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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
721
|
B
|
Passwords
|
PROGRAMMING
| 1,100
|
[
"implementation",
"math",
"sortings",
"strings"
] | null | null |
Vanya is managed to enter his favourite site Codehorses. Vanya uses *n* distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password *k* times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
|
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next *n* lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his *n* passwords.
|
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
|
[
"5 2\ncba\nabc\nbb1\nabC\nABC\nabc\n",
"4 100\n11\n22\n1\n2\n22\n"
] |
[
"1 15\n",
"3 4\n"
] |
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
| 1,000
|
[
{
"input": "5 2\ncba\nabc\nbb1\nabC\nABC\nabc",
"output": "1 15"
},
{
"input": "4 100\n11\n22\n1\n2\n22",
"output": "3 4"
},
{
"input": "1 1\na1\na1",
"output": "1 1"
},
{
"input": "1 100\na1\na1",
"output": "1 1"
},
{
"input": "2 1\nabc\nAbc\nAbc",
"output": "1 7"
},
{
"input": "2 2\nabc\nAbc\nabc",
"output": "1 2"
},
{
"input": "2 1\nab\nabc\nab",
"output": "1 1"
},
{
"input": "2 2\nab\nabc\nab",
"output": "1 1"
},
{
"input": "2 1\nab\nabc\nabc",
"output": "7 7"
},
{
"input": "2 2\nab\nabc\nabc",
"output": "2 2"
},
{
"input": "10 3\nOIbV1igi\no\nZS\nQM\n9woLzI\nWreboD\nQ7yl\nA5Rb\nS9Lno72TkP\nfT97o\no",
"output": "1 1"
},
{
"input": "10 3\nHJZNMsT\nLaPcH2C\nlrhqIO\n9cxw\noTC1XwjW\nGHL9Ul6\nUyIs\nPuzwgR4ZKa\nyIByoKR5\nd3QA\nPuzwgR4ZKa",
"output": "25 25"
},
{
"input": "20 5\nvSyC787KlIL8kZ2Uv5sw\nWKWOP\n7i8J3E8EByIq\nNW2VyGweL\nmyR2sRNu\nmXusPP0\nf4jgGxra\n4wHRzRhOCpEt\npPz9kybGb\nOtSpePCRoG5nkjZ2VxRy\nwHYsSttWbJkg\nKBOP9\nQfiOiFyHPPsw3GHo8J8\nxB8\nqCpehZEeEhdq\niOLjICK6\nQ91\nHmCsfMGTFKoFFnv238c\nJKjhg\ngkEUh\nKBOP9",
"output": "3 11"
},
{
"input": "15 2\nw6S9WyU\nMVh\nkgUhQHW\nhGQNOF\nUuym\n7rGQA\nBM8vLPRB\n9E\nDs32U\no\nz1aV2C5T\n8\nzSXjrqQ\n1FO\n3kIt\nBM8vLPRB",
"output": "44 50"
},
{
"input": "20 2\ni\n5Rp6\nE4vsr\nSY\nORXx\nh13C\nk6tzC\ne\nN\nKQf4C\nWZcdL\ndiA3v\n0InQT\nuJkAr\nGCamp\nBuIRd\nY\nM\nxZYx7\n0a5A\nWZcdL",
"output": "36 65"
},
{
"input": "20 2\naWLQ6\nSgQ9r\nHcPdj\n2BNaO\n3TjNb\nnvwFM\nqsKt7\nFnb6N\nLoc0p\njxuLq\nBKAjf\nEKgZB\nBfOSa\nsMIvr\nuIWcR\nIura3\nLAqSf\ntXq3G\n8rQ8I\n8otAO\nsMIvr",
"output": "1 65"
},
{
"input": "20 15\n0ZpQugVlN7\nm0SlKGnohN\nRFXTqhNGcn\n1qm2ZbB\nQXtJWdf78P\nbc2vH\nP21dty2Z1P\nm2c71LFhCk\n23EuP1Dvh3\nanwri5RhQN\n55v6HYv288\n1u5uKOjM5r\n6vg0GC1\nDAPYiA3ns1\nUZaaJ3Gmnk\nwB44x7V4Zi\n4hgB2oyU8P\npYFQpy8gGK\ndbz\nBv\n55v6HYv288",
"output": "6 25"
},
{
"input": "3 1\na\nb\naa\naa",
"output": "13 13"
},
{
"input": "6 3\nab\nac\nad\nabc\nabd\nabe\nabc",
"output": "9 11"
},
{
"input": "4 2\n1\n2\n11\n22\n22",
"output": "8 9"
},
{
"input": "2 1\n1\n12\n12",
"output": "7 7"
},
{
"input": "3 1\nab\nabc\nabd\nabc",
"output": "7 13"
},
{
"input": "2 1\na\nab\nab",
"output": "7 7"
},
{
"input": "5 2\na\nb\nc\nab\naa\naa",
"output": "9 15"
},
{
"input": "6 1\n1\n2\n11\n22\n111\n2222\n22",
"output": "13 19"
},
{
"input": "3 1\n1\n2\n11\n11",
"output": "13 13"
},
{
"input": "10 4\na\nb\nc\nd\ne\nf\nab\ncd\nac\nad\nac",
"output": "12 20"
},
{
"input": "4 2\na\nb\nc\nd\na",
"output": "1 9"
},
{
"input": "4 1\n1\n2\n3\n4\n4",
"output": "1 19"
},
{
"input": "5 1\na\nb\nc\nd\nef\nef",
"output": "25 25"
},
{
"input": "6 4\n1\n2\n22\n33\n44\n555\n555",
"output": "11 11"
},
{
"input": "5 2\na\nb\nc\nd\nab\nab",
"output": "15 15"
},
{
"input": "6 2\n1\n2\n3\n4\n5\n23\n23",
"output": "16 16"
},
{
"input": "4 2\na\nb\naa\nbb\naa",
"output": "8 9"
},
{
"input": "5 4\na\nbb\ncc\ndd\nee\nbb",
"output": "2 10"
},
{
"input": "4 1\na\nb\nc\nab\nab",
"output": "19 19"
},
{
"input": "7 100\na\nb\nc\nd\ne\ng\nab\nab",
"output": "7 7"
},
{
"input": "6 1\na\nb\nc\nd\ne\naa\naa",
"output": "31 31"
},
{
"input": "4 1\na\nas\nasd\nasde\nasde",
"output": "19 19"
},
{
"input": "5 2\n1\n2\n3\n11\n22\n22",
"output": "9 15"
},
{
"input": "10 2\na\nb\nc\nd\nee\nff\ngg\nhh\nii\njj\nii",
"output": "15 30"
},
{
"input": "3 1\na\nab\nbc\nab",
"output": "7 13"
},
{
"input": "6 4\na\nb\nc\nbb\nbc\ncc\ncc",
"output": "4 11"
}
] | 1,636,516,007
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 46
| 4,505,600
|
first_line = input().split()
cases = int(first_line[0])
limit = int(first_line[1])
full_list = [input() for i in range(cases+1)]
def password_time(full_list, limit):
real_password = full_list[-1]
all_tries = full_list[:-1]
shorter = [i for i in all_tries if len(i) < len(real_password)]
same_length = [i for i in all_tries if len(i) == len(real_password)]
best_case_tries = len(shorter) + 1
best_case_total = best_case_tries + ((best_case_tries // limit) * 5)
worst_case_tries = len(same_length) + len(shorter)
worst_case_total = worst_case_tries + ((worst_case_tries // limit) * 5)
print(str(best_case_total) + ' ' + str(worst_case_total))
password_time(full_list, limit)
|
Title: Passwords
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya is managed to enter his favourite site Codehorses. Vanya uses *n* distinct passwords for sites at all, however he can't remember which one exactly he specified during Codehorses registration.
Vanya will enter passwords in order of non-decreasing their lengths, and he will enter passwords of same length in arbitrary order. Just when Vanya will have entered the correct password, he is immediately authorized on the site. Vanya will not enter any password twice.
Entering any passwords takes one second for Vanya. But if Vanya will enter wrong password *k* times, then he is able to make the next try only 5 seconds after that. Vanya makes each try immediately, that is, at each moment when Vanya is able to enter password, he is doing that.
Determine how many seconds will Vanya need to enter Codehorses in the best case for him (if he spends minimum possible number of second) and in the worst case (if he spends maximum possible amount of seconds).
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of Vanya's passwords and the number of failed tries, after which the access to the site is blocked for 5 seconds.
The next *n* lines contains passwords, one per line — pairwise distinct non-empty strings consisting of latin letters and digits. Each password length does not exceed 100 characters.
The last line of the input contains the Vanya's Codehorses password. It is guaranteed that the Vanya's Codehorses password is equal to some of his *n* passwords.
Output Specification:
Print two integers — time (in seconds), Vanya needs to be authorized to Codehorses in the best case for him and in the worst case respectively.
Demo Input:
['5 2\ncba\nabc\nbb1\nabC\nABC\nabc\n', '4 100\n11\n22\n1\n2\n22\n']
Demo Output:
['1 15\n', '3 4\n']
Note:
Consider the first sample case. As soon as all passwords have the same length, Vanya can enter the right password at the first try as well as at the last try. If he enters it at the first try, he spends exactly 1 second. Thus in the best case the answer is 1. If, at the other hand, he enters it at the last try, he enters another 4 passwords before. He spends 2 seconds to enter first 2 passwords, then he waits 5 seconds as soon as he made 2 wrong tries. Then he spends 2 more seconds to enter 2 wrong passwords, again waits 5 seconds and, finally, enters the correct password spending 1 more second. In summary in the worst case he is able to be authorized in 15 seconds.
Consider the second sample case. There is no way of entering passwords and get the access to the site blocked. As soon as the required password has length of 2, Vanya enters all passwords of length 1 anyway, spending 2 seconds for that. Then, in the best case, he immediately enters the correct password and the answer for the best case is 3, but in the worst case he enters wrong password of length 2 and only then the right one, spending 4 seconds at all.
|
```python
first_line = input().split()
cases = int(first_line[0])
limit = int(first_line[1])
full_list = [input() for i in range(cases+1)]
def password_time(full_list, limit):
real_password = full_list[-1]
all_tries = full_list[:-1]
shorter = [i for i in all_tries if len(i) < len(real_password)]
same_length = [i for i in all_tries if len(i) == len(real_password)]
best_case_tries = len(shorter) + 1
best_case_total = best_case_tries + ((best_case_tries // limit) * 5)
worst_case_tries = len(same_length) + len(shorter)
worst_case_total = worst_case_tries + ((worst_case_tries // limit) * 5)
print(str(best_case_total) + ' ' + str(worst_case_total))
password_time(full_list, limit)
```
| 0
|
|
424
|
A
|
Squats
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Pasha has many hamsters and he makes them work out. Today, *n* hamsters (*n* is even) came to work out. The hamsters lined up and each hamster either sat down or stood up.
For another exercise, Pasha needs exactly hamsters to stand up and the other hamsters to sit down. In one minute, Pasha can make some hamster ether sit down or stand up. How many minutes will he need to get what he wants if he acts optimally well?
|
The first line contains integer *n* (2<=≤<=*n*<=≤<=200; *n* is even). The next line contains *n* characters without spaces. These characters describe the hamsters' position: the *i*-th character equals 'X', if the *i*-th hamster in the row is standing, and 'x', if he is sitting.
|
In the first line, print a single integer — the minimum required number of minutes. In the second line, print a string that describes the hamsters' position after Pasha makes the required changes. If there are multiple optimal positions, print any of them.
|
[
"4\nxxXx\n",
"2\nXX\n",
"6\nxXXxXx\n"
] |
[
"1\nXxXx\n",
"1\nxX\n",
"0\nxXXxXx\n"
] |
none
| 500
|
[
{
"input": "4\nxxXx",
"output": "1\nXxXx"
},
{
"input": "2\nXX",
"output": "1\nxX"
},
{
"input": "6\nxXXxXx",
"output": "0\nxXXxXx"
},
{
"input": "4\nxXXX",
"output": "1\nxxXX"
},
{
"input": "2\nXx",
"output": "0\nXx"
},
{
"input": "22\nXXxXXxxXxXxXXXXXXXXXxx",
"output": "4\nxxxxxxxXxXxXXXXXXXXXxx"
},
{
"input": "30\nXXxXxxXXXXxxXXxxXXxxxxXxxXXXxx",
"output": "0\nXXxXxxXXXXxxXXxxXXxxxxXxxXXXxx"
},
{
"input": "104\nxxXxXxxXXXxxXxXxxXXXxxxXxxXXXxxXXXxXxXxXXxxXxxxxxXXXXxXXXXxXXXxxxXxxxxxxxXxxXxXXxxXXXXxXXXxxXXXXXXXXXxXX",
"output": "4\nxxxxxxxxxXxxXxXxxXXXxxxXxxXXXxxXXXxXxXxXXxxXxxxxxXXXXxXXXXxXXXxxxXxxxxxxxXxxXxXXxxXXXXxXXXxxXXXXXXXXXxXX"
},
{
"input": "78\nxxxXxxXxXxxXxxxxxXxXXXxXXXXxxxxxXxXXXxxXxXXXxxxxXxxXXXxxxxxxxxXXXXxXxXXxXXXxXX",
"output": "3\nXXXXxxXxXxxXxxxxxXxXXXxXXXXxxxxxXxXXXxxXxXXXxxxxXxxXXXxxxxxxxxXXXXxXxXXxXXXxXX"
},
{
"input": "200\nxxXXxxXXxXxxXxxXxXxxXxXxXxXxxxxxXXxXXxxXXXXxXXXxXXxXxXxxxxXxxXXXxxxXxXxxxXxxXXxXxXxxxxxxxXxxXxXxxXxXXXxxXxXXXXxxXxxxXxXXXXXXxXxXXxxxxXxxxXxxxXxXXXxXxXXXXxXXxxxXxXXxxXXxxxXxXxXXxXXXxXxXxxxXXxxxxXXxXXXX",
"output": "4\nXXXXXXXXxXxxXxxXxXxxXxXxXxXxxxxxXXxXXxxXXXXxXXXxXXxXxXxxxxXxxXXXxxxXxXxxxXxxXXxXxXxxxxxxxXxxXxXxxXxXXXxxXxXXXXxxXxxxXxXXXXXXxXxXXxxxxXxxxXxxxXxXXXxXxXXXXxXXxxxXxXXxxXXxxxXxXxXXxXXXxXxXxxxXXxxxxXXxXXXX"
},
{
"input": "198\nxXxxXxxXxxXXxXxXxXxxXXXxxXxxxxXXXXxxXxxxxXXXXxXxXXxxxXXXXXXXxXXXxxxxXXxXXxXxXXxxxxXxXXXXXXxXxxXxXxxxXxXXXXxxXXxxXxxxXXxXxXXxXxXXxXXXXxxxxxXxXXxxxXxXXXXxXxXXxxXxXXxXxXXxxxXxXXXXxXxxXxXXXxxxxXxXXXXxXx",
"output": "5\nxxxxxxxxxxxxxXxXxXxxXXXxxXxxxxXXXXxxXxxxxXXXXxXxXXxxxXXXXXXXxXXXxxxxXXxXXxXxXXxxxxXxXXXXXXxXxxXxXxxxXxXXXXxxXXxxXxxxXXxXxXXxXxXXxXXXXxxxxxXxXXxxxXxXXXXxXxXXxxXxXXxXxXXxxxXxXXXXxXxxXxXXXxxxxXxXXXXxXx"
},
{
"input": "200\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
"output": "100\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
},
{
"input": "198\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
"output": "99\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
},
{
"input": "200\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX",
"output": "100\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX"
},
{
"input": "198\nXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX",
"output": "99\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX"
},
{
"input": "2\nxx",
"output": "1\nXx"
},
{
"input": "2\nXx",
"output": "0\nXx"
},
{
"input": "2\nxX",
"output": "0\nxX"
},
{
"input": "4\nXXXX",
"output": "2\nxxXX"
},
{
"input": "4\nxxxx",
"output": "2\nXXxx"
},
{
"input": "4\nxxXX",
"output": "0\nxxXX"
},
{
"input": "4\nXXxx",
"output": "0\nXXxx"
},
{
"input": "4\nxXxx",
"output": "1\nXXxx"
},
{
"input": "4\nXxxx",
"output": "1\nXXxx"
},
{
"input": "4\nxxxX",
"output": "1\nXxxX"
},
{
"input": "4\nxxXx",
"output": "1\nXxXx"
},
{
"input": "4\nXXXx",
"output": "1\nxXXx"
},
{
"input": "4\nxXXX",
"output": "1\nxxXX"
},
{
"input": "4\nXxXX",
"output": "1\nxxXX"
},
{
"input": "4\nXXxX",
"output": "1\nxXxX"
},
{
"input": "4\nXxXx",
"output": "0\nXxXx"
},
{
"input": "6\nxXXxXX",
"output": "1\nxxXxXX"
}
] | 1,490,719,970
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 62
| 4,608,000
|
ham_count = int(input())
order = input()
final_order = order
small_count = 0
big_count = 0
for i in order:
if i == 'x':
small_count = small_count + 1
else:
big_count = big_count + 1
if small_count > big_count:
diff = int((small_count - big_count) / 2)
print(diff)
for i, index in enumerate(order):
if diff == 0:
break
elif index == 'x':
final_order = final_order[:i] + 'X' + final_order[i + 1:]
diff = diff - 1
print(final_order)
elif big_count > small_count:
diff = int((big_count - small_count) / 2)
print(diff)
for i, index in enumerate(order):
if diff == 0:
break
elif index == 'X':
final_order = final_order[:i] + 'x' + final_order[i + 1:]
diff = diff - 1
print(final_order)
else:
print(0)
print(order)
|
Title: Squats
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha has many hamsters and he makes them work out. Today, *n* hamsters (*n* is even) came to work out. The hamsters lined up and each hamster either sat down or stood up.
For another exercise, Pasha needs exactly hamsters to stand up and the other hamsters to sit down. In one minute, Pasha can make some hamster ether sit down or stand up. How many minutes will he need to get what he wants if he acts optimally well?
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=200; *n* is even). The next line contains *n* characters without spaces. These characters describe the hamsters' position: the *i*-th character equals 'X', if the *i*-th hamster in the row is standing, and 'x', if he is sitting.
Output Specification:
In the first line, print a single integer — the minimum required number of minutes. In the second line, print a string that describes the hamsters' position after Pasha makes the required changes. If there are multiple optimal positions, print any of them.
Demo Input:
['4\nxxXx\n', '2\nXX\n', '6\nxXXxXx\n']
Demo Output:
['1\nXxXx\n', '1\nxX\n', '0\nxXXxXx\n']
Note:
none
|
```python
ham_count = int(input())
order = input()
final_order = order
small_count = 0
big_count = 0
for i in order:
if i == 'x':
small_count = small_count + 1
else:
big_count = big_count + 1
if small_count > big_count:
diff = int((small_count - big_count) / 2)
print(diff)
for i, index in enumerate(order):
if diff == 0:
break
elif index == 'x':
final_order = final_order[:i] + 'X' + final_order[i + 1:]
diff = diff - 1
print(final_order)
elif big_count > small_count:
diff = int((big_count - small_count) / 2)
print(diff)
for i, index in enumerate(order):
if diff == 0:
break
elif index == 'X':
final_order = final_order[:i] + 'x' + final_order[i + 1:]
diff = diff - 1
print(final_order)
else:
print(0)
print(order)
```
| 3
|
|
242
|
B
|
Big Segment
|
PROGRAMMING
| 1,100
|
[
"implementation",
"sortings"
] | null | null |
A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*].
You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1.
Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment.
It is guaranteed that no two segments coincide.
|
Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1.
The segments are numbered starting from 1 in the order in which they appear in the input.
|
[
"3\n1 1\n2 2\n3 3\n",
"6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n"
] |
[
"-1\n",
"3\n"
] |
none
| 1,000
|
[
{
"input": "3\n1 1\n2 2\n3 3",
"output": "-1"
},
{
"input": "6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10",
"output": "3"
},
{
"input": "4\n1 5\n2 2\n2 4\n2 5",
"output": "1"
},
{
"input": "5\n3 3\n1 3\n2 2\n2 3\n1 2",
"output": "2"
},
{
"input": "7\n7 7\n8 8\n3 7\n1 6\n1 7\n4 7\n2 8",
"output": "-1"
},
{
"input": "3\n2 5\n3 4\n2 3",
"output": "1"
},
{
"input": "16\n15 15\n8 12\n6 9\n15 16\n8 14\n3 12\n7 19\n9 13\n5 16\n9 17\n10 15\n9 14\n9 9\n18 19\n5 15\n6 19",
"output": "-1"
},
{
"input": "9\n1 10\n7 8\n6 7\n1 4\n5 9\n2 8\n3 10\n1 1\n2 3",
"output": "1"
},
{
"input": "1\n1 100000",
"output": "1"
},
{
"input": "6\n2 2\n3 3\n3 5\n4 5\n1 1\n1 5",
"output": "6"
},
{
"input": "33\n2 18\n4 14\n2 16\n10 12\n4 6\n9 17\n2 8\n4 12\n8 20\n1 10\n11 14\n11 17\n8 15\n3 16\n3 4\n6 9\n6 19\n4 17\n17 19\n6 16\n3 12\n1 7\n6 20\n8 16\n12 19\n1 3\n12 18\n6 11\n7 20\n16 18\n4 15\n3 15\n15 19",
"output": "-1"
},
{
"input": "34\n3 8\n5 9\n2 9\n1 4\n3 7\n3 3\n8 9\n6 10\n4 7\n6 7\n5 8\n5 10\n1 5\n8 8\n2 5\n3 5\n7 7\n2 8\n4 5\n1 1\n7 9\n5 6\n2 3\n1 2\n2 4\n8 10\n7 8\n1 3\n4 8\n9 10\n1 7\n10 10\n2 2\n1 8",
"output": "-1"
},
{
"input": "55\n3 4\n6 8\n9 10\n3 9\n9 9\n2 5\n4 8\n3 8\n8 10\n1 1\n4 9\n10 10\n6 6\n8 8\n1 8\n5 5\n4 5\n5 9\n2 2\n3 10\n4 6\n3 6\n1 6\n1 7\n6 10\n2 6\n3 7\n2 4\n4 4\n5 10\n1 4\n2 9\n1 3\n7 9\n7 8\n1 9\n1 10\n2 8\n8 9\n6 7\n1 2\n6 9\n7 7\n4 7\n3 3\n2 7\n4 10\n7 10\n2 3\n2 10\n5 7\n3 5\n5 8\n1 5\n5 6",
"output": "37"
},
{
"input": "1\n999999999 1000000000",
"output": "1"
},
{
"input": "3\n1 20\n2 22\n3 18",
"output": "-1"
},
{
"input": "1\n1000000000 1000000000",
"output": "1"
},
{
"input": "2\n100001 100008\n100005 100006",
"output": "1"
},
{
"input": "1\n1000000 10000000",
"output": "1"
},
{
"input": "3\n3 6\n2 4\n1 5",
"output": "-1"
},
{
"input": "2\n3 5\n1 2",
"output": "-1"
}
] | 1,594,672,241
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 0
| 2,000
| 6,963,200
|
nums = int(input())
first = []
second = []
for i in range(nums):
d = [int(i) for i in input().split()]
first.append(d[0])
second.append(d[1])
answer = -1
i = 0
while i < nums:
if first[i] == min(first) and second[i] == max(second):
answer = i+1
break
print(answer)
|
Title: Big Segment
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A coordinate line has *n* segments, the *i*-th segment starts at the position *l**i* and ends at the position *r**i*. We will denote such a segment as [*l**i*,<=*r**i*].
You have suggested that one of the defined segments covers all others. In other words, there is such segment in the given set, which contains all other ones. Now you want to test your assumption. Find in the given set the segment which covers all other segments, and print its number. If such a segment doesn't exist, print -1.
Formally we will assume that segment [*a*,<=*b*] covers segment [*c*,<=*d*], if they meet this condition *a*<=≤<=*c*<=≤<=*d*<=≤<=*b*.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of segments. Next *n* lines contain the descriptions of the segments. The *i*-th line contains two space-separated integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=109) — the borders of the *i*-th segment.
It is guaranteed that no two segments coincide.
Output Specification:
Print a single integer — the number of the segment that covers all other segments in the set. If there's no solution, print -1.
The segments are numbered starting from 1 in the order in which they appear in the input.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '6\n1 5\n2 3\n1 10\n7 10\n7 7\n10 10\n']
Demo Output:
['-1\n', '3\n']
Note:
none
|
```python
nums = int(input())
first = []
second = []
for i in range(nums):
d = [int(i) for i in input().split()]
first.append(d[0])
second.append(d[1])
answer = -1
i = 0
while i < nums:
if first[i] == min(first) and second[i] == max(second):
answer = i+1
break
print(answer)
```
| 0
|
|
884
|
B
|
Japanese Crosswords Strike Back
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
|
The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding.
|
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
|
[
"2 4\n1 3\n",
"3 10\n3 3 2\n",
"2 10\n1 3\n"
] |
[
"NO\n",
"YES\n",
"NO\n"
] |
none
| 0
|
[
{
"input": "2 4\n1 3",
"output": "NO"
},
{
"input": "3 10\n3 3 2",
"output": "YES"
},
{
"input": "2 10\n1 3",
"output": "NO"
},
{
"input": "1 1\n1",
"output": "YES"
},
{
"input": "1 10\n10",
"output": "YES"
},
{
"input": "1 10000\n10000",
"output": "YES"
},
{
"input": "10 1\n5 78 3 87 4 9 5 8 9 1235",
"output": "NO"
},
{
"input": "3 12\n3 3 3",
"output": "NO"
},
{
"input": "3 9\n2 2 2",
"output": "NO"
},
{
"input": "2 5\n1 1",
"output": "NO"
},
{
"input": "1 2\n1",
"output": "NO"
},
{
"input": "3 13\n3 3 3",
"output": "NO"
},
{
"input": "3 6\n1 1 1",
"output": "NO"
},
{
"input": "1 6\n5",
"output": "NO"
},
{
"input": "3 11\n3 3 2",
"output": "NO"
},
{
"input": "2 6\n1 3",
"output": "NO"
},
{
"input": "3 10\n2 2 2",
"output": "NO"
},
{
"input": "3 8\n2 1 1",
"output": "NO"
},
{
"input": "1 5\n2",
"output": "NO"
},
{
"input": "1 3\n1",
"output": "NO"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "NO"
},
{
"input": "2 10\n4 4",
"output": "NO"
},
{
"input": "2 8\n2 3",
"output": "NO"
},
{
"input": "2 4\n1 1",
"output": "NO"
},
{
"input": "3 10\n1 2 4",
"output": "NO"
},
{
"input": "3 10\n2 1 3",
"output": "NO"
},
{
"input": "2 6\n1 2",
"output": "NO"
},
{
"input": "3 4\n1 1 1",
"output": "NO"
},
{
"input": "3 11\n1 2 4",
"output": "NO"
},
{
"input": "3 12\n3 3 2",
"output": "NO"
},
{
"input": "4 9\n1 1 1 1",
"output": "NO"
},
{
"input": "1 10\n9",
"output": "NO"
},
{
"input": "1 7\n5",
"output": "NO"
},
{
"input": "2 5\n1 2",
"output": "NO"
},
{
"input": "3 8\n1 1 2",
"output": "NO"
},
{
"input": "3 7\n1 1 1",
"output": "NO"
},
{
"input": "1 10\n1",
"output": "NO"
},
{
"input": "3 10\n2 3 4",
"output": "NO"
},
{
"input": "3 9\n1 2 3",
"output": "NO"
},
{
"input": "3 9\n3 3 2",
"output": "NO"
},
{
"input": "3 6\n3 3 2",
"output": "NO"
},
{
"input": "1 1\n3",
"output": "NO"
},
{
"input": "1 3\n2",
"output": "NO"
},
{
"input": "3 10\n3 3 3",
"output": "NO"
},
{
"input": "3 5\n1 1 1",
"output": "YES"
},
{
"input": "2 1\n100 100",
"output": "NO"
},
{
"input": "3 3\n3 3 3",
"output": "NO"
},
{
"input": "4 17\n3 3 9 1",
"output": "NO"
},
{
"input": "4 1660\n1505 13 37 100",
"output": "NO"
},
{
"input": "3 5\n3 3 2",
"output": "NO"
},
{
"input": "4 10\n3 3 2 5",
"output": "NO"
},
{
"input": "5 5\n5 5 5 5 5",
"output": "NO"
},
{
"input": "1 1\n2",
"output": "NO"
},
{
"input": "5 10\n1 2 2 4 5",
"output": "NO"
},
{
"input": "2 1\n1 1",
"output": "NO"
}
] | 1,509,546,825
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 66
| 108
| 7,372,800
|
n,x=map(int,input().split())
l=list(map(int,input().split()))
s=0
for i in range(n):
s+=l[i]
if s+n-1==x:
print('yes')
else : print('no')
|
Title: Japanese Crosswords Strike Back
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A one-dimensional Japanese crossword can be represented as a binary string of length *x*. An encoding of this crossword is an array *a* of size *n*, where *n* is the number of segments formed completely of 1's, and *a**i* is the length of *i*-th segment. No two segments touch or intersect.
For example:
- If *x*<==<=6 and the crossword is 111011, then its encoding is an array {3,<=2}; - If *x*<==<=8 and the crossword is 01101010, then its encoding is an array {2,<=1,<=1}; - If *x*<==<=5 and the crossword is 11111, then its encoding is an array {5}; - If *x*<==<=5 and the crossword is 00000, then its encoding is an empty array.
Mishka wants to create a new one-dimensional Japanese crossword. He has already picked the length and the encoding for this crossword. And now he needs to check if there is exactly one crossword such that its length and encoding are equal to the length and encoding he picked. Help him to check it!
Input Specification:
The first line contains two integer numbers *n* and *x* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*x*<=≤<=109) — the number of elements in the encoding and the length of the crossword Mishka picked.
The second line contains *n* integer numbers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=10000) — the encoding.
Output Specification:
Print YES if there exists exaclty one crossword with chosen length and encoding. Otherwise, print NO.
Demo Input:
['2 4\n1 3\n', '3 10\n3 3 2\n', '2 10\n1 3\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none
|
```python
n,x=map(int,input().split())
l=list(map(int,input().split()))
s=0
for i in range(n):
s+=l[i]
if s+n-1==x:
print('yes')
else : print('no')
```
| 3
|
|
224
|
A
|
Parallelepiped
|
PROGRAMMING
| 1,100
|
[
"brute force",
"geometry",
"math"
] | null | null |
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
|
The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.
|
Print a single number — the sum of all edges of the parallelepiped.
|
[
"1 1 1\n",
"4 6 6\n"
] |
[
"12\n",
"28\n"
] |
In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3.
| 500
|
[
{
"input": "1 1 1",
"output": "12"
},
{
"input": "4 6 6",
"output": "28"
},
{
"input": "20 10 50",
"output": "68"
},
{
"input": "9 4 36",
"output": "56"
},
{
"input": "324 9 36",
"output": "184"
},
{
"input": "1333 93 129",
"output": "308"
},
{
"input": "1022 584 112",
"output": "380"
},
{
"input": "66 174 319",
"output": "184"
},
{
"input": "912 276 1748",
"output": "444"
},
{
"input": "65 156 60",
"output": "120"
},
{
"input": "1 10000 10000",
"output": "40008"
},
{
"input": "1485 55 27",
"output": "332"
},
{
"input": "152 108 4104",
"output": "528"
},
{
"input": "1656 6900 1350",
"output": "740"
},
{
"input": "12 14 42",
"output": "60"
},
{
"input": "615 18 1230",
"output": "856"
},
{
"input": "680 60 408",
"output": "336"
},
{
"input": "644 966 6",
"output": "1308"
},
{
"input": "1 432 432",
"output": "1736"
},
{
"input": "2239 2239 1",
"output": "8964"
},
{
"input": "4106 8212 2",
"output": "16436"
},
{
"input": "10000 10000 10000",
"output": "1200"
},
{
"input": "3623 3623 1",
"output": "14500"
},
{
"input": "9801 9801 9801",
"output": "1188"
},
{
"input": "10000 1 10000",
"output": "40008"
},
{
"input": "9 9 9",
"output": "36"
},
{
"input": "9801 9702 9702",
"output": "1184"
}
] | 1,623,218,554
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 218
| 0
|
#%%
import math
ins = list(map(int, input().split(" ")))
ins[2] = int(math.sqrt(ins[2] * ins[1] / ins[0]))
ins[1] /= ins[2]
ins[0] /= ins[1]
ins[0], ins[1] = int(ins[0]), int(ins[1])
print(4 * sum(ins))
|
Title: Parallelepiped
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.
Input Specification:
The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive (<=><=0) and do not exceed 104. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.
Output Specification:
Print a single number — the sum of all edges of the parallelepiped.
Demo Input:
['1 1 1\n', '4 6 6\n']
Demo Output:
['12\n', '28\n']
Note:
In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3.
|
```python
#%%
import math
ins = list(map(int, input().split(" ")))
ins[2] = int(math.sqrt(ins[2] * ins[1] / ins[0]))
ins[1] /= ins[2]
ins[0] /= ins[1]
ins[0], ins[1] = int(ins[0]), int(ins[1])
print(4 * sum(ins))
```
| 3
|
|
277
|
A
|
Learning Languages
|
PROGRAMMING
| 1,400
|
[
"dfs and similar",
"dsu"
] | null | null |
The "BerCorp" company has got *n* employees. These employees can use *m* approved official languages for the formal correspondence. The languages are numbered with integers from 1 to *m*. For each employee we have the list of languages, which he knows. This list could be empty, i. e. an employee may know no official languages. But the employees are willing to learn any number of official languages, as long as the company pays their lessons. A study course in one language for one employee costs 1 berdollar.
Find the minimum sum of money the company needs to spend so as any employee could correspond to any other one (their correspondence can be indirect, i. e. other employees can help out translating).
|
The first line contains two integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of employees and the number of languages.
Then *n* lines follow — each employee's language list. At the beginning of the *i*-th line is integer *k**i* (0<=≤<=*k**i*<=≤<=*m*) — the number of languages the *i*-th employee knows. Next, the *i*-th line contains *k**i* integers — *a**ij* (1<=≤<=*a**ij*<=≤<=*m*) — the identifiers of languages the *i*-th employee knows. It is guaranteed that all the identifiers in one list are distinct. Note that an employee may know zero languages.
The numbers in the lines are separated by single spaces.
|
Print a single integer — the minimum amount of money to pay so that in the end every employee could write a letter to every other one (other employees can help out translating).
|
[
"5 5\n1 2\n2 2 3\n2 3 4\n2 4 5\n1 5\n",
"8 7\n0\n3 1 2 3\n1 1\n2 5 4\n2 6 7\n1 3\n2 7 4\n1 1\n",
"2 2\n1 2\n0\n"
] |
[
"0\n",
"2\n",
"1\n"
] |
In the second sample the employee 1 can learn language 2, and employee 8 can learn language 4.
In the third sample employee 2 must learn language 2.
| 500
|
[
{
"input": "5 5\n1 2\n2 2 3\n2 3 4\n2 4 5\n1 5",
"output": "0"
},
{
"input": "8 7\n0\n3 1 2 3\n1 1\n2 5 4\n2 6 7\n1 3\n2 7 4\n1 1",
"output": "2"
},
{
"input": "2 2\n1 2\n0",
"output": "1"
},
{
"input": "2 2\n0\n0",
"output": "2"
},
{
"input": "5 5\n1 3\n0\n0\n2 4 1\n0",
"output": "4"
},
{
"input": "6 2\n0\n0\n2 1 2\n1 1\n1 1\n0",
"output": "3"
},
{
"input": "7 3\n3 1 3 2\n3 2 1 3\n2 2 3\n1 1\n2 2 3\n3 3 2 1\n3 2 3 1",
"output": "0"
},
{
"input": "8 4\n0\n0\n4 2 3 1 4\n4 2 1 4 3\n3 4 3 1\n1 2\n2 4 1\n2 4 2",
"output": "2"
},
{
"input": "10 10\n5 7 5 2 8 1\n7 10 6 9 5 8 2 4\n2 2 7\n5 8 6 9 10 1\n2 9 5\n3 6 5 2\n6 5 8 7 9 10 4\n0\n1 1\n2 8 6",
"output": "1"
},
{
"input": "11 42\n4 20 26 9 24\n14 34 7 28 32 12 15 26 4 10 38 21 20 8 11\n4 21 8 36 6\n11 32 1 39 11 21 10 25 17 26 15 4\n2 8 12\n2 21 31\n8 17 10 3 39 32 30 5 15\n20 24 20 38 17 4 7 21 19 32 28 31 22 30 37 10 5 33 2 13 9\n7 38 34 42 27 20 11 6\n3 40 3 39\n14 39 40 4 30 33 8 36 28 14 23 16 7 25 9",
"output": "0"
},
{
"input": "100 100\n1 33\n0\n2 35 4\n2 40 78\n1 69\n0\n1 2\n0\n0\n2 81 34\n1 45\n0\n1 87\n1 50\n1 97\n0\n2 100 59\n0\n0\n0\n1 45\n0\n0\n0\n1 6\n1 54\n0\n0\n0\n4 79 96 52 84\n1 86\n0\n0\n0\n0\n0\n0\n0\n0\n3 90 2 80\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1 2\n0\n1 64\n0\n1 78\n1 82\n0\n0\n0\n0\n0\n1 6\n0\n0\n2 47 57\n1 95\n0\n2 91 79\n0\n1 27\n0\n1 74\n0\n0\n1 14\n0\n0\n2 90 19\n0\n1 10\n0\n0\n0\n0\n0\n1 57\n0\n2 28 50\n0\n0\n0\n1 47\n0\n0\n1 14\n0\n1 84\n1 1\n0\n0",
"output": "87"
},
{
"input": "2 2\n2 1 2\n2 1 2",
"output": "0"
},
{
"input": "2 2\n2 1 2\n1 1",
"output": "0"
},
{
"input": "2 2\n1 2\n1 1",
"output": "1"
},
{
"input": "3 100\n0\n0\n0",
"output": "3"
},
{
"input": "100 100\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "100"
},
{
"input": "3 3\n0\n0\n0",
"output": "3"
}
] | 1,610,215,708
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 186
| 0
|
import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
input = sys.stdin.readline
def dfs(i, grp):
if vis[i]:
return
vis[i] = grp
langs = employees[i]
for lang in langs:
for emp in langs_d[lang]:
dfs(emp, grp)
n, m = map(int, input().split())
employees = []
langs_d = {}
for i in range(n):
langs = list(map(int, input().split()))[1:]
employees.append(langs)
for lang in langs:
if lang in langs_d:
langs_d[lang].append(i)
else:
langs_d[lang] = [i]
vis = [0]*n
grp = 1
for i in range(n):
if vis[i]:
continue
dfs(i, grp)
grp += 1
print(max(vis)-1)
|
Title: Learning Languages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The "BerCorp" company has got *n* employees. These employees can use *m* approved official languages for the formal correspondence. The languages are numbered with integers from 1 to *m*. For each employee we have the list of languages, which he knows. This list could be empty, i. e. an employee may know no official languages. But the employees are willing to learn any number of official languages, as long as the company pays their lessons. A study course in one language for one employee costs 1 berdollar.
Find the minimum sum of money the company needs to spend so as any employee could correspond to any other one (their correspondence can be indirect, i. e. other employees can help out translating).
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of employees and the number of languages.
Then *n* lines follow — each employee's language list. At the beginning of the *i*-th line is integer *k**i* (0<=≤<=*k**i*<=≤<=*m*) — the number of languages the *i*-th employee knows. Next, the *i*-th line contains *k**i* integers — *a**ij* (1<=≤<=*a**ij*<=≤<=*m*) — the identifiers of languages the *i*-th employee knows. It is guaranteed that all the identifiers in one list are distinct. Note that an employee may know zero languages.
The numbers in the lines are separated by single spaces.
Output Specification:
Print a single integer — the minimum amount of money to pay so that in the end every employee could write a letter to every other one (other employees can help out translating).
Demo Input:
['5 5\n1 2\n2 2 3\n2 3 4\n2 4 5\n1 5\n', '8 7\n0\n3 1 2 3\n1 1\n2 5 4\n2 6 7\n1 3\n2 7 4\n1 1\n', '2 2\n1 2\n0\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the second sample the employee 1 can learn language 2, and employee 8 can learn language 4.
In the third sample employee 2 must learn language 2.
|
```python
import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
input = sys.stdin.readline
def dfs(i, grp):
if vis[i]:
return
vis[i] = grp
langs = employees[i]
for lang in langs:
for emp in langs_d[lang]:
dfs(emp, grp)
n, m = map(int, input().split())
employees = []
langs_d = {}
for i in range(n):
langs = list(map(int, input().split()))[1:]
employees.append(langs)
for lang in langs:
if lang in langs_d:
langs_d[lang].append(i)
else:
langs_d[lang] = [i]
vis = [0]*n
grp = 1
for i in range(n):
if vis[i]:
continue
dfs(i, grp)
grp += 1
print(max(vis)-1)
```
| 0
|
|
37
|
A
|
Towers
|
PROGRAMMING
| 1,000
|
[
"sortings"
] |
A. Towers
|
2
|
256
|
Little Vasya has received a young builder’s kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same.
Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible.
|
The first line contains an integer *N* (1<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000.
|
In one line output two numbers — the height of the largest tower and their total number. Remember that Vasya should use all the bars.
|
[
"3\n1 2 3\n",
"4\n6 5 6 7\n"
] |
[
"1 3\n",
"2 3\n"
] |
none
| 500
|
[
{
"input": "3\n1 2 3",
"output": "1 3"
},
{
"input": "4\n6 5 6 7",
"output": "2 3"
},
{
"input": "4\n3 2 1 1",
"output": "2 3"
},
{
"input": "4\n1 2 3 3",
"output": "2 3"
},
{
"input": "3\n20 22 36",
"output": "1 3"
},
{
"input": "25\n47 30 94 41 45 20 96 51 110 129 24 116 9 47 32 82 105 114 116 75 154 151 70 42 162",
"output": "2 23"
},
{
"input": "45\n802 664 442 318 318 827 417 878 711 291 231 414 807 553 657 392 279 202 386 606 465 655 658 112 887 15 25 502 95 44 679 775 942 609 209 871 31 234 4 231 150 110 22 823 193",
"output": "2 43"
},
{
"input": "63\n93 180 116 7 8 179 268 279 136 94 221 153 264 190 278 19 19 63 153 26 158 225 25 49 89 218 111 149 255 225 197 122 243 80 3 224 107 178 202 17 53 92 69 42 228 24 81 205 95 8 265 82 228 156 127 241 172 159 106 60 67 155 111",
"output": "2 57"
},
{
"input": "83\n246 535 994 33 390 927 321 97 223 922 812 705 79 80 977 457 476 636 511 137 6 360 815 319 717 674 368 551 714 628 278 713 761 553 184 414 623 753 428 214 581 115 439 61 677 216 772 592 187 603 658 310 439 559 870 376 109 321 189 337 277 26 70 734 796 907 979 693 570 227 345 650 737 633 701 914 134 403 972 940 371 6 642",
"output": "2 80"
},
{
"input": "105\n246 57 12 204 165 123 246 68 191 310 3 152 386 333 374 257 158 104 333 50 80 290 8 340 101 76 221 316 388 289 138 359 316 26 93 290 105 178 81 195 41 196 218 180 244 292 187 97 315 323 174 119 248 239 92 312 31 2 101 180 307 170 338 314 163 281 217 31 142 238 280 190 190 156 70 74 329 113 151 8 141 313 366 40 253 116 168 124 135 230 294 266 353 389 371 359 195 200 183 237 93 102 315 118 188",
"output": "2 92"
},
{
"input": "123\n112 277 170 247 252 115 157 293 256 143 196 90 12 164 164 42 8 223 167 109 175 232 239 111 148 51 9 254 93 32 268 162 231 91 47 162 161 191 195 145 247 292 129 199 230 94 144 217 18 205 176 20 143 198 121 243 211 262 230 277 195 255 108 290 220 275 158 2 286 200 60 267 278 207 123 150 123 116 131 13 12 226 33 244 30 275 263 45 158 192 254 149 242 176 62 224 221 288 250 160 155 225 132 143 276 293 218 145 197 175 33 129 79 206 210 192 222 262 190 52 274 243 233",
"output": "3 101"
},
{
"input": "5\n5 5 5 5 5",
"output": "5 1"
},
{
"input": "3\n1000 1000 1000",
"output": "3 1"
},
{
"input": "1\n1000",
"output": "1 1"
},
{
"input": "1\n1",
"output": "1 1"
},
{
"input": "5\n1 1000 1000 1000 1000",
"output": "4 2"
},
{
"input": "5\n1000 1000 1000 8 7",
"output": "3 3"
}
] | 1,522,308,707
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 61
| 156
| 8,089,600
|
n = int(input())
a = [int(i) for i in input().split()]
b = dict()
for i in a:
b[i] = b.get(i, 0) + 1
print(max(b.values()), len(b.keys()))
|
Title: Towers
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Vasya has received a young builder’s kit. The kit consists of several wooden bars, the lengths of all of them are known. The bars can be put one on the top of the other if their lengths are the same.
Vasya wants to construct the minimal number of towers from the bars. Help Vasya to use the bars in the best way possible.
Input Specification:
The first line contains an integer *N* (1<=≤<=*N*<=≤<=1000) — the number of bars at Vasya’s disposal. The second line contains *N* space-separated integers *l**i* — the lengths of the bars. All the lengths are natural numbers not exceeding 1000.
Output Specification:
In one line output two numbers — the height of the largest tower and their total number. Remember that Vasya should use all the bars.
Demo Input:
['3\n1 2 3\n', '4\n6 5 6 7\n']
Demo Output:
['1 3\n', '2 3\n']
Note:
none
|
```python
n = int(input())
a = [int(i) for i in input().split()]
b = dict()
for i in a:
b[i] = b.get(i, 0) + 1
print(max(b.values()), len(b.keys()))
```
| 3.945932
|
910
|
A
|
The Way to Home
|
PROGRAMMING
| 800
|
[
"dfs and similar",
"dp",
"greedy",
"implementation"
] | null | null |
A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.
|
The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.
|
If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.
|
[
"8 4\n10010101\n",
"4 2\n1001\n",
"8 4\n11100101\n",
"12 3\n101111100101\n"
] |
[
"2\n",
"-1\n",
"3\n",
"4\n"
] |
In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.
| 500
|
[
{
"input": "8 4\n10010101",
"output": "2"
},
{
"input": "4 2\n1001",
"output": "-1"
},
{
"input": "8 4\n11100101",
"output": "3"
},
{
"input": "12 3\n101111100101",
"output": "4"
},
{
"input": "5 4\n11011",
"output": "1"
},
{
"input": "5 4\n10001",
"output": "1"
},
{
"input": "10 7\n1101111011",
"output": "2"
},
{
"input": "10 9\n1110000101",
"output": "1"
},
{
"input": "10 9\n1100000001",
"output": "1"
},
{
"input": "20 5\n11111111110111101001",
"output": "4"
},
{
"input": "20 11\n11100000111000011011",
"output": "2"
},
{
"input": "20 19\n10100000000000000001",
"output": "1"
},
{
"input": "50 13\n10011010100010100111010000010000000000010100000101",
"output": "5"
},
{
"input": "50 8\n11010100000011001100001100010001110000101100110011",
"output": "8"
},
{
"input": "99 4\n111111111111111111111111111111111111111111111111111111111011111111111111111111111111111111111111111",
"output": "25"
},
{
"input": "99 98\n100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "1"
},
{
"input": "100 5\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "20"
},
{
"input": "100 4\n1111111111111111111111111111111111111111111111111111111111111111111111111111110111111111111111111111",
"output": "25"
},
{
"input": "100 4\n1111111111111111111111111111111111111111111111111111111111111101111111011111111111111111111111111111",
"output": "25"
},
{
"input": "100 3\n1111110111111111111111111111111111111111101111111111111111111111111101111111111111111111111111111111",
"output": "34"
},
{
"input": "100 8\n1111111111101110111111111111111111111111111111111111111111111111111111110011111111111111011111111111",
"output": "13"
},
{
"input": "100 7\n1011111111111111111011101111111011111101111111111101111011110111111111111111111111110111111011111111",
"output": "15"
},
{
"input": "100 9\n1101111110111110101111111111111111011001110111011101011111111111010101111111100011011111111010111111",
"output": "12"
},
{
"input": "100 6\n1011111011111111111011010110011001010101111110111111000111011011111110101101110110101111110000100111",
"output": "18"
},
{
"input": "100 7\n1110001111101001110011111111111101111101101001010001101000101100000101101101011111111101101000100001",
"output": "16"
},
{
"input": "100 11\n1000010100011100011011100000010011001111011110100100001011010100011011111001101101110110010110001101",
"output": "10"
},
{
"input": "100 9\n1001001110000011100100000001000110111101101010101001000101001010011001101100110011011110110011011111",
"output": "13"
},
{
"input": "100 7\n1010100001110101111011000111000001110100100110110001110110011010100001100100001110111100110000101001",
"output": "18"
},
{
"input": "100 10\n1110110000000110000000101110100000111000001011100000100110010001110111001010101000011000000001011011",
"output": "12"
},
{
"input": "100 13\n1000000100000000100011000010010000101010011110000000001000011000110100001000010001100000011001011001",
"output": "9"
},
{
"input": "100 11\n1000000000100000010000100001000100000000010000100100000000100100001000000001011000110001000000000101",
"output": "12"
},
{
"input": "100 22\n1000100000001010000000000000000001000000100000000000000000010000000000001000000000000000000100000001",
"output": "7"
},
{
"input": "100 48\n1000000000000000011000000000000000000000000000000001100000000000000000000000000000000000000000000001",
"output": "3"
},
{
"input": "100 48\n1000000000000000000000100000000000000000000000000000000000000000000001000000000000000000100000000001",
"output": "3"
},
{
"input": "100 75\n1000000100000000000000000000000000000000000000000000000000000000000000000000000001000000000000000001",
"output": "3"
},
{
"input": "100 73\n1000000000000000000000000000000100000000000000000000000000000000000000000000000000000000000000000001",
"output": "2"
},
{
"input": "100 99\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "1"
},
{
"input": "100 1\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "99"
},
{
"input": "100 2\n1111111111111111111111111111111110111111111111111111111111111111111111111111111111111111111111111111",
"output": "50"
},
{
"input": "100 1\n1111111111111111011111111111111111111111111111111111111111111111111101111111111111111111111111111111",
"output": "-1"
},
{
"input": "100 3\n1111111111111111111111111101111111111111111111111011111111111111111111111111111011111111111111111111",
"output": "33"
},
{
"input": "100 1\n1101111111111111111111101111111111111111111111111111111111111011111111101111101111111111111111111111",
"output": "-1"
},
{
"input": "100 6\n1111111111111111111111101111111101011110001111111111111111110111111111111111111111111110010111111111",
"output": "17"
},
{
"input": "100 2\n1111111101111010110111011011110111101111111011111101010101011111011111111111111011111001101111101111",
"output": "-1"
},
{
"input": "100 8\n1100110101111001101001111000111100110100011110111011001011111110000110101000001110111011100111011011",
"output": "14"
},
{
"input": "100 10\n1000111110100000001001101100000010011100010101001100010011111001001101111110110111101111001010001101",
"output": "11"
},
{
"input": "100 7\n1110000011010001110101011010000011110001000000011101110111010110001000011101111010010001101111110001",
"output": "-1"
},
{
"input": "100 3\n1111010001000001011011000011001111000100101000101101000010111101111000010000011110110011001101010111",
"output": "-1"
},
{
"input": "100 9\n1101010101101100010111011000010100001010000101010011001001100010110110000000010000101000000001101101",
"output": "13"
},
{
"input": "100 14\n1010100000000000010101000010001100000000000011100010000001000001011010001110001010100000100001101101",
"output": "9"
},
{
"input": "100 13\n1000000001101001110000010000011001000000000000001010000000100001001010000000000000000100010000000001",
"output": "-1"
},
{
"input": "100 18\n1000000000000000110000000000000000010000000001000001000001000000000100000000000010000000000000000001",
"output": "-1"
},
{
"input": "100 32\n1000000000000000000000000001000000000000000000000101000000000000000000000000000000000001000000000001",
"output": "-1"
},
{
"input": "100 79\n1000000001000000000101000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "2"
},
{
"input": "100 41\n1000000000000000000000000000000000010000000000000000000000000000000000000000100000000000000000000001",
"output": "3"
},
{
"input": "100 82\n1000000000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "2"
},
{
"input": "100 96\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "-1"
},
{
"input": "43 30\n1001000001111111010100100100110101011101101",
"output": "2"
},
{
"input": "7 1\n1111111",
"output": "6"
},
{
"input": "9 3\n101000001",
"output": "-1"
},
{
"input": "10 3\n1100000001",
"output": "-1"
},
{
"input": "8 2\n10000101",
"output": "-1"
},
{
"input": "2 1\n11",
"output": "1"
}
] | 1,596,442,730
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 109
| 20,172,800
|
import sys, os.path #=========================================================================
if(os.path.exists('in.txt')):sys.stdin = open("in.txt","r");sys.stdout = open("out.txt","w")#=
#=============================================================================================
n, x = map(int, input().split())
s = input()
i = 0
ans = 0
no_ans = False
while(i < n):
p = i
ans += 1
i += x
if(i >= n):break
if(s[i] != "1"):
while(s[i] != "1"):
i -= 1
if(i <= p):
no_ans = 1
break
if(no_ans):break
if(no_ans):print(-1)
else: print(ans)
|
Title: The Way to Home
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.
Input Specification:
The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.
Output Specification:
If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.
Demo Input:
['8 4\n10010101\n', '4 2\n1001\n', '8 4\n11100101\n', '12 3\n101111100101\n']
Demo Output:
['2\n', '-1\n', '3\n', '4\n']
Note:
In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.
|
```python
import sys, os.path #=========================================================================
if(os.path.exists('in.txt')):sys.stdin = open("in.txt","r");sys.stdout = open("out.txt","w")#=
#=============================================================================================
n, x = map(int, input().split())
s = input()
i = 0
ans = 0
no_ans = False
while(i < n):
p = i
ans += 1
i += x
if(i >= n):break
if(s[i] != "1"):
while(s[i] != "1"):
i -= 1
if(i <= p):
no_ans = 1
break
if(no_ans):break
if(no_ans):print(-1)
else: print(ans)
```
| 0
|
|
520
|
A
|
Pangram
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
|
Output "YES", if the string is a pangram and "NO" otherwise.
|
[
"12\ntoosmallword\n",
"35\nTheQuickBrownFoxJumpsOverTheLazyDog\n"
] |
[
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "12\ntoosmallword",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLazyDog",
"output": "YES"
},
{
"input": "1\na",
"output": "NO"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "YES"
},
{
"input": "26\nABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "YES"
},
{
"input": "48\nthereisasyetinsufficientdataforameaningfulanswer",
"output": "NO"
},
{
"input": "30\nToBeOrNotToBeThatIsTheQuestion",
"output": "NO"
},
{
"input": "30\njackdawslovemybigsphinxofquarz",
"output": "NO"
},
{
"input": "31\nTHEFIVEBOXINGWIZARDSJUMPQUICKLY",
"output": "YES"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "26\nMGJYIZDKsbhpVeNFlquRTcWoAx",
"output": "YES"
},
{
"input": "26\nfWMOhAPsbIVtyUEZrGNQXDklCJ",
"output": "YES"
},
{
"input": "26\nngPMVFSThiRCwLEuyOAbKxQzDJ",
"output": "YES"
},
{
"input": "25\nnxYTzLFwzNolAumjgcAboyxAj",
"output": "NO"
},
{
"input": "26\npRWdodGdxUESvcScPGbUoooZsC",
"output": "NO"
},
{
"input": "66\nBovdMlDzTaqKllZILFVfxbLGsRnzmtVVTmqiIDTYrossLEPlmsPrkUYtWEsGHVOnFj",
"output": "NO"
},
{
"input": "100\nmKtsiDRJypUieHIkvJaMFkwaKxcCIbBszZQLIyPpCDCjhNpAnYFngLjRpnKWpKWtGnwoSteeZXuFHWQxxxOpFlNeYTwKocsXuCoa",
"output": "YES"
},
{
"input": "26\nEoqxUbsLjPytUHMiFnvcGWZdRK",
"output": "NO"
},
{
"input": "26\nvCUFRKElZOnjmXGylWQaHDiPst",
"output": "NO"
},
{
"input": "26\nWtrPuaHdXLKJMsnvQfgOiJZBEY",
"output": "NO"
},
{
"input": "26\npGiFluRteQwkaVoPszJyNBChxM",
"output": "NO"
},
{
"input": "26\ncTUpqjPmANrdbzSFhlWIoKxgVY",
"output": "NO"
},
{
"input": "26\nLndjgvAEuICHKxPwqYztosrmBN",
"output": "NO"
},
{
"input": "26\nMdaXJrCipnOZLykfqHWEStevbU",
"output": "NO"
},
{
"input": "26\nEjDWsVxfKTqGXRnUMOLYcIzPba",
"output": "NO"
},
{
"input": "26\nxKwzRMpunYaqsdfaBgJcVElTHo",
"output": "NO"
},
{
"input": "26\nnRYUQsTwCPLZkgshfEXvBdoiMa",
"output": "NO"
},
{
"input": "26\nHNCQPfJutyAlDGsvRxZWMEbIdO",
"output": "NO"
},
{
"input": "26\nDaHJIpvKznQcmUyWsTGObXRFDe",
"output": "NO"
},
{
"input": "26\nkqvAnFAiRhzlJbtyuWedXSPcOG",
"output": "NO"
},
{
"input": "26\nhlrvgdwsIOyjcmUZXtAKEqoBpF",
"output": "NO"
},
{
"input": "26\njLfXXiMhBTcAwQVReGnpKzdsYu",
"output": "NO"
},
{
"input": "26\nlNMcVuwItjxRBGAekjhyDsQOzf",
"output": "NO"
},
{
"input": "26\nRkSwbNoYldUGtAZvpFMcxhIJFE",
"output": "NO"
},
{
"input": "26\nDqspXZJTuONYieKgaHLMBwfVSC",
"output": "NO"
},
{
"input": "26\necOyUkqNljFHRVXtIpWabGMLDz",
"output": "NO"
},
{
"input": "26\nEKAvqZhBnPmVCDRlgWJfOusxYI",
"output": "NO"
},
{
"input": "26\naLbgqeYchKdMrsZxIPFvTOWNjA",
"output": "NO"
},
{
"input": "26\nxfpBLsndiqtacOCHGmeWUjRkYz",
"output": "NO"
},
{
"input": "26\nXsbRKtqleZPNIVCdfUhyagAomJ",
"output": "NO"
},
{
"input": "26\nAmVtbrwquEthZcjKPLiyDgSoNF",
"output": "NO"
},
{
"input": "26\nOhvXDcwqAUmSEPRZGnjFLiKtNB",
"output": "NO"
},
{
"input": "26\nEKWJqCFLRmstxVBdYuinpbhaOg",
"output": "NO"
},
{
"input": "26\nmnbvcxxlkjhgfdsapoiuytrewq",
"output": "NO"
},
{
"input": "26\naAbcdefghijklmnopqrstuvwxy",
"output": "NO"
},
{
"input": "30\nABCDEFGHTYRIOPLabcdefghtyriopl",
"output": "NO"
},
{
"input": "25\nabcdefghijklmnopqrstuvwxy",
"output": "NO"
},
{
"input": "26\nabcdefhijklmnopqrstVxyzABC",
"output": "NO"
},
{
"input": "25\nqwertyuiopasdfghjklxcvbnm",
"output": "NO"
},
{
"input": "34\nTheQuickBrownFoxJumpsOverTheLayDog",
"output": "NO"
},
{
"input": "26\nabcdefghigklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxyA",
"output": "NO"
},
{
"input": "50\nqazwsxedcrfvtgbyhnujmikolQWERTYUIOASDFGHJKLZXCVBNM",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLasyDog",
"output": "NO"
},
{
"input": "25\nbcdefghijklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "38\nAbCdEfGhIjKlMnOpQrStVwXyZzzzzzzaaaaaaa",
"output": "NO"
},
{
"input": "26\nabcdefghiklmnopqrstvxyzABC",
"output": "NO"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxzZ",
"output": "NO"
},
{
"input": "50\nabcdefghijklmnopqrstuvwxyABCDEFGHIJKLMNOPQRSTUVWXY",
"output": "NO"
}
] | 1,695,800,040
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 59
| 46
| 5,632,000
|
n = int(input())
s = input().lower()
a = set(list(s))
if n >= 26 and len(a) == 26:
print("YES")
else:
print("NO")
|
Title: Pangram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
Output Specification:
Output "YES", if the string is a pangram and "NO" otherwise.
Demo Input:
['12\ntoosmallword\n', '35\nTheQuickBrownFoxJumpsOverTheLazyDog\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none
|
```python
n = int(input())
s = input().lower()
a = set(list(s))
if n >= 26 and len(a) == 26:
print("YES")
else:
print("NO")
```
| 3
|
|
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"
}
] | 1,689,068,496
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 35
| 124
| 0
|
n, m = map(int, input().split())
rectSize = n*m
dominoCount = rectSize//2
print(dominoCount)
|
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
n, m = map(int, input().split())
rectSize = n*m
dominoCount = rectSize//2
print(dominoCount)
```
| 3.969
|
612
|
A
|
The Text Splitting
|
PROGRAMMING
| 1,300
|
[
"brute force",
"implementation",
"strings"
] | null | null |
You are given the string *s* of length *n* and the numbers *p*,<=*q*. Split the string *s* to pieces of length *p* and *q*.
For example, the string "Hello" for *p*<==<=2, *q*<==<=3 can be split to the two strings "Hel" and "lo" or to the two strings "He" and "llo".
Note it is allowed to split the string *s* to the strings only of length *p* or to the strings only of length *q* (see the second sample test).
|
The first line contains three positive integers *n*,<=*p*,<=*q* (1<=≤<=*p*,<=*q*<=≤<=*n*<=≤<=100).
The second line contains the string *s* consists of lowercase and uppercase latin letters and digits.
|
If it's impossible to split the string *s* to the strings of length *p* and *q* print the only number "-1".
Otherwise in the first line print integer *k* — the number of strings in partition of *s*.
Each of the next *k* lines should contain the strings in partition. Each string should be of the length *p* or *q*. The string should be in order of their appearing in string *s* — from left to right.
