 2 26 54 47 42 31 31 81 21 21 65 40 82 53 60 71 75 33 96 98 6 22 95 12 5 48 18 27 58 62 5 96 36 75",
"output": "7"
},
{
"input": "8 8\n8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "3 8\n4 3 2",
"output": "2"
},
{
"input": "3 8\n2 4 2",
"output": "2"
},
{
"input": "3 6\n1 3 2",
"output": "2"
},
{
"input": "3 6\n3 2 5",
"output": "2"
},
{
"input": "3 8\n4 2 1",
"output": "2"
},
{
"input": "5 6\n2 3 5 1 2",
"output": "2"
},
{
"input": "2 6\n5 3",
"output": "2"
},
{
"input": "4 12\n6 4 3 1",
"output": "2"
},
{
"input": "3 18\n1 9 6",
"output": "2"
},
{
"input": "3 9\n3 2 1",
"output": "3"
},
{
"input": "3 6\n5 3 2",
"output": "2"
},
{
"input": "2 10\n5 2",
"output": "2"
},
{
"input": "2 18\n6 3",
"output": "3"
},
{
"input": "4 12\n1 2 12 3",
"output": "1"
},
{
"input": "3 7\n3 2 1",
"output": "7"
},
{
"input": "3 6\n3 2 1",
"output": "2"
},
{
"input": "5 10\n5 4 3 2 1",
"output": "2"
},
{
"input": "5 16\n8 4 2 1 7",
"output": "2"
},
{
"input": "6 7\n6 5 4 3 7 1",
"output": "1"
},
{
"input": "2 6\n3 2",
"output": "2"
},
{
"input": "2 4\n4 1",
"output": "1"
},
{
"input": "6 8\n2 4 1 3 5 7",
"output": "2"
},
{
"input": "6 8\n6 5 4 3 2 1",
"output": "2"
},
{
"input": "6 15\n5 2 3 6 4 3",
"output": "3"
},
{
"input": "4 8\n2 4 8 1",
"output": "1"
},
{
"input": "2 5\n5 1",
"output": "1"
},
{
"input": "4 18\n3 1 1 2",
"output": "6"
},
{
"input": "2 1\n2 1",
"output": "1"
},
{
"input": "3 10\n2 10 5",
"output": "1"
},
{
"input": "5 12\n12 4 4 4 3",
"output": "1"
},
{
"input": "3 6\n6 3 2",
"output": "1"
},
{
"input": "2 2\n2 1",
"output": "1"
},
{
"input": "3 18\n1 9 3",
"output": "2"
},
{
"input": "3 8\n7 2 4",
"output": "2"
},
{
"input": "2 100\n99 1",
"output": "100"
},
{
"input": "4 12\n1 3 4 2",
"output": "3"
},
{
"input": "3 6\n2 3 1",
"output": "2"
},
{
"input": "4 6\n3 2 5 12",
"output": "2"
},
{
"input": "4 97\n97 1 50 10",
"output": "1"
},
{
"input": "3 12\n1 12 2",
"output": "1"
},
{
"input": "4 12\n1 4 3 2",
"output": "3"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "3 19\n7 1 1",
"output": "19"
},
{
"input": "5 12\n12 4 3 4 4",
"output": "1"
},
{
"input": "3 8\n8 4 2",
"output": "1"
},
{
"input": "3 3\n3 2 1",
"output": "1"
},
{
"input": "5 6\n3 2 4 2 2",
"output": "2"
},
{
"input": "2 16\n8 4",
"output": "2"
},
{
"input": "3 6\n10 2 3",
"output": "2"
},
{
"input": "5 3\n2 4 5 3 6",
"output": "1"
},
{
"input": "11 99\n1 2 3 6 5 4 7 8 99 33 66",
"output": "1"
},
{
"input": "3 12\n3 12 2",
"output": "1"
},
{
"input": "5 25\n24 5 15 25 23",
"output": "1"
},
{
"input": "2 4\n8 1",
"output": "4"
},
{
"input": "4 100\n2 50 4 1",
"output": "2"
},
{
"input": "3 28\n7 14 1",
"output": "2"
},
{
"input": "4 8\n2 8 4 1",
"output": "1"
},
{
"input": "4 6\n6 1 2 3",
"output": "1"
},
{
"input": "2 12\n4 3",
"output": "3"
},
{
"input": "4 12\n1 2 4 3",
"output": "3"
},
{
"input": "5 12\n2 3 12 6 4",
"output": "1"
},
{
"input": "4 4\n1 2 2 4",
"output": "1"
},
{
"input": "3 6\n2 3 2",
"output": "2"
},
{
"input": "4 21\n21 20 21 2",
"output": "1"
},
{
"input": "3 8\n3 4 2",
"output": "2"
},
{
"input": "1 25\n25",
"output": "1"
},
{
"input": "99 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99",
"output": "1"
},
{
"input": "98 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98",
"output": "1"
},
{
"input": "79 12\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79",
"output": "1"
},
{
"input": "4 32\n1 1 1 1",
"output": "32"
},
{
"input": "1 100\n1",
"output": "100"
},
{
"input": "2 100\n7 1",
"output": "100"
},
{
"input": "7 24\n1 3 6 4 5 2 7",
"output": "4"
},
{
"input": "6 87\n1 2 8 4 5 7",
"output": "87"
},
{
"input": "1 88\n1",
"output": "88"
},
{
"input": "1 89\n1",
"output": "89"
}
] | 1,530,324,971
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 83
| 124
| 0
|
[_,k]=list(map(int,input().split(' ')))
a=list(map(int,input().split(' ')))
for i in sorted(a,reverse=True):
if k%i==0:
break
print(k//i)
|
Title: Garden
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding.
Input Specification:
The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.
Output Specification:
Print one integer number — the minimum number of hours required to water the garden.
Demo Input:
['3 6\n2 3 5\n', '6 7\n1 2 3 4 5 6\n']
Demo Output:
['2\n', '7\n']
Note:
In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1.
|
```python
[_,k]=list(map(int,input().split(' ')))
a=list(map(int,input().split(' ')))
for i in sorted(a,reverse=True):
if k%i==0:
break
print(k//i)
```
| 3
|
|
1,009
|
A
|
Game Shopping
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Maxim wants to buy some games at the local game shop. There are $n$ games in the shop, the $i$-th game costs $c_i$.
Maxim has a wallet which can be represented as an array of integers. His wallet contains $m$ bills, the $j$-th bill has value $a_j$.
Games in the shop are ordered from left to right, Maxim tries to buy every game in that order.
When Maxim stands at the position $i$ in the shop, he takes the first bill from his wallet (if his wallet is empty then he proceeds to the next position immediately) and tries to buy the $i$-th game using this bill. After Maxim tried to buy the $n$-th game, he leaves the shop.
Maxim buys the $i$-th game if and only if the value of the first bill (which he takes) from his wallet is greater or equal to the cost of the $i$-th game. If he successfully buys the $i$-th game, the first bill from his wallet disappears and the next bill becomes first. Otherwise Maxim leaves the first bill in his wallet (this bill still remains the first one) and proceeds to the next game.
For example, for array $c = [2, 4, 5, 2, 4]$ and array $a = [5, 3, 4, 6]$ the following process takes place: Maxim buys the first game using the first bill (its value is $5$), the bill disappears, after that the second bill (with value $3$) becomes the first one in Maxim's wallet, then Maxim doesn't buy the second game because $c_2 > a_2$, the same with the third game, then he buys the fourth game using the bill of value $a_2$ (the third bill becomes the first one in Maxim's wallet) and buys the fifth game using the bill of value $a_3$.
Your task is to get the number of games Maxim will buy.
|
The first line of the input contains two integers $n$ and $m$ ($1 \le n, m \le 1000$) — the number of games and the number of bills in Maxim's wallet.
The second line of the input contains $n$ integers $c_1, c_2, \dots, c_n$ ($1 \le c_i \le 1000$), where $c_i$ is the cost of the $i$-th game.
The third line of the input contains $m$ integers $a_1, a_2, \dots, a_m$ ($1 \le a_j \le 1000$), where $a_j$ is the value of the $j$-th bill from the Maxim's wallet.
|
Print a single integer — the number of games Maxim will buy.
|
[
"5 4\n2 4 5 2 4\n5 3 4 6\n",
"5 2\n20 40 50 20 40\n19 20\n",
"6 4\n4 8 15 16 23 42\n1000 1000 1000 1000\n"
] |
[
"3\n",
"0\n",
"4\n"
] |
The first example is described in the problem statement.
In the second example Maxim cannot buy any game because the value of the first bill in his wallet is smaller than the cost of any game in the shop.
In the third example the values of the bills in Maxim's wallet are large enough to buy any game he encounter until he runs out of bills in his wallet.
| 0
|
[
{
"input": "5 4\n2 4 5 2 4\n5 3 4 6",
"output": "3"
},
{
"input": "5 2\n20 40 50 20 40\n19 20",
"output": "0"
},
{
"input": "6 4\n4 8 15 16 23 42\n1000 1000 1000 1000",
"output": "4"
},
{
"input": "5 1\n1 1 1 1 1\n5",
"output": "1"
},
{
"input": "5 1\n10 1 1 1 1\n1000",
"output": "1"
},
{
"input": "5 1\n100 100 100 100 100\n100",
"output": "1"
},
{
"input": "2 1\n2 1\n1",
"output": "1"
},
{
"input": "2 3\n3 1\n2 4 2",
"output": "1"
},
{
"input": "1 5\n4\n1 4 3 3 2",
"output": "0"
},
{
"input": "5 3\n4 2 3 1 1\n2 1 3",
"output": "3"
},
{
"input": "3 5\n5 2 5\n1 4 1 4 2",
"output": "0"
},
{
"input": "7 3\n9 7 10 2 1 1 1\n8 9 6",
"output": "3"
},
{
"input": "5 3\n2 5 3 3 2\n2 5 3",
"output": "3"
}
] | 1,694,205,275
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 46
| 0
|
n,m=map(int,input().split())
l1=list(map(int,input().split()))
l2=list(map(int,input().split()))
i1,i2=0,0
while(i1!=n and i2!=m):
if l1[i1]<=l2[i2]:
i2+=1
i1+=1
print(i2)
|
Title: Game Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Maxim wants to buy some games at the local game shop. There are $n$ games in the shop, the $i$-th game costs $c_i$.
Maxim has a wallet which can be represented as an array of integers. His wallet contains $m$ bills, the $j$-th bill has value $a_j$.
Games in the shop are ordered from left to right, Maxim tries to buy every game in that order.
When Maxim stands at the position $i$ in the shop, he takes the first bill from his wallet (if his wallet is empty then he proceeds to the next position immediately) and tries to buy the $i$-th game using this bill. After Maxim tried to buy the $n$-th game, he leaves the shop.
Maxim buys the $i$-th game if and only if the value of the first bill (which he takes) from his wallet is greater or equal to the cost of the $i$-th game. If he successfully buys the $i$-th game, the first bill from his wallet disappears and the next bill becomes first. Otherwise Maxim leaves the first bill in his wallet (this bill still remains the first one) and proceeds to the next game.
For example, for array $c = [2, 4, 5, 2, 4]$ and array $a = [5, 3, 4, 6]$ the following process takes place: Maxim buys the first game using the first bill (its value is $5$), the bill disappears, after that the second bill (with value $3$) becomes the first one in Maxim's wallet, then Maxim doesn't buy the second game because $c_2 > a_2$, the same with the third game, then he buys the fourth game using the bill of value $a_2$ (the third bill becomes the first one in Maxim's wallet) and buys the fifth game using the bill of value $a_3$.
Your task is to get the number of games Maxim will buy.
Input Specification:
The first line of the input contains two integers $n$ and $m$ ($1 \le n, m \le 1000$) — the number of games and the number of bills in Maxim's wallet.
The second line of the input contains $n$ integers $c_1, c_2, \dots, c_n$ ($1 \le c_i \le 1000$), where $c_i$ is the cost of the $i$-th game.
The third line of the input contains $m$ integers $a_1, a_2, \dots, a_m$ ($1 \le a_j \le 1000$), where $a_j$ is the value of the $j$-th bill from the Maxim's wallet.
Output Specification:
Print a single integer — the number of games Maxim will buy.
Demo Input:
['5 4\n2 4 5 2 4\n5 3 4 6\n', '5 2\n20 40 50 20 40\n19 20\n', '6 4\n4 8 15 16 23 42\n1000 1000 1000 1000\n']
Demo Output:
['3\n', '0\n', '4\n']
Note:
The first example is described in the problem statement.
In the second example Maxim cannot buy any game because the value of the first bill in his wallet is smaller than the cost of any game in the shop.