If there are several solutions print any of them.
|
[
"5 2 3\nHello\n",
"10 9 5\nCodeforces\n",
"6 4 5\nPrivet\n",
"8 1 1\nabacabac\n"
] |
[
"2\nHe\nllo\n",
"2\nCodef\norces\n",
"-1\n",
"8\na\nb\na\nc\na\nb\na\nc\n"
] |
none
| 0
|
[
{
"input": "5 2 3\nHello",
"output": "2\nHe\nllo"
},
{
"input": "10 9 5\nCodeforces",
"output": "2\nCodef\norces"
},
{
"input": "6 4 5\nPrivet",
"output": "-1"
},
{
"input": "8 1 1\nabacabac",
"output": "8\na\nb\na\nc\na\nb\na\nc"
},
{
"input": "1 1 1\n1",
"output": "1\n1"
},
{
"input": "10 8 1\nuTl9w4lcdo",
"output": "10\nu\nT\nl\n9\nw\n4\nl\nc\nd\no"
},
{
"input": "20 6 4\nfmFRpk2NrzSvnQC9gB61",
"output": "5\nfmFR\npk2N\nrzSv\nnQC9\ngB61"
},
{
"input": "30 23 6\nWXDjl9kitaDTY673R5xyTlbL9gqeQ6",
"output": "5\nWXDjl9\nkitaDT\nY673R5\nxyTlbL\n9gqeQ6"
},
{
"input": "40 14 3\nSOHBIkWEv7ScrkHgMtFFxP9G7JQLYXFoH1sJDAde",
"output": "6\nSOHBIkWEv7Scrk\nHgMtFFxP9G7JQL\nYXF\noH1\nsJD\nAde"
},
{
"input": "50 16 3\nXCgVJUu4aMQ7HMxZjNxe3XARNiahK303g9y7NV8oN6tWdyXrlu",
"output": "8\nXCgVJUu4aMQ7HMxZ\njNxe3XARNiahK303\ng9y\n7NV\n8oN\n6tW\ndyX\nrlu"
},
{
"input": "60 52 8\nhae0PYwXcW2ziQCOSci5VaElHLZCZI81ULSHgpyG3fuZaP0fHjN4hCKogONj",
"output": "2\nhae0PYwXcW2ziQCOSci5VaElHLZCZI81ULSHgpyG3fuZaP0fHjN4\nhCKogONj"
},
{
"input": "70 50 5\n1BH1ECq7hjzooQOZdbiYHTAgATcP5mxI7kLI9rqA9AriWc9kE5KoLa1zmuTDFsd2ClAPPY",
"output": "14\n1BH1E\nCq7hj\nzooQO\nZdbiY\nHTAgA\nTcP5m\nxI7kL\nI9rqA\n9AriW\nc9kE5\nKoLa1\nzmuTD\nFsd2C\nlAPPY"
},
{
"input": "80 51 8\no2mpu1FCofuiLQb472qczCNHfVzz5TfJtVMrzgN3ff7FwlAY0fQ0ROhWmIX2bggodORNA76bHMjA5yyc",
"output": "10\no2mpu1FC\nofuiLQb4\n72qczCNH\nfVzz5TfJ\ntVMrzgN3\nff7FwlAY\n0fQ0ROhW\nmIX2bggo\ndORNA76b\nHMjA5yyc"
},
{
"input": "90 12 7\nclcImtsw176FFOA6OHGFxtEfEyhFh5bH4iktV0Y8onIcn0soTwiiHUFRWC6Ow36tT5bsQjgrVSTcB8fAVoe7dJIWkE",
"output": "10\nclcImtsw176F\nFOA6OHGFxtEf\nEyhFh5bH4ikt\nV0Y8onIcn0so\nTwiiHUF\nRWC6Ow3\n6tT5bsQ\njgrVSTc\nB8fAVoe\n7dJIWkE"
},
{
"input": "100 25 5\n2SRB9mRpXMRND5zQjeRxc4GhUBlEQSmLgnUtB9xTKoC5QM9uptc8dKwB88XRJy02r7edEtN2C6D60EjzK1EHPJcWNj6fbF8kECeB",
"output": "20\n2SRB9\nmRpXM\nRND5z\nQjeRx\nc4GhU\nBlEQS\nmLgnU\ntB9xT\nKoC5Q\nM9upt\nc8dKw\nB88XR\nJy02r\n7edEt\nN2C6D\n60Ejz\nK1EHP\nJcWNj\n6fbF8\nkECeB"
},
{
"input": "100 97 74\nxL8yd8lENYnXZs28xleyci4SxqsjZqkYzkEbQXfLQ4l4gKf9QQ9xjBjeZ0f9xQySf5psDUDkJEtPLsa62n4CLc6lF6E2yEqvt4EJ",
"output": "-1"
},
{
"input": "51 25 11\nwpk5wqrB6d3qE1slUrzJwMFafnnOu8aESlvTEb7Pp42FDG2iGQn",
"output": "-1"
},
{
"input": "70 13 37\nfzL91QIJvNoZRP4A9aNRT2GTksd8jEb1713pnWFaCGKHQ1oYvlTHXIl95lqyZRKJ1UPYvT",
"output": "-1"
},
{
"input": "10 3 1\nXQ2vXLPShy",
"output": "10\nX\nQ\n2\nv\nX\nL\nP\nS\nh\ny"
},
{
"input": "4 2 3\naaaa",
"output": "2\naa\naa"
},
{
"input": "100 1 1\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "100\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb\nb"
},
{
"input": "99 2 4\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "11 2 3\nhavanahavan",
"output": "4\nha\nvan\naha\nvan"
},
{
"input": "100 2 2\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "50\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa\naa"
},
{
"input": "17 3 5\ngopstopmipodoshli",
"output": "5\ngop\nsto\npmi\npod\noshli"
},
{
"input": "5 4 3\nfoyku",
"output": "-1"
},
{
"input": "99 2 2\n123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789",
"output": "-1"
},
{
"input": "99 2 2\nrecursionishellrecursionishellrecursionishellrecursionishellrecursionishellrecursionishelldontuseit",
"output": "-1"
},
{
"input": "11 2 3\nqibwnnvqqgo",
"output": "4\nqi\nbwn\nnvq\nqgo"
},
{
"input": "4 4 3\nhhhh",
"output": "1\nhhhh"
},
{
"input": "99 2 2\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "99 2 5\nhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh",
"output": "21\nhh\nhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh\nhhhhh"
},
{
"input": "10 5 9\nCodeforces",
"output": "2\nCodef\norces"
},
{
"input": "10 5 9\naaaaaaaaaa",
"output": "2\naaaaa\naaaaa"
},
{
"input": "11 3 2\nmlmqpohwtsf",
"output": "5\nmlm\nqp\noh\nwt\nsf"
},
{
"input": "3 3 2\nzyx",
"output": "1\nzyx"
},
{
"input": "100 3 3\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "4 2 3\nzyxw",
"output": "2\nzy\nxw"
},
{
"input": "3 2 3\nejt",
"output": "1\nejt"
},
{
"input": "5 2 4\nzyxwv",
"output": "-1"
},
{
"input": "100 1 1\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "100\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na\na"
},
{
"input": "100 5 4\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "25\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa\naaaa"
},
{
"input": "3 2 2\nzyx",
"output": "-1"
},
{
"input": "99 2 2\nhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh",
"output": "-1"
},
{
"input": "26 8 9\nabcabcabcabcabcabcabcabcab",
"output": "3\nabcabcab\ncabcabcab\ncabcabcab"
},
{
"input": "6 3 5\naaaaaa",
"output": "2\naaa\naaa"
},
{
"input": "3 2 3\nzyx",
"output": "1\nzyx"
},
{
"input": "5 5 2\naaaaa",
"output": "1\naaaaa"
},
{
"input": "4 3 2\nzyxw",
"output": "2\nzy\nxw"
},
{
"input": "5 4 3\nzyxwv",
"output": "-1"
},
{
"input": "95 3 29\nabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab",
"output": "23\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabc\nabcabcabcabcabcabcabcabcabcab"
},
{
"input": "3 2 2\naaa",
"output": "-1"
},
{
"input": "91 62 3\nfjzhkfwzoabaauvbkuzaahkozofaophaafhfpuhobufawkzbavaazwavwppfwapkapaofbfjwaavajojgjguahphofj",
"output": "-1"
},
{
"input": "99 2 2\nabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc",
"output": "-1"
},
{
"input": "56 13 5\nabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcab",
"output": "8\nabcabcabcabca\nbcabcabcabcab\ncabca\nbcabc\nabcab\ncabca\nbcabc\nabcab"
},
{
"input": "79 7 31\nabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabca",
"output": "-1"
},
{
"input": "92 79 6\nxlvplpckwnhmctoethhslkcyashqtsoeltriddglfwtgkfvkvgytygbcyohrvcxvosdioqvackxiuifmkgdngvbbudcb",
"output": "-1"
},
{
"input": "48 16 13\nibhfinipihcbsqnvtgsbkobepmwymlyfmlfgblvhlfhyojsy",
"output": "3\nibhfinipihcbsqnv\ntgsbkobepmwymlyf\nmlfgblvhlfhyojsy"
},
{
"input": "16 3 7\naaaaaaaaaaaaaaaa",
"output": "4\naaa\naaa\naaa\naaaaaaa"
},
{
"input": "11 10 3\naaaaaaaaaaa",
"output": "-1"
},
{
"input": "11 8 8\naaaaaaaaaaa",
"output": "-1"
},
{
"input": "11 7 3\naaaaaaaaaaa",
"output": "-1"
},
{
"input": "41 3 4\nabcabcabcabcabcabcabcabcabcabcabcabcabcab",
"output": "11\nabc\nabc\nabc\nabca\nbcab\ncabc\nabca\nbcab\ncabc\nabca\nbcab"
},
{
"input": "11 3 2\naaaaaaaaaaa",
"output": "5\naaa\naa\naa\naa\naa"
},
{
"input": "14 9 4\nabcdefghijklmn",
"output": "-1"
},
{
"input": "9 9 5\n123456789",
"output": "1\n123456789"
},
{
"input": "92 10 41\nmeotryyfneonmnrvfnhqlehlxtvpracifpadcofecvbikoitrlgeftiqofpvacgocrdiquhatlqosqvtduenaqkwrnnw",
"output": "3\nmeotryyfne\nonmnrvfnhqlehlxtvpracifpadcofecvbikoitrlg\neftiqofpvacgocrdiquhatlqosqvtduenaqkwrnnw"
},
{
"input": "17 16 3\ndxyhgtsxtuyljmclj",
"output": "-1"
},
{
"input": "82 13 30\nfmtwumakkejtolxqxrnydhqoufwtdwldfxcfjrndauqcarhbwmdwxsxfbqjsfspuxobywhcrvlndsdmkqd",
"output": "5\nfmtwumakkejto\nlxqxrnydhqouf\nwtdwldfxcfjrn\ndauqcarhbwmdw\nxsxfbqjsfspuxobywhcrvlndsdmkqd"
},
{
"input": "95 3 3\nihnfqcswushyoirjxlxxnwqtwtaowounkumxukwpacxwatimhhhoggqwkkspcplvyndfukbxickcixidgxkjtnpkoeiwlor",
"output": "-1"
},
{
"input": "7 5 3\nzyxwvut",
"output": "-1"
},
{
"input": "17 16 4\nctvfhkiakagcilrdw",
"output": "-1"
}
] | 1,538,019,125
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 93
| 0
|
import sys
n, p, q = [int(x) for x in input().split()]
possible, possible2, possiblePQ = True, True, True
subs = list()
s = input()
sArr = list(s)
if n % p == 0:
possible = False
possible2 = False
pCount = 0
aux = ""
for s1 in sArr:
aux += s1
pCount += 1
if pCount == p:
subs.append(aux)
pCount = 0
aux = ""
if n % q == 0 and possible:
possiblePQ = False
qCount = 0
aux = ""
for s1 in sArr:
aux += s1
qCount += 1
if qCount == q:
subs.append(aux)
qCount = 0
aux = ""
if possible2 and possiblePQ:
letter = p
sliced = False
while s != "":
if sliced is False:
if len(s) >= letter:
y = s[0:letter]
subs.append(y)
s = s[letter:]
else:
letter = q
sliced = True
else:
if len(subs) != 0:
if len(s) % q == 0:
y = s[0:letter]
subs.append(y)
s = s[letter:]
else:
s = subs.pop(len(subs) - 1) + s
else:
print(-1)
sys.exit(1)
if len(subs) != 0:
print(len(subs))
for i in (range(0, len(subs), 1)):
print(subs[i])
sys.exit(1)
print(len(subs))
for i in range(0, len(subs), 1):
print(subs[i])
|
Title: The Text Splitting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given the string *s* of length *n* and the numbers *p*,<=*q*. Split the string *s* to pieces of length *p* and *q*.
For example, the string "Hello" for *p*<==<=2, *q*<==<=3 can be split to the two strings "Hel" and "lo" or to the two strings "He" and "llo".
Note it is allowed to split the string *s* to the strings only of length *p* or to the strings only of length *q* (see the second sample test).
Input Specification:
The first line contains three positive integers *n*,<=*p*,<=*q* (1<=≤<=*p*,<=*q*<=≤<=*n*<=≤<=100).
The second line contains the string *s* consists of lowercase and uppercase latin letters and digits.
Output Specification:
If it's impossible to split the string *s* to the strings of length *p* and *q* print the only number "-1".
Otherwise in the first line print integer *k* — the number of strings in partition of *s*.
Each of the next *k* lines should contain the strings in partition. Each string should be of the length *p* or *q*. The string should be in order of their appearing in string *s* — from left to right.
If there are several solutions print any of them.
Demo Input:
['5 2 3\nHello\n', '10 9 5\nCodeforces\n', '6 4 5\nPrivet\n', '8 1 1\nabacabac\n']
Demo Output:
['2\nHe\nllo\n', '2\nCodef\norces\n', '-1\n', '8\na\nb\na\nc\na\nb\na\nc\n']
Note:
none
|
```python
import sys
n, p, q = [int(x) for x in input().split()]
possible, possible2, possiblePQ = True, True, True
subs = list()
s = input()
sArr = list(s)
if n % p == 0:
possible = False
possible2 = False
pCount = 0
aux = ""
for s1 in sArr:
aux += s1
pCount += 1
if pCount == p:
subs.append(aux)
pCount = 0
aux = ""
if n % q == 0 and possible:
possiblePQ = False
qCount = 0
aux = ""
for s1 in sArr:
aux += s1
qCount += 1
if qCount == q:
subs.append(aux)
qCount = 0
aux = ""
if possible2 and possiblePQ:
letter = p
sliced = False
while s != "":
if sliced is False:
if len(s) >= letter:
y = s[0:letter]
subs.append(y)
s = s[letter:]
else:
letter = q
sliced = True
else:
if len(subs) != 0:
if len(s) % q == 0:
y = s[0:letter]
subs.append(y)
s = s[letter:]
else:
s = subs.pop(len(subs) - 1) + s
else:
print(-1)
sys.exit(1)
if len(subs) != 0:
print(len(subs))
for i in (range(0, len(subs), 1)):
print(subs[i])
sys.exit(1)
print(len(subs))
for i in range(0, len(subs), 1):
print(subs[i])
```
| -1
|
|
996
|
A
|
Hit the Lottery
|
PROGRAMMING
| 800
|
[
"dp",
"greedy"
] | null | null |
Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
|
The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).
|
Output the minimum number of bills that Allen could receive.
|
[
"125\n",
"43\n",
"1000000000\n"
] |
[
"3\n",
"5\n",
"10000000\n"
] |
In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills.
| 500
|
[
{
"input": "125",
"output": "3"
},
{
"input": "43",
"output": "5"
},
{
"input": "1000000000",
"output": "10000000"
},
{
"input": "4",
"output": "4"
},
{
"input": "5",
"output": "1"
},
{
"input": "1",
"output": "1"
},
{
"input": "74",
"output": "8"
},
{
"input": "31",
"output": "3"
},
{
"input": "59",
"output": "8"
},
{
"input": "79",
"output": "9"
},
{
"input": "7",
"output": "3"
},
{
"input": "55",
"output": "4"
},
{
"input": "40",
"output": "2"
},
{
"input": "719",
"output": "13"
},
{
"input": "847",
"output": "13"
},
{
"input": "225",
"output": "4"
},
{
"input": "4704",
"output": "51"
},
{
"input": "1132",
"output": "15"
},
{
"input": "7811",
"output": "80"
},
{
"input": "7981",
"output": "84"
},
{
"input": "82655",
"output": "830"
},
{
"input": "6364",
"output": "70"
},
{
"input": "74611",
"output": "748"
},
{
"input": "45391",
"output": "459"
},
{
"input": "620448",
"output": "6210"
},
{
"input": "265145",
"output": "2654"
},
{
"input": "671704",
"output": "6721"
},
{
"input": "365173",
"output": "3658"
},
{
"input": "7130872",
"output": "71314"
},
{
"input": "9628747",
"output": "96292"
},
{
"input": "8898399",
"output": "88993"
},
{
"input": "9497953",
"output": "94985"
},
{
"input": "19070947",
"output": "190714"
},
{
"input": "20185520",
"output": "201856"
},
{
"input": "91402248",
"output": "914028"
},
{
"input": "27035533",
"output": "270360"
},
{
"input": "22717528",
"output": "227180"
},
{
"input": "24403439",
"output": "244041"
},
{
"input": "163565555",
"output": "1635659"
},
{
"input": "152541908",
"output": "1525423"
},
{
"input": "668439837",
"output": "6684403"
},
{
"input": "20",
"output": "1"
},
{
"input": "999999999",
"output": "10000009"
},
{
"input": "10",
"output": "1"
},
{
"input": "7",
"output": "3"
},
{
"input": "6",
"output": "2"
},
{
"input": "99999999",
"output": "1000009"
},
{
"input": "1000001",
"output": "10001"
},
{
"input": "2521",
"output": "27"
}
] | 1,684,312,202
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 46
| 0
|
x = int(input())
list1 = [100, 20, 10, 5, 1]
c = 0
for item in list1:
c += x // item
x %= item
print(c)
|
Title: Hit the Lottery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input Specification:
The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).
Output Specification:
Output the minimum number of bills that Allen could receive.
Demo Input:
['125\n', '43\n', '1000000000\n']
Demo Output:
['3\n', '5\n', '10000000\n']
Note:
In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills.
|
```python
x = int(input())
list1 = [100, 20, 10, 5, 1]
c = 0
for item in list1:
c += x // item
x %= item
print(c)
```
| 3
|
|
26
|
B
|
Regular Bracket Sequence
|
PROGRAMMING
| 1,400
|
[
"greedy"
] |
B. Regular Bracket Sequence
|
5
|
256
|
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
|
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
|
Output the maximum possible length of a regular bracket sequence.
|
[
"(()))(\n",
"((()())\n"
] |
[
"4\n",
"6\n"
] |
none
| 1,000
|
[
{
"input": "(()))(",
"output": "4"
},
{
"input": "((()())",
"output": "6"
},
{
"input": "(",
"output": "0"
},
{
"input": ")",
"output": "0"
},
{
"input": ")(()(",
"output": "2"
},
{
"input": "))))))(",
"output": "0"
},
{
"input": "()()(()(((",
"output": "6"
},
{
"input": "()))(()((((()(())",
"output": "10"
},
{
"input": "())))((()())())))))())",
"output": "14"
},
{
"input": ")))((((())(()((()((((()()())((",
"output": "16"
},
{
"input": "))()()((()()))())()(((((((())((((((((())()()((())(",
"output": "32"
},
{
"input": "))())))))))())))))()()))()()))))())))))()))))))))))))(()))())(()))))(()))))())))((((()()))))()))()))",
"output": "48"
}
] | 1,591,830,274
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 38
| 1,184
| 6,246,400
|
from math import *
s=list(input())
n=len(s)
c,k=0,0
ans=0
for i in range(n):
if s[i]=='(':
c=c+1
else:
k=k+1
if (c-k)<0:
if s[i]==')':
k=k-1
print(2*k)
|
Title: Regular Bracket Sequence
Time Limit: 5 seconds
Memory Limit: 256 megabytes
Problem Description:
A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters «+» and «1» into this sequence. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.
One day Johnny got bracket sequence. He decided to remove some of the brackets from it in order to obtain a regular bracket sequence. What is the maximum length of a regular bracket sequence which can be obtained?
Input Specification:
Input consists of a single line with non-empty string of «(» and «)» characters. Its length does not exceed 106.
Output Specification:
Output the maximum possible length of a regular bracket sequence.
Demo Input:
['(()))(\n', '((()())\n']
Demo Output:
['4\n', '6\n']
Note:
none
|
```python
from math import *
s=list(input())
n=len(s)
c,k=0,0
ans=0
for i in range(n):
if s[i]=='(':
c=c+1
else:
k=k+1
if (c-k)<0:
if s[i]==')':
k=k-1
print(2*k)
```
| 3.869965
|
820
|
A
|
Mister B and Book Reading
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages.
At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day.
Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time.
Help Mister B to calculate how many days he needed to finish the book.
|
First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading.
|
Print one integer — the number of days Mister B needed to finish the book.
|
[
"5 5 10 5 4\n",
"12 4 12 4 1\n",
"15 1 100 0 0\n"
] |
[
"1\n",
"3\n",
"15\n"
] |
In the first sample test the book contains 5 pages, so Mister B read it right at the first day.
In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book.
In third sample test every day Mister B read 1 page of the book, so he finished in 15 days.
| 500
|
[
{
"input": "5 5 10 5 4",
"output": "1"
},
{
"input": "12 4 12 4 1",
"output": "3"
},
{
"input": "15 1 100 0 0",
"output": "15"
},
{
"input": "1 1 1 0 0",
"output": "1"
},
{
"input": "1000 999 1000 1000 998",
"output": "2"
},
{
"input": "1000 2 2 5 1",
"output": "999"
},
{
"input": "1000 1 1 1000 0",
"output": "1000"
},
{
"input": "737 41 74 12 11",
"output": "13"
},
{
"input": "1000 1000 1000 0 999",
"output": "1"
},
{
"input": "765 12 105 5 7",
"output": "17"
},
{
"input": "15 2 2 1000 0",
"output": "8"
},
{
"input": "1000 1 1000 1000 0",
"output": "2"
},
{
"input": "20 3 7 1 2",
"output": "6"
},
{
"input": "1000 500 500 1000 499",
"output": "501"
},
{
"input": "1 1000 1000 1000 0",
"output": "1"
},
{
"input": "1000 2 1000 56 0",
"output": "7"
},
{
"input": "1000 2 1000 802 0",
"output": "3"
},
{
"input": "16 1 8 2 0",
"output": "4"
},
{
"input": "20 6 10 2 2",
"output": "3"
},
{
"input": "8 2 12 4 1",
"output": "3"
},
{
"input": "8 6 13 2 5",
"output": "2"
},
{
"input": "70 4 20 87 0",
"output": "5"
},
{
"input": "97 8 13 234 5",
"output": "13"
},
{
"input": "16 4 23 8 3",
"output": "3"
},
{
"input": "65 7 22 7 4",
"output": "5"
},
{
"input": "93 10 18 11 7",
"output": "9"
},
{
"input": "86 13 19 15 9",
"output": "9"
},
{
"input": "333 17 50 10 16",
"output": "12"
},
{
"input": "881 16 55 10 12",
"output": "23"
},
{
"input": "528 11 84 3 9",
"output": "19"
},
{
"input": "896 2 184 8 1",
"output": "16"
},
{
"input": "236 10 930 9 8",
"output": "8"
},
{
"input": "784 1 550 14 0",
"output": "12"
},
{
"input": "506 1 10 4 0",
"output": "53"
},
{
"input": "460 1 3 2 0",
"output": "154"
},
{
"input": "701 1 3 1 0",
"output": "235"
},
{
"input": "100 49 50 1000 2",
"output": "3"
},
{
"input": "100 1 100 100 0",
"output": "2"
},
{
"input": "12 1 4 2 0",
"output": "4"
},
{
"input": "22 10 12 0 0",
"output": "3"
},
{
"input": "20 10 15 1 4",
"output": "3"
},
{
"input": "1000 5 10 1 4",
"output": "169"
},
{
"input": "1000 1 1000 1 0",
"output": "45"
},
{
"input": "4 1 2 2 0",
"output": "3"
},
{
"input": "1 5 5 1 1",
"output": "1"
},
{
"input": "19 10 11 0 2",
"output": "3"
},
{
"input": "1 2 3 0 0",
"output": "1"
},
{
"input": "10 1 4 10 0",
"output": "4"
},
{
"input": "20 3 100 1 1",
"output": "5"
},
{
"input": "1000 5 9 5 0",
"output": "112"
},
{
"input": "1 11 12 0 10",
"output": "1"
},
{
"input": "1 1 1 1 0",
"output": "1"
},
{
"input": "1000 1 20 1 0",
"output": "60"
},
{
"input": "9 1 4 2 0",
"output": "4"
},
{
"input": "129 2 3 4 0",
"output": "44"
},
{
"input": "4 2 2 0 1",
"output": "3"
},
{
"input": "1000 1 10 100 0",
"output": "101"
},
{
"input": "100 1 100 1 0",
"output": "14"
},
{
"input": "8 3 4 2 0",
"output": "3"
},
{
"input": "20 1 6 4 0",
"output": "5"
},
{
"input": "8 2 4 2 0",
"output": "3"
},
{
"input": "11 5 6 7 2",
"output": "3"
},
{
"input": "100 120 130 120 0",
"output": "1"
},
{
"input": "7 1 4 1 0",
"output": "4"
},
{
"input": "5 3 10 0 2",
"output": "3"
},
{
"input": "5 2 2 0 0",
"output": "3"
},
{
"input": "1000 10 1000 10 0",
"output": "14"
},
{
"input": "25 3 50 4 2",
"output": "4"
},
{
"input": "9 10 10 10 9",
"output": "1"
},
{
"input": "17 10 12 6 5",
"output": "2"
},
{
"input": "15 5 10 3 0",
"output": "3"
},
{
"input": "8 3 5 1 0",
"output": "3"
},
{
"input": "19 1 12 5 0",
"output": "4"
},
{
"input": "1000 10 1000 1 0",
"output": "37"
},
{
"input": "100 1 2 1000 0",
"output": "51"
},
{
"input": "20 10 11 1000 9",
"output": "6"
},
{
"input": "16 2 100 1 1",
"output": "5"
},
{
"input": "18 10 13 2 5",
"output": "3"
},
{
"input": "12 3 5 3 1",
"output": "4"
},
{
"input": "17 3 11 2 0",
"output": "4"
},
{
"input": "4 2 100 1 1",
"output": "2"
},
{
"input": "7 4 5 2 3",
"output": "3"
},
{
"input": "100 1 2 2 0",
"output": "51"
},
{
"input": "50 4 5 5 0",
"output": "11"
},
{
"input": "1 2 2 0 1",
"output": "1"
},
{
"input": "1000 2 3 10 1",
"output": "500"
},
{
"input": "500 10 500 1000 0",
"output": "2"
},
{
"input": "1000 4 12 1 0",
"output": "87"
},
{
"input": "18 10 13 1 5",
"output": "3"
},
{
"input": "7 3 6 2 2",
"output": "3"
},
{
"input": "15 5 100 1 2",
"output": "4"
},
{
"input": "100 1 10 1 0",
"output": "15"
},
{
"input": "8 2 7 5 1",
"output": "2"
},
{
"input": "11 2 4 1 1",
"output": "5"
},
{
"input": "1000 500 900 100 300",
"output": "3"
},
{
"input": "7 1 2 5 0",
"output": "4"
},
{
"input": "7 3 5 3 2",
"output": "3"
},
{
"input": "7 3 10 2 1",
"output": "2"
},
{
"input": "1000 501 510 1 499",
"output": "50"
},
{
"input": "1000 1 1000 2 0",
"output": "32"
},
{
"input": "1 5 5 0 0",
"output": "1"
},
{
"input": "18 10 15 1 5",
"output": "3"
},
{
"input": "100 4 1000 1 2",
"output": "13"
},
{
"input": "20 2 40 1 1",
"output": "6"
},
{
"input": "1 11 1000 100 1",
"output": "1"
},
{
"input": "6 4 4 1 2",
"output": "2"
},
{
"input": "8 3 5 3 1",
"output": "3"
},
{
"input": "10 5 7 1 2",
"output": "3"
},
{
"input": "400 100 198 1 99",
"output": "25"
},
{
"input": "3 1 2 5 0",
"output": "2"
}
] | 1,498,840,083
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 110
| 77
| 5,529,600
|
c, v0, v1, a, l = map(int, input().split())
if (c <= v0):
print(1)
else:
mp = v0
cur = v0
cnt = 1
while (mp < c):
cur = min(cur + a, v1)
mp += cur - l
cnt += 1
print(cnt)
|
Title: Mister B and Book Reading
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages.
At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day.
Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time.
Help Mister B to calculate how many days he needed to finish the book.
Input Specification:
First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading.
Output Specification:
Print one integer — the number of days Mister B needed to finish the book.
Demo Input:
['5 5 10 5 4\n', '12 4 12 4 1\n', '15 1 100 0 0\n']
Demo Output:
['1\n', '3\n', '15\n']
Note:
In the first sample test the book contains 5 pages, so Mister B read it right at the first day.
In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book.
In third sample test every day Mister B read 1 page of the book, so he finished in 15 days.
|
```python
c, v0, v1, a, l = map(int, input().split())
if (c <= v0):
print(1)
else:
mp = v0
cur = v0
cnt = 1
while (mp < c):
cur = min(cur + a, v1)
mp += cur - l
cnt += 1
print(cnt)
```
| 3
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,581,100,006
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 216
| 0
|
import sys
n=int(input())# your code goes here
c=[]
s=0
for i in range(n):
a,b,d=map(int,input().split())
d1=a+b+d
c.append(d1)
s=s+c[i]
if(a==0 or b==0 or c==0):
print("NO")
sys.exit()
if(s==0):
print("YES")
else:
print("NO")
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
import sys
n=int(input())# your code goes here
c=[]
s=0
for i in range(n):
a,b,d=map(int,input().split())
d1=a+b+d
c.append(d1)
s=s+c[i]
if(a==0 or b==0 or c==0):
print("NO")
sys.exit()
if(s==0):
print("YES")
else:
print("NO")
```
| 0
|
735
|
B
|
Urbanization
|
PROGRAMMING
| 1,100
|
[
"greedy",
"number theory",
"sortings"
] | null | null |
Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are *n* people who plan to move to the cities. The wealth of the *i* of them is equal to *a**i*. Authorities plan to build two cities, first for *n*1 people and second for *n*2 people. Of course, each of *n* candidates can settle in only one of the cities. Thus, first some subset of candidates of size *n*1 settle in the first city and then some subset of size *n*2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home.
To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth *a**i* among all its residents divided by the number of them (*n*1 or *n*2 depending on the city). The division should be done in real numbers without any rounding.
Please, help authorities find the optimal way to pick residents for two cities.
|
The first line of the input contains three integers *n*, *n*1 and *n*2 (1<=≤<=*n*,<=*n*1,<=*n*2<=≤<=100<=000, *n*1<=+<=*n*2<=≤<=*n*) — the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), the *i*-th of them is equal to the wealth of the *i*-th candidate.
|
Print one real value — the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
|
[
"2 1 1\n1 5\n",
"4 2 1\n1 4 2 3\n"
] |
[
"6.00000000\n",
"6.50000000\n"
] |
In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second.
In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (*a*<sub class="lower-index">3</sub> + *a*<sub class="lower-index">4</sub>) / 2 + *a*<sub class="lower-index">2</sub> = (3 + 2) / 2 + 4 = 6.5
| 1,000
|
[
{
"input": "2 1 1\n1 5",
"output": "6.00000000"
},
{
"input": "4 2 1\n1 4 2 3",
"output": "6.50000000"
},
{
"input": "3 1 2\n1 2 3",
"output": "4.50000000"
},
{
"input": "10 4 6\n3 5 7 9 12 25 67 69 83 96",
"output": "88.91666667"
},
{
"input": "19 7 12\n1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 100000 100000",
"output": "47052.10714286"
},
{
"input": "100 9 6\n109 711 40 95 935 48 228 253 308 726 816 534 252 8 966 363 162 508 84 83 807 506 748 178 45 30 106 108 764 698 825 198 336 353 158 790 64 262 403 334 577 571 742 541 946 602 279 621 910 776 421 886 29 133 114 394 762 965 339 263 750 530 49 80 124 31 322 292 27 590 960 278 111 932 849 491 561 744 469 511 106 271 156 160 836 363 149 473 457 543 976 809 490 29 85 626 265 88 995 946",
"output": "1849.66666667"
},
{
"input": "69 6 63\n53475 22876 79144 6335 33763 79104 65441 45527 65847 94406 74670 43529 75330 19403 67629 56187 57949 23071 64910 54409 55348 18056 855 24961 50565 6622 26467 33989 22660 79469 41246 13965 79706 14422 16075 93378 81313 48173 13470 97348 2346 27452 59427 29925 29847 73823 32021 10988 24609 98855 90919 45939 17203 8439 43007 40138 55693 30314 71734 33458 66850 4011 20089 20546 92090 50842 78859 62756 40177",
"output": "135712.88888889"
},
{
"input": "69 6 9\n2612 17461 69001 33130 10662 85485 88195 45974 16712 81365 67119 87797 15559 20197 74716 92979 97268 49466 68603 48351 99905 35606 54242 98603 68232 54398 82637 49647 38979 46171 54680 23334 15892 92186 69670 29711 67999 2220 32317 717 70667 68262 86760 55720 97158 61122 7251 138 21022 27197 12691 59331 13576 66999 38332 13574 83484 66646 17704 33065 98583 80259 64631 16745 69431 40747 82089 82788 32739",
"output": "183129.44444444"
}
] | 1,582,045,923
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 155
| 7,475,200
|
n,n1,n2=map(int,input().split())
a=list(map(int,input().split()))
a.sort(reverse=True)
b=[]
c=[]
if n1<n2:
p1,q1=0,n1
p2,q2=n1,(n1+n2)
else:
p1,q1=n2,(n1+n2)
p2,q2=0,n2
for i in range(p1,q1):
b.append(a[i])
for i in range(p2,q2):
c.append(a[i])
x=sum(b)/n1+sum(c)/n2
print('%.8f'%x)
|
Title: Urbanization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are *n* people who plan to move to the cities. The wealth of the *i* of them is equal to *a**i*. Authorities plan to build two cities, first for *n*1 people and second for *n*2 people. Of course, each of *n* candidates can settle in only one of the cities. Thus, first some subset of candidates of size *n*1 settle in the first city and then some subset of size *n*2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home.
To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth *a**i* among all its residents divided by the number of them (*n*1 or *n*2 depending on the city). The division should be done in real numbers without any rounding.
Please, help authorities find the optimal way to pick residents for two cities.
Input Specification:
The first line of the input contains three integers *n*, *n*1 and *n*2 (1<=≤<=*n*,<=*n*1,<=*n*2<=≤<=100<=000, *n*1<=+<=*n*2<=≤<=*n*) — the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), the *i*-th of them is equal to the wealth of the *i*-th candidate.
Output Specification:
Print one real value — the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
Demo Input:
['2 1 1\n1 5\n', '4 2 1\n1 4 2 3\n']
Demo Output:
['6.00000000\n', '6.50000000\n']
Note:
In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second.
In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (*a*<sub class="lower-index">3</sub> + *a*<sub class="lower-index">4</sub>) / 2 + *a*<sub class="lower-index">2</sub> = (3 + 2) / 2 + 4 = 6.5
|
```python
n,n1,n2=map(int,input().split())
a=list(map(int,input().split()))
a.sort(reverse=True)
b=[]
c=[]
if n1<n2:
p1,q1=0,n1
p2,q2=n1,(n1+n2)
else:
p1,q1=n2,(n1+n2)
p2,q2=0,n2
for i in range(p1,q1):
b.append(a[i])
for i in range(p2,q2):
c.append(a[i])
x=sum(b)/n1+sum(c)/n2
print('%.8f'%x)
```
| 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"
}
] | 1,490,907,331
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 62
| 5,529,600
|
x = input()
x = int(x,2)
print(x)
count = 0
while x >1:
if x%2 == 0:
x/=2
count+=1
elif x%2!=0:
x+=1
count+=1
print(count)
|
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
x = input()
x = int(x,2)
print(x)
count = 0
while x >1:
if x%2 == 0:
x/=2
count+=1
elif x%2!=0:
x+=1
count+=1
print(count)
```
| 0
|
297
|
B
|
Fish Weight
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"greedy"
] | null | null |
It is known that there are *k* fish species in the polar ocean, numbered from 1 to *k*. They are sorted by non-decreasing order of their weight, which is a positive number. Let the weight of the *i*-th type of fish be *w**i*, then 0<=<<=*w*1<=≤<=*w*2<=≤<=...<=≤<=*w**k* holds.
Polar bears Alice and Bob each have caught some fish, and they are guessing who has the larger sum of weight of the fish he/she's caught. Given the type of the fish they've caught, determine whether it is possible that the fish caught by Alice has a strictly larger total weight than Bob's. In other words, does there exist a sequence of weights *w**i* (not necessary integers), such that the fish caught by Alice has a strictly larger total weight?
|
The first line contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=105,<=1<=≤<=*k*<=≤<=109) — the number of fish caught by Alice and Bob respectively, and the number of fish species.
The second line contains *n* integers each from 1 to *k*, the list of fish type caught by Alice. The third line contains *m* integers each from 1 to *k*, the list of fish type caught by Bob.
Note that one may have caught more than one fish for a same species.
|
Output "YES" (without quotes) if it is possible, and "NO" (without quotes) otherwise.
|
[
"3 3 3\n2 2 2\n1 1 3\n",
"4 7 9\n5 2 7 3\n3 5 2 7 3 8 7\n"
] |
[
"YES\n",
"NO\n"
] |
In the first sample, if *w*<sub class="lower-index">1</sub> = 1, *w*<sub class="lower-index">2</sub> = 2, *w*<sub class="lower-index">3</sub> = 2.5, then Alice has a total of 2 + 2 + 2 = 6 weight units, while Bob only has 1 + 1 + 2.5 = 4.5.
In the second sample, the fish that Alice caught is a subset of Bob's. Therefore, the total weight of Bob’s fish is always not less than the total weight of Alice’s fish.
| 500
|
[
{
"input": "3 3 3\n2 2 2\n1 1 3",
"output": "YES"
},
{
"input": "4 7 9\n5 2 7 3\n3 5 2 7 3 8 7",
"output": "NO"
},
{
"input": "5 5 10\n8 2 8 5 9\n9 1 7 5 1",
"output": "YES"
},
{
"input": "7 7 10\n8 2 8 10 6 9 10\n2 4 9 5 6 2 5",
"output": "YES"
},
{
"input": "15 15 10\n4 5 9 1 4 6 4 1 4 3 7 9 9 2 6\n6 6 7 7 2 9 1 6 10 9 7 10 7 10 9",
"output": "NO"
},
{
"input": "25 25 10\n10 6 2 1 9 7 2 5 6 9 2 3 2 8 5 8 2 9 10 8 9 7 7 4 8\n6 2 10 4 7 9 3 2 4 5 1 8 6 9 8 6 9 8 4 8 7 9 10 2 8",
"output": "NO"
},
{
"input": "50 100 10\n10 9 10 5 5 2 2 6 4 8 9 1 6 3 9 7 8 3 8 5 6 6 5 7 2 10 3 6 8 1 8 8 9 5 10 1 5 10 9 4 7 8 10 3 3 4 7 8 6 3\n5 3 2 6 4 10 2 3 1 8 8 10 1 1 4 3 9 2 9 9 8 8 7 9 4 1 1 10 5 6 3 7 2 10 2 3 3 3 7 4 1 3 1 6 7 6 1 9 1 7 6 8 6 1 3 3 3 4 3 6 7 8 2 5 4 1 4 8 3 9 7 4 10 5 3 6 3 1 4 10 3 6 1 8 4 6 10 9 6 2 8 3 7 5 3 4 10 9 1 4",
"output": "NO"
},
{
"input": "100 50 10\n7 8 7 1 6 7 9 2 4 6 7 7 3 9 4 5 1 7 8 10 4 1 3 6 8 10 4 6 6 1 6 6 7 4 10 3 10 1 3 2 10 6 9 9 5 2 9 2 9 8 10 2 10 3 3 2 3 8 6 2 7 10 7 2 7 2 8 9 6 2 5 4 4 5 3 3 9 10 9 4 9 3 9 5 3 6 6 1 3 6 10 3 10 2 6 10 10 10 4 8\n3 3 2 9 4 4 10 2 7 3 3 2 6 3 3 4 7 4 1 2 3 8 1 6 7 7 2 10 1 1 1 5 7 7 5 1 6 8 7 5 3 7 4 6 10 5 5 5 1 9",
"output": "YES"
},
{
"input": "2 2 1000000000\n398981840 446967516\n477651114 577011341",
"output": "NO"
},
{
"input": "1 1 1\n1\n1",
"output": "NO"
},
{
"input": "1 1 1000000000\n502700350\n502700349",
"output": "YES"
},
{
"input": "1 1 1000000000\n406009709\n406009709",
"output": "NO"
},
{
"input": "2 1 1000000000\n699573624 308238132\n308238132",
"output": "YES"
},
{
"input": "10 10 10\n2 10 8 1 10 4 6 1 3 7\n8 1 1 5 7 1 9 10 2 3",
"output": "YES"
},
{
"input": "5 4 5\n1 2 2 3 4\n1 3 4 5",
"output": "YES"
}
] | 1,658,737,072
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
n, m, k = list(map(int, input().split()))
# fishs = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
if n > m:
print('YES')
else:
a.sort(reverse=True)
b.sort(reverse=True)
for i, val in enumerate(a):
print(val, b[i])
if val > b[i]:
print('True')
print('No')
|
Title: Fish Weight
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It is known that there are *k* fish species in the polar ocean, numbered from 1 to *k*. They are sorted by non-decreasing order of their weight, which is a positive number. Let the weight of the *i*-th type of fish be *w**i*, then 0<=<<=*w*1<=≤<=*w*2<=≤<=...<=≤<=*w**k* holds.
Polar bears Alice and Bob each have caught some fish, and they are guessing who has the larger sum of weight of the fish he/she's caught. Given the type of the fish they've caught, determine whether it is possible that the fish caught by Alice has a strictly larger total weight than Bob's. In other words, does there exist a sequence of weights *w**i* (not necessary integers), such that the fish caught by Alice has a strictly larger total weight?
Input Specification:
The first line contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=105,<=1<=≤<=*k*<=≤<=109) — the number of fish caught by Alice and Bob respectively, and the number of fish species.
The second line contains *n* integers each from 1 to *k*, the list of fish type caught by Alice. The third line contains *m* integers each from 1 to *k*, the list of fish type caught by Bob.
Note that one may have caught more than one fish for a same species.
Output Specification:
Output "YES" (without quotes) if it is possible, and "NO" (without quotes) otherwise.
Demo Input:
['3 3 3\n2 2 2\n1 1 3\n', '4 7 9\n5 2 7 3\n3 5 2 7 3 8 7\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, if *w*<sub class="lower-index">1</sub> = 1, *w*<sub class="lower-index">2</sub> = 2, *w*<sub class="lower-index">3</sub> = 2.5, then Alice has a total of 2 + 2 + 2 = 6 weight units, while Bob only has 1 + 1 + 2.5 = 4.5.
In the second sample, the fish that Alice caught is a subset of Bob's. Therefore, the total weight of Bob’s fish is always not less than the total weight of Alice’s fish.
|
```python
n, m, k = list(map(int, input().split()))
# fishs = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
if n > m:
print('YES')
else:
a.sort(reverse=True)
b.sort(reverse=True)
for i, val in enumerate(a):
print(val, b[i])
if val > b[i]:
print('True')
print('No')
```
| 0
|
|
608
|
B
|
Hamming Distance Sum
|
PROGRAMMING
| 1,500
|
[
"combinatorics",
"strings"
] | null | null |
Genos needs your help. He was asked to solve the following programming problem by Saitama:
The length of some string *s* is denoted |*s*|. The Hamming distance between two strings *s* and *t* of equal length is defined as , where *s**i* is the *i*-th character of *s* and *t**i* is the *i*-th character of *t*. For example, the Hamming distance between string "0011" and string "0110" is |0<=-<=0|<=+<=|0<=-<=1|<=+<=|1<=-<=1|<=+<=|1<=-<=0|<==<=0<=+<=1<=+<=0<=+<=1<==<=2.
Given two binary strings *a* and *b*, find the sum of the Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
|
The first line of the input contains binary string *a* (1<=≤<=|*a*|<=≤<=200<=000).
The second line of the input contains binary string *b* (|*a*|<=≤<=|*b*|<=≤<=200<=000).
Both strings are guaranteed to consist of characters '0' and '1' only.
|
Print a single integer — the sum of Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
|
[
"01\n00111\n",
"0011\n0110\n"
] |
[
"3\n",
"2\n"
] |
For the first sample case, there are four contiguous substrings of *b* of length |*a*|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.
The second sample case is described in the statement.
| 1,000
|
[
{
"input": "01\n00111",
"output": "3"
},
{
"input": "0011\n0110",
"output": "2"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "0\n1",
"output": "1"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1001101001101110101101000\n01111000010011111111110010001101000100011110101111",
"output": "321"
},
{
"input": "1110010001000101001011111\n00011011000000100001010000010100110011010001111010",
"output": "316"
}
] | 1,644,142,917
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 8
| 2,000
| 614,400
|
a = input()
b = input()
answer = 0
for a_i in range(len(a)):
for b_i in range(a_i,len(b)-len(a)+a_i+1):
if a[a_i]!=b[b_i]:
answer += 1
print(answer)
|
Title: Hamming Distance Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Genos needs your help. He was asked to solve the following programming problem by Saitama:
The length of some string *s* is denoted |*s*|. The Hamming distance between two strings *s* and *t* of equal length is defined as , where *s**i* is the *i*-th character of *s* and *t**i* is the *i*-th character of *t*. For example, the Hamming distance between string "0011" and string "0110" is |0<=-<=0|<=+<=|0<=-<=1|<=+<=|1<=-<=1|<=+<=|1<=-<=0|<==<=0<=+<=1<=+<=0<=+<=1<==<=2.
Given two binary strings *a* and *b*, find the sum of the Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
Input Specification:
The first line of the input contains binary string *a* (1<=≤<=|*a*|<=≤<=200<=000).
The second line of the input contains binary string *b* (|*a*|<=≤<=|*b*|<=≤<=200<=000).
Both strings are guaranteed to consist of characters '0' and '1' only.
Output Specification:
Print a single integer — the sum of Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
Demo Input:
['01\n00111\n', '0011\n0110\n']
Demo Output:
['3\n', '2\n']
Note:
For the first sample case, there are four contiguous substrings of *b* of length |*a*|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.
The second sample case is described in the statement.
|
```python
a = input()
b = input()
answer = 0
for a_i in range(len(a)):
for b_i in range(a_i,len(b)-len(a)+a_i+1):
if a[a_i]!=b[b_i]:
answer += 1
print(answer)
```
| 0
|
|
616
|
A
|
Comparing Two Long Integers
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
You are given two very long integers *a*,<=*b* (leading zeroes are allowed). You should check what number *a* or *b* is greater or determine that they are equal.
The input size is very large so don't use the reading of symbols one by one. Instead of that use the reading of a whole line or token.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. Don't use the function input() in Python2 instead of it use the function raw_input().
|
The first line contains a non-negative integer *a*.
The second line contains a non-negative integer *b*.
The numbers *a*,<=*b* may contain leading zeroes. Each of them contains no more than 106 digits.
|
Print the symbol "<" if *a*<=<<=*b* and the symbol ">" if *a*<=><=*b*. If the numbers are equal print the symbol "=".
|
[
"9\n10\n",
"11\n10\n",
"00012345\n12345\n",
"0123\n9\n",
"0123\n111\n"
] |
[
"<\n",
">\n",
"=\n",
">\n",
">\n"
] |
none
| 0
|
[
{
"input": "9\n10",
"output": "<"
},
{
"input": "11\n10",
"output": ">"
},
{
"input": "00012345\n12345",
"output": "="
},
{
"input": "0123\n9",
"output": ">"
},
{
"input": "0123\n111",
"output": ">"
},
{
"input": "9\n9",
"output": "="
},
{
"input": "0\n0000",
"output": "="
},
{
"input": "1213121\n1213121",
"output": "="
},
{
"input": "8631749422082281871941140403034638286979613893271246118706788645620907151504874585597378422393911017\n1460175633701201615285047975806206470993708143873675499262156511814213451040881275819636625899967479",
"output": ">"
},
{
"input": "6421902501252475186372406731932548506197390793597574544727433297197476846519276598727359617092494798\n8",
"output": ">"
},
{
"input": "9\n3549746075165939381145061479392284958612916596558639332310874529760172204736013341477640605383578772",
"output": "<"
},
{
"input": "11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "="
},
{
"input": "0000000001\n2",
"output": "<"
},
{
"input": "1000000000000000000000000000000000\n1000000000000000000000000000000001",
"output": "<"
},
{
"input": "123456123456123456123456123456123456123456123456123456123456123456\n123456123456123456123456123456123456123456123456123456123456123456123456123456",
"output": "<"
},
{
"input": "1111111111111111111111111111111111111111\n2222222222222222222222222222222222222222",
"output": "<"
},
{
"input": "123456789999999\n123456789999999",
"output": "="
},
{
"input": "111111111111111111111111111111\n222222222222222222222222222222",
"output": "<"
},
{
"input": "1111111111111111111111111111111111111111111111111111111111111111111111\n1111111111111111111111111111111111111111111111111111111111111111111111",
"output": "="
},
{
"input": "587345873489573457357834\n47957438573458347574375348",
"output": "<"
},
{
"input": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111\n33333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333",
"output": "<"
},
{
"input": "11111111111111111111111111111111111\n44444444444444444444444444444444444",
"output": "<"
},
{
"input": "11111111111111111111111111111111111\n22222222222222222222222222222222222",
"output": "<"
},
{
"input": "9999999999999999999999999999999999999999999999999999999999999999999\n99999999999999999999999999999999999999999999999999999999999999999999999999999999999999",
"output": "<"
},
{
"input": "1\n2",
"output": "<"
},
{
"input": "9\n0",
"output": ">"
},
{
"input": "222222222222222222222222222222222222222222222222222222222\n22222222222222222222222222222222222222222222222222222222222",
"output": "<"
},
{
"input": "66646464222222222222222222222222222222222222222222222222222222222222222\n111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "<"
},
{
"input": "222222222222222222222222222222222222222222222222222\n111111111111111111111111111111111111111111111111111111111111111",
"output": "<"
},
{
"input": "11111111111111111111111111111111111111\n44444444444444444444444444444444444444",
"output": "<"
},
{
"input": "01\n2",
"output": "<"
},
{
"input": "00\n01",
"output": "<"
},
{
"input": "99999999999999999999999999999999999999999999999\n99999999999999999999999999999999999999999999999",
"output": "="
},
{
"input": "43278947323248843213443272432\n793439250984509434324323453435435",
"output": "<"
},
{
"input": "0\n1",
"output": "<"
},
{
"input": "010\n011",
"output": "<"
},
{
"input": "999999999999999999999999999999999999999999999999\n999999999999999999999999999999999999999999999999",
"output": "="
},
{
"input": "0001001\n0001010",
"output": "<"
},
{
"input": "1111111111111111111111111111111111111111111111111111111111111\n1111111111111111111111111111111111111111111111111111111111111",
"output": "="
},
{
"input": "00000\n00",
"output": "="
},
{
"input": "999999999999999999999999999\n999999999999999999999999999",
"output": "="
},
{
"input": "999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999\n999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999",
"output": "="
},
{
"input": "001\n000000000010",
"output": "<"
},
{
"input": "01\n10",
"output": "<"
},
{
"input": "555555555555555555555555555555555555555555555555555555555555\n555555555555555555555555555555555555555555555555555555555555",
"output": "="
},
{
"input": "5555555555555555555555555555555555555555555555555\n5555555555555555555555555555555555555555555555555",
"output": "="
},
{
"input": "01\n02",
"output": "<"
},
{
"input": "001111\n0001111",
"output": "="
},
{
"input": "55555555555555555555555555555555555555555555555555\n55555555555555555555555555555555555555555555555555",
"output": "="
},
{
"input": "1029301293019283091283091283091280391283\n1029301293019283091283091283091280391283",
"output": "="
},
{
"input": "001\n2",
"output": "<"
},
{
"input": "000000000\n000000000",
"output": "="
},
{
"input": "000000\n10",
"output": "<"
},
{
"input": "000000000000000\n001",
"output": "<"
},
{
"input": "0000001\n2",
"output": "<"
},
{
"input": "0000\n123",
"output": "<"
},
{
"input": "951\n960",
"output": "<"
},
{
"input": "002\n0001",
"output": ">"
},
{
"input": "0000001\n01",
"output": "="
},
{
"input": "99999999999999999999999999999999999999999999999999999999999999\n99999999999999999999999999999999999999999999999999999999999999",
"output": "="
},
{
"input": "12345678901234567890123456789012345678901234567890123456789012\n12345678901234567890123456789012345678901234567890123456789012",
"output": "="
},
{
"input": "02\n01",
"output": ">"
},
{
"input": "00000111111\n00000110111",
"output": ">"
},
{
"input": "0123\n123",
"output": "="
},
{
"input": "123771237912798378912\n91239712798379812897389123123123123",
"output": "<"
},
{
"input": "00001\n002",
"output": "<"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000\n000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "="
},
{
"input": "000000001\n00002",
"output": "<"
},
{
"input": "00002\n00003",
"output": "<"
},
{
"input": "000123456\n123457",
"output": "<"
},
{
"input": "01\n00",
"output": ">"
},
{
"input": "00\n0",
"output": "="
},
{
"input": "10\n11",
"output": "<"
},
{
"input": "0011\n12",
"output": "<"
},
{
"input": "00\n1",
"output": "<"
},
{
"input": "0\n0",
"output": "="
},
{
"input": "00\n10",
"output": "<"
},
{
"input": "011\n10",
"output": ">"
},
{
"input": "00011111111111111111111111111111111111000000000000000000000000000000000000000000000000000210000000000000000000000000000000000000000011000\n11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111112091",
"output": "<"
},
{
"input": "0000001\n00",
"output": ">"
},
{
"input": "01\n1",
"output": "="
},
{
"input": "010\n001",
"output": ">"
},
{
"input": "100\n111",
"output": "<"
},
{
"input": "1\n0",
"output": ">"
},
{
"input": "000000\n000000000000000000000",
"output": "="
},
{
"input": "010101\n010101",
"output": "="
},
{
"input": "00000000000000000001111111111111111111111111111111111111111111111111111111\n11111111111111111111111",
"output": ">"
},
{
"input": "0000000\n0",
"output": "="
},
{
"input": "187923712738712879387912839182381\n871279397127389781927389718923789178923897123",
"output": "<"
},
{
"input": "0010\n030",
"output": "<"
}
] | 1,623,125,631
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 8
| 109
| 0
|
a = int(input())
b = int(input())
m = pow(10,16)
if int(a)%m>int(b)%m:
print(">")
elif int(a)%m<int(b)%m:
print("<")
elif int(a)%m==int(b)%m:
print("=")
|
Title: Comparing Two Long Integers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two very long integers *a*,<=*b* (leading zeroes are allowed). You should check what number *a* or *b* is greater or determine that they are equal.
The input size is very large so don't use the reading of symbols one by one. Instead of that use the reading of a whole line or token.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. Don't use the function input() in Python2 instead of it use the function raw_input().
Input Specification:
The first line contains a non-negative integer *a*.
The second line contains a non-negative integer *b*.
The numbers *a*,<=*b* may contain leading zeroes. Each of them contains no more than 106 digits.
Output Specification:
Print the symbol "<" if *a*<=<<=*b* and the symbol ">" if *a*<=><=*b*. If the numbers are equal print the symbol "=".
Demo Input:
['9\n10\n', '11\n10\n', '00012345\n12345\n', '0123\n9\n', '0123\n111\n']
Demo Output:
['<\n', '>\n', '=\n', '>\n', '>\n']
Note:
none
|
```python
a = int(input())
b = int(input())
m = pow(10,16)
if int(a)%m>int(b)%m:
print(">")
elif int(a)%m<int(b)%m:
print("<")
elif int(a)%m==int(b)%m:
print("=")
```
| 0
|
|
191
|
D
|
Metro Scheme
|
PROGRAMMING
| 2,700
|
[
"graphs",
"greedy"
] | null | null |
Berland is very concerned with privacy, so almost all plans and blueprints are secret. However, a spy of the neighboring state managed to steal the Bertown subway scheme.
The Bertown Subway has *n* stations, numbered from 1 to *n*, and *m* bidirectional tunnels connecting them. All Bertown Subway consists of lines. To be more precise, there are two types of lines: circular and radial.
A radial line is a sequence of stations *v*1,<=...,<=*v**k* (*k*<=><=1), where stations *v**i* and *v**i*<=+<=1 (*i*<=<<=*k*) are connected by a tunnel and no station occurs in the line more than once (*v**i*<=≠<=*v**j* for *i*<=≠<=*j*).
A loop line is a series of stations, *v*1,<=...,<=*v**k* (*k*<=><=2), where stations *v**i* и *v**i*<=+<=1 are connected by a tunnel. In addition, stations *v*1 and *v**k* are also connected by a tunnel. No station is occurs in the loop line more than once.
Note that a single station can be passed by any number of lines.
According to Berland standards, there can't be more than one tunnel between two stations and each tunnel belongs to exactly one line. Naturally, each line has at least one tunnel. Between any two stations there is the way along the subway tunnels. In addition, in terms of graph theory, a subway is a vertex cactus: if we consider the subway as a graph in which the stations are the vertexes and the edges are tunnels, then each vertex lies on no more than one simple cycle.
Unfortunately, scheme, stolen by the spy, had only the stations and the tunnels. It was impossible to determine to which line every tunnel corresponds. But to sabotage successfully, the spy needs to know what minimum and maximum number of lines may be in the Bertown subway.
Help him!
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=105, 0<=≤<=*m*<=≤<=3·105) — the number of stations and the number of tunnels, correspondingly.
Each of the next *m* lines contain two integers — the numbers of stations connected by the corresponding tunnel. The stations are numbered with integers from 1 to *n*.
It is guaranteed that the graph that corresponds to the subway has no multiple edges or loops, it is connected and it is a vertex cactus.
|
Print two numbers — the minimum and maximum number of lines correspondingly.
|
[
"3 3\n1 2\n2 3\n3 1\n",
"8 8\n1 2\n2 3\n3 4\n4 5\n6 4\n4 7\n7 2\n2 8\n",
"6 6\n1 2\n2 3\n2 5\n5 6\n3 4\n3 5\n"
] |
[
"1 3\n",
"2 8\n",
"3 6\n"
] |
The subway scheme with minimum possible number of lines for the second sample is:
| 2,000
|
[] | 1,623,006,207
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 186
| 0
|
n = int(input())
l = list(map(int, input().strip().split()))[:n]
r = False
for i in range(len(l)):
if l[l[l[i]-1]-1]-1 == i:
r = True
break
if r:
print("YES")
else:
print("NO")
|
Title: Metro Scheme
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Berland is very concerned with privacy, so almost all plans and blueprints are secret. However, a spy of the neighboring state managed to steal the Bertown subway scheme.
The Bertown Subway has *n* stations, numbered from 1 to *n*, and *m* bidirectional tunnels connecting them. All Bertown Subway consists of lines. To be more precise, there are two types of lines: circular and radial.
A radial line is a sequence of stations *v*1,<=...,<=*v**k* (*k*<=><=1), where stations *v**i* and *v**i*<=+<=1 (*i*<=<<=*k*) are connected by a tunnel and no station occurs in the line more than once (*v**i*<=≠<=*v**j* for *i*<=≠<=*j*).
A loop line is a series of stations, *v*1,<=...,<=*v**k* (*k*<=><=2), where stations *v**i* и *v**i*<=+<=1 are connected by a tunnel. In addition, stations *v*1 and *v**k* are also connected by a tunnel. No station is occurs in the loop line more than once.
Note that a single station can be passed by any number of lines.
According to Berland standards, there can't be more than one tunnel between two stations and each tunnel belongs to exactly one line. Naturally, each line has at least one tunnel. Between any two stations there is the way along the subway tunnels. In addition, in terms of graph theory, a subway is a vertex cactus: if we consider the subway as a graph in which the stations are the vertexes and the edges are tunnels, then each vertex lies on no more than one simple cycle.
Unfortunately, scheme, stolen by the spy, had only the stations and the tunnels. It was impossible to determine to which line every tunnel corresponds. But to sabotage successfully, the spy needs to know what minimum and maximum number of lines may be in the Bertown subway.
Help him!
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=105, 0<=≤<=*m*<=≤<=3·105) — the number of stations and the number of tunnels, correspondingly.
Each of the next *m* lines contain two integers — the numbers of stations connected by the corresponding tunnel. The stations are numbered with integers from 1 to *n*.
It is guaranteed that the graph that corresponds to the subway has no multiple edges or loops, it is connected and it is a vertex cactus.
Output Specification:
Print two numbers — the minimum and maximum number of lines correspondingly.
Demo Input:
['3 3\n1 2\n2 3\n3 1\n', '8 8\n1 2\n2 3\n3 4\n4 5\n6 4\n4 7\n7 2\n2 8\n', '6 6\n1 2\n2 3\n2 5\n5 6\n3 4\n3 5\n']
Demo Output:
['1 3\n', '2 8\n', '3 6\n']
Note:
The subway scheme with minimum possible number of lines for the second sample is:
|
```python
n = int(input())
l = list(map(int, input().strip().split()))[:n]
r = False
for i in range(len(l)):
if l[l[l[i]-1]-1]-1 == i:
r = True
break
if r:
print("YES")
else:
print("NO")
```
| -1
|
|
808
|
C
|
Tea Party
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"greedy",
"sortings"
] | null | null |
Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=≤<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
|
The first line contains two integer numbers *n* and *w* (1<=≤<=*n*<=≤<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
|
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
|
[
"2 10\n8 7\n",
"4 4\n1 1 1 1\n",
"3 10\n9 8 10\n"
] |
[
"6 4 \n",
"1 1 1 1 \n",
"-1\n"
] |
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.
| 0
|
[
{
"input": "2 10\n8 7",
"output": "6 4 "
},
{
"input": "4 4\n1 1 1 1",
"output": "1 1 1 1 "
},
{
"input": "3 10\n9 8 10",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "1 "
},
{
"input": "1 1\n2",
"output": "1 "
},
{
"input": "1 10\n20",
"output": "10 "
},
{
"input": "3 10\n8 4 8",
"output": "4 2 4 "
},
{
"input": "3 100\n37 26 37",
"output": "37 26 37 "
},
{
"input": "3 60\n43 23 24",
"output": "36 12 12 "
},
{
"input": "20 14\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 8\n1 2 1 2 1 1 1 2 1 1 1 2 1 1 2 1 1 1 2 2",
"output": "-1"
},
{
"input": "50 1113\n25 21 23 37 28 23 19 25 5 12 3 11 46 50 13 50 7 1 8 40 4 6 34 27 11 39 45 31 10 12 48 2 19 37 47 45 30 24 21 42 36 14 31 30 31 50 6 3 33 49",
"output": "13 11 12 37 28 12 10 18 3 6 2 6 46 50 7 50 4 1 4 40 2 3 34 27 6 39 45 31 5 6 48 1 10 37 47 45 30 12 11 42 36 7 31 30 31 50 3 2 33 49 "
},
{
"input": "50 440\n14 69 33 38 83 65 21 66 89 3 93 60 31 16 61 20 42 64 13 1 50 50 74 58 67 61 52 22 69 68 18 33 28 59 4 8 96 32 84 85 87 87 61 89 2 47 15 64 88 18",
"output": "-1"
},
{
"input": "100 640\n82 51 81 14 37 17 78 92 64 15 8 86 89 8 87 77 66 10 15 12 100 25 92 47 21 78 20 63 13 49 41 36 41 79 16 87 87 69 3 76 80 60 100 49 70 59 72 8 38 71 45 97 71 14 76 54 81 4 59 46 39 29 92 3 49 22 53 99 59 52 74 31 92 43 42 23 44 9 82 47 7 40 12 9 3 55 37 85 46 22 84 52 98 41 21 77 63 17 62 91",
"output": "-1"
},
{
"input": "100 82\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"
},
{
"input": "100 55\n1 1 1 1 2 1 1 1 1 1 2 2 1 1 2 1 2 1 1 1 2 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 2 2 1 1 1 2 1 2 2 1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 1 1 1 1 2 2 2 2 2 2 2 1 1 1 2 1 2 1",
"output": "-1"
},
{
"input": "30 50\n3 1 2 4 1 2 2 4 3 4 4 3 3 3 3 5 3 2 5 4 3 3 5 3 3 5 4 5 3 5",
"output": "-1"
},
{
"input": "40 100\n3 3 3 3 4 1 1 1 1 1 2 2 1 3 1 2 3 2 1 2 2 2 1 4 2 2 3 3 3 2 4 6 4 4 3 2 2 2 4 5",
"output": "3 3 3 3 4 1 1 1 1 1 2 2 1 3 1 2 3 2 1 2 2 2 1 4 2 2 3 3 3 2 4 6 4 4 3 2 2 2 4 5 "
},
{
"input": "100 10000\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": "100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 "
},
{
"input": "2 5\n3 4",
"output": "2 3 "
},
{
"input": "2 6\n2 6",
"output": "1 5 "
},
{
"input": "23 855\n5 63 94 57 38 84 77 79 83 36 47 31 60 79 75 48 88 17 46 33 23 15 27",
"output": "3 32 94 29 19 84 39 72 83 18 24 16 30 79 38 24 88 9 23 17 12 8 14 "
},
{
"input": "52 2615\n73 78 70 92 94 74 46 19 55 20 70 3 1 42 68 10 66 80 1 31 65 19 73 74 56 35 53 38 92 35 65 81 6 98 74 51 27 49 76 19 86 76 5 60 14 75 64 99 43 7 36 79",
"output": "73 78 70 92 94 74 46 10 55 10 70 2 1 42 68 5 66 80 1 16 65 10 73 74 56 18 53 38 92 30 65 81 3 98 74 51 14 49 76 10 86 76 3 60 7 75 64 99 43 4 36 79 "
},
{
"input": "11 287\n34 30 69 86 22 53 11 91 62 44 5",
"output": "17 15 35 43 11 27 6 77 31 22 3 "
},
{
"input": "55 1645\n60 53 21 20 87 48 10 21 76 35 52 41 82 86 93 11 93 86 34 15 37 63 57 3 57 57 32 8 55 25 29 38 46 22 13 87 27 35 40 83 5 7 6 18 88 25 4 59 95 62 31 93 98 50 62",
"output": "30 27 11 10 82 24 5 11 38 18 26 21 41 43 93 6 93 43 17 8 19 32 29 2 29 29 16 4 28 13 15 19 23 11 7 87 14 18 20 42 3 4 3 9 88 13 2 30 95 31 16 93 98 25 31 "
},
{
"input": "71 3512\n97 46 76 95 81 96 99 83 10 50 19 18 73 5 41 60 12 73 60 31 21 64 88 61 43 57 61 19 75 35 41 85 12 59 32 47 37 43 35 92 90 47 3 98 21 18 61 79 39 86 74 8 52 33 39 27 93 54 35 38 96 36 83 51 97 10 8 66 75 87 68",
"output": "97 46 76 95 81 96 99 83 5 50 10 9 73 3 41 60 6 73 60 16 11 64 88 61 43 57 61 10 75 18 41 85 6 59 16 47 19 43 18 92 90 47 2 98 11 9 61 79 20 86 74 4 52 17 21 14 93 54 18 19 96 18 83 51 97 5 4 66 75 87 68 "
},
{
"input": "100 2633\n99 50 64 81 75 73 26 31 31 36 95 12 100 2 70 72 78 56 76 23 94 8 91 1 39 82 97 67 64 25 71 90 48 34 31 46 64 37 46 50 99 93 14 56 1 89 95 89 50 52 12 58 43 65 45 88 90 14 38 19 6 15 91 67 43 48 82 20 11 48 33 20 39 52 73 5 25 84 26 54 42 56 10 28 9 63 60 98 30 1 25 74 86 56 85 9 12 94 80 95",
"output": "50 25 32 41 38 37 13 16 16 18 48 6 61 1 35 36 39 28 38 12 47 4 46 1 20 41 49 34 32 13 36 45 24 17 16 23 32 19 23 25 50 47 7 28 1 45 48 45 25 26 6 29 22 33 23 44 45 7 19 10 3 8 46 34 22 24 41 10 6 24 17 10 20 26 37 3 13 42 13 27 21 28 5 14 5 32 30 49 15 1 13 37 43 28 43 5 6 47 40 48 "
},
{
"input": "71 1899\n23 55 58 87 69 85 100 21 19 72 81 68 20 25 29 92 18 74 89 70 53 7 78 57 41 79 64 87 63 76 95 84 1 28 32 1 79 34 77 17 71 61 35 31 62 92 69 99 60 26 2 18 61 9 27 77 82 6 30 65 52 3 51 43 13 77 41 59 19 29 86",
"output": "12 28 29 44 35 43 95 11 10 36 41 34 10 13 15 46 9 37 45 35 27 4 39 29 21 40 32 44 32 38 48 42 1 14 16 1 40 17 39 9 36 31 18 16 31 46 35 50 30 13 1 9 31 5 14 39 41 3 15 33 26 2 26 22 7 39 21 30 10 15 43 "
},
{
"input": "10 21\n3 3 3 3 4 3 3 3 3 3",
"output": "2 2 2 2 3 2 2 2 2 2 "
}
] | 1,625,336,690
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 61
| 0
|
n,w = map(int,input().split())
l = list(map(int,input().split()))
ans = []
for i in range(n):
ans.append(l[i]//2+l[i]%2)
w-=l[i]//2+l[i]%2
l[i] = (l[i],i)
if w <0:
print(-1)
else:
l = list(reversed(list(sorted(l))))
for i in range(n):
w,ans[l[i][1]] = max(0,ans[l[i][1]]-l[i][0]),min(l[i][0],ans[l[i][1]]+w)
if(w==0):
break
for i in range(n):
print(ans[i],end = " ")
|
Title: Tea Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp invited all his friends to the tea party to celebrate the holiday. He has *n* cups, one for each of his *n* friends, with volumes *a*1,<=*a*2,<=...,<=*a**n*. His teapot stores *w* milliliters of tea (*w*<=≤<=*a*1<=+<=*a*2<=+<=...<=+<=*a**n*). Polycarp wants to pour tea in cups in such a way that:
- Every cup will contain tea for at least half of its volume - Every cup will contain integer number of milliliters of tea - All the tea from the teapot will be poured into cups - All friends will be satisfied.