In the third example the values of the bills in Maxim's wallet are large enough to buy any game he encounter until he runs out of bills in his wallet.
|
```python
n,m=map(int,input().split())
l1=list(map(int,input().split()))
l2=list(map(int,input().split()))
i1,i2=0,0
while(i1!=n and i2!=m):
if l1[i1]<=l2[i2]:
i2+=1
i1+=1
print(i2)
```
| 3
|
|
296
|
A
|
Yaroslav and Permutations
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
|
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
|
[
"1\n1\n",
"3\n1 1 2\n",
"4\n7 7 7 7\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs.
| 500
|
[
{
"input": "1\n1",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "YES"
},
{
"input": "4\n7 7 7 7",
"output": "NO"
},
{
"input": "4\n479 170 465 146",
"output": "YES"
},
{
"input": "5\n996 437 605 996 293",
"output": "YES"
},
{
"input": "6\n727 539 896 668 36 896",
"output": "YES"
},
{
"input": "7\n674 712 674 674 674 674 674",
"output": "NO"
},
{
"input": "8\n742 742 742 742 742 289 742 742",
"output": "NO"
},
{
"input": "9\n730 351 806 806 806 630 85 757 967",
"output": "YES"
},
{
"input": "10\n324 539 83 440 834 640 440 440 440 440",
"output": "YES"
},
{
"input": "7\n925 830 925 98 987 162 356",
"output": "YES"
},
{
"input": "68\n575 32 53 351 151 942 725 967 431 108 192 8 338 458 288 754 384 946 910 210 759 222 589 423 947 507 31 414 169 901 592 763 656 411 360 625 538 549 484 596 42 603 351 292 837 375 21 597 22 349 200 669 485 282 735 54 1000 419 939 901 789 128 468 729 894 649 484 808",
"output": "YES"
},
{
"input": "22\n618 814 515 310 617 936 452 601 250 520 557 799 304 225 9 845 610 990 703 196 486 94",
"output": "YES"
},
{
"input": "44\n459 581 449 449 449 449 449 449 449 623 449 449 449 449 449 449 449 449 889 449 203 273 329 449 449 449 449 449 449 845 882 323 22 449 449 893 449 449 449 449 449 870 449 402",
"output": "NO"
},
{
"input": "90\n424 3 586 183 286 89 427 618 758 833 933 170 155 722 190 977 330 369 693 426 556 435 550 442 513 146 61 719 754 140 424 280 997 688 530 550 438 867 950 194 196 298 417 287 106 489 283 456 735 115 702 317 672 787 264 314 356 186 54 913 809 833 946 314 757 322 559 647 983 482 145 197 223 130 162 536 451 174 467 45 660 293 440 254 25 155 511 746 650 187",
"output": "YES"
},
{
"input": "14\n959 203 478 315 788 788 373 834 488 519 774 764 193 103",
"output": "YES"
},
{
"input": "81\n544 528 528 528 528 4 506 528 32 528 528 528 528 528 528 528 528 975 528 528 528 528 528 528 528 528 528 528 528 528 528 20 528 528 528 528 528 528 528 528 852 528 528 120 528 528 61 11 528 528 528 228 528 165 883 528 488 475 628 528 528 528 528 528 528 597 528 528 528 528 528 528 528 528 528 528 528 412 528 521 925",
"output": "NO"
},
{
"input": "89\n354 356 352 355 355 355 352 354 354 352 355 356 355 352 354 356 354 355 355 354 353 352 352 355 355 356 352 352 353 356 352 353 354 352 355 352 353 353 353 354 353 354 354 353 356 353 353 354 354 354 354 353 352 353 355 356 356 352 356 354 353 352 355 354 356 356 356 354 354 356 354 355 354 355 353 352 354 355 352 355 355 354 356 353 353 352 356 352 353",
"output": "YES"
},
{
"input": "71\n284 284 285 285 285 284 285 284 284 285 284 285 284 284 285 284 285 285 285 285 284 284 285 285 284 284 284 285 284 285 284 285 285 284 284 284 285 284 284 285 285 285 284 284 285 284 285 285 284 285 285 284 285 284 284 284 285 285 284 285 284 285 285 285 285 284 284 285 285 284 285",
"output": "NO"
},
{
"input": "28\n602 216 214 825 814 760 814 28 76 814 814 288 814 814 222 707 11 490 814 543 914 705 814 751 976 814 814 99",
"output": "YES"
},
{
"input": "48\n546 547 914 263 986 945 914 914 509 871 324 914 153 571 914 914 914 528 970 566 544 914 914 914 410 914 914 589 609 222 914 889 691 844 621 68 914 36 914 39 630 749 914 258 945 914 727 26",
"output": "YES"
},
{
"input": "56\n516 76 516 197 516 427 174 516 706 813 94 37 516 815 516 516 937 483 16 516 842 516 638 691 516 635 516 516 453 263 516 516 635 257 125 214 29 81 516 51 362 516 677 516 903 516 949 654 221 924 516 879 516 516 972 516",
"output": "YES"
},
{
"input": "46\n314 723 314 314 314 235 314 314 314 314 270 314 59 972 314 216 816 40 314 314 314 314 314 314 314 381 314 314 314 314 314 314 314 789 314 957 114 942 314 314 29 314 314 72 314 314",
"output": "NO"
},
{
"input": "72\n169 169 169 599 694 81 250 529 865 406 817 169 667 169 965 169 169 663 65 169 903 169 942 763 169 807 169 603 169 169 13 169 169 810 169 291 169 169 169 169 169 169 169 713 169 440 169 169 169 169 169 480 169 169 867 169 169 169 169 169 169 169 169 393 169 169 459 169 99 169 601 800",
"output": "NO"
},
{
"input": "100\n317 316 317 316 317 316 317 316 317 316 316 317 317 316 317 316 316 316 317 316 317 317 316 317 316 316 316 316 316 316 317 316 317 317 317 317 317 317 316 316 316 317 316 317 316 317 316 317 317 316 317 316 317 317 316 317 316 317 316 317 316 316 316 317 317 317 317 317 316 317 317 316 316 316 316 317 317 316 317 316 316 316 316 316 316 317 316 316 317 317 317 317 317 317 317 317 317 316 316 317",
"output": "NO"
},
{
"input": "100\n510 510 510 162 969 32 510 511 510 510 911 183 496 875 903 461 510 510 123 578 510 510 510 510 510 755 510 673 510 510 763 510 510 909 510 435 487 959 807 510 368 788 557 448 284 332 510 949 510 510 777 112 857 926 487 510 510 510 678 510 510 197 829 427 698 704 409 509 510 238 314 851 510 651 510 455 682 510 714 635 973 510 443 878 510 510 510 591 510 24 596 510 43 183 510 510 671 652 214 784",
"output": "YES"
},
{
"input": "100\n476 477 474 476 476 475 473 476 474 475 473 477 476 476 474 476 474 475 476 477 473 473 473 474 474 476 473 473 476 476 475 476 473 474 473 473 477 475 475 475 476 475 477 477 477 476 475 475 475 473 476 477 475 476 477 473 474 477 473 475 476 476 474 477 476 474 473 477 473 475 477 473 476 474 477 473 475 477 473 476 476 475 476 475 474 473 477 473 475 473 477 473 473 474 475 473 477 476 477 474",
"output": "YES"
},
{
"input": "100\n498 498 498 498 498 499 498 499 499 499 498 498 498 498 499 498 499 499 498 499 498 498 498 499 499 499 498 498 499 499 498 498 498 499 498 499 498 498 498 499 498 499 498 498 498 498 499 498 498 499 498 498 499 498 499 499 498 499 499 499 498 498 498 498 499 498 499 498 499 499 499 499 498 498 499 499 498 499 499 498 498 499 499 498 498 499 499 499 498 498 499 498 498 498 499 499 499 498 498 499",
"output": "NO"
},
{
"input": "100\n858 53 816 816 816 816 816 816 816 181 816 816 816 816 579 879 816 948 171 816 816 150 866 816 816 816 897 816 816 816 816 816 816 706 816 539 816 816 816 816 816 816 423 487 816 615 254 816 816 816 816 83 816 816 816 816 816 816 816 816 816 816 816 136 775 999 816 816 816 644 816 816 816 816 927 816 802 816 856 816 816 816 816 816 816 816 816 816 816 700 816 816 816 816 982 477 816 891 806 816",
"output": "NO"
},
{
"input": "100\n167 169 169 167 169 169 167 167 167 167 168 166 170 170 169 170 170 170 169 168 166 167 170 169 167 169 168 169 166 170 166 167 170 166 166 167 169 166 166 169 166 167 168 168 170 167 168 166 168 170 167 168 167 169 169 166 168 167 170 168 167 169 168 169 166 168 168 169 169 166 170 168 167 169 170 168 167 169 168 167 168 168 166 169 170 170 166 166 167 170 167 168 167 167 169 169 166 166 169 167",
"output": "YES"
},
{
"input": "100\n1000 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",
"output": "NO"
},
{
"input": "99\n1000 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",
"output": "NO"
},
{
"input": "100\n1000 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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "YES"
},
{
"input": "99\n1000 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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "YES"
},
{
"input": "2\n1 1",
"output": "NO"
},
{
"input": "1\n1000",
"output": "YES"
},
{
"input": "12\n2 2 4 4 4 4 6 6 6 6 6 6",
"output": "YES"
}
] | 1,575,567,783
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 17
| 248
| 0
|
v=[]
n=[]
s=0
z=int(input())
x=list(map(int,input().split()))
for c in x:
if c not in v:
v.append(c)
if z>1 and len(v)==1:
s=99999
for b in v:
n.append(x.count(b))
#print('n=',n)
m=max(n)
n.remove(m)
#print('nnnnnnn=',n)
for a in n:
s=s+a
#print('s=',s)
if s==m-1 or z//2<len(v) or s==m :
print("YES")
else:
print('NO')
|
Title: Yaroslav and Permutations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
Output Specification:
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
Demo Input:
['1\n1\n', '3\n1 1 2\n', '4\n7 7 7 7\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs.
|
```python
v=[]
n=[]
s=0
z=int(input())
x=list(map(int,input().split()))
for c in x:
if c not in v:
v.append(c)
if z>1 and len(v)==1:
s=99999
for b in v:
n.append(x.count(b))
#print('n=',n)
m=max(n)
n.remove(m)
#print('nnnnnnn=',n)
for a in n:
s=s+a
#print('s=',s)
if s==m-1 or z//2<len(v) or s==m :
print("YES")
else:
print('NO')
```
| 0
|
|
348
|
C
|
Subset Sums
|
PROGRAMMING
| 2,500
|
[
"brute force",
"data structures"
] | null | null |
You are given an array *a*1,<=*a*2,<=...,<=*a**n* and *m* sets *S*1,<=*S*2,<=...,<=*S**m* of indices of elements of this array. Let's denote *S**k*<==<={*S**k*,<=*i*} (1<=≤<=*i*<=≤<=|*S**k*|). In other words, *S**k*,<=*i* is some element from set *S**k*.
In this problem you have to answer *q* queries of the two types:
1. Find the sum of elements with indices from set *S**k*: . The query format is "? k". 1. Add number *x* to all elements at indices from set *S**k*: *a**S**k*,<=*i* is replaced by *a**S**k*,<=*i*<=+<=*x* for all *i* (1<=≤<=*i*<=≤<=|*S**k*|). The query format is "+ k x".
After each first type query print the required sum.
|
The first line contains integers *n*,<=*m*,<=*q* (1<=≤<=*n*,<=*m*,<=*q*<=≤<=105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=108) — elements of array *a*.
Each of the following *m* lines describes one set of indices. The *k*-th line first contains a positive integer, representing the number of elements in set (|*S**k*|), then follow |*S**k*| distinct integers *S**k*,<=1,<=*S**k*,<=2,<=...,<=*S**k*,<=|*S**k*| (1<=≤<=*S**k*,<=*i*<=≤<=*n*) — elements of set *S**k*.
The next *q* lines contain queries. Each query looks like either "? k" or "+ k x" and sits on a single line. For all queries the following limits are held: 1<=≤<=*k*<=≤<=*m*, |*x*|<=≤<=108. The queries are given in order they need to be answered.