Friend with cup *i* won't be satisfied, if there exists such cup *j* that cup *i* contains less tea than cup *j* but *a**i*<=><=*a**j*.
For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.
Input Specification:
The first line contains two integer numbers *n* and *w* (1<=≤<=*n*<=≤<=100, ).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.
If it's impossible to pour all the tea and satisfy all conditions then output -1.
Demo Input:
['2 10\n8 7\n', '4 4\n1 1 1 1\n', '3 10\n9 8 10\n']
Demo Output:
['6 4 \n', '1 1 1 1 \n', '-1\n']
Note:
In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.
|
```python
n,w = map(int,input().split())
l = list(map(int,input().split()))
ans = []
for i in range(n):
ans.append(l[i]//2+l[i]%2)
w-=l[i]//2+l[i]%2
l[i] = (l[i],i)
if w <0:
print(-1)
else:
l = list(reversed(list(sorted(l))))
for i in range(n):
w,ans[l[i][1]] = max(0,ans[l[i][1]]-l[i][0]),min(l[i][0],ans[l[i][1]]+w)
if(w==0):
break
for i in range(n):
print(ans[i],end = " ")
```
| 0
|
|
253
|
A
|
Boys and Girls
|
PROGRAMMING
| 1,100
|
[
"greedy"
] | null | null |
There are *n* boys and *m* girls studying in the class. They should stand in a line so that boys and girls alternated there as much as possible. Let's assume that positions in the line are indexed from left to right by numbers from 1 to *n*<=+<=*m*. Then the number of integers *i* (1<=≤<=*i*<=<<=*n*<=+<=*m*) such that positions with indexes *i* and *i*<=+<=1 contain children of different genders (position *i* has a girl and position *i*<=+<=1 has a boy or vice versa) must be as large as possible.
Help the children and tell them how to form the line.
|
The single line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), separated by a space.
|
Print a line of *n*<=+<=*m* characters. Print on the *i*-th position of the line character "B", if the *i*-th position of your arrangement should have a boy and "G", if it should have a girl.
Of course, the number of characters "B" should equal *n* and the number of characters "G" should equal *m*. If there are multiple optimal solutions, print any of them.
|
[
"3 3\n",
"4 2\n"
] |
[
"GBGBGB\n",
"BGBGBB\n"
] |
In the first sample another possible answer is BGBGBG.
In the second sample answer BBGBGB is also optimal.
| 500
|
[
{
"input": "3 3",
"output": "GBGBGB"
},
{
"input": "4 2",
"output": "BGBGBB"
},
{
"input": "5 5",
"output": "GBGBGBGBGB"
},
{
"input": "6 4",
"output": "BGBGBGBGBB"
},
{
"input": "100 1",
"output": "BGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "76 48",
"output": "BGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "100 90",
"output": "BGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBBBBBBBBBB"
},
{
"input": "90 100",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGGGGGGGGGG"
},
{
"input": "1 98",
"output": "GBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "1 100",
"output": "GBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "56 98",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "89 89",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGB"
},
{
"input": "18 94",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "84 27",
"output": "BGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "1 1",
"output": "GB"
},
{
"input": "1 2",
"output": "GBG"
},
{
"input": "2 1",
"output": "BGB"
},
{
"input": "1 34",
"output": "GBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "46 2",
"output": "BGBGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "99 3",
"output": "BGBGBGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "10 100",
"output": "GBGBGBGBGBGBGBGBGBGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "100 100",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGB"
},
{
"input": "1 4",
"output": "GBGGG"
}
] | 1,678,712,614
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 154
| 2,867,200
|
import sys
b, g = map(int, input().split())
smaller = min(b, g)
while smaller > 0:
if b > g:
print("BG", end="")
else:
print("GB", end="")
smaller -= 1
for i in range(max(b, g) - min(b, g)):
if b > g:
print("B", end = "")
else:
print("G", end = "")
|
Title: Boys and Girls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* boys and *m* girls studying in the class. They should stand in a line so that boys and girls alternated there as much as possible. Let's assume that positions in the line are indexed from left to right by numbers from 1 to *n*<=+<=*m*. Then the number of integers *i* (1<=≤<=*i*<=<<=*n*<=+<=*m*) such that positions with indexes *i* and *i*<=+<=1 contain children of different genders (position *i* has a girl and position *i*<=+<=1 has a boy or vice versa) must be as large as possible.
Help the children and tell them how to form the line.
Input Specification:
The single line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), separated by a space.
Output Specification:
Print a line of *n*<=+<=*m* characters. Print on the *i*-th position of the line character "B", if the *i*-th position of your arrangement should have a boy and "G", if it should have a girl.
Of course, the number of characters "B" should equal *n* and the number of characters "G" should equal *m*. If there are multiple optimal solutions, print any of them.
Demo Input:
['3 3\n', '4 2\n']
Demo Output:
['GBGBGB\n', 'BGBGBB\n']
Note:
In the first sample another possible answer is BGBGBG.
In the second sample answer BBGBGB is also optimal.
|
```python
import sys
b, g = map(int, input().split())
smaller = min(b, g)
while smaller > 0:
if b > g:
print("BG", end="")
else:
print("GB", end="")
smaller -= 1
for i in range(max(b, g) - min(b, g)):
if b > g:
print("B", end = "")
else:
print("G", end = "")
```
| -1
|
|
58
|
A
|
Chat room
|
PROGRAMMING
| 1,000
|
[
"greedy",
"strings"
] |
A. Chat room
|
1
|
256
|
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
|
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
|
If Vasya managed to say hello, print "YES", otherwise print "NO".
|
[
"ahhellllloou\n",
"hlelo\n"
] |
[
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,663,228,130
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 46
| 0
|
s=list(input())
count=1
for x in list("hello"):
if x in s:
del s[0:s.index(x)+1]
count*=1
else:
count*=0
print("YES" if count==1 else "NO")
|
Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none
|
```python
s=list(input())
count=1
for x in list("hello"):
if x in s:
del s[0:s.index(x)+1]
count*=1
else:
count*=0
print("YES" if count==1 else "NO")
```
| 3.977
|
461
|
A
|
Appleman and Toastman
|
PROGRAMMING
| 1,200
|
[
"greedy",
"sortings"
] | null | null |
Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman.
|
Print a single integer — the largest possible score.
|
[
"3\n3 1 5\n",
"1\n10\n"
] |
[
"26\n",
"10\n"
] |
Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions.
| 500
|
[
{
"input": "3\n3 1 5",
"output": "26"
},
{
"input": "1\n10",
"output": "10"
},
{
"input": "10\n8 10 2 5 6 2 4 7 2 1",
"output": "376"
},
{
"input": "10\n171308 397870 724672 431255 228496 892002 542924 718337 888642 161821",
"output": "40204082"
},
{
"input": "10\n1 2 2 2 4 5 6 7 8 10",
"output": "376"
},
{
"input": "10\n161821 171308 228496 397870 431255 542924 718337 724672 888642 892002",
"output": "40204082"
},
{
"input": "1\n397870",
"output": "397870"
},
{
"input": "1\n1000000",
"output": "1000000"
},
{
"input": "10\n10 8 7 6 5 4 2 2 2 1",
"output": "376"
},
{
"input": "10\n892002 888642 724672 718337 542924 431255 397870 228496 171308 161821",
"output": "40204082"
},
{
"input": "10\n5 2 6 10 10 10 10 2 2 5",
"output": "485"
},
{
"input": "10\n431255 724672 228496 397870 397870 397870 397870 724672 888642 431255",
"output": "36742665"
},
{
"input": "10\n2 2 2 5 5 6 10 10 10 10",
"output": "485"
},
{
"input": "10\n228496 397870 397870 397870 397870 431255 431255 724672 724672 888642",
"output": "36742665"
},
{
"input": "10\n10 10 10 10 6 5 5 2 2 2",
"output": "485"
},
{
"input": "10\n888642 724672 724672 431255 431255 397870 397870 397870 397870 228496",
"output": "36742665"
},
{
"input": "10\n10 10 10 10 10 10 10 10 10 10",
"output": "640"
},
{
"input": "10\n1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000",
"output": "64000000"
},
{
"input": "1\n397870",
"output": "397870"
},
{
"input": "2\n1 2",
"output": "6"
},
{
"input": "2\n2 3",
"output": "10"
},
{
"input": "2\n1 1",
"output": "4"
}
] | 1,480,649,279
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 46
| 0
|
n=int(input())
nums=input().split()
nums.sort()
num=[int(i) for i in nums]
x=0
while len(num)>1:
x=x+sum(num)
x=x+num[0]
del num[0]
x=x+num[0]
print(x)
|
Title: Appleman and Toastman
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman.
Output Specification:
Print a single integer — the largest possible score.
Demo Input:
['3\n3 1 5\n', '1\n10\n']
Demo Output:
['26\n', '10\n']
Note:
Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions.
|
```python
n=int(input())
nums=input().split()
nums.sort()
num=[int(i) for i in nums]
x=0
while len(num)>1:
x=x+sum(num)
x=x+num[0]
del num[0]
x=x+num[0]
print(x)
```
| 0
|
|
863
|
B
|
Kayaking
|
PROGRAMMING
| 1,500
|
[
"brute force",
"greedy",
"sortings"
] | null | null |
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability!
|
The first line contains one number *n* (2<=≤<=*n*<=≤<=50).
The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000).
|
Print minimum possible total instability.
|
[
"2\n1 2 3 4\n",
"4\n1 3 4 6 3 4 100 200\n"
] |
[
"1\n",
"5\n"
] |
none
| 0
|
[
{
"input": "2\n1 2 3 4",
"output": "1"
},
{
"input": "4\n1 3 4 6 3 4 100 200",
"output": "5"
},
{
"input": "3\n305 139 205 406 530 206",
"output": "102"
},
{
"input": "3\n610 750 778 6 361 407",
"output": "74"
},
{
"input": "5\n97 166 126 164 154 98 221 7 51 47",
"output": "35"
},
{
"input": "50\n1 1 2 2 1 3 2 2 1 1 1 1 2 3 3 1 2 1 3 3 2 1 2 3 1 1 2 1 3 1 3 1 3 3 3 1 1 1 3 3 2 2 2 2 3 2 2 2 2 3 1 3 3 3 3 1 3 3 1 3 3 3 3 2 3 1 3 3 1 1 1 3 1 2 2 2 1 1 1 3 1 2 3 2 1 3 3 2 2 1 3 1 3 1 2 2 1 2 3 2",
"output": "0"
},
{
"input": "50\n5 5 5 5 4 2 2 3 2 2 4 1 5 5 1 2 4 2 4 2 5 2 2 2 2 3 2 4 2 5 5 4 3 1 2 3 3 5 4 2 2 5 2 4 5 5 4 4 1 5 5 3 2 2 5 1 3 3 2 4 4 5 1 2 3 4 4 1 3 3 3 5 1 2 4 4 4 4 2 5 2 5 3 2 4 5 5 2 1 1 2 4 5 3 2 1 2 4 4 4",
"output": "1"
},
{
"input": "50\n499 780 837 984 481 526 944 482 862 136 265 605 5 631 974 967 574 293 969 467 573 845 102 224 17 873 648 120 694 996 244 313 404 129 899 583 541 314 525 496 443 857 297 78 575 2 430 137 387 319 382 651 594 411 845 746 18 232 6 289 889 81 174 175 805 1000 799 950 475 713 951 685 729 925 262 447 139 217 788 514 658 572 784 185 112 636 10 251 621 218 210 89 597 553 430 532 264 11 160 476",
"output": "368"
},
{
"input": "50\n873 838 288 87 889 364 720 410 565 651 577 356 740 99 549 592 994 385 777 435 486 118 887 440 749 533 356 790 413 681 267 496 475 317 88 660 374 186 61 437 729 860 880 538 277 301 667 180 60 393 955 540 896 241 362 146 74 680 734 767 851 337 751 860 542 735 444 793 340 259 495 903 743 961 964 966 87 275 22 776 368 701 835 732 810 735 267 988 352 647 924 183 1 924 217 944 322 252 758 597",
"output": "393"
},
{
"input": "50\n297 787 34 268 439 629 600 398 425 833 721 908 830 636 64 509 420 647 499 675 427 599 396 119 798 742 577 355 22 847 389 574 766 453 196 772 808 261 106 844 726 975 173 992 874 89 775 616 678 52 69 591 181 573 258 381 665 301 589 379 362 146 790 842 765 100 229 916 938 97 340 793 758 177 736 396 247 562 571 92 923 861 165 748 345 703 431 930 101 761 862 595 505 393 126 846 431 103 596 21",
"output": "387"
},
{
"input": "50\n721 631 587 746 692 406 583 90 388 16 161 948 921 70 387 426 39 398 517 724 879 377 906 502 359 950 798 408 846 718 911 845 57 886 9 668 537 632 344 762 19 193 658 447 870 173 98 156 592 519 183 539 274 393 962 615 551 626 148 183 769 763 829 120 796 761 14 744 537 231 696 284 581 688 611 826 703 145 224 600 965 613 791 275 984 375 402 281 851 580 992 8 816 454 35 532 347 250 242 637",
"output": "376"
},
{
"input": "50\n849 475 37 120 754 183 758 374 543 198 896 691 11 607 198 343 761 660 239 669 628 259 223 182 216 158 20 565 454 884 137 923 156 22 310 77 267 707 582 169 120 308 439 309 59 152 206 696 210 177 296 887 559 22 154 553 142 247 491 692 473 572 461 206 532 319 503 164 328 365 541 366 300 392 486 257 863 432 877 404 520 69 418 99 519 239 374 927 601 103 226 316 423 219 240 26 455 101 184 61",
"output": "351"
},
{
"input": "3\n1 2 10 11 100 100",
"output": "1"
},
{
"input": "17\n814 744 145 886 751 1000 272 914 270 529 467 164 410 369 123 424 991 12 702 582 561 858 746 950 598 393 606 498 648 686 455 873 728 858",
"output": "318"
},
{
"input": "45\n476 103 187 696 463 457 588 632 763 77 391 721 95 124 378 812 980 193 694 898 859 572 721 274 605 264 929 615 257 918 42 493 1 3 697 349 990 800 82 535 382 816 943 735 11 272 562 323 653 370 766 332 666 130 704 604 645 717 267 255 37 470 925 941 376 611 332 758 504 40 477 263 708 434 38 596 650 990 714 662 572 467 949 799 648 581 545 828 508 636",
"output": "355"
},
{
"input": "2\n55 5 25 51",
"output": "4"
},
{
"input": "25\n89 50 640 463 858 301 522 241 923 378 892 822 550 17 42 66 706 779 657 840 273 222 444 459 94 925 437 159 182 727 92 851 742 215 653 891 782 533 29 128 133 883 317 475 165 994 802 434 744 973",
"output": "348"
},
{
"input": "4\n35 48 71 44 78 79 57 48",
"output": "10"
},
{
"input": "3\n58 89 73 15 5 47",
"output": "21"
},
{
"input": "2\n1 20 99 100",
"output": "1"
}
] | 1,512,475,259
| 1,859
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 5,529,600
|
n=int(input())
a=sorted(list(map(int,input().split())))
b=[]
for i in range(1,2*n):
b.append(abs(a[i]-a[i-1]))
b.sort()
print(b)
print(sum(b)-b[-1]-b[-2])
|
Title: Kayaking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability!
Input Specification:
The first line contains one number *n* (2<=≤<=*n*<=≤<=50).
The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000).
Output Specification:
Print minimum possible total instability.
Demo Input:
['2\n1 2 3 4\n', '4\n1 3 4 6 3 4 100 200\n']
Demo Output:
['1\n', '5\n']
Note:
none
|
```python
n=int(input())
a=sorted(list(map(int,input().split())))
b=[]
for i in range(1,2*n):
b.append(abs(a[i]-a[i-1]))
b.sort()
print(b)
print(sum(b)-b[-1]-b[-2])
```
| 0
|
|
584
|
A
|
Olesya and Rodion
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.
|
The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by.
|
Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
|
[
"3 2\n"
] |
[
"712"
] |
none
| 500
|
[
{
"input": "3 2",
"output": "222"
},
{
"input": "2 2",
"output": "22"
},
{
"input": "4 3",
"output": "3333"
},
{
"input": "5 3",
"output": "33333"
},
{
"input": "10 7",
"output": "7777777777"
},
{
"input": "2 9",
"output": "99"
},
{
"input": "18 8",
"output": "888888888888888888"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 10",
"output": "-1"
},
{
"input": "100 5",
"output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555"
},
{
"input": "10 2",
"output": "2222222222"
},
{
"input": "18 10",
"output": "111111111111111110"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "7 6",
"output": "6666666"
},
{
"input": "4 4",
"output": "4444"
},
{
"input": "14 7",
"output": "77777777777777"
},
{
"input": "3 8",
"output": "888"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "2 8",
"output": "88"
},
{
"input": "3 8",
"output": "888"
},
{
"input": "4 3",
"output": "3333"
},
{
"input": "5 9",
"output": "99999"
},
{
"input": "4 8",
"output": "8888"
},
{
"input": "3 4",
"output": "444"
},
{
"input": "9 4",
"output": "444444444"
},
{
"input": "8 10",
"output": "11111110"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "20 3",
"output": "33333333333333333333"
},
{
"input": "15 10",
"output": "111111111111110"
},
{
"input": "31 4",
"output": "4444444444444444444444444444444"
},
{
"input": "18 9",
"output": "999999999999999999"
},
{
"input": "72 4",
"output": "444444444444444444444444444444444444444444444444444444444444444444444444"
},
{
"input": "76 8",
"output": "8888888888888888888888888888888888888888888888888888888888888888888888888888"
},
{
"input": "12 5",
"output": "555555555555"
},
{
"input": "54 5",
"output": "555555555555555555555555555555555555555555555555555555"
},
{
"input": "96 10",
"output": "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "15 9",
"output": "999999999999999"
},
{
"input": "100 2",
"output": "2222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222"
},
{
"input": "99 3",
"output": "333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333"
},
{
"input": "98 4",
"output": "44444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444"
},
{
"input": "97 5",
"output": "5555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555"
},
{
"input": "100 6",
"output": "6666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666"
},
{
"input": "99 7",
"output": "777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "98 8",
"output": "88888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888"
},
{
"input": "97 9",
"output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"
},
{
"input": "100 10",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "3 10",
"output": "110"
},
{
"input": "2 4",
"output": "44"
},
{
"input": "2 10",
"output": "10"
},
{
"input": "10 10",
"output": "1111111110"
},
{
"input": "4 10",
"output": "1110"
},
{
"input": "25 10",
"output": "1111111111111111111111110"
},
{
"input": "50 10",
"output": "11111111111111111111111111111111111111111111111110"
},
{
"input": "5 10",
"output": "11110"
},
{
"input": "99 10",
"output": "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "100 10",
"output": "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111110"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "20 10",
"output": "11111111111111111110"
},
{
"input": "99 7",
"output": "777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
},
{
"input": "89 5",
"output": "55555555555555555555555555555555555555555555555555555555555555555555555555555555555555555"
},
{
"input": "6 10",
"output": "111110"
},
{
"input": "3 8",
"output": "888"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "100 7",
"output": "7777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777"
}
] | 1,696,923,545
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 8
| 31
| 0
|
n,t=map(int,input().split())
i=10**(n-1)
while i < 10**n:
if i % t == 0:
break
else:
i+=1
print(i)
|
Title: Olesya and Rodion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.
Input Specification:
The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by.
Output Specification:
Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Demo Input:
['3 2\n']
Demo Output:
['712']
Note:
none
|
```python
n,t=map(int,input().split())
i=10**(n-1)
while i < 10**n:
if i % t == 0:
break
else:
i+=1
print(i)
```
| 0
|
|
32
|
B
|
Borze
|
PROGRAMMING
| 800
|
[
"expression parsing",
"implementation"
] |
B. Borze
|
2
|
256
|
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
|
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
|
Output the decoded ternary number. It can have leading zeroes.
|
[
".-.--\n",
"--.\n",
"-..-.--\n"
] |
[
"012",
"20",
"1012"
] |
none
| 1,000
|
[
{
"input": ".-.--",
"output": "012"
},
{
"input": "--.",
"output": "20"
},
{
"input": "-..-.--",
"output": "1012"
},
{
"input": "---..",
"output": "210"
},
{
"input": "..--.---..",
"output": "0020210"
},
{
"input": "-.....----.",
"output": "10000220"
},
{
"input": ".",
"output": "0"
},
{
"input": "-.",
"output": "1"
},
{
"input": "--",
"output": "2"
},
{
"input": "..",
"output": "00"
},
{
"input": "--.",
"output": "20"
},
{
"input": ".--.",
"output": "020"
},
{
"input": ".-.-..",
"output": "0110"
},
{
"input": "----.-.",
"output": "2201"
},
{
"input": "-..--.-.",
"output": "10201"
},
{
"input": "..--..--.",
"output": "0020020"
},
{
"input": "-.-.---.--..-..-.-.-..-..-.--.",
"output": "112120010111010120"
},
{
"input": "---.-.-.------..-..-..-..-.-..-.--.-.-..-.-.-----..-.-.",
"output": "21112220010101011012011011221011"
},
{
"input": "-.-..--.-.-.-.-.-..-.-.-.---------.--.---..--...--.-----.-.-.-...--.-.-.---.------.--..-.--.-----.-...-..------",
"output": "11020111110111222212021020002022111100201121222020012022110010222"
},
{
"input": "-.-..-.--.---..---.-..---.-...-.-.----..-.---.-.---..-.--.---.-.-------.---.--....----.-.---.---.---.----.-----..---.-.-.-.-----.--.-------.-..",
"output": "110120210211021100112200121121012021122212120000220121212122022102111122120222110"
},
{
"input": ".-..-.-.---.-----.--.---...-.--.-.-....-..",
"output": "01011212212021001201100010"
},
{
"input": ".------.-.---..--...-..-..-.-.-.--.--.-..-.--...-.-.---.-.-.------..--..-.---..----.-..-.--.---.-.----.-.---...-.-.-.-----.-.-.---.---.-.....-.-...-----.-...-.---.-..-.-----.--...---.-.-..-.--.-.---..",
"output": "022201210200010101112020101200011211122200200121022010120211220121001112211121211000011002211001211012212000211101201210"
},
{
"input": ".-.--.---.-----.-.-----.-.-..-----..-..----..--.-.--.----..---.---..-.-.-----..-------.----..----.-..---...-----..-..-----...-..-.-.-----....---..---..-.-----...-.--...--.-.---.-.-.-.-.-...---..----.",
"output": "01202122112211102210102200201202200212101122102221220022010210022101022100101122100021021012210012000201211111100210220"
},
{
"input": "..-.-.-.---.-.-.-..-.-..-.-.---.-------.---..-----.---....-.---.--.--.-.---.---------.-..---.-.-.--..---.---.-.---.-.-..-.-..-.-.-.----.--.-....--------.-.---..----.------.-.-.--.--.-----.-----.----",
"output": "0011121111011011212221210221210001212020121222211021112002121121110110111220201000222201210220222011202022122122"
},
{
"input": "-..-------.------.-..--.-.-..--.-.-..-----..-.-.-..-..-..--.---..-----..---..-..--.-..-.-.---...-.....-------.---.-----.-...-.-...-.-.---.---.-----.--.--...-.--..-.-..-...-.-.-.-.---..---.-..-.-.-.-..",
"output": "102221222010201102011022101110101020210221021010201011210010000222121221100110011212122120200012001101001111210211011110"
},
{
"input": ".-.----.-.--..-.-.-.-..----..-.-...--.-.---.---.-------..-.--..-......--.------.--.----.--...-.--.--..-----..-.....--.--.-.-.------..--------.----------..-.---.----.---.-..--..-.....-..------.--.",
"output": "012201200111102200110020121212221012001000002022202022020001202002210100002020112220022220222220012122021102001000010222020"
},
{
"input": "------.-----.-....--.-.----.-.---.-.-..---.-.---.-----..-...-.-.---..-.-.-..-.-.-...-.-.-.----..--.------.----.-..-.--...-.-------...-.-..-.-.--.--.---..--..--------.--.-.-.---.-.-.-...----.--..-.--..",
"output": "222022110002012201211102112122101001121011101110011122002022202201012000122210011011202021020022220201121111002202001200"
},
{
"input": "-.---...----...--.--...-.--.----",
"output": "121002200020200012022"
},
{
"input": "--.--.--.---.--.-.---.-.-..-..--.-..---.-.....-..---.-----.--...-.-.-------.-.--.-.----.-..-.------.",
"output": "202020212012111010201021100001021221200011222112012201012220"
}
] | 1,646,109,808
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 62
| 0
|
t=int(input())
list=[]
for i in range(t):
li=[]
lis1=[]
lis2=[]
for i in range(t):
i=int(input())
li.append(i)
li2=li.sort()
for i in range(1,t//2):
lis1.append(i)
for i in range(t):
lis2.append(i)
if max(lis1)==li2[-1]and max(lis2)==li2[t-2]:
list.append("YES")
else:list.append("NO")
print(list)
|
Title: Borze
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
Input Specification:
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
Output Specification:
Output the decoded ternary number. It can have leading zeroes.
Demo Input:
['.-.--\n', '--.\n', '-..-.--\n']
Demo Output:
['012', '20', '1012']
Note:
none
|
```python
t=int(input())
list=[]
for i in range(t):
li=[]
lis1=[]
lis2=[]
for i in range(t):
i=int(input())
li.append(i)
li2=li.sort()
for i in range(1,t//2):
lis1.append(i)
for i in range(t):
lis2.append(i)
if max(lis1)==li2[-1]and max(lis2)==li2[t-2]:
list.append("YES")
else:list.append("NO")
print(list)
```
| -1
|
507
|
A
|
Amr and Music
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.
Amr has *n* instruments, it takes *a**i* days to learn *i*-th instrument. Being busy, Amr dedicated *k* days to learn how to play the maximum possible number of instruments.
Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.
|
The first line contains two numbers *n*, *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=10<=000), the number of instruments and number of days respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100), representing number of days required to learn the *i*-th instrument.
|
In the first line output one integer *m* representing the maximum number of instruments Amr can learn.
In the second line output *m* space-separated integers: the indices of instruments to be learnt. You may output indices in any order.
if there are multiple optimal solutions output any. It is not necessary to use all days for studying.
|
[
"4 10\n4 3 1 2\n",
"5 6\n4 3 1 1 2\n",
"1 3\n4\n"
] |
[
"4\n1 2 3 4",
"3\n1 3 4",
"0\n"
] |
In the first test Amr can learn all 4 instruments.
In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.
In the third test Amr doesn't have enough time to learn the only presented instrument.
| 500
|
[
{
"input": "4 10\n4 3 1 2",
"output": "4\n1 2 3 4"
},
{
"input": "5 6\n4 3 1 1 2",
"output": "3\n3 4 5"
},
{
"input": "1 3\n4",
"output": "0"
},
{
"input": "2 100\n100 100",
"output": "1\n1"
},
{
"input": "3 150\n50 50 50",
"output": "3\n1 2 3"
},
{
"input": "4 0\n100 100 100 100",
"output": "0"
},
{
"input": "100 7567\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": "75\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"
},
{
"input": "68 3250\n95 84 67 7 82 75 100 39 31 45 69 100 8 97 13 58 74 40 88 69 35 91 94 28 62 85 51 97 37 15 87 51 24 96 89 49 53 54 35 17 23 54 51 91 94 18 26 92 79 63 23 37 98 43 16 44 82 25 100 59 97 3 60 92 76 58 56 50",
"output": "60\n1 2 3 4 5 6 8 9 10 11 13 15 16 17 18 19 20 21 22 23 24 25 26 27 29 30 31 32 33 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 54 55 56 57 58 60 62 63 64 65 66 67 68"
},
{
"input": "100 10000\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": "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"
},
{
"input": "25 1293\n96 13 7 2 81 72 39 45 5 88 47 23 60 81 54 46 63 52 41 57 2 87 90 28 93",
"output": "25\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"
},
{
"input": "98 7454\n71 57 94 76 52 90 76 81 67 60 99 88 98 61 73 61 80 91 88 93 53 55 88 64 71 55 81 76 52 63 87 99 84 66 65 52 83 99 92 62 95 81 90 67 64 57 80 80 67 75 77 58 71 85 97 50 97 55 52 59 55 96 57 53 85 100 95 95 74 51 78 88 66 98 97 86 94 81 56 64 61 57 67 95 85 82 85 60 76 95 69 95 76 91 74 100 69 76",
"output": "98\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"
},
{
"input": "5 249\n96 13 7 2 81",
"output": "5\n1 2 3 4 5"
},
{
"input": "61 3331\n12 63 99 56 57 70 53 21 41 82 97 63 42 91 18 84 99 78 85 89 6 63 76 28 33 78 100 46 78 78 32 13 11 12 73 50 34 60 12 73 9 19 88 100 28 51 50 45 51 10 78 38 25 22 8 40 71 55 56 83 44",
"output": "61\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"
},
{
"input": "99 10000\n42 88 21 63 59 38 23 100 86 37 57 86 11 22 19 89 6 19 15 64 18 77 83 29 14 26 80 73 8 51 14 19 9 98 81 96 47 77 22 19 86 71 91 61 84 8 80 28 6 25 33 95 96 21 57 92 96 57 31 88 38 32 70 19 25 67 29 78 18 90 37 50 62 33 49 16 47 39 9 33 88 69 69 29 14 66 75 76 41 98 40 52 65 25 33 47 39 24 80",
"output": "99\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"
},
{
"input": "89 4910\n44 9 31 70 85 72 55 9 85 84 63 43 92 85 10 34 83 28 73 45 62 7 34 52 89 58 24 10 28 6 72 45 57 36 71 34 26 24 38 59 5 15 48 82 58 99 8 77 49 84 14 58 29 46 88 50 13 7 58 23 40 63 96 23 46 31 17 8 59 93 12 76 69 20 43 44 91 78 68 94 37 27 100 65 40 25 52 30 97",
"output": "89\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"
},
{
"input": "40 2110\n91 18 52 22 26 67 59 10 55 43 97 78 20 81 99 36 33 12 86 32 82 87 70 63 48 48 45 94 78 23 77 15 68 17 71 54 44 98 54 8",
"output": "39\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 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"
},
{
"input": "27 1480\n38 95 9 36 21 70 19 89 35 46 7 31 88 25 10 72 81 32 65 83 68 57 50 20 73 42 12",
"output": "27\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"
},
{
"input": "57 2937\n84 73 23 62 93 64 23 17 53 100 47 67 52 53 90 58 19 84 33 69 46 47 50 28 73 74 40 42 92 70 32 29 57 52 23 82 42 32 46 83 45 87 40 58 50 51 48 37 57 52 78 26 21 54 16 66 93",
"output": "55\n1 2 3 4 5 6 7 8 9 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"
},
{
"input": "6 41\n6 8 9 8 9 8",
"output": "5\n1 2 3 4 6"
},
{
"input": "9 95\n9 11 12 11 12 11 8 11 10",
"output": "9\n1 2 3 4 5 6 7 8 9"
},
{
"input": "89 6512\n80 87 61 91 85 51 58 69 79 57 81 67 74 55 88 70 77 61 55 81 56 76 79 67 92 52 54 73 67 72 81 54 72 81 65 88 83 57 83 92 62 66 63 58 61 66 92 77 73 66 71 85 92 73 82 65 76 64 58 62 64 51 90 59 79 70 86 89 86 51 72 61 60 71 52 74 58 72 77 91 91 60 76 56 64 55 61 81 52",
"output": "89\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"
},
{
"input": "5 29\n6 3 7 2 1",
"output": "5\n1 2 3 4 5"
},
{
"input": "5 49\n16 13 7 2 1",
"output": "5\n1 2 3 4 5"
},
{
"input": "6 84\n16 21 25 6 17 16",
"output": "5\n1 2 4 5 6"
},
{
"input": "4 9\n7 4 2 1",
"output": "3\n2 3 4"
},
{
"input": "50 2500\n50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50 50",
"output": "50\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"
},
{
"input": "100 10000\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": "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"
},
{
"input": "100 100\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": "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"
},
{
"input": "96 514\n6 3 7 2 1 2 9 5 5 8 7 3 10 1 4 6 3 2 1 7 2 7 10 8 3 8 10 4 8 8 2 5 3 2 1 4 4 8 4 3 3 7 4 4 2 7 8 3 9 2 2 6 3 4 8 6 7 5 4 3 10 7 6 5 10 1 7 10 7 7 8 2 1 2 3 10 9 8 8 2 7 1 2 7 10 1 2 2 3 8 6 2 9 6 9 6",
"output": "96\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"
},
{
"input": "47 350\n6 1 9 12 8 8 11 4 4 8 8 3 3 2 12 7 7 7 12 2 9 1 5 10 6 1 5 2 6 3 9 13 8 3 10 10 10 10 6 9 10 10 8 5 12 11 3",
"output": "47\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"
},
{
"input": "100 200\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": "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"
},
{
"input": "2 10000\n1 1",
"output": "2\n1 2"
},
{
"input": "1 2\n1",
"output": "1\n1"
},
{
"input": "1 3\n2",
"output": "1\n1"
},
{
"input": "34 4964\n37 27 90 83 36 59 80 7 28 41 97 72 64 8 40 30 76 4 92 51 52 44 42 13 38 64 60 66 47 93 30 35 71 71",
"output": "34\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"
},
{
"input": "2 2\n1 10",
"output": "1\n1"
},
{
"input": "2 5\n1 1",
"output": "2\n1 2"
},
{
"input": "1 4\n3",
"output": "1\n1"
},
{
"input": "4 384\n1 2 3 4",
"output": "4\n1 2 3 4"
}
] | 1,664,540,437
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 39
| 62
| 0
|
n, k = list(map(int, input().split()))
values = [(value, pos) for pos, value in enumerate(map(int, input().split()), start=1)]
values.sort()
x,ans = 0, []
while (x < n) and (values[x][0]) <= k:
ans.append(values[x][1])
k -= values[x][0]
x += 1
print(len(ans))
print(*ans)
|
Title: Amr and Music
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.
Amr has *n* instruments, it takes *a**i* days to learn *i*-th instrument. Being busy, Amr dedicated *k* days to learn how to play the maximum possible number of instruments.
Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.
Input Specification:
The first line contains two numbers *n*, *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=10<=000), the number of instruments and number of days respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100), representing number of days required to learn the *i*-th instrument.
Output Specification:
In the first line output one integer *m* representing the maximum number of instruments Amr can learn.
In the second line output *m* space-separated integers: the indices of instruments to be learnt. You may output indices in any order.
if there are multiple optimal solutions output any. It is not necessary to use all days for studying.
Demo Input:
['4 10\n4 3 1 2\n', '5 6\n4 3 1 1 2\n', '1 3\n4\n']
Demo Output:
['4\n1 2 3 4', '3\n1 3 4', '0\n']
Note:
In the first test Amr can learn all 4 instruments.
In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.
In the third test Amr doesn't have enough time to learn the only presented instrument.
|
```python
n, k = list(map(int, input().split()))
values = [(value, pos) for pos, value in enumerate(map(int, input().split()), start=1)]
values.sort()
x,ans = 0, []
while (x < n) and (values[x][0]) <= k:
ans.append(values[x][1])
k -= values[x][0]
x += 1
print(len(ans))
print(*ans)
```
| 3
|
|
622
|
A
|
Infinite Sequence
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Consider the infinite sequence of integers: 1,<=1,<=2,<=1,<=2,<=3,<=1,<=2,<=3,<=4,<=1,<=2,<=3,<=4,<=5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the *n*-th position of the sequence.
|
The only line contains integer *n* (1<=≤<=*n*<=≤<=1014) — the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
|
Print the element in the *n*-th position of the sequence (the elements are numerated from one).
|
[
"3\n",
"5\n",
"10\n",
"55\n",
"56\n"
] |
[
"2\n",
"2\n",
"4\n",
"10\n",
"1\n"
] |
none
| 0
|
[
{
"input": "3",
"output": "2"
},
{
"input": "5",
"output": "2"
},
{
"input": "10",
"output": "4"
},
{
"input": "55",
"output": "10"
},
{
"input": "56",
"output": "1"
},
{
"input": "1000000000000",
"output": "88209"
},
{
"input": "847194127849",
"output": "255708"
},
{
"input": "294719472984",
"output": "593358"
},
{
"input": "999999911791",
"output": "1414213"
},
{
"input": "999999911792",
"output": "1"
},
{
"input": "100000000000000",
"output": "1749820"
},
{
"input": "1",
"output": "1"
},
{
"input": "99993",
"output": "312"
},
{
"input": "99994",
"output": "313"
},
{
"input": "99995",
"output": "314"
},
{
"input": "99990",
"output": "309"
},
{
"input": "2",
"output": "1"
},
{
"input": "99991",
"output": "310"
},
{
"input": "99992",
"output": "311"
},
{
"input": "99996",
"output": "315"
},
{
"input": "99997",
"output": "316"
},
{
"input": "99998",
"output": "317"
},
{
"input": "99999",
"output": "318"
},
{
"input": "1021",
"output": "31"
},
{
"input": "4",
"output": "1"
},
{
"input": "23",
"output": "2"
},
{
"input": "9994",
"output": "124"
},
{
"input": "99939",
"output": "258"
},
{
"input": "99999998250180",
"output": "14142135"
},
{
"input": "6",
"output": "3"
},
{
"input": "8",
"output": "2"
},
{
"input": "35",
"output": "7"
},
{
"input": "100",
"output": "9"
},
{
"input": "10101010",
"output": "745"
},
{
"input": "103",
"output": "12"
},
{
"input": "102",
"output": "11"
},
{
"input": "101",
"output": "10"
}
] | 1,455,641,373
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
import sys
# sys.stdin = open('input.txt')
# sys.stdout = open('output.txt', 'w')
def main():
n = int(input())
tl, tr = 0, n
while tr - tl > 1:
tm = tl + (tr - tl) / 2
s = (1 + tm) * tm / 2
if s >= n:
tr = tm
else:
tl = tm
s = (1 + tr) * tr / 2
s -= tr
n -= s
print(n)
main()
|
Title: Infinite Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Consider the infinite sequence of integers: 1,<=1,<=2,<=1,<=2,<=3,<=1,<=2,<=3,<=4,<=1,<=2,<=3,<=4,<=5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the *n*-th position of the sequence.
Input Specification:
The only line contains integer *n* (1<=≤<=*n*<=≤<=1014) — the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output Specification:
Print the element in the *n*-th position of the sequence (the elements are numerated from one).
Demo Input:
['3\n', '5\n', '10\n', '55\n', '56\n']
Demo Output:
['2\n', '2\n', '4\n', '10\n', '1\n']
Note:
none
|
```python
import sys
# sys.stdin = open('input.txt')
# sys.stdout = open('output.txt', 'w')
def main():
n = int(input())
tl, tr = 0, n
while tr - tl > 1:
tm = tl + (tr - tl) / 2
s = (1 + tm) * tm / 2
if s >= n:
tr = tm
else:
tl = tm
s = (1 + tr) * tr / 2
s -= tr
n -= s
print(n)
main()
```
| 0
|
|
697
|
B
|
Barnicle
|
PROGRAMMING
| 1,400
|
[
"brute force",
"implementation",
"math",
"strings"
] | null | null |
Barney is standing in a bar and starring at a pretty girl. He wants to shoot her with his heart arrow but he needs to know the distance between him and the girl to make his shot accurate.
Barney asked the bar tender Carl about this distance value, but Carl was so busy talking to the customers so he wrote the distance value (it's a real number) on a napkin. The problem is that he wrote it in scientific notation. The scientific notation of some real number *x* is the notation of form *AeB*, where *A* is a real number and *B* is an integer and *x*<==<=*A*<=×<=10*B* is true. In our case *A* is between 0 and 9 and *B* is non-negative.
Barney doesn't know anything about scientific notation (as well as anything scientific at all). So he asked you to tell him the distance value in usual decimal representation with minimal number of digits after the decimal point (and no decimal point if it is an integer). See the output format for better understanding.
|
The first and only line of input contains a single string of form *a*.*deb* where *a*, *d* and *b* are integers and *e* is usual character 'e' (0<=≤<=*a*<=≤<=9,<=0<=≤<=*d*<=<<=10100,<=0<=≤<=*b*<=≤<=100) — the scientific notation of the desired distance value.
*a* and *b* contain no leading zeros and *d* contains no trailing zeros (but may be equal to 0). Also, *b* can not be non-zero if *a* is zero.
|
Print the only real number *x* (the desired distance value) in the only line in its decimal notation.
Thus if *x* is an integer, print it's integer value without decimal part and decimal point and without leading zeroes.
Otherwise print *x* in a form of *p*.*q* such that *p* is an integer that have no leading zeroes (but may be equal to zero), and *q* is an integer that have no trailing zeroes (and may not be equal to zero).
|
[
"8.549e2\n",
"8.549e3\n",
"0.33e0\n"
] |
[
"854.9\n",
"8549\n",
"0.33\n"
] |
none
| 1,000
|
[
{
"input": "8.549e2",
"output": "854.9"
},
{
"input": "8.549e3",
"output": "8549"
},
{
"input": "0.33e0",
"output": "0.33"
},
{
"input": "1.31e1",
"output": "13.1"
},
{
"input": "1.038e0",
"output": "1.038"
},
{
"input": "8.25983e5",
"output": "825983"
},
{
"input": "8.77056e6",
"output": "8770560"
},
{
"input": "4.28522890224373996236468418851564462623381500262405e30",
"output": "4285228902243739962364684188515.64462623381500262405"
},
{
"input": "4.09336275522154223604344399571355118601483591618747e85",
"output": "40933627552215422360434439957135511860148359161874700000000000000000000000000000000000"
},
{
"input": "2.0629094807595491132306264747042243928486303384791951220362096240931158821630792563855724946791054152e85",
"output": "20629094807595491132306264747042243928486303384791951220362096240931158821630792563855.724946791054152"
},
{
"input": "0.7e0",
"output": "0.7"
},
{
"input": "0.75e0",
"output": "0.75"
},
{
"input": "0.3299209894804593859495773277850971828150469972132991597085582244596065712639531451e0",
"output": "0.3299209894804593859495773277850971828150469972132991597085582244596065712639531451"
},
{
"input": "0.1438410315232821898580886049593487999249997483354329425897344341660326482795266134253882860655873197e0",
"output": "0.1438410315232821898580886049593487999249997483354329425897344341660326482795266134253882860655873197"
},
{
"input": "1.7282220592677586155528202123627915992640276211396528871e0",
"output": "1.7282220592677586155528202123627915992640276211396528871"
},
{
"input": "1.91641639840522198229453882518758458881136053577016034847369545687354908120008812644841021662133251e89",
"output": "191641639840522198229453882518758458881136053577016034847369545687354908120008812644841021.662133251"
},
{
"input": "7.0e100",
"output": "70000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "1.7390193766535948887334396973270576641602486903095355363287177932797263236084900516267835886881779051e100",
"output": "17390193766535948887334396973270576641602486903095355363287177932797263236084900516267835886881779051"
},
{
"input": "4.6329496401734172195e50",
"output": "463294964017341721950000000000000000000000000000000"
},
{
"input": "2.806303180541991592302230754797823269634e39",
"output": "2806303180541991592302230754797823269634"
},
{
"input": "5.8743505652112692964508303637002e64",
"output": "58743505652112692964508303637002000000000000000000000000000000000"
},
{
"input": "6.8778661934058405217475274375560252344373481358834598914724956711e31",
"output": "68778661934058405217475274375560.252344373481358834598914724956711"
},
{
"input": "9.4e100",
"output": "94000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "3.2371070627618799335840070613481911588919091676203766004638236894609230433739617153911544972468224113e50",
"output": "323710706276187993358400706134819115889190916762037.66004638236894609230433739617153911544972468224113"
},
{
"input": "4.8133196117786711780806656271869913331127534865038175322117213586960112955982462632332925275690064929e0",
"output": "4.8133196117786711780806656271869913331127534865038175322117213586960112955982462632332925275690064929"
},
{
"input": "7.7060200967648284035308242369118752594772564843152902469146249303976625961451358536989314351204406625e1",
"output": "77.060200967648284035308242369118752594772564843152902469146249303976625961451358536989314351204406625"
},
{
"input": "8.1089882894234341219420177467603732503076124872188628349726911362800974096687340341040683238197289136e31",
"output": "81089882894234341219420177467603.732503076124872188628349726911362800974096687340341040683238197289136"
},
{
"input": "9.6576660076120385279859051742522204516365367878315639937449558670629833997839913220859648564428655877e99",
"output": "9657666007612038527985905174252220451636536787831563993744955867062983399783991322085964856442865587.7"
},
{
"input": "0.0e0",
"output": "0"
},
{
"input": "1.0e0",
"output": "1"
},
{
"input": "8.0e0",
"output": "8"
},
{
"input": "3.0e0",
"output": "3"
},
{
"input": "4.0e0",
"output": "4"
},
{
"input": "2.0e0",
"output": "2"
},
{
"input": "9.0e0",
"output": "9"
},
{
"input": "0.888888e0",
"output": "0.888888"
},
{
"input": "9.99999999999999999999999999999999999999999999999999999999999999999999999999999999e100",
"output": "99999999999999999999999999999999999999999999999999999999999999999999999999999999900000000000000000000"
},
{
"input": "5.0e0",
"output": "5"
},
{
"input": "1.0e10",
"output": "10000000000"
},
{
"input": "1.0e5",
"output": "100000"
},
{
"input": "6.0e0",
"output": "6"
},
{
"input": "1.1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111e1",
"output": "11.111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111"
}
] | 1,651,834,540
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 42
| 62
| 409,600
|
import re
s = input()
d = re.findall(".\d+e", s)[0][1:-1] + "0" * 100
a = re.findall("\d+.", s)[0][:-1]
b = int(re.findall("e\d+", s)[0][1:])
r = a + d[:b] + "." + d[b:]
r = r.replace("0", ' ').strip().replace(' ', '0')
if r[0] == '.':
r = "0" + r
if r[-1] == '.':
r = r[:-1]
print(r)
|
Title: Barnicle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Barney is standing in a bar and starring at a pretty girl. He wants to shoot her with his heart arrow but he needs to know the distance between him and the girl to make his shot accurate.
Barney asked the bar tender Carl about this distance value, but Carl was so busy talking to the customers so he wrote the distance value (it's a real number) on a napkin. The problem is that he wrote it in scientific notation. The scientific notation of some real number *x* is the notation of form *AeB*, where *A* is a real number and *B* is an integer and *x*<==<=*A*<=×<=10*B* is true. In our case *A* is between 0 and 9 and *B* is non-negative.
Barney doesn't know anything about scientific notation (as well as anything scientific at all). So he asked you to tell him the distance value in usual decimal representation with minimal number of digits after the decimal point (and no decimal point if it is an integer). See the output format for better understanding.
Input Specification:
The first and only line of input contains a single string of form *a*.*deb* where *a*, *d* and *b* are integers and *e* is usual character 'e' (0<=≤<=*a*<=≤<=9,<=0<=≤<=*d*<=<<=10100,<=0<=≤<=*b*<=≤<=100) — the scientific notation of the desired distance value.
*a* and *b* contain no leading zeros and *d* contains no trailing zeros (but may be equal to 0). Also, *b* can not be non-zero if *a* is zero.
Output Specification:
Print the only real number *x* (the desired distance value) in the only line in its decimal notation.
Thus if *x* is an integer, print it's integer value without decimal part and decimal point and without leading zeroes.
Otherwise print *x* in a form of *p*.*q* such that *p* is an integer that have no leading zeroes (but may be equal to zero), and *q* is an integer that have no trailing zeroes (and may not be equal to zero).
Demo Input:
['8.549e2\n', '8.549e3\n', '0.33e0\n']
Demo Output:
['854.9\n', '8549\n', '0.33\n']
Note:
none
|
```python
import re
s = input()
d = re.findall(".\d+e", s)[0][1:-1] + "0" * 100
a = re.findall("\d+.", s)[0][:-1]
b = int(re.findall("e\d+", s)[0][1:])
r = a + d[:b] + "." + d[b:]
r = r.replace("0", ' ').strip().replace(' ', '0')
if r[0] == '.':
r = "0" + r
if r[-1] == '.':
r = r[:-1]
print(r)
```
| 3
|
|
20
|
C
|
Dijkstra?
|
PROGRAMMING
| 1,900
|
[
"graphs",
"shortest paths"
] |
C. Dijkstra?
|
1
|
64
|
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
|
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
|
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
|
[
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n",
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n"
] |
[
"1 4 3 5 ",
"1 4 3 5 "
] |
none
| 1,500
|
[
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "2 1\n1 2 1",
"output": "1 2 "
},
{
"input": "3 1\n1 2 1",
"output": "-1"
},
{
"input": "3 3\n1 2 1\n1 3 2\n2 3 1",
"output": "1 3 "
},
{
"input": "10 10\n1 5 12\n2 4 140\n2 10 149\n3 6 154\n3 7 9\n3 8 226\n3 10 132\n4 10 55\n5 8 33\n7 8 173",
"output": "1 5 8 7 3 10 "
},
{
"input": "10 10\n1 5 178\n1 8 221\n2 7 92\n2 8 159\n3 5 55\n3 6 179\n3 10 237\n4 8 205\n5 6 191\n8 10 157",
"output": "1 8 10 "
},
{
"input": "10 10\n1 4 200\n1 9 197\n3 4 79\n3 5 213\n3 6 149\n5 8 3\n5 9 189\n6 7 130\n6 9 51\n8 10 135",
"output": "1 9 5 8 10 "
},
{
"input": "10 10\n1 4 201\n2 3 238\n3 4 40\n3 6 231\n3 8 45\n4 5 227\n4 6 58\n4 9 55\n5 7 14\n6 10 242",
"output": "1 4 6 10 "
}
] | 1,568,934,999
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 30
| 1,000
| 45,772,800
|
import heapq
n = 0
graph = []
dist = {}
prev = {}
def dijk(t):
global graph
global dist
global prev
seen = set()
q = [(0, 1)]
heapq.heapify(q)
dist = {}
dist[1] = 0
while len(q) != 0:
d, u = heapq.heappop(q)
if d == float('inf') or u == t:
return
if u in seen:
continue
seen.add(u)
for v, v_len in graph[u].items():
d_ = v_len + d
if v not in dist:
dist[v] = float("inf")
if d_ < dist[v]:
dist[v] = d_
prev[v] = u
if v not in seen:
heapq.heappush(q, (d_, v))
def main():
global n
n, m = map(int, input().split())
for i in range(n+1):
graph.append({})
for i in range(m):
u,v,w = map(int, input().split())
if u == v:
continue
if v in graph[u]:
graph[u][v] = min(graph[u][v], w)
else:
graph[u][v] = w
graph[v][u] = graph[u][v]
dijk(n)
if n not in prev:
print(-1)
return
p = prev[n]
l = [n]
while p != 1:
l.append(p)
p = prev[p]
l.append(1)
print(' '.join(map(str, l[::-1])))
main()
|
Title: Dijkstra?
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
Output Specification:
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
Demo Input:
['5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n', '5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n']
Demo Output:
['1 4 3 5 ', '1 4 3 5 ']
Note:
none
|
```python
import heapq
n = 0
graph = []
dist = {}
prev = {}
def dijk(t):
global graph
global dist
global prev
seen = set()
q = [(0, 1)]
heapq.heapify(q)
dist = {}
dist[1] = 0
while len(q) != 0:
d, u = heapq.heappop(q)
if d == float('inf') or u == t:
return
if u in seen:
continue
seen.add(u)
for v, v_len in graph[u].items():
d_ = v_len + d
if v not in dist:
dist[v] = float("inf")
if d_ < dist[v]:
dist[v] = d_
prev[v] = u
if v not in seen:
heapq.heappush(q, (d_, v))
def main():
global n
n, m = map(int, input().split())
for i in range(n+1):
graph.append({})
for i in range(m):
u,v,w = map(int, input().split())
if u == v:
continue
if v in graph[u]:
graph[u][v] = min(graph[u][v], w)
else:
graph[u][v] = w
graph[v][u] = graph[u][v]
dijk(n)
if n not in prev:
print(-1)
return
p = prev[n]
l = [n]
while p != 1:
l.append(p)
p = prev[p]
l.append(1)
print(' '.join(map(str, l[::-1])))
main()
```
| 0
|
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"
}
] | 1,699,740,574
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 15
| 46
| 0
|
k, n, w = input().split()
k = int(k)
n = int(n)
w = int(w)
w = w*(w+1)/2
w = w*k
if n < w:
print(int(w-n))
else:
print(0)
|
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 = input().split()
k = int(k)
n = int(n)
w = int(w)
w = w*(w+1)/2
w = w*k
if n < w:
print(int(w-n))
else:
print(0)
```
| 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"
}
] | 1,532,929,163
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 4
| 216
| 0
|
a=str(input())
c=True
k=0
for i in range(len(a)):
if a[i].isupper():
c
k+=1
else:
n=len(a)
b=n-k
if b>k:
print(a.lower())
elif b<k:
print(a.upper())
else:
print(a.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=str(input())
c=True
k=0
for i in range(len(a)):
if a[i].isupper():
c
k+=1
else:
n=len(a)
b=n-k
if b>k:
print(a.lower())
elif b<k:
print(a.upper())
else:
print(a.lower())
```
| -1
|
475
|
D
|
CGCDSSQ
|
PROGRAMMING
| 2,000
|
[
"brute force",
"data structures",
"math"
] | null | null |
Given a sequence of integers *a*1,<=...,<=*a**n* and *q* queries *x*1,<=...,<=*x**q* on it. For each query *x**i* you have to count the number of pairs (*l*,<=*r*) such that 1<=≤<=*l*<=≤<=*r*<=≤<=*n* and *gcd*(*a**l*,<=*a**l*<=+<=1,<=...,<=*a**r*)<==<=*x**i*.
is a greatest common divisor of *v*1,<=*v*2,<=...,<=*v**n*, that is equal to a largest positive integer that divides all *v**i*.
|
The first line of the input contains integer *n*, (1<=≤<=*n*<=≤<=105), denoting the length of the sequence. The next line contains *n* space separated integers *a*1,<=...,<=*a**n*, (1<=≤<=*a**i*<=≤<=109).
The third line of the input contains integer *q*, (1<=≤<=*q*<=≤<=3<=×<=105), denoting the number of queries. Then follows *q* lines, each contain an integer *x**i*, (1<=≤<=*x**i*<=≤<=109).
|
For each query print the result in a separate line.
|
[
"3\n2 6 3\n5\n1\n2\n3\n4\n6\n",
"7\n10 20 3 15 1000 60 16\n10\n1\n2\n3\n4\n5\n6\n10\n20\n60\n1000\n"
] |
[
"1\n2\n2\n0\n1\n",
"14\n0\n2\n2\n2\n0\n2\n2\n1\n1\n"
] |
none
| 2,000
|
[
{
"input": "3\n2 6 3\n5\n1\n2\n3\n4\n6",
"output": "1\n2\n2\n0\n1"
},
{
"input": "7\n10 20 3 15 1000 60 16\n10\n1\n2\n3\n4\n5\n6\n10\n20\n60\n1000",
"output": "14\n0\n2\n2\n2\n0\n2\n2\n1\n1"
},
{
"input": "10\n2 2 4 3 2 4 4 2 4 2\n104\n3\n3\n1\n4\n1\n1\n4\n1\n1\n3\n1\n1\n4\n1\n1\n1\n4\n3\n1\n1\n4\n1\n1\n1\n1\n1\n4\n1\n1\n1\n4\n1\n1\n4\n1\n1\n1\n1\n1\n4\n4\n1\n3\n1\n4\n1\n1\n1\n4\n1\n2\n4\n1\n4\n1\n4\n1\n4\n3\n1\n2\n2\n4\n2\n1\n1\n2\n4\n4\n1\n2\n3\n1\n1\n4\n4\n4\n4\n4\n2\n2\n4\n1\n1\n1\n1\n4\n2\n1\n1\n4\n1\n4\n3\n4\n4\n1\n2\n3\n1\n2\n1\n3\n1",
"output": "1\n1\n27\n5\n27\n27\n5\n27\n27\n1\n27\n27\n5\n27\n27\n27\n5\n1\n27\n27\n5\n27\n27\n27\n27\n27\n5\n27\n27\n27\n5\n27\n27\n5\n27\n27\n27\n27\n27\n5\n5\n27\n1\n27\n5\n27\n27\n27\n5\n27\n22\n5\n27\n5\n27\n5\n27\n5\n1\n27\n22\n22\n5\n22\n27\n27\n22\n5\n5\n27\n22\n1\n27\n27\n5\n5\n5\n5\n5\n22\n22\n5\n27\n27\n27\n27\n5\n22\n27\n27\n5\n27\n5\n1\n5\n5\n27\n22\n1\n27\n22\n27\n1\n27"
}
] | 1,634,470,320
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 67
| 2,000
| 21,401,600
|
from collections import defaultdict
def gcd(a, b):
while b:
a %= b
a, b = b, a
return a
def test():
n = int(input())
nums = list(map(int, input().split()))
d = defaultdict(int)
ans = defaultdict(int)
d[nums[0]] = 1
ans[nums[0]] = 1
for i in range(1, n):
t = defaultdict(int)
t[nums[i]] = 1
ans[nums[i]] += 1
for k, v in d.items():
g = gcd(k, nums[i])
ans[g] += v
t[g] += v
d = t
num_queries = int(input())
for i in range(num_queries):
print(ans[int(input())])
if __name__ == '__main__': test()
|
Title: CGCDSSQ
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Given a sequence of integers *a*1,<=...,<=*a**n* and *q* queries *x*1,<=...,<=*x**q* on it. For each query *x**i* you have to count the number of pairs (*l*,<=*r*) such that 1<=≤<=*l*<=≤<=*r*<=≤<=*n* and *gcd*(*a**l*,<=*a**l*<=+<=1,<=...,<=*a**r*)<==<=*x**i*.
is a greatest common divisor of *v*1,<=*v*2,<=...,<=*v**n*, that is equal to a largest positive integer that divides all *v**i*.
Input Specification:
The first line of the input contains integer *n*, (1<=≤<=*n*<=≤<=105), denoting the length of the sequence. The next line contains *n* space separated integers *a*1,<=...,<=*a**n*, (1<=≤<=*a**i*<=≤<=109).
The third line of the input contains integer *q*, (1<=≤<=*q*<=≤<=3<=×<=105), denoting the number of queries. Then follows *q* lines, each contain an integer *x**i*, (1<=≤<=*x**i*<=≤<=109).
Output Specification:
For each query print the result in a separate line.
Demo Input:
['3\n2 6 3\n5\n1\n2\n3\n4\n6\n', '7\n10 20 3 15 1000 60 16\n10\n1\n2\n3\n4\n5\n6\n10\n20\n60\n1000\n']
Demo Output:
['1\n2\n2\n0\n1\n', '14\n0\n2\n2\n2\n0\n2\n2\n1\n1\n']
Note:
none
|
```python
from collections import defaultdict
def gcd(a, b):
while b:
a %= b
a, b = b, a
return a
def test():
n = int(input())
nums = list(map(int, input().split()))
d = defaultdict(int)
ans = defaultdict(int)
d[nums[0]] = 1
ans[nums[0]] = 1
for i in range(1, n):
t = defaultdict(int)
t[nums[i]] = 1
ans[nums[i]] += 1
for k, v in d.items():
g = gcd(k, nums[i])
ans[g] += v
t[g] += v
d = t
num_queries = int(input())
for i in range(num_queries):
print(ans[int(input())])
if __name__ == '__main__': test()
```
| 0
|
|
658
|
A
|
Bear and Reverse Radewoosh
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order.
There will be *n* problems. The *i*-th problem has initial score *p**i* and it takes exactly *t**i* minutes to solve it. Problems are sorted by difficulty — it's guaranteed that *p**i*<=<<=*p**i*<=+<=1 and *t**i*<=<<=*t**i*<=+<=1.
A constant *c* is given too, representing the speed of loosing points. Then, submitting the *i*-th problem at time *x* (*x* minutes after the start of the contest) gives *max*(0,<= *p**i*<=-<=*c*·*x*) points.
Limak is going to solve problems in order 1,<=2,<=...,<=*n* (sorted increasingly by *p**i*). Radewoosh is going to solve them in order *n*,<=*n*<=-<=1,<=...,<=1 (sorted decreasingly by *p**i*). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie.
You may assume that the duration of the competition is greater or equal than the sum of all *t**i*. That means both Limak and Radewoosh will accept all *n* problems.
|
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=50,<=1<=≤<=*c*<=≤<=1000) — the number of problems and the constant representing the speed of loosing points.
The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=1000,<=*p**i*<=<<=*p**i*<=+<=1) — initial scores.
The third line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000,<=*t**i*<=<<=*t**i*<=+<=1) where *t**i* denotes the number of minutes one needs to solve the *i*-th problem.
|
Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points.
|
[
"3 2\n50 85 250\n10 15 25\n",
"3 6\n50 85 250\n10 15 25\n",
"8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76\n"
] |
[
"Limak\n",
"Radewoosh\n",
"Tie\n"
] |
In the first sample, there are 3 problems. Limak solves them as follows:
1. Limak spends 10 minutes on the 1-st problem and he gets 50 - *c*·10 = 50 - 2·10 = 30 points. 1. Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points. 1. He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points.
So, Limak got 30 + 35 + 150 = 215 points.
Radewoosh solves problem in the reversed order:
1. Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points. 1. He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem. 1. He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets *max*(0, 50 - 2·50) = *max*(0, - 50) = 0 points.
Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins.
In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway.
In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4.
| 500
|
[
{
"input": "3 2\n50 85 250\n10 15 25",
"output": "Limak"
},
{
"input": "3 6\n50 85 250\n10 15 25",
"output": "Radewoosh"
},
{
"input": "8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76",
"output": "Tie"
},
{
"input": "4 1\n3 5 6 9\n1 2 4 8",
"output": "Limak"
},
{
"input": "4 1\n1 3 6 10\n1 5 7 8",
"output": "Radewoosh"
},
{
"input": "4 1\n2 4 5 10\n2 3 9 10",
"output": "Tie"
},
{
"input": "18 4\n68 97 121 132 146 277 312 395 407 431 458 461 595 634 751 855 871 994\n1 2 3 4 9 10 13 21 22 29 31 34 37 38 39 41 48 49",
"output": "Radewoosh"
},
{
"input": "50 1\n5 14 18 73 137 187 195 197 212 226 235 251 262 278 287 304 310 322 342 379 393 420 442 444 448 472 483 485 508 515 517 523 559 585 618 627 636 646 666 682 703 707 780 853 937 951 959 989 991 992\n30 84 113 173 199 220 235 261 266 277 300 306 310 312 347 356 394 396 397 409 414 424 446 462 468 487 507 517 537 566 594 643 656 660 662 668 706 708 773 774 779 805 820 827 868 896 929 942 961 995",
"output": "Tie"
},
{
"input": "4 1\n4 6 9 10\n2 3 4 5",
"output": "Radewoosh"
},
{
"input": "4 1\n4 6 9 10\n3 4 5 7",
"output": "Radewoosh"
},
{
"input": "4 1\n1 6 7 10\n2 7 8 10",
"output": "Tie"
},
{
"input": "4 1\n4 5 7 9\n1 4 5 8",
"output": "Limak"
},
{
"input": "50 1\n6 17 44 82 94 127 134 156 187 211 212 252 256 292 294 303 352 355 379 380 398 409 424 434 480 524 584 594 631 714 745 756 777 778 789 793 799 821 841 849 859 878 879 895 925 932 944 952 958 990\n15 16 40 42 45 71 99 100 117 120 174 181 186 204 221 268 289 332 376 394 403 409 411 444 471 487 499 539 541 551 567 589 619 623 639 669 689 722 735 776 794 822 830 840 847 907 917 927 936 988",
"output": "Radewoosh"
},
{
"input": "50 10\n25 49 52 73 104 117 127 136 149 164 171 184 226 251 257 258 286 324 337 341 386 390 428 453 464 470 492 517 543 565 609 634 636 660 678 693 710 714 729 736 739 749 781 836 866 875 956 960 977 979\n2 4 7 10 11 22 24 26 27 28 31 35 37 38 42 44 45 46 52 53 55 56 57 59 60 61 64 66 67 68 69 71 75 76 77 78 79 81 83 85 86 87 89 90 92 93 94 98 99 100",
"output": "Limak"
},
{
"input": "50 10\n11 15 25 71 77 83 95 108 143 150 182 183 198 203 213 223 279 280 346 348 350 355 375 376 412 413 415 432 470 545 553 562 589 595 607 633 635 637 688 719 747 767 771 799 842 883 905 924 942 944\n1 3 5 6 7 10 11 12 13 14 15 16 19 20 21 23 25 32 35 36 37 38 40 41 42 43 47 50 51 54 55 56 57 58 59 60 62 63 64 65 66 68 69 70 71 72 73 75 78 80",
"output": "Radewoosh"
},
{
"input": "32 6\n25 77 141 148 157 159 192 196 198 244 245 255 332 392 414 457 466 524 575 603 629 700 738 782 838 841 845 847 870 945 984 985\n1 2 4 5 8 9 10 12 13 14 15 16 17 18 20 21 22 23 24 26 28 31 38 39 40 41 42 43 45 47 48 49",
"output": "Radewoosh"
},
{
"input": "5 1\n256 275 469 671 842\n7 9 14 17 26",
"output": "Limak"
},
{
"input": "2 1000\n1 2\n1 2",
"output": "Tie"
},
{
"input": "3 1\n1 50 809\n2 8 800",
"output": "Limak"
},
{
"input": "1 13\n866\n10",
"output": "Tie"
},
{
"input": "15 1\n9 11 66 128 199 323 376 386 393 555 585 718 935 960 971\n3 11 14 19 20 21 24 26 32 38 40 42 44 47 50",
"output": "Limak"
},
{
"input": "1 10\n546\n45",
"output": "Tie"
},
{
"input": "50 20\n21 43 51 99 117 119 158 167 175 190 196 244 250 316 335 375 391 403 423 428 451 457 460 480 487 522 539 559 566 584 598 602 604 616 626 666 675 730 771 787 828 841 861 867 886 889 898 970 986 991\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",
"output": "Limak"
},
{
"input": "50 21\n13 20 22 38 62 84 118 135 141 152 170 175 194 218 227 229 232 253 260 263 278 313 329 357 396 402 422 452 454 533 575 576 580 594 624 644 653 671 676 759 789 811 816 823 831 833 856 924 933 987\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",
"output": "Tie"
},
{
"input": "1 36\n312\n42",
"output": "Tie"
},
{
"input": "1 1000\n1\n1000",
"output": "Tie"
},
{
"input": "1 1\n1000\n1",
"output": "Tie"
},
{
"input": "50 35\n9 17 28 107 136 152 169 174 186 188 201 262 291 312 324 330 341 358 385 386 393 397 425 431 479 498 502 523 530 540 542 554 578 588 622 623 684 696 709 722 784 819 836 845 850 932 945 969 983 984\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",
"output": "Tie"
},
{
"input": "50 20\n12 113 116 120 138 156 167 183 185 194 211 228 234 261 278 287 310 317 346 361 364 397 424 470 496 522 527 536 611 648 668 704 707 712 717 752 761 766 815 828 832 864 872 885 889 901 904 929 982 993\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",
"output": "Limak"
}
] | 1,586,504,202
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 78
| 307,200
|
#Codeforce 658A
n,c=(int(v) for v in input().split())
score=[int(u) for u in input().split()]
time=[int(t) for t in input().split()]
Limak=0
Radewoosh=0
tL=0
tR=0
for i in range(n):
tL += time[i]
Limak += max(0,(score[i] - c*tL))
tR += time[n-1-i]
Radewoosh += max(0,(score[n-1-i] - c*tR))
if Limak > Radewoosh:
print("Limak")
elif Radewoosh > Limak:
print("Radewoos")
else:
print("Tie")
|
Title: Bear and Reverse Radewoosh
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak and Radewoosh are going to compete against each other in the upcoming algorithmic contest. They are equally skilled but they won't solve problems in the same order.