It is guaranteed that the sum of sizes of all sets *S**k* doesn't exceed 105.
|
After each first type query print the required sum on a single line.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
[
"5 3 5\n5 -5 5 1 -4\n2 1 2\n4 2 1 4 5\n2 2 5\n? 2\n+ 3 4\n? 1\n+ 2 1\n? 2\n"
] |
[
"-3\n4\n9\n"
] |
none
| 1,500
|
[
{
"input": "5 3 5\n5 -5 5 1 -4\n2 1 2\n4 2 1 4 5\n2 2 5\n? 2\n+ 3 4\n? 1\n+ 2 1\n? 2",
"output": "-3\n4\n9"
},
{
"input": "10 10 10\n0 0 0 1 -5 8 7 2 9 -2\n8 1 9 7 8 2 5 10 3\n2 7 10\n8 8 7 1 4 2 5 10 9\n2 7 5\n8 9 3 6 4 8 1 2 10\n2 9 2\n8 4 2 9 1 10 7 5 3\n2 3 6\n8 2 8 5 9 4 1 6 10\n2 5 2\n? 1\n? 3\n? 5\n? 4\n+ 9 0\n+ 2 -5\n+ 4 7\n+ 5 -1\n? 2\n? 7",
"output": "11\n12\n18\n2\n1\n8"
},
{
"input": "1 1 1\n0\n1 1\n? 1",
"output": "0"
},
{
"input": "1 1 1\n100000000\n1 1\n? 1",
"output": "100000000"
}
] | 1,646,162,613
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 154
| 2,048,000
|
n = int(input())
arr2 = input()
arr2 = arr2.split(' ')
arr = list(map(int,arr2))
fin = []
def sum(arr, n):
total = 1 << n
for i in range(total):
s = 0
for j in range(n):
if ((i & (1 << j)) != 0):
s += arr[j]
fin.append(s)
sum(arr, n);
fin.sort()
fin.reverse()
for i in range(len(fin)):
print(fin[i])
|
Title: Subset Sums
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a*1,<=*a*2,<=...,<=*a**n* and *m* sets *S*1,<=*S*2,<=...,<=*S**m* of indices of elements of this array. Let's denote *S**k*<==<={*S**k*,<=*i*} (1<=≤<=*i*<=≤<=|*S**k*|). In other words, *S**k*,<=*i* is some element from set *S**k*.
In this problem you have to answer *q* queries of the two types:
1. Find the sum of elements with indices from set *S**k*: . The query format is "? k". 1. Add number *x* to all elements at indices from set *S**k*: *a**S**k*,<=*i* is replaced by *a**S**k*,<=*i*<=+<=*x* for all *i* (1<=≤<=*i*<=≤<=|*S**k*|). The query format is "+ k x".
After each first type query print the required sum.
Input Specification:
The first line contains integers *n*,<=*m*,<=*q* (1<=≤<=*n*,<=*m*,<=*q*<=≤<=105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=108) — elements of array *a*.
Each of the following *m* lines describes one set of indices. The *k*-th line first contains a positive integer, representing the number of elements in set (|*S**k*|), then follow |*S**k*| distinct integers *S**k*,<=1,<=*S**k*,<=2,<=...,<=*S**k*,<=|*S**k*| (1<=≤<=*S**k*,<=*i*<=≤<=*n*) — elements of set *S**k*.
The next *q* lines contain queries. Each query looks like either "? k" or "+ k x" and sits on a single line. For all queries the following limits are held: 1<=≤<=*k*<=≤<=*m*, |*x*|<=≤<=108. The queries are given in order they need to be answered.
It is guaranteed that the sum of sizes of all sets *S**k* doesn't exceed 105.
Output Specification:
After each first type query print the required sum on a single line.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['5 3 5\n5 -5 5 1 -4\n2 1 2\n4 2 1 4 5\n2 2 5\n? 2\n+ 3 4\n? 1\n+ 2 1\n? 2\n']
Demo Output:
['-3\n4\n9\n']
Note:
none
|
```python
n = int(input())
arr2 = input()
arr2 = arr2.split(' ')
arr = list(map(int,arr2))
fin = []
def sum(arr, n):
total = 1 << n
for i in range(total):
s = 0
for j in range(n):
if ((i & (1 << j)) != 0):
s += arr[j]
fin.append(s)
sum(arr, n);
fin.sort()
fin.reverse()
for i in range(len(fin)):
print(fin[i])
```
| -1
|
|
546
|
A
|
Soldier and Bananas
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation",
"math"
] | null | null |
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
|
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=≤<=<=*k*,<=*w*<=<=≤<=<=1000, 0<=≤<=*n*<=≤<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
|
Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
|
[
"3 17 4\n"
] |
[
"13"
] |
none
| 500
|
[
{
"input": "3 17 4",
"output": "13"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "1 5 6",
"output": "16"
},
{
"input": "1 1000000000 1",
"output": "0"
},
{
"input": "1000 0 1000",
"output": "500500000"
},
{
"input": "859 453892 543",
"output": "126416972"
},
{
"input": "1000 1000000000 1000",
"output": "0"
},
{
"input": "1000 500500000 1000",
"output": "0"
},
{
"input": "1000 500500001 1000",
"output": "0"
},
{
"input": "1000 500499999 1000",
"output": "1"
},
{
"input": "634 87973 214",
"output": "14497197"
},
{
"input": "432 10000 241",
"output": "12587552"
},
{
"input": "111 111111111 111",
"output": "0"
},
{
"input": "20 43 3",
"output": "77"
}
] | 1,695,113,436
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 15
| 31
| 0
|
k,n,w = map(int,input().split())
print(max(0,k*w*(w+1)//2-n))
|
Title: Soldier and Bananas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
Input Specification:
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=≤<=<=*k*,<=*w*<=<=≤<=<=1000, 0<=≤<=*n*<=≤<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
Output Specification:
Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
Demo Input:
['3 17 4\n']
Demo Output:
['13']
Note:
none
|
```python
k,n,w = map(int,input().split())
print(max(0,k*w*(w+1)//2-n))
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
The last stage of Football World Cup is played using the play-off system.
There are *n* teams left in this stage, they are enumerated from 1 to *n*. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids *a* and *b* can meet.
|
The only line contains three integers *n*, *a* and *b* (2<=≤<=*n*<=≤<=256, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that *n* is such that in each round an even number of team advance, and that *a* and *b* are not equal.
|
In the only line print "Final!" (without quotes), if teams *a* and *b* can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams *a* and *b* can meet. The round are enumerated from 1.
|
[
"4 1 2\n",
"8 2 6\n",
"8 7 5\n"
] |
[
"1\n",
"Final!\n",
"2\n"
] |
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round.
| 0
|
[
{
"input": "4 1 2",
"output": "1"
},
{
"input": "8 2 6",
"output": "Final!"
},
{
"input": "8 7 5",
"output": "2"
},
{
"input": "128 30 98",
"output": "Final!"
},
{
"input": "256 128 256",
"output": "Final!"
},
{
"input": "256 2 127",
"output": "7"
},
{
"input": "2 1 2",
"output": "Final!"
},
{
"input": "2 2 1",
"output": "Final!"
},
{
"input": "4 1 3",
"output": "Final!"
},
{
"input": "4 1 4",
"output": "Final!"
},
{
"input": "4 2 1",
"output": "1"
},
{
"input": "4 2 3",
"output": "Final!"
},
{
"input": "4 2 4",
"output": "Final!"
},
{
"input": "4 3 1",
"output": "Final!"
},
{
"input": "4 3 2",
"output": "Final!"
},
{
"input": "4 3 4",
"output": "1"
},
{
"input": "4 4 1",
"output": "Final!"
},
{
"input": "4 4 2",
"output": "Final!"
},
{
"input": "4 4 3",
"output": "1"
},
{
"input": "8 8 7",
"output": "1"
},
{
"input": "8 8 5",
"output": "2"
},
{
"input": "8 8 1",
"output": "Final!"
},
{
"input": "16 4 3",
"output": "1"
},
{
"input": "16 2 4",
"output": "2"
},
{
"input": "16 14 11",
"output": "3"
},
{
"input": "16 3 11",
"output": "Final!"
},
{
"input": "32 10 9",
"output": "1"
},
{
"input": "32 25 28",
"output": "2"
},
{
"input": "32 22 18",
"output": "3"
},
{
"input": "32 17 25",
"output": "4"
},
{
"input": "32 18 3",
"output": "Final!"
},
{
"input": "64 40 39",
"output": "1"
},
{
"input": "64 60 58",
"output": "2"
},
{
"input": "64 34 37",
"output": "3"
},
{
"input": "64 26 24",
"output": "4"
},
{
"input": "64 50 43",
"output": "5"
},
{
"input": "64 17 42",
"output": "Final!"
},
{
"input": "128 116 115",
"output": "1"
},
{
"input": "128 35 33",
"output": "2"
},
{
"input": "128 61 59",
"output": "3"
},
{
"input": "128 116 123",
"output": "4"
},
{
"input": "128 17 15",
"output": "5"
},
{
"input": "128 124 77",
"output": "6"
},
{
"input": "128 4 80",
"output": "Final!"
},
{
"input": "256 224 223",
"output": "1"
},
{
"input": "256 24 22",
"output": "2"
},
{
"input": "256 199 196",
"output": "3"
},
{
"input": "256 148 159",
"output": "4"
},
{
"input": "256 178 166",
"output": "5"
},
{
"input": "256 75 97",
"output": "6"
},
{
"input": "256 185 200",
"output": "7"
},
{
"input": "256 3 238",
"output": "Final!"
},
{
"input": "256 128 129",
"output": "Final!"
},
{
"input": "256 255 129",
"output": "7"
},
{
"input": "256 255 128",
"output": "Final!"
},
{
"input": "256 129 256",
"output": "7"
},
{
"input": "128 98 69",
"output": "6"
},
{
"input": "128 47 83",
"output": "Final!"
},
{
"input": "16 2 3",
"output": "2"
},
{
"input": "64 32 30",
"output": "2"
},
{
"input": "8 4 5",
"output": "Final!"
},
{
"input": "8 7 8",
"output": "1"
},
{
"input": "8 2 3",
"output": "2"
},
{
"input": "8 2 5",
"output": "Final!"
}
] | 1,520,154,229
| 1,429
|
Python 3
|
OK
|
TESTS
| 64
| 77
| 5,632,000
|
n, a, b = map(int, input().split())
for i in range(8):
a = (a + 1) // 2
b = (b + 1) // 2
if 2 ** (i + 1) == n:
print('Final!')
break
elif a == b:
print(i + 1)
break
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The last stage of Football World Cup is played using the play-off system.
There are *n* teams left in this stage, they are enumerated from 1 to *n*. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids *a* and *b* can meet.
Input Specification:
The only line contains three integers *n*, *a* and *b* (2<=≤<=*n*<=≤<=256, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that *n* is such that in each round an even number of team advance, and that *a* and *b* are not equal.
Output Specification:
In the only line print "Final!" (without quotes), if teams *a* and *b* can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams *a* and *b* can meet. The round are enumerated from 1.
Demo Input:
['4 1 2\n', '8 2 6\n', '8 7 5\n']
Demo Output:
['1\n', 'Final!\n', '2\n']
Note:
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round.
|
```python
n, a, b = map(int, input().split())
for i in range(8):
a = (a + 1) // 2
b = (b + 1) // 2
if 2 ** (i + 1) == n:
print('Final!')