There will be *n* problems. The *i*-th problem has initial score *p**i* and it takes exactly *t**i* minutes to solve it. Problems are sorted by difficulty — it's guaranteed that *p**i*<=<<=*p**i*<=+<=1 and *t**i*<=<<=*t**i*<=+<=1.
A constant *c* is given too, representing the speed of loosing points. Then, submitting the *i*-th problem at time *x* (*x* minutes after the start of the contest) gives *max*(0,<= *p**i*<=-<=*c*·*x*) points.
Limak is going to solve problems in order 1,<=2,<=...,<=*n* (sorted increasingly by *p**i*). Radewoosh is going to solve them in order *n*,<=*n*<=-<=1,<=...,<=1 (sorted decreasingly by *p**i*). Your task is to predict the outcome — print the name of the winner (person who gets more points at the end) or a word "Tie" in case of a tie.
You may assume that the duration of the competition is greater or equal than the sum of all *t**i*. That means both Limak and Radewoosh will accept all *n* problems.
Input Specification:
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=50,<=1<=≤<=*c*<=≤<=1000) — the number of problems and the constant representing the speed of loosing points.
The second line contains *n* integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=1000,<=*p**i*<=<<=*p**i*<=+<=1) — initial scores.
The third line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=1000,<=*t**i*<=<<=*t**i*<=+<=1) where *t**i* denotes the number of minutes one needs to solve the *i*-th problem.
Output Specification:
Print "Limak" (without quotes) if Limak will get more points in total. Print "Radewoosh" (without quotes) if Radewoosh will get more points in total. Print "Tie" (without quotes) if Limak and Radewoosh will get the same total number of points.
Demo Input:
['3 2\n50 85 250\n10 15 25\n', '3 6\n50 85 250\n10 15 25\n', '8 1\n10 20 30 40 50 60 70 80\n8 10 58 63 71 72 75 76\n']
Demo Output:
['Limak\n', 'Radewoosh\n', 'Tie\n']
Note:
In the first sample, there are 3 problems. Limak solves them as follows:
1. Limak spends 10 minutes on the 1-st problem and he gets 50 - *c*·10 = 50 - 2·10 = 30 points. 1. Limak spends 15 minutes on the 2-nd problem so he submits it 10 + 15 = 25 minutes after the start of the contest. For the 2-nd problem he gets 85 - 2·25 = 35 points. 1. He spends 25 minutes on the 3-rd problem so he submits it 10 + 15 + 25 = 50 minutes after the start. For this problem he gets 250 - 2·50 = 150 points.
So, Limak got 30 + 35 + 150 = 215 points.
Radewoosh solves problem in the reversed order:
1. Radewoosh solves 3-rd problem after 25 minutes so he gets 250 - 2·25 = 200 points. 1. He spends 15 minutes on the 2-nd problem so he submits it 25 + 15 = 40 minutes after the start. He gets 85 - 2·40 = 5 points for this problem. 1. He spends 10 minutes on the 1-st problem so he submits it 25 + 15 + 10 = 50 minutes after the start. He gets *max*(0, 50 - 2·50) = *max*(0, - 50) = 0 points.
Radewoosh got 200 + 5 + 0 = 205 points in total. Limak has 215 points so Limak wins.
In the second sample, Limak will get 0 points for each problem and Radewoosh will first solve the hardest problem and he will get 250 - 6·25 = 100 points for that. Radewoosh will get 0 points for other two problems but he is the winner anyway.
In the third sample, Limak will get 2 points for the 1-st problem and 2 points for the 2-nd problem. Radewoosh will get 4 points for the 8-th problem. They won't get points for other problems and thus there is a tie because 2 + 2 = 4.
|
```python
#Codeforce 658A
n,c=(int(v) for v in input().split())
score=[int(u) for u in input().split()]
time=[int(t) for t in input().split()]
Limak=0
Radewoosh=0
tL=0
tR=0
for i in range(n):
tL += time[i]
Limak += max(0,(score[i] - c*tL))
tR += time[n-1-i]
Radewoosh += max(0,(score[n-1-i] - c*tR))
if Limak > Radewoosh:
print("Limak")
elif Radewoosh > Limak:
print("Radewoos")
else:
print("Tie")
```
| 0
|
|
514
|
A
|
Chewbaсca and Number
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation"
] | null | null |
Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero.
|
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca.
|
Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes.
|
[
"27\n",
"4545\n"
] |
[
"22\n",
"4444\n"
] |
none
| 500
|
[
{
"input": "27",
"output": "22"
},
{
"input": "4545",
"output": "4444"
},
{
"input": "1",
"output": "1"
},
{
"input": "9",
"output": "9"
},
{
"input": "8772",
"output": "1222"
},
{
"input": "81",
"output": "11"
},
{
"input": "71723447",
"output": "21223442"
},
{
"input": "91730629",
"output": "91230320"
},
{
"input": "420062703497",
"output": "420032203402"
},
{
"input": "332711047202",
"output": "332211042202"
},
{
"input": "3395184971407775",
"output": "3304114021402224"
},
{
"input": "8464062628894325",
"output": "1434032321104324"
},
{
"input": "164324828731963982",
"output": "134324121231033012"
},
{
"input": "384979173822804784",
"output": "314020123122104214"
},
{
"input": "41312150450968417",
"output": "41312140440031412"
},
{
"input": "2156",
"output": "2143"
},
{
"input": "1932",
"output": "1032"
},
{
"input": "5902",
"output": "4002"
},
{
"input": "5728",
"output": "4221"
},
{
"input": "8537",
"output": "1432"
},
{
"input": "55403857",
"output": "44403142"
},
{
"input": "270739",
"output": "220230"
},
{
"input": "28746918",
"output": "21243011"
},
{
"input": "10279211",
"output": "10220211"
},
{
"input": "40289679",
"output": "40210320"
},
{
"input": "545203238506",
"output": "444203231403"
},
{
"input": "461117063340",
"output": "431112033340"
},
{
"input": "658492686568",
"output": "341402313431"
},
{
"input": "857373361868",
"output": "142323331131"
},
{
"input": "429325660016",
"output": "420324330013"
},
{
"input": "9894448650287940",
"output": "9104441340212040"
},
{
"input": "6354510839296263",
"output": "3344410130203233"
},
{
"input": "6873575462224593",
"output": "3123424432224403"
},
{
"input": "4237951492601449",
"output": "4232041402301440"
},
{
"input": "2680352384836991",
"output": "2310342314133001"
},
{
"input": "606187734191890310",
"output": "303112234101100310"
},
{
"input": "351499943576823355",
"output": "341400043423123344"
},
{
"input": "180593481782177068",
"output": "110403411212122031"
},
{
"input": "999999999999999999",
"output": "900000000000000000"
},
{
"input": "1000000000000000000",
"output": "1000000000000000000"
},
{
"input": "9999",
"output": "9000"
},
{
"input": "99",
"output": "90"
},
{
"input": "9991",
"output": "9001"
}
] | 1,687,913,012
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 46
| 0
|
a=list(map(int, list(input())))
for i in range(len(a)):
if a[i]>4:
a[i]=9-a[i]
if a[0]==0:
a[0]=9
print(''.join(map(str, a)))
|
Title: Chewbaсca and Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero.
Input Specification:
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca.
Output Specification:
Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes.
Demo Input:
['27\n', '4545\n']
Demo Output:
['22\n', '4444\n']
Note:
none
|
```python
a=list(map(int, list(input())))
for i in range(len(a)):
if a[i]>4:
a[i]=9-a[i]
if a[0]==0:
a[0]=9
print(''.join(map(str, a)))
```
| 3
|
|
222
|
A
|
Shooshuns and Sequence
|
PROGRAMMING
| 1,200
|
[
"brute force",
"implementation"
] | null | null |
One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
|
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found.
|
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
|
[
"3 2\n3 1 1\n",
"3 1\n3 1 1\n"
] |
[
"1\n",
"-1\n"
] |
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
| 500
|
[
{
"input": "3 2\n3 1 1",
"output": "1"
},
{
"input": "3 1\n3 1 1",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "0"
},
{
"input": "2 1\n2 1",
"output": "-1"
},
{
"input": "4 4\n1 2 3 4",
"output": "3"
},
{
"input": "4 3\n1 2 3 4",
"output": "-1"
},
{
"input": "5 3\n2 1 1 1 1",
"output": "1"
},
{
"input": "5 3\n1 1 1 1 1",
"output": "0"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "0"
},
{
"input": "10 1\n1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "10 1\n1 2 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 1\n2 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 2\n2 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "10 2\n1 2 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "10 3\n3 2 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10 1\n1 2 3 4 5 6 7 8 9 10",
"output": "-1"
},
{
"input": "10 10\n1 2 3 4 5 6 7 8 9 10",
"output": "9"
},
{
"input": "20 10\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 11\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "10"
},
{
"input": "20 9\n1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 1 1 1 1 1",
"output": "-1"
},
{
"input": "20 10\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "20 20\n20 1 19 2 18 3 17 4 16 5 15 6 14 7 13 8 12 9 11 10",
"output": "19"
}
] | 1,623,500,052
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 92
| 0
|
n, k = map(int, input().split())
arr = list(map(int, input().split()))
count = 0
arr2 = arr.copy()
while True:
count += 1
arr.append(arr[k-1])
del arr[0]
if len(set(arr)) == 1:
print(count)
break
if arr == arr2:
print(-1)
break
|
Title: Shooshuns and Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input Specification:
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found.
Output Specification:
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
Demo Input:
['3 2\n3 1 1\n', '3 1\n3 1 1\n']
Demo Output:
['1\n', '-1\n']
Note:
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1.
|
```python
n, k = map(int, input().split())
arr = list(map(int, input().split()))
count = 0
arr2 = arr.copy()
while True:
count += 1
arr.append(arr[k-1])
del arr[0]
if len(set(arr)) == 1:
print(count)
break
if arr == arr2:
print(-1)
break
```
| 0
|
|
192
|
B
|
Walking in the Rain
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] | null | null |
In Berland the opposition is going to arrange mass walking on the boulevard. The boulevard consists of *n* tiles that are lain in a row and are numbered from 1 to *n* from right to left. The opposition should start walking on the tile number 1 and the finish on the tile number *n*. During the walk it is allowed to move from right to left between adjacent tiles in a row, and jump over a tile. More formally, if you are standing on the tile number *i* (*i*<=<<=*n*<=-<=1), you can reach the tiles number *i*<=+<=1 or the tile number *i*<=+<=2 from it (if you stand on the tile number *n*<=-<=1, you can only reach tile number *n*). We can assume that all the opposition movements occur instantaneously.
In order to thwart an opposition rally, the Berland bloody regime organized the rain. The tiles on the boulevard are of poor quality and they are rapidly destroyed in the rain. We know that the *i*-th tile is destroyed after *a**i* days of rain (on day *a**i* tile isn't destroyed yet, and on day *a**i*<=+<=1 it is already destroyed). Of course, no one is allowed to walk on the destroyed tiles! So the walk of the opposition is considered thwarted, if either the tile number 1 is broken, or the tile number *n* is broken, or it is impossible to reach the tile number *n* from the tile number 1 if we can walk on undestroyed tiles.
The opposition wants to gather more supporters for their walk. Therefore, the more time they have to pack, the better. Help the opposition to calculate how much time they still have and tell us for how many days the walk from the tile number 1 to the tile number *n* will be possible.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=103) — the boulevard's length in tiles.
The second line contains *n* space-separated integers *a**i* — the number of days after which the *i*-th tile gets destroyed (1<=≤<=*a**i*<=≤<=103).
|
Print a single number — the sought number of days.
|
[
"4\n10 3 5 10\n",
"5\n10 2 8 3 5\n"
] |
[
"5\n",
"5\n"
] |
In the first sample the second tile gets destroyed after day three, and the only path left is 1 → 3 → 4. After day five there is a two-tile gap between the first and the last tile, you can't jump over it.
In the second sample path 1 → 3 → 5 is available up to day five, inclusive. On day six the last tile is destroyed and the walk is thwarted.
| 1,000
|
[
{
"input": "4\n10 3 5 10",
"output": "5"
},
{
"input": "5\n10 2 8 3 5",
"output": "5"
},
{
"input": "10\n10 3 1 6 7 1 3 3 8 1",
"output": "1"
},
{
"input": "10\n26 72 10 52 2 5 61 2 39 64",
"output": "5"
},
{
"input": "100\n8 2 1 2 8 3 5 8 5 1 9 3 4 1 5 6 4 2 9 10 6 10 10 3 9 4 10 5 3 1 5 10 7 6 8 10 2 6 4 4 2 2 10 7 2 7 3 2 6 3 6 4 7 6 2 5 5 8 6 9 5 2 7 5 8 6 5 8 10 6 10 8 5 3 1 10 6 1 7 5 1 8 10 5 1 3 10 7 10 5 7 1 4 3 8 6 3 4 9 6",
"output": "2"
},
{
"input": "100\n10 2 8 7 5 1 5 4 9 2 7 9 3 5 6 2 3 6 10 1 2 7 1 4 8 8 6 1 7 8 8 1 5 8 1 2 7 4 10 7 3 1 2 5 8 1 1 4 9 7 7 4 7 3 8 8 7 1 5 1 6 9 8 8 1 10 4 4 7 7 10 9 5 1 1 3 6 2 6 3 6 4 9 8 2 9 6 2 7 8 10 9 9 6 3 5 3 1 4 8",
"output": "1"
},
{
"input": "100\n21 57 14 6 58 61 37 54 43 22 90 90 90 14 10 97 47 43 19 66 96 58 88 92 22 62 99 97 15 36 58 93 44 42 45 38 41 21 16 30 66 92 39 70 1 73 83 27 63 21 20 84 30 30 30 77 93 30 62 96 33 34 28 59 48 89 68 62 50 16 18 19 42 42 80 58 31 59 40 81 92 26 28 47 26 8 8 74 86 80 88 82 98 27 41 97 11 91 42 67",
"output": "8"
},
{
"input": "100\n37 75 11 81 60 33 17 80 37 77 26 86 31 78 59 23 92 38 8 15 30 91 99 75 79 34 78 80 19 51 48 48 61 74 59 30 26 2 71 74 48 42 42 81 20 55 49 69 60 10 53 2 21 44 10 18 45 64 21 18 5 62 3 34 52 72 16 28 70 31 93 5 21 69 21 90 31 90 91 79 54 94 77 27 97 4 74 9 29 29 81 5 33 81 75 37 61 73 57 75",
"output": "15"
},
{
"input": "100\n190 544 642 723 577 689 757 509 165 193 396 972 742 367 83 294 404 308 683 399 551 770 564 721 465 839 379 68 687 554 821 719 304 533 146 180 596 713 546 743 949 100 458 735 17 525 568 907 957 670 914 374 347 801 227 884 284 444 686 410 127 508 504 273 624 213 873 658 336 79 819 938 3 722 649 368 733 747 577 746 940 308 970 963 145 487 102 559 790 243 609 77 552 565 151 492 726 448 393 837",
"output": "180"
},
{
"input": "100\n606 358 399 589 724 454 741 183 571 244 984 867 828 232 189 821 642 855 220 839 585 203 135 305 970 503 362 658 491 562 706 62 721 465 560 880 833 646 365 23 679 549 317 834 583 947 134 253 250 768 343 996 541 163 355 925 336 874 997 632 498 529 932 487 415 391 766 224 364 790 486 512 183 458 343 751 633 126 688 536 845 380 423 447 904 779 520 843 977 392 406 147 888 520 886 179 176 129 8 750",
"output": "129"
},
{
"input": "5\n3 2 3 4 2",
"output": "2"
},
{
"input": "5\n4 8 9 10 6",
"output": "4"
},
{
"input": "5\n2 21 6 5 9",
"output": "2"
},
{
"input": "5\n34 39 30 37 35",
"output": "34"
},
{
"input": "5\n14 67 15 28 21",
"output": "14"
},
{
"input": "5\n243 238 138 146 140",
"output": "140"
},
{
"input": "5\n46 123 210 119 195",
"output": "46"
},
{
"input": "5\n725 444 477 661 761",
"output": "477"
},
{
"input": "10\n2 2 3 4 4 1 5 3 1 2",
"output": "2"
},
{
"input": "10\n1 10 1 10 1 1 7 8 6 7",
"output": "1"
},
{
"input": "10\n5 17 8 1 10 20 9 18 12 20",
"output": "5"
},
{
"input": "10\n18 11 23 7 9 10 28 29 46 21",
"output": "9"
},
{
"input": "10\n2 17 53 94 95 57 36 47 68 48",
"output": "2"
},
{
"input": "10\n93 231 176 168 177 222 22 137 110 4",
"output": "4"
},
{
"input": "10\n499 173 45 141 425 276 96 290 428 95",
"output": "95"
},
{
"input": "10\n201 186 897 279 703 376 238 93 253 316",
"output": "201"
},
{
"input": "25\n3 2 3 2 2 2 3 4 5 1 1 4 1 2 1 3 5 5 3 5 1 2 4 1 3",
"output": "1"
},
{
"input": "25\n9 9 1 9 10 5 6 4 6 1 5 2 2 1 2 8 4 6 5 7 1 10 5 4 9",
"output": "2"
},
{
"input": "25\n2 17 21 4 13 6 14 18 17 1 16 13 24 4 12 7 8 16 9 25 25 9 11 20 18",
"output": "2"
},
{
"input": "25\n38 30 9 35 33 48 8 4 49 2 39 19 34 35 47 49 33 4 23 5 42 35 49 11 30",
"output": "8"
},
{
"input": "25\n75 34 77 68 60 38 76 89 35 68 28 36 96 63 43 12 9 4 37 75 88 30 11 58 35",
"output": "9"
},
{
"input": "25\n108 3 144 140 239 105 59 126 224 181 147 102 94 201 68 121 167 94 60 130 64 162 45 95 235",
"output": "94"
},
{
"input": "25\n220 93 216 467 134 408 132 220 292 11 363 404 282 253 141 313 310 356 214 256 380 81 42 128 363",
"output": "81"
},
{
"input": "25\n371 884 75 465 891 510 471 52 382 829 514 610 660 642 179 108 41 818 346 106 738 993 706 574 623",
"output": "108"
},
{
"input": "50\n1 2 1 3 2 5 2 2 2 3 4 4 4 3 3 4 1 2 3 1 5 4 1 2 2 1 5 3 2 2 1 5 4 5 2 5 4 1 1 3 5 2 1 4 5 5 1 5 5 5",
"output": "1"
},
{
"input": "50\n2 4 9 8 1 3 7 1 2 3 8 9 8 8 5 2 10 5 8 1 3 1 8 2 3 7 9 10 2 9 9 7 3 8 6 10 6 5 4 8 1 1 5 6 8 9 5 9 5 3",
"output": "1"
},
{
"input": "50\n22 9 5 3 24 21 25 13 17 21 14 8 22 18 2 3 22 9 10 11 25 22 5 10 16 7 15 3 2 13 2 12 9 24 3 14 2 18 3 22 8 2 19 6 16 4 5 20 10 12",
"output": "3"
},
{
"input": "50\n14 4 20 37 50 46 19 20 25 47 10 6 34 12 41 47 9 22 28 41 34 47 40 12 42 9 4 15 15 27 8 38 9 4 17 8 13 47 7 9 38 30 48 50 7 41 34 23 11 16",
"output": "9"
},
{
"input": "50\n69 9 97 15 22 69 27 7 23 84 73 74 60 94 43 98 13 4 63 49 7 31 93 23 6 75 32 63 49 32 99 43 68 48 16 54 20 38 40 65 34 28 21 55 79 50 2 18 22 95",
"output": "13"
},
{
"input": "50\n50 122 117 195 42 178 153 194 7 89 142 40 158 230 213 104 179 56 244 196 85 159 167 19 157 20 230 201 152 98 250 242 10 52 96 242 139 181 90 107 178 52 196 79 23 61 212 47 97 97",
"output": "50"
},
{
"input": "50\n354 268 292 215 187 232 35 38 179 79 108 491 346 384 345 103 14 260 148 322 459 238 220 493 374 237 474 148 21 221 88 377 289 121 201 198 490 117 382 454 359 390 346 456 294 325 130 306 484 83",
"output": "38"
},
{
"input": "50\n94 634 27 328 629 967 728 177 379 908 801 715 787 192 427 48 559 923 841 6 759 335 251 172 193 593 456 780 647 638 750 881 206 129 278 744 91 49 523 248 286 549 593 451 216 753 471 325 870 16",
"output": "16"
},
{
"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": "1"
},
{
"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": "2"
},
{
"input": "100\n14 7 6 21 12 5 22 23 2 9 8 1 9 2 20 2 24 7 14 24 8 19 15 19 10 24 9 4 21 12 3 21 9 16 9 22 18 4 17 19 19 9 6 1 13 15 23 3 14 3 7 15 17 10 7 24 4 18 21 14 25 20 19 19 14 25 24 21 16 10 2 16 1 21 1 24 13 7 13 20 12 20 2 16 3 6 6 2 19 9 16 4 1 2 7 18 15 14 10 22",
"output": "2"
},
{
"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": "1"
},
{
"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": "5"
},
{
"input": "100\n26 171 37 63 189 202 180 210 179 131 43 33 227 5 211 130 105 23 229 48 174 48 182 68 174 146 200 166 246 116 106 86 72 206 216 207 70 148 83 149 94 64 142 8 241 211 27 190 58 116 113 96 210 237 73 240 180 110 34 115 167 4 42 30 162 114 74 131 34 206 174 168 216 101 216 149 212 172 180 220 123 201 25 116 42 143 105 40 30 123 174 220 57 238 145 222 105 184 131 162",
"output": "26"
},
{
"input": "100\n182 9 8 332 494 108 117 203 43 473 451 426 119 408 342 84 88 35 383 84 48 69 31 54 347 363 342 69 422 489 194 16 55 171 71 355 116 142 181 246 275 402 155 282 160 179 240 448 49 101 42 499 434 258 21 327 95 376 38 422 68 381 170 372 427 149 38 48 400 224 246 438 62 43 280 40 108 385 351 379 224 311 66 125 300 41 372 358 5 221 223 341 201 261 455 165 74 379 214 10",
"output": "9"
},
{
"input": "100\n836 969 196 706 812 64 743 262 667 27 227 730 50 510 374 915 124 527 778 528 175 151 439 994 835 87 197 91 121 243 534 634 4 410 936 6 979 227 745 734 492 792 209 95 602 446 299 533 376 595 971 879 36 126 528 759 116 499 571 664 787 820 870 838 604 240 334 872 477 415 57 689 870 690 304 122 487 191 253 610 301 348 358 806 828 911 8 320 414 172 268 867 978 205 812 60 845 395 406 155",
"output": "121"
},
{
"input": "250\n5 3 5 1 3 5 3 4 4 3 1 5 2 2 1 1 5 2 3 3 2 5 4 3 2 4 2 3 5 4 1 2 3 5 2 2 5 4 1 3 3 5 4 4 4 4 4 2 4 2 3 5 1 4 3 3 2 3 5 3 3 4 4 2 3 1 3 4 1 4 5 4 1 2 3 4 1 5 3 3 2 3 5 4 2 5 2 2 3 5 4 3 5 4 2 1 4 1 4 1 1 3 5 1 1 2 1 3 4 5 4 3 2 5 1 3 5 1 1 3 3 5 1 4 5 1 2 1 1 5 5 3 5 1 4 1 4 4 4 4 4 1 4 3 4 5 4 1 2 2 5 2 2 4 2 3 5 3 5 5 3 3 2 2 2 1 1 4 4 4 2 1 4 5 3 1 5 4 4 5 5 5 3 3 5 2 1 4 5 4 1 1 1 5 3 5 2 3 3 2 1 3 4 1 4 1 5 3 1 2 5 5 2 1 4 4 2 3 5 2 4 1 3 4 5 5 4 3 2 2 3 2 4 2 5 3 5 5 1 5 3 2 2 4 2 5 5 5 2 5",
"output": "1"
},
{
"input": "1\n987",
"output": "987"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "5\n2 5 5 5 5",
"output": "2"
},
{
"input": "1\n500",
"output": "500"
}
] | 1,586,858,645
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 85
| 312
| 0
|
from sys import stdin,stdout
ii1 = lambda: int(stdin.readline().strip())
is1 = lambda: stdin.readline().strip()
iia = lambda: list(map(int, stdin.readline().strip().split()))
isa = lambda: stdin.readline().strip().split()
mod = 1000000007
n = ii1()
arr = iia()
m = float("inf")
for i in range(n):
if i == 0 or i == n - 1:
m = min(m, arr[i])
if i < n - 2:
m = min(m,max(arr[i + 1], arr[i + 2]))
elif i<n-1:
m = min(m, arr[i + 1])
print(m)
|
Title: Walking in the Rain
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland the opposition is going to arrange mass walking on the boulevard. The boulevard consists of *n* tiles that are lain in a row and are numbered from 1 to *n* from right to left. The opposition should start walking on the tile number 1 and the finish on the tile number *n*. During the walk it is allowed to move from right to left between adjacent tiles in a row, and jump over a tile. More formally, if you are standing on the tile number *i* (*i*<=<<=*n*<=-<=1), you can reach the tiles number *i*<=+<=1 or the tile number *i*<=+<=2 from it (if you stand on the tile number *n*<=-<=1, you can only reach tile number *n*). We can assume that all the opposition movements occur instantaneously.
In order to thwart an opposition rally, the Berland bloody regime organized the rain. The tiles on the boulevard are of poor quality and they are rapidly destroyed in the rain. We know that the *i*-th tile is destroyed after *a**i* days of rain (on day *a**i* tile isn't destroyed yet, and on day *a**i*<=+<=1 it is already destroyed). Of course, no one is allowed to walk on the destroyed tiles! So the walk of the opposition is considered thwarted, if either the tile number 1 is broken, or the tile number *n* is broken, or it is impossible to reach the tile number *n* from the tile number 1 if we can walk on undestroyed tiles.
The opposition wants to gather more supporters for their walk. Therefore, the more time they have to pack, the better. Help the opposition to calculate how much time they still have and tell us for how many days the walk from the tile number 1 to the tile number *n* will be possible.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=103) — the boulevard's length in tiles.
The second line contains *n* space-separated integers *a**i* — the number of days after which the *i*-th tile gets destroyed (1<=≤<=*a**i*<=≤<=103).
Output Specification:
Print a single number — the sought number of days.
Demo Input:
['4\n10 3 5 10\n', '5\n10 2 8 3 5\n']
Demo Output:
['5\n', '5\n']
Note:
In the first sample the second tile gets destroyed after day three, and the only path left is 1 → 3 → 4. After day five there is a two-tile gap between the first and the last tile, you can't jump over it.
In the second sample path 1 → 3 → 5 is available up to day five, inclusive. On day six the last tile is destroyed and the walk is thwarted.
|
```python
from sys import stdin,stdout
ii1 = lambda: int(stdin.readline().strip())
is1 = lambda: stdin.readline().strip()
iia = lambda: list(map(int, stdin.readline().strip().split()))
isa = lambda: stdin.readline().strip().split()
mod = 1000000007
n = ii1()
arr = iia()
m = float("inf")
for i in range(n):
if i == 0 or i == n - 1:
m = min(m, arr[i])
if i < n - 2:
m = min(m,max(arr[i + 1], arr[i + 2]))
elif i<n-1:
m = min(m, arr[i + 1])
print(m)
```
| 3
|
|
106
|
C
|
Buns
|
PROGRAMMING
| 1,700
|
[
"dp"
] |
C. Buns
|
2
|
256
|
Lavrenty, a baker, is going to make several buns with stuffings and sell them.
Lavrenty has *n* grams of dough as well as *m* different stuffing types. The stuffing types are numerated from 1 to *m*. Lavrenty knows that he has *a**i* grams left of the *i*-th stuffing. It takes exactly *b**i* grams of stuffing *i* and *c**i* grams of dough to cook a bun with the *i*-th stuffing. Such bun can be sold for *d**i* tugriks.
Also he can make buns without stuffings. Each of such buns requires *c*0 grams of dough and it can be sold for *d*0 tugriks. So Lavrenty can cook any number of buns with different stuffings or without it unless he runs out of dough and the stuffings. Lavrenty throws away all excess material left after baking.
Find the maximum number of tugriks Lavrenty can earn.
|
The first line contains 4 integers *n*, *m*, *c*0 and *d*0 (1<=≤<=*n*<=≤<=1000, 1<=≤<=*m*<=≤<=10, 1<=≤<=*c*0,<=*d*0<=≤<=100). Each of the following *m* lines contains 4 integers. The *i*-th line contains numbers *a**i*, *b**i*, *c**i* and *d**i* (1<=≤<=*a**i*,<=*b**i*,<=*c**i*,<=*d**i*<=≤<=100).
|
Print the only number — the maximum number of tugriks Lavrenty can earn.
|
[
"10 2 2 1\n7 3 2 100\n12 3 1 10\n",
"100 1 25 50\n15 5 20 10\n"
] |
[
"241",
"200"
] |
To get the maximum number of tugriks in the first sample, you need to cook 2 buns with stuffing 1, 4 buns with stuffing 2 and a bun without any stuffing.
In the second sample Lavrenty should cook 4 buns without stuffings.
| 1,500
|
[
{
"input": "10 2 2 1\n7 3 2 100\n12 3 1 10",
"output": "241"
},
{
"input": "100 1 25 50\n15 5 20 10",
"output": "200"
},
{
"input": "10 1 5 2\n100 1 2 3",
"output": "15"
},
{
"input": "10 1 5 11\n3 1 3 8",
"output": "24"
},
{
"input": "10 2 11 5\n100 1 3 10\n100 1 2 4",
"output": "30"
},
{
"input": "5 8 6 5\n1 2 5 4\n1 2 6 7\n1 2 3 5\n1 2 1 6\n1 2 8 3\n1 2 2 4\n1 2 5 6\n1 2 7 7",
"output": "0"
},
{
"input": "300 4 100 2\n10 1 24 5\n10 1 25 6\n10 1 26 7\n10 1 27 8",
"output": "87"
},
{
"input": "1 1 1 1\n1 1 1 1",
"output": "1"
},
{
"input": "2 1 2 1\n1 2 1 1",
"output": "1"
},
{
"input": "10 2 13 100\n20 1 3 10\n20 1 2 6",
"output": "32"
},
{
"input": "100 5 8 80\n25 8 2 70\n27 6 7 30\n26 1 6 5\n7 1 1 86\n18 8 4 54",
"output": "1670"
},
{
"input": "150 8 3 46\n39 4 10 25\n31 17 8 70\n37 2 13 1\n29 17 17 59\n54 20 5 39\n53 14 10 23\n50 12 16 41\n8 2 6 61",
"output": "2300"
},
{
"input": "231 10 9 30\n98 11 5 17\n59 13 1 47\n83 1 7 2\n42 21 1 6\n50 16 2 9\n44 10 5 31\n12 20 8 9\n61 23 7 2\n85 18 2 19\n82 25 10 20",
"output": "1065"
},
{
"input": "345 10 5 45\n1 23 14 55\n51 26 15 11\n65 4 16 36\n81 14 13 25\n8 9 13 60\n43 4 7 59\n85 11 14 35\n82 13 5 49\n85 28 15 3\n51 21 18 53",
"output": "3129"
},
{
"input": "401 10 2 82\n17 9 14 48\n79 4 3 38\n1 2 6 31\n45 2 9 60\n45 2 4 50\n6 1 3 36\n3 1 19 37\n78 3 8 33\n59 8 19 19\n65 10 2 61",
"output": "16400"
},
{
"input": "777 10 23 20\n50 90 86 69\n33 90 59 73\n79 26 35 31\n57 48 97 4\n5 10 48 87\n35 99 33 34\n7 32 54 35\n56 25 10 38\n5 3 89 76\n13 33 91 66",
"output": "734"
},
{
"input": "990 10 7 20\n38 82 14 69\n5 66 51 5\n11 26 91 11\n29 12 73 96\n93 82 48 59\n19 15 5 50\n15 36 6 63\n16 57 94 90\n45 3 57 72\n61 41 47 18",
"output": "2850"
},
{
"input": "1000 10 51 56\n2 62 82 65\n37 90 87 97\n11 94 47 95\n49 24 97 24\n33 38 40 31\n27 15 17 66\n91 80 34 71\n60 93 42 94\n9 35 73 68\n93 65 83 58",
"output": "1145"
},
{
"input": "1000 10 1 53\n63 1 1 58\n58 1 2 28\n100 1 1 25\n61 1 1 90\n96 2 2 50\n19 2 1 90\n7 2 1 30\n90 1 2 5\n34 2 1 12\n3 2 1 96",
"output": "55948"
},
{
"input": "1000 10 1 65\n77 1 1 36\n74 1 1 41\n96 1 1 38\n48 1 1 35\n1 1 1 54\n42 1 1 67\n26 1 1 23\n43 1 1 89\n82 1 1 7\n45 1 1 63",
"output": "66116"
},
{
"input": "1000 10 1 87\n100 1 1 38\n100 1 1 45\n100 1 1 73\n100 1 1 89\n100 1 1 38\n100 1 1 13\n100 1 1 93\n100 1 1 89\n100 1 1 71\n100 1 1 29",
"output": "88000"
},
{
"input": "1000 10 1 7\n100 1 1 89\n100 1 1 38\n100 1 1 13\n100 1 1 93\n100 1 1 89\n100 1 1 38\n100 1 1 45\n100 1 1 73\n100 1 1 71\n100 1 1 29",
"output": "57800"
},
{
"input": "1000 10 1 100\n100 1 1 100\n100 1 1 100\n100 1 1 100\n100 1 1 100\n100 1 1 100\n100 1 1 100\n100 1 1 100\n100 1 1 100\n100 1 1 100\n100 1 1 100",
"output": "100000"
},
{
"input": "99 10 100 100\n100 1 100 100\n100 1 100 100\n100 1 100 100\n100 1 100 100\n100 1 100 100\n100 1 100 100\n100 1 100 100\n100 1 100 100\n100 1 100 100\n100 1 100 100",
"output": "0"
},
{
"input": "1000 10 100 75\n100 97 100 95\n100 64 100 78\n100 82 100 35\n100 51 100 64\n100 67 100 25\n100 79 100 33\n100 65 100 85\n100 99 100 78\n100 53 100 74\n100 87 100 73",
"output": "786"
},
{
"input": "999 10 5 100\n100 1 10 100\n100 1 10 100\n100 1 10 100\n100 1 10 100\n100 1 10 100\n100 1 10 100\n100 1 10 100\n100 1 10 100\n100 1 10 100\n100 1 10 100",
"output": "19900"
},
{
"input": "1000 10 50 100\n7 1 80 100\n5 1 37 100\n9 1 25 100\n7 1 17 100\n6 1 10 100\n5 1 15 100\n6 1 13 100\n2 1 14 100\n4 1 17 100\n3 1 32 100",
"output": "4800"
},
{
"input": "1000 10 1 1\n1 2 1 97\n1 2 1 95\n1 2 1 99\n1 2 1 98\n1 2 1 93\n1 2 1 91\n1 2 1 90\n1 2 1 94\n1 2 1 92\n1 2 1 99",
"output": "1000"
},
{
"input": "1 10 1 97\n1 1 1 98\n1 1 1 99\n1 1 1 76\n1 1 1 89\n1 1 1 64\n1 1 1 83\n1 1 1 72\n1 1 1 66\n1 1 1 54\n1 1 1 73",
"output": "99"
},
{
"input": "3 10 10 98\n10 5 5 97\n6 7 1 56\n23 10 5 78\n40 36 4 35\n30 50 1 30\n60 56 8 35\n70 90 2 17\n10 11 3 68\n1 2 17 70\n13 4 8 19",
"output": "0"
},
{
"input": "1000 1 23 76\n74 22 14 5",
"output": "3268"
},
{
"input": "1000 2 95 56\n58 54 66 61\n61 14 67 65",
"output": "713"
},
{
"input": "1000 3 67 88\n90 86 66 17\n97 38 63 17\n55 78 39 51",
"output": "1232"
},
{
"input": "1000 4 91 20\n74 18 18 73\n33 10 59 21\n7 42 87 79\n9 100 77 100",
"output": "515"
},
{
"input": "1000 5 63 52\n6 98 18 77\n17 34 3 73\n59 6 35 7\n61 16 85 64\n73 62 40 11",
"output": "804"
},
{
"input": "1000 6 87 32\n90 30 70 33\n53 6 99 77\n59 22 83 35\n65 32 93 28\n85 50 60 7\n15 15 5 82",
"output": "771"
},
{
"input": "1000 7 59 64\n22 62 70 89\n37 78 43 29\n11 86 83 63\n17 48 1 92\n97 38 80 55\n15 3 89 42\n87 80 62 35",
"output": "1024"
},
{
"input": "1000 8 31 96\n6 94 70 93\n73 2 39 33\n63 50 31 91\n21 64 9 56\n61 26 100 51\n67 39 21 50\n79 4 2 71\n100 9 18 86",
"output": "4609"
},
{
"input": "1000 9 55 28\n38 74 22 49\n9 74 83 85\n63 66 79 19\n25 32 17 20\n73 62 20 47\n19 27 53 58\n71 80 94 7\n56 69 62 98\n49 7 65 76",
"output": "831"
},
{
"input": "1000 10 67 55\n10 21 31 19\n95 29 53 1\n55 53 19 18\n26 88 19 94\n31 1 45 50\n70 38 33 93\n2 12 7 95\n54 37 81 31\n65 32 63 16\n93 66 98 38",
"output": "1161"
},
{
"input": "1000 10 37 38\n65 27 78 14\n16 70 78 66\n93 86 91 43\n95 6 72 86\n72 59 94 36\n66 58 96 40\n41 72 64 4\n26 47 69 13\n85 2 52 15\n34 62 16 79",
"output": "1156"
},
{
"input": "1000 10 58 21\n73 85 73 10\n38 60 55 31\n32 66 62 16\n63 76 73 78\n61 17 92 70\n61 79 11 87\n27 31 21 62\n47 9 4 94\n4 71 42 61\n76 5 35 72",
"output": "1823"
},
{
"input": "12 2 100 1\n100 1 9 10\n100 1 4 4",
"output": "12"
},
{
"input": "1 1 1 10\n100 100 1 100",
"output": "100"
},
{
"input": "10 3 5 1\n100 1 3 7\n100 1 2 5\n1 1 1 10",
"output": "32"
},
{
"input": "10 3 5 1\n100 1 3 7\n100 1 2 5\n1 1 1 10",
"output": "32"
},
{
"input": "1000 10 1 1\n100 1 1 1\n100 1 1 1\n100 1 1 1\n100 1 1 1\n100 1 1 1\n100 1 1 1\n100 1 1 1\n100 1 1 1\n100 1 1 1\n100 1 1 1",
"output": "1000"
},
{
"input": "10 2 100 1\n4 4 5 7\n6 2 3 4",
"output": "12"
},
{
"input": "8 2 10 10\n5 5 5 15\n50 5 4 8",
"output": "16"
},
{
"input": "8 2 10 10\n5 5 5 15\n50 5 4 8",
"output": "16"
},
{
"input": "4 1 2 4\n10 1 3 7",
"output": "8"
},
{
"input": "4 1 2 4\n10 1 3 7",
"output": "8"
},
{
"input": "10 2 5 1\n100 1 2 5\n100 1 3 8",
"output": "26"
},
{
"input": "1000 10 10 10\n100 1 1 1\n100 1 1 2\n100 1 2 1\n100 1 2 2\n100 1 1 1\n100 1 2 3\n100 1 3 2\n100 1 3 3\n100 1 1 3\n100 1 3 1",
"output": "1400"
},
{
"input": "10 3 5 1\n100 1 3 7\n100 1 2 5\n1 1 1 10",
"output": "32"
}
] | 1,598,192,103
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 55
| 1,620
| 6,963,200
|
n,m,c,d=map(int,input().split())
ar=[0]*1005
for i in range(c,n+1):
ar[i]=ar[i-c]+d
for i in range(m):
a,b,c,d=map(int,input().split())
for j in range(a//b):
e=n
while(e>=c):
ar[e]=max(ar[e],ar[e-c]+d)
e-=1
print(ar[n])
|
Title: Buns
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Lavrenty, a baker, is going to make several buns with stuffings and sell them.
Lavrenty has *n* grams of dough as well as *m* different stuffing types. The stuffing types are numerated from 1 to *m*. Lavrenty knows that he has *a**i* grams left of the *i*-th stuffing. It takes exactly *b**i* grams of stuffing *i* and *c**i* grams of dough to cook a bun with the *i*-th stuffing. Such bun can be sold for *d**i* tugriks.
Also he can make buns without stuffings. Each of such buns requires *c*0 grams of dough and it can be sold for *d*0 tugriks. So Lavrenty can cook any number of buns with different stuffings or without it unless he runs out of dough and the stuffings. Lavrenty throws away all excess material left after baking.
Find the maximum number of tugriks Lavrenty can earn.
Input Specification:
The first line contains 4 integers *n*, *m*, *c*0 and *d*0 (1<=≤<=*n*<=≤<=1000, 1<=≤<=*m*<=≤<=10, 1<=≤<=*c*0,<=*d*0<=≤<=100). Each of the following *m* lines contains 4 integers. The *i*-th line contains numbers *a**i*, *b**i*, *c**i* and *d**i* (1<=≤<=*a**i*,<=*b**i*,<=*c**i*,<=*d**i*<=≤<=100).
Output Specification:
Print the only number — the maximum number of tugriks Lavrenty can earn.
Demo Input:
['10 2 2 1\n7 3 2 100\n12 3 1 10\n', '100 1 25 50\n15 5 20 10\n']
Demo Output:
['241', '200']
Note:
To get the maximum number of tugriks in the first sample, you need to cook 2 buns with stuffing 1, 4 buns with stuffing 2 and a bun without any stuffing.
In the second sample Lavrenty should cook 4 buns without stuffings.
|
```python
n,m,c,d=map(int,input().split())
ar=[0]*1005
for i in range(c,n+1):
ar[i]=ar[i-c]+d
for i in range(m):
a,b,c,d=map(int,input().split())
for j in range(a//b):
e=n
while(e>=c):
ar[e]=max(ar[e],ar[e-c]+d)
e-=1
print(ar[n])
```
| 3.58203
|
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"
}
] | 1,684,259,176
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
m, n = map(int, input().split())
max_d = (m * n) // 2
print(max_d)
|
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())
max_d = (m * n) // 2
print(max_d)
```
| 3.977
|
265
|
B
|
Roadside Trees (Simplified Edition)
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation"
] | null | null |
Squirrel Liss loves nuts. There are *n* trees (numbered 1 to *n* from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree *i* is *h**i*. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree. - Eat a nut on the top of the current tree. - Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height *h* of the tree *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), she jumps to height *h* of the tree *i*<=+<=1. This action can't be performed if *h*<=><=*h**i*<=+<=1.
Compute the minimal time (in seconds) required to eat all nuts.
|
The first line contains an integer *n* (1<=<=≤<=<=*n*<=≤<=105) — the number of trees.
Next *n* lines contains the height of trees: *i*-th line contains an integer *h**i* (1<=≤<=*h**i*<=≤<=104) — the height of the tree with the number *i*.
|
Print a single integer — the minimal time required to eat all nuts in seconds.
|
[
"2\n1\n2\n",
"5\n2\n1\n2\n1\n1\n"
] |
[
"5\n",
"14\n"
] |
none
| 1,000
|
[
{
"input": "2\n1\n2",
"output": "5"
},
{
"input": "5\n2\n1\n2\n1\n1",
"output": "14"
},
{
"input": "1\n1",
"output": "2"
}
] | 1,665,905,000
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 15
| 1,184
| 9,011,200
|
n = int(input())
# l = [int(x) for x in input().split()]
# print(l)
l = []
for i in range(n):
b = int(input())
l.append(b)
x = 2*n-1+l[0]
for i in range(1,n):
x = x + abs(l[i]-l[i-1])
print(x)
|
Title: Roadside Trees (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Squirrel Liss loves nuts. There are *n* trees (numbered 1 to *n* from west to east) along a street and there is a delicious nut on the top of each tree. The height of the tree *i* is *h**i*. Liss wants to eat all nuts.
Now Liss is on the root of the tree with the number 1. In one second Liss can perform one of the following actions:
- Walk up or down one unit on a tree. - Eat a nut on the top of the current tree. - Jump to the next tree. In this action the height of Liss doesn't change. More formally, when Liss is at height *h* of the tree *i* (1<=≤<=*i*<=≤<=*n*<=-<=1), she jumps to height *h* of the tree *i*<=+<=1. This action can't be performed if *h*<=><=*h**i*<=+<=1.
Compute the minimal time (in seconds) required to eat all nuts.
Input Specification:
The first line contains an integer *n* (1<=<=≤<=<=*n*<=≤<=105) — the number of trees.
Next *n* lines contains the height of trees: *i*-th line contains an integer *h**i* (1<=≤<=*h**i*<=≤<=104) — the height of the tree with the number *i*.
Output Specification:
Print a single integer — the minimal time required to eat all nuts in seconds.
Demo Input:
['2\n1\n2\n', '5\n2\n1\n2\n1\n1\n']
Demo Output:
['5\n', '14\n']
Note:
none
|
```python
n = int(input())
# l = [int(x) for x in input().split()]
# print(l)
l = []
for i in range(n):
b = int(input())
l.append(b)
x = 2*n-1+l[0]
for i in range(1,n):
x = x + abs(l[i]-l[i-1])
print(x)
```
| 3
|
|
228
|
A
|
Is your horseshoe on the other hoof?
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
|
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
|
Print a single integer — the minimum number of horseshoes Valera needs to buy.
|
[
"1 7 3 3\n",
"7 7 7 7\n"
] |
[
"1\n",
"3\n"
] |
none
| 500
|
[
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 586975278",
"output": "1"
},
{
"input": "259504825 377489979 588153796 377489979",
"output": "1"
},
{
"input": "652588203 931100304 931100304 652588203",
"output": "2"
},
{
"input": "391958720 651507265 391958720 651507265",
"output": "2"
},
{
"input": "90793237 90793237 90793237 90793237",
"output": "3"
},
{
"input": "551651653 551651653 551651653 551651653",
"output": "3"
},
{
"input": "156630260 609654355 668943582 973622757",
"output": "0"
},
{
"input": "17061017 110313588 434481173 796661222",
"output": "0"
},
{
"input": "24975422 256716298 337790533 690960249",
"output": "0"
},
{
"input": "255635360 732742923 798648949 883146723",
"output": "0"
},
{
"input": "133315691 265159773 734556507 265159773",
"output": "1"
},
{
"input": "28442865 741657755 978106882 978106882",
"output": "1"
},
{
"input": "131245479 174845575 497483467 131245479",
"output": "1"
},
{
"input": "139159884 616215581 958341883 616215581",
"output": "1"
},
{
"input": "147784432 947653080 947653080 947653080",
"output": "2"
},
{
"input": "94055790 756126496 756126496 94055790",
"output": "2"
},
{
"input": "240458500 511952208 240458500 511952208",
"output": "2"
},
{
"input": "681828506 972810624 972810624 681828506",
"output": "2"
},
{
"input": "454961014 454961014 454961014 454961014",
"output": "3"
},
{
"input": "915819430 915819430 915819430 915819430",
"output": "3"
},
{
"input": "671645142 671645142 671645142 671645142",
"output": "3"
},
{
"input": "132503558 132503558 132503558 132503558",
"output": "3"
},
{
"input": "5 5 999999 6",
"output": "1"
},
{
"input": "1 1 2 5",
"output": "1"
},
{
"input": "2 1 2 3",
"output": "1"
},
{
"input": "1 1 3 5",
"output": "1"
},
{
"input": "1 1 3 3",
"output": "2"
},
{
"input": "2 2 2 1",
"output": "2"
},
{
"input": "3 1 1 1",
"output": "2"
},
{
"input": "1 2 2 2",
"output": "2"
}
] | 1,698,606,631
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 92
| 0
|
a = list(map(int,input().split()))
b = set()
for i in a:
b.add(i)
print(4 - len(b))
|
Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none
|
```python
a = list(map(int,input().split()))
b = set()
for i in a:
b.add(i)
print(4 - len(b))
```
| 3
|
|
465
|
B
|
Inbox (100500)
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Over time, Alexey's mail box got littered with too many letters. Some of them are read, while others are unread.
Alexey's mail program can either show a list of all letters or show the content of a single letter. As soon as the program shows the content of an unread letter, it becomes read letter (if the program shows the content of a read letter nothing happens). In one click he can do any of the following operations:
- Move from the list of letters to the content of any single letter.- Return to the list of letters from single letter viewing mode.- In single letter viewing mode, move to the next or to the previous letter in the list. You cannot move from the first letter to the previous one or from the last letter to the next one.
The program cannot delete the letters from the list or rearrange them.
Alexey wants to read all the unread letters and go watch football. Now he is viewing the list of all letters and for each letter he can see if it is read or unread. What minimum number of operations does Alexey need to perform to read all unread letters?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of letters in the mailbox.
The second line contains *n* space-separated integers (zeros and ones) — the state of the letter list. The *i*-th number equals either 1, if the *i*-th number is unread, or 0, if the *i*-th letter is read.
|
Print a single number — the minimum number of operations needed to make all the letters read.
|
[
"5\n0 1 0 1 0\n",
"5\n1 1 0 0 1\n",
"2\n0 0\n"
] |
[
"3\n",
"4\n",
"0\n"
] |
In the first sample Alexey needs three operations to cope with the task: open the second letter, move to the third one, move to the fourth one.
In the second sample the action plan: open the first letter, move to the second letter, return to the list, open the fifth letter.
In the third sample all letters are already read.
| 1,000
|
[
{
"input": "5\n0 1 0 1 0",
"output": "3"
},
{
"input": "5\n1 1 0 0 1",
"output": "4"
},
{
"input": "2\n0 0",
"output": "0"
},
{
"input": "9\n1 0 1 0 1 0 1 0 1",
"output": "9"
},
{
"input": "5\n1 1 1 1 1",
"output": "5"
},
{
"input": "14\n0 0 1 1 1 0 1 1 1 0 1 1 1 0",
"output": "11"
},
{
"input": "23\n1 1 1 0 1 1 0 1 1 0 1 1 1 0 1 1 0 1 1 0 1 1 1",
"output": "23"
},
{
"input": "27\n0 1 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",
"output": "25"
},
{
"input": "10\n1 0 0 0 0 1 0 0 0 1",
"output": "5"
},
{
"input": "10\n1 0 0 1 0 0 1 1 0 1",
"output": "8"
},
{
"input": "27\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",
"output": "0"
},
{
"input": "39\n1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1",
"output": "39"
},
{
"input": "48\n1 0 1 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 1 0 0 1 0 1 0 0 1 0 0 1 0 0 1 0 1 0 1 0 0 1 0 0 1",
"output": "39"
},
{
"input": "71\n0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0 0 0 0 0",
"output": "59"
},
{
"input": "99\n1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1",
"output": "99"
},
{
"input": "99\n1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1",
"output": "99"
},
{
"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": "100"
},
{
"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 0 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 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "99"
},
{
"input": "193\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 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "1"
},
{
"input": "213\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 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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",
"output": "5"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "5\n0 0 0 0 1",
"output": "1"
},
{
"input": "12\n0 1 1 0 1 1 0 1 1 0 0 0",
"output": "8"
},
{
"input": "3\n1 0 0",
"output": "1"
},
{
"input": "5\n0 0 1 0 0",
"output": "1"
},
{
"input": "3\n0 1 0",
"output": "1"
},
{
"input": "6\n1 1 0 0 0 0",
"output": "2"
},
{
"input": "4\n0 1 0 0",
"output": "1"
},
{
"input": "5\n1 1 1 0 0",
"output": "3"
},
{
"input": "4\n1 0 0 0",
"output": "1"
},
{
"input": "6\n1 1 1 1 0 0",
"output": "4"
}
] | 1,633,958,777
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 44
| 77
| 6,963,200
|
n = int(input())
letters = list(map(int, input().split()))
steps = 0 if letters[0] == 0 else 2
for i in range(1,n):
if letters[i] == 1:
steps += 2
if letters[i - 1] == 0:
pass
else:
steps -= 1
else:
pass
if steps == 0:
print(0)
else:
print(steps - 1)
|
Title: Inbox (100500)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Over time, Alexey's mail box got littered with too many letters. Some of them are read, while others are unread.
Alexey's mail program can either show a list of all letters or show the content of a single letter. As soon as the program shows the content of an unread letter, it becomes read letter (if the program shows the content of a read letter nothing happens). In one click he can do any of the following operations:
- Move from the list of letters to the content of any single letter.- Return to the list of letters from single letter viewing mode.- In single letter viewing mode, move to the next or to the previous letter in the list. You cannot move from the first letter to the previous one or from the last letter to the next one.
The program cannot delete the letters from the list or rearrange them.
Alexey wants to read all the unread letters and go watch football. Now he is viewing the list of all letters and for each letter he can see if it is read or unread. What minimum number of operations does Alexey need to perform to read all unread letters?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of letters in the mailbox.
The second line contains *n* space-separated integers (zeros and ones) — the state of the letter list. The *i*-th number equals either 1, if the *i*-th number is unread, or 0, if the *i*-th letter is read.
Output Specification:
Print a single number — the minimum number of operations needed to make all the letters read.
Demo Input:
['5\n0 1 0 1 0\n', '5\n1 1 0 0 1\n', '2\n0 0\n']
Demo Output:
['3\n', '4\n', '0\n']
Note:
In the first sample Alexey needs three operations to cope with the task: open the second letter, move to the third one, move to the fourth one.
In the second sample the action plan: open the first letter, move to the second letter, return to the list, open the fifth letter.
In the third sample all letters are already read.
|
```python
n = int(input())
letters = list(map(int, input().split()))
steps = 0 if letters[0] == 0 else 2
for i in range(1,n):
if letters[i] == 1:
steps += 2
if letters[i - 1] == 0:
pass
else:
steps -= 1
else:
pass
if steps == 0:
print(0)
else:
print(steps - 1)
```
| 3
|
|
832
|
A
|
Sasha and Sticks
|
PROGRAMMING
| 800
|
[
"games",
"math"
] | null | null |
It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn.
|
If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower).
|
[
"1 1\n",
"10 4\n"
] |
[
"YES\n",
"NO\n"
] |
In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win.
| 500
|
[
{
"input": "1 1",
"output": "YES"
},
{
"input": "10 4",
"output": "NO"
},
{
"input": "251656215122324104 164397544865601257",
"output": "YES"
},
{
"input": "963577813436662285 206326039287271924",
"output": "NO"
},
{
"input": "1000000000000000000 1",
"output": "NO"
},
{
"input": "253308697183523656 25332878317796706",
"output": "YES"
},
{
"input": "669038685745448997 501718093668307460",
"output": "YES"
},
{
"input": "116453141993601660 87060381463547965",
"output": "YES"
},
{
"input": "766959657 370931668",
"output": "NO"
},
{
"input": "255787422422806632 146884995820359999",
"output": "YES"
},
{
"input": "502007866464507926 71266379084204128",
"output": "YES"
},
{
"input": "257439908778973480 64157133126869976",
"output": "NO"
},
{
"input": "232709385 91708542",
"output": "NO"
},
{
"input": "252482458300407528 89907711721009125",
"output": "NO"
},
{
"input": "6 2",
"output": "YES"
},
{
"input": "6 3",
"output": "NO"
},
{
"input": "6 4",
"output": "YES"
},
{
"input": "6 5",
"output": "YES"
},
{
"input": "6 6",
"output": "YES"
},
{
"input": "258266151957056904 30153168463725364",
"output": "NO"
},
{
"input": "83504367885565783 52285355047292458",
"output": "YES"
},
{
"input": "545668929424440387 508692735816921376",
"output": "YES"
},
{
"input": "547321411485639939 36665750286082900",
"output": "NO"
},
{
"input": "548973893546839491 183137237979822911",
"output": "NO"
},
{
"input": "544068082 193116851",
"output": "NO"
},
{
"input": "871412474 749817171",
"output": "YES"
},
{
"input": "999999999 1247",
"output": "NO"
},
{
"input": "851941088 712987048",
"output": "YES"
},
{
"input": "559922900 418944886",
"output": "YES"
},
{
"input": "293908937 37520518",
"output": "YES"
},
{
"input": "650075786 130049650",
"output": "NO"
},
{
"input": "1000000000 1000000000",
"output": "YES"
},
{
"input": "548147654663723363 107422751713800746",
"output": "YES"
},
{
"input": "828159210 131819483",
"output": "NO"
},
{
"input": "6242634 4110365",
"output": "YES"
},
{
"input": "458601973 245084155",
"output": "YES"
},
{
"input": "349593257 18089089",
"output": "YES"
},
{
"input": "814768821 312514745",
"output": "NO"
},
{
"input": "697884949 626323363",
"output": "YES"
},
{
"input": "667011589 54866795",
"output": "NO"
},
{
"input": "1000000000000000000 2",
"output": "NO"
},
{
"input": "1000000000000000000 3",
"output": "YES"
},
{
"input": "1000000000000000000 4",
"output": "NO"
},
{
"input": "999999999999999 1",
"output": "YES"
},
{
"input": "17 4",
"output": "NO"
},
{
"input": "2 2",
"output": "YES"
},
{
"input": "1000000000000000 2",
"output": "NO"
},
{
"input": "12 4",
"output": "YES"
},
{
"input": "6 1",
"output": "NO"
},
{
"input": "2 1",
"output": "NO"
},
{
"input": "10000000005 1",
"output": "YES"
},
{
"input": "10000000000000009 2",
"output": "NO"
},
{
"input": "12457895452123 1",
"output": "YES"
},
{
"input": "999999999999999999 9",
"output": "YES"
},
{
"input": "1000000000000 3",
"output": "YES"
},
{
"input": "13099714659575475 6549849616514894",
"output": "NO"
},
{
"input": "100000000000000001 1",
"output": "YES"
},
{
"input": "825175814723458 324",
"output": "YES"
},
{
"input": "20 4",
"output": "YES"
},
{
"input": "100000176877 4",
"output": "YES"
},
{
"input": "100000 3",
"output": "YES"
},
{
"input": "946744073709551614 10",
"output": "YES"
}
] | 1,632,303,027
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 77
| 6,758,400
|
n, k = map(int, input().split())
if n == k:
print("YES")
elif n % k * 2 > k or n % k == 0:
print("YES")
else:
print("NO")
|
Title: Sasha and Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn.
Output Specification:
If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower).
Demo Input:
['1 1\n', '10 4\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win.
|
```python
n, k = map(int, input().split())
if n == k:
print("YES")
elif n % k * 2 > k or n % k == 0:
print("YES")
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"
}
] | 1,644,035,715
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
import math
m,n = map(int,input().split())
prod = m*n
no_of_dominoes = prod//2
print(no_of_dominoes)
|
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
import math
m,n = map(int,input().split())
prod = m*n
no_of_dominoes = prod//2
print(no_of_dominoes)
```
| 3.977
|
197
|
A
|
Plate Game
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"games",
"math"
] | null | null |
You've got a rectangular table with length *a* and width *b* and the infinite number of plates of radius *r*. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
|
A single line contains three space-separated integers *a*, *b*, *r* (1<=≤<=*a*,<=*b*,<=*r*<=≤<=100) — the table sides and the plates' radius, correspondingly.
|
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
|
[
"5 5 2\n",
"6 7 4\n"
] |
[
"First\n",
"Second\n"
] |
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
| 1,000
|
[
{
"input": "5 5 2",
"output": "First"
},
{
"input": "6 7 4",
"output": "Second"
},
{
"input": "100 100 1",
"output": "First"
},
{
"input": "1 1 100",
"output": "Second"
},
{
"input": "13 7 3",
"output": "First"
},
{
"input": "23 7 3",
"output": "First"
},
{
"input": "9 9 2",
"output": "First"
},
{
"input": "13 13 2",
"output": "First"
},
{
"input": "21 21 10",
"output": "First"
},
{
"input": "20 21 10",
"output": "First"
},
{
"input": "20 20 10",
"output": "First"
},
{
"input": "9 13 2",
"output": "First"
},
{
"input": "19 7 3",
"output": "First"
},
{
"input": "19 19 10",
"output": "Second"
},
{
"input": "19 20 10",
"output": "Second"
},
{
"input": "19 21 10",
"output": "Second"
},
{
"input": "1 100 1",
"output": "Second"
},
{
"input": "2 100 1",
"output": "First"
},
{
"input": "3 100 1",
"output": "First"
},
{
"input": "100 100 49",
"output": "First"
},
{
"input": "100 100 50",
"output": "First"
},
{
"input": "100 100 51",
"output": "Second"
},
{
"input": "100 99 50",
"output": "Second"
},
{
"input": "4 10 5",
"output": "Second"
},
{
"input": "8 11 2",
"output": "First"
},
{
"input": "3 12 5",
"output": "Second"
},
{
"input": "14 15 5",
"output": "First"
},
{
"input": "61 2 3",
"output": "Second"
},
{
"input": "82 20 5",
"output": "First"
},
{
"input": "16 80 10",
"output": "Second"
},
{
"input": "2 1 20",
"output": "Second"
},
{
"input": "78 82 5",
"output": "First"
},
{
"input": "8 55 7",
"output": "Second"
},
{
"input": "75 55 43",
"output": "Second"
},
{
"input": "34 43 70",
"output": "Second"
},
{
"input": "86 74 36",
"output": "First"
},
{
"input": "86 74 37",
"output": "First"
},
{
"input": "86 74 38",
"output": "Second"
},
{
"input": "24 70 11",
"output": "First"
},
{
"input": "24 70 12",
"output": "First"
},
{
"input": "24 70 13",
"output": "Second"
},
{
"input": "78 95 38",
"output": "First"
},
{
"input": "78 95 39",
"output": "First"
},
{
"input": "78 95 40",
"output": "Second"
},
{
"input": "88 43 21",
"output": "First"
},
{
"input": "88 43 22",
"output": "Second"
},
{
"input": "88 43 23",
"output": "Second"
},
{
"input": "30 40 14",
"output": "First"
},
{
"input": "30 40 15",
"output": "First"
},
{
"input": "30 40 16",
"output": "Second"
},
{
"input": "2 5 2",
"output": "Second"
},
{
"input": "5 100 3",
"output": "Second"
},
{
"input": "44 58 5",
"output": "First"
},
{
"input": "4 4 6",
"output": "Second"
},
{
"input": "10 20 6",
"output": "Second"
},
{
"input": "100 1 1",
"output": "Second"
},
{
"input": "60 60 1",
"output": "First"
},
{
"input": "100 1 2",
"output": "Second"
},
{
"input": "2 4 2",
"output": "Second"
},
{
"input": "10 90 11",
"output": "Second"
},
{
"input": "20 5 6",
"output": "Second"
},
{
"input": "1 44 2",
"output": "Second"
},
{
"input": "10 5 5",
"output": "Second"
},
{
"input": "5 100 4",
"output": "Second"
},
{
"input": "99 99 50",
"output": "Second"
},
{
"input": "1 100 2",
"output": "Second"
},
{
"input": "100 20 12",
"output": "Second"
},
{
"input": "10 2 4",
"output": "Second"
},
{
"input": "1 50 2",
"output": "Second"
},
{
"input": "10 4 3",
"output": "Second"
},
{
"input": "74 1 1",
"output": "Second"
},
{
"input": "6 6 1",
"output": "First"
},
{
"input": "10 10 1",
"output": "First"
},
{
"input": "21 41 5",
"output": "First"
},
{
"input": "13 1 2",
"output": "Second"
},
{
"input": "1 100 3",
"output": "Second"
},
{
"input": "1 64 2",
"output": "Second"
},
{
"input": "3 4 1",
"output": "First"
},
{
"input": "15 15 1",
"output": "First"
},
{
"input": "15 16 1",
"output": "First"
},
{
"input": "16 15 1",
"output": "First"
},
{
"input": "16 16 1",
"output": "First"
},
{
"input": "15 15 2",
"output": "First"
},
{
"input": "15 16 2",
"output": "First"
},
{
"input": "16 15 2",
"output": "First"
},
{
"input": "16 16 2",
"output": "First"
},
{
"input": "15 15 3",
"output": "First"
},
{
"input": "15 16 3",
"output": "First"
},
{
"input": "16 15 3",
"output": "First"
},
{
"input": "16 16 3",
"output": "First"
},
{
"input": "15 17 3",
"output": "First"
},
{
"input": "16 17 3",
"output": "First"
},
{
"input": "17 17 3",
"output": "First"
},
{
"input": "17 15 3",
"output": "First"
},
{
"input": "17 16 3",
"output": "First"
}
] | 1,530,472,129
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 186
| 0
|
abr = list(map(int, input().strip().split()))
a = abr[0]
b = abr[1]
d = 2 * abr[2]
circle_num = (a // d) * (b // d)
if circle_num % 2 == 0:
print("Second")
else:
print("First")
|
Title: Plate Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a rectangular table with length *a* and width *b* and the infinite number of plates of radius *r*. Two players play the following game: they take turns to put the plates on the table so that the plates don't lie on each other (but they can touch each other), and so that any point on any plate is located within the table's border. During the game one cannot move the plates that already lie on the table. The player who cannot make another move loses. Determine which player wins, the one who moves first or the one who moves second, provided that both players play optimally well.
Input Specification:
A single line contains three space-separated integers *a*, *b*, *r* (1<=≤<=*a*,<=*b*,<=*r*<=≤<=100) — the table sides and the plates' radius, correspondingly.
Output Specification:
If wins the player who moves first, print "First" (without the quotes). Otherwise print "Second" (without the quotes).
Demo Input:
['5 5 2\n', '6 7 4\n']
Demo Output:
['First\n', 'Second\n']
Note:
In the first sample the table has place for only one plate. The first player puts a plate on the table, the second player can't do that and loses.
In the second sample the table is so small that it doesn't have enough place even for one plate. So the first player loses without making a single move.
|
```python
abr = list(map(int, input().strip().split()))
a = abr[0]
b = abr[1]
d = 2 * abr[2]
circle_num = (a // d) * (b // d)
if circle_num % 2 == 0:
print("Second")
else:
print("First")
```
| 0
|
|
231
|
A
|
Team
|
PROGRAMMING
| 800
|
[
"brute force",
"greedy"
] | null | null |
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
|
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
|
Print a single integer — the number of problems the friends will implement on the contest.
|
[
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] |
[
"2\n",
"1\n"
] |
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
| 500
|
[
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,697,889,463
| 2,147,483,647
|
PyPy 3-64
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
n = int(input())
c =0
for i in range(n):
a,b,c = int(input().split())
if a + b +c >=2 :
c++
print(c)
|
Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
|
```python
n = int(input())
c =0
for i in range(n):
a,b,c = int(input().split())
if a + b +c >=2 :
c++
print(c)
```
| -1
|
|
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"
}
] | 1,642,434,481
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 35
| 184
| 0
|
m,n=map(int,input().split())
if m%2==0 and n%2==0:#2*4
print((m//2)*n)
elif m%2!=0 and n%2==0:#3*4
print(m*(n//2))
elif m%2==0 and n%2!=0:
print(n*(m//2))
else:
print((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%2==0 and n%2==0:#2*4
print((m//2)*n)
elif m%2!=0 and n%2==0:#3*4
print(m*(n//2))
elif m%2==0 and n%2!=0:
print(n*(m//2))
else:
print((m*n)//2)
```
| 3.954
|
16
|
C
|
Monitor
|
PROGRAMMING
| 1,800
|
[
"binary search",
"number theory"
] |
C. Monitor
|
0
|
64
|
Reca company makes monitors, the most popular of their models is AB999 with the screen size *a*<=×<=*b* centimeters. Because of some production peculiarities a screen parameters are integer numbers. Recently the screen sides ratio *x*:<=*y* became popular with users. That's why the company wants to reduce monitor AB999 size so that its screen sides ratio becomes *x*:<=*y*, at the same time they want its total area to be maximal of all possible variants. Your task is to find the screen parameters of the reduced size model, or find out that such a reduction can't be performed.
|
The first line of the input contains 4 integers — *a*, *b*, *x* and *y* (1<=≤<=*a*,<=*b*,<=*x*,<=*y*<=≤<=2·109).
|
If the answer exists, output 2 positive integers — screen parameters of the reduced size model. Output 0 0 otherwise.
|
[
"800 600 4 3\n",
"1920 1200 16 9\n",
"1 1 1 2\n"
] |
[
"800 600\n",
"1920 1080\n",
"0 0\n"
] |
none
| 0
|
[
{
"input": "800 600 4 3",
"output": "800 600"
},
{
"input": "1920 1200 16 9",
"output": "1920 1080"
},
{
"input": "1 1 1 2",
"output": "0 0"
},
{
"input": "1002105126 227379125 179460772 1295256518",
"output": "0 0"
},
{
"input": "625166755 843062051 1463070160 1958300154",
"output": "0 0"
},
{
"input": "248228385 1458744978 824699604 1589655888",
"output": "206174901 397413972"
},
{
"input": "186329049 1221011622 90104472 1769702163",
"output": "60069648 1179801442"
},
{
"input": "511020182 242192314 394753578 198572007",
"output": "394753578 198572007"
},
{
"input": "134081812 857875240 82707261 667398699",
"output": "105411215 850606185"
},
{
"input": "721746595 799202881 143676564 380427290",
"output": "287353128 760854580"
},
{
"input": "912724694 1268739154 440710604 387545692",
"output": "881421208 775091384"
},
{
"input": "1103702793 1095784840 788679477 432619528",
"output": "788679477 432619528"
},
{
"input": "548893795 861438648 131329677 177735812",
"output": "525318708 710943248"
},
{
"input": "652586118 1793536161 127888702 397268645",
"output": "511554808 1589074580"
},
{
"input": "756278440 578150025 96644319 26752094",
"output": "676510233 187264658"
},
{
"input": "859970763 1510247537 37524734 97452508",
"output": "562871010 1461787620"
},
{
"input": "547278097 1977241684 51768282 183174370",
"output": "543566961 1923330885"
},
{
"input": "62256611 453071697 240966 206678",
"output": "62169228 53322924"
},
{
"input": "1979767797 878430446 5812753 3794880",
"output": "1342745943 876617280"
},
{
"input": "1143276347 1875662241 178868040 116042960",
"output": "1140283755 739773870"
},
{
"input": "435954880 1740366589 19415065 185502270",
"output": "182099920 1739883360"
},
{
"input": "664035593 983601098 4966148 2852768",
"output": "664032908 381448928"
},
{
"input": "1461963719 350925487 135888396 83344296",
"output": "572153868 350918568"
},
{
"input": "754199095 348965411 161206703 67014029",
"output": "754119492 313489356"
},
{
"input": "166102153 494841162 14166516 76948872",
"output": "91096406 494812252"
},
{
"input": "1243276346 1975662240 38441120 291740200",
"output": "259477560 1969246350"
},
{
"input": "535954879 1840366588 26278959 73433046",
"output": "535849118 1497358892"
},
{
"input": "764035592 1083601097 1192390 7267738",
"output": "177777265 1083570463"
},
{
"input": "1561963718 450925486 475523188 136236856",
"output": "1561914768 447486816"
},
{
"input": "854199094 448965410 364102983 125971431",
"output": "853687785 295356745"
},
{
"input": "266102152 594841161 15854566 13392106",
"output": "266043102 224722482"
},
{
"input": "1 1 2 1",
"output": "0 0"
},
{
"input": "2000000000 2000000000 1 1999999999",
"output": "1 1999999999"
},
{
"input": "2000000000 2000000000 1999999999 1",
"output": "1999999999 1"
},
{
"input": "2000000000 2000000000 2 1999999999",
"output": "2 1999999999"
},
{
"input": "1000000000 1000000000 999999999 2",
"output": "999999999 2"
},
{
"input": "2000000000 2000000000 1999999999 2",
"output": "1999999999 2"
},
{
"input": "2000000000 2000000000 1999999999 1999999998",
"output": "1999999999 1999999998"
},
{
"input": "2000000000 2000000000 1999999998 1999999999",
"output": "1999999998 1999999999"
}
] | 1,502,163,339
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 3
| 500
| 4,608,000
|
s = input()
a = []
a = s.split()
for i in range(len(a)):
a[i] = int(a[i])
while a[0]/a[2] != a[1]/a[3]:
if a[0]/a[2] < a[1]/a[3]:
a[1] = int((a[0]/a[2])*a[3])
else:
a[0] = int((a[1]/a[3])*a[2])
print(a[0], ' ', a[1])
|
Title: Monitor
Time Limit: 0 seconds
Memory Limit: 64 megabytes
Problem Description:
Reca company makes monitors, the most popular of their models is AB999 with the screen size *a*<=×<=*b* centimeters. Because of some production peculiarities a screen parameters are integer numbers. Recently the screen sides ratio *x*:<=*y* became popular with users. That's why the company wants to reduce monitor AB999 size so that its screen sides ratio becomes *x*:<=*y*, at the same time they want its total area to be maximal of all possible variants. Your task is to find the screen parameters of the reduced size model, or find out that such a reduction can't be performed.
Input Specification:
The first line of the input contains 4 integers — *a*, *b*, *x* and *y* (1<=≤<=*a*,<=*b*,<=*x*,<=*y*<=≤<=2·109).
Output Specification:
If the answer exists, output 2 positive integers — screen parameters of the reduced size model. Output 0 0 otherwise.
Demo Input:
['800 600 4 3\n', '1920 1200 16 9\n', '1 1 1 2\n']
Demo Output:
['800 600\n', '1920 1080\n', '0 0\n']
Note:
none
|
```python
s = input()
a = []
a = s.split()
for i in range(len(a)):
a[i] = int(a[i])
while a[0]/a[2] != a[1]/a[3]:
if a[0]/a[2] < a[1]/a[3]:
a[1] = int((a[0]/a[2])*a[3])
else:
a[0] = int((a[1]/a[3])*a[2])
print(a[0], ' ', a[1])
```
| 0
|
284
|
A
|
Cows and Primitive Roots
|
PROGRAMMING
| 1,400
|
[
"implementation",
"math",
"number theory"
] | null | null |
The cows have just learned what a primitive root is! Given a prime *p*, a primitive root is an integer *x* (1<=≤<=*x*<=<<=*p*) such that none of integers *x*<=-<=1,<=*x*2<=-<=1,<=...,<=*x**p*<=-<=2<=-<=1 are divisible by *p*, but *x**p*<=-<=1<=-<=1 is.
Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime *p*, help the cows find the number of primitive roots .
|
The input contains a single line containing an integer *p* (2<=≤<=*p*<=<<=2000). It is guaranteed that *p* is a prime.
|
Output on a single line the number of primitive roots .
|
[
"3\n",
"5\n"
] |
[
"1\n",
"2\n"
] |
The only primitive root <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3722298ba062e95b18705d1253eb4e5d31e3b2d1.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.
The primitive roots <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1d85c6a17ef1c42b53cf94d00bc49a7ac458fd58.png" style="max-width: 100.0%;max-height: 100.0%;"/> are 2 and 3.
| 500
|
[
{
"input": "3",
"output": "1"
},
{
"input": "5",
"output": "2"
},
{
"input": "7",
"output": "2"
},
{
"input": "11",
"output": "4"
},
{
"input": "17",
"output": "8"
},
{
"input": "19",
"output": "6"
},
{
"input": "1583",
"output": "672"
},
{
"input": "997",
"output": "328"
},
{
"input": "2",
"output": "1"
},
{
"input": "1987",
"output": "660"
},
{
"input": "1993",
"output": "656"
},
{
"input": "1997",
"output": "996"
},
{
"input": "1999",
"output": "648"
},
{
"input": "1297",
"output": "432"
},
{
"input": "163",
"output": "54"
},
{
"input": "1259",
"output": "576"
},
{
"input": "821",
"output": "320"
},
{
"input": "29",
"output": "12"
},
{
"input": "43",
"output": "12"
},
{
"input": "1949",
"output": "972"
},
{
"input": "1879",
"output": "624"
},
{
"input": "1759",
"output": "584"
},
{
"input": "1597",
"output": "432"
},
{
"input": "1619",
"output": "808"
},
{
"input": "1277",
"output": "560"
},
{
"input": "1097",
"output": "544"
},
{
"input": "953",
"output": "384"
},
{
"input": "751",
"output": "200"
},
{
"input": "607",
"output": "200"
},
{
"input": "509",
"output": "252"
},
{
"input": "431",
"output": "168"
},
{
"input": "317",
"output": "156"
},
{
"input": "211",
"output": "48"
},
{
"input": "127",
"output": "36"
},
{
"input": "101",
"output": "40"
},
{
"input": "31",
"output": "8"
}
] | 1,699,536,751
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 62
| 0
|
p = int(input())
total = 0
for x in range(2, p):
if (x ** ((p - 1) / 2)) % p == p - 1:
total += 1
print(total)
|
Title: Cows and Primitive Roots
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The cows have just learned what a primitive root is! Given a prime *p*, a primitive root is an integer *x* (1<=≤<=*x*<=<<=*p*) such that none of integers *x*<=-<=1,<=*x*2<=-<=1,<=...,<=*x**p*<=-<=2<=-<=1 are divisible by *p*, but *x**p*<=-<=1<=-<=1 is.
Unfortunately, computing primitive roots can be time consuming, so the cows need your help. Given a prime *p*, help the cows find the number of primitive roots .
Input Specification:
The input contains a single line containing an integer *p* (2<=≤<=*p*<=<<=2000). It is guaranteed that *p* is a prime.
Output Specification:
Output on a single line the number of primitive roots .
Demo Input:
['3\n', '5\n']
Demo Output:
['1\n', '2\n']
Note:
The only primitive root <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3722298ba062e95b18705d1253eb4e5d31e3b2d1.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.
The primitive roots <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1d85c6a17ef1c42b53cf94d00bc49a7ac458fd58.png" style="max-width: 100.0%;max-height: 100.0%;"/> are 2 and 3.
|
```python
p = int(input())
total = 0
for x in range(2, p):
if (x ** ((p - 1) / 2)) % p == p - 1:
total += 1
print(total)
```
| 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"
}
] | 1,696,586,739
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 38
| 248
| 13,824,000
|
n, t = map(int, input().split())
book_times = list(map(int, input().split()))
left = 0 # Left pointer initially points to the first book
current_time = 0 # Initialize the current time spent reading books
max_books = 0 # Initialize the maximum number of books read to 0
for right in range(n):
current_time += book_times[right]
# If the current time exceeds the available time, move the left pointer
while current_time > t:
current_time -= book_times[left]
left += 1
# Update the maximum number of books read
max_books = max(max_books, right - left + 1)
print(max_books)
|
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())
book_times = list(map(int, input().split()))
left = 0 # Left pointer initially points to the first book
current_time = 0 # Initialize the current time spent reading books
max_books = 0 # Initialize the maximum number of books read to 0
for right in range(n):
current_time += book_times[right]
# If the current time exceeds the available time, move the left pointer
while current_time > t:
current_time -= book_times[left]
left += 1
# Update the maximum number of books read
max_books = max(max_books, right - left + 1)
print(max_books)
```
| 3
|
|
765
|
C
|
Table Tennis Game 2
|
PROGRAMMING
| 1,200
|
[
"math"
] | null | null |
Misha and Vanya have played several table tennis sets. Each set consists of several serves, each serve is won by one of the players, he receives one point and the loser receives nothing. Once one of the players scores exactly *k* points, the score is reset and a new set begins.
Across all the sets Misha scored *a* points in total, and Vanya scored *b* points. Given this information, determine the maximum number of sets they could have played, or that the situation is impossible.
Note that the game consisted of several complete sets.
|
The first line contains three space-separated integers *k*, *a* and *b* (1<=≤<=*k*<=≤<=109, 0<=≤<=*a*,<=*b*<=≤<=109, *a*<=+<=*b*<=><=0).
|
If the situation is impossible, print a single number -1. Otherwise, print the maximum possible number of sets.
|
[
"11 11 5\n",
"11 2 3\n"
] |
[
"1\n",
"-1\n"
] |
Note that the rules of the game in this problem differ from the real table tennis game, for example, the rule of "balance" (the winning player has to be at least two points ahead to win a set) has no power within the present problem.
| 1,250
|
[
{
"input": "11 11 5",
"output": "1"
},
{
"input": "11 2 3",
"output": "-1"
},
{
"input": "1 5 9",
"output": "14"
},
{
"input": "2 3 3",
"output": "2"
},
{
"input": "1 1000000000 1000000000",
"output": "2000000000"
},
{
"input": "2 3 5",
"output": "3"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "2"
},
{
"input": "1 0 1",
"output": "1"
},
{
"input": "101 99 97",
"output": "-1"
},
{
"input": "1000000000 0 1",
"output": "-1"
},
{
"input": "137 137 136",
"output": "1"
},
{
"input": "255 255 255",
"output": "2"
},
{
"input": "1 0 1000000000",
"output": "1000000000"
},
{
"input": "123 456 789",
"output": "9"
},
{
"input": "666666 6666666 666665",
"output": "-1"
},
{
"input": "1000000000 999999999 999999999",
"output": "-1"
},
{
"input": "100000000 100000001 99999999",
"output": "-1"
},
{
"input": "3 2 1000000000",
"output": "-1"
},
{
"input": "999999999 1000000000 999999998",
"output": "-1"
},
{
"input": "12938621 192872393 102739134",
"output": "21"
},
{
"input": "666666666 1230983 666666666",
"output": "1"
},
{
"input": "123456789 123456789 123456787",
"output": "1"
},
{
"input": "5 6 0",
"output": "-1"
},
{
"input": "11 0 12",
"output": "-1"
},
{
"input": "2 11 0",
"output": "-1"
},
{
"input": "2 1 0",
"output": "-1"
},
{
"input": "10 11 12",
"output": "2"
},
{
"input": "11 12 5",
"output": "-1"
},
{
"input": "11 12 3",
"output": "-1"
},
{
"input": "11 15 4",
"output": "-1"
},
{
"input": "2 3 1",
"output": "-1"
},
{
"input": "11 12 0",
"output": "-1"
},
{
"input": "11 13 2",
"output": "-1"
},
{
"input": "11 23 22",
"output": "4"
},
{
"input": "10 21 0",
"output": "-1"
},
{
"input": "11 23 1",
"output": "-1"
},
{
"input": "11 10 12",
"output": "-1"
},
{
"input": "11 1 12",
"output": "-1"
},
{
"input": "11 5 12",
"output": "-1"
},
{
"input": "11 8 12",
"output": "-1"
},
{
"input": "11 12 1",
"output": "-1"
},
{
"input": "5 4 6",
"output": "-1"
},
{
"input": "10 1 22",
"output": "-1"
},
{
"input": "2 3 0",
"output": "-1"
},
{
"input": "11 23 2",
"output": "-1"
},
{
"input": "2 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "11 0 15",
"output": "-1"
},
{
"input": "11 5 0",
"output": "-1"
},
{
"input": "11 5 15",
"output": "-1"
},
{
"input": "10 0 13",
"output": "-1"
},
{
"input": "4 7 0",
"output": "-1"
},
{
"input": "10 2 8",
"output": "-1"
},
{
"input": "11 5 22",
"output": "2"
},
{
"input": "11 13 0",
"output": "-1"
},
{
"input": "2 0 3",
"output": "-1"
},
{
"input": "10 10 0",
"output": "1"
},
{
"input": "10 11 10",
"output": "2"
},
{
"input": "3 5 4",
"output": "2"
},
{
"input": "11 22 3",
"output": "2"
},
{
"input": "11 12 10",
"output": "-1"
},
{
"input": "10 2 13",
"output": "-1"
},
{
"input": "5 6 1",
"output": "-1"
},
{
"input": "10 21 5",
"output": "-1"
},
{
"input": "10 11 9",
"output": "-1"
},
{
"input": "10 17 7",
"output": "-1"
},
{
"input": "3 4 1",
"output": "-1"
},
{
"input": "4 5 3",
"output": "-1"
},
{
"input": "11 3 23",
"output": "-1"
},
{
"input": "11 3 12",
"output": "-1"
},
{
"input": "2 5 0",
"output": "-1"
},
{
"input": "10 21 2",
"output": "-1"
},
{
"input": "5 1 6",
"output": "-1"
},
{
"input": "10 11 0",
"output": "-1"
},
{
"input": "10 9 11",
"output": "-1"
},
{
"input": "7 10 5",
"output": "-1"
},
{
"input": "5 7 2",
"output": "-1"
},
{
"input": "6 5 7",
"output": "-1"
},
{
"input": "11 16 2",
"output": "-1"
},
{
"input": "11 1000000000 10",
"output": "-1"
},
{
"input": "10 2 21",
"output": "-1"
},
{
"input": "10 15 1",
"output": "-1"
},
{
"input": "5 2 8",
"output": "-1"
},
{
"input": "11 10000000 10",
"output": "-1"
},
{
"input": "10 1 101",
"output": "-1"
},
{
"input": "20 24 2",
"output": "-1"
},
{
"input": "11 24 0",
"output": "-1"
},
{
"input": "11 17 4",
"output": "-1"
},
{
"input": "11 13 1",
"output": "-1"
},
{
"input": "10 11 2",
"output": "-1"
},
{
"input": "11 23 3",
"output": "-1"
},
{
"input": "10 99 0",
"output": "-1"
},
{
"input": "6 7 4",
"output": "-1"
},
{
"input": "11 1 22",
"output": "2"
},
{
"input": "11 2 13",
"output": "-1"
},
{
"input": "2 1 3",
"output": "-1"
},
{
"input": "11 6 18",
"output": "-1"
},
{
"input": "11 122 4",
"output": "-1"
},
{
"input": "11 21 10",
"output": "-1"
},
{
"input": "3 2 4",
"output": "-1"
},
{
"input": "9 11 2",
"output": "-1"
},
{
"input": "11 0 7",
"output": "-1"
},
{
"input": "5 9 4",
"output": "-1"
},
{
"input": "100 105 5",
"output": "-1"
},
{
"input": "11 15 0",
"output": "-1"
},
{
"input": "5 6 4",
"output": "-1"
},
{
"input": "3 4 2",
"output": "-1"
},
{
"input": "2 9 0",
"output": "-1"
},
{
"input": "11 13 11",
"output": "2"
},
{
"input": "11 15 5",
"output": "-1"
},
{
"input": "11 4 15",
"output": "-1"
},
{
"input": "10 1 0",
"output": "-1"
},
{
"input": "11 16 8",
"output": "-1"
},
{
"input": "10 43 0",
"output": "-1"
},
{
"input": "11 13 5",
"output": "-1"
},
{
"input": "11 22 0",
"output": "2"
},
{
"input": "5 6 3",
"output": "-1"
},
{
"input": "2 1 11",
"output": "-1"
},
{
"input": "4 5 1",
"output": "-1"
},
{
"input": "11 23 0",
"output": "-1"
},
{
"input": "11 4 12",
"output": "-1"
},
{
"input": "12 13 1",
"output": "-1"
},
{
"input": "10 19 9",
"output": "-1"
},
{
"input": "3 7 2",
"output": "-1"
},
{
"input": "12 18 0",
"output": "-1"
},
{
"input": "11 25 3",
"output": "-1"
},
{
"input": "11 23 5",
"output": "-1"
},
{
"input": "2 1 5",
"output": "-1"
},
{
"input": "2 0 5",
"output": "-1"
},
{
"input": "11 24 1",
"output": "-1"
},
{
"input": "10 11 4",
"output": "-1"
},
{
"input": "2 0 1",
"output": "-1"
},
{
"input": "10 0 21",
"output": "-1"
},
{
"input": "3 0 7",
"output": "-1"
},
{
"input": "18 11 21",
"output": "-1"
},
{
"input": "3 7 0",
"output": "-1"
},
{
"input": "5 11 0",
"output": "-1"
},
{
"input": "11 5 13",
"output": "-1"
},
{
"input": "11 9 34",
"output": "-1"
},
{
"input": "11 13 9",
"output": "-1"
},
{
"input": "10 0 22",
"output": "-1"
},
{
"input": "5 1 12",
"output": "-1"
},
{
"input": "11 2 12",
"output": "-1"
},
{
"input": "11 9 12",
"output": "-1"
},
{
"input": "11 24 2",
"output": "-1"
},
{
"input": "11 23 6",
"output": "-1"
},
{
"input": "11 20 4",
"output": "-1"
},
{
"input": "2 5 1",
"output": "-1"
},
{
"input": "120 132 133",
"output": "2"
},
{
"input": "11 111 4",
"output": "-1"
},
{
"input": "10 7 11",
"output": "-1"
},
{
"input": "6 13 0",
"output": "-1"
},
{
"input": "5 11 1",
"output": "-1"
},
{
"input": "11 5 27",
"output": "-1"
},
{
"input": "11 15 3",
"output": "-1"
},
{
"input": "11 0 13",
"output": "-1"
},
{
"input": "11 13 10",
"output": "-1"
},
{
"input": "11 25 5",
"output": "-1"
},
{
"input": "4 3 5",
"output": "-1"
},
{
"input": "100 199 100",
"output": "2"
},
{
"input": "11 2 22",
"output": "2"
},
{
"input": "10 20 2",
"output": "2"
},
{
"input": "5 5 0",
"output": "1"
},
{
"input": "10 11 1",
"output": "-1"
},
{
"input": "11 12 2",
"output": "-1"
},
{
"input": "5 16 3",
"output": "-1"
},
{
"input": "12 14 1",
"output": "-1"
},
{
"input": "10 22 2",
"output": "-1"
},
{
"input": "2 4 0",
"output": "2"
},
{
"input": "11 34 7",
"output": "-1"
},
{
"input": "6 13 1",
"output": "-1"
},
{
"input": "11 0 23",
"output": "-1"
},
{
"input": "20 21 19",
"output": "-1"
},
{
"input": "11 33 22",
"output": "5"
},
{
"input": "10 4 41",
"output": "-1"
},
{
"input": "3 4 0",
"output": "-1"
},
{
"input": "11 15 7",
"output": "-1"
},
{
"input": "5 0 6",
"output": "-1"
},
{
"input": "11 3 22",
"output": "2"
},
{
"input": "2 6 0",
"output": "3"
},
{
"input": "10 11 11",
"output": "2"
},
{
"input": "11 33 0",
"output": "3"
},
{
"input": "4 6 2",
"output": "-1"
},
{
"input": "11 76 2",
"output": "-1"
},
{
"input": "7 9 4",
"output": "-1"
},
{
"input": "10 43 1",
"output": "-1"
},
{
"input": "22 25 5",
"output": "-1"
},
{
"input": "3 5 2",
"output": "-1"
},
{
"input": "11 1 24",
"output": "-1"
},
{
"input": "12 25 3",
"output": "-1"
},
{
"input": "11 0 22",
"output": "2"
},
{
"input": "4 2 5",
"output": "-1"
},
{
"input": "11 13 3",
"output": "-1"
},
{
"input": "11 12 9",
"output": "-1"
},
{
"input": "11 35 1",
"output": "-1"
},
{
"input": "5 3 6",
"output": "-1"
},
{
"input": "5 11 4",
"output": "-1"
},
{
"input": "12 8 14",
"output": "-1"
},
{
"input": "10 12 9",
"output": "-1"
},
{
"input": "11 12 13",
"output": "2"
},
{
"input": "11 15 2",
"output": "-1"
},
{
"input": "11 23 4",
"output": "-1"
},
{
"input": "5 3 11",
"output": "-1"
},
{
"input": "6 13 2",
"output": "-1"
},
{
"input": "4 1 0",
"output": "-1"
},
{
"input": "11 32 10",
"output": "-1"
},
{
"input": "2 11 1",
"output": "-1"
},
{
"input": "10 11 7",
"output": "-1"
},
{
"input": "11 26 0",
"output": "-1"
},
{
"input": "100 205 5",
"output": "-1"
},
{
"input": "4 0 2",
"output": "-1"
},
{
"input": "10 11 8",
"output": "-1"
},
{
"input": "11 22 5",
"output": "2"
},
{
"input": "4 0 5",
"output": "-1"
},
{
"input": "11 87 22",
"output": "9"
},
{
"input": "4 8 0",
"output": "2"
},
{
"input": "9 8 17",
"output": "-1"
},
{
"input": "10 20 0",
"output": "2"
},
{
"input": "10 9 19",
"output": "-1"
},
{
"input": "12 2 13",
"output": "-1"
},
{
"input": "11 24 5",
"output": "-1"
},
{
"input": "10 1 11",
"output": "-1"
},
{
"input": "4 0 9",
"output": "-1"
},
{
"input": "3 0 1",
"output": "-1"
},
{
"input": "11 12 4",
"output": "-1"
},
{
"input": "3 8 2",
"output": "-1"
},
{
"input": "11 17 10",
"output": "-1"
},
{
"input": "6 1 13",
"output": "-1"
},
{
"input": "11 25 0",
"output": "-1"
},
{
"input": "12 0 13",
"output": "-1"
},
{
"input": "10 5 20",
"output": "2"
},
{
"input": "11 89 2",
"output": "-1"
},
{
"input": "2 4 1",
"output": "2"
},
{
"input": "10 31 0",
"output": "-1"
},
{
"input": "11 34 1",
"output": "-1"
},
{
"input": "999 6693 8331",
"output": "14"
},
{
"input": "10 55 1",
"output": "-1"
},
{
"input": "11 12 8",
"output": "-1"
},
{
"input": "1 9 22",
"output": "31"
},
{
"input": "7572 9186 895",
"output": "-1"
},
{
"input": "3 2 11",
"output": "-1"
},
{
"input": "2 1 4",
"output": "2"
},
{
"input": "11 10 19",
"output": "-1"
},
{
"input": "100 199 99",
"output": "-1"
},
{
"input": "2537 8926 1523",
"output": "-1"
},
{
"input": "11 0 5",
"output": "-1"
},
{
"input": "5 1 11",
"output": "-1"
},
{
"input": "12 13 5",
"output": "-1"
},
{
"input": "10 12 0",
"output": "-1"
},
{
"input": "5 4 7",
"output": "-1"
},
{
"input": "12 25 1",
"output": "-1"
},
{
"input": "7 9 0",
"output": "-1"
},
{
"input": "4 15 0",
"output": "-1"
},
{
"input": "5 11 2",
"output": "-1"
},
{
"input": "11 58 3",
"output": "-1"
},
{
"input": "10 11 5",
"output": "-1"
},
{
"input": "10 3 1003",
"output": "-1"
},
{
"input": "11 14 0",
"output": "-1"
},
{
"input": "11 8 23",
"output": "-1"
},
{
"input": "11 22 4",
"output": "2"
},
{
"input": "5 2 0",
"output": "-1"
},
{
"input": "11 1 35",
"output": "-1"
},
{
"input": "12 25 5",
"output": "-1"
},
{
"input": "11 37 4",
"output": "-1"
},
{
"input": "20 42 3",
"output": "-1"
},
{
"input": "10 12 2",
"output": "-1"
},
{
"input": "4 1 5",
"output": "-1"
},
{
"input": "5 16 0",
"output": "-1"
},
{
"input": "11 13 7",
"output": "-1"
},
{
"input": "5 0 3",
"output": "-1"
},
{
"input": "11 2 11",
"output": "1"
},
{
"input": "3 2 2000",
"output": "-1"
},
{
"input": "3 0 4241",
"output": "-1"
},
{
"input": "2 12 11",
"output": "11"
},
{
"input": "10 2 82",
"output": "-1"
},
{
"input": "11 0 11",
"output": "1"
},
{
"input": "11 20 20",
"output": "2"
},
{
"input": "5 0 10",
"output": "2"
},
{
"input": "11 1 0",
"output": "-1"
},
{
"input": "11 13 13",
"output": "2"
}
] | 1,487,098,641
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
k, a, b = map(int, input().split())
if a < k and b < k:
print(-1)
else if a < k and b % k != 0:
print(-1)
else if b < k and a % k != 0:
print(-1)
else:
print(a // k + b // k)
|
Title: Table Tennis Game 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha and Vanya have played several table tennis sets. Each set consists of several serves, each serve is won by one of the players, he receives one point and the loser receives nothing. Once one of the players scores exactly *k* points, the score is reset and a new set begins.
Across all the sets Misha scored *a* points in total, and Vanya scored *b* points. Given this information, determine the maximum number of sets they could have played, or that the situation is impossible.
Note that the game consisted of several complete sets.
Input Specification:
The first line contains three space-separated integers *k*, *a* and *b* (1<=≤<=*k*<=≤<=109, 0<=≤<=*a*,<=*b*<=≤<=109, *a*<=+<=*b*<=><=0).
Output Specification:
If the situation is impossible, print a single number -1. Otherwise, print the maximum possible number of sets.
Demo Input:
['11 11 5\n', '11 2 3\n']
Demo Output:
['1\n', '-1\n']
Note:
Note that the rules of the game in this problem differ from the real table tennis game, for example, the rule of "balance" (the winning player has to be at least two points ahead to win a set) has no power within the present problem.
|
```python
k, a, b = map(int, input().split())
if a < k and b < k:
print(-1)
else if a < k and b % k != 0:
print(-1)
else if b < k and a % k != 0:
print(-1)
else:
print(a // k + b // k)
```
| -1
|
|
508
|
A
|
Pasha and Pixels
|
PROGRAMMING
| 1,100
|
[
"brute force"
] | null | null |
Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.
Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of *n* row with *m* pixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2<=×<=2 square consisting of black pixels is formed.
Pasha has made a plan of *k* moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers *i* and *j*, denoting respectively the row and the column of the pixel to be colored on the current move.
Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2<=×<=2 square consisting of black pixels is formed.
|
The first line of the input contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform.
The next *k* lines contain Pasha's moves in the order he makes them. Each line contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*j*<=≤<=*m*), representing the row number and column number of the pixel that was painted during a move.
|
If Pasha loses, print the number of the move when the 2<=×<=2 square consisting of black pixels is formed.
If Pasha doesn't lose, that is, no 2<=×<=2 square consisting of black pixels is formed during the given *k* moves, print 0.
|
[
"2 2 4\n1 1\n1 2\n2 1\n2 2\n",
"2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1\n",
"5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2\n"
] |
[
"4\n",
"5\n",
"0\n"
] |
none
| 500
|
[
{
"input": "2 2 4\n1 1\n1 2\n2 1\n2 2",
"output": "4"
},
{
"input": "2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1",
"output": "5"
},
{
"input": "5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2",
"output": "0"
},
{
"input": "3 3 11\n2 1\n3 1\n1 1\n1 3\n1 2\n2 3\n3 3\n3 2\n2 2\n1 3\n3 3",
"output": "9"
},
{
"input": "2 2 5\n1 1\n2 1\n2 1\n1 2\n2 2",
"output": "5"
},
{
"input": "518 518 10\n37 97\n47 278\n17 467\n158 66\n483 351\n83 123\n285 219\n513 187\n380 75\n304 352",
"output": "0"
},
{
"input": "1 1 5\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "0"
},
{
"input": "1 5 5\n1 1\n1 2\n1 3\n1 4\n1 5",
"output": "0"
},
{
"input": "5 1 5\n1 1\n2 1\n3 1\n4 1\n5 1",
"output": "0"
},
{
"input": "1 1 1\n1 1",
"output": "0"
},
{
"input": "10 10 4\n5 9\n6 9\n6 10\n5 10",
"output": "4"
},
{
"input": "1000 1000 4\n999 999\n999 1000\n1000 999\n1000 1000",
"output": "4"
},
{
"input": "2 3 5\n2 3\n1 3\n1 2\n1 1\n2 2",
"output": "5"
},
{
"input": "1000 1000 4\n1000 1000\n999 999\n1000 999\n999 1000",
"output": "4"
}
] | 1,697,685,814
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 12
| 2,000
| 14,028,800
|
list1=input().split()
step_record=set()
steps=[]#第几步走了啥
for z in range(int(list1[-1])):
step=input().split()
x,y=int(step[0]),int(step[1])
step_record.add((x, y))
steps.append((x,y))
sign=[]
for a,b in step_record:
if (a-1,b-1) in step_record and (a,b-1) in step_record and (a-1,b) in step_record:
sign.append(max(steps.index((a, b)), steps.index((a - 1, b - 1)), steps.index((a, b - 1)), steps.index((a - 1, b))))
if not sign:
print(0)
else:
print(min(sign)+1)
|
Title: Pasha and Pixels
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.
Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of *n* row with *m* pixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2<=×<=2 square consisting of black pixels is formed.
Pasha has made a plan of *k* moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers *i* and *j*, denoting respectively the row and the column of the pixel to be colored on the current move.
Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2<=×<=2 square consisting of black pixels is formed.
Input Specification:
The first line of the input contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform.
The next *k* lines contain Pasha's moves in the order he makes them. Each line contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*j*<=≤<=*m*), representing the row number and column number of the pixel that was painted during a move.
Output Specification:
If Pasha loses, print the number of the move when the 2<=×<=2 square consisting of black pixels is formed.
If Pasha doesn't lose, that is, no 2<=×<=2 square consisting of black pixels is formed during the given *k* moves, print 0.
Demo Input:
['2 2 4\n1 1\n1 2\n2 1\n2 2\n', '2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1\n', '5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2\n']
Demo Output:
['4\n', '5\n', '0\n']
Note:
none
|
```python
list1=input().split()
step_record=set()
steps=[]#第几步走了啥
for z in range(int(list1[-1])):
step=input().split()
x,y=int(step[0]),int(step[1])
step_record.add((x, y))
steps.append((x,y))
sign=[]
for a,b in step_record:
if (a-1,b-1) in step_record and (a,b-1) in step_record and (a-1,b) in step_record:
sign.append(max(steps.index((a, b)), steps.index((a - 1, b - 1)), steps.index((a, b - 1)), steps.index((a - 1, b))))
if not sign:
print(0)
else:
print(min(sign)+1)
```
| 0
|
|
839
|
C
|
Journey
|
PROGRAMMING
| 1,500
|
[
"dfs and similar",
"dp",
"graphs",
"probabilities",
"trees"
] | null | null |
There are *n* cities and *n*<=-<=1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link [https://en.wikipedia.org/wiki/Expected_value](https://en.wikipedia.org/wiki/Expected_value).
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100000) — number of cities.
Then *n*<=-<=1 lines follow. The *i*-th line of these lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*) — the cities connected by the *i*-th road.
It is guaranteed that one can reach any city from any other by the roads.
|
Print a number — the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
|
[
"4\n1 2\n1 3\n2 4\n",
"5\n1 2\n1 3\n3 4\n2 5\n"
] |
[
"1.500000000000000\n",
"2.000000000000000\n"
] |
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
| 1,500
|
[
{
"input": "4\n1 2\n1 3\n2 4",
"output": "1.500000000000000"
},
{
"input": "5\n1 2\n1 3\n3 4\n2 5",
"output": "2.000000000000000"
},
{
"input": "70\n1 25\n57 1\n18 1\n65 1\n38 1\n1 41\n1 5\n1 69\n1 3\n31 1\n1 8\n1 9\n53 1\n70 1\n45 1\n1 24\n1 42\n1 30\n1 12\n1 37\n64 1\n1 28\n1 58\n1 22\n11 1\n1 4\n1 27\n1 16\n1 21\n54 1\n1 51\n1 43\n29 1\n56 1\n1 39\n32 1\n1 15\n1 17\n1 19\n1 40\n36 1\n48 1\n63 1\n1 7\n1 47\n1 13\n1 46\n60 1\n1 6\n23 1\n20 1\n1 52\n2 1\n26 1\n1 59\n1 66\n10 1\n1 62\n1 68\n1 55\n50 1\n33 1\n44 1\n1 34\n1 35\n1 61\n14 1\n67 1\n49 1",
"output": "1.000000000000000"
},
{
"input": "10\n8 6\n9 10\n8 7\n1 4\n1 8\n9 5\n9 8\n2 5\n3 1",
"output": "1.500000000000000"
},
{
"input": "1",
"output": "0.000000000000000"
}
] | 1,609,173,771
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 3
| 108
| 1,228,800
|
class Node:
def __init__(self, name):
self.name = name
self.neighs = set()
self.expect = 0
self.visited = False
def connect(self, node):
self.neighs.add(node)
def read_graph_from_input():
nodes = {}
n = int(input())
for i in range(1, n+1):
nodes[i] = Node(i)
for _ in range(0, n-1):
i, j = map(int, input().split())
nodes[i].connect(nodes[j])
nodes[j].connect(nodes[i])
return nodes[1]
def get_math_exp_for_node(node):
node.visited = True
not_visited_neighs = set(filter(lambda x: not x.visited, node.neighs))
if len(not_visited_neighs) == 0:
return 0
expect = 0
for neigh in not_visited_neighs:
expect += get_math_exp_for_node(neigh)
return expect / len(not_visited_neighs) + 1
def main():
root = read_graph_from_input()
print(get_math_exp_for_node(root))
if __name__ == '__main__':
main()
|
Title: Journey
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cities and *n*<=-<=1 roads in the Seven Kingdoms, each road connects two cities and we can reach any city from any other by the roads.
Theon and Yara Greyjoy are on a horse in the first city, they are starting traveling through the roads. But the weather is foggy, so they can’t see where the horse brings them. When the horse reaches a city (including the first one), it goes to one of the cities connected to the current city. But it is a strange horse, it only goes to cities in which they weren't before. In each such city, the horse goes with equal probabilities and it stops when there are no such cities.
Let the length of each road be 1. The journey starts in the city 1. What is the expected length (expected value of length) of their journey? You can read about expected (average) value by the link [https://en.wikipedia.org/wiki/Expected_value](https://en.wikipedia.org/wiki/Expected_value).
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100000) — number of cities.
Then *n*<=-<=1 lines follow. The *i*-th line of these lines contains two integers *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*, *u**i*<=≠<=*v**i*) — the cities connected by the *i*-th road.
It is guaranteed that one can reach any city from any other by the roads.
Output Specification:
Print a number — the expected length of their journey. The journey starts in the city 1.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
Demo Input:
['4\n1 2\n1 3\n2 4\n', '5\n1 2\n1 3\n3 4\n2 5\n']
Demo Output:
['1.500000000000000\n', '2.000000000000000\n']
Note:
In the first sample, their journey may end in cities 3 or 4 with equal probability. The distance to city 3 is 1 and to city 4 is 2, so the expected length is 1.5.
In the second sample, their journey may end in city 4 or 5. The distance to the both cities is 2, so the expected length is 2.
|
```python
class Node:
def __init__(self, name):
self.name = name
self.neighs = set()
self.expect = 0
self.visited = False
def connect(self, node):
self.neighs.add(node)
def read_graph_from_input():
nodes = {}
n = int(input())
for i in range(1, n+1):
nodes[i] = Node(i)
for _ in range(0, n-1):
i, j = map(int, input().split())
nodes[i].connect(nodes[j])
nodes[j].connect(nodes[i])
return nodes[1]
def get_math_exp_for_node(node):
node.visited = True
not_visited_neighs = set(filter(lambda x: not x.visited, node.neighs))
if len(not_visited_neighs) == 0:
return 0
expect = 0
for neigh in not_visited_neighs:
expect += get_math_exp_for_node(neigh)
return expect / len(not_visited_neighs) + 1
def main():
root = read_graph_from_input()
print(get_math_exp_for_node(root))
if __name__ == '__main__':
main()
```
| -1
|
|
231
|
A
|
Team
|
PROGRAMMING
| 800
|
[
"brute force",
"greedy"
] | null | null |
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
|
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
|
Print a single integer — the number of problems the friends will implement on the contest.
|
[
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] |
[
"2\n",
"1\n"
] |
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
| 500
|
[
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
{
"input": "10\n0 1 0\n0 1 0\n1 1 0\n1 0 0\n0 0 1\n0 1 1\n1 1 1\n1 1 0\n0 0 0\n0 0 0",
"output": "4"
},
{
"input": "15\n0 1 0\n1 0 0\n1 1 0\n1 1 1\n0 1 0\n0 0 1\n1 0 1\n1 0 1\n1 0 1\n0 0 0\n1 1 1\n1 1 0\n0 1 1\n1 1 0\n1 1 1",
"output": "10"
},
{
"input": "50\n0 0 0\n0 1 1\n1 1 1\n0 1 0\n1 0 1\n1 1 1\n0 0 1\n1 0 0\n1 1 0\n1 0 1\n0 1 0\n0 0 1\n1 1 0\n0 1 0\n1 1 0\n0 0 0\n1 1 1\n1 0 1\n0 0 1\n1 1 0\n1 1 1\n0 1 1\n1 1 0\n0 0 0\n0 0 0\n1 1 1\n0 0 0\n1 1 1\n0 1 1\n0 0 1\n0 0 0\n0 0 0\n1 1 0\n1 1 0\n1 0 1\n1 0 0\n1 0 1\n1 0 1\n0 1 1\n1 1 0\n1 1 0\n0 1 0\n1 0 1\n0 0 0\n0 0 0\n0 0 0\n0 0 1\n1 1 1\n0 1 1\n1 0 1",
"output": "29"
},
{
"input": "1\n1 1 1",
"output": "1"
},
{
"input": "8\n0 0 0\n0 0 1\n0 0 0\n0 1 1\n1 0 0\n1 0 1\n1 1 0\n1 1 1",
"output": "4"
},
{
"input": "16\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1",
"output": "16"
}
] | 1,697,902,780
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 92
| 0
|
n = int(input())
sum = 0
for i in range(0,n,1):
a, b, c = input().split()
a = int(a)
b = int(b)
c = int(c)
if (a + b + c) >= 2:
sum += 1
print(sum)
|
Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution.
|
```python
n = int(input())
sum = 0
for i in range(0,n,1):
a, b, c = input().split()
a = int(a)
b = int(b)
c = int(c)
if (a + b + c) >= 2:
sum += 1
print(sum)
```
| 3
|
|
609
|
A
|
USB Flash Drives
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"sortings"
] | null | null |
Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes.
Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives.
|
The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives.
The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file.
Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes.
It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*.
|
Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives.
|
[
"3\n5\n2\n1\n3\n",
"3\n6\n2\n3\n2\n",
"2\n5\n5\n10\n"
] |
[
"2\n",
"3\n",
"1\n"
] |
In the first example Sean needs only two USB flash drives — the first and the third.
In the second example Sean needs all three USB flash drives.
In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second.
| 0
|
[
{
"input": "3\n5\n2\n1\n3",
"output": "2"
},
{
"input": "3\n6\n2\n3\n2",
"output": "3"
},
{
"input": "2\n5\n5\n10",
"output": "1"
},
{
"input": "5\n16\n8\n1\n3\n4\n9",
"output": "2"
},
{
"input": "10\n121\n10\n37\n74\n56\n42\n39\n6\n68\n8\n100",
"output": "2"
},
{
"input": "12\n4773\n325\n377\n192\n780\n881\n816\n839\n223\n215\n125\n952\n8",
"output": "7"
},
{
"input": "15\n7758\n182\n272\n763\n910\n24\n359\n583\n890\n735\n819\n66\n992\n440\n496\n227",
"output": "15"
},
{
"input": "30\n70\n6\n2\n10\n4\n7\n10\n5\n1\n8\n10\n4\n3\n5\n9\n3\n6\n6\n4\n2\n6\n5\n10\n1\n9\n7\n2\n1\n10\n7\n5",
"output": "8"
},
{
"input": "40\n15705\n702\n722\n105\n873\n417\n477\n794\n300\n869\n496\n572\n232\n456\n298\n473\n584\n486\n713\n934\n121\n303\n956\n934\n840\n358\n201\n861\n497\n131\n312\n957\n96\n914\n509\n60\n300\n722\n658\n820\n103",
"output": "21"
},
{
"input": "50\n18239\n300\n151\n770\n9\n200\n52\n247\n753\n523\n263\n744\n463\n540\n244\n608\n569\n771\n32\n425\n777\n624\n761\n628\n124\n405\n396\n726\n626\n679\n237\n229\n49\n512\n18\n671\n290\n768\n632\n739\n18\n136\n413\n117\n83\n413\n452\n767\n664\n203\n404",
"output": "31"
},
{
"input": "70\n149\n5\n3\n3\n4\n6\n1\n2\n9\n8\n3\n1\n8\n4\n4\n3\n6\n10\n7\n1\n10\n8\n4\n9\n3\n8\n3\n2\n5\n1\n8\n6\n9\n10\n4\n8\n6\n9\n9\n9\n3\n4\n2\n2\n5\n8\n9\n1\n10\n3\n4\n3\n1\n9\n3\n5\n1\n3\n7\n6\n9\n8\n9\n1\n7\n4\n4\n2\n3\n5\n7",
"output": "17"
},
{
"input": "70\n2731\n26\n75\n86\n94\n37\n25\n32\n35\n92\n1\n51\n73\n53\n66\n16\n80\n15\n81\n100\n87\n55\n48\n30\n71\n39\n87\n77\n25\n70\n22\n75\n23\n97\n16\n75\n95\n61\n61\n28\n10\n78\n54\n80\n51\n25\n24\n90\n58\n4\n77\n40\n54\n53\n47\n62\n30\n38\n71\n97\n71\n60\n58\n1\n21\n15\n55\n99\n34\n88\n99",
"output": "35"
},
{
"input": "70\n28625\n34\n132\n181\n232\n593\n413\n862\n887\n808\n18\n35\n89\n356\n640\n339\n280\n975\n82\n345\n398\n948\n372\n91\n755\n75\n153\n948\n603\n35\n694\n722\n293\n363\n884\n264\n813\n175\n169\n646\n138\n449\n488\n828\n417\n134\n84\n763\n288\n845\n801\n556\n972\n332\n564\n934\n699\n842\n942\n644\n203\n406\n140\n37\n9\n423\n546\n675\n491\n113\n587",
"output": "45"
},
{
"input": "80\n248\n3\n9\n4\n5\n10\n7\n2\n6\n2\n2\n8\n2\n1\n3\n7\n9\n2\n8\n4\n4\n8\n5\n4\n4\n10\n2\n1\n4\n8\n4\n10\n1\n2\n10\n2\n3\n3\n1\n1\n8\n9\n5\n10\n2\n8\n10\n5\n3\n6\n1\n7\n8\n9\n10\n5\n10\n10\n2\n10\n1\n2\n4\n1\n9\n4\n7\n10\n8\n5\n8\n1\n4\n2\n2\n3\n9\n9\n9\n10\n6",
"output": "27"
},
{
"input": "80\n2993\n18\n14\n73\n38\n14\n73\n77\n18\n81\n6\n96\n65\n77\n86\n76\n8\n16\n81\n83\n83\n34\n69\n58\n15\n19\n1\n16\n57\n95\n35\n5\n49\n8\n15\n47\n84\n99\n94\n93\n55\n43\n47\n51\n61\n57\n13\n7\n92\n14\n4\n83\n100\n60\n75\n41\n95\n74\n40\n1\n4\n95\n68\n59\n65\n15\n15\n75\n85\n46\n77\n26\n30\n51\n64\n75\n40\n22\n88\n68\n24",
"output": "38"
},
{
"input": "80\n37947\n117\n569\n702\n272\n573\n629\n90\n337\n673\n589\n576\n205\n11\n284\n645\n719\n777\n271\n567\n466\n251\n402\n3\n97\n288\n699\n208\n173\n530\n782\n266\n395\n957\n159\n463\n43\n316\n603\n197\n386\n132\n799\n778\n905\n784\n71\n851\n963\n883\n705\n454\n275\n425\n727\n223\n4\n870\n833\n431\n463\n85\n505\n800\n41\n954\n981\n242\n578\n336\n48\n858\n702\n349\n929\n646\n528\n993\n506\n274\n227",
"output": "70"
},
{
"input": "90\n413\n5\n8\n10\n7\n5\n7\n5\n7\n1\n7\n8\n4\n3\n9\n4\n1\n10\n3\n1\n10\n9\n3\n1\n8\n4\n7\n5\n2\n9\n3\n10\n10\n3\n6\n3\n3\n10\n7\n5\n1\n1\n2\n4\n8\n2\n5\n5\n3\n9\n5\n5\n3\n10\n2\n3\n8\n5\n9\n1\n3\n6\n5\n9\n2\n3\n7\n10\n3\n4\n4\n1\n5\n9\n2\n6\n9\n1\n1\n9\n9\n7\n7\n7\n8\n4\n5\n3\n4\n6\n9",
"output": "59"
},
{
"input": "90\n4226\n33\n43\n83\n46\n75\n14\n88\n36\n8\n25\n47\n4\n96\n19\n33\n49\n65\n17\n59\n72\n1\n55\n94\n92\n27\n33\n39\n14\n62\n79\n12\n89\n22\n86\n13\n19\n77\n53\n96\n74\n24\n25\n17\n64\n71\n81\n87\n52\n72\n55\n49\n74\n36\n65\n86\n91\n33\n61\n97\n38\n87\n61\n14\n73\n95\n43\n67\n42\n67\n22\n12\n62\n32\n96\n24\n49\n82\n46\n89\n36\n75\n91\n11\n10\n9\n33\n86\n28\n75\n39",
"output": "64"
},
{
"input": "90\n40579\n448\n977\n607\n745\n268\n826\n479\n59\n330\n609\n43\n301\n970\n726\n172\n632\n600\n181\n712\n195\n491\n312\n849\n722\n679\n682\n780\n131\n404\n293\n387\n567\n660\n54\n339\n111\n833\n612\n911\n869\n356\n884\n635\n126\n639\n712\n473\n663\n773\n435\n32\n973\n484\n662\n464\n699\n274\n919\n95\n904\n253\n589\n543\n454\n250\n349\n237\n829\n511\n536\n36\n45\n152\n626\n384\n199\n877\n941\n84\n781\n115\n20\n52\n726\n751\n920\n291\n571\n6\n199",
"output": "64"
},
{
"input": "100\n66\n7\n9\n10\n5\n2\n8\n6\n5\n4\n10\n10\n6\n5\n2\n2\n1\n1\n5\n8\n7\n8\n10\n5\n6\n6\n5\n9\n9\n6\n3\n8\n7\n10\n5\n9\n6\n7\n3\n5\n8\n6\n8\n9\n1\n1\n1\n2\n4\n5\n5\n1\n1\n2\n6\n7\n1\n5\n8\n7\n2\n1\n7\n10\n9\n10\n2\n4\n10\n4\n10\n10\n5\n3\n9\n1\n2\n1\n10\n5\n1\n7\n4\n4\n5\n7\n6\n10\n4\n7\n3\n4\n3\n6\n2\n5\n2\n4\n9\n5\n3",
"output": "7"
},
{
"input": "100\n4862\n20\n47\n85\n47\n76\n38\n48\n93\n91\n81\n31\n51\n23\n60\n59\n3\n73\n72\n57\n67\n54\n9\n42\n5\n32\n46\n72\n79\n95\n61\n79\n88\n33\n52\n97\n10\n3\n20\n79\n82\n93\n90\n38\n80\n18\n21\n43\n60\n73\n34\n75\n65\n10\n84\n100\n29\n94\n56\n22\n59\n95\n46\n22\n57\n69\n67\n90\n11\n10\n61\n27\n2\n48\n69\n86\n91\n69\n76\n36\n71\n18\n54\n90\n74\n69\n50\n46\n8\n5\n41\n96\n5\n14\n55\n85\n39\n6\n79\n75\n87",
"output": "70"
},
{
"input": "100\n45570\n14\n881\n678\n687\n993\n413\n760\n451\n426\n787\n503\n343\n234\n530\n294\n725\n941\n524\n574\n441\n798\n399\n360\n609\n376\n525\n229\n995\n478\n347\n47\n23\n468\n525\n749\n601\n235\n89\n995\n489\n1\n239\n415\n122\n671\n128\n357\n886\n401\n964\n212\n968\n210\n130\n871\n360\n661\n844\n414\n187\n21\n824\n266\n713\n126\n496\n916\n37\n193\n755\n894\n641\n300\n170\n176\n383\n488\n627\n61\n897\n33\n242\n419\n881\n698\n107\n391\n418\n774\n905\n87\n5\n896\n835\n318\n373\n916\n393\n91\n460",
"output": "78"
},
{
"input": "100\n522\n1\n5\n2\n4\n2\n6\n3\n4\n2\n10\n10\n6\n7\n9\n7\n1\n7\n2\n5\n3\n1\n5\n2\n3\n5\n1\n7\n10\n10\n4\n4\n10\n9\n10\n6\n2\n8\n2\n6\n10\n9\n2\n7\n5\n9\n4\n6\n10\n7\n3\n1\n1\n9\n5\n10\n9\n2\n8\n3\n7\n5\n4\n7\n5\n9\n10\n6\n2\n9\n2\n5\n10\n1\n7\n7\n10\n5\n6\n2\n9\n4\n7\n10\n10\n8\n3\n4\n9\n3\n6\n9\n10\n2\n9\n9\n3\n4\n1\n10\n2",
"output": "74"
},
{
"input": "100\n32294\n414\n116\n131\n649\n130\n476\n630\n605\n213\n117\n757\n42\n109\n85\n127\n635\n629\n994\n410\n764\n204\n161\n231\n577\n116\n936\n537\n565\n571\n317\n722\n819\n229\n284\n487\n649\n304\n628\n727\n816\n854\n91\n111\n549\n87\n374\n417\n3\n868\n882\n168\n743\n77\n534\n781\n75\n956\n910\n734\n507\n568\n802\n946\n891\n659\n116\n678\n375\n380\n430\n627\n873\n350\n930\n285\n6\n183\n96\n517\n81\n794\n235\n360\n551\n6\n28\n799\n226\n996\n894\n981\n551\n60\n40\n460\n479\n161\n318\n952\n433",
"output": "42"
},
{
"input": "100\n178\n71\n23\n84\n98\n8\n14\n4\n42\n56\n83\n87\n28\n22\n32\n50\n5\n96\n90\n1\n59\n74\n56\n96\n77\n88\n71\n38\n62\n36\n85\n1\n97\n98\n98\n32\n99\n42\n6\n81\n20\n49\n57\n71\n66\n9\n45\n41\n29\n28\n32\n68\n38\n29\n35\n29\n19\n27\n76\n85\n68\n68\n41\n32\n78\n72\n38\n19\n55\n83\n83\n25\n46\n62\n48\n26\n53\n14\n39\n31\n94\n84\n22\n39\n34\n96\n63\n37\n42\n6\n78\n76\n64\n16\n26\n6\n79\n53\n24\n29\n63",
"output": "2"
},
{
"input": "100\n885\n226\n266\n321\n72\n719\n29\n121\n533\n85\n672\n225\n830\n783\n822\n30\n791\n618\n166\n487\n922\n434\n814\n473\n5\n741\n947\n910\n305\n998\n49\n945\n588\n868\n809\n803\n168\n280\n614\n434\n634\n538\n591\n437\n540\n445\n313\n177\n171\n799\n778\n55\n617\n554\n583\n611\n12\n94\n599\n182\n765\n556\n965\n542\n35\n460\n177\n313\n485\n744\n384\n21\n52\n879\n792\n411\n614\n811\n565\n695\n428\n587\n631\n794\n461\n258\n193\n696\n936\n646\n756\n267\n55\n690\n730\n742\n734\n988\n235\n762\n440",
"output": "1"
},
{
"input": "100\n29\n9\n2\n10\n8\n6\n7\n7\n3\n3\n10\n4\n5\n2\n5\n1\n6\n3\n2\n5\n10\n10\n9\n1\n4\n5\n2\n2\n3\n1\n2\n2\n9\n6\n9\n7\n8\n8\n1\n5\n5\n3\n1\n5\n6\n1\n9\n2\n3\n8\n10\n8\n3\n2\n7\n1\n2\n1\n2\n8\n10\n5\n2\n3\n1\n10\n7\n1\n7\n4\n9\n6\n6\n4\n7\n1\n2\n7\n7\n9\n9\n7\n10\n4\n10\n8\n2\n1\n5\n5\n10\n5\n8\n1\n5\n6\n5\n1\n5\n6\n8",
"output": "3"
},
{
"input": "100\n644\n94\n69\n43\n36\n54\n93\n30\n74\n56\n95\n70\n49\n11\n36\n57\n30\n59\n3\n52\n59\n90\n82\n39\n67\n32\n8\n80\n64\n8\n65\n51\n48\n89\n90\n35\n4\n54\n66\n96\n68\n90\n30\n4\n13\n97\n41\n90\n85\n17\n45\n94\n31\n58\n4\n39\n76\n95\n92\n59\n67\n46\n96\n55\n82\n64\n20\n20\n83\n46\n37\n15\n60\n37\n79\n45\n47\n63\n73\n76\n31\n52\n36\n32\n49\n26\n61\n91\n31\n25\n62\n90\n65\n65\n5\n94\n7\n15\n97\n88\n68",
"output": "7"
},
{
"input": "100\n1756\n98\n229\n158\n281\n16\n169\n149\n239\n235\n182\n147\n215\n49\n270\n194\n242\n295\n289\n249\n19\n12\n144\n157\n92\n270\n122\n212\n97\n152\n14\n42\n12\n198\n98\n295\n154\n229\n191\n294\n5\n156\n43\n185\n184\n20\n125\n23\n10\n257\n244\n264\n79\n46\n277\n13\n22\n97\n212\n77\n293\n20\n51\n17\n109\n37\n68\n117\n51\n248\n10\n149\n179\n192\n239\n161\n13\n173\n297\n73\n43\n109\n288\n198\n81\n70\n254\n187\n277\n1\n295\n113\n95\n291\n293\n119\n205\n191\n37\n34\n116",
"output": "6"
},
{
"input": "100\n20562\n721\n452\n11\n703\n376\n183\n197\n203\n406\n642\n346\n446\n256\n760\n201\n360\n702\n707\n388\n779\n653\n610\n497\n768\n670\n134\n780\n306\n661\n180\n259\n256\n362\n6\n121\n415\n747\n170\n67\n439\n728\n193\n622\n481\n38\n225\n343\n303\n253\n436\n305\n68\n794\n247\n291\n600\n750\n188\n199\n757\n28\n776\n749\n253\n351\n53\n629\n129\n578\n209\n89\n651\n262\n638\n353\n469\n31\n144\n460\n176\n535\n562\n366\n639\n234\n577\n364\n761\n617\n303\n450\n778\n311\n289\n221\n274\n239\n626\n194\n36",
"output": "30"
},
{
"input": "100\n32630\n548\n21\n756\n138\n56\n719\n97\n86\n599\n531\n464\n137\n737\n239\n626\n179\n170\n271\n335\n178\n298\n597\n616\n104\n190\n216\n333\n606\n621\n310\n70\n286\n593\n655\n472\n556\n77\n625\n744\n395\n487\n762\n316\n469\n258\n475\n596\n651\n180\n737\n101\n511\n708\n163\n713\n475\n557\n649\n705\n583\n97\n447\n119\n91\n372\n194\n394\n667\n384\n635\n611\n746\n758\n613\n54\n508\n405\n188\n570\n254\n28\n599\n767\n115\n737\n194\n148\n686\n705\n236\n762\n613\n770\n633\n401\n773\n448\n274\n141\n718",
"output": "52"
},
{
"input": "1\n1\n1",
"output": "1"
}
] | 1,577,453,729
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 34
| 140
| 0
|
n = int(input())
m = int(input())
l = [0] * n
for i in range(n):
l[i] = int(input())
l.sort(reverse = True)
ans = 0
temp = 0
for i in l:
temp += i
ans -= -1
if temp >= m:
print(ans)
break
|
Title: USB Flash Drives
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes.
Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives.
Input Specification:
The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives.
The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file.
Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes.
It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*.
Output Specification:
Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives.
Demo Input:
['3\n5\n2\n1\n3\n', '3\n6\n2\n3\n2\n', '2\n5\n5\n10\n']
Demo Output:
['2\n', '3\n', '1\n']
Note:
In the first example Sean needs only two USB flash drives — the first and the third.
In the second example Sean needs all three USB flash drives.
In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second.
|
```python
n = int(input())
m = int(input())
l = [0] * n
for i in range(n):
l[i] = int(input())
l.sort(reverse = True)
ans = 0
temp = 0
for i in l:
temp += i
ans -= -1
if temp >= m:
print(ans)
break
```
| 3
|
|
1,006
|
C
|
Three Parts of the Array
|
PROGRAMMING
| 1,200
|
[
"binary search",
"data structures",
"two pointers"
] | null | null |
You are given an array $d_1, d_2, \dots, d_n$ consisting of $n$ integer numbers.
Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.
Let the sum of elements of the first part be $sum_1$, the sum of elements of the second part be $sum_2$ and the sum of elements of the third part be $sum_3$. Among all possible ways to split the array you have to choose a way such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
More formally, if the first part of the array contains $a$ elements, the second part of the array contains $b$ elements and the third part contains $c$ elements, then:
$$sum_1 = \sum\limits_{1 \le i \le a}d_i,$$ $$sum_2 = \sum\limits_{a + 1 \le i \le a + b}d_i,$$ $$sum_3 = \sum\limits_{a + b + 1 \le i \le a + b + c}d_i.$$
The sum of an empty array is $0$.
Your task is to find a way to split the array such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
|
The first line of the input contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $d$.
The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^9$) — the elements of the array $d$.
|
Print a single integer — the maximum possible value of $sum_1$, considering that the condition $sum_1 = sum_3$ must be met.
Obviously, at least one valid way to split the array exists (use $a=c=0$ and $b=n$).
|
[
"5\n1 3 1 1 4\n",
"5\n1 3 2 1 4\n",
"3\n4 1 2\n"
] |
[
"5\n",
"4\n",
"0\n"
] |
In the first example there is only one possible splitting which maximizes $sum_1$: $[1, 3, 1], [~], [1, 4]$.
In the second example the only way to have $sum_1=4$ is: $[1, 3], [2, 1], [4]$.
In the third example there is only one way to split the array: $[~], [4, 1, 2], [~]$.
| 0
|
[
{
"input": "5\n1 3 1 1 4",
"output": "5"
},
{
"input": "5\n1 3 2 1 4",
"output": "4"
},
{
"input": "3\n4 1 2",
"output": "0"
},
{
"input": "1\n1000000000",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "5\n1 3 5 4 5",
"output": "9"
}
] | 1,619,154,677
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 280
| 16,793,600
|
n = int(input())
a = [int(i) for i in input().split()]
l = 0
r = len(a)-1
sum1 = 0
sum3 = 0
max = 0
while l<=r:
if sum1 + a[l] > sum3 +a[r]:
sum3 += a[r]
r-=1
else :
sum1 += a[l]
l+=1
if sum1 == sum3:
max = sum1
print(max)
|
Title: Three Parts of the Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array $d_1, d_2, \dots, d_n$ consisting of $n$ integer numbers.
Your task is to split this array into three parts (some of which may be empty) in such a way that each element of the array belongs to exactly one of the three parts, and each of the parts forms a consecutive contiguous subsegment (possibly, empty) of the original array.
Let the sum of elements of the first part be $sum_1$, the sum of elements of the second part be $sum_2$ and the sum of elements of the third part be $sum_3$. Among all possible ways to split the array you have to choose a way such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
More formally, if the first part of the array contains $a$ elements, the second part of the array contains $b$ elements and the third part contains $c$ elements, then:
$$sum_1 = \sum\limits_{1 \le i \le a}d_i,$$ $$sum_2 = \sum\limits_{a + 1 \le i \le a + b}d_i,$$ $$sum_3 = \sum\limits_{a + b + 1 \le i \le a + b + c}d_i.$$
The sum of an empty array is $0$.
Your task is to find a way to split the array such that $sum_1 = sum_3$ and $sum_1$ is maximum possible.
Input Specification:
The first line of the input contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in the array $d$.
The second line of the input contains $n$ integers $d_1, d_2, \dots, d_n$ ($1 \le d_i \le 10^9$) — the elements of the array $d$.
Output Specification:
Print a single integer — the maximum possible value of $sum_1$, considering that the condition $sum_1 = sum_3$ must be met.
Obviously, at least one valid way to split the array exists (use $a=c=0$ and $b=n$).
Demo Input:
['5\n1 3 1 1 4\n', '5\n1 3 2 1 4\n', '3\n4 1 2\n']
Demo Output:
['5\n', '4\n', '0\n']
Note:
In the first example there is only one possible splitting which maximizes $sum_1$: $[1, 3, 1], [~], [1, 4]$.
In the second example the only way to have $sum_1=4$ is: $[1, 3], [2, 1], [4]$.