break
elif a == b:
print(i + 1)
break
```
| 3
|
|
75
|
A
|
Life Without Zeros
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
A. Life Without Zeros
|
2
|
256
|
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
|
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
|
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
|
[
"101\n102\n",
"105\n106\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "101\n102",
"output": "YES"
},
{
"input": "105\n106",
"output": "NO"
},
{
"input": "544\n397",
"output": "YES"
},
{
"input": "822\n280",
"output": "NO"
},
{
"input": "101\n413",
"output": "NO"
},
{
"input": "309\n139",
"output": "NO"
},
{
"input": "693\n970",
"output": "NO"
},
{
"input": "981\n1",
"output": "YES"
},
{
"input": "352\n276",
"output": "YES"
},
{
"input": "164\n691",
"output": "YES"
},
{
"input": "110036\n43",
"output": "YES"
},
{
"input": "100\n1000",
"output": "NO"
},
{
"input": "1000000000\n1000000000",
"output": "YES"
},
{
"input": "999999999\n999999999",
"output": "YES"
},
{
"input": "6\n4",
"output": "NO"
},
{
"input": "123456\n876543",
"output": "YES"
},
{
"input": "1234567\n9876543",
"output": "NO"
},
{
"input": "1111111\n1119111",
"output": "NO"
},
{
"input": "10502405\n40403403",
"output": "YES"
},
{
"input": "1\n1",
"output": "YES"
},
{
"input": "5\n4",
"output": "YES"
},
{
"input": "1\n6",
"output": "YES"
},
{
"input": "40\n17",
"output": "NO"
},
{
"input": "35\n81",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "2081\n7450",
"output": "NO"
},
{
"input": "84355694\n373629503",
"output": "NO"
},
{
"input": "885992048\n510468672",
"output": "NO"
},
{
"input": "540144752\n647307841",
"output": "NO"
},
{
"input": "341781108\n784147010",
"output": "NO"
},
{
"input": "960440938\n572334653",
"output": "NO"
},
{
"input": "657414510\n768771903",
"output": "NO"
},
{
"input": "501871728\n725074574",
"output": "NO"
},
{
"input": "968\n851",
"output": "YES"
},
{
"input": "943\n939",
"output": "YES"
},
{
"input": "347\n908",
"output": "NO"
},
{
"input": "611\n41",
"output": "YES"
},
{
"input": "737\n289",
"output": "NO"
},
{
"input": "168\n571",
"output": "YES"
},
{
"input": "868\n463",
"output": "YES"
},
{
"input": "815732007\n913882008",
"output": "NO"
},
{
"input": "459039901\n453033850",
"output": "NO"
},
{
"input": "975305404\n484601256",
"output": "NO"
}
] | 1,595,216,026
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 218
| 20,172,800
|
a=input()
b=input()
c=''
for i in list(a):
if i== '0':
pass
else:
c+=i
d=''
for i in list(b):
if i=='0':
pass
else:
d+=i
j=str(int(a)+int(b))
l=''
for i in list(j):
if i=='0' :
pass
else:
l+=i
if ((int(a)+int(b)) ==( int(l) )):
print("YES")
else:
print("NO")
|
Title: Life Without Zeros
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
Input Specification:
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
Output Specification:
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
Demo Input:
['101\n102\n', '105\n106\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
a=input()
b=input()
c=''
for i in list(a):
if i== '0':
pass
else:
c+=i
d=''
for i in list(b):
if i=='0':
pass
else:
d+=i
j=str(int(a)+int(b))
l=''
for i in list(j):
if i=='0' :
pass
else:
l+=i
if ((int(a)+int(b)) ==( int(l) )):
print("YES")
else:
print("NO")
```
| 0
|
495
|
B
|
Modular Equations
|
PROGRAMMING
| 1,600
|
[
"math",
"number theory"
] | null | null |
Last week, Hamed learned about a new type of equations in his math class called Modular Equations. Lets define *i* modulo *j* as the remainder of division of *i* by *j* and denote it by . A Modular Equation, as Hamed's teacher described, is an equation of the form in which *a* and *b* are two non-negative integers and *x* is a variable. We call a positive integer *x* for which a solution of our equation.
Hamed didn't pay much attention to the class since he was watching a movie. He only managed to understand the definitions of these equations.
Now he wants to write his math exercises but since he has no idea how to do that, he asked you for help. He has told you all he knows about Modular Equations and asked you to write a program which given two numbers *a* and *b* determines how many answers the Modular Equation has.
|
In the only line of the input two space-separated integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=109) are given.
|
If there is an infinite number of answers to our equation, print "infinity" (without the quotes). Otherwise print the number of solutions of the Modular Equation .
|
[
"21 5\n",
"9435152 272\n",
"10 10\n"
] |
[
"2\n",
"282\n",
"infinity\n"
] |
In the first sample the answers of the Modular Equation are 8 and 16 since <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/6f5ff39ebd209bf990adaf91f4b82f9687097224.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 1,000
|
[
{
"input": "21 5",
"output": "2"
},
{
"input": "9435152 272",
"output": "282"
},
{
"input": "10 10",
"output": "infinity"
},
{
"input": "0 1000000000",
"output": "0"
},
{
"input": "11 2",
"output": "2"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "0 0",
"output": "infinity"
},
{
"input": "121 0",
"output": "3"
},
{
"input": "772930485 686893955",
"output": "0"
},
{
"input": "257424 24",
"output": "127"
},
{
"input": "295138437 589952171",
"output": "0"
},
{
"input": "223093836 966",
"output": "399"
},
{
"input": "233758336 10665466",
"output": "13"
},
{
"input": "223092887 17",
"output": "500"
},
{
"input": "223094728 1858",
"output": "371"
},
{
"input": "223092899 29",
"output": "495"
},
{
"input": "997920 0",
"output": "240"
},
{
"input": "887043 3",
"output": "213"
},
{
"input": "124 24",
"output": "3"
},
{
"input": "982901 101",
"output": "193"
},
{
"input": "357987 35",
"output": "45"
},
{
"input": "954374 1030",
"output": "32"
},
{
"input": "49106 46",
"output": "15"
},
{
"input": "325508499 119510657",
"output": "1"
},
{
"input": "89768760 885778845",
"output": "0"
},
{
"input": "944387968 700818251",
"output": "0"
},
{
"input": "12 3",
"output": "1"
},
{
"input": "1000000000 1",
"output": "19"
},
{
"input": "923456789 3",
"output": "14"
},
{
"input": "1000000000 6",
"output": "6"
},
{
"input": "1000000000 333333300",
"output": "2"
},
{
"input": "5 2",
"output": "1"
},
{
"input": "1 10",
"output": "0"
},
{
"input": "15 3",
"output": "3"
},
{
"input": "2 0",
"output": "2"
},
{
"input": "77 75",
"output": "0"
},
{
"input": "1000000000 1000000000",
"output": "infinity"
}
] | 1,434,056,313
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 62
| 0
|
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import time
#
(a, b) = (int(i) for i in input().split())
start = time.time()
ans = 0
if ( a == b):
ans = "infinity"
elif (a > b):
i = 1
max = (a-b) ** 0.5
while(i <= max ):
if (divmod(a-b, i)[1] == 0):
if i > b:
ans += 1
buf = (a-b)//i
if buf > b and buf != i:
ans += 1
i += 1
print(ans)
finish = time.time()
#print(finish - start)
|
Title: Modular Equations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Last week, Hamed learned about a new type of equations in his math class called Modular Equations. Lets define *i* modulo *j* as the remainder of division of *i* by *j* and denote it by . A Modular Equation, as Hamed's teacher described, is an equation of the form in which *a* and *b* are two non-negative integers and *x* is a variable. We call a positive integer *x* for which a solution of our equation.
Hamed didn't pay much attention to the class since he was watching a movie. He only managed to understand the definitions of these equations.
Now he wants to write his math exercises but since he has no idea how to do that, he asked you for help. He has told you all he knows about Modular Equations and asked you to write a program which given two numbers *a* and *b* determines how many answers the Modular Equation has.
Input Specification:
In the only line of the input two space-separated integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=109) are given.
Output Specification:
If there is an infinite number of answers to our equation, print "infinity" (without the quotes). Otherwise print the number of solutions of the Modular Equation .
Demo Input:
['21 5\n', '9435152 272\n', '10 10\n']
Demo Output:
['2\n', '282\n', 'infinity\n']
Note:
In the first sample the answers of the Modular Equation are 8 and 16 since <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/6f5ff39ebd209bf990adaf91f4b82f9687097224.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import time
#
(a, b) = (int(i) for i in input().split())
start = time.time()
ans = 0
if ( a == b):
ans = "infinity"
elif (a > b):
i = 1
max = (a-b) ** 0.5
while(i <= max ):
if (divmod(a-b, i)[1] == 0):
if i > b:
ans += 1
buf = (a-b)//i
if buf > b and buf != i:
ans += 1
i += 1
print(ans)
finish = time.time()
#print(finish - start)
```
| 3
|
|
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"
}
] | 1,671,455,261
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 0
|
n,m,a=map(int,input().split())
if n%a==0:
len=int(n/a)
else:
len=int(n/a+1)
if m%a==0:
bre=int(m/a)
else:
bre=int(m/a+1)
print(len+bre)
|
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 n%a==0:
len=int(n/a)
else:
len=int(n/a+1)
if m%a==0:
bre=int(m/a)
else:
bre=int(m/a+1)
print(len+bre)
```
| 0
|
279
|
B
|
Books
|
PROGRAMMING
| 1,400
|
[
"binary search",
"brute force",
"implementation",
"two pointers"
] | null | null |
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
|
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
|
Print a single integer — the maximum number of books Valera can read.
|
[
"4 5\n3 1 2 1\n",
"3 3\n2 2 3\n"
] |
[
"3\n",
"1\n"
] |
none
| 1,000
|
[
{
"input": "4 5\n3 1 2 1",
"output": "3"
},
{
"input": "3 3\n2 2 3",
"output": "1"
},
{
"input": "1 3\n5",
"output": "0"
},
{
"input": "1 10\n4",
"output": "1"
},
{
"input": "2 10\n6 4",
"output": "2"
},
{
"input": "6 10\n2 3 4 2 1 1",
"output": "4"
},
{
"input": "7 13\n6 8 14 9 4 11 10",
"output": "2"
},
{
"input": "10 15\n10 9 1 1 5 10 5 3 7 2",
"output": "3"
},
{
"input": "20 30\n8 1 2 6 9 4 1 9 9 10 4 7 8 9 5 7 1 8 7 4",
"output": "6"
},
{
"input": "30 60\n16 13 22 38 13 35 17 17 20 38 12 19 9 22 20 3 35 34 34 21 35 40 22 3 27 19 12 4 8 19",
"output": "4"
},
{
"input": "100 100\n75 92 18 6 81 67 7 92 100 65 82 32 50 67 85 31 80 91 84 63 39 52 92 81 1 98 24 12 43 48 17 86 51 72 48 95 45 50 12 66 19 79 49 89 34 1 97 75 20 33 96 27 42 23 73 71 93 1 85 19 66 14 17 61 20 39 36 33 42 61 56 64 23 91 80 99 40 74 13 18 98 85 74 39 62 84 46 74 50 23 38 11 79 14 9 25 66 100 25 52",
"output": "3"
},
{
"input": "10 1\n4418 7528 8170 1736 1317 3205 8183 4995 8039 4708",
"output": "0"
},
{
"input": "50 2\n124 214 63 73 996 760 38 571 451 300 970 1 706 937 837 494 619 88 851 411 957 990 842 613 821 649 627 34 693 678 734 116 816 985 705 940 499 493 922 967 854 439 112 644 961 438 189 572 655 550",
"output": "1"
}
] | 1,679,350,591
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 8
| 2,000
| 5,939,200
|
n, t = map(int, input().split())
a = list(map(int, input().split()))
maxx = 0
for i in range(n):
for j in range(i, n):
sres = a[i:j + 1]
if sum(sres) <= t:
maxx = max(maxx, len(sres))
print(maxx)
|
Title: Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
Output Specification:
Print a single integer — the maximum number of books Valera can read.
Demo Input:
['4 5\n3 1 2 1\n', '3 3\n2 2 3\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
n, t = map(int, input().split())
a = list(map(int, input().split()))
maxx = 0
for i in range(n):
for j in range(i, n):
sres = a[i:j + 1]
if sum(sres) <= t:
maxx = max(maxx, len(sres))
print(maxx)
```
| 0
|
|
258
|
A
|
Little Elephant and Bits
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
|
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
|
In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.
|
[
"101\n",
"110010\n"
] |
[
"11\n",
"11010\n"
] |
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>.
| 500
|
[
{
"input": "101",
"output": "11"
},
{
"input": "110010",
"output": "11010"
},
{
"input": "10000",
"output": "1000"
},
{
"input": "1111111110",
"output": "111111111"
},
{
"input": "10100101011110101",
"output": "1100101011110101"
},
{
"input": "111010010111",
"output": "11110010111"
},
{
"input": "11110111011100000000",
"output": "1111111011100000000"
},
{
"input": "11110010010100001110110101110011110110100111101",
"output": "1111010010100001110110101110011110110100111101"
},
{
"input": "1001011111010010100111111",
"output": "101011111010010100111111"
},
{
"input": "1111111111",
"output": "111111111"
},
{
"input": "1111111111111111111100111101001110110111111000001111110101001101001110011000001011001111111000110101",
"output": "111111111111111111110111101001110110111111000001111110101001101001110011000001011001111111000110101"
},
{
"input": "11010110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100",
"output": "1110110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100"
},
{
"input": "11111111111111111111111110110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011",
"output": "1111111111111111111111111110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011"
},
{
"input": "11100010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011",
"output": "1110010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011"
},
{
"input": "11",
"output": "1"
},
{
"input": "111",
"output": "11"
},
{
"input": "111111",
"output": "11111"
},
{
"input": "11111",
"output": "1111"
},
{
"input": "1111",
"output": "111"
}
] | 1,556,831,130
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 248
| 409,600
|
s = str(input())
if "0" not in s:
print(s[0:-1])
exit()
ind = s.index("0")
s = s[0 : ind] + s[1 + ind::]
print(s)
|
Title: Little Elephant and Bits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
Input Specification:
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
Output Specification:
In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.