In the third example there is only one way to split the array: $[~], [4, 1, 2], [~]$.
|
```python
n = int(input())
a = [int(i) for i in input().split()]
l = 0
r = len(a)-1
sum1 = 0
sum3 = 0
max = 0
while l<=r:
if sum1 + a[l] > sum3 +a[r]:
sum3 += a[r]
r-=1
else :
sum1 += a[l]
l+=1
if sum1 == sum3:
max = sum1
print(max)
```
| 3
|
|
397
|
A
|
On Segment's Own Points
|
PROGRAMMING
| 0
|
[
"implementation"
] | null | null |
Our old friend Alexey has finally entered the University of City N — the Berland capital. Alexey expected his father to get him a place to live in but his father said it was high time for Alexey to practice some financial independence. So, Alexey is living in a dorm.
The dorm has exactly one straight dryer — a 100 centimeter long rope to hang clothes on. The dryer has got a coordinate system installed: the leftmost end of the dryer has coordinate 0, and the opposite end has coordinate 100. Overall, the university has *n* students. Dean's office allows *i*-th student to use the segment (*l**i*,<=*r**i*) of the dryer. However, the dean's office actions are contradictory and now one part of the dryer can belong to multiple students!
Alexey don't like when someone touch his clothes. That's why he want make it impossible to someone clothes touch his ones. So Alexey wonders: what is the total length of the parts of the dryer that he may use in a such way that clothes of the others (*n*<=-<=1) students aren't drying there. Help him! Note that Alexey, as the most respected student, has number 1.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100). The (*i*<=+<=1)-th line contains integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=100) — the endpoints of the corresponding segment for the *i*-th student.
|
On a single line print a single number *k*, equal to the sum of lengths of the parts of the dryer which are inside Alexey's segment and are outside all other segments.
|
[
"3\n0 5\n2 8\n1 6\n",
"3\n0 10\n1 5\n7 15\n"
] |
[
"1\n",
"3\n"
] |
Note that it's not important are clothes drying on the touching segments (e.g. (0, 1) and (1, 2)) considered to be touching or not because you need to find the length of segments.
In the first test sample Alexey may use the only segment (0, 1). In such case his clothes will not touch clothes on the segments (1, 6) and (2, 8). The length of segment (0, 1) is 1.
In the second test sample Alexey may dry his clothes on segments (0, 1) and (5, 7). Overall length of these segments is 3.
| 500
|
[
{
"input": "3\n0 5\n2 8\n1 6",
"output": "1"
},
{
"input": "3\n0 10\n1 5\n7 15",
"output": "3"
},
{
"input": "1\n0 100",
"output": "100"
},
{
"input": "2\n1 9\n1 9",
"output": "0"
},
{
"input": "2\n1 9\n5 10",
"output": "4"
},
{
"input": "2\n1 9\n3 5",
"output": "6"
},
{
"input": "2\n3 5\n1 9",
"output": "0"
},
{
"input": "10\n43 80\n39 75\n26 71\n4 17\n11 57\n31 42\n1 62\n9 19\n27 76\n34 53",
"output": "4"
},
{
"input": "50\n33 35\n98 99\n1 2\n4 6\n17 18\n63 66\n29 30\n35 37\n44 45\n73 75\n4 5\n39 40\n92 93\n96 97\n23 27\n49 50\n2 3\n60 61\n43 44\n69 70\n7 8\n45 46\n21 22\n85 86\n48 49\n41 43\n70 71\n10 11\n27 28\n71 72\n6 7\n15 16\n46 47\n89 91\n54 55\n19 21\n86 87\n37 38\n77 82\n84 85\n54 55\n93 94\n45 46\n37 38\n75 76\n22 23\n50 52\n38 39\n1 2\n66 67",
"output": "2"
},
{
"input": "2\n1 5\n7 9",
"output": "4"
},
{
"input": "2\n1 5\n3 5",
"output": "2"
},
{
"input": "2\n1 5\n1 2",
"output": "3"
},
{
"input": "5\n5 10\n5 10\n5 10\n5 10\n5 10",
"output": "0"
},
{
"input": "6\n1 99\n33 94\n68 69\n3 35\n93 94\n5 98",
"output": "3"
},
{
"input": "11\n2 98\n63 97\n4 33\n12 34\n34 65\n23 31\n43 54\n82 99\n15 84\n23 52\n4 50",
"output": "2"
},
{
"input": "10\n95 96\n19 20\n72 73\n1 2\n25 26\n48 49\n90 91\n22 23\n16 17\n16 17",
"output": "1"
},
{
"input": "11\n1 100\n63 97\n4 33\n12 34\n34 65\n23 31\n43 54\n82 99\n15 84\n23 52\n4 50",
"output": "4"
},
{
"input": "21\n0 100\n81 90\n11 68\n18 23\n75 78\n45 86\n37 58\n15 21\n40 98\n53 100\n10 70\n14 75\n1 92\n23 81\n13 66\n93 100\n6 34\n22 87\n27 84\n15 63\n54 91",
"output": "1"
},
{
"input": "10\n60 66\n5 14\n1 3\n55 56\n70 87\n34 35\n16 21\n23 24\n30 31\n25 27",
"output": "6"
},
{
"input": "40\n29 31\n22 23\n59 60\n70 71\n42 43\n13 15\n11 12\n64 65\n1 2\n62 63\n54 56\n8 9\n2 3\n53 54\n27 28\n48 49\n72 73\n17 18\n46 47\n18 19\n43 44\n39 40\n83 84\n63 64\n52 53\n33 34\n3 4\n24 25\n74 75\n0 1\n61 62\n68 69\n80 81\n5 6\n36 37\n81 82\n50 51\n66 67\n69 70\n20 21",
"output": "2"
},
{
"input": "15\n22 31\n0 4\n31 40\n77 80\n81 83\n11 13\n59 61\n53 59\n51 53\n87 88\n14 22\n43 45\n8 10\n45 47\n68 71",
"output": "9"
},
{
"input": "31\n0 100\n2 97\n8 94\n9 94\n14 94\n15 93\n15 90\n17 88\n19 88\n19 87\n20 86\n25 86\n30 85\n32 85\n35 82\n35 81\n36 80\n37 78\n38 74\n38 74\n39 71\n40 69\n40 68\n41 65\n43 62\n44 62\n45 61\n45 59\n46 57\n49 54\n50 52",
"output": "5"
},
{
"input": "21\n0 97\n46 59\n64 95\n3 16\n86 95\n55 71\n51 77\n26 28\n47 88\n30 40\n26 34\n2 12\n9 10\n4 19\n35 36\n41 92\n1 16\n41 78\n56 81\n23 35\n40 68",
"output": "7"
},
{
"input": "27\n0 97\n7 9\n6 9\n12 33\n12 26\n15 27\n10 46\n33 50\n31 47\n15 38\n12 44\n21 35\n24 37\n51 52\n65 67\n58 63\n53 60\n63 68\n57 63\n60 68\n55 58\n74 80\n70 75\n89 90\n81 85\n93 99\n93 98",
"output": "19"
},
{
"input": "20\n23 24\n22 23\n84 86\n6 10\n40 45\n11 13\n24 27\n81 82\n53 58\n87 90\n14 15\n49 50\n70 75\n75 78\n98 100\n66 68\n18 21\n1 2\n92 93\n34 37",
"output": "1"
},
{
"input": "11\n2 100\n34 65\n4 50\n63 97\n82 99\n43 54\n23 52\n4 33\n15 84\n23 31\n12 34",
"output": "3"
},
{
"input": "60\n73 75\n6 7\n69 70\n15 16\n54 55\n66 67\n7 8\n39 40\n38 39\n37 38\n1 2\n46 47\n7 8\n21 22\n23 27\n15 16\n45 46\n37 38\n60 61\n4 6\n63 66\n10 11\n33 35\n43 44\n2 3\n4 6\n10 11\n93 94\n45 46\n7 8\n44 45\n41 43\n35 37\n17 18\n48 49\n89 91\n27 28\n46 47\n71 72\n1 2\n75 76\n49 50\n84 85\n17 18\n98 99\n54 55\n46 47\n19 21\n77 82\n29 30\n4 5\n70 71\n85 86\n96 97\n86 87\n92 93\n22 23\n50 52\n44 45\n63 66",
"output": "2"
},
{
"input": "40\n47 48\n42 44\n92 94\n15 17\n20 22\n11 13\n37 39\n6 8\n39 40\n35 37\n21 22\n41 42\n77 78\n76 78\n69 71\n17 19\n18 19\n17 18\n84 85\n9 10\n11 12\n51 52\n99 100\n7 8\n97 99\n22 23\n60 62\n7 8\n67 69\n20 22\n13 14\n89 91\n15 17\n12 13\n56 57\n37 39\n29 30\n24 26\n37 38\n25 27",
"output": "1"
},
{
"input": "10\n28 36\n18 26\n28 35\n95 100\n68 72\n41 42\n76 84\n99 100\n6 8\n58 60",
"output": "1"
},
{
"input": "20\n69 72\n88 92\n77 80\n64 69\n66 67\n79 81\n91 96\n78 83\n81 86\n11 12\n48 53\n22 23\n81 84\n89 92\n56 60\n1 4\n1 5\n60 62\n20 23\n63 66",
"output": "3"
},
{
"input": "71\n1 99\n11 69\n86 92\n7 49\n31 70\n42 53\n48 81\n86 96\n36 91\n19 38\n39 91\n41 64\n87 93\n83 97\n40 41\n3 32\n15 18\n58 65\n22 32\n1 71\n58 86\n64 77\n15 69\n4 34\n42 89\n9 66\n15 18\n58 65\n59 96\n39 89\n19 38\n6 63\n26 73\n29 47\n55 88\n5 78\n41 74\n48 81\n20 71\n59 96\n42 49\n4 69\n41 74\n87 93\n0 65\n2 34\n15 18\n10 56\n55 88\n33 56\n42 89\n86 92\n42 81\n65 82\n5 78\n13 52\n32 85\n7 65\n59 96\n4 65\n46 69\n10 56\n42 89\n4 69\n0 65\n32 35\n5 78\n32 75\n42 53\n55 59\n64 77",
"output": "2"
},
{
"input": "1\n1 2",
"output": "1"
}
] | 1,578,668,655
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 23
| 109
| 307,200
|
def function(dryer, owner):
l, r = owner
count = 0
for i in range(l, r):
if i + 1 <= 100:
if dryer[i] == 1 and dryer[i + 1] == 0:
count += 1
elif dryer[i] == 0 and dryer[i + 1] == 1:
count += 1
elif dryer[i] == 0 and dryer[i + 1] == 0:
count += 1
return count
def main():
n = int(input().strip())
owner = None
dryer = [0] * 101
for i in range(n):
l, r = map(int, input().strip().split())
if owner is None:
owner = l, r
else:
for j in range(l, r + 1):
dryer[j] = 1
l = function(dryer, owner)
print(l)
if __name__ == '__main__':
main()
|
Title: On Segment's Own Points
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our old friend Alexey has finally entered the University of City N — the Berland capital. Alexey expected his father to get him a place to live in but his father said it was high time for Alexey to practice some financial independence. So, Alexey is living in a dorm.
The dorm has exactly one straight dryer — a 100 centimeter long rope to hang clothes on. The dryer has got a coordinate system installed: the leftmost end of the dryer has coordinate 0, and the opposite end has coordinate 100. Overall, the university has *n* students. Dean's office allows *i*-th student to use the segment (*l**i*,<=*r**i*) of the dryer. However, the dean's office actions are contradictory and now one part of the dryer can belong to multiple students!
Alexey don't like when someone touch his clothes. That's why he want make it impossible to someone clothes touch his ones. So Alexey wonders: what is the total length of the parts of the dryer that he may use in a such way that clothes of the others (*n*<=-<=1) students aren't drying there. Help him! Note that Alexey, as the most respected student, has number 1.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100). The (*i*<=+<=1)-th line contains integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=100) — the endpoints of the corresponding segment for the *i*-th student.
Output Specification:
On a single line print a single number *k*, equal to the sum of lengths of the parts of the dryer which are inside Alexey's segment and are outside all other segments.
Demo Input:
['3\n0 5\n2 8\n1 6\n', '3\n0 10\n1 5\n7 15\n']
Demo Output:
['1\n', '3\n']
Note:
Note that it's not important are clothes drying on the touching segments (e.g. (0, 1) and (1, 2)) considered to be touching or not because you need to find the length of segments.
In the first test sample Alexey may use the only segment (0, 1). In such case his clothes will not touch clothes on the segments (1, 6) and (2, 8). The length of segment (0, 1) is 1.
In the second test sample Alexey may dry his clothes on segments (0, 1) and (5, 7). Overall length of these segments is 3.
|
```python
def function(dryer, owner):
l, r = owner
count = 0
for i in range(l, r):
if i + 1 <= 100:
if dryer[i] == 1 and dryer[i + 1] == 0:
count += 1
elif dryer[i] == 0 and dryer[i + 1] == 1:
count += 1
elif dryer[i] == 0 and dryer[i + 1] == 0:
count += 1
return count
def main():
n = int(input().strip())
owner = None
dryer = [0] * 101
for i in range(n):
l, r = map(int, input().strip().split())
if owner is None:
owner = l, r
else:
for j in range(l, r + 1):
dryer[j] = 1
l = function(dryer, owner)
print(l)
if __name__ == '__main__':
main()
```
| 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"
}
] | 1,686,845,837
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
n, m = map(int, input().split())
ans = (n * m) // 2
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
n, m = map(int, input().split())
ans = (n * m) // 2
print(ans)
```
| 3.977
|
115
|
A
|
Party
|
PROGRAMMING
| 900
|
[
"dfs and similar",
"graphs",
"trees"
] | null | null |
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
|
Print a single integer denoting the minimum number of groups that will be formed in the party.
|
[
"5\n-1\n1\n2\n1\n-1\n"
] |
[
"3\n"
] |
For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5
| 500
|
[
{
"input": "5\n-1\n1\n2\n1\n-1",
"output": "3"
},
{
"input": "4\n-1\n1\n2\n3",
"output": "4"
},
{
"input": "12\n-1\n1\n2\n3\n-1\n5\n6\n7\n-1\n9\n10\n11",
"output": "4"
},
{
"input": "6\n-1\n-1\n2\n3\n1\n1",
"output": "3"
},
{
"input": "3\n-1\n1\n1",
"output": "2"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n2\n-1",
"output": "2"
},
{
"input": "2\n-1\n-1",
"output": "1"
},
{
"input": "3\n2\n-1\n1",
"output": "3"
},
{
"input": "3\n-1\n-1\n-1",
"output": "1"
},
{
"input": "5\n4\n5\n1\n-1\n4",
"output": "3"
},
{
"input": "12\n-1\n1\n1\n1\n1\n1\n3\n4\n3\n3\n4\n7",
"output": "4"
},
{
"input": "12\n-1\n-1\n1\n-1\n1\n1\n5\n11\n8\n6\n6\n4",
"output": "5"
},
{
"input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n2\n-1\n-1\n-1",
"output": "2"
},
{
"input": "12\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1",
"output": "1"
},
{
"input": "12\n3\n4\n2\n8\n7\n1\n10\n12\n5\n-1\n9\n11",
"output": "12"
},
{
"input": "12\n5\n6\n7\n1\n-1\n9\n12\n4\n8\n-1\n3\n2",
"output": "11"
},
{
"input": "12\n-1\n9\n11\n6\n6\n-1\n6\n3\n8\n6\n1\n6",
"output": "6"
},
{
"input": "12\n7\n8\n4\n12\n7\n9\n-1\n-1\n-1\n8\n6\n-1",
"output": "3"
},
{
"input": "12\n-1\n10\n-1\n1\n-1\n5\n9\n12\n-1\n-1\n3\n-1",
"output": "2"
},
{
"input": "12\n-1\n7\n9\n12\n1\n7\n-1\n-1\n8\n5\n4\n-1",
"output": "3"
},
{
"input": "12\n11\n11\n8\n9\n1\n1\n2\n-1\n10\n3\n-1\n8",
"output": "5"
},
{
"input": "12\n-1\n8\n9\n-1\n4\n2\n11\n1\n-1\n6\n-1\n10",
"output": "6"
},
{
"input": "12\n7\n4\n4\n-1\n6\n7\n9\n7\n-1\n7\n4\n1",
"output": "4"
},
{
"input": "12\n-1\n6\n10\n6\n1\n11\n5\n9\n11\n5\n1\n9",
"output": "4"
},
{
"input": "12\n8\n6\n6\n8\n6\n-1\n6\n6\n8\n6\n8\n8",
"output": "3"
},
{
"input": "12\n3\n8\n9\n12\n2\n8\n11\n12\n2\n9\n-1\n11",
"output": "7"
},
{
"input": "7\n7\n6\n-1\n-1\n4\n4\n-1",
"output": "3"
},
{
"input": "77\n53\n-1\n48\n-1\n51\n-1\n54\n43\n70\n19\n64\n31\n45\n58\n6\n61\n41\n56\n34\n-1\n54\n-1\n66\n65\n74\n16\n38\n1\n2\n61\n-1\n34\n-1\n-1\n7\n19\n57\n13\n61\n42\n-1\n34\n65\n16\n31\n61\n6\n5\n34\n15\n-1\n13\n4\n22\n74\n7\n-1\n70\n38\n61\n45\n53\n46\n22\n-1\n65\n34\n19\n43\n48\n33\n58\n36\n-1\n47\n24\n34",
"output": "6"
},
{
"input": "5\n2\n3\n4\n5\n-1",
"output": "5"
}
] | 1,670,571,469
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 154
| 0
|
n = int(input())
a = [0] * 1001
ans = 0
for i in range(n):
p = int(input())
if p != -1 and a[p] == 0:
ans += 1
print(ans)
|
Title: Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
Output Specification:
Print a single integer denoting the minimum number of groups that will be formed in the party.
Demo Input:
['5\n-1\n1\n2\n1\n-1\n']
Demo Output:
['3\n']
Note:
For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5
|
```python
n = int(input())
a = [0] * 1001
ans = 0
for i in range(n):
p = int(input())
if p != -1 and a[p] == 0:
ans += 1
print(ans)
```
| 0
|
|
611
|
A
|
New Year and Days
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.
Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.
Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.
Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him.
|
The only line of the input is in one of the following two formats:
- "*x* of week" where *x* (1<=≤<=*x*<=≤<=7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday. - "*x* of month" where *x* (1<=≤<=*x*<=≤<=31) denotes the day of the month.
|
Print one integer — the number of candies Limak will save in the year 2016.
|
[
"4 of week\n",
"30 of month\n"
] |
[
"52\n",
"11\n"
] |
Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – [https://en.wikipedia.org/wiki/Gregorian_calendar](https://en.wikipedia.org/wiki/Gregorian_calendar). The week starts with Monday.
In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.
In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total.
| 500
|
[
{
"input": "4 of week",
"output": "52"
},
{
"input": "30 of month",
"output": "11"
},
{
"input": "17 of month",
"output": "12"
},
{
"input": "31 of month",
"output": "7"
},
{
"input": "6 of week",
"output": "53"
},
{
"input": "1 of week",
"output": "52"
},
{
"input": "2 of week",
"output": "52"
},
{
"input": "3 of week",
"output": "52"
},
{
"input": "5 of week",
"output": "53"
},
{
"input": "7 of week",
"output": "52"
},
{
"input": "1 of month",
"output": "12"
},
{
"input": "2 of month",
"output": "12"
},
{
"input": "3 of month",
"output": "12"
},
{
"input": "4 of month",
"output": "12"
},
{
"input": "5 of month",
"output": "12"
},
{
"input": "6 of month",
"output": "12"
},
{
"input": "7 of month",
"output": "12"
},
{
"input": "8 of month",
"output": "12"
},
{
"input": "9 of month",
"output": "12"
},
{
"input": "10 of month",
"output": "12"
},
{
"input": "11 of month",
"output": "12"
},
{
"input": "12 of month",
"output": "12"
},
{
"input": "13 of month",
"output": "12"
},
{
"input": "14 of month",
"output": "12"
},
{
"input": "15 of month",
"output": "12"
},
{
"input": "16 of month",
"output": "12"
},
{
"input": "18 of month",
"output": "12"
},
{
"input": "19 of month",
"output": "12"
},
{
"input": "20 of month",
"output": "12"
},
{
"input": "21 of month",
"output": "12"
},
{
"input": "22 of month",
"output": "12"
},
{
"input": "23 of month",
"output": "12"
},
{
"input": "24 of month",
"output": "12"
},
{
"input": "25 of month",
"output": "12"
},
{
"input": "26 of month",
"output": "12"
},
{
"input": "27 of month",
"output": "12"
},
{
"input": "28 of month",
"output": "12"
},
{
"input": "29 of month",
"output": "12"
}
] | 1,540,305,269
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 124
| 0
|
x,y,s=input().split()
x=int(x)
cnt=0
if s=="month":
if x==31:
cnt=6
elif x==30:
cnt=11
else:
cnt=12
else:
if x==5:
cnt=53
else:
cnt=52
print(cnt)
|
Title: New Year and Days
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.
Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.
Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.
Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him.
Input Specification:
The only line of the input is in one of the following two formats:
- "*x* of week" where *x* (1<=≤<=*x*<=≤<=7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday. - "*x* of month" where *x* (1<=≤<=*x*<=≤<=31) denotes the day of the month.
Output Specification:
Print one integer — the number of candies Limak will save in the year 2016.
Demo Input:
['4 of week\n', '30 of month\n']
Demo Output:
['52\n', '11\n']
Note:
Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – [https://en.wikipedia.org/wiki/Gregorian_calendar](https://en.wikipedia.org/wiki/Gregorian_calendar). The week starts with Monday.
In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.
In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total.
|
```python
x,y,s=input().split()
x=int(x)
cnt=0
if s=="month":
if x==31:
cnt=6
elif x==30:
cnt=11
else:
cnt=12
else:
if x==5:
cnt=53
else:
cnt=52
print(cnt)
```
| 0
|
|
120
|
E
|
Put Knight!
|
PROGRAMMING
| 1,400
|
[
"games",
"math"
] | null | null |
Petya and Gena play a very interesting game "Put a Knight!" on a chessboard *n*<=×<=*n* in size. In this game they take turns to put chess pieces called "knights" on the board so that no two knights could threat each other. A knight located in square (*r*,<=*c*) can threat squares (*r*<=-<=1,<=*c*<=+<=2), (*r*<=-<=1,<=*c*<=-<=2), (*r*<=+<=1,<=*c*<=+<=2), (*r*<=+<=1,<=*c*<=-<=2), (*r*<=-<=2,<=*c*<=+<=1), (*r*<=-<=2,<=*c*<=-<=1), (*r*<=+<=2,<=*c*<=+<=1) and (*r*<=+<=2,<=*c*<=-<=1) (some of the squares may be located outside the chessboard). The player who can't put a new knight during his move loses. Determine which player wins considering that both players play optimally well and Petya starts.
|
The first line contains integer *T* (1<=≤<=*T*<=≤<=100) — the number of boards, for which you should determine the winning player. Next *T* lines contain *T* integers *n**i* (1<=≤<=*n**i*<=≤<=10000) — the sizes of the chessboards.
|
For each *n**i*<=×<=*n**i* board print on a single line "0" if Petya wins considering both players play optimally well. Otherwise, print "1".
|
[
"2\n2\n1\n"
] |
[
"1\n0\n"
] |
none
| 0
|
[
{
"input": "2\n2\n1",
"output": "1\n0"
},
{
"input": "10\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10",
"output": "0\n1\n0\n1\n0\n1\n0\n1\n0\n1"
},
{
"input": "15\n10\n4\n7\n8\n9\n6\n2\n1\n3\n1\n5\n2\n3\n4\n5",
"output": "1\n1\n0\n1\n0\n1\n1\n0\n0\n0\n0\n1\n0\n1\n0"
},
{
"input": "6\n10\n7\n10\n8\n5\n1",
"output": "1\n0\n1\n1\n0\n0"
},
{
"input": "100\n5\n6\n8\n7\n5\n7\n10\n2\n8\n3\n10\n3\n7\n3\n2\n7\n10\n3\n7\n3\n9\n5\n1\n1\n1\n5\n7\n5\n4\n8\n7\n3\n2\n10\n5\n10\n1\n10\n5\n2\n10\n6\n4\n10\n7\n6\n10\n8\n8\n5\n5\n7\n5\n7\n8\n6\n7\n8\n5\n8\n7\n9\n1\n1\n1\n5\n10\n6\n3\n3\n2\n7\n5\n2\n4\n4\n10\n1\n5\n2\n9\n1\n9\n9\n8\n2\n6\n9\n8\n2\n6\n2\n1\n10\n10\n8\n9\n7\n8\n8",
"output": "0\n1\n1\n0\n0\n0\n1\n1\n1\n0\n1\n0\n0\n0\n1\n0\n1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n1\n1\n0\n0\n1\n1\n0\n1\n0\n1\n0\n1\n1\n1\n1\n1\n0\n1\n1\n1\n1\n0\n0\n0\n0\n0\n1\n1\n0\n1\n0\n1\n0\n0\n0\n0\n0\n0\n1\n1\n0\n0\n1\n0\n0\n1\n1\n1\n1\n0\n0\n1\n0\n0\n0\n0\n1\n1\n1\n0\n1\n1\n1\n1\n0\n1\n1\n1\n0\n0\n1\n1"
},
{
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}
] | 1,609,088,282
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
#include <bits/stdc++.h>
using namespace std;
int main(){freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);int t,n;cin>>t;while(t--){cin>>n;cout<<(n+1)%2<<"\n";}}
|
Title: Put Knight!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya and Gena play a very interesting game "Put a Knight!" on a chessboard *n*<=×<=*n* in size. In this game they take turns to put chess pieces called "knights" on the board so that no two knights could threat each other. A knight located in square (*r*,<=*c*) can threat squares (*r*<=-<=1,<=*c*<=+<=2), (*r*<=-<=1,<=*c*<=-<=2), (*r*<=+<=1,<=*c*<=+<=2), (*r*<=+<=1,<=*c*<=-<=2), (*r*<=-<=2,<=*c*<=+<=1), (*r*<=-<=2,<=*c*<=-<=1), (*r*<=+<=2,<=*c*<=+<=1) and (*r*<=+<=2,<=*c*<=-<=1) (some of the squares may be located outside the chessboard). The player who can't put a new knight during his move loses. Determine which player wins considering that both players play optimally well and Petya starts.
Input Specification:
The first line contains integer *T* (1<=≤<=*T*<=≤<=100) — the number of boards, for which you should determine the winning player. Next *T* lines contain *T* integers *n**i* (1<=≤<=*n**i*<=≤<=10000) — the sizes of the chessboards.
Output Specification:
For each *n**i*<=×<=*n**i* board print on a single line "0" if Petya wins considering both players play optimally well. Otherwise, print "1".
Demo Input:
['2\n2\n1\n']
Demo Output:
['1\n0\n']
Note:
none
|
```python
#include <bits/stdc++.h>
using namespace std;
int main(){freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);int t,n;cin>>t;while(t--){cin>>n;cout<<(n+1)%2<<"\n";}}
```
| -1
|
|
750
|
A
|
New Year and Hurry
|
PROGRAMMING
| 800
|
[
"binary search",
"brute force",
"implementation",
"math"
] | null | null |
Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem.
Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first.
How many problems can Limak solve if he wants to make it to the party?
|
The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.
|
Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.
|
[
"3 222\n",
"4 190\n",
"7 1\n"
] |
[
"2\n",
"4\n",
"7\n"
] |
In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2.
In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight.
In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.
| 500
|
[
{
"input": "3 222",
"output": "2"
},
{
"input": "4 190",
"output": "4"
},
{
"input": "7 1",
"output": "7"
},
{
"input": "10 135",
"output": "6"
},
{
"input": "10 136",
"output": "5"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 240",
"output": "0"
},
{
"input": "10 1",
"output": "9"
},
{
"input": "10 240",
"output": "0"
},
{
"input": "9 240",
"output": "0"
},
{
"input": "9 1",
"output": "9"
},
{
"input": "9 235",
"output": "1"
},
{
"input": "9 236",
"output": "0"
},
{
"input": "5 225",
"output": "2"
},
{
"input": "5 226",
"output": "1"
},
{
"input": "4 210",
"output": "3"
},
{
"input": "4 211",
"output": "2"
},
{
"input": "4 191",
"output": "3"
},
{
"input": "10 165",
"output": "5"
},
{
"input": "10 166",
"output": "4"
},
{
"input": "8 100",
"output": "7"
},
{
"input": "8 101",
"output": "6"
},
{
"input": "8 60",
"output": "8"
},
{
"input": "8 61",
"output": "7"
},
{
"input": "10 15",
"output": "9"
},
{
"input": "10 16",
"output": "8"
},
{
"input": "4 100",
"output": "4"
},
{
"input": "4 101",
"output": "4"
},
{
"input": "7 167",
"output": "4"
},
{
"input": "10 164",
"output": "5"
},
{
"input": "9 170",
"output": "4"
},
{
"input": "8 160",
"output": "5"
},
{
"input": "1 100",
"output": "1"
},
{
"input": "8 123",
"output": "6"
},
{
"input": "2 99",
"output": "2"
},
{
"input": "10 88",
"output": "7"
},
{
"input": "1 235",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "4 240",
"output": "0"
},
{
"input": "1 55",
"output": "1"
},
{
"input": "1 240",
"output": "0"
},
{
"input": "3 240",
"output": "0"
},
{
"input": "10 240",
"output": "0"
},
{
"input": "2 240",
"output": "0"
},
{
"input": "10 1",
"output": "9"
},
{
"input": "9 1",
"output": "9"
},
{
"input": "2 236",
"output": "0"
},
{
"input": "10 2",
"output": "9"
},
{
"input": "3 239",
"output": "0"
},
{
"input": "1 237",
"output": "0"
},
{
"input": "7 8",
"output": "7"
},
{
"input": "10 235",
"output": "1"
}
] | 1,691,091,270
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 52
| 46
| 0
|
def max_problems(n, k):
total_time = k
max_time_solving_problems = 240 - total_time
problems_solved = 0
for i in range(1, n + 1):
times = 5 * i
if max_time_solving_problems >= times:
max_time_solving_problems -= times
problems_solved += 1
else:
break
return problems_solved
n, k = map(int, input().split())
print(max_problems(n, k))
|
Title: New Year and Hurry
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem.
Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first.
How many problems can Limak solve if he wants to make it to the party?
Input Specification:
The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.
Output Specification:
Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.
Demo Input:
['3 222\n', '4 190\n', '7 1\n']
Demo Output:
['2\n', '4\n', '7\n']
Note:
In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2.
In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight.
In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems.
|
```python
def max_problems(n, k):
total_time = k
max_time_solving_problems = 240 - total_time
problems_solved = 0
for i in range(1, n + 1):
times = 5 * i
if max_time_solving_problems >= times:
max_time_solving_problems -= times
problems_solved += 1
else:
break
return problems_solved
n, k = map(int, input().split())
print(max_problems(n, k))
```
| 3
|
|
570
|
B
|
Simple Game
|
PROGRAMMING
| 1,300
|
[
"constructive algorithms",
"games",
"greedy",
"implementation",
"math"
] | null | null |
One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to *n*. Let's assume that Misha chose number *m*, and Andrew chose number *a*.
Then, by using a random generator they choose a random integer *c* in the range between 1 and *n* (any integer from 1 to *n* is chosen with the same probability), after which the winner is the player, whose number was closer to *c*. The boys agreed that if *m* and *a* are located on the same distance from *c*, Misha wins.
Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number *n*. You need to determine which value of *a* Andrew must choose, so that the probability of his victory is the highest possible.
More formally, you need to find such integer *a* (1<=≤<=*a*<=≤<=*n*), that the probability that is maximal, where *c* is the equiprobably chosen integer from 1 to *n* (inclusive).
|
The first line contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the range of numbers in the game, and the number selected by Misha respectively.
|
Print a single number — such value *a*, that probability that Andrew wins is the highest. If there are multiple such values, print the minimum of them.
|
[
"3 1\n",
"4 3\n"
] |
[
"2",
"2"
] |
In the first sample test: Andrew wins if *c* is equal to 2 or 3. The probability that Andrew wins is 2 / 3. If Andrew chooses *a* = 3, the probability of winning will be 1 / 3. If *a* = 1, the probability of winning is 0.
In the second sample test: Andrew wins if *c* is equal to 1 and 2. The probability that Andrew wins is 1 / 2. For other choices of *a* the probability of winning is less.
| 1,000
|
[
{
"input": "3 1",
"output": "2"
},
{
"input": "4 3",
"output": "2"
},
{
"input": "5 5",
"output": "4"
},
{
"input": "10 5",
"output": "6"
},
{
"input": "20 13",
"output": "12"
},
{
"input": "51 1",
"output": "2"
},
{
"input": "100 50",
"output": "51"
},
{
"input": "100 51",
"output": "50"
},
{
"input": "100 49",
"output": "50"
},
{
"input": "1000000000 1000000000",
"output": "999999999"
},
{
"input": "1000000000 1",
"output": "2"
},
{
"input": "1000000000 100000000",
"output": "100000001"
},
{
"input": "1000000000 500000000",
"output": "500000001"
},
{
"input": "1000000000 123124",
"output": "123125"
},
{
"input": "12412523 125123",
"output": "125124"
},
{
"input": "54645723 432423",
"output": "432424"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "262833325 131416663",
"output": "131416662"
},
{
"input": "477667530 238833766",
"output": "238833765"
},
{
"input": "692501734 346250868",
"output": "346250867"
},
{
"input": "907335939 453667970",
"output": "453667969"
},
{
"input": "746085224 373042613",
"output": "373042612"
},
{
"input": "189520699 94760350",
"output": "94760349"
},
{
"input": "404354904 202177453",
"output": "202177452"
},
{
"input": "619189108 309594555",
"output": "309594554"
},
{
"input": "81813292 40906647",
"output": "40906646"
},
{
"input": "296647497 148323750",
"output": "148323749"
},
{
"input": "511481701 255740851",
"output": "255740850"
},
{
"input": "726315905 363157953",
"output": "363157952"
},
{
"input": "496110970 201868357",
"output": "201868358"
},
{
"input": "710945175 173165570",
"output": "173165571"
},
{
"input": "925779379 720443954",
"output": "720443953"
},
{
"input": "140613583 93171580",
"output": "93171579"
},
{
"input": "355447788 85890184",
"output": "85890185"
},
{
"input": "570281992 291648263",
"output": "291648262"
},
{
"input": "541904957 459371829",
"output": "459371828"
},
{
"input": "756739161 125332525",
"output": "125332526"
},
{
"input": "971573366 216791157",
"output": "216791158"
},
{
"input": "186407570 160453970",
"output": "160453969"
},
{
"input": "401241775 170032078",
"output": "170032079"
},
{
"input": "616075979 207073797",
"output": "207073798"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 1",
"output": "2"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "3 1",
"output": "2"
},
{
"input": "3 2",
"output": "1"
},
{
"input": "3 3",
"output": "2"
},
{
"input": "4 1",
"output": "2"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "4 3",
"output": "2"
},
{
"input": "4 4",
"output": "3"
},
{
"input": "5 1",
"output": "2"
},
{
"input": "5 2",
"output": "3"
},
{
"input": "5 3",
"output": "2"
},
{
"input": "5 4",
"output": "3"
},
{
"input": "5 5",
"output": "4"
},
{
"input": "3 2",
"output": "1"
},
{
"input": "7 4",
"output": "3"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "7 3",
"output": "4"
}
] | 1,579,820,035
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 16
| 171
| 0
|
n, m = map(int, input().split())
range1, range2 = n - m, m - 1
print(m + 1 if range1 > range2 else m - 1)
|
Title: Simple Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to *n*. Let's assume that Misha chose number *m*, and Andrew chose number *a*.
Then, by using a random generator they choose a random integer *c* in the range between 1 and *n* (any integer from 1 to *n* is chosen with the same probability), after which the winner is the player, whose number was closer to *c*. The boys agreed that if *m* and *a* are located on the same distance from *c*, Misha wins.
Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number *n*. You need to determine which value of *a* Andrew must choose, so that the probability of his victory is the highest possible.
More formally, you need to find such integer *a* (1<=≤<=*a*<=≤<=*n*), that the probability that is maximal, where *c* is the equiprobably chosen integer from 1 to *n* (inclusive).
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=109) — the range of numbers in the game, and the number selected by Misha respectively.
Output Specification:
Print a single number — such value *a*, that probability that Andrew wins is the highest. If there are multiple such values, print the minimum of them.
Demo Input:
['3 1\n', '4 3\n']
Demo Output:
['2', '2']
Note:
In the first sample test: Andrew wins if *c* is equal to 2 or 3. The probability that Andrew wins is 2 / 3. If Andrew chooses *a* = 3, the probability of winning will be 1 / 3. If *a* = 1, the probability of winning is 0.
In the second sample test: Andrew wins if *c* is equal to 1 and 2. The probability that Andrew wins is 1 / 2. For other choices of *a* the probability of winning is less.
|
```python
n, m = map(int, input().split())
range1, range2 = n - m, m - 1
print(m + 1 if range1 > range2 else m - 1)
```
| 0
|
|
723
|
A
|
The New Year: Meeting Friends
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"sortings"
] | null | null |
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
|
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
|
Print one integer — the minimum total distance the friends need to travel in order to meet together.
|
[
"7 1 4\n",
"30 20 10\n"
] |
[
"6\n",
"20\n"
] |
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
| 500
|
[
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"input": "71 85 88",
"output": "17"
},
{
"input": "30 38 99",
"output": "69"
},
{
"input": "23 82 95",
"output": "72"
},
{
"input": "22 41 47",
"output": "25"
},
{
"input": "9 94 77",
"output": "85"
},
{
"input": "1 53 51",
"output": "52"
},
{
"input": "25 97 93",
"output": "72"
},
{
"input": "42 53 51",
"output": "11"
},
{
"input": "81 96 94",
"output": "15"
},
{
"input": "21 5 93",
"output": "88"
},
{
"input": "50 13 75",
"output": "62"
},
{
"input": "41 28 98",
"output": "70"
},
{
"input": "69 46 82",
"output": "36"
},
{
"input": "87 28 89",
"output": "61"
},
{
"input": "44 45 40",
"output": "5"
},
{
"input": "86 97 68",
"output": "29"
},
{
"input": "43 92 30",
"output": "62"
},
{
"input": "16 70 1",
"output": "69"
},
{
"input": "40 46 19",
"output": "27"
},
{
"input": "71 38 56",
"output": "33"
},
{
"input": "82 21 80",
"output": "61"
},
{
"input": "75 8 35",
"output": "67"
},
{
"input": "75 24 28",
"output": "51"
},
{
"input": "78 23 56",
"output": "55"
},
{
"input": "85 31 10",
"output": "75"
},
{
"input": "76 50 9",
"output": "67"
},
{
"input": "95 37 34",
"output": "61"
},
{
"input": "84 61 35",
"output": "49"
},
{
"input": "87 85 37",
"output": "50"
},
{
"input": "1 3 2",
"output": "2"
},
{
"input": "4 2 6",
"output": "4"
},
{
"input": "6 9 3",
"output": "6"
},
{
"input": "12 4 8",
"output": "8"
},
{
"input": "15 10 5",
"output": "10"
},
{
"input": "1 50 17",
"output": "49"
},
{
"input": "10 5 15",
"output": "10"
},
{
"input": "8 1 9",
"output": "8"
},
{
"input": "3 5 4",
"output": "2"
},
{
"input": "2 1 3",
"output": "2"
},
{
"input": "1 8 2",
"output": "7"
},
{
"input": "1 100 2",
"output": "99"
},
{
"input": "1 4 6",
"output": "5"
}
] | 1,685,166,896
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 46
| 0
|
a, b, c = map(int, input().split())
sr = sorted([a, b, c])[1]
print((sorted([a, b, c])[2]-sr) + (sr-sorted([a, b, c])[0]))
|
Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integer — the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
|
```python
a, b, c = map(int, input().split())
sr = sorted([a, b, c])[1]
print((sorted([a, b, c])[2]-sr) + (sr-sorted([a, b, c])[0]))
```
| 3
|
|
577
|
A
|
Multiplication Table
|
PROGRAMMING
| 1,000
|
[
"implementation",
"number theory"
] | null | null |
Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
|
The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table.
|
Print a single number: the number of times *x* occurs in the table.
|
[
"10 5\n",
"6 12\n",
"5 13\n"
] |
[
"2\n",
"4\n",
"0\n"
] |
A table for the second sample test is given below. The occurrences of number 12 are marked bold.
| 500
|
[
{
"input": "10 5",
"output": "2"
},
{
"input": "6 12",
"output": "4"
},
{
"input": "5 13",
"output": "0"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "1 1000000000",
"output": "0"
},
{
"input": "100000 1000000000",
"output": "16"
},
{
"input": "100000 362880",
"output": "154"
},
{
"input": "1 4",
"output": "0"
},
{
"input": "9 12",
"output": "4"
},
{
"input": "10 123",
"output": "0"
},
{
"input": "9551 975275379",
"output": "0"
},
{
"input": "17286 948615687",
"output": "0"
},
{
"input": "58942 936593001",
"output": "0"
},
{
"input": "50000 989460910",
"output": "4"
},
{
"input": "22741 989460910",
"output": "0"
},
{
"input": "22740 989460910",
"output": "0"
},
{
"input": "100000 989460910",
"output": "4"
},
{
"input": "100000 98280",
"output": "128"
},
{
"input": "100000 997920",
"output": "222"
},
{
"input": "100000 720720",
"output": "226"
},
{
"input": "100000 2162160",
"output": "282"
},
{
"input": "100000 4324320",
"output": "320"
},
{
"input": "100000 8648640",
"output": "348"
},
{
"input": "100000 183783600",
"output": "438"
},
{
"input": "100000 551350800",
"output": "392"
},
{
"input": "40000 551350800",
"output": "150"
},
{
"input": "20000 400000000",
"output": "1"
},
{
"input": "19999 400000000",
"output": "0"
},
{
"input": "19999 399960001",
"output": "1"
},
{
"input": "31621 999887641",
"output": "1"
},
{
"input": "31622 999887641",
"output": "1"
},
{
"input": "31620 999887641",
"output": "0"
},
{
"input": "100000 999887641",
"output": "3"
},
{
"input": "100000 25",
"output": "3"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "100000 3628800",
"output": "220"
},
{
"input": "100000 39916800",
"output": "328"
},
{
"input": "100000 479001600",
"output": "254"
},
{
"input": "4 9",
"output": "1"
},
{
"input": "2 6",
"output": "0"
},
{
"input": "20 100",
"output": "3"
},
{
"input": "10 3",
"output": "2"
},
{
"input": "4 4",
"output": "3"
},
{
"input": "2 4",
"output": "1"
},
{
"input": "89874 1",
"output": "1"
}
] | 1,643,471,304
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
from math import sqrt
n, x = [int(i) for i in input().split()]
ans = 0
for i in range(1, int(sqrt(x))+1):
if x%i == 0:
q1 = i
q2 = x//i
print(q1, q2, ans)
if q1 <= n and q2 <= n and q1 != q2: ans += 2
if q1 <= n and q2 <= n and q1 == q2: ans += 1
print(ans)
|
Title: Multiplication Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
Input Specification:
The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table.
Output Specification:
Print a single number: the number of times *x* occurs in the table.
Demo Input:
['10 5\n', '6 12\n', '5 13\n']
Demo Output:
['2\n', '4\n', '0\n']
Note:
A table for the second sample test is given below. The occurrences of number 12 are marked bold.
|
```python
from math import sqrt
n, x = [int(i) for i in input().split()]
ans = 0
for i in range(1, int(sqrt(x))+1):
if x%i == 0:
q1 = i
q2 = x//i
print(q1, q2, ans)
if q1 <= n and q2 <= n and q1 != q2: ans += 2
if q1 <= n and q2 <= n and q1 == q2: ans += 1
print(ans)
```
| 0
|
|
540
|
A
|
Combination Lock
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock.
The second line contains a string of *n* digits — the original state of the disks.
The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.
|
Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.
|
[
"5\n82195\n64723\n"
] |
[
"13\n"
] |
In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 500
|
[
{
"input": "5\n82195\n64723",
"output": "13"
},
{
"input": "12\n102021090898\n010212908089",
"output": "16"
},
{
"input": "1\n8\n1",
"output": "3"
},
{
"input": "2\n83\n57",
"output": "7"
},
{
"input": "10\n0728592530\n1362615763",
"output": "27"
},
{
"input": "100\n4176196363694273682807653052945037727131821799902563705176501742060696655282954944720643131654235909\n3459912084922154505910287499879975659298239371519889866585472674423008837878123067103005344986554746",
"output": "245"
},
{
"input": "1\n8\n1",
"output": "3"
},
{
"input": "2\n83\n57",
"output": "7"
},
{
"input": "3\n607\n684",
"output": "5"
},
{
"input": "4\n0809\n0636",
"output": "8"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8037856825987124762280548\n9519431339078678836940020",
"output": "72"
},
{
"input": "125\n23269567683904664184142384849516523616863461607751021071772615078579713054027902974007001544768640273491193035874486891541257\n47635110303703399505805044019026243695451609639556649012447370081552870340011971572363458960190590266459684717415349529509024",
"output": "305"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8037856825987124762285484\n9519431339078678836940202",
"output": "74"
},
{
"input": "125\n23269567689466418414238845152168634610771021717726157879713054270294007001544768647391193035874486891412573389247025830678706\n47635110307339950580504010224954516093956649124473708152870340117152363458960190596659684717415349529090241694059599629136831",
"output": "357"
},
{
"input": "5\n84284\n08941",
"output": "16"
},
{
"input": "25\n8378525987476228048406972\n9194339078883694020217816",
"output": "55"
},
{
"input": "125\n23269576839046618414238484916523616863461607750210717761078579713054027902974007015447686027349193035874486891541257338624472\n47635103037033950580504401926243695451609639556490124437081552870340011971572363489601905026645984717415349529509024169604599",
"output": "274"
},
{
"input": "1\n0\n0",
"output": "0"
},
{
"input": "1\n7\n7",
"output": "0"
},
{
"input": "1\n0\n5",
"output": "5"
},
{
"input": "1\n2\n7",
"output": "5"
},
{
"input": "1\n7\n9",
"output": "2"
},
{
"input": "1\n9\n7",
"output": "2"
},
{
"input": "1\n2\n9",
"output": "3"
},
{
"input": "1\n9\n2",
"output": "3"
},
{
"input": "25\n3164978461316464614169874\n9413979197249127496597357",
"output": "66"
},
{
"input": "4\n9999\n9999",
"output": "0"
},
{
"input": "2\n11\n11",
"output": "0"
}
] | 1,611,417,595
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 43
| 62
| 0
|
n = int(input())
expected = int(input())
final = int(input())
ans = 0
for _ in range(n):
a, b = expected % 10, final % 10
n = abs(a - b)
ans += min(n, 10 - n)
expected //= 10
final //= 10
print(ans)
|
Title: Combination Lock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock.
The second line contains a string of *n* digits — the original state of the disks.
The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.
Output Specification:
Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.
Demo Input:
['5\n82195\n64723\n']
Demo Output:
['13\n']
Note:
In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
n = int(input())
expected = int(input())
final = int(input())
ans = 0
for _ in range(n):
a, b = expected % 10, final % 10
n = abs(a - b)
ans += min(n, 10 - n)
expected //= 10
final //= 10
print(ans)
```
| 3
|
|
381
|
A
|
Sereja and Dima
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"two pointers"
] | null | null |
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
|
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
|
[
"4\n4 1 2 10\n",
"7\n1 2 3 4 5 6 7\n"
] |
[
"12 5\n",
"16 12\n"
] |
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
| 500
|
[
{
"input": "4\n4 1 2 10",
"output": "12 5"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "16 12"
},
{
"input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13",
"output": "613 418"
},
{
"input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24",
"output": "644 500"
},
{
"input": "1\n3",
"output": "3 0"
},
{
"input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646",
"output": "6848 6568"
},
{
"input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727",
"output": "9562 9561"
},
{
"input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12",
"output": "315 315"
},
{
"input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304",
"output": "3238 2222"
},
{
"input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325",
"output": "5246 4864"
},
{
"input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71",
"output": "8147 7807"
},
{
"input": "1\n1",
"output": "1 0"
}
] | 1,672,927,674
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 46
| 0
|
n = int(input())
l = list(map(int, input().split()))
s = 0
d = 0
c = 0
while True:
if c%2 == 0:
s += max(l[0], l[-1])
else:
d += max(l[0], l[-1])
c -= 1
l.remove(max(l[0], l[-1]))
if len(l) == 0:
break
print(s,d)
|
Title: Sereja and Dima
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output Specification:
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Demo Input:
['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n']
Demo Output:
['12 5\n', '16 12\n']
Note:
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
|
```python
n = int(input())
l = list(map(int, input().split()))
s = 0
d = 0
c = 0
while True:
if c%2 == 0:
s += max(l[0], l[-1])
else:
d += max(l[0], l[-1])
c -= 1
l.remove(max(l[0], l[-1]))
if len(l) == 0:
break
print(s,d)
```
| 3
|
|
239
|
A
|
Two Bags of Potatoes
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation",
"math"
] | null | null |
Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order.
|
The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105).
|
Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1.
|
[
"10 1 10\n",
"10 6 40\n"
] |
[
"-1\n",
"2 8 14 20 26 \n"
] |
none
| 500
|
[
{
"input": "10 1 10",
"output": "-1"
},
{
"input": "10 6 40",
"output": "2 8 14 20 26 "
},
{
"input": "10 1 20",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "1 10000 1000000000",
"output": "9999 19999 29999 39999 49999 59999 69999 79999 89999 99999 109999 119999 129999 139999 149999 159999 169999 179999 189999 199999 209999 219999 229999 239999 249999 259999 269999 279999 289999 299999 309999 319999 329999 339999 349999 359999 369999 379999 389999 399999 409999 419999 429999 439999 449999 459999 469999 479999 489999 499999 509999 519999 529999 539999 549999 559999 569999 579999 589999 599999 609999 619999 629999 639999 649999 659999 669999 679999 689999 699999 709999 719999 729999 739999 7499..."
},
{
"input": "84817 1 33457",
"output": "-1"
},
{
"input": "21 37 99",
"output": "16 53 "
},
{
"input": "78 7 15",
"output": "-1"
},
{
"input": "74 17 27",
"output": "-1"
},
{
"input": "79 23 43",
"output": "-1"
},
{
"input": "32 33 3",
"output": "-1"
},
{
"input": "55 49 44",
"output": "-1"
},
{
"input": "64 59 404",
"output": "54 113 172 231 290 "
},
{
"input": "61 69 820",
"output": "8 77 146 215 284 353 422 491 560 629 698 "
},
{
"input": "17 28 532",
"output": "11 39 67 95 123 151 179 207 235 263 291 319 347 375 403 431 459 487 515 "
},
{
"input": "46592 52 232",
"output": "-1"
},
{
"input": "1541 58 648",
"output": "-1"
},
{
"input": "15946 76 360",
"output": "-1"
},
{
"input": "30351 86 424",
"output": "-1"
},
{
"input": "1 2 37493",
"output": "1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 101 103 105 107 109 111 113 115 117 119 121 123 125 127 129 131 133 135 137 139 141 143 145 147 149 151 153 155 157 159 161 163 165 167 169 171 173 175 177 179 181 183 185 187 189 191 193 195 197 199 201 203 205 207 209 211 213 215 217 219 221 223 225 227 229 231 233 235 237 239 241 243 245 247 249 251 253 255 257 259 261 263 265 267 269 271 273 275 277 279 281 28..."
},
{
"input": "1 3 27764",
"output": "2 5 8 11 14 17 20 23 26 29 32 35 38 41 44 47 50 53 56 59 62 65 68 71 74 77 80 83 86 89 92 95 98 101 104 107 110 113 116 119 122 125 128 131 134 137 140 143 146 149 152 155 158 161 164 167 170 173 176 179 182 185 188 191 194 197 200 203 206 209 212 215 218 221 224 227 230 233 236 239 242 245 248 251 254 257 260 263 266 269 272 275 278 281 284 287 290 293 296 299 302 305 308 311 314 317 320 323 326 329 332 335 338 341 344 347 350 353 356 359 362 365 368 371 374 377 380 383 386 389 392 395 398 401 404 407 410..."
},
{
"input": "10 4 9174",
"output": "2 6 10 14 18 22 26 30 34 38 42 46 50 54 58 62 66 70 74 78 82 86 90 94 98 102 106 110 114 118 122 126 130 134 138 142 146 150 154 158 162 166 170 174 178 182 186 190 194 198 202 206 210 214 218 222 226 230 234 238 242 246 250 254 258 262 266 270 274 278 282 286 290 294 298 302 306 310 314 318 322 326 330 334 338 342 346 350 354 358 362 366 370 374 378 382 386 390 394 398 402 406 410 414 418 422 426 430 434 438 442 446 450 454 458 462 466 470 474 478 482 486 490 494 498 502 506 510 514 518 522 526 530 534 53..."
},
{
"input": "33 7 4971",
"output": "2 9 16 23 30 37 44 51 58 65 72 79 86 93 100 107 114 121 128 135 142 149 156 163 170 177 184 191 198 205 212 219 226 233 240 247 254 261 268 275 282 289 296 303 310 317 324 331 338 345 352 359 366 373 380 387 394 401 408 415 422 429 436 443 450 457 464 471 478 485 492 499 506 513 520 527 534 541 548 555 562 569 576 583 590 597 604 611 618 625 632 639 646 653 660 667 674 681 688 695 702 709 716 723 730 737 744 751 758 765 772 779 786 793 800 807 814 821 828 835 842 849 856 863 870 877 884 891 898 905 912 919..."
},
{
"input": "981 1 3387",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "386 1 2747",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "123 2 50000",
"output": "1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 101 103 105 107 109 111 113 115 117 119 121 123 125 127 129 131 133 135 137 139 141 143 145 147 149 151 153 155 157 159 161 163 165 167 169 171 173 175 177 179 181 183 185 187 189 191 193 195 197 199 201 203 205 207 209 211 213 215 217 219 221 223 225 227 229 231 233 235 237 239 241 243 245 247 249 251 253 255 257 259 261 263 265 267 269 271 273 275 277 279 281 28..."
},
{
"input": "3123 100 10000000",
"output": "77 177 277 377 477 577 677 777 877 977 1077 1177 1277 1377 1477 1577 1677 1777 1877 1977 2077 2177 2277 2377 2477 2577 2677 2777 2877 2977 3077 3177 3277 3377 3477 3577 3677 3777 3877 3977 4077 4177 4277 4377 4477 4577 4677 4777 4877 4977 5077 5177 5277 5377 5477 5577 5677 5777 5877 5977 6077 6177 6277 6377 6477 6577 6677 6777 6877 6977 7077 7177 7277 7377 7477 7577 7677 7777 7877 7977 8077 8177 8277 8377 8477 8577 8677 8777 8877 8977 9077 9177 9277 9377 9477 9577 9677 9777 9877 9977 10077 10177 10277 1037..."
},
{
"input": "2 10000 1000000000",
"output": "9998 19998 29998 39998 49998 59998 69998 79998 89998 99998 109998 119998 129998 139998 149998 159998 169998 179998 189998 199998 209998 219998 229998 239998 249998 259998 269998 279998 289998 299998 309998 319998 329998 339998 349998 359998 369998 379998 389998 399998 409998 419998 429998 439998 449998 459998 469998 479998 489998 499998 509998 519998 529998 539998 549998 559998 569998 579998 589998 599998 609998 619998 629998 639998 649998 659998 669998 679998 689998 699998 709998 719998 729998 739998 7499..."
},
{
"input": "3 10000 1000000000",
"output": "9997 19997 29997 39997 49997 59997 69997 79997 89997 99997 109997 119997 129997 139997 149997 159997 169997 179997 189997 199997 209997 219997 229997 239997 249997 259997 269997 279997 289997 299997 309997 319997 329997 339997 349997 359997 369997 379997 389997 399997 409997 419997 429997 439997 449997 459997 469997 479997 489997 499997 509997 519997 529997 539997 549997 559997 569997 579997 589997 599997 609997 619997 629997 639997 649997 659997 669997 679997 689997 699997 709997 719997 729997 739997 7499..."
},
{
"input": "12312223 10000 1000000000",
"output": "7777 17777 27777 37777 47777 57777 67777 77777 87777 97777 107777 117777 127777 137777 147777 157777 167777 177777 187777 197777 207777 217777 227777 237777 247777 257777 267777 277777 287777 297777 307777 317777 327777 337777 347777 357777 367777 377777 387777 397777 407777 417777 427777 437777 447777 457777 467777 477777 487777 497777 507777 517777 527777 537777 547777 557777 567777 577777 587777 597777 607777 617777 627777 637777 647777 657777 667777 677777 687777 697777 707777 717777 727777 737777 7477..."
},
{
"input": "500000000 1000000000 1000000000",
"output": "500000000 "
},
{
"input": "1 1000000000 1000000000",
"output": "999999999 "
},
{
"input": "10 6 11",
"output": "-1"
},
{
"input": "2 100 10",
"output": "-1"
},
{
"input": "1 100000007 1000000000",
"output": "100000006 200000013 300000020 400000027 500000034 600000041 700000048 800000055 900000062 "
},
{
"input": "1 999999999 1000000000",
"output": "999999998 "
},
{
"input": "100000000 1000000000 1000000000",
"output": "900000000 "
},
{
"input": "11 2 12",
"output": "1 "
},
{
"input": "31 10 39",
"output": "-1"
},
{
"input": "48 6 50",
"output": "-1"
},
{
"input": "500000000 500000000 1000000000",
"output": "500000000 "
},
{
"input": "1 1000000000 999999999",
"output": "-1"
},
{
"input": "4 2 10",
"output": "2 4 6 "
},
{
"input": "1000000000 1 1",
"output": "-1"
},
{
"input": "1000000000 1 100000",
"output": "-1"
},
{
"input": "1000000000 1 10",
"output": "-1"
},
{
"input": "10 5 14",
"output": "-1"
},
{
"input": "500000000 499999999 1000000000",
"output": "499999998 "
},
{
"input": "1 999999997 1000000000",
"output": "999999996 "
}
] | 1,614,077,311
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 92
| 307,200
|
def isDiv(y,n,k):
res=[]
temp=y
while temp<=n:
if temp%k==0:
res.append(temp)
temp+=1
return res
y,k,n=map(int,input().split())
if y==n:
print(-1)
else:
res=isDiv(y,n,k)
ans=set()
for i in range(len(res)):
ans.add((res[i]-y))
# print(res[i]-y,end=" ")
# ans2=sorted(ans)
for i in ans:
print(i,end=" ")
|
Title: Two Bags of Potatoes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order.
Input Specification:
The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105).
Output Specification:
Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1.
Demo Input:
['10 1 10\n', '10 6 40\n']
Demo Output:
['-1\n', '2 8 14 20 26 \n']
Note:
none
|
```python
def isDiv(y,n,k):
res=[]
temp=y
while temp<=n:
if temp%k==0:
res.append(temp)
temp+=1
return res
y,k,n=map(int,input().split())
if y==n:
print(-1)
else:
res=isDiv(y,n,k)
ans=set()
for i in range(len(res)):
ans.add((res[i]-y))
# print(res[i]-y,end=" ")
# ans2=sorted(ans)
for i in ans:
print(i,end=" ")
```
| 0
|
|
754
|
C
|
Vladik and chat
|
PROGRAMMING
| 2,200
|
[
"brute force",
"constructive algorithms",
"dp",
"implementation",
"strings"
] | null | null |
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download *t* chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
|
The first line contains single integer *t* (1<=≤<=*t*<=≤<=10) — the number of chats. The *t* chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of users in the chat.
The next line contains *n* space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer *m* (1<=≤<=*m*<=≤<=100) — the number of messages in the chat. The next *m* line contain the messages in the following formats, one per line:
- <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat. - <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
|
Print the information about the *t* chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print *m* messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
|
[
"1\n2\nVladik netman\n2\n?: Hello, Vladik!\n?: Hi\n",
"1\n2\nnetman vladik\n3\nnetman:how are you?\n?:wrong message\nvladik:im fine\n",
"2\n3\nnetman vladik Fedosik\n2\n?: users are netman, vladik, Fedosik\nvladik: something wrong with this chat\n4\nnetman tigerrrrr banany2001 klinchuh\n4\n?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?\nklinchuh: yes, coach!\n?: yes, netman\nbanany2001: yes of course.\n"
] |
[
"netman: Hello, Vladik!\nVladik: Hi\n",
"Impossible\n",
"Impossible\nnetman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?\nklinchuh: yes, coach!\ntigerrrrr: yes, netman\nbanany2001: yes of course.\n"
] |
none
| 1,500
|
[
{
"input": "1\n2\nVladik netman\n2\n?: Hello, Vladik!\n?: Hi",
"output": "netman: Hello, Vladik!\nVladik: Hi"
},
{
"input": "1\n2\nnetman vladik\n3\nnetman:how are you?\n?:wrong message\nvladik:im fine",
"output": "Impossible"
},
{
"input": "2\n3\nnetman vladik Fedosik\n2\n?: users are netman, vladik, Fedosik\nvladik: something wrong with this chat\n4\nnetman tigerrrrr banany2001 klinchuh\n4\n?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?\nklinchuh: yes, coach!\n?: yes, netman\nbanany2001: yes of course.",
"output": "Impossible\nnetman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?\nklinchuh: yes, coach!\ntigerrrrr: yes, netman\nbanany2001: yes of course."
},
{
"input": "1\n1\nb\n1\nb:lala!",
"output": "b:lala!"
},
{
"input": "1\n1\nb\n1\n?:lala b!",
"output": "Impossible"
},
{
"input": "1\n1\nb\n2\n?:lala hhe!\nb:wat?",
"output": "Impossible"
},
{
"input": "1\n3\nA B C\n3\nA: HI\n?: HI\nB: HI",
"output": "A: HI\nC: HI\nB: HI"
}
] | 1,566,993,375
| 4,515
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 124
| 0
|
t = int(input())
for g in range(t):
n = int(input())
s = input().split()
orig = s
q = int(input())
x = []
for i in range(q):
x += [input().split(':')]
last = "xxxxxxxxx"
for i in range(len(x)):
if(x[i][0]=='?'):
for j in s:
if(j not in x[i][1] and j != last and i+1 < len(x) and j != x[i+1][0]):
x[i][0] = j
break
last = x[i][0]
if(x[-1][0] == '?'):
for j in s:
if j not in x[-1][1] and j != x[-2][0]:
x[-1][0] = j
break
count = 0
for i in x:
if(i[0] in s and i[0] != '?'):
count += 1
if(count == len(x)):
for i in x:
index = 0
while(index < len(i[1]) and i[1][index]==' '):
index += 1
print(i[0]+": "+i[1][index:])
else:
print("Impossible")
|
Title: Vladik and chat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Vladik discovered a new entertainment — coding bots for social networks. He would like to use machine learning in his bots so now he want to prepare some learning data for them.
At first, he need to download *t* chats. Vladik coded a script which should have downloaded the chats, however, something went wrong. In particular, some of the messages have no information of their sender. It is known that if a person sends several messages in a row, they all are merged into a single message. It means that there could not be two or more messages in a row with the same sender. Moreover, a sender never mention himself in his messages.
Vladik wants to recover senders of all the messages so that each two neighboring messages will have different senders and no sender will mention himself in his messages.
He has no idea of how to do this, and asks you for help. Help Vladik to recover senders in each of the chats!
Input Specification:
The first line contains single integer *t* (1<=≤<=*t*<=≤<=10) — the number of chats. The *t* chats follow. Each chat is given in the following format.
The first line of each chat description contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of users in the chat.
The next line contains *n* space-separated distinct usernames. Each username consists of lowercase and uppercase English letters and digits. The usernames can't start with a digit. Two usernames are different even if they differ only with letters' case. The length of username is positive and doesn't exceed 10 characters.
The next line contains single integer *m* (1<=≤<=*m*<=≤<=100) — the number of messages in the chat. The next *m* line contain the messages in the following formats, one per line:
- <username>:<text> — the format of a message with known sender. The username should appear in the list of usernames of the chat. - <?>:<text> — the format of a message with unknown sender.
The text of a message can consist of lowercase and uppercase English letter, digits, characters '.' (dot), ',' (comma), '!' (exclamation mark), '?' (question mark) and ' ' (space). The text doesn't contain trailing spaces. The length of the text is positive and doesn't exceed 100 characters.
We say that a text mention a user if his username appears in the text as a word. In other words, the username appears in a such a position that the two characters before and after its appearance either do not exist or are not English letters or digits. For example, the text "Vasya, masha13 and Kate!" can mention users "Vasya", "masha13", "and" and "Kate", but not "masha".
It is guaranteed that in each chat no known sender mention himself in his messages and there are no two neighboring messages with the same known sender.
Output Specification:
Print the information about the *t* chats in the following format:
If it is not possible to recover senders, print single line "Impossible" for this chat. Otherwise print *m* messages in the following format:
<username>:<text>
If there are multiple answers, print any of them.
Demo Input:
['1\n2\nVladik netman\n2\n?: Hello, Vladik!\n?: Hi\n', '1\n2\nnetman vladik\n3\nnetman:how are you?\n?:wrong message\nvladik:im fine\n', '2\n3\nnetman vladik Fedosik\n2\n?: users are netman, vladik, Fedosik\nvladik: something wrong with this chat\n4\nnetman tigerrrrr banany2001 klinchuh\n4\n?: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?\nklinchuh: yes, coach!\n?: yes, netman\nbanany2001: yes of course.\n']
Demo Output:
['netman: Hello, Vladik!\nVladik: Hi\n', 'Impossible\n', 'Impossible\nnetman: tigerrrrr, banany2001, klinchuh, my favourite team ever, are you ready?\nklinchuh: yes, coach!\ntigerrrrr: yes, netman\nbanany2001: yes of course.\n']
Note:
none
|
```python
t = int(input())
for g in range(t):
n = int(input())
s = input().split()
orig = s
q = int(input())
x = []
for i in range(q):
x += [input().split(':')]
last = "xxxxxxxxx"
for i in range(len(x)):
if(x[i][0]=='?'):
for j in s:
if(j not in x[i][1] and j != last and i+1 < len(x) and j != x[i+1][0]):
x[i][0] = j
break
last = x[i][0]
if(x[-1][0] == '?'):
for j in s:
if j not in x[-1][1] and j != x[-2][0]:
x[-1][0] = j
break
count = 0
for i in x:
if(i[0] in s and i[0] != '?'):
count += 1
if(count == len(x)):
for i in x:
index = 0
while(index < len(i[1]) and i[1][index]==' '):
index += 1
print(i[0]+": "+i[1][index:])
else:
print("Impossible")
```
| 0
|
|
416
|
E
|
President's Path
|
PROGRAMMING
| 2,500
|
[
"dp",
"graphs",
"shortest paths"
] | null | null |
Good old Berland has *n* cities and *m* roads. Each road connects a pair of distinct cities and is bidirectional. Between any pair of cities, there is at most one road. For each road, we know its length.