Demo Input:
['101\n', '110010\n']
Demo Output:
['11\n', '11010\n']
Note:
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>.
|
```python
s = str(input())
if "0" not in s:
print(s[0:-1])
exit()
ind = s.index("0")
s = s[0 : ind] + s[1 + ind::]
print(s)
```
| 3
|
|
344
|
A
|
Magnets
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
|
On the single line of the output print the number of groups of magnets.
|
[
"6\n10\n10\n10\n01\n10\n10\n",
"4\n01\n01\n10\n10\n"
] |
[
"3\n",
"2\n"
] |
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets.
| 500
|
[
{
"input": "6\n10\n10\n10\n01\n10\n10",
"output": "3"
},
{
"input": "4\n01\n01\n10\n10",
"output": "2"
},
{
"input": "1\n10",
"output": "1"
},
{
"input": "2\n01\n10",
"output": "2"
},
{
"input": "2\n10\n10",
"output": "1"
},
{
"input": "3\n10\n01\n10",
"output": "3"
},
{
"input": "1\n01",
"output": "1"
},
{
"input": "2\n01\n01",
"output": "1"
},
{
"input": "2\n10\n01",
"output": "2"
},
{
"input": "3\n01\n01\n01",
"output": "1"
},
{
"input": "3\n10\n10\n01",
"output": "2"
},
{
"input": "3\n01\n10\n10",
"output": "2"
},
{
"input": "115\n10\n10\n10\n10\n01\n01\n10\n10\n10\n01\n01\n10\n01\n01\n10\n10\n10\n01\n10\n01\n10\n10\n01\n01\n10\n10\n10\n10\n01\n10\n01\n01\n10\n10\n10\n10\n01\n10\n10\n10\n01\n10\n01\n10\n10\n10\n10\n01\n01\n01\n10\n10\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n01\n10\n01\n10\n10\n01\n01\n01\n10\n01\n01\n10\n10\n01\n01\n01\n01\n01\n10\n01\n10\n01\n10\n01\n01\n01\n10\n01\n10\n10\n01\n10\n10\n01\n01\n01\n10\n10\n10\n10\n10\n10\n10\n10",
"output": "55"
}
] | 1,693,279,019
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 310
| 7,680,000
|
number_of_magnets = int(input())
_list = []
for i in range(number_of_magnets):
x = input()
_list.append(x)
number_of_groups = 1
for i in range(len(_list)):
if i != len(_list) - 1:
if _list[i] != _list[i+1]:
number_of_groups += 1
print(number_of_groups)
|
Title: Magnets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
Output Specification:
On the single line of the output print the number of groups of magnets.
Demo Input:
['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n']
Demo Output:
['3\n', '2\n']
Note:
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets.
|
```python
number_of_magnets = int(input())
_list = []
for i in range(number_of_magnets):
x = input()
_list.append(x)
number_of_groups = 1
for i in range(len(_list)):
if i != len(_list) - 1:
if _list[i] != _list[i+1]:
number_of_groups += 1
print(number_of_groups)
```
| 3
|
|
424
|
B
|
Megacity
|
PROGRAMMING
| 1,200
|
[
"binary search",
"greedy",
"implementation",
"sortings"
] | null | null |
The administration of the Tomsk Region firmly believes that it's time to become a megacity (that is, get population of one million). Instead of improving the demographic situation, they decided to achieve its goal by expanding the boundaries of the city.
The city of Tomsk can be represented as point on the plane with coordinates (0; 0). The city is surrounded with *n* other locations, the *i*-th one has coordinates (*x**i*, *y**i*) with the population of *k**i* people. You can widen the city boundaries to a circle of radius *r*. In such case all locations inside the circle and on its border are included into the city.
Your goal is to write a program that will determine the minimum radius *r*, to which is necessary to expand the boundaries of Tomsk, so that it becomes a megacity.
|
The first line of the input contains two integers *n* and *s* (1<=≤<=*n*<=≤<=103; 1<=≤<=*s*<=<<=106) — the number of locatons around Tomsk city and the population of the city. Then *n* lines follow. The *i*-th line contains three integers — the *x**i* and *y**i* coordinate values of the *i*-th location and the number *k**i* of people in it (1<=≤<=*k**i*<=<<=106). Each coordinate is an integer and doesn't exceed 104 in its absolute value.
It is guaranteed that no two locations are at the same point and no location is at point (0; 0).
|
In the output, print "-1" (without the quotes), if Tomsk won't be able to become a megacity. Otherwise, in the first line print a single real number — the minimum radius of the circle that the city needs to expand to in order to become a megacity.
The answer is considered correct if the absolute or relative error don't exceed 10<=-<=6.
|
[
"4 999998\n1 1 1\n2 2 1\n3 3 1\n2 -2 1\n",
"4 999998\n1 1 2\n2 2 1\n3 3 1\n2 -2 1\n",
"2 1\n1 1 999997\n2 2 1\n"
] |
[
"2.8284271\n",
"1.4142136\n",
"-1"
] |
none
| 1,000
|
[
{
"input": "4 999998\n1 1 1\n2 2 1\n3 3 1\n2 -2 1",
"output": "2.8284271"
},
{
"input": "4 999998\n1 1 2\n2 2 1\n3 3 1\n2 -2 1",
"output": "1.4142136"
},
{
"input": "2 1\n1 1 999997\n2 2 1",
"output": "-1"
},
{
"input": "4 999998\n3 3 10\n-3 3 10\n3 -3 10\n-3 -3 10",
"output": "4.2426407"
},
{
"input": "15 95473\n-9 6 199715\n0 -8 110607\n0 2 6621\n-3 -2 59894\n-10 -8 175440\n-2 0 25814\n10 -4 68131\n7 1 9971\n6 7 821\n6 5 20208\n6 2 68468\n0 7 37427\n1 -3 13337\n-10 7 113041\n-6 -2 44028",
"output": "12.8062485"
},
{
"input": "20 93350\n13 -28 486\n26 -26 48487\n5 -23 143368\n-23 -25 10371\n-2 -7 75193\n0 -8 3\n-6 -11 5015\n-19 -18 315278\n28 -15 45801\n21 8 4590\n-4 -28 12926\n-16 17 9405\n-28 -23 222092\n1 -10 1857\n14 -28 35170\n-4 -22 22036\n-2 -10 1260\n-1 12 375745\n-19 -24 38845\n10 -25 9256",
"output": "26.1725047"
},
{
"input": "30 505231\n-18 16 88130\n-10 16 15693\n16 -32 660\n-27 17 19042\n30 -37 6680\n36 19 299674\n-45 21 3300\n11 27 76\n-49 -34 28649\n-1 11 31401\n25 42 20858\n-40 6 455660\n-29 43 105001\n-38 10 6042\n19 -45 65551\n20 -9 148533\n-5 -24 393442\n-43 2 8577\n-39 18 97059\n12 28 39189\n35 23 28178\n40 -34 51687\n23 41 219028\n21 -44 927\n47 8 13206\n33 41 97342\n10 18 24895\n0 12 288\n0 -44 1065\n-25 43 44231",
"output": "24.5153013"
},
{
"input": "2 500000\n936 1000 500000\n961 976 500000",
"output": "1369.7065379"
},
{
"input": "10 764008\n959 32 23049\n-513 797 38979\n-603 -838 24916\n598 -430 25414\n-280 -624 18714\n330 891 21296\n-347 -68 27466\n650 -842 30125\n-314 889 35394\n275 969 5711",
"output": "1063.7029661"
},
{
"input": "30 295830\n1 -4 24773\n4 3 26175\n-2 -3 14789\n2 -1 46618\n-2 -2 52997\n-3 0 517\n-2 0 18173\n-4 -3 54465\n2 4 63579\n4 -4 41821\n2 2 11018\n0 4 42856\n0 -1 51885\n-3 4 57137\n3 0 4688\n0 2 60137\n-4 4 33484\n-1 3 66196\n3 -1 53634\n0 -2 41630\n-2 1 54606\n2 -2 2978\n2 -3 23733\n1 -2 35248\n-3 -3 15124\n-2 -4 26518\n4 0 28151\n4 -1 18348\n3 3 16914\n-4 2 26013",
"output": "4.4721360"
},
{
"input": "10 511500\n-5129 -3858 76954\n1296 1130 36126\n1219 6732 102003\n-8026 -178 4150\n-3261 1342 105429\n7965 -3013 62561\n5607 8963 53539\n-9044 -3999 16509\n1406 4103 115667\n-3716 2522 110626",
"output": "6841.4753526"
},
{
"input": "20 39342\n2 0 36476\n-3 1 136925\n1 3 31234\n0 -3 23785\n-1 3 77700\n-3 -1 50490\n-1 -3 13965\n-3 2 121093\n3 0 118933\n-3 0 125552\n-3 3 54779\n-2 0 96250\n1 2 142643\n2 2 23848\n0 2 29845\n0 -2 80462\n-1 1 91852\n-1 2 26526\n0 -1 136272\n1 1 108999",
"output": "3.0000000"
},
{
"input": "2 1\n1 0 1\n0 1 999999",
"output": "1.0000000"
},
{
"input": "2 999997\n1 1 1\n1 2 1",
"output": "-1"
}
] | 1,524,493,480
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 54
| 155
| 8,089,600
|
import queue
import math
if __name__ == '__main__':
n, s = map(int, input().split())
pq = queue.PriorityQueue()
for i in range(n):
x, y, k = map(int, input().split())
r = math.sqrt(x**2 + y**2)
pq.put((r, k))
r = 0
while pq.empty() == False and s < 1000000:
r, k = pq.get()
s += k
if s < 1000000:
print(-1)
else:
print("{0:.7f}".format(r))
|
Title: Megacity
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The administration of the Tomsk Region firmly believes that it's time to become a megacity (that is, get population of one million). Instead of improving the demographic situation, they decided to achieve its goal by expanding the boundaries of the city.
The city of Tomsk can be represented as point on the plane with coordinates (0; 0). The city is surrounded with *n* other locations, the *i*-th one has coordinates (*x**i*, *y**i*) with the population of *k**i* people. You can widen the city boundaries to a circle of radius *r*. In such case all locations inside the circle and on its border are included into the city.
Your goal is to write a program that will determine the minimum radius *r*, to which is necessary to expand the boundaries of Tomsk, so that it becomes a megacity.
Input Specification:
The first line of the input contains two integers *n* and *s* (1<=≤<=*n*<=≤<=103; 1<=≤<=*s*<=<<=106) — the number of locatons around Tomsk city and the population of the city. Then *n* lines follow. The *i*-th line contains three integers — the *x**i* and *y**i* coordinate values of the *i*-th location and the number *k**i* of people in it (1<=≤<=*k**i*<=<<=106). Each coordinate is an integer and doesn't exceed 104 in its absolute value.
It is guaranteed that no two locations are at the same point and no location is at point (0; 0).
Output Specification:
In the output, print "-1" (without the quotes), if Tomsk won't be able to become a megacity. Otherwise, in the first line print a single real number — the minimum radius of the circle that the city needs to expand to in order to become a megacity.
The answer is considered correct if the absolute or relative error don't exceed 10<=-<=6.
Demo Input:
['4 999998\n1 1 1\n2 2 1\n3 3 1\n2 -2 1\n', '4 999998\n1 1 2\n2 2 1\n3 3 1\n2 -2 1\n', '2 1\n1 1 999997\n2 2 1\n']
Demo Output:
['2.8284271\n', '1.4142136\n', '-1']
Note:
none
|
```python
import queue
import math
if __name__ == '__main__':
n, s = map(int, input().split())
pq = queue.PriorityQueue()
for i in range(n):
x, y, k = map(int, input().split())
r = math.sqrt(x**2 + y**2)
pq.put((r, k))
r = 0
while pq.empty() == False and s < 1000000:
r, k = pq.get()
s += k
if s < 1000000:
print(-1)
else:
print("{0:.7f}".format(r))
```
| 3
|
|
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"
}
] | 1,652,387,255
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 20
| 31
| 0
|
import math
arr = input().split()
m = int(arr[0])
n = int(arr[1])
a = int(arr[2])
def theatreSquare(m, n, a):
topSide = math.ceil(n/a)
leftSide = math.ceil(m/a)
return topSide*leftSide
print(theatreSquare(m,n,a))
|
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
import math
arr = input().split()
m = int(arr[0])
n = int(arr[1])
a = int(arr[2])
def theatreSquare(m, n, a):
topSide = math.ceil(n/a)
leftSide = math.ceil(m/a)
return topSide*leftSide
print(theatreSquare(m,n,a))
```
| 3.9845
|
109
|
A
|
Lucky Sum of Digits
|
PROGRAMMING
| 1,000
|
[
"brute force",
"implementation"
] |
A. Lucky Sum of Digits
|
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.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.
|
The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number.
|
Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.
|
[
"11\n",
"10\n"
] |
[
"47\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "11",
"output": "47"
},
{
"input": "10",
"output": "-1"
},
{
"input": "64",
"output": "4477777777"
},
{
"input": "1",
"output": "-1"
},
{
"input": "4",
"output": "4"
},
{
"input": "7",
"output": "7"
},
{
"input": "12",
"output": "444"
},
{
"input": "1000000",
"output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "47",
"output": "44477777"
},
{
"input": "100",
"output": "4444777777777777"
},
{
"input": "700",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "485",
"output": "44447777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "111",
"output": "444447777777777777"
},
{
"input": "85",
"output": "4477777777777"
},
{
"input": "114",
"output": "444477777777777777"
},
{
"input": "474",
"output": "444777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "74",
"output": "47777777777"
},
{
"input": "1000",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "1024",
"output": "4444777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "4444",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "45784",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "10000",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "9854",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "186",
"output": "477777777777777777777777777"
},
{
"input": "10416",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "10417",
"output": "4477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "3840",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "100000",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "9876",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "99999",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "777777",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "854759",
"output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "11000",
"output": "4444447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "18951",
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},
{
"input": "999999",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "888887",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999998",
"output": "4444477777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "40008",
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},
{
"input": "10691",
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},
{
"input": "5",
"output": "-1"
},
{
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"output": "-1"
},
{
"input": "9",
"output": "-1"
},
{
"input": "8",
"output": "44"
},
{
"input": "2",
"output": "-1"
},
{
"input": "3",
"output": "-1"
},
{
"input": "999997",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999996",
"output": "4777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999990",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "999980",
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},
{
"input": "800000",
"output": "4447777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
},
{
"input": "980000",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777..."