We also know that the President will soon ride along the Berland roads from city *s* to city *t*. Naturally, he will choose one of the shortest paths from *s* to *t*, but nobody can say for sure which path he will choose.
The Minister for Transport is really afraid that the President might get upset by the state of the roads in the country. That is the reason he is planning to repair the roads in the possible President's path.
Making the budget for such an event is not an easy task. For all possible distinct pairs *s*,<=*t* (*s*<=<<=*t*) find the number of roads that lie on at least one shortest path from *s* to *t*.
|
The first line of the input contains integers *n*,<=*m* (2<=≤<=*n*<=≤<=500, 0<=≤<=*m*<=≤<=*n*·(*n*<=-<=1)<=/<=2) — the number of cities and roads, correspondingly. Then *m* lines follow, containing the road descriptions, one description per line. Each description contains three integers *x**i*,<=*y**i*,<=*l**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*,<=*x**i*<=≠<=*y**i*,<=1<=≤<=*l**i*<=≤<=106), where *x**i*,<=*y**i* are the numbers of the cities connected by the *i*-th road and *l**i* is its length.
|
Print the sequence of integers *c*12,<=*c*13,<=...,<=*c*1*n*,<=*c*23,<=*c*24,<=...,<=*c*2*n*,<=...,<=*c**n*<=-<=1,<=*n*, where *c**st* is the number of roads that can lie on the shortest path from *s* to *t*. Print the elements of sequence *c* in the described order. If the pair of cities *s* and *t* don't have a path between them, then *c**st*<==<=0.
|
[
"5 6\n1 2 1\n2 3 1\n3 4 1\n4 1 1\n2 4 2\n4 5 4\n"
] |
[
"1 4 1 2 1 5 6 1 2 1 "
] |
none
| 2,500
|
[] | 1,689,194,739
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
print("_RANDOM_GUESS_1689194739.4809332")# 1689194739.480979
|
Title: President's Path
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Good old Berland has *n* cities and *m* roads. Each road connects a pair of distinct cities and is bidirectional. Between any pair of cities, there is at most one road. For each road, we know its length.
We also know that the President will soon ride along the Berland roads from city *s* to city *t*. Naturally, he will choose one of the shortest paths from *s* to *t*, but nobody can say for sure which path he will choose.
The Minister for Transport is really afraid that the President might get upset by the state of the roads in the country. That is the reason he is planning to repair the roads in the possible President's path.
Making the budget for such an event is not an easy task. For all possible distinct pairs *s*,<=*t* (*s*<=<<=*t*) find the number of roads that lie on at least one shortest path from *s* to *t*.
Input Specification:
The first line of the input contains integers *n*,<=*m* (2<=≤<=*n*<=≤<=500, 0<=≤<=*m*<=≤<=*n*·(*n*<=-<=1)<=/<=2) — the number of cities and roads, correspondingly. Then *m* lines follow, containing the road descriptions, one description per line. Each description contains three integers *x**i*,<=*y**i*,<=*l**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*,<=*x**i*<=≠<=*y**i*,<=1<=≤<=*l**i*<=≤<=106), where *x**i*,<=*y**i* are the numbers of the cities connected by the *i*-th road and *l**i* is its length.
Output Specification:
Print the sequence of integers *c*12,<=*c*13,<=...,<=*c*1*n*,<=*c*23,<=*c*24,<=...,<=*c*2*n*,<=...,<=*c**n*<=-<=1,<=*n*, where *c**st* is the number of roads that can lie on the shortest path from *s* to *t*. Print the elements of sequence *c* in the described order. If the pair of cities *s* and *t* don't have a path between them, then *c**st*<==<=0.
Demo Input:
['5 6\n1 2 1\n2 3 1\n3 4 1\n4 1 1\n2 4 2\n4 5 4\n']
Demo Output:
['1 4 1 2 1 5 6 1 2 1 ']
Note:
none
|
```python
print("_RANDOM_GUESS_1689194739.4809332")# 1689194739.480979
```
| 0
|
|
869
|
B
|
The Eternal Immortality
|
PROGRAMMING
| 1,100
|
[
"math"
] | null | null |
Even if the world is full of counterfeits, I still regard it as wonderful.
Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this.
The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1.
Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer.
As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge.
|
The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018).
|
Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.
|
[
"2 4\n",
"0 10\n",
"107 109\n"
] |
[
"2\n",
"0\n",
"2\n"
] |
In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2;
In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0;
In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.
| 1,000
|
[
{
"input": "2 4",
"output": "2"
},
{
"input": "0 10",
"output": "0"
},
{
"input": "107 109",
"output": "2"
},
{
"input": "10 13",
"output": "6"
},
{
"input": "998244355 998244359",
"output": "4"
},
{
"input": "999999999000000000 1000000000000000000",
"output": "0"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "3 15",
"output": "0"
},
{
"input": "24 26",
"output": "0"
},
{
"input": "14 60",
"output": "0"
},
{
"input": "11 79",
"output": "0"
},
{
"input": "1230 1232",
"output": "2"
},
{
"input": "2633 2634",
"output": "4"
},
{
"input": "535 536",
"output": "6"
},
{
"input": "344319135 396746843",
"output": "0"
},
{
"input": "696667767 696667767",
"output": "1"
},
{
"input": "419530302 610096911",
"output": "0"
},
{
"input": "238965115 821731161",
"output": "0"
},
{
"input": "414626436 728903812",
"output": "0"
},
{
"input": "274410639 293308324",
"output": "0"
},
{
"input": "650636673091305697 650636673091305702",
"output": "0"
},
{
"input": "651240548333620923 651240548333620924",
"output": "4"
},
{
"input": "500000000000000000 1000000000000000000",
"output": "0"
},
{
"input": "999999999999999999 1000000000000000000",
"output": "0"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "1"
},
{
"input": "0 4",
"output": "4"
},
{
"input": "50000000062000007 50000000062000011",
"output": "0"
},
{
"input": "0 0",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "0 2",
"output": "2"
},
{
"input": "10000000000012 10000000000015",
"output": "0"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "12 23",
"output": "0"
},
{
"input": "0 11",
"output": "0"
},
{
"input": "11111234567890 11111234567898",
"output": "0"
},
{
"input": "0 3",
"output": "6"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999999999999999997 999999999999999999",
"output": "2"
},
{
"input": "4 5",
"output": "5"
},
{
"input": "0 1",
"output": "1"
},
{
"input": "101 1002",
"output": "0"
},
{
"input": "0 100000000000000001",
"output": "0"
},
{
"input": "99999999999999997 99999999999999999",
"output": "2"
},
{
"input": "14 15",
"output": "5"
},
{
"input": "8 19",
"output": "0"
},
{
"input": "12 22",
"output": "0"
},
{
"input": "999999999999996 999999999999999",
"output": "4"
},
{
"input": "1 3",
"output": "6"
},
{
"input": "124 125",
"output": "5"
},
{
"input": "11 32",
"output": "0"
},
{
"input": "0 5",
"output": "0"
},
{
"input": "0 999999",
"output": "0"
},
{
"input": "151151151515 151151151526",
"output": "0"
},
{
"input": "6 107",
"output": "0"
},
{
"input": "5 16",
"output": "0"
},
{
"input": "7 16",
"output": "0"
},
{
"input": "6 19",
"output": "0"
},
{
"input": "11113111111111 13111111111111",
"output": "0"
},
{
"input": "1 1000",
"output": "0"
},
{
"input": "24 25",
"output": "5"
},
{
"input": "0 100000000000",
"output": "0"
},
{
"input": "1 22",
"output": "0"
},
{
"input": "999999999999999996 999999999999999999",
"output": "4"
}
] | 1,561,893,771
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 15
| 140
| 0
|
a, b = [int(i) for i in input().split()]
a += 1
ans = a % 10
a+=1
while ans != 0 and a <= b:
last = a % 10
ans = (ans * last) % 10
a += 1
print(ans)
|
Title: The Eternal Immortality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Even if the world is full of counterfeits, I still regard it as wonderful.
Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this.
The phoenix has a rather long lifespan, and reincarnates itself once every *a*! years. Here *a*! denotes the factorial of integer *a*, that is, *a*!<==<=1<=×<=2<=×<=...<=×<=*a*. Specifically, 0!<==<=1.
Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of *b*! years, that is, . Note that when *b*<=≥<=*a* this value is always integer.
As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge.
Input Specification:
The first and only line of input contains two space-separated integers *a* and *b* (0<=≤<=*a*<=≤<=*b*<=≤<=1018).
Output Specification:
Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.
Demo Input:
['2 4\n', '0 10\n', '107 109\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/99c47ca8b182f097e38094d12f0c06ce0b081b76.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2;
In the second example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/9642ef11a23e7c5a3f3c2b1255c1b1b3533802a4.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 0;
In the third example, the last digit of <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/844938cef52ee264c183246d2a9df05cca94dc60.png" style="max-width: 100.0%;max-height: 100.0%;"/> is 2.
|
```python
a, b = [int(i) for i in input().split()]
a += 1
ans = a % 10
a+=1
while ans != 0 and a <= b:
last = a % 10
ans = (ans * last) % 10
a += 1
print(ans)
```
| 0
|
|
275
|
A
|
Lights Out
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.
|
The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed.
|
Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".
|
[
"1 0 0\n0 0 0\n0 0 1\n",
"1 0 1\n8 8 8\n2 0 3\n"
] |
[
"001\n010\n100\n",
"010\n011\n100\n"
] |
none
| 500
|
[
{
"input": "1 0 0\n0 0 0\n0 0 1",
"output": "001\n010\n100"
},
{
"input": "1 0 1\n8 8 8\n2 0 3",
"output": "010\n011\n100"
},
{
"input": "13 85 77\n25 50 45\n65 79 9",
"output": "000\n010\n000"
},
{
"input": "96 95 5\n8 84 74\n67 31 61",
"output": "011\n011\n101"
},
{
"input": "24 54 37\n60 63 6\n1 84 26",
"output": "110\n101\n011"
},
{
"input": "23 10 40\n15 6 40\n92 80 77",
"output": "101\n100\n000"
},
{
"input": "62 74 80\n95 74 93\n2 47 95",
"output": "010\n001\n110"
},
{
"input": "80 83 48\n26 0 66\n47 76 37",
"output": "000\n000\n010"
},
{
"input": "32 15 65\n7 54 36\n5 51 3",
"output": "111\n101\n001"
},
{
"input": "22 97 12\n71 8 24\n100 21 64",
"output": "100\n001\n100"
},
{
"input": "46 37 13\n87 0 50\n90 8 55",
"output": "111\n011\n000"
},
{
"input": "57 43 58\n20 82 83\n66 16 52",
"output": "111\n010\n110"
},
{
"input": "45 56 93\n47 51 59\n18 51 63",
"output": "101\n011\n100"
},
{
"input": "47 66 67\n14 1 37\n27 81 69",
"output": "001\n001\n110"
},
{
"input": "26 69 69\n85 18 23\n14 22 74",
"output": "110\n001\n010"
},
{
"input": "10 70 65\n94 27 25\n74 66 30",
"output": "111\n010\n100"
},
{
"input": "97 1 74\n15 99 1\n88 68 86",
"output": "001\n011\n000"
},
{
"input": "36 48 42\n45 41 66\n26 64 1",
"output": "001\n111\n010"
},
{
"input": "52 81 97\n29 77 71\n66 11 2",
"output": "100\n100\n111"
},
{
"input": "18 66 33\n19 49 49\n48 46 26",
"output": "011\n100\n000"
},
{
"input": "68 79 52\n51 39 100\n29 14 26",
"output": "110\n000\n111"
},
{
"input": "91 69 77\n91 26 64\n91 88 57",
"output": "001\n011\n110"
},
{
"input": "16 69 64\n48 21 80\n81 51 51",
"output": "010\n101\n111"
},
{
"input": "96 14 2\n100 18 12\n65 34 89",
"output": "111\n010\n010"
},
{
"input": "93 95 90\n8 59 42\n53 13 19",
"output": "100\n001\n111"
},
{
"input": "71 84 18\n100 19 67\n9 76 15",
"output": "010\n010\n001"
},
{
"input": "38 93 85\n21 88 64\n4 96 25",
"output": "111\n011\n000"
},
{
"input": "75 20 20\n60 5 78\n77 4 69",
"output": "011\n001\n000"
},
{
"input": "65 70 96\n19 6 83\n33 37 82",
"output": "100\n000\n011"
},
{
"input": "11 13 60\n17 13 46\n42 21 39",
"output": "000\n011\n101"
},
{
"input": "0 0 0\n0 0 0\n0 0 0",
"output": "111\n111\n111"
},
{
"input": "0 0 0\n0 1 0\n0 0 0",
"output": "101\n000\n101"
},
{
"input": "0 0 0\n0 0 0\n0 0 1",
"output": "111\n110\n100"
}
] | 1,594,651,866
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 33
| 155
| 20,172,800
|
a=[]
b=[[1,1,1],[1,1,1],[1,1,1]]
for i in range(3):
row=list(map(int,input().split()))
a.append(row)
for i in range(3):
for j in range(3):
if a[i][j]%2!=0 and a[i][j]!=0:
b[i][j]^=1
if j+1<3:
b[i][j+1]^=1
if j - 1 >=0:
b[i][j - 1] ^= 1
if i - 1 >=0:
b[i-1][j] ^= 1
if i + 1 <3:
b[i+1][j] ^= 1
for i in range(3):
for j in range(3):
print(b[i][j],end="")
print(end="\n")
|
Title: Lights Out
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.
Input Specification:
The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed.
Output Specification:
Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".
Demo Input:
['1 0 0\n0 0 0\n0 0 1\n', '1 0 1\n8 8 8\n2 0 3\n']
Demo Output:
['001\n010\n100\n', '010\n011\n100\n']
Note:
none
|
```python
a=[]
b=[[1,1,1],[1,1,1],[1,1,1]]
for i in range(3):
row=list(map(int,input().split()))
a.append(row)
for i in range(3):
for j in range(3):
if a[i][j]%2!=0 and a[i][j]!=0:
b[i][j]^=1
if j+1<3:
b[i][j+1]^=1
if j - 1 >=0:
b[i][j - 1] ^= 1
if i - 1 >=0:
b[i-1][j] ^= 1
if i + 1 <3:
b[i+1][j] ^= 1
for i in range(3):
for j in range(3):
print(b[i][j],end="")
print(end="\n")
```
| 3
|
|
181
|
A
|
Series of Crimes
|
PROGRAMMING
| 800
|
[
"brute force",
"geometry",
"implementation"
] | null | null |
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
|
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
|
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
|
[
"3 2\n.*\n..\n**\n",
"3 3\n*.*\n*..\n...\n"
] |
[
"1 1\n",
"2 3\n"
] |
none
| 500
|
[
{
"input": "3 2\n.*\n..\n**",
"output": "1 1"
},
{
"input": "2 5\n*....\n*...*",
"output": "1 5"
},
{
"input": "7 2\n..\n**\n..\n..\n..\n..\n.*",
"output": "7 1"
},
{
"input": "7 2\n*.\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "2 10\n*......*..\n.......*..",
"output": "2 1"
},
{
"input": "10 3\n*..\n...\n...\n...\n...\n...\n...\n...\n...\n**.",
"output": "1 2"
},
{
"input": "100 2\n*.\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n..\n**",
"output": "1 2"
},
{
"input": "99 3\n**.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\n...\n.*.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "99 3\n**.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n*..",
"output": "99 2"
},
{
"input": "98 3\n...\n.*.\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n...\n.**",
"output": "2 3"
},
{
"input": "2 100\n...................................................................................................*\n*..................................................................................................*",
"output": "1 1"
},
{
"input": "2 2\n*.\n**",
"output": "1 2"
},
{
"input": "2 2\n**\n.*",
"output": "2 1"
},
{
"input": "2 3\n*.*\n..*",
"output": "2 1"
}
] | 1,668,390,215
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 36
| 124
| 1,536,000
|
num = input()
num = list(num)+[' ']
n = ''
list_store = []
# creating values
for o in num:
if o != ' ':
n = str(n) + o
n = int(n)
if o == ' ':
list_store = list_store + [n]
n = ''
rows = list_store[0]
columns = list_store[1]
i = 1
grid = {}
location_list = []
# creating a grid
while i <= rows:
column_structure = input()
grid[i] = column_structure
i = i + 1
# assigning co-ordinates to char's
for k in grid:
z = 1
while z <= columns:
index = z - 1
if grid[k][index] == '*':
location_list = location_list + [[k, z]]
z = z + 1
# Checking the four conditions
if location_list[0][1] - location_list[1][1] == 0 :
unkown_location = [location_list[0][0], location_list[2][1]]
elif location_list[0][1] - location_list[2][1] == 0 and location_list[0][0] - location_list[1][0] != 0:
unkown_location = [location_list[0][0], location_list[1][1]]
elif location_list[0][1] - location_list[2][1] == 0 and location_list[0][0] - location_list[1][0] == 0:
unkown_location = [location_list[2][0], location_list[1][1]]
else:
unkown_location = [location_list[2][0], location_list[0][1]]
for f in unkown_location:
print(f, end=' ')
|
Title: Series of Crimes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland capital is shaken with three bold crimes committed by the Pihsters, a notorious criminal gang.
The Berland capital's map is represented by an *n*<=×<=*m* rectangular table. Each cell of the table on the map represents some districts of the capital.
The capital's main detective Polycarpus took a map and marked there the districts where the first three robberies had been committed as asterisks. Deduction tells Polycarpus that the fourth robbery will be committed in such district, that all four robbed districts will form the vertices of some rectangle, parallel to the sides of the map.
Polycarpus is good at deduction but he's hopeless at math. So he asked you to find the district where the fourth robbery will be committed.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*,<=*m*<=≤<=100) — the number of rows and columns in the table, correspondingly.
Each of the next *n* lines contains *m* characters — the description of the capital's map. Each character can either be a "." (dot), or an "*" (asterisk). A character equals "*" if the corresponding district has been robbed. Otherwise, it equals ".".
It is guaranteed that the map has exactly three characters "*" and we can always find the fourth district that meets the problem requirements.
Output Specification:
Print two integers — the number of the row and the number of the column of the city district that is the fourth one to be robbed. The rows are numbered starting from one from top to bottom and the columns are numbered starting from one from left to right.
Demo Input:
['3 2\n.*\n..\n**\n', '3 3\n*.*\n*..\n...\n']
Demo Output:
['1 1\n', '2 3\n']
Note:
none
|
```python
num = input()
num = list(num)+[' ']
n = ''
list_store = []
# creating values
for o in num:
if o != ' ':
n = str(n) + o
n = int(n)
if o == ' ':
list_store = list_store + [n]
n = ''
rows = list_store[0]
columns = list_store[1]
i = 1
grid = {}
location_list = []
# creating a grid
while i <= rows:
column_structure = input()
grid[i] = column_structure
i = i + 1
# assigning co-ordinates to char's
for k in grid:
z = 1
while z <= columns:
index = z - 1
if grid[k][index] == '*':
location_list = location_list + [[k, z]]
z = z + 1
# Checking the four conditions
if location_list[0][1] - location_list[1][1] == 0 :
unkown_location = [location_list[0][0], location_list[2][1]]
elif location_list[0][1] - location_list[2][1] == 0 and location_list[0][0] - location_list[1][0] != 0:
unkown_location = [location_list[0][0], location_list[1][1]]
elif location_list[0][1] - location_list[2][1] == 0 and location_list[0][0] - location_list[1][0] == 0:
unkown_location = [location_list[2][0], location_list[1][1]]
else:
unkown_location = [location_list[2][0], location_list[0][1]]
for f in unkown_location:
print(f, end=' ')
```
| 3
|
|
433
|
B
|
Kuriyama Mirai's Stones
|
PROGRAMMING
| 1,200
|
[
"dp",
"implementation",
"sortings"
] | null | null |
Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
|
Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
|
[
"6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n",
"4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n"
] |
[
"24\n9\n28\n",
"10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n"
] |
Please note that the answers to the questions may overflow 32-bit integer type.
| 1,500
|
[
{
"input": "6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6",
"output": "24\n9\n28"
},
{
"input": "4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2",
"output": "10\n15\n5\n15\n5\n5\n2\n12\n3\n5"
},
{
"input": "4\n2 2 3 6\n9\n2 2 3\n1 1 3\n2 2 3\n2 2 3\n2 2 2\n1 1 3\n1 1 3\n2 1 4\n1 1 2",
"output": "5\n7\n5\n5\n2\n7\n7\n13\n4"
},
{
"input": "18\n26 46 56 18 78 88 86 93 13 77 21 84 59 61 5 74 72 52\n25\n1 10 10\n1 9 13\n2 13 17\n1 8 14\n2 2 6\n1 12 16\n2 15 17\n2 3 6\n1 3 13\n2 8 9\n2 17 17\n1 17 17\n2 5 10\n2 1 18\n1 4 16\n1 1 13\n1 1 8\n2 7 11\n2 6 12\n1 5 9\n1 4 5\n2 7 15\n1 8 8\n1 8 14\n1 3 7",
"output": "77\n254\n413\n408\n124\n283\n258\n111\n673\n115\n88\n72\n300\n1009\n757\n745\n491\n300\n420\n358\n96\n613\n93\n408\n326"
},
{
"input": "56\n43 100 44 66 65 11 26 75 96 77 5 15 75 96 11 44 11 97 75 53 33 26 32 33 90 26 68 72 5 44 53 26 33 88 68 25 84 21 25 92 1 84 21 66 94 35 76 51 11 95 67 4 61 3 34 18\n27\n1 20 38\n1 11 46\n2 42 53\n1 8 11\n2 11 42\n2 35 39\n2 37 41\n1 48 51\n1 32 51\n1 36 40\n1 31 56\n1 18 38\n2 9 51\n1 7 48\n1 15 52\n1 27 31\n2 5 19\n2 35 50\n1 31 34\n1 2 7\n2 15 33\n2 46 47\n1 26 28\n2 3 29\n1 23 45\n2 29 55\n1 14 29",
"output": "880\n1727\n1026\n253\n1429\n335\n350\n224\n1063\n247\n1236\n1052\n2215\n2128\n1840\n242\n278\n1223\n200\n312\n722\n168\n166\n662\n1151\n2028\n772"
},
{
"input": "18\n38 93 48 14 69 85 26 47 71 11 57 9 38 65 72 78 52 47\n38\n2 10 12\n1 6 18\n2 2 2\n1 3 15\n2 1 16\n2 5 13\n1 9 17\n1 2 15\n2 5 17\n1 15 15\n2 4 11\n2 3 4\n2 2 5\n2 1 17\n2 6 16\n2 8 16\n2 8 14\n1 9 12\n2 8 13\n2 1 14\n2 5 13\n1 2 3\n1 9 14\n2 12 15\n2 3 3\n2 9 13\n2 4 12\n2 11 14\n2 6 16\n1 8 14\n1 12 15\n2 3 4\n1 3 5\n2 4 14\n1 6 6\n2 7 14\n2 7 18\n1 8 12",
"output": "174\n658\n11\n612\n742\n461\n453\n705\n767\n72\n353\n40\n89\n827\n644\n559\n409\n148\n338\n592\n461\n141\n251\n277\n14\n291\n418\n262\n644\n298\n184\n40\n131\n558\n85\n456\n784\n195"
},
{
"input": "1\n2\n10\n1 1 1\n1 1 1\n2 1 1\n1 1 1\n1 1 1\n1 1 1\n1 1 1\n2 1 1\n1 1 1\n1 1 1",
"output": "2\n2\n2\n2\n2\n2\n2\n2\n2\n2"
},
{
"input": "2\n1 5\n8\n2 1 2\n1 1 1\n1 1 2\n1 1 1\n2 2 2\n2 1 2\n1 1 1\n1 2 2",
"output": "6\n1\n6\n1\n5\n6\n1\n5"
},
{
"input": "8\n2 6 4 6 8 4 7 7\n9\n2 6 8\n1 2 3\n2 3 4\n2 1 2\n2 4 5\n1 3 8\n2 4 4\n2 2 4\n1 2 4",
"output": "22\n10\n10\n6\n12\n36\n6\n14\n16"
},
{
"input": "4\n1 1 2 8\n9\n1 2 4\n1 2 4\n1 1 2\n2 3 4\n2 1 1\n1 2 3\n2 3 4\n1 1 4\n1 1 4",
"output": "11\n11\n2\n10\n1\n3\n10\n12\n12"
},
{
"input": "5\n8 8 1 10 7\n5\n1 2 2\n1 1 4\n1 3 3\n1 2 2\n1 2 5",
"output": "8\n27\n1\n8\n26"
},
{
"input": "10\n10 3 3 2 1 2 7 1 1 1\n4\n2 10 10\n2 6 6\n1 5 10\n1 6 8",
"output": "10\n2\n13\n10"
},
{
"input": "6\n6 6 3 8 5 4\n3\n2 2 3\n2 1 5\n1 1 5",
"output": "9\n24\n28"
},
{
"input": "1\n1\n2\n1 1 1\n2 1 1",
"output": "1\n1"
}
] | 1,648,814,515
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 46
| 1,170
| 10,035,200
|
n = int(input())
v_list = list(map(int, input().split()))
m = int(input())
v_list_sorted = sorted(v_list)
prefix_sum = [[0,v_list[0]],[0,v_list_sorted[0]]]
for i in range(1, n):
sum_val = prefix_sum[0][i] + v_list[i]
prefix_sum[0].append(sum_val)
for i in range(1, n):
sum_val = prefix_sum[1][i] + v_list_sorted[i]
prefix_sum[1].append(sum_val)
for i in range(m):
type_question, l, r = map(int, input().split())
print(str(prefix_sum[type_question-1][r] - prefix_sum[type_question-1][l-1]))
|
Title: Kuriyama Mirai's Stones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output Specification:
Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Demo Input:
['6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n', '4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n']
Demo Output:
['24\n9\n28\n', '10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n']
Note:
Please note that the answers to the questions may overflow 32-bit integer type.
|
```python
n = int(input())
v_list = list(map(int, input().split()))
m = int(input())
v_list_sorted = sorted(v_list)
prefix_sum = [[0,v_list[0]],[0,v_list_sorted[0]]]
for i in range(1, n):
sum_val = prefix_sum[0][i] + v_list[i]
prefix_sum[0].append(sum_val)
for i in range(1, n):
sum_val = prefix_sum[1][i] + v_list_sorted[i]
prefix_sum[1].append(sum_val)
for i in range(m):
type_question, l, r = map(int, input().split())
print(str(prefix_sum[type_question-1][r] - prefix_sum[type_question-1][l-1]))
```
| 3
|
|
427
|
A
|
Police Recruits
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
|
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
|
Print a single integer, the number of crimes which will go untreated.
|
[
"3\n-1 -1 1\n",
"8\n1 -1 1 -1 -1 1 1 1\n",
"11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n"
] |
[
"2\n",
"1\n",
"8\n"
] |
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
| 500
|
[
{
"input": "3\n-1 -1 1",
"output": "2"
},
{
"input": "8\n1 -1 1 -1 -1 1 1 1",
"output": "1"
},
{
"input": "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1",
"output": "8"
},
{
"input": "7\n-1 -1 1 1 -1 -1 1",
"output": "2"
},
{
"input": "21\n-1 -1 -1 -1 -1 3 2 -1 6 -1 -1 2 1 -1 2 2 1 6 5 -1 5",
"output": "5"
},
{
"input": "98\n-1 -1 1 -1 -1 -1 -1 1 -1 -1 1 -1 -1 1 -1 1 1 1 -1 1 1 1 1 1 -1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 1 -1 1 -1 1 1 1 1 1 1 1 -1 -1 1 1 -1 -1 -1 -1 -1 -1 -1 -1 -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": "13"
},
{
"input": "3\n-1 5 4",
"output": "1"
},
{
"input": "146\n4 -1 -1 -1 -1 -1 -1 -1 -1 -1 4 -1 3 -1 3 -1 -1 1 4 -1 2 -1 -1 3 -1 -1 -1 4 1 -1 -1 -1 4 -1 -1 -1 -1 -1 -1 3 2 -1 3 2 4 5 2 4 1 5 -1 -1 2 -1 -1 1 -1 5 3 -1 1 2 2 3 5 3 -1 -1 3 -1 -1 3 5 5 -1 -1 5 -1 4 4 1 -1 -1 -1 2 1 -1 -1 -1 2 5 3 -1 -1 -1 3 -1 5 4 -1 1 -1 -1 3 -1 -1 3 1 1 2 -1 -1 -1 1 3 1 -1 2 -1 -1 5 5 -1 -1 3 4 5 1 -1 2 -1 -1 -1 3 -1 5 3 2 -1 2 -1 -1 5 -1 3 -1",
"output": "5"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n-1",
"output": "1"
},
{
"input": "2\n1 -1",
"output": "0"
},
{
"input": "2\n-1 1",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n-1 -1",
"output": "2"
},
{
"input": "4\n10 -1 -1 -1",
"output": "0"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "4\n-1 -1 1 1",
"output": "2"
}
] | 1,683,492,759
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 63
| 62
| 10,547,200
|
n = int(input())
x = input().split()
x = list(map(int, x))
police = 0
pr = 0
for i in range(n):
if x[i] != -1:
police += x[i]
elif x[i] == -1:
pr += 1
if police > 0:
pr -= 1
police -= 1
print(pr)
|
Title: Police Recruits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
Input Specification:
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
Output Specification:
Print a single integer, the number of crimes which will go untreated.
Demo Input:
['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n']
Demo Output:
['2\n', '1\n', '8\n']
Note:
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated.
|
```python
n = int(input())
x = input().split()
x = list(map(int, x))
police = 0
pr = 0
for i in range(n):
if x[i] != -1:
police += x[i]
elif x[i] == -1:
pr += 1
if police > 0:
pr -= 1
police -= 1
print(pr)
```
| 3
|
|
802
|
G
|
Fake News (easy)
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...
|
The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z).
|
Output YES if the string *s* contains heidi as a subsequence and NO otherwise.
|
[
"abcheaibcdi\n",
"hiedi\n"
] |
[
"YES",
"NO"
] |
A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*.
| 0
|
[
{
"input": "abcheaibcdi",
"output": "YES"
},
{
"input": "hiedi",
"output": "NO"
},
{
"input": "ihied",
"output": "NO"
},
{
"input": "diehi",
"output": "NO"
},
{
"input": "deiih",
"output": "NO"
},
{
"input": "iheid",
"output": "NO"
},
{
"input": "eihdi",
"output": "NO"
},
{
"input": "ehdii",
"output": "NO"
},
{
"input": "edhii",
"output": "NO"
},
{
"input": "deiih",
"output": "NO"
},
{
"input": "ehdii",
"output": "NO"
},
{
"input": "eufyajkssayhjhqcwxmctecaeepjwmfoscqprpcxsqfwnlgzsmmuwuoruantipholrauvxydfvftwfzhnckxswussvlidcojiciflpvkcxkkcmmvtfvxrkwcpeelwsuzqgamamdtdgzscmikvojfvqehblmjczkvtdeymgertgkwfwfukafqlfdhtedcctixhyetdypswgagrpyto",
"output": "YES"
},
{
"input": "arfbvxgdvqzuloojjrwoyqqbxamxybaqltfimofulusfebodjkwwrgwcppkwiodtpjaraglyplgerrpqjkpoggjmfxhwtqrijpijrcyxnoodvwpyjfpvqaoazllbrpzananbrvvybboedidtuvqquklkpeflfaltukjhzjgiofombhbmqbihgtapswykfvlgdoapjqntvqsaohmbvnphvyyhvhavslamczuqifxnwknkaenqmlvetrqogqxmlptgrmqvxzdxdmwobjesmgxckpmawtioavwdngyiwkzypfnxcovwzdohshwlavwsthdssiadhiwmhpvgkrbezm",
"output": "YES"
},
{
"input": "zcectngbqnejjjtsfrluummmqabzqbyccshjqbrjthzhlbmzjfxugvjouwhumsgrnopiyakfadjnbsesamhynsbfbfunupwbxvohfmpwlcpxhovwpfpciclatgmiufwdvtsqrsdcymvkldpnhfeisrzhyhhlkwdzthgprvkpyldeysvbmcibqkpudyrraqdlxpjecvwcvuiklcrsbgvqasmxmtxqzmawcjtozioqlfflinnxpeexbzloaeqjvglbdeufultpjqexvjjjkzemtzuzmxvawilcqdrcjzpqyhtwfphuonzwkotthsaxrmwtnlmcdylxqcfffyndqeouztluqwlhnkkvzwcfiscikv",
"output": "YES"
},
{
"input": "plqaykgovxkvsiahdbglktdlhcqwelxxmtlyymrsyubxdskvyjkrowvcbpdofpjqspsrgpakdczletxujzlsegepzleipiyycpinzxgwjsgslnxsotouddgfcybozfpjhhocpybfjbaywsehbcfrayvancbrumdfngqytnhihyxnlvilrqyhnxeckprqafofelospffhtwguzjbbjlzbqrtiielbvzutzgpqxosiaqznndgobcluuqlhmffiowkjdlkokehtjdyjvmxsiyxureflmdomerfekxdvtitvwzmdsdzplkpbtafxqfpudnhfqpoiwvjnylanunmagoweobdvfjgepbsymfutrjarlxclhgavpytiiqwvojrptofuvlohzeguxdsrihsbucelhhuedltnnjgzxwyblbqvnoliiydfinzlogbvucwykryzcyibnniggbkdkdcdgcsbvvnavtyhtkanrblpvomvjs",
"output": "YES"
},
{
"input": "fbldqzggeunkpwcfirxanmntbfrudijltoertsdvcvcmbwodbibsrxendzebvxwydpasaqnisrijctsuatihxxygbeovhxjdptdcppkvfytdpjspvrannxavmkmisqtygntxkdlousdypyfkrpzapysfpdbyprufwzhunlsfugojddkmxzinatiwfxdqmgyrnjnxvrclhxyuwxtshoqdjptmeecvgmrlvuwqtmnfnfeeiwcavwnqmyustawbjodzwsqmnjxhpqmgpysierlwbbdzcwprpsexyvreewcmlbvaiytjlxdqdaqftefdlmtmmjcwvfejshymhnouoshdzqcwzxpzupkbcievodzqkqvyjuuxxwepxjalvkzufnveji",
"output": "YES"
},
{
"input": "htsyljgoelbbuipivuzrhmfpkgderqpoprlxdpasxhpmxvaztccldtmujjzjmcpdvsdghzpretlsyyiljhjznseaacruriufswuvizwwuvdioazophhyytvbiogttnnouauxllbdn",
"output": "YES"
},
{
"input": "ikmxzqdzxqlvgeojsnhqzciujslwjyzzexnregabdqztpplosdakimjxmuqccbnwvzbajoiqgdobccwnrwmixohrbdarhoeeelzbpigiybtesybwefpcfx",
"output": "YES"
},
{
"input": "bpvbpjvbdfiodsmahxpcubjxdykesubnypalhypantshkjffmxjmelblqnjdmtaltneuyudyevkgedkqrdmrfeemgpghwrifcwincfixokfgurhqbcfzeajrgkgpwqwsepudxulywowwxzdxkumsicsvnzfxspmjpaixgejeaoyoibegosqoyoydmphfpbutrrewyjecowjckvpcceoamtfbitdneuwqfvnagswlskmsmkhmxyfsrpqwhxzocyffiumcy",
"output": "YES"
},
{
"input": "vllsexwrazvlfvhvrtqeohvzzresjdiuhomfpgqcxpqdevplecuaepixhlijatxzegciizpvyvxuembiplwklahlqibykfideysjygagjbgqkbhdhkatddcwlxboinfuomnpc",
"output": "YES"
},
{
"input": "pnjdwpxmvfoqkjtbhquqcuredrkwqzzfjmdvpnbqtypzdovemhhclkvigjvtprrpzbrbcbatkucaqteuciuozytsptvsskkeplaxdaqmjkmef",
"output": "NO"
},
{
"input": "jpwfhvlxvsdhtuozvlmnfiotrgapgjxtcsgcjnodcztupysvvvmjpzqkpommadppdrykuqkcpzojcwvlogvkddedwbggkrhuvtsvdiokehlkdlnukcufjvqxnikcdawvexxwffxtriqbdmkahxdtygodzohwtdmmuvmatdkvweqvaehaxiefpevkvqpyxsrhtmgjsdfcwzqobibeduooldrmglbinrepmunizheqzvgqvpdskhxfidxfnbisyizhepwyrcykcmjxnkyfjgrqlkixcvysa",
"output": "YES"
},
{
"input": "aftcrvuumeqbfvaqlltscnuhkpcifrrhnutjinxdhhdbzvizlrapzjdatuaynoplgjketupgaejciosofuhcgcjdcucarfvtsofgubtphijciswsvidnvpztlaarydkeqxzwdhfbmullkimerukusbrdnnujviydldrwhdfllsjtziwfeaiqotbiprespmxjulnyunkdtcghrzvhtcychkwatqqmladxpvmvlkzscthylbzkpgwlzfjqwarqvdeyngekqvrhrftpxnkfcibbowvnqdkulcdydspcubwlgoyinpnzgidbgunparnueddzwtzdiavbprbbg",
"output": "YES"
},
{
"input": "oagjghsidigeh",
"output": "NO"
},
{
"input": "chdhzpfzabupskiusjoefrwmjmqkbmdgboicnszkhdrlegeqjsldurmbshijadlwsycselhlnudndpdhcnhruhhvsgbthpruiqfirxkhpqhzhqdfpyozolbionodypfcqfeqbkcgmqkizgeyyelzeoothexcoaahedgrvoemqcwccbvoeqawqeuusyjxmgjkpfwcdttfmwunzuwvsihliexlzygqcgpbdiawfvqukikhbjerjkyhpcknlndaystrgsinghlmekbvhntcpypmchcwoglsmwwdulqneuabuuuvtyrnjxfcgoothalwkzzfxakneusezgnnepkpipzromqubraiggqndliz",
"output": "YES"
},
{
"input": "lgirxqkrkgjcutpqitmffvbujcljkqardlalyigxorscczuzikoylcxenryhskoavymexysvmhbsvhtycjlmzhijpuvcjshyfeycvvcfyzytzoyvxajpqdjtfiatnvxnyeqtfcagfftafllhhjhplbdsrfpctkqpinpdfrtlzyjllfbeffputywcckupyslkbbzpgcnxgbmhtqeqqehpdaokkjtatrhyiuusjhwgiiiikxpzdueasemosmmccoakafgvxduwiuflovhhfhffgnnjhoperhhjtvocpqytjxkmrknnknqeglffhfuplopmktykxuvcmbwpoeisrlyyhdpxfvzseucofyhziuiikihpqheqdyzwigeaqzhxzvporgisxgvhyicqyejovqloibhbunsvsunpvmdckkbuokitdzleilfwutcvuuytpupizinfjrzhxudsmjcjyfcpfgthujjowdwtgbvi",
"output": "YES"
},
{
"input": "uuehrvufgerqbzyzksmqnewacotuimawhlbycdbsmhshrsbqwybbkwjwsrkwptvlbbwjiivqugzrxxwgidrcrhrwsmwgeoleptfamzefgaeyxouxocrpvomjrazmxrnffdwrrmblgdiabdncvfougtmjgvvazasnygdrigbsrieoonirlivfyodvulouslxosswgpdexuldmkdbpdlgutiotvxjyecbrsvbmqxrlcpcipjjncduyqtohlzybvlemmfdeubihwlwqglkgjvnwrbgydcpwklmjeewqklmqdbajqgrpnynaxfvxjzgibqerxyhnxenrmcdqaaeksbzyrcaepozqpetaurlhjuxxhwppuhgoihxdxbmxeiahyaqkbknktlzkheaarjoqqrsyeducvoygwalgarldcdlqogfvsncejssmx",
"output": "YES"
},
{
"input": "iiopulfjxoitgiusqrhgbkiyzinphjtclodbkkydetylvuimkhdkklmyoacmekdvjpuxcrvqnjhqhhbfenlpzpwijtykqziocilvtpqhxuyrphdlamawjuzgjwiebkqyrzyqgtkcrhntjxqmcgkrqcslogjegfrivzidfedeegwbbsopvxvdoididlpypjogxaodtueebbwuwilacunqpozczcgrpaxxrtogpxgfkudtxchoravrrdtimhenwmnafxaruuojduxxglefejycfcyxllfimkszmbrhcwpnwjqgwvzysmlaaifdxfjjpgjmcksiigpanappjdloiolojmcqbnpnjjzaufdpjaknylmyvolhwypygwepmqwpiglpcnpypnudhdzpdvgzosyjthzcwtytxq",
"output": "YES"
}
] | 1,674,494,024
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 46
| 0
|
s = input()
count = 0
temp = "heidi"
for i in s:
if i == temp[count]:
count += 1
if count == 5:
break
print('YES' if count==5 else "NO")
|
Title: Fake News (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...
Input Specification:
The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z).
Output Specification:
Output YES if the string *s* contains heidi as a subsequence and NO otherwise.
Demo Input:
['abcheaibcdi\n', 'hiedi\n']
Demo Output:
['YES', 'NO']
Note:
A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*.
|
```python
s = input()
count = 0
temp = "heidi"
for i in s:
if i == temp[count]:
count += 1
if count == 5:
break
print('YES' if count==5 else "NO")
```
| 3
|
|
448
|
A
|
Rewards
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Bizon the Champion is called the Champion for a reason.
Bizon the Champion has recently got a present — a new glass cupboard with *n* shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has *a*1 first prize cups, *a*2 second prize cups and *a*3 third prize cups. Besides, he has *b*1 first prize medals, *b*2 second prize medals and *b*3 third prize medals.
Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules:
- any shelf cannot contain both cups and medals at the same time; - no shelf can contain more than five cups; - no shelf can have more than ten medals.
Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled.
|
The first line contains integers *a*1, *a*2 and *a*3 (0<=≤<=*a*1,<=*a*2,<=*a*3<=≤<=100). The second line contains integers *b*1, *b*2 and *b*3 (0<=≤<=*b*1,<=*b*2,<=*b*3<=≤<=100). The third line contains integer *n* (1<=≤<=*n*<=≤<=100).
The numbers in the lines are separated by single spaces.
|
Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes).
|
[
"1 1 1\n1 1 1\n4\n",
"1 1 3\n2 3 4\n2\n",
"1 0 0\n1 0 0\n1\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "1 1 1\n1 1 1\n4",
"output": "YES"
},
{
"input": "1 1 3\n2 3 4\n2",
"output": "YES"
},
{
"input": "1 0 0\n1 0 0\n1",
"output": "NO"
},
{
"input": "0 0 0\n0 0 0\n1",
"output": "YES"
},
{
"input": "100 100 100\n100 100 100\n100",
"output": "YES"
},
{
"input": "100 100 100\n100 100 100\n1",
"output": "NO"
},
{
"input": "1 10 100\n100 10 1\n20",
"output": "NO"
},
{
"input": "1 1 1\n0 0 0\n1",
"output": "YES"
},
{
"input": "0 0 0\n1 1 1\n1",
"output": "YES"
},
{
"input": "5 5 5\n0 0 0\n2",
"output": "NO"
},
{
"input": "0 0 0\n10 10 10\n2",
"output": "NO"
},
{
"input": "21 61 39\n63 58 69\n44",
"output": "YES"
},
{
"input": "18 95 4\n7 1 75\n46",
"output": "YES"
},
{
"input": "64 27 81\n72 35 23\n48",
"output": "YES"
},
{
"input": "6 6 6\n11 11 11\n7",
"output": "NO"
},
{
"input": "1 2 3\n2 4 6\n3",
"output": "NO"
},
{
"input": "1 2 3\n2 4 6\n4",
"output": "YES"
},
{
"input": "99 99 99\n99 99 99\n89",
"output": "NO"
},
{
"input": "5 0 0\n15 0 0\n2",
"output": "NO"
},
{
"input": "10 10 10\n0 0 0\n1",
"output": "NO"
},
{
"input": "1 1 1\n1 1 1\n15",
"output": "YES"
},
{
"input": "2 3 5\n2 3 5\n2",
"output": "NO"
},
{
"input": "2 2 2\n3 3 5\n3",
"output": "NO"
},
{
"input": "1 2 2\n2 4 4\n1",
"output": "NO"
},
{
"input": "1 2 3\n1 5 5\n2",
"output": "NO"
}
] | 1,635,703,214
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 46
| 4,505,600
|
a1,a2,a3 = map(int,input().split())
b1,b2,b3 = map(int,input().split())
pol = int(input())
med = a1+a2+a3
cub = b1+b2+b3
polm = med//5
polc = cub//10
if (med % 5) != 0 :
polm = polm+1
if (cub % 10) != 0 :
polc = polc+1
if (polc + polm) > pol :
print('NO')
else :
print('YES')
# Sun Oct 31 2021 18:00:13 GMT+0000 (Coordinated Universal Time)
|
Title: Rewards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bizon the Champion is called the Champion for a reason.
Bizon the Champion has recently got a present — a new glass cupboard with *n* shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has *a*1 first prize cups, *a*2 second prize cups and *a*3 third prize cups. Besides, he has *b*1 first prize medals, *b*2 second prize medals and *b*3 third prize medals.
Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules:
- any shelf cannot contain both cups and medals at the same time; - no shelf can contain more than five cups; - no shelf can have more than ten medals.
Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled.
Input Specification:
The first line contains integers *a*1, *a*2 and *a*3 (0<=≤<=*a*1,<=*a*2,<=*a*3<=≤<=100). The second line contains integers *b*1, *b*2 and *b*3 (0<=≤<=*b*1,<=*b*2,<=*b*3<=≤<=100). The third line contains integer *n* (1<=≤<=*n*<=≤<=100).
The numbers in the lines are separated by single spaces.
Output Specification:
Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes).
Demo Input:
['1 1 1\n1 1 1\n4\n', '1 1 3\n2 3 4\n2\n', '1 0 0\n1 0 0\n1\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none
|
```python
a1,a2,a3 = map(int,input().split())
b1,b2,b3 = map(int,input().split())
pol = int(input())
med = a1+a2+a3
cub = b1+b2+b3
polm = med//5
polc = cub//10
if (med % 5) != 0 :
polm = polm+1
if (cub % 10) != 0 :
polc = polc+1
if (polc + polm) > pol :
print('NO')
else :
print('YES')
# Sun Oct 31 2021 18:00:13 GMT+0000 (Coordinated Universal Time)
```
| 3
|
|
604
|
A
|
Uncowed Forces
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point values of 500, 1000, 1500, 2000, and 2500, respectively. Despite the challenging tasks, Kevin was uncowed and bulldozed through all of them, distinguishing himself from the herd as the best cowmputer scientist in all of Bovinia. Kevin knows his submission time for each problem, the number of wrong submissions that he made on each problem, and his total numbers of successful and unsuccessful hacks. Because Codeforces scoring is complicated, Kevin wants you to write a program to compute his final score.
Codeforces scores are computed as follows: If the maximum point value of a problem is *x*, and Kevin submitted correctly at minute *m* but made *w* wrong submissions, then his score on that problem is . His total score is equal to the sum of his scores for each problem. In addition, Kevin's total score gets increased by 100 points for each successful hack, but gets decreased by 50 points for each unsuccessful hack.
All arithmetic operations are performed with absolute precision and no rounding. It is guaranteed that Kevin's final score is an integer.
|
The first line of the input contains five space-separated integers *m*1, *m*2, *m*3, *m*4, *m*5, where *m**i* (0<=≤<=*m**i*<=≤<=119) is the time of Kevin's last submission for problem *i*. His last submission is always correct and gets accepted.
The second line contains five space-separated integers *w*1, *w*2, *w*3, *w*4, *w*5, where *w**i* (0<=≤<=*w**i*<=≤<=10) is Kevin's number of wrong submissions on problem *i*.
The last line contains two space-separated integers *h**s* and *h**u* (0<=≤<=*h**s*,<=*h**u*<=≤<=20), denoting the Kevin's numbers of successful and unsuccessful hacks, respectively.
|
Print a single integer, the value of Kevin's final score.
|
[
"20 40 60 80 100\n0 1 2 3 4\n1 0\n",
"119 119 119 119 119\n0 0 0 0 0\n10 0\n"
] |
[
"4900\n",
"4930\n"
] |
In the second sample, Kevin takes 119 minutes on all of the problems. Therefore, he gets <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/42158dc2bc78cd21fa679530ae9ef8b9ea298d15.png" style="max-width: 100.0%;max-height: 100.0%;"/> of the points on each problem. So his score from solving problems is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/fdf392d8508500b57f8057ac0c4c892ab5f925a2.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Adding in 10·100 = 1000 points from hacks, his total score becomes 3930 + 1000 = 4930.
| 500
|
[
{
"input": "20 40 60 80 100\n0 1 2 3 4\n1 0",
"output": "4900"
},
{
"input": "119 119 119 119 119\n0 0 0 0 0\n10 0",
"output": "4930"
},
{
"input": "3 6 13 38 60\n6 10 10 3 8\n9 9",
"output": "5088"
},
{
"input": "21 44 11 68 75\n6 2 4 8 4\n2 8",
"output": "4522"
},
{
"input": "16 112 50 114 68\n1 4 8 4 9\n19 11",
"output": "5178"
},
{
"input": "55 66 75 44 47\n6 0 6 6 10\n19 0",
"output": "6414"
},
{
"input": "47 11 88 5 110\n6 10 4 2 3\n10 6",
"output": "5188"
},
{
"input": "5 44 61 103 92\n9 0 10 4 8\n15 7",
"output": "4914"
},
{
"input": "115 53 96 62 110\n7 8 1 7 9\n7 16",
"output": "3416"
},
{
"input": "102 83 26 6 11\n3 4 1 8 3\n17 14",
"output": "6704"
},
{
"input": "36 102 73 101 19\n5 9 2 2 6\n4 13",
"output": "4292"
},
{
"input": "40 115 93 107 113\n5 7 2 6 8\n6 17",
"output": "2876"
},
{
"input": "53 34 53 107 81\n4 3 1 10 8\n7 7",
"output": "4324"
},
{
"input": "113 37 4 84 66\n2 0 10 3 0\n20 19",
"output": "6070"
},
{
"input": "10 53 101 62 1\n8 0 9 7 9\n0 11",
"output": "4032"
},
{
"input": "45 45 75 36 76\n6 2 2 0 0\n8 17",
"output": "5222"
},
{
"input": "47 16 44 78 111\n7 9 8 0 2\n1 19",
"output": "3288"
},
{
"input": "7 54 39 102 31\n6 0 2 10 1\n18 3",
"output": "6610"
},
{
"input": "0 46 86 72 40\n1 5 5 5 9\n6 5",
"output": "4924"
},
{
"input": "114 4 45 78 113\n0 4 8 10 2\n10 12",
"output": "4432"
},
{
"input": "56 56 96 105 107\n4 9 10 4 8\n2 1",
"output": "3104"
},
{
"input": "113 107 59 50 56\n3 7 10 6 3\n10 12",
"output": "4586"
},
{
"input": "96 104 9 94 84\n6 10 7 8 3\n14 11",
"output": "4754"
},
{
"input": "98 15 116 43 55\n4 3 0 9 3\n10 7",
"output": "5400"
},
{
"input": "0 26 99 108 35\n0 4 3 0 10\n9 5",
"output": "5388"
},
{
"input": "89 24 51 49 84\n5 6 2 2 9\n2 14",
"output": "4066"
},
{
"input": "57 51 76 45 96\n1 0 4 3 6\n12 15",
"output": "5156"
},
{
"input": "79 112 37 36 116\n2 8 4 7 5\n4 12",
"output": "3872"
},
{
"input": "71 42 60 20 7\n7 1 1 10 6\n1 7",
"output": "5242"
},
{
"input": "86 10 66 80 55\n0 2 5 10 5\n15 6",
"output": "5802"
},
{
"input": "66 109 22 22 62\n3 1 5 4 5\n10 5",
"output": "5854"
},
{
"input": "97 17 43 84 58\n2 8 3 8 6\n10 7",
"output": "5028"
},
{
"input": "109 83 5 114 104\n6 0 3 9 5\n5 2",
"output": "4386"
},
{
"input": "94 18 24 91 105\n2 0 7 10 3\n1 4",
"output": "4118"
},
{
"input": "64 17 86 59 45\n8 0 10 2 2\n4 4",
"output": "5144"
},
{
"input": "70 84 31 57 2\n7 0 0 2 7\n12 5",
"output": "6652"
},
{
"input": "98 118 117 86 4\n2 10 9 7 5\n11 15",
"output": "4476"
},
{
"input": "103 110 101 97 70\n4 2 1 0 5\n7 5",
"output": "4678"
},
{
"input": "78 96 6 97 62\n7 7 9 2 9\n10 3",
"output": "4868"
},
{
"input": "95 28 3 31 115\n1 9 0 7 3\n10 13",
"output": "5132"
},
{
"input": "45 17 116 58 3\n8 8 7 6 4\n3 19",
"output": "3992"
},
{
"input": "19 12 0 113 77\n3 0 10 9 2\n8 6",
"output": "5040"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0",
"output": "7500"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n20 0",
"output": "9500"
},
{
"input": "119 119 119 119 119\n10 10 10 10 10\n0 20",
"output": "1310"
},
{
"input": "0 0 0 0 0\n10 10 10 10 10\n0 20",
"output": "4150"
},
{
"input": "119 0 0 0 0\n10 0 0 0 0\n5 5",
"output": "7400"
},
{
"input": "0 119 0 0 0\n0 10 0 0 0\n5 5",
"output": "7050"
},
{
"input": "0 0 119 0 0\n0 0 10 0 0\n0 0",
"output": "6450"
},
{
"input": "0 0 0 119 0\n0 0 0 10 0\n5 5",
"output": "6350"
},
{
"input": "0 0 0 0 119\n0 0 0 0 10\n5 5",
"output": "6060"
},
{
"input": "119 0 0 0 0\n2 0 0 0 0\n5 5",
"output": "7412"
},
{
"input": "0 119 0 0 0\n0 2 0 0 0\n5 5",
"output": "7174"
},
{
"input": "0 0 119 0 0\n0 0 2 0 0\n5 5",
"output": "6936"
},
{
"input": "0 0 0 119 0\n0 0 0 2 0\n5 5",
"output": "6698"
},
{
"input": "0 0 0 0 119\n0 0 0 0 2\n5 5",
"output": "6460"
},
{
"input": "119 0 0 0 0\n0 0 0 0 0\n4 9",
"output": "7212"
}
] | 1,448,984,609
| 509
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 0
| 62
| 1,126,400
|
from fractions import Fraction
m = list(map(Fraction, input().split()))
w = list(map(Fraction, input().split()))
h1,h2 = map(Fraction, input().split())
ball = [500,1000,1500,2000,2500]
score = Fraction(0)
for i in range(5):
score += max(Fraction(3,10)*ball[i],(1-Fraction(m[i],250)*ball[i])-50*w[i])
score += h1*100-50*h2
print(int(score))
|
Title: Uncowed Forces
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kevin Sun has just finished competing in Codeforces Round #334! The round was 120 minutes long and featured five problems with maximum point values of 500, 1000, 1500, 2000, and 2500, respectively. Despite the challenging tasks, Kevin was uncowed and bulldozed through all of them, distinguishing himself from the herd as the best cowmputer scientist in all of Bovinia. Kevin knows his submission time for each problem, the number of wrong submissions that he made on each problem, and his total numbers of successful and unsuccessful hacks. Because Codeforces scoring is complicated, Kevin wants you to write a program to compute his final score.
Codeforces scores are computed as follows: If the maximum point value of a problem is *x*, and Kevin submitted correctly at minute *m* but made *w* wrong submissions, then his score on that problem is . His total score is equal to the sum of his scores for each problem. In addition, Kevin's total score gets increased by 100 points for each successful hack, but gets decreased by 50 points for each unsuccessful hack.
All arithmetic operations are performed with absolute precision and no rounding. It is guaranteed that Kevin's final score is an integer.
Input Specification:
The first line of the input contains five space-separated integers *m*1, *m*2, *m*3, *m*4, *m*5, where *m**i* (0<=≤<=*m**i*<=≤<=119) is the time of Kevin's last submission for problem *i*. His last submission is always correct and gets accepted.
The second line contains five space-separated integers *w*1, *w*2, *w*3, *w*4, *w*5, where *w**i* (0<=≤<=*w**i*<=≤<=10) is Kevin's number of wrong submissions on problem *i*.
The last line contains two space-separated integers *h**s* and *h**u* (0<=≤<=*h**s*,<=*h**u*<=≤<=20), denoting the Kevin's numbers of successful and unsuccessful hacks, respectively.
Output Specification:
Print a single integer, the value of Kevin's final score.
Demo Input:
['20 40 60 80 100\n0 1 2 3 4\n1 0\n', '119 119 119 119 119\n0 0 0 0 0\n10 0\n']
Demo Output:
['4900\n', '4930\n']
Note:
In the second sample, Kevin takes 119 minutes on all of the problems. Therefore, he gets <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/42158dc2bc78cd21fa679530ae9ef8b9ea298d15.png" style="max-width: 100.0%;max-height: 100.0%;"/> of the points on each problem. So his score from solving problems is <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/fdf392d8508500b57f8057ac0c4c892ab5f925a2.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Adding in 10·100 = 1000 points from hacks, his total score becomes 3930 + 1000 = 4930.
|
```python
from fractions import Fraction
m = list(map(Fraction, input().split()))
w = list(map(Fraction, input().split()))
h1,h2 = map(Fraction, input().split())
ball = [500,1000,1500,2000,2500]
score = Fraction(0)
for i in range(5):
score += max(Fraction(3,10)*ball[i],(1-Fraction(m[i],250)*ball[i])-50*w[i])
score += h1*100-50*h2
print(int(score))
```
| 0
|
|
508
|
B
|
Anton and currency you all know
|
PROGRAMMING
| 1,300
|
[
"greedy",
"math",
"strings"
] | null | null |
Berland, 2016. The exchange rate of currency you all know against the burle has increased so much that to simplify the calculations, its fractional part was neglected and the exchange rate is now assumed to be an integer.
Reliable sources have informed the financier Anton of some information about the exchange rate of currency you all know against the burle for tomorrow. Now Anton knows that tomorrow the exchange rate will be an even number, which can be obtained from the present rate by swapping exactly two distinct digits in it. Of all the possible values that meet these conditions, the exchange rate for tomorrow will be the maximum possible. It is guaranteed that today the exchange rate is an odd positive integer *n*. Help Anton to determine the exchange rate of currency you all know for tomorrow!
|
The first line contains an odd positive integer *n* — the exchange rate of currency you all know for today. The length of number *n*'s representation is within range from 2 to 105, inclusive. The representation of *n* doesn't contain any leading zeroes.
|
If the information about tomorrow's exchange rate is inconsistent, that is, there is no integer that meets the condition, print <=-<=1.
Otherwise, print the exchange rate of currency you all know against the burle for tomorrow. This should be the maximum possible number of those that are even and that are obtained from today's exchange rate by swapping exactly two digits. Exchange rate representation should not contain leading zeroes.
|
[
"527\n",
"4573\n",
"1357997531\n"
] |
[
"572\n",
"3574\n",
"-1\n"
] |
none
| 1,000
|
[
{
"input": "527",
"output": "572"
},
{
"input": "4573",
"output": "3574"
},
{
"input": "1357997531",
"output": "-1"
},
{
"input": "444443",
"output": "444434"
},
{
"input": "22227",
"output": "72222"
},
{
"input": "24683",
"output": "34682"
},
{
"input": "11",
"output": "-1"
},
{
"input": "1435678543",
"output": "1435678534"
},
{
"input": "4250769",
"output": "9250764"
},
{
"input": "4052769",
"output": "9052764"
},
{
"input": "5685341",
"output": "5685314"
},
{
"input": "1111111111111111231",
"output": "1111111111111111132"
},
{
"input": "333333332379",
"output": "333333339372"
},
{
"input": "85",
"output": "58"
},
{
"input": "7700016673",
"output": "7730016670"
},
{
"input": "35451519805848712272404365322858764249299938505103",
"output": "35451519835848712272404365322858764249299938505100"
},
{
"input": "4314752277691991627730686134692292422155753465948025897701703862445837045929984759093775762579123919",
"output": "9314752277691991627730686134692292422155753465948025897701703862445837045929984759093775762579123914"
},
{
"input": "21",
"output": "12"
},
{
"input": "101",
"output": "110"
},
{
"input": "503147",
"output": "573140"
},
{
"input": "333333333333333333333",
"output": "-1"
},
{
"input": "55555555555555555555555",
"output": "-1"
},
{
"input": "99",
"output": "-1"
},
{
"input": "23759",
"output": "93752"
},
{
"input": "235749",
"output": "935742"
},
{
"input": "435729",
"output": "935724"
},
{
"input": "8623",
"output": "8632"
},
{
"input": "109",
"output": "190"
},
{
"input": "20000000000000000000001",
"output": "21000000000000000000000"
},
{
"input": "1001",
"output": "1100"
}
] | 1,663,702,011
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 58
| 93
| 1,433,600
|
from collections import defaultdict, deque
from functools import lru_cache
from heapq import heappush, heappop
from typing import Counter
from bisect import bisect_right, bisect_left
import math
hpop = heappop
hpush = heappush
def solution():
num = list(input())
# to days is odd
# so change the last one with even number
# can't be equal
# if am less than than definitly pass
# if no I better be the last one
# for that start from the back
# if no all make it the negative one
for i in range(len(num)-1):
if int(num[i]) % 2 == 0 and num[i] < num[-1]:
num[i],num[-1] = num[-1],num[i]
return print("".join(num))
for i in range(len(num)-1)[::-1]:
if int(num[i]) % 2 == 0:
num[i],num[-1] = num[-1],num[i]
return print("".join(num))
print(-1)
def main():
t = 1
#t = int(input())
for _ in range(t):
solution()
#import sys
#import threading
#sys.setrecursionlimit(1 << 30)
#threading.stack_size(1 << 27)
#thread = threading.Thread(target=main)
#thread.start(); thread.join()
main()
"""
num = int(input())
arr = list(map(int, input().split()))
a,b = map(int, input().split())
graph = defaultdict(list)
for i in range(#)
graph[a].append(b)
graph[b].append(a)
MOD = 10**9 + 7
for di,dj in [(0,1),(1,0),(0,-1),(-1,0)]:
ni = i + di
nj = j + dj
if not (0<= ni < len(grid) and 0<= nj < len(grid[0])):
continue
def gcd(a,b):
if a < b: a,b = b,a
if b == 0: return a
return gcd(a%b,b)
"""
|
Title: Anton and currency you all know
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Berland, 2016. The exchange rate of currency you all know against the burle has increased so much that to simplify the calculations, its fractional part was neglected and the exchange rate is now assumed to be an integer.
Reliable sources have informed the financier Anton of some information about the exchange rate of currency you all know against the burle for tomorrow. Now Anton knows that tomorrow the exchange rate will be an even number, which can be obtained from the present rate by swapping exactly two distinct digits in it. Of all the possible values that meet these conditions, the exchange rate for tomorrow will be the maximum possible. It is guaranteed that today the exchange rate is an odd positive integer *n*. Help Anton to determine the exchange rate of currency you all know for tomorrow!
Input Specification:
The first line contains an odd positive integer *n* — the exchange rate of currency you all know for today. The length of number *n*'s representation is within range from 2 to 105, inclusive. The representation of *n* doesn't contain any leading zeroes.
Output Specification:
If the information about tomorrow's exchange rate is inconsistent, that is, there is no integer that meets the condition, print <=-<=1.
Otherwise, print the exchange rate of currency you all know against the burle for tomorrow. This should be the maximum possible number of those that are even and that are obtained from today's exchange rate by swapping exactly two digits. Exchange rate representation should not contain leading zeroes.