}
] | 1,560,627,252
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 2
| 2,000
| 2,048,000
|
x = int(input())
def som2(y):
sum = 0
while(y != 0):
r = y%10
y = y//10
if((r == 4) or (r == 7)):
sum = sum+r
else:
sum = 0
break
return sum
i = 3
u = 0
while(u == 0):
an = som2(i)
if(an == 0):
i = i+1
else:
if(an == x):
u = 1
elif(an > x):
u = 2
else:
i = i+1
if(u == 1):
print(i)
else:
print(-1)
|
Title: Lucky Sum of Digits
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.
Petya wonders eagerly what minimum lucky number has the sum of digits equal to *n*. Help him cope with the task.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the sum of digits of the required lucky number.
Output Specification:
Print on the single line the result — the minimum lucky number, whose sum of digits equals *n*. If such number does not exist, print -1.
Demo Input:
['11\n', '10\n']
Demo Output:
['47\n', '-1\n']
Note:
none
|
```python
x = int(input())
def som2(y):
sum = 0
while(y != 0):
r = y%10
y = y//10
if((r == 4) or (r == 7)):
sum = sum+r
else:
sum = 0
break
return sum
i = 3
u = 0
while(u == 0):
an = som2(i)
if(an == 0):
i = i+1
else:
if(an == x):
u = 1
elif(an > x):
u = 2
else:
i = i+1
if(u == 1):
print(i)
else:
print(-1)
```
| 0
|
9
|
A
|
Die Roll
|
PROGRAMMING
| 800
|
[
"math",
"probabilities"
] |
A. Die Roll
|
1
|
64
|
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
|
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
|
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
|
[
"4 2\n"
] |
[
"1/2\n"
] |
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
| 0
|
[
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,643,019,527
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 36
| 186
| 0
|
a,b = map(int,input().split()) ; c = 7-(max(a,b)) ; d = (str(c)+'/'+str(6))
if d == '3/6': print('1/2')
elif d == '2/6': print('1/3')
elif d == '4/6': print('2/3')
elif d == '5/6': print('5/6')
elif d == '6/6': print('1/1')
elif d == '1/6': print('1/6')
|
Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
|
```python
a,b = map(int,input().split()) ; c = 7-(max(a,b)) ; d = (str(c)+'/'+str(6))
if d == '3/6': print('1/2')
elif d == '2/6': print('1/3')
elif d == '4/6': print('2/3')
elif d == '5/6': print('5/6')
elif d == '6/6': print('1/1')
elif d == '1/6': print('1/6')
```
| 3.907
|
78
|
C
|
Beaver Game
|
PROGRAMMING
| 2,000
|
[
"dp",
"games",
"number theory"
] |
C. Beaver Game
|
1
|
256
|
Two beavers, Timur and Marsel, play the following game.
There are *n* logs, each of exactly *m* meters in length. The beavers move in turns. For each move a beaver chooses a log and gnaws it into some number (more than one) of equal parts, the length of each one is expressed by an integer and is no less than *k* meters. Each resulting part is also a log which can be gnawed in future by any beaver. The beaver that can't make a move loses. Thus, the other beaver wins.
Timur makes the first move. The players play in the optimal way. Determine the winner.
|
The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=109).
|
Print "Timur", if Timur wins, or "Marsel", if Marsel wins. You should print everything without the quotes.
|
[
"1 15 4\n",
"4 9 5\n"
] |
[
"Timur",
"Marsel"
] |
In the first sample the beavers only have one log, of 15 meters in length. Timur moves first. The only move he can do is to split the log into 3 parts each 5 meters in length. Then Marsel moves but he can't split any of the resulting logs, as *k* = 4. Thus, the winner is Timur.
In the second example the beavers have 4 logs 9 meters in length. Timur can't split any of them, so that the resulting parts possessed the length of not less than 5 meters, that's why he loses instantly.
| 1,500
|
[
{
"input": "1 15 4",
"output": "Timur"
},
{
"input": "4 9 5",
"output": "Marsel"
},
{
"input": "14 30 9",
"output": "Marsel"
},
{
"input": "81 180 53",
"output": "Timur"
},
{
"input": "225 187 20",
"output": "Marsel"
},
{
"input": "501 840 11",
"output": "Timur"
},
{
"input": "3 11 6",
"output": "Marsel"
},
{
"input": "1 2 1",
"output": "Timur"
},
{
"input": "1 1 2",
"output": "Marsel"
},
{
"input": "1 1 1",
"output": "Marsel"
},
{
"input": "1 2 2",
"output": "Marsel"
},
{
"input": "2 1 1",
"output": "Marsel"
},
{
"input": "2 1 2",
"output": "Marsel"
},
{
"input": "2 2 1",
"output": "Marsel"
},
{
"input": "2 2 2",
"output": "Marsel"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "Marsel"
},
{
"input": "999999999 735134400 1",
"output": "Timur"
},
{
"input": "999999999 735134400 100",
"output": "Timur"
},
{
"input": "787965358 999999986 2",
"output": "Marsel"
},
{
"input": "588462355 999999986 2",
"output": "Timur"
},
{
"input": "994427144 999999986 10",
"output": "Marsel"
},
{
"input": "36948983 999999986 10",
"output": "Timur"
},
{
"input": "431525232 999999986 250000000",
"output": "Marsel"
},
{
"input": "502291145 999999986 250000000",
"output": "Timur"
},
{
"input": "612116696 999999986 499999993",
"output": "Marsel"
},
{
"input": "559551717 999999986 499999993",
"output": "Timur"
},
{
"input": "844774784 999999894 30",
"output": "Marsel"
},
{
"input": "776970029 999999894 30",
"output": "Timur"
},
{
"input": "726014914 999950443 31643",
"output": "Marsel"
},
{
"input": "350735191 999950443 31643",
"output": "Timur"
},
{
"input": "114514534 999950443 31601",
"output": "Marsel"
},
{
"input": "689283015 999950443 31601",
"output": "Timur"
},
{
"input": "585863414 999002449 31607",
"output": "Marsel"
},
{
"input": "271824601 999002449 31607",
"output": "Timur"
},
{
"input": "57784860 999002449 1",
"output": "Marsel"
},
{
"input": "844093323 999002449 1",
"output": "Timur"
},
{
"input": "623425842 999002449 10",
"output": "Marsel"
},
{
"input": "283831003 999002449 10",
"output": "Timur"
},
{
"input": "884219706 735134400 1",
"output": "Marsel"
},
{
"input": "519313933 735134400 1",
"output": "Timur"
},
{
"input": "483982088 735134400 2",
"output": "Marsel"
},
{
"input": "966249765 735134400 2",
"output": "Timur"
},
{
"input": "362305942 735134400 367567201",
"output": "Marsel"
},
{
"input": "311659875 735134400 367567201",
"output": "Marsel"
},
{
"input": "717978658 735134400 100000000",
"output": "Marsel"
},
{
"input": "367890897 735134400 100000000",
"output": "Timur"
},
{
"input": "52049820 735134400 200000000",
"output": "Marsel"
},
{
"input": "395681265 735134400 200000000",
"output": "Timur"
},
{
"input": "195335332 735134400 300000000",
"output": "Marsel"
},
{
"input": "2240305 735134400 300000000",
"output": "Timur"
},
{
"input": "166548792 901800900 30030",
"output": "Marsel"
},
{
"input": "708577575 901800900 30030",
"output": "Timur"
},
{
"input": "304908040 901800900 901800900",
"output": "Marsel"
},
{
"input": "799852313 901800900 901800900",
"output": "Marsel"
},
{
"input": "624970498 901800900 901800901",
"output": "Marsel"
},
{
"input": "373293701 901800900 901800901",
"output": "Marsel"
},
{
"input": "1608018 999999937 1",
"output": "Marsel"
},
{
"input": "830273749 999999937 1",
"output": "Timur"
},
{
"input": "859437048 999999937 2",
"output": "Marsel"
},
{
"input": "672396661 999999937 2",
"output": "Marsel"
},
{
"input": "763677180 999999937 111",
"output": "Marsel"
},
{
"input": "605998637 999999937 111",
"output": "Marsel"
},
{
"input": "938316524 999999937 1000000000",
"output": "Marsel"
},
{
"input": "885233939 999999937 1000000000",
"output": "Marsel"
},
{
"input": "522619786 1000000000 500000000",
"output": "Marsel"
},
{
"input": "780954779 1000000000 500000000",
"output": "Timur"
},
{
"input": "568419816 1000000000 500000001",
"output": "Marsel"
},
{
"input": "300872687 1000000000 500000001",
"output": "Marsel"
},
{
"input": "628290298 1000000000 1250",
"output": "Marsel"
},
{
"input": "56021277 1000000000 1250",
"output": "Timur"
},
{
"input": "538604320 536870912 268435456",
"output": "Marsel"
},
{
"input": "719320775 536870912 268435456",
"output": "Timur"
},
{
"input": "911637544 536870912 134217728",
"output": "Marsel"
},
{
"input": "104077283 536870912 134217728",
"output": "Timur"
},
{
"input": "167283896 935821597 191101",
"output": "Marsel"
},
{
"input": "753528595 935821597 191101",
"output": "Timur"
},
{
"input": "907605112 935821597 11274959",
"output": "Marsel"
},
{
"input": "327082339 935821597 11274959",
"output": "Timur"
},
{
"input": "316344442 935821597 22824917",
"output": "Marsel"
},
{
"input": "244672063 935821597 22824917",
"output": "Timur"
},
{
"input": "1000000000 1000000000 36",
"output": "Marsel"
},
{
"input": "1 1000000000 1000000000",
"output": "Marsel"
},
{
"input": "2 10 1",
"output": "Marsel"
},
{
"input": "1 999999991 488888888",
"output": "Marsel"
},
{
"input": "4 9 4",
"output": "Marsel"
},
{
"input": "1 3 2",
"output": "Marsel"
},
{
"input": "1 1000000000 1",
"output": "Timur"
},
{
"input": "1 999999938 999999937",
"output": "Marsel"
},
{
"input": "1 7 1",
"output": "Timur"
},
{
"input": "101 1 1",
"output": "Marsel"
},
{
"input": "999999999 1000000000 1000000000",
"output": "Marsel"
},
{
"input": "1 200000 100000",
"output": "Timur"
},
{
"input": "1 9 3",
"output": "Timur"
}
] | 1,552,977,970
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 7
| 124
| 0
|
from sys import exit
def IsPrime(n):
d = 2
if n != 1:
while n % d != 0:
d += 1
return d == n
p = False
n, m, k = [int(i) for i in input().split()]
for i in range(m // 2, k - 1, -1):
if m % i == 0:
p = True
if p==False:
print('Marsel')
quit()
if IsPrime(m):
print('Marsel')
elif k >= m:
print('Marsel')
elif m % 2 == 0 or m % 2 != 0:
if n % 2 != 0:
print('Timur')
else:
print('Marsel')
|
Title: Beaver Game
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Two beavers, Timur and Marsel, play the following game.
There are *n* logs, each of exactly *m* meters in length. The beavers move in turns. For each move a beaver chooses a log and gnaws it into some number (more than one) of equal parts, the length of each one is expressed by an integer and is no less than *k* meters. Each resulting part is also a log which can be gnawed in future by any beaver. The beaver that can't make a move loses. Thus, the other beaver wins.
Timur makes the first move. The players play in the optimal way. Determine the winner.
Input Specification:
The first line contains three integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=109).
Output Specification:
Print "Timur", if Timur wins, or "Marsel", if Marsel wins. You should print everything without the quotes.
Demo Input:
['1 15 4\n', '4 9 5\n']
Demo Output:
['Timur', 'Marsel']
Note:
In the first sample the beavers only have one log, of 15 meters in length. Timur moves first. The only move he can do is to split the log into 3 parts each 5 meters in length. Then Marsel moves but he can't split any of the resulting logs, as *k* = 4. Thus, the winner is Timur.
In the second example the beavers have 4 logs 9 meters in length. Timur can't split any of them, so that the resulting parts possessed the length of not less than 5 meters, that's why he loses instantly.
|
```python
from sys import exit
def IsPrime(n):
d = 2
if n != 1:
while n % d != 0:
d += 1
return d == n
p = False
n, m, k = [int(i) for i in input().split()]
for i in range(m // 2, k - 1, -1):
if m % i == 0:
p = True
if p==False:
print('Marsel')
quit()
if IsPrime(m):
print('Marsel')
elif k >= m:
print('Marsel')
elif m % 2 == 0 or m % 2 != 0:
if n % 2 != 0:
print('Timur')
else:
print('Marsel')
```
| 0
|
937
|
A
|
Olympiad
|
PROGRAMMING
| 800
|
[
"implementation",
"sortings"
] | null | null |
The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score.
|
Print a single integer — the desired number of ways.
|
[
"4\n1 3 3 2\n",
"3\n1 1 1\n",
"4\n42 0 0 42\n"
] |
[
"3\n",
"1\n",
"1\n"
] |
There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything.
| 500
|
[
{
"input": "4\n1 3 3 2",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "4\n42 0 0 42",
"output": "1"
},
{
"input": "10\n1 0 1 0 1 0 0 0 0 1",
"output": "1"
},
{
"input": "10\n572 471 540 163 50 30 561 510 43 200",
"output": "10"
},
{
"input": "100\n122 575 426 445 172 81 247 429 97 202 175 325 382 384 417 356 132 502 328 537 57 339 518 211 479 306 140 168 268 16 140 263 593 249 391 310 555 468 231 180 157 18 334 328 276 155 21 280 322 545 111 267 467 274 291 304 235 34 365 180 21 95 501 552 325 331 302 353 296 22 289 399 7 466 32 302 568 333 75 192 284 10 94 128 154 512 9 480 243 521 551 492 420 197 207 125 367 117 438 600",
"output": "94"
},
{
"input": "100\n600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600",
"output": "1"
},
{
"input": "78\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12",
"output": "13"
},
{
"input": "34\n220 387 408 343 184 447 197 307 337 414 251 319 426 322 347 242 208 412 188 185 241 235 216 259 331 372 322 284 444 384 214 297 389 391",
"output": "33"
},
{
"input": "100\n1 2 1 0 3 0 2 0 0 1 2 0 1 3 0 3 3 1 3 0 0 2 1 2 2 1 3 3 3 3 3 2 0 0 2 1 2 3 2 3 0 1 1 3 3 2 0 3 1 0 2 2 2 1 2 3 2 1 0 3 0 2 0 3 0 2 1 0 3 1 0 2 2 1 3 1 3 0 2 3 3 1 1 3 1 3 0 3 2 0 2 3 3 0 2 0 2 0 1 3",
"output": "3"
},
{
"input": "100\n572 471 540 163 50 30 561 510 43 200 213 387 500 424 113 487 357 333 294 337 435 202 447 494 485 465 161 344 470 559 104 356 393 207 224 213 511 514 60 386 149 216 392 229 429 173 165 401 395 150 127 579 344 390 529 296 225 425 318 79 465 447 177 110 367 212 459 270 41 500 277 567 125 436 178 9 214 342 203 112 144 24 79 155 495 556 40 549 463 281 241 316 2 246 1 396 510 293 332 55",
"output": "93"
},
{
"input": "99\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12 2 3 9 3 7 13 7 13 0 11 8 12 2 5 9 4 0 6 6 2 13",
"output": "13"
},
{
"input": "99\n1 0 1 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 0 0 0 1 1 0 0 0 0 0 1",
"output": "1"
},
{
"input": "99\n21 74 25 44 71 80 46 28 96 1 74 24 81 83 16 55 31 1 27 36 56 38 17 10 78 5 39 67 67 15 39 62 92 48 90 9 54 67 30 79 56 17 33 27 75 54 20 79 21 44 10 66 66 73 90 3 34 33 64 79 20 94 0 51 24 30 1 52 95 21 88 98 6 65 31 1 67 32 74 91 83 9 93 27 53 11 8 79 42 20 50 91 19 96 6 24 66 16 37",
"output": "61"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "2\n0 600",
"output": "1"
},
{
"input": "4\n1 1 1 2",
"output": "2"
},
{
"input": "4\n0 0 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "2\n0 5",
"output": "1"
},
{
"input": "5\n1 0 0 1 2",
"output": "2"
}
] | 1,632,352,006
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 31
| 0
|
list = []
n =int(input())
for i in range(0,n):
o = int(input())
if o != 0 and o not in list:
list.append(o)
print(len(list))
|
Title: Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score.
Output Specification:
Print a single integer — the desired number of ways.
Demo Input:
['4\n1 3 3 2\n', '3\n1 1 1\n', '4\n42 0 0 42\n']
Demo Output:
['3\n', '1\n', '1\n']
Note:
There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything.
|
```python
list = []
n =int(input())
for i in range(0,n):
o = int(input())
if o != 0 and o not in list:
list.append(o)
print(len(list))
```
| -1
|
|
2
|
A
|
Winner
|
PROGRAMMING
| 1,500
|
[
"hashing",
"implementation"
] |
A. Winner
|
1
|
64
|
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
|
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
|
Print the name of the winner.
|
[
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] |
[
"andrew\n",
"andrew\n"
] |
none
| 0
|
[
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},
{
"input": "7\nksjuuerbnlklcfdjeyq 312\ndthjlkrvvbyahttifpdewvyslsh -983\nksjuuerbnlklcfdjeyq 268\ndthjlkrvvbyahttifpdewvyslsh 788\nksjuuerbnlklcfdjeyq -79\nksjuuerbnlklcfdjeyq -593\nksjuuerbnlklcfdjeyq 734",
"output": "ksjuuerbnlklcfdjeyq"
},
{
"input": "12\natrtthfpcvishmqbakprquvnejr 185\natrtthfpcvishmqbakprquvnejr -699\natrtthfpcvishmqbakprquvnejr -911\natrtthfpcvishmqbakprquvnejr -220\nfcgslzkicjrpbqaifgweyzreajjfdo 132\nfcgslzkicjrpbqaifgweyzreajjfdo -242\nm 177\nm -549\natrtthfpcvishmqbakprquvnejr -242\nm 38\natrtthfpcvishmqbakprquvnejr -761\nfcgslzkicjrpbqaifgweyzreajjfdo 879",
"output": "fcgslzkicjrpbqaifgweyzreajjfdo"
},
{
"input": "15\naawtvezfntstrcpgbzjbf 681\nzhahpvqiptvksnbjkdvmknb -74\naawtvezfntstrcpgbzjbf 661\njpdwmyke 474\naawtvezfntstrcpgbzjbf -547\naawtvezfntstrcpgbzjbf 600\nzhahpvqiptvksnbjkdvmknb -11\njpdwmyke 711\nbjmj 652\naawtvezfntstrcpgbzjbf -1000\naawtvezfntstrcpgbzjbf -171\nbjmj -302\naawtvezfntstrcpgbzjbf 961\nzhahpvqiptvksnbjkdvmknb 848\nbjmj -735",
"output": "aawtvezfntstrcpgbzjbf"
},
{
"input": "17\nqdplghhx -649\nivhgbxiv 424\nivhgbxiv -72\nivhgbxiv 479\nugowfouebkelargxylsxn 83\nugowfouebkelargxylsxn -840\nivhgbxiv 382\nqdplghhx -904\nivhgbxiv -710\nugowfouebkelargxylsxn -838\nqdplghhx -974\nqdplghhx 571\nivhgbxiv -145\nugowfouebkelargxylsxn -788\nivhgbxiv 209\nivhgbxiv 420\nqdplghhx -158",
"output": "ivhgbxiv"
}
] | 1,667,432,308
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 9
| 92
| 0
|
from sys import stdin
input = stdin.readline
from collections import defaultdict
scores_by_name = defaultdict(int)
winner = None
max_score = float('-inf')
for l in range(int(input())):
name, sc = map(str, input().split())
sc = int(sc)
scores_by_name[name] += sc
if len(scores_by_name) == 1:
winner = name
max_score = max(max_score, scores_by_name[name])
else:
if scores_by_name[name] > max_score:
winner = name
max_score = scores_by_name[name]
print(winner)
|
Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none
|
```python
from sys import stdin
input = stdin.readline
from collections import defaultdict
scores_by_name = defaultdict(int)
winner = None
max_score = float('-inf')
for l in range(int(input())):
name, sc = map(str, input().split())
sc = int(sc)
scores_by_name[name] += sc
if len(scores_by_name) == 1:
winner = name
max_score = max(max_score, scores_by_name[name])
else:
if scores_by_name[name] > max_score:
winner = name
max_score = scores_by_name[name]
print(winner)
```
| 0
|
366
|
B
|
Dima and To-do List
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
You helped Dima to have a great weekend, but it's time to work. Naturally, Dima, as all other men who have girlfriends, does everything wrong.
Inna and Dima are now in one room. Inna tells Dima off for everything he does in her presence. After Inna tells him off for something, she goes to another room, walks there in circles muttering about how useless her sweetheart is. During that time Dima has time to peacefully complete *k*<=-<=1 tasks. Then Inna returns and tells Dima off for the next task he does in her presence and goes to another room again. It continues until Dima is through with his tasks.
Overall, Dima has *n* tasks to do, each task has a unique number from 1 to *n*. Dima loves order, so he does tasks consecutively, starting from some task. For example, if Dima has 6 tasks to do in total, then, if he starts from the 5-th task, the order is like that: first Dima does the 5-th task, then the 6-th one, then the 1-st one, then the 2-nd one, then the 3-rd one, then the 4-th one.
Inna tells Dima off (only lovingly and appropriately!) so often and systematically that he's very well learned the power with which she tells him off for each task. Help Dima choose the first task so that in total he gets told off with as little power as possible.
|
The first line of the input contains two integers *n*,<=*k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=103), where *a**i* is the power Inna tells Dima off with if she is present in the room while he is doing the *i*-th task.
It is guaranteed that *n* is divisible by *k*.
|
In a single line print the number of the task Dima should start with to get told off with as little power as possible. If there are multiple solutions, print the one with the minimum number of the first task to do.
|
[
"6 2\n3 2 1 6 5 4\n",
"10 5\n1 3 5 7 9 9 4 1 8 5\n"
] |
[
"1\n",
"3\n"
] |
Explanation of the first example.
If Dima starts from the first task, Inna tells him off with power 3, then Dima can do one more task (as *k* = 2), then Inna tells him off for the third task with power 1, then she tells him off for the fifth task with power 5. Thus, Dima gets told off with total power 3 + 1 + 5 = 9. If Dima started from the second task, for example, then Inna would tell him off for tasks 2, 4 and 6 with power 2 + 6 + 4 = 12.
Explanation of the second example.
In the second example *k* = 5, thus, Dima manages to complete 4 tasks in-between the telling off sessions. Thus, Inna tells Dima off for tasks number 1 and 6 (if he starts from 1 or 6), 2 and 7 (if he starts from 2 or 7) and so on. The optimal answer is to start from task 3 or 8, 3 has a smaller number, so the answer is 3.
| 1,000
|
[
{
"input": "6 2\n3 2 1 6 5 4",
"output": "1"
},
{
"input": "10 5\n1 3 5 7 9 9 4 1 8 5",
"output": "3"
},
{
"input": "20 4\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "10 10\n8 4 5 7 6 9 2 2 3 5",
"output": "7"
},
{
"input": "50 10\n1 2 3 4 5 6 7 8 9 10 10 1 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "2 1\n1 1",
"output": "1"
},
{
"input": "4 2\n2 1 1 3",
"output": "1"
},
{
"input": "15 5\n5 5 5 5 5 1 2 3 4 5 1 2 3 4 5",
"output": "1"
},
{
"input": "20 10\n3 3 3 3 3 3 3 3 2 3 3 3 3 3 3 3 3 3 6 4",
"output": "1"
}
] | 1,642,958,560
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 512,000
|
import math
def main_function():
n, k = [int(i) for i in input().split(" ")]
a = [int(i) for i in input().split(" ")]
positions = n // k
divided = [0 for i in range(k + 1)]
indexes = [(i + 1) % k for i in range(len(a))]
#print(indexes)
for i in range(len(a)):
j = i + 1
remainder = j % k
if remainder == 0:
remainder = k
divided[remainder] += a[i]
if n <= k:
print(a.index(min(a)) + 1)
else:
#print(divided)
min_val = 100000000000000000
index = 0
for j in range(0, k):
i = k - j
print(i)
if min_val > divided[i]:
index = i
print(index)
main_function()
|
Title: Dima and To-do List
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You helped Dima to have a great weekend, but it's time to work. Naturally, Dima, as all other men who have girlfriends, does everything wrong.
Inna and Dima are now in one room. Inna tells Dima off for everything he does in her presence. After Inna tells him off for something, she goes to another room, walks there in circles muttering about how useless her sweetheart is. During that time Dima has time to peacefully complete *k*<=-<=1 tasks. Then Inna returns and tells Dima off for the next task he does in her presence and goes to another room again. It continues until Dima is through with his tasks.
Overall, Dima has *n* tasks to do, each task has a unique number from 1 to *n*. Dima loves order, so he does tasks consecutively, starting from some task. For example, if Dima has 6 tasks to do in total, then, if he starts from the 5-th task, the order is like that: first Dima does the 5-th task, then the 6-th one, then the 1-st one, then the 2-nd one, then the 3-rd one, then the 4-th one.
Inna tells Dima off (only lovingly and appropriately!) so often and systematically that he's very well learned the power with which she tells him off for each task. Help Dima choose the first task so that in total he gets told off with as little power as possible.
Input Specification:
The first line of the input contains two integers *n*,<=*k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=103), where *a**i* is the power Inna tells Dima off with if she is present in the room while he is doing the *i*-th task.
It is guaranteed that *n* is divisible by *k*.
Output Specification:
In a single line print the number of the task Dima should start with to get told off with as little power as possible. If there are multiple solutions, print the one with the minimum number of the first task to do.
Demo Input:
['6 2\n3 2 1 6 5 4\n', '10 5\n1 3 5 7 9 9 4 1 8 5\n']
Demo Output:
['1\n', '3\n']
Note:
Explanation of the first example.
If Dima starts from the first task, Inna tells him off with power 3, then Dima can do one more task (as *k* = 2), then Inna tells him off for the third task with power 1, then she tells him off for the fifth task with power 5. Thus, Dima gets told off with total power 3 + 1 + 5 = 9. If Dima started from the second task, for example, then Inna would tell him off for tasks 2, 4 and 6 with power 2 + 6 + 4 = 12.
Explanation of the second example.
In the second example *k* = 5, thus, Dima manages to complete 4 tasks in-between the telling off sessions. Thus, Inna tells Dima off for tasks number 1 and 6 (if he starts from 1 or 6), 2 and 7 (if he starts from 2 or 7) and so on. The optimal answer is to start from task 3 or 8, 3 has a smaller number, so the answer is 3.
|
```python
import math
def main_function():
n, k = [int(i) for i in input().split(" ")]
a = [int(i) for i in input().split(" ")]
positions = n // k
divided = [0 for i in range(k + 1)]
indexes = [(i + 1) % k for i in range(len(a))]
#print(indexes)
for i in range(len(a)):
j = i + 1
remainder = j % k
if remainder == 0:
remainder = k
divided[remainder] += a[i]
if n <= k:
print(a.index(min(a)) + 1)
else:
#print(divided)
min_val = 100000000000000000
index = 0
for j in range(0, k):
i = k - j
print(i)
if min_val > divided[i]:
index = i
print(index)
main_function()
```
| 0
|
|
420
|
A
|
Start Up
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
|
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
|
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
|
[
"AHA\n",
"Z\n",
"XO\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "AHA",
"output": "YES"
},
{
"input": "Z",
"output": "NO"
},
{
"input": "XO",
"output": "NO"
},
{
"input": "AAA",
"output": "YES"
},
{
"input": "AHHA",
"output": "YES"
},
{
"input": "BAB",
"output": "NO"
},
{
"input": "OMMMAAMMMO",
"output": "YES"
},
{
"input": "YYHUIUGYI",
"output": "NO"
},
{
"input": "TT",
"output": "YES"
},
{
"input": "UUU",
"output": "YES"
},
{
"input": "WYYW",
"output": "YES"
},
{
"input": "MITIM",
"output": "YES"
},
{
"input": "VO",
"output": "NO"
},
{
"input": "WWS",
"output": "NO"
},
{
"input": "VIYMAXXAVM",
"output": "NO"
},
{
"input": "OVWIHIWVYXMVAAAATOXWOIUUHYXHIHHVUIOOXWHOXTUUMUUVHVWWYUTIAUAITAOMHXWMTTOIVMIVOTHOVOIOHYHAOXWAUVWAVIVM",
"output": "NO"
},
{
"input": "CC",
"output": "NO"
},
{
"input": "QOQ",
"output": "NO"
},
{
"input": "AEEA",
"output": "NO"
},
{
"input": "OQQQO",
"output": "NO"
},
{
"input": "HNCMEEMCNH",
"output": "NO"
},
{
"input": "QDPINBMCRFWXPDBFGOZVVOCEMJRUCTOADEWEGTVBVBFWWRPGYEEYGPRWWFBVBVTGEWEDAOTCURJMECOVVZOGFBDPXWFRCMBNIPDQ",
"output": "NO"
},
{
"input": "A",
"output": "YES"
},
{
"input": "B",
"output": "NO"
},
{
"input": "C",
"output": "NO"
},
{
"input": "D",
"output": "NO"
},
{
"input": "E",
"output": "NO"
},
{
"input": "F",
"output": "NO"
},
{
"input": "G",
"output": "NO"
},
{
"input": "H",
"output": "YES"
},
{
"input": "I",
"output": "YES"
},
{
"input": "J",
"output": "NO"
},
{
"input": "K",
"output": "NO"
},
{
"input": "L",
"output": "NO"
},
{
"input": "M",
"output": "YES"
},
{
"input": "N",
"output": "NO"
},
{
"input": "O",
"output": "YES"
},
{
"input": "P",
"output": "NO"
},
{
"input": "Q",
"output": "NO"
},
{
"input": "R",
"output": "NO"
},
{
"input": "S",
"output": "NO"
},
{
"input": "T",
"output": "YES"
},
{
"input": "U",
"output": "YES"
},
{
"input": "V",
"output": "YES"
},
{
"input": "W",
"output": "YES"
},
{
"input": "X",
"output": "YES"
},
{
"input": "Y",
"output": "YES"
},
{
"input": "JL",
"output": "NO"
},
{
"input": "AAAKTAAA",
"output": "NO"
},
{
"input": "AKA",
"output": "NO"
},
{
"input": "AAJAA",
"output": "NO"
},
{
"input": "ABA",
"output": "NO"
},
{
"input": "AAAAAABAAAAAA",
"output": "NO"
},
{
"input": "ZZ",
"output": "NO"
},
{
"input": "ADA",
"output": "NO"
},
{
"input": "N",
"output": "NO"
},
{
"input": "P",
"output": "NO"
},
{
"input": "LAL",
"output": "NO"
},
{
"input": "AABAA",
"output": "NO"
},
{
"input": "AZA",
"output": "NO"
},
{
"input": "V",
"output": "YES"
},
{
"input": "SSS",
"output": "NO"
},
{
"input": "NNN",
"output": "NO"
},
{
"input": "S",
"output": "NO"
},
{
"input": "I",
"output": "YES"
},
{
"input": "SS",
"output": "NO"
},
{
"input": "E",
"output": "NO"
}
] | 1,608,014,123
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 80
| 140
| 1,843,200
|
s = str(input())
output = "YES"
for j in range(len(s)):
if s[j] in ['A', 'H', 'I', 'M', 'O', 'T', 'U', 'V', 'W', 'X', 'Y']:
continue
else:
output = "NO"
break
p1, p2 = 0, len(s) - 1
while p1 < p2:
if s[p1] != s[p2]:
output = "NO"
break
p1 += 1
p2 -= 1
print(output)
|
Title: Start Up
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, a start up by two students of a state university of city F gained incredible popularity. Now it's time to start a new company. But what do we call it?
The market analysts came up with a very smart plan: the name of the company should be identical to its reflection in a mirror! In other words, if we write out the name of the company on a piece of paper in a line (horizontally, from left to right) with large English letters, then put this piece of paper in front of the mirror, then the reflection of the name in the mirror should perfectly match the line written on the piece of paper.
There are many suggestions for the company name, so coming up to the mirror with a piece of paper for each name wouldn't be sensible. The founders of the company decided to automatize this process. They asked you to write a program that can, given a word, determine whether the word is a 'mirror' word or not.
Input Specification:
The first line contains a non-empty name that needs to be checked. The name contains at most 105 large English letters. The name will be written with the next sans serif font:
Output Specification:
Print 'YES' (without the quotes), if the given name matches its mirror reflection. Otherwise, print 'NO' (without the quotes).
Demo Input:
['AHA\n', 'Z\n', 'XO\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
s = str(input())
output = "YES"
for j in range(len(s)):
if s[j] in ['A', 'H', 'I', 'M', 'O', 'T', 'U', 'V', 'W', 'X', 'Y']:
continue
else:
output = "NO"
break
p1, p2 = 0, len(s) - 1
while p1 < p2:
if s[p1] != s[p2]:
output = "NO"
break
p1 += 1
p2 -= 1
print(output)
```
| 3
|
|
122
|
A
|
Lucky Division
|
PROGRAMMING
| 1,000
|
[
"brute force",
"number theory"
] | null | null |
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
|
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
|
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
|
[
"47\n",
"16\n",
"78\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
| 500
|
[
{
"input": "47",
"output": "YES"
},
{
"input": "16",
"output": "YES"
},
{
"input": "78",
"output": "NO"
},
{
"input": "48",
"output": "YES"
},
{
"input": "100",
"output": "YES"
},
{
"input": "107",
"output": "NO"
},
{
"input": "77",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "480",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "49",
"output": "YES"
},
{
"input": "56",
"output": "YES"
},
{
"input": "124",
"output": "YES"
},
{
"input": "1000",
"output": "YES"
},
{
"input": "999",
"output": "NO"
},
{
"input": "298",
"output": "NO"
},
{
"input": "274",
"output": "NO"
},
{
"input": "998",
"output": "NO"
},
{
"input": "42",
"output": "YES"
},
{
"input": "788",
"output": "YES"
},
{
"input": "70",
"output": "YES"
},
{
"input": "444",
"output": "YES"
},
{
"input": "777",
"output": "YES"
},
{
"input": "799",
"output": "YES"
},
{
"input": "25",
"output": "NO"
},
{
"input": "882",
"output": "YES"
},
{
"input": "88",
"output": "YES"
},
{
"input": "11",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "7",
"output": "YES"
},
{
"input": "8",
"output": "YES"
},
{
"input": "94",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "141",
"output": "YES"
}
] | 1,690,352,074
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 60
| 0
|
a = int(input())
if a % 4 == 0 or a % 7 == 7: print('YES')
else: print('NO')
|
Title: Lucky Division
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
Output Specification:
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['47\n', '16\n', '78\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
|
```python
a = int(input())
if a % 4 == 0 or a % 7 == 7: print('YES')
else: print('NO')
```
| 0
|
|
270
|
A
|
Fancy Fence
|
PROGRAMMING
| 1,100
|
[
"geometry",
"implementation",
"math"
] | null | null |
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
|
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
|
[
"3\n30\n60\n90\n"
] |
[
"NO\nYES\nYES\n"
] |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
| 500
|
[
{
"input": "3\n30\n60\n90",
"output": "NO\nYES\nYES"
},
{
"input": "6\n1\n2\n3\n170\n179\n25",
"output": "NO\nNO\nNO\nYES\nYES\nNO"
}
] | 1,632,923,028
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 6,758,400
|
lp= [360]
for i in range(3,181):
if 180%i==0:
lp.append(i)
n=int(input())
for i in range(n):
d=int(input())
print("YES" if d in lp and d>=60 else "NO")
|
Title: Fancy Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
Input Specification:
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
Output Specification:
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
Demo Input:
['3\n30\n60\n90\n']
Demo Output:
['NO\nYES\nYES\n']
Note:
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
```python
lp= [360]
for i in range(3,181):
if 180%i==0:
lp.append(i)
n=int(input())
for i in range(n):
d=int(input())
print("YES" if d in lp and d>=60 else "NO")
```
| 0
|
Subsets and Splits
Successful Python Submissions
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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.