Demo Input:
['527\n', '4573\n', '1357997531\n']
Demo Output:
['572\n', '3574\n', '-1\n']
Note:
none
|
```python
from collections import defaultdict, deque
from functools import lru_cache
from heapq import heappush, heappop
from typing import Counter
from bisect import bisect_right, bisect_left
import math
hpop = heappop
hpush = heappush
def solution():
num = list(input())
# to days is odd
# so change the last one with even number
# can't be equal
# if am less than than definitly pass
# if no I better be the last one
# for that start from the back
# if no all make it the negative one
for i in range(len(num)-1):
if int(num[i]) % 2 == 0 and num[i] < num[-1]:
num[i],num[-1] = num[-1],num[i]
return print("".join(num))
for i in range(len(num)-1)[::-1]:
if int(num[i]) % 2 == 0:
num[i],num[-1] = num[-1],num[i]
return print("".join(num))
print(-1)
def main():
t = 1
#t = int(input())
for _ in range(t):
solution()
#import sys
#import threading
#sys.setrecursionlimit(1 << 30)
#threading.stack_size(1 << 27)
#thread = threading.Thread(target=main)
#thread.start(); thread.join()
main()
"""
num = int(input())
arr = list(map(int, input().split()))
a,b = map(int, input().split())
graph = defaultdict(list)
for i in range(#)
graph[a].append(b)
graph[b].append(a)
MOD = 10**9 + 7
for di,dj in [(0,1),(1,0),(0,-1),(-1,0)]:
ni = i + di
nj = j + dj
if not (0<= ni < len(grid) and 0<= nj < len(grid[0])):
continue
def gcd(a,b):
if a < b: a,b = b,a
if b == 0: return a
return gcd(a%b,b)
"""
```
| 3
|
|
47
|
B
|
Coins
|
PROGRAMMING
| 1,200
|
[
"implementation"
] |
B. Coins
|
2
|
256
|
One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal.
|
The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B.
|
It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights.
|
[
"A>B\nC<B\nA>C\n",
"A<B\nB>C\nC>A\n"
] |
[
"CBA",
"ACB"
] |
none
| 1,000
|
[
{
"input": "A>B\nC<B\nA>C",
"output": "CBA"
},
{
"input": "A<B\nB>C\nC>A",
"output": "ACB"
},
{
"input": "A<C\nB<A\nB>C",
"output": "Impossible"
},
{
"input": "A<B\nA<C\nB>C",
"output": "ACB"
},
{
"input": "B>A\nC<B\nC>A",
"output": "ACB"
},
{
"input": "A>B\nB>C\nC<A",
"output": "CBA"
},
{
"input": "A>C\nA>B\nB<C",
"output": "BCA"
},
{
"input": "C<B\nB>A\nA<C",
"output": "ACB"
},
{
"input": "C<B\nA>B\nC<A",
"output": "CBA"
},
{
"input": "C>B\nB>A\nA<C",
"output": "ABC"
},
{
"input": "C<B\nB<A\nC>A",
"output": "Impossible"
},
{
"input": "B<C\nC<A\nA>B",
"output": "BCA"
},
{
"input": "A>B\nC<B\nC<A",
"output": "CBA"
},
{
"input": "B>A\nC>B\nA>C",
"output": "Impossible"
},
{
"input": "B<A\nC>B\nC>A",
"output": "BAC"
},
{
"input": "A<B\nC>B\nA<C",
"output": "ABC"
},
{
"input": "A<B\nC<A\nB<C",
"output": "Impossible"
},
{
"input": "A>C\nC<B\nB>A",
"output": "CAB"
},
{
"input": "C>A\nA<B\nB>C",
"output": "ACB"
},
{
"input": "C>A\nC<B\nB>A",
"output": "ACB"
},
{
"input": "B>C\nB>A\nA<C",
"output": "ACB"
},
{
"input": "C<B\nC<A\nB<A",
"output": "CBA"
},
{
"input": "A<C\nA<B\nB>C",
"output": "ACB"
},
{
"input": "B>A\nA>C\nB>C",
"output": "CAB"
},
{
"input": "B<A\nA<C\nC<B",
"output": "Impossible"
},
{
"input": "A<C\nB>C\nA>B",
"output": "Impossible"
},
{
"input": "B>A\nC<A\nC>B",
"output": "Impossible"
},
{
"input": "A>C\nC>B\nB<A",
"output": "BCA"
},
{
"input": "B<C\nB<A\nA>C",
"output": "BCA"
},
{
"input": "A>B\nC>B\nA<C",
"output": "BAC"
},
{
"input": "C<B\nC<A\nB<A",
"output": "CBA"
},
{
"input": "A<C\nA>B\nB>C",
"output": "Impossible"
},
{
"input": "B>A\nB>C\nA<C",
"output": "ACB"
},
{
"input": "B>C\nC<A\nB<A",
"output": "CBA"
},
{
"input": "C>A\nB>A\nB>C",
"output": "ACB"
},
{
"input": "B<A\nB>C\nA<C",
"output": "Impossible"
},
{
"input": "B<C\nA<B\nC>A",
"output": "ABC"
},
{
"input": "C<B\nB>A\nA>C",
"output": "CAB"
},
{
"input": "A>B\nC>A\nB<C",
"output": "BAC"
},
{
"input": "A>B\nB>C\nC<A",
"output": "CBA"
},
{
"input": "B>C\nC>A\nA>B",
"output": "Impossible"
},
{
"input": "B<A\nB>C\nC<A",
"output": "CBA"
},
{
"input": "C>B\nB>A\nC>A",
"output": "ABC"
},
{
"input": "A>C\nA>B\nB>C",
"output": "CBA"
},
{
"input": "B<C\nA>B\nA<C",
"output": "BAC"
},
{
"input": "C>A\nC<B\nB>A",
"output": "ACB"
},
{
"input": "C>B\nA<B\nC<A",
"output": "Impossible"
},
{
"input": "A<C\nC<B\nA>B",
"output": "Impossible"
},
{
"input": "B>C\nA<B\nA<C",
"output": "ACB"
},
{
"input": "C>A\nA<B\nC>B",
"output": "ABC"
}
] | 1,559,835,611
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 50
| 218
| 0
|
weight=[]
for i in range(3):
weight.append(input())
d={'A':0,'B':0,'C' :0}
for comp in weight:
if comp[1] =='>':
d[comp[0]] += 1
else:
d[comp[2]] += 1
if len(set(d.values())) == 3:
print("".join(sorted(d.keys(),key = lambda x : d[x])))
else:
print("Impossible")
|
Title: Coins
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal.
Input Specification:
The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B.
Output Specification:
It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights.
Demo Input:
['A>B\nC<B\nA>C\n', 'A<B\nB>C\nC>A\n']
Demo Output:
['CBA', 'ACB']
Note:
none
|
```python
weight=[]
for i in range(3):
weight.append(input())
d={'A':0,'B':0,'C' :0}
for comp in weight:
if comp[1] =='>':
d[comp[0]] += 1
else:
d[comp[2]] += 1
if len(set(d.values())) == 3:
print("".join(sorted(d.keys(),key = lambda x : d[x])))
else:
print("Impossible")
```
| 3.9455
|
31
|
A
|
Worms Evolution
|
PROGRAMMING
| 1,200
|
[
"implementation"
] |
A. Worms Evolution
|
2
|
256
|
Professor Vasechkin is studying evolution of worms. Recently he put forward hypotheses that all worms evolve by division. There are *n* forms of worms. Worms of these forms have lengths *a*1, *a*2, ..., *a**n*. To prove his theory, professor needs to find 3 different forms that the length of the first form is equal to sum of lengths of the other two forms. Help him to do this.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of worm's forms. The second line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000) — lengths of worms of each form.
|
Output 3 distinct integers *i* *j* *k* (1<=≤<=*i*,<=*j*,<=*k*<=≤<=*n*) — such indexes of worm's forms that *a**i*<==<=*a**j*<=+<=*a**k*. If there is no such triple, output -1. If there are several solutions, output any of them. It possible that *a**j*<==<=*a**k*.
|
[
"5\n1 2 3 5 7\n",
"5\n1 8 1 5 1\n"
] |
[
"3 2 1\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "5\n1 2 3 5 7",
"output": "3 2 1"
},
{
"input": "5\n1 8 1 5 1",
"output": "-1"
},
{
"input": "4\n303 872 764 401",
"output": "-1"
},
{
"input": "6\n86 402 133 524 405 610",
"output": "6 4 1"
},
{
"input": "8\n217 779 418 895 996 473 3 22",
"output": "5 2 1"
},
{
"input": "10\n858 972 670 15 662 114 33 273 53 310",
"output": "2 6 1"
},
{
"input": "100\n611 697 572 770 603 870 128 245 49 904 468 982 788 943 549 288 668 796 803 515 999 735 912 49 298 80 412 841 494 434 543 298 17 571 271 105 70 313 178 755 194 279 585 766 412 164 907 841 776 556 731 268 735 880 176 267 287 65 239 588 155 658 821 47 783 595 585 69 226 906 429 161 999 148 7 484 362 585 952 365 92 749 904 525 307 626 883 367 450 755 564 950 728 724 69 106 119 157 96 290",
"output": "1 38 25"
},
{
"input": "100\n713 572 318 890 577 657 646 146 373 783 392 229 455 871 20 593 573 336 26 381 280 916 907 732 820 713 111 840 570 446 184 711 481 399 788 647 492 15 40 530 549 506 719 782 126 20 778 996 712 761 9 74 812 418 488 175 103 585 900 3 604 521 109 513 145 708 990 361 682 827 791 22 596 780 596 385 450 643 158 496 876 975 319 783 654 895 891 361 397 81 682 899 347 623 809 557 435 279 513 438",
"output": "1 63 61"
},
{
"input": "100\n156 822 179 298 981 82 610 345 373 378 895 734 768 15 78 335 764 608 932 297 717 553 916 367 425 447 361 195 66 70 901 236 905 744 919 564 296 610 963 628 840 52 100 750 345 308 37 687 192 704 101 815 10 990 216 358 823 546 578 821 706 148 182 582 421 482 829 425 121 337 500 301 402 868 66 935 625 527 746 585 308 523 488 914 608 709 875 252 151 781 447 2 756 176 976 302 450 35 680 791",
"output": "1 98 69"
},
{
"input": "100\n54 947 785 838 359 647 92 445 48 465 323 486 101 86 607 31 860 420 709 432 435 372 272 37 903 814 309 197 638 58 259 822 793 564 309 22 522 907 101 853 486 824 614 734 630 452 166 532 256 499 470 9 933 452 256 450 7 26 916 406 257 285 895 117 59 369 424 133 16 417 352 440 806 236 478 34 889 469 540 806 172 296 73 655 261 792 868 380 204 454 330 53 136 629 236 850 134 560 264 291",
"output": "2 29 27"
},
{
"input": "99\n175 269 828 129 499 890 127 263 995 807 508 289 996 226 437 320 365 642 757 22 190 8 345 499 834 713 962 889 336 171 608 492 320 257 472 801 176 325 301 306 198 729 933 4 640 322 226 317 567 586 249 237 202 633 287 128 911 654 719 988 420 855 361 574 716 899 317 356 581 440 284 982 541 111 439 29 37 560 961 224 478 906 319 416 736 603 808 87 762 697 392 713 19 459 262 238 239 599 997",
"output": "1 44 30"
},
{
"input": "98\n443 719 559 672 16 69 529 632 953 999 725 431 54 22 346 968 558 696 48 669 963 129 257 712 39 870 498 595 45 821 344 925 179 388 792 346 755 213 423 365 344 659 824 356 773 637 628 897 841 155 243 536 951 361 192 105 418 431 635 596 150 162 145 548 473 531 750 306 377 354 450 975 79 743 656 733 440 940 19 139 237 346 276 227 64 799 479 633 199 17 796 362 517 234 729 62 995 535",
"output": "2 70 40"
},
{
"input": "97\n359 522 938 862 181 600 283 1000 910 191 590 220 761 818 903 264 751 751 987 316 737 898 168 925 244 674 34 950 754 472 81 6 37 520 112 891 981 454 897 424 489 238 363 709 906 951 677 828 114 373 589 835 52 89 97 435 277 560 551 204 879 469 928 523 231 163 183 609 821 915 615 969 616 23 874 437 844 321 78 53 643 786 585 38 744 347 150 179 988 985 200 11 15 9 547 886 752",
"output": "1 23 10"
},
{
"input": "4\n303 872 764 401",
"output": "-1"
},
{
"input": "100\n328 397 235 453 188 254 879 225 423 36 384 296 486 592 231 849 856 255 213 898 234 800 701 529 951 693 507 326 15 905 618 348 967 927 28 979 752 850 343 35 84 302 36 390 482 826 249 918 91 289 973 457 557 348 365 239 709 565 320 560 153 130 647 708 483 469 788 473 322 844 830 562 611 961 397 673 69 960 74 703 369 968 382 451 328 160 211 230 566 208 7 545 293 73 806 375 157 410 303 58",
"output": "1 79 6"
},
{
"input": "33\n52 145 137 734 180 847 178 286 716 134 181 630 358 764 593 762 785 28 1 468 189 540 764 485 165 656 114 58 628 108 605 584 257",
"output": "8 30 7"
},
{
"input": "57\n75 291 309 68 444 654 985 158 514 204 116 918 374 806 176 31 49 455 269 66 722 713 164 818 317 295 546 564 134 641 28 13 987 478 146 219 213 940 289 173 157 666 168 391 392 71 870 477 446 988 414 568 964 684 409 671 454",
"output": "2 41 29"
},
{
"input": "88\n327 644 942 738 84 118 981 686 530 404 137 197 434 16 693 183 423 325 410 345 941 329 7 106 79 867 584 358 533 675 192 718 641 329 900 768 404 301 101 538 954 590 401 954 447 14 559 337 756 586 934 367 538 928 945 936 770 641 488 579 206 869 902 139 216 446 723 150 829 205 373 578 357 368 960 40 121 206 503 385 521 161 501 694 138 370 709 308",
"output": "1 77 61"
},
{
"input": "100\n804 510 266 304 788 625 862 888 408 82 414 470 777 991 729 229 933 406 601 1 596 720 608 706 432 361 527 548 59 548 474 515 4 991 263 568 681 24 117 563 576 587 281 643 904 521 891 106 842 884 943 54 605 815 504 757 311 374 335 192 447 652 633 410 455 402 382 150 432 836 413 819 669 875 638 925 217 805 632 520 605 266 728 795 162 222 603 159 284 790 914 443 775 97 789 606 859 13 851 47",
"output": "1 77 42"
},
{
"input": "100\n449 649 615 713 64 385 927 466 138 126 143 886 80 199 208 43 196 694 92 89 264 180 617 970 191 196 910 150 275 89 693 190 191 99 542 342 45 592 114 56 451 170 64 589 176 102 308 92 402 153 414 675 352 157 69 150 91 288 163 121 816 184 20 234 836 12 593 150 793 439 540 93 99 663 186 125 349 247 476 106 77 523 215 7 363 278 441 745 337 25 148 384 15 915 108 211 240 58 23 408",
"output": "1 6 5"
},
{
"input": "90\n881 436 52 308 97 261 153 931 670 538 702 156 114 445 154 685 452 76 966 790 93 42 547 65 736 364 136 489 719 322 239 628 696 735 55 703 622 375 100 188 804 341 546 474 484 446 729 290 974 301 602 225 996 244 488 983 882 460 962 754 395 617 61 640 534 292 158 375 632 902 420 979 379 38 100 67 963 928 190 456 545 571 45 716 153 68 844 2 102 116",
"output": "1 14 2"
},
{
"input": "80\n313 674 262 240 697 146 391 221 793 504 896 818 92 899 86 370 341 339 306 887 937 570 830 683 729 519 240 833 656 847 427 958 435 704 853 230 758 347 660 575 843 293 649 396 437 787 654 599 35 103 779 783 447 379 444 585 902 713 791 150 851 228 306 721 996 471 617 403 102 168 197 741 877 481 968 545 331 715 236 654",
"output": "1 13 8"
},
{
"input": "70\n745 264 471 171 946 32 277 511 269 469 89 831 69 2 369 407 583 602 646 633 429 747 113 302 722 321 344 824 241 372 263 287 822 24 652 758 246 967 219 313 882 597 752 965 389 775 227 556 95 904 308 340 899 514 400 187 275 318 621 546 659 488 199 154 811 1 725 79 925 82",
"output": "1 63 60"
},
{
"input": "60\n176 502 680 102 546 917 516 801 392 435 635 492 398 456 653 444 472 513 634 378 273 276 44 920 68 124 800 167 825 250 452 264 561 344 98 933 381 939 426 51 568 548 206 887 342 763 151 514 156 354 486 546 998 649 356 438 295 570 450 589",
"output": "2 26 20"
},
{
"input": "50\n608 92 889 33 146 803 402 91 868 400 828 505 375 558 584 129 361 776 974 123 765 804 326 186 61 927 904 511 762 775 640 593 300 664 897 461 869 911 986 789 607 500 309 457 294 104 724 471 216 155",
"output": "3 25 11"
},
{
"input": "40\n40 330 98 612 747 336 640 381 991 366 22 167 352 12 868 166 603 40 313 869 609 981 609 804 54 729 8 854 347 300 828 922 39 633 695 988 4 530 545 176",
"output": "5 10 8"
},
{
"input": "30\n471 920 308 544 347 222 878 671 467 332 215 180 681 114 151 203 492 951 653 614 453 510 540 422 399 532 113 198 932 825",
"output": "2 21 9"
},
{
"input": "20\n551 158 517 475 595 108 764 961 590 297 761 841 659 568 82 888 733 214 993 359",
"output": "3 20 2"
},
{
"input": "10\n983 748 726 406 196 993 2 251 66 263",
"output": "-1"
},
{
"input": "9\n933 266 457 863 768 257 594 136 145",
"output": "-1"
},
{
"input": "8\n537 198 48 771 944 868 700 163",
"output": "7 8 1"
},
{
"input": "7\n140 779 639 679 768 479 158",
"output": "2 3 1"
},
{
"input": "6\n744 359 230 586 944 442",
"output": "-1"
},
{
"input": "5\n700 939 173 494 120",
"output": "-1"
},
{
"input": "4\n303 872 764 401",
"output": "-1"
},
{
"input": "3\n907 452 355",
"output": "-1"
},
{
"input": "3\n963 630 333",
"output": "1 3 2"
},
{
"input": "3\n2 2 4",
"output": "3 2 1"
},
{
"input": "3\n2 4 100",
"output": "-1"
}
] | 1,549,553,270
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 216
| 0
|
n = int(input())
container = list(map(int, input().split()))
ans = "-1"
needBreak = False
for i in range(len(container)):
for j in range(len(container)):
if j == i:
continue
for k in range(len(container)):
if k == j or k == i:
continue
if container[j] + container[k] == container[i]:
ans = str(i + 1) + str(j + 1) + str(k + 1)
needBreak = True
break
if needBreak:
break
if needBreak:
break
print(' '.join(ans))
|
Title: Worms Evolution
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Professor Vasechkin is studying evolution of worms. Recently he put forward hypotheses that all worms evolve by division. There are *n* forms of worms. Worms of these forms have lengths *a*1, *a*2, ..., *a**n*. To prove his theory, professor needs to find 3 different forms that the length of the first form is equal to sum of lengths of the other two forms. Help him to do this.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of worm's forms. The second line contains *n* space-separated integers *a**i* (1<=≤<=*a**i*<=≤<=1000) — lengths of worms of each form.
Output Specification:
Output 3 distinct integers *i* *j* *k* (1<=≤<=*i*,<=*j*,<=*k*<=≤<=*n*) — such indexes of worm's forms that *a**i*<==<=*a**j*<=+<=*a**k*. If there is no such triple, output -1. If there are several solutions, output any of them. It possible that *a**j*<==<=*a**k*.
Demo Input:
['5\n1 2 3 5 7\n', '5\n1 8 1 5 1\n']
Demo Output:
['3 2 1\n', '-1\n']
Note:
none
|
```python
n = int(input())
container = list(map(int, input().split()))
ans = "-1"
needBreak = False
for i in range(len(container)):
for j in range(len(container)):
if j == i:
continue
for k in range(len(container)):
if k == j or k == i:
continue
if container[j] + container[k] == container[i]:
ans = str(i + 1) + str(j + 1) + str(k + 1)
needBreak = True
break
if needBreak:
break
if needBreak:
break
print(' '.join(ans))
```
| 0
|
863
|
B
|
Kayaking
|
PROGRAMMING
| 1,500
|
[
"brute force",
"greedy",
"sortings"
] | null | null |
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability!
|
The first line contains one number *n* (2<=≤<=*n*<=≤<=50).
The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000).
|
Print minimum possible total instability.
|
[
"2\n1 2 3 4\n",
"4\n1 3 4 6 3 4 100 200\n"
] |
[
"1\n",
"5\n"
] |
none
| 0
|
[
{
"input": "2\n1 2 3 4",
"output": "1"
},
{
"input": "4\n1 3 4 6 3 4 100 200",
"output": "5"
},
{
"input": "3\n305 139 205 406 530 206",
"output": "102"
},
{
"input": "3\n610 750 778 6 361 407",
"output": "74"
},
{
"input": "5\n97 166 126 164 154 98 221 7 51 47",
"output": "35"
},
{
"input": "50\n1 1 2 2 1 3 2 2 1 1 1 1 2 3 3 1 2 1 3 3 2 1 2 3 1 1 2 1 3 1 3 1 3 3 3 1 1 1 3 3 2 2 2 2 3 2 2 2 2 3 1 3 3 3 3 1 3 3 1 3 3 3 3 2 3 1 3 3 1 1 1 3 1 2 2 2 1 1 1 3 1 2 3 2 1 3 3 2 2 1 3 1 3 1 2 2 1 2 3 2",
"output": "0"
},
{
"input": "50\n5 5 5 5 4 2 2 3 2 2 4 1 5 5 1 2 4 2 4 2 5 2 2 2 2 3 2 4 2 5 5 4 3 1 2 3 3 5 4 2 2 5 2 4 5 5 4 4 1 5 5 3 2 2 5 1 3 3 2 4 4 5 1 2 3 4 4 1 3 3 3 5 1 2 4 4 4 4 2 5 2 5 3 2 4 5 5 2 1 1 2 4 5 3 2 1 2 4 4 4",
"output": "1"
},
{
"input": "50\n499 780 837 984 481 526 944 482 862 136 265 605 5 631 974 967 574 293 969 467 573 845 102 224 17 873 648 120 694 996 244 313 404 129 899 583 541 314 525 496 443 857 297 78 575 2 430 137 387 319 382 651 594 411 845 746 18 232 6 289 889 81 174 175 805 1000 799 950 475 713 951 685 729 925 262 447 139 217 788 514 658 572 784 185 112 636 10 251 621 218 210 89 597 553 430 532 264 11 160 476",
"output": "368"
},
{
"input": "50\n873 838 288 87 889 364 720 410 565 651 577 356 740 99 549 592 994 385 777 435 486 118 887 440 749 533 356 790 413 681 267 496 475 317 88 660 374 186 61 437 729 860 880 538 277 301 667 180 60 393 955 540 896 241 362 146 74 680 734 767 851 337 751 860 542 735 444 793 340 259 495 903 743 961 964 966 87 275 22 776 368 701 835 732 810 735 267 988 352 647 924 183 1 924 217 944 322 252 758 597",
"output": "393"
},
{
"input": "50\n297 787 34 268 439 629 600 398 425 833 721 908 830 636 64 509 420 647 499 675 427 599 396 119 798 742 577 355 22 847 389 574 766 453 196 772 808 261 106 844 726 975 173 992 874 89 775 616 678 52 69 591 181 573 258 381 665 301 589 379 362 146 790 842 765 100 229 916 938 97 340 793 758 177 736 396 247 562 571 92 923 861 165 748 345 703 431 930 101 761 862 595 505 393 126 846 431 103 596 21",
"output": "387"
},
{
"input": "50\n721 631 587 746 692 406 583 90 388 16 161 948 921 70 387 426 39 398 517 724 879 377 906 502 359 950 798 408 846 718 911 845 57 886 9 668 537 632 344 762 19 193 658 447 870 173 98 156 592 519 183 539 274 393 962 615 551 626 148 183 769 763 829 120 796 761 14 744 537 231 696 284 581 688 611 826 703 145 224 600 965 613 791 275 984 375 402 281 851 580 992 8 816 454 35 532 347 250 242 637",
"output": "376"
},
{
"input": "50\n849 475 37 120 754 183 758 374 543 198 896 691 11 607 198 343 761 660 239 669 628 259 223 182 216 158 20 565 454 884 137 923 156 22 310 77 267 707 582 169 120 308 439 309 59 152 206 696 210 177 296 887 559 22 154 553 142 247 491 692 473 572 461 206 532 319 503 164 328 365 541 366 300 392 486 257 863 432 877 404 520 69 418 99 519 239 374 927 601 103 226 316 423 219 240 26 455 101 184 61",
"output": "351"
},
{
"input": "3\n1 2 10 11 100 100",
"output": "1"
},
{
"input": "17\n814 744 145 886 751 1000 272 914 270 529 467 164 410 369 123 424 991 12 702 582 561 858 746 950 598 393 606 498 648 686 455 873 728 858",
"output": "318"
},
{
"input": "45\n476 103 187 696 463 457 588 632 763 77 391 721 95 124 378 812 980 193 694 898 859 572 721 274 605 264 929 615 257 918 42 493 1 3 697 349 990 800 82 535 382 816 943 735 11 272 562 323 653 370 766 332 666 130 704 604 645 717 267 255 37 470 925 941 376 611 332 758 504 40 477 263 708 434 38 596 650 990 714 662 572 467 949 799 648 581 545 828 508 636",
"output": "355"
},
{
"input": "2\n55 5 25 51",
"output": "4"
},
{
"input": "25\n89 50 640 463 858 301 522 241 923 378 892 822 550 17 42 66 706 779 657 840 273 222 444 459 94 925 437 159 182 727 92 851 742 215 653 891 782 533 29 128 133 883 317 475 165 994 802 434 744 973",
"output": "348"
},
{
"input": "4\n35 48 71 44 78 79 57 48",
"output": "10"
},
{
"input": "3\n58 89 73 15 5 47",
"output": "21"
},
{
"input": "2\n1 20 99 100",
"output": "1"
}
] | 1,674,116,350
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 77
| 0
|
n = int(input())
weights = list(map(int, input().split()))
weights.sort()
instability = 0
weight = 0
while weight < (2*n)-2:
instability += abs(weights[weight] - weights[weight+1])
weight += 2
print(instability)
|
Title: Kayaking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vadim is really keen on travelling. Recently he heard about kayaking activity near his town and became very excited about it, so he joined a party of kayakers.
Now the party is ready to start its journey, but firstly they have to choose kayaks. There are 2·*n* people in the group (including Vadim), and they have exactly *n*<=-<=1 tandem kayaks (each of which, obviously, can carry two people) and 2 single kayaks. *i*-th person's weight is *w**i*, and weight is an important matter in kayaking — if the difference between the weights of two people that sit in the same tandem kayak is too large, then it can crash. And, of course, people want to distribute their seats in kayaks in order to minimize the chances that kayaks will crash.
Formally, the instability of a single kayak is always 0, and the instability of a tandem kayak is the absolute difference between weights of the people that are in this kayak. Instability of the whole journey is the total instability of all kayaks.
Help the party to determine minimum possible total instability!
Input Specification:
The first line contains one number *n* (2<=≤<=*n*<=≤<=50).
The second line contains 2·*n* integer numbers *w*1, *w*2, ..., *w*2*n*, where *w**i* is weight of person *i* (1<=≤<=*w**i*<=≤<=1000).
Output Specification:
Print minimum possible total instability.
Demo Input:
['2\n1 2 3 4\n', '4\n1 3 4 6 3 4 100 200\n']
Demo Output:
['1\n', '5\n']
Note:
none
|
```python
n = int(input())
weights = list(map(int, input().split()))
weights.sort()
instability = 0
weight = 0
while weight < (2*n)-2:
instability += abs(weights[weight] - weights[weight+1])
weight += 2
print(instability)
```
| 0
|
|
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"
}
] | 1,698,083,145
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 15
| 0
|
def toggle_case(s):
if not s.isalpha():
return s
return s.lower() if s.isupper() else s.upper()
def apply_rule(word):
if word.isupper() or word[1:].isupper():
return ''.join(toggle_case(c) for c in word)
return word
print(apply_rule("hELLO")) # hello
print(apply_rule("HTTP")) # http
print(apply_rule("z")) # Z
print(apply_rule("Hello")) # Hello
|
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 toggle_case(s):
if not s.isalpha():
return s
return s.lower() if s.isupper() else s.upper()
def apply_rule(word):
if word.isupper() or word[1:].isupper():
return ''.join(toggle_case(c) for c in word)
return word
print(apply_rule("hELLO")) # hello
print(apply_rule("HTTP")) # http
print(apply_rule("z")) # Z
print(apply_rule("Hello")) # Hello
```
| 0
|
|
588
|
B
|
Duff in Love
|
PROGRAMMING
| 1,300
|
[
"math"
] | null | null |
Duff is in love with lovely numbers! A positive integer *x* is called lovely if and only if there is no such positive integer *a*<=><=1 such that *a*2 is a divisor of *x*.
Malek has a number store! In his store, he has only divisors of positive integer *n* (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.
Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.
|
The first and only line of input contains one integer, *n* (1<=≤<=*n*<=≤<=1012).
|
Print the answer in one line.
|
[
"10\n",
"12\n"
] |
[
"10\n",
"6\n"
] |
In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.
In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 2<sup class="upper-index">2</sup>, so 12 is not lovely, while 6 is indeed lovely.
| 1,000
|
[
{
"input": "10",
"output": "10"
},
{
"input": "12",
"output": "6"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "2"
},
{
"input": "4",
"output": "2"
},
{
"input": "8",
"output": "2"
},
{
"input": "3",
"output": "3"
},
{
"input": "31",
"output": "31"
},
{
"input": "97",
"output": "97"
},
{
"input": "1000000000000",
"output": "10"
},
{
"input": "15",
"output": "15"
},
{
"input": "894",
"output": "894"
},
{
"input": "271",
"output": "271"
},
{
"input": "2457",
"output": "273"
},
{
"input": "2829",
"output": "2829"
},
{
"input": "5000",
"output": "10"
},
{
"input": "20",
"output": "10"
},
{
"input": "68",
"output": "34"
},
{
"input": "3096",
"output": "258"
},
{
"input": "1024",
"output": "2"
},
{
"input": "1048576",
"output": "2"
},
{
"input": "413933789280",
"output": "25870861830"
},
{
"input": "817634153013",
"output": "817634153013"
},
{
"input": "56517269141",
"output": "56517269141"
},
{
"input": "30707328551",
"output": "30707328551"
},
{
"input": "279564127218",
"output": "10354226934"
},
{
"input": "491159577042",
"output": "18191095446"
},
{
"input": "734337660466",
"output": "734337660466"
},
{
"input": "808453785117",
"output": "808453785117"
},
{
"input": "55926835837",
"output": "55926835837"
},
{
"input": "294809951965",
"output": "294809951965"
},
{
"input": "537988035389",
"output": "76855433627"
},
{
"input": "822722434952",
"output": "205680608738"
},
{
"input": "699511759613",
"output": "699511759613"
},
{
"input": "942689843037",
"output": "104743315893"
},
{
"input": "663634158717",
"output": "663634158717"
},
{
"input": "213612977250",
"output": "11730"
},
{
"input": "999999999989",
"output": "999999999989"
},
{
"input": "999999999988",
"output": "499999999994"
},
{
"input": "87178291200",
"output": "30030"
},
{
"input": "927668721948",
"output": "463834360974"
},
{
"input": "562436815639",
"output": "37927"
},
{
"input": "302981118597",
"output": "35853"
},
{
"input": "5",
"output": "5"
},
{
"input": "9",
"output": "3"
},
{
"input": "36",
"output": "6"
},
{
"input": "2231",
"output": "2231"
},
{
"input": "27648",
"output": "6"
},
{
"input": "40320",
"output": "210"
},
{
"input": "648000",
"output": "30"
},
{
"input": "999966000289",
"output": "999983"
},
{
"input": "999985999949",
"output": "999985999949"
},
{
"input": "991921850317",
"output": "9973"
}
] | 1,607,805,972
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 16
| 140
| 0
|
from math import sqrt
def func(n):
x=n
div=0
ans = 1
i = 2
lim=int(sqrt(n))+1
while n > 1:
flag = 0
if i>=lim and div:
return x
while n % i == 0:
n = n // i
flag = 1
div=1
if flag:
ans *= i
i += 1
return ans
def main():
n = int(input())
print(func(n))
if __name__ == '__main__':
main()
|
Title: Duff in Love
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Duff is in love with lovely numbers! A positive integer *x* is called lovely if and only if there is no such positive integer *a*<=><=1 such that *a*2 is a divisor of *x*.
Malek has a number store! In his store, he has only divisors of positive integer *n* (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.
Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.
Input Specification:
The first and only line of input contains one integer, *n* (1<=≤<=*n*<=≤<=1012).
Output Specification:
Print the answer in one line.
Demo Input:
['10\n', '12\n']
Demo Output:
['10\n', '6\n']
Note:
In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.
In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 2<sup class="upper-index">2</sup>, so 12 is not lovely, while 6 is indeed lovely.
|
```python
from math import sqrt
def func(n):
x=n
div=0
ans = 1
i = 2
lim=int(sqrt(n))+1
while n > 1:
flag = 0
if i>=lim and div:
return x
while n % i == 0:
n = n // i
flag = 1
div=1
if flag:
ans *= i
i += 1
return ans
def main():
n = int(input())
print(func(n))
if __name__ == '__main__':
main()
```
| 0
|
|
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"
}
] | 1,651,318,172
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
x=input()
x=x.split(" ")
if 2*int(x[0])+2*int(x[1])<=int(x[0])+int(x[1])+int(x[2]):
print(2*int(x[0])+2*int(x[1]))
else:
print(int(x[0])+int(x[1])+int(x[2]))
|
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=input()
x=x.split(" ")
if 2*int(x[0])+2*int(x[1])<=int(x[0])+int(x[1])+int(x[2]):
print(2*int(x[0])+2*int(x[1]))
else:
print(int(x[0])+int(x[1])+int(x[2]))
```
| 0
|
|
776
|
B
|
Sherlock and his girlfriend
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"number theory"
] | null | null |
Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry.
He bought *n* pieces of jewelry. The *i*-th piece has price equal to *i*<=+<=1, that is, the prices of the jewelry are 2,<=3,<=4,<=... *n*<=+<=1.
Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used.
Help Sherlock complete this trivial task.
|
The only line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of jewelry pieces.
|
The first line of output should contain a single integer *k*, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints.
The next line should consist of *n* space-separated integers (between 1 and *k*) that specify the color of each piece in the order of increasing price.
If there are multiple ways to color the pieces using *k* colors, you can output any of them.
|
[
"3\n",
"4\n"
] |
[
"2\n1 1 2 ",
"2\n2 1 1 2\n"
] |
In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2 respectively.
In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct.
| 1,000
|
[
{
"input": "3",
"output": "2\n1 1 2 "
},
{
"input": "4",
"output": "2\n1 1 2 1 "
},
{
"input": "17",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 "
},
{
"input": "25",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 "
},
{
"input": "85",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 "
},
{
"input": "105",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 "
},
{
"input": "123",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 "
},
{
"input": "452",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "641",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "293",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "733",
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{
"input": "13736",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "29594",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "19283",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "15346",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "41794",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "99998",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "100000",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "2",
"output": "1\n1 1 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "2",
"output": "1\n1 1 "
},
{
"input": "100000",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
},
{
"input": "99971",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 1 2 2 2 2 ..."
}
] | 1,662,134,002
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 140
| 4,812,800
|
from collections import deque
from math import *
import sys
import random
from bisect import *
from functools import reduce
from sys import stdin
import copy
"""
for _ in range(int(input())):
n,x = map(int,input().split())
arr = list(map(int,input().split()))
ans = set()
for i in arr:
if (i*x) in ans:
ans.remove(i*x)
elif i>=x and (i/x)%1==0 and (i//x) in ans:
ans.remove(i//x)
ans.add(i)
print(ans)
for _ in range(int(input())):
n = int(input())
s = input()
if n == 1:
ans = 1
else:
ans = 1 if ((s[1] == s[0] ))else 0
i = 1
while i<n:
if s[i]!=s[i-1]:
prev = s[i-1]
ans+=2
pika = 0
while i<n and s[i] != prev:
i+=1
pika += 1
if i == n:
ans+=pika
else:
ans+=(2*(pika+1))
else:
ans+=1
i+=1
print(ans)
"""
n = int(input())
a = [1 for i in range(n+1)]
i = 2
if n == 1:
print(1)
print(1)
elif n==2:
print(1)
print(*[1,1])
elif n==3:
print(2)
print(*[1,1,2])
else:
while i*i<=n:
j = 1
while i*j<=n:
a[i*j]=j+1
j+=1
i+=1
print(len(set(a))-1)
print(*a[1:])
|
Title: Sherlock and his girlfriend
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry.
He bought *n* pieces of jewelry. The *i*-th piece has price equal to *i*<=+<=1, that is, the prices of the jewelry are 2,<=3,<=4,<=... *n*<=+<=1.
Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used.
Help Sherlock complete this trivial task.
Input Specification:
The only line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of jewelry pieces.
Output Specification:
The first line of output should contain a single integer *k*, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints.
The next line should consist of *n* space-separated integers (between 1 and *k*) that specify the color of each piece in the order of increasing price.
If there are multiple ways to color the pieces using *k* colors, you can output any of them.
Demo Input:
['3\n', '4\n']
Demo Output:
['2\n1 1 2 ', '2\n2 1 1 2\n']
Note:
In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2 respectively.
In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct.
|
```python
from collections import deque
from math import *
import sys
import random
from bisect import *
from functools import reduce
from sys import stdin
import copy
"""
for _ in range(int(input())):
n,x = map(int,input().split())
arr = list(map(int,input().split()))
ans = set()
for i in arr:
if (i*x) in ans:
ans.remove(i*x)
elif i>=x and (i/x)%1==0 and (i//x) in ans:
ans.remove(i//x)
ans.add(i)
print(ans)
for _ in range(int(input())):
n = int(input())
s = input()
if n == 1:
ans = 1
else:
ans = 1 if ((s[1] == s[0] ))else 0
i = 1
while i<n:
if s[i]!=s[i-1]:
prev = s[i-1]
ans+=2
pika = 0
while i<n and s[i] != prev:
i+=1
pika += 1
if i == n:
ans+=pika
else:
ans+=(2*(pika+1))
else:
ans+=1
i+=1
print(ans)
"""
n = int(input())
a = [1 for i in range(n+1)]
i = 2
if n == 1:
print(1)
print(1)
elif n==2:
print(1)
print(*[1,1])
elif n==3:
print(2)
print(*[1,1,2])
else:
while i*i<=n:
j = 1
while i*j<=n:
a[i*j]=j+1
j+=1
i+=1
print(len(set(a))-1)
print(*a[1:])
```
| 0
|
|
69
|
A
|
Young Physicist
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] |
A. Young Physicist
|
2
|
256
|
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
|
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
|
[
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] |
[
"NO",
"YES"
] |
none
| 500
|
[
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,602,149,046
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 80
| 280
| 0
|
n=int(input())
tx,ty,tz=0,0,0
for i in range(n):
s=list(map(int,input().split()))
tx+=s[0]
ty+=s[1]
tz+=s[2]
print(["NO","YES"][(tx+ty+tz)==0])
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n=int(input())
tx,ty,tz=0,0,0
for i in range(n):
s=list(map(int,input().split()))
tx+=s[0]
ty+=s[1]
tz+=s[2]
print(["NO","YES"][(tx+ty+tz)==0])
```
| 0
|
468
|
C
|
Hack it!
|
PROGRAMMING
| 2,500
|
[
"binary search",
"constructive algorithms",
"math"
] | null | null |
Little X has met the following problem recently.
Let's define *f*(*x*) as the sum of digits in decimal representation of number *x* (for example, *f*(1234)<==<=1<=+<=2<=+<=3<=+<=4). You are to calculate
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
|
The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1018).
|
Print two integers: *l*,<=*r* (1<=≤<=*l*<=≤<=*r*<=<<=10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
|
[
"46\n",
"126444381000032\n"
] |
[
"1 10\n",
"2333333 2333333333333\n"
] |
none
| 1,500
|
[
{
"input": "46",
"output": "1 10"
},
{
"input": "126444381000032",
"output": "2333333 2333333333333"
},
{
"input": "69645082595",
"output": "613752823618441225798858488535 713259406474207764329704856394"
},
{
"input": "70602205995",
"output": "11 249221334020432074498656960922"
},
{
"input": "33898130785",
"output": "9 558855506346909386939077840182"
},
{
"input": "58929554039",
"output": "22 855783114773435710171914224422"
},
{
"input": "81696185182",
"output": "499118531974994927425925323518 956291458400902769638235161661"
},
{
"input": "1",
"output": "149268802942315027273202513064 277551734280589260570057105889"
},
{
"input": "2",
"output": "119692200833686078608961312319 629363568954685219494592939495"
},
{
"input": "3",
"output": "2 302254410562920936884653943506"
},
{
"input": "4",
"output": "284378053387469023431537894255 317250990020830090421009164911"
},
{
"input": "5",
"output": "2 62668056583245293799710157951"
},
{
"input": "6",
"output": "3 93810188780011787541394067841"
},
{
"input": "7",
"output": "2 834286447477504059026206246185"
},
{
"input": "8",
"output": "3 257583347960907690857477857197"
},
{
"input": "10",
"output": "3 163048811987317819669274448265"
},
{
"input": "11",
"output": "3 919618203693907154039906935669"
},
{
"input": "12",
"output": "448221703341269567451520778454 698029790336105644790102859494"
},
{
"input": "43",
"output": "9 172412961300207091437973214327"
},
{
"input": "36",
"output": "8 619355518777647869838990701242"
},
{
"input": "65",
"output": "6 709024330418134127413755925068"
},
{
"input": "43",
"output": "7 669540448846929747909766131221"
},
{
"input": "23",
"output": "2 104579054315773428039906118259"
},
{
"input": "100",
"output": "15 324437778467489559125023403167"
},
{
"input": "10000",
"output": "2 936791129091842315790163514642"
},
{
"input": "1000000",
"output": "18 369591628030718549289473454545"
},
{
"input": "100000000",
"output": "7 870405265198051697453938746950"
},
{
"input": "10000000000",
"output": "20 972749766921651560604778558599"
},
{
"input": "1000000000000",
"output": "6 68997070398311657294228230677"
},
{
"input": "100000000000000",
"output": "249537318528661282822184562278 397003438246047829818181818181"
},
{
"input": "10000000000000000",
"output": "10 778165727326620883431915444624"
},
{
"input": "1000000000000000000",
"output": "408256298986776744812953390000 824018301451167837914299999999"
},
{
"input": "450000000000000000",
"output": "2 357722688084551093593033993033"
},
{
"input": "432022",
"output": "3 333556238531076799985515487090"
},
{
"input": "428033",
"output": "22 730314748425770554502599499142"
},
{
"input": "776930",
"output": "20 521232359366297130685112811874"
},
{
"input": "329824",
"output": "308969571112207311167474021348 745620588073413831210052337999"
},
{
"input": "85058261498638",
"output": "16 931187081941564769324316582547"
},
{
"input": "2130909834463",
"output": "21 895378349209612431051450316022"
},
{
"input": "3427089130241",
"output": "10 676758114393938690602742889714"
},
{
"input": "22881472397923",
"output": "174523915446146844994793303441 429392837423394397373605399524"
},
{
"input": "756499070280135900",
"output": "17 414254565210363110239866979636"
},
{
"input": "348919756917359012",
"output": "15 148711424846749040017925495268"
},
{
"input": "789136710974630947",
"output": "13 214740524663440316532540770911"
},
{
"input": "32851678509090784",
"output": "15 728787006985867306455350840047"
},
{
"input": "219760524792138559",
"output": "8 51530430692061277616906746431"
},
{
"input": "1000000007",
"output": "19 96093944063249823335491170999"
},
{
"input": "999999874000003969",
"output": "306259936121099507549028331741 983025349899931537481252156006"
},
{
"input": "999999999999999989",
"output": "140766388309929586494194507367 705096656792775954903252198639"
},
{
"input": "999999999999999983",
"output": "4 365946710276937286091309116340"
},
{
"input": "999999999999999999",
"output": "26 983966655749419565978291923474"
}
] | 1,411,521,236
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 31
| 0
|
m = int(input())
x,t=10**100-1,m-100*45*10**99%m
pirntf(t,t+x)
|
Title: Hack it!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little X has met the following problem recently.
Let's define *f*(*x*) as the sum of digits in decimal representation of number *x* (for example, *f*(1234)<==<=1<=+<=2<=+<=3<=+<=4). You are to calculate
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
Input Specification:
The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1018).
Output Specification:
Print two integers: *l*,<=*r* (1<=≤<=*l*<=≤<=*r*<=<<=10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
Demo Input:
['46\n', '126444381000032\n']
Demo Output:
['1 10\n', '2333333 2333333333333\n']
Note:
none
|
```python
m = int(input())
x,t=10**100-1,m-100*45*10**99%m
pirntf(t,t+x)
```
| -1
|
|
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"
}
] | 1,673,111,762
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 124
| 0
|
m,n = map(int,input().split())
a = m*n
p = a/2
print(p)
|
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 = m*n
p = a/2
print(p)
```
| 0
|
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",
"output": "28 29"
},
{
"input": "98",
"output": "28 28"
},
{
"input": "97",
"output": "27 28"
},
{
"input": "96",
"output": "26 28"
},
{
"input": "95",
"output": "26 28"
},
{
"input": "94",
"output": "26 28"
},
{
"input": "93",
"output": "26 28"
},
{
"input": "92",
"output": "26 27"
},
{
"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"
},
{
"input": "984",
"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"
},
{
"input": "9996",
"output": "2856 2856"
},
{
"input": "9995",
"output": "2855 2856"
},
{
"input": "9994",
"output": "2854 2856"
},
{
"input": "9993",
"output": "2854 2856"
},
{
"input": "9992",
"output": "2854 2856"
},
{
"input": "9991",
"output": "2854 2856"
},
{
"input": "9990",
"output": "2854 2855"
},
{
"input": "9989",
"output": "2854 2854"
},
{
"input": "9988",
"output": "2853 2854"
},
{
"input": "9987",
"output": "2852 2854"
},
{
"input": "9986",
"output": "2852 2854"
},
{
"input": "9985",
"output": "2852 2854"
},
{
"input": "9984",
"output": "2852 2854"
},
{
"input": "9983",
"output": "2852 2853"
},
{
"input": "9982",
"output": "2852 2852"
},
{
"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"
},
{
"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"
},
{
"input": "234118",
"output": "66890 66892"
},
{
"input": "234117",
"output": "66890 66892"
},
{
"input": "234116",
"output": "66890 66891"
},
{
"input": "234115",
"output": "66890 66890"
},
{
"input": "234114",
"output": "66889 66890"
},
{
"input": "234113",
"output": "66888 66890"
},
{
"input": "234112",
"output": "66888 66890"
},
{
"input": "234111",
"output": "66888 66890"
},
{
"input": "234110",
"output": "66888 66890"
},
{
"input": "234109",
"output": "66888 66889"
},
{
"input": "234108",
"output": "66888 66888"
},
{
"input": "234107",
"output": "66887 66888"
},
{
"input": "234106",
"output": "66886 66888"
},
{
"input": "234105",
"output": "66886 66888"
},
{
"input": "234104",
"output": "66886 66888"
},
{
"input": "234103",
"output": "66886 66888"
},
{
"input": "868531",
"output": "248151 248152"
},
{
"input": "868530",
"output": "248150 248152"
},
{
"input": "868529",
"output": "248150 248152"
},
{
"input": "868528",
"output": "248150 248152"
},
{
"input": "868527",
"output": "248150 248152"
},
{
"input": "868526",
"output": "248150 248151"
},
{
"input": "868525",
"output": "248150 248150"
},
{
"input": "868524",
"output": "248149 248150"
},
{
"input": "868523",
"output": "248148 248150"
},
{
"input": "868522",
"output": "248148 248150"
},
{
"input": "868521",
"output": "248148 248150"
},
{
"input": "868520",
"output": "248148 248150"
},
{
"input": "868519",
"output": "248148 248149"
},
{
"input": "868518",
"output": "248148 248148"
},
{
"input": "868517",
"output": "248147 248148"
},
{
"input": "868516",
"output": "248146 248148"
},
{
"input": "868515",
"output": "248146 248148"
},
{
"input": "868514",
"output": "248146 248148"
},
{
"input": "868513",
"output": "248146 248148"
},
{
"input": "868512",
"output": "248146 248147"
},
{
"input": "868511",
"output": "248146 248146"
},
{
"input": "123413",
"output": "35260 35262"
},
{
"input": "123412",
"output": "35260 35262"
},
{
"input": "123411",
"output": "35260 35261"
},
{
"input": "123410",
"output": "35260 35260"
},
{
"input": "123409",
"output": "35259 35260"
},
{
"input": "123408",
"output": "35258 35260"
},
{
"input": "123407",
"output": "35258 35260"
},
{
"input": "123406",
"output": "35258 35260"
},
{
"input": "123405",
"output": "35258 35260"
},
{
"input": "123404",
"output": "35258 35259"
},
{
"input": "123403",
"output": "35258 35258"
},
{
"input": "123402",
"output": "35257 35258"
},
{
"input": "123401",
"output": "35256 35258"
},
{
"input": "123400",
"output": "35256 35258"
},
{
"input": "123399",
"output": "35256 35258"
},
{
"input": "123398",
"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"
}
] | 1,691,080,998
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
n = int(input())
min_days =max(0,(n%7)-5)
max_days =min(n%7,2)
print(min_days,max_days)
|
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 =max(0,(n%7)-5)
max_days =min(n%7,2)
print(min_days,max_days)
```
| 0
|
|
117
|
C
|
Cycle
|
PROGRAMMING
| 2,000
|
[
"dfs and similar",
"graphs"
] | null | null |
A tournament is a directed graph without self-loops in which every pair of vertexes is connected by exactly one directed edge. That is, for any two vertexes *u* and *v* (*u*<=≠<=*v*) exists either an edge going from *u* to *v*, or an edge from *v* to *u*.
You are given a tournament consisting of *n* vertexes. Your task is to find there a cycle of length three.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=5000). Next *n* lines contain the adjacency matrix *A* of the graph (without spaces). *A**i*,<=*j*<==<=1 if the graph has an edge going from vertex *i* to vertex *j*, otherwise *A**i*,<=*j*<==<=0. *A**i*,<=*j* stands for the *j*-th character in the *i*-th line.
It is guaranteed that the given graph is a tournament, that is, *A**i*,<=*i*<==<=0,<=*A**i*,<=*j*<=≠<=*A**j*,<=*i* (1<=≤<=*i*,<=*j*<=≤<=*n*,<=*i*<=≠<=*j*).
|
Print three distinct vertexes of the graph *a*1, *a*2, *a*3 (1<=≤<=*a**i*<=≤<=*n*), such that *A**a*1,<=*a*2<==<=*A**a*2,<=*a*3<==<=*A**a*3,<=*a*1<==<=1, or "-1", if a cycle whose length equals three does not exist.
If there are several solutions, print any of them.
|
[
"5\n00100\n10000\n01001\n11101\n11000\n",
"5\n01111\n00000\n01000\n01100\n01110\n"
] |
[
"1 3 2 ",
"-1\n"
] |
none
| 1,500
|
[
{
"input": "5\n00100\n10000\n01001\n11101\n11000",
"output": "1 3 2 "
},
{
"input": "5\n01111\n00000\n01000\n01100\n01110",
"output": "-1"
},
{
"input": "5\n01000\n00101\n10010\n11001\n10100",
"output": "1 2 3 "
},
{
"input": "5\n00110\n10110\n00011\n00000\n11010",
"output": "1 3 5 "
},
{
"input": "10\n0011000010\n1011001101\n0000101100\n0010101010\n1100000100\n1111101100\n1000100000\n1001001011\n0110111001\n1011111000",
"output": "1 3 5 "
},
{
"input": "10\n0111001000\n0011111000\n0000110110\n0010101110\n1000011001\n1001000010\n0010010101\n1100110000\n1100101100\n1111010110",
"output": "1 3 5 "
},
{
"input": "10\n0101111011\n0001111111\n1100011110\n0010011000\n0011000110\n0000101011\n0000100000\n1001011011\n0001001000\n0011101010",
"output": "1 4 3 "
},
{
"input": "10\n0000010011\n1001001111\n1100001110\n1010010011\n1111011000\n0110000001\n1001010100\n1001110000\n0000111101\n0010101100",
"output": "1 6 2 "
},
{
"input": "10\n0000000000\n1001100111\n1101101111\n1000000011\n1001000111\n1111101111\n1101100111\n1001000011\n1000000001\n1000000000",
"output": "-1"
},
{
"input": "1\n0",
"output": "-1"
},
{
"input": "2\n00\n10",
"output": "-1"
},
{
"input": "3\n001\n100\n010",
"output": "1 3 2 "
},
{
"input": "3\n010\n001\n100",
"output": "1 2 3 "
},
{
"input": "2\n01\n00",
"output": "-1"
},
{
"input": "3\n011\n000\n010",
"output": "-1"
},
{
"input": "4\n0000\n1010\n1001\n1100",
"output": "2 3 4 "
},
{
"input": "5\n01111\n00111\n00010\n00001\n00100",
"output": "3 4 5 "
}
] | 1,621,341,014
| 2,147,483,647
|
PyPy 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
/*
Auther: ghoshashis545 Ashis Ghosh 😎
College: Jalpaiguri Govt Enggineering College
*/
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#include <bits/stdc++.h>
#include<algorithm>
#include<string>
using namespace std;
#define ll int
#define int long long
#define ld long double
#define ff first
#define ss string
#define se second
#define sp(x) cout << fixed << setprecision(x)
#define endl "\n"
#define ub upper_bound
#define lb lower_bound
#define vi vector<int>
#define ipair pair<int,int>
#define vpi vector<ipair>
#define clr(a,x) memset(a,x,sizeof(a))
#define alt(v) v.begin(),v.end()
#define ralt(v) v.rbegin(),v.rend()
#define pb emplace_back
#define mp make_pair
#define PI 3.1415926535897932384626
#define cntb(x) __builtin_popcount(x)
#define fab(i,a,b) for(int i=(a);i<(b);i++)
#define fba(i,a,b) for(int i=(b);i>=(a);i--)
bool ispoweroftwo(int n){return n&(!(n&(n-1)));}
int mod=1000000007;
// int mod=998244353;
int dx[] = {1,0,-1,0};
int dy[] = {0,-1,0,1};
bool test = 0;
const int inf = 1e18;
const int N = 5e3+5;
int n;
string s;
vector<int>adj[N],vis(N);
vector<int>res;
int dfs(int u,int st,int sz){
if(sz == 0)
return u == st;
for(auto it : adj[u]){
if(dfs(it,st,sz-1)){
res.push_back(it);
return 1;
}
}
return 0;
}
void solve(int tc = 0)
{
cin >> n;
for(int i = 0; i < n; ++i){
cin >>s;
for(int j = 0; j < n; ++j)
if(s[j] == '1')
adj[i+1].push_back(j+1);
}
for(int i = 1; i <= n; ++i){
if(dfs(i,i,3)){
for(auto it : res)
cout<<it<<" ";
cout<<"\n";
return;
}
}
cout<<"-1\n";
}
signed main()
{
ios_base::sync_with_stdio(false);cin.tie(NULL);
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
int t=1;
if(test)
cin>>t;
for(int i = 1; i <= t; ++i){
// cout<<"Case #"<<i<<": ";
solve(i);
}
return 0;
}
|
Title: Cycle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A tournament is a directed graph without self-loops in which every pair of vertexes is connected by exactly one directed edge. That is, for any two vertexes *u* and *v* (*u*<=≠<=*v*) exists either an edge going from *u* to *v*, or an edge from *v* to *u*.
You are given a tournament consisting of *n* vertexes. Your task is to find there a cycle of length three.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=5000). Next *n* lines contain the adjacency matrix *A* of the graph (without spaces). *A**i*,<=*j*<==<=1 if the graph has an edge going from vertex *i* to vertex *j*, otherwise *A**i*,<=*j*<==<=0. *A**i*,<=*j* stands for the *j*-th character in the *i*-th line.
It is guaranteed that the given graph is a tournament, that is, *A**i*,<=*i*<==<=0,<=*A**i*,<=*j*<=≠<=*A**j*,<=*i* (1<=≤<=*i*,<=*j*<=≤<=*n*,<=*i*<=≠<=*j*).
Output Specification:
Print three distinct vertexes of the graph *a*1, *a*2, *a*3 (1<=≤<=*a**i*<=≤<=*n*), such that *A**a*1,<=*a*2<==<=*A**a*2,<=*a*3<==<=*A**a*3,<=*a*1<==<=1, or "-1", if a cycle whose length equals three does not exist.
If there are several solutions, print any of them.
Demo Input:
['5\n00100\n10000\n01001\n11101\n11000\n', '5\n01111\n00000\n01000\n01100\n01110\n']
Demo Output:
['1 3 2 ', '-1\n']
Note:
none
|
```python
/*
Auther: ghoshashis545 Ashis Ghosh 😎
College: Jalpaiguri Govt Enggineering College
*/
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#include <bits/stdc++.h>
#include<algorithm>
#include<string>
using namespace std;
#define ll int
#define int long long
#define ld long double
#define ff first
#define ss string
#define se second
#define sp(x) cout << fixed << setprecision(x)
#define endl "\n"
#define ub upper_bound
#define lb lower_bound
#define vi vector<int>
#define ipair pair<int,int>
#define vpi vector<ipair>
#define clr(a,x) memset(a,x,sizeof(a))
#define alt(v) v.begin(),v.end()
#define ralt(v) v.rbegin(),v.rend()
#define pb emplace_back
#define mp make_pair
#define PI 3.1415926535897932384626
#define cntb(x) __builtin_popcount(x)
#define fab(i,a,b) for(int i=(a);i<(b);i++)
#define fba(i,a,b) for(int i=(b);i>=(a);i--)
bool ispoweroftwo(int n){return n&(!(n&(n-1)));}
int mod=1000000007;
// int mod=998244353;
int dx[] = {1,0,-1,0};
int dy[] = {0,-1,0,1};
bool test = 0;
const int inf = 1e18;
const int N = 5e3+5;
int n;
string s;
vector<int>adj[N],vis(N);
vector<int>res;
int dfs(int u,int st,int sz){
if(sz == 0)
return u == st;
for(auto it : adj[u]){
if(dfs(it,st,sz-1)){
res.push_back(it);
return 1;
}
}
return 0;
}
void solve(int tc = 0)
{
cin >> n;
for(int i = 0; i < n; ++i){
cin >>s;
for(int j = 0; j < n; ++j)
if(s[j] == '1')
adj[i+1].push_back(j+1);
}
for(int i = 1; i <= n; ++i){
if(dfs(i,i,3)){
for(auto it : res)
cout<<it<<" ";
cout<<"\n";
return;
}
}
cout<<"-1\n";
}
signed main()
{
ios_base::sync_with_stdio(false);cin.tie(NULL);
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
int t=1;
if(test)
cin>>t;
for(int i = 1; i <= t; ++i){
// cout<<"Case #"<<i<<": ";
solve(i);
}
return 0;
}
```
| -1
|
|
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"
}
] | 1,644,349,875
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 92
| 0
|
x= (input())
print((int(x[0])) * int(x[2]) // 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
x= (input())
print((int(x[0])) * int(x[2]) // 2)
```
| 0
|
852
|
I
|
Dating
|
PROGRAMMING
| 2,300
|
[
"brute force",
"dfs and similar",
"graphs",
"trees"
] | null | null |
This story is happening in a town named BubbleLand. There are *n* houses in BubbleLand. In each of these *n* houses lives a boy or a girl. People there really love numbers and everyone has their favorite number *f*. That means that the boy or girl that lives in the *i*-th house has favorite number equal to *f**i*.
The houses are numerated with numbers 1 to *n*.
The houses are connected with *n*<=-<=1 bidirectional roads and you can travel from any house to any other house in the town. There is exactly one path between every pair of houses.
A new dating had agency opened their offices in this mysterious town and the citizens were very excited. They immediately sent *q* questions to the agency and each question was of the following format:
- *a* *b* — asking how many ways are there to choose a couple (boy and girl) that have the same favorite number and live in one of the houses on the unique path from house *a* to house *b*.
Help the dating agency to answer the questions and grow their business.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105), the number of houses in the town.
The second line contains *n* integers, where the *i*-th number is 1 if a boy lives in the *i*-th house or 0 if a girl lives in *i*-th house.
The third line contains *n* integers, where the *i*-th number represents the favorite number *f**i* (1<=≤<=*f**i*<=≤<=109) of the girl or boy that lives in the *i*-th house.
The next *n*<=-<=1 lines contain information about the roads and the *i*-th line contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*) which means that there exists road between those two houses. It is guaranteed that it's possible to reach any house from any other.
The following line contains an integer *q* (1<=≤<=*q*<=≤<=105), the number of queries.
Each of the following *q* lines represents a question and consists of two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=*n*).
|
For each of the *q* questions output a single number, the answer to the citizens question.
|
[
"7\n1 0 0 1 0 1 0\n9 2 9 2 2 9 9\n2 6\n1 2\n4 2\n6 5\n3 6\n7 4\n2\n1 3\n7 5\n"
] |
[
"2\n3\n"
] |
In the first question from house 1 to house 3, the potential couples are (1, 3) and (6, 3).
In the second question from house 7 to house 5, the potential couples are (7, 6), (4, 2) and (4, 5).
| 0
|
[] | 1,689,249,462
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
print("_RANDOM_GUESS_1689249462.505291")# 1689249462.505329
|
Title: Dating
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This story is happening in a town named BubbleLand. There are *n* houses in BubbleLand. In each of these *n* houses lives a boy or a girl. People there really love numbers and everyone has their favorite number *f*. That means that the boy or girl that lives in the *i*-th house has favorite number equal to *f**i*.
The houses are numerated with numbers 1 to *n*.
The houses are connected with *n*<=-<=1 bidirectional roads and you can travel from any house to any other house in the town. There is exactly one path between every pair of houses.
A new dating had agency opened their offices in this mysterious town and the citizens were very excited. They immediately sent *q* questions to the agency and each question was of the following format:
- *a* *b* — asking how many ways are there to choose a couple (boy and girl) that have the same favorite number and live in one of the houses on the unique path from house *a* to house *b*.
Help the dating agency to answer the questions and grow their business.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105), the number of houses in the town.
The second line contains *n* integers, where the *i*-th number is 1 if a boy lives in the *i*-th house or 0 if a girl lives in *i*-th house.
The third line contains *n* integers, where the *i*-th number represents the favorite number *f**i* (1<=≤<=*f**i*<=≤<=109) of the girl or boy that lives in the *i*-th house.
The next *n*<=-<=1 lines contain information about the roads and the *i*-th line contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*) which means that there exists road between those two houses. It is guaranteed that it's possible to reach any house from any other.
The following line contains an integer *q* (1<=≤<=*q*<=≤<=105), the number of queries.
Each of the following *q* lines represents a question and consists of two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=*n*).
Output Specification:
For each of the *q* questions output a single number, the answer to the citizens question.
Demo Input:
['7\n1 0 0 1 0 1 0\n9 2 9 2 2 9 9\n2 6\n1 2\n4 2\n6 5\n3 6\n7 4\n2\n1 3\n7 5\n']
Demo Output:
['2\n3\n']
Note:
In the first question from house 1 to house 3, the potential couples are (1, 3) and (6, 3).
In the second question from house 7 to house 5, the potential couples are (7, 6), (4, 2) and (4, 5).
|
```python
print("_RANDOM_GUESS_1689249462.505291")# 1689249462.505329
```
| 0
|
|
886
|
B
|
Vlad and Cafes
|
PROGRAMMING
| 1,000
|
[] | null | null |
Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.
First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe.
|
In first line there is one integer *n* (1<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted.
|
Print one integer — index of the cafe that Vlad hasn't visited for as long as possible.
|
[
"5\n1 3 2 1 2\n",
"6\n2 1 2 2 4 1\n"
] |
[
"3\n",
"2\n"
] |
In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.
In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes.
| 1,000
|
[
{
"input": "5\n1 3 2 1 2",
"output": "3"
},
{
"input": "6\n2 1 2 2 4 1",
"output": "2"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n200000",
"output": "200000"
},
{
"input": "2\n2018 2017",
"output": "2018"
},
{
"input": "5\n100 1000 1000 1000 1000",
"output": "100"
},
{
"input": "8\n200000 1 200000 1 200000 1 200000 2",
"output": "1"
},
{
"input": "5\n20000 1 2 3 4",
"output": "20000"
},
{
"input": "2\n200000 1",
"output": "200000"
},
{
"input": "3\n2 100 2",
"output": "100"
},
{
"input": "2\n10 1",
"output": "10"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "3\n5 5 5",
"output": "5"
},
{
"input": "2\n5 1",
"output": "5"
}
] | 1,510,507,785
| 5,085
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
PRETESTS
| 4
| 2,000
| 11,571,200
|
n = int(input())
a = input().split()
b = []
for i in range (n):
b.append(int(0))
for i in range (n):
for j in range (n):
if a[i]==a[j]:
b[i] = j
m = int(1000)
for i in range (n):
if b[i]<m:
m = b[i]
k = i
print(a[k])
|
Title: Vlad and Cafes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vlad likes to eat in cafes very much. During his life, he has visited cafes *n* times. Unfortunately, Vlad started to feel that his last visits are not any different from each other. To fix that Vlad had a small research.
First of all, Vlad assigned individual indices to all cafes. Then, he wrote down indices of cafes he visited in a row, in order of visiting them. Now, Vlad wants to find such a cafe that his last visit to that cafe was before his last visits to every other cafe. In other words, he wants to find such a cafe that he hasn't been there for as long as possible. Help Vlad to find that cafe.
Input Specification:
In first line there is one integer *n* (1<=≤<=*n*<=≤<=2·105) — number of cafes indices written by Vlad.
In second line, *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2·105) are written — indices of cafes in order of being visited by Vlad. Vlad could visit some cafes more than once. Note that in numeration, some indices could be omitted.
Output Specification:
Print one integer — index of the cafe that Vlad hasn't visited for as long as possible.
Demo Input:
['5\n1 3 2 1 2\n', '6\n2 1 2 2 4 1\n']
Demo Output:
['3\n', '2\n']
Note:
In first test, there are three cafes, and the last visits to cafes with indices 1 and 2 were after the last visit to cafe with index 3; so this cafe is the answer.
In second test case, there are also three cafes, but with indices 1, 2 and 4. Cafes with indices 1 and 4 were visited after the last visit of cafe with index 2, so the answer is 2. Note that Vlad could omit some numbers while numerating the cafes.
|
```python
n = int(input())
a = input().split()
b = []
for i in range (n):
b.append(int(0))
for i in range (n):
for j in range (n):
if a[i]==a[j]:
b[i] = j
m = int(1000)
for i in range (n):
if b[i]<m:
m = b[i]
k = i
print(a[k])
```
| 0
|
|
849
|
B
|
Tell Your World
|
PROGRAMMING
| 1,600
|
[
"brute force",
"geometry"
] | null | null |
Connect the countless points with lines, till we reach the faraway yonder.
There are *n* points on a coordinate plane, the *i*-th of which being (*i*,<=*y**i*).
Determine whether it's possible to draw two parallel and non-overlapping lines, such that every point in the set lies on exactly one of them, and each of them passes through at least one point in the set.
|
The first line of input contains a positive integer *n* (3<=≤<=*n*<=≤<=1<=000) — the number of points.
The second line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (<=-<=109<=≤<=*y**i*<=≤<=109) — the vertical coordinates of each point.
|
Output "Yes" (without quotes) if it's possible to fulfill the requirements, and "No" otherwise.
You can print each letter in any case (upper or lower).
|
[
"5\n7 5 8 6 9\n",
"5\n-1 -2 0 0 -5\n",
"5\n5 4 3 2 1\n",
"5\n1000000000 0 0 0 0\n"
] |
[
"Yes\n",
"No\n",
"No\n",
"Yes\n"
] |
In the first example, there are five points: (1, 7), (2, 5), (3, 8), (4, 6) and (5, 9). It's possible to draw a line that passes through points 1, 3, 5, and another one that passes through points 2, 4 and is parallel to the first one.
In the second example, while it's possible to draw two lines that cover all points, they cannot be made parallel.
In the third example, it's impossible to satisfy both requirements at the same time.
| 1,000
|
[
{
"input": "5\n7 5 8 6 9",
"output": "Yes"
},
{
"input": "5\n-1 -2 0 0 -5",
"output": "No"
},
{
"input": "5\n5 4 3 2 1",
"output": "No"
},
{
"input": "5\n1000000000 0 0 0 0",
"output": "Yes"
},
{
"input": "5\n1000000000 1 0 -999999999 -1000000000",
"output": "Yes"
},
{
"input": "3\n998 244 353",
"output": "Yes"
},
{
"input": "3\n-1000000000 0 1000000000",
"output": "No"
},
{
"input": "5\n-1 -1 -1 -1 1",
"output": "Yes"
},
{
"input": "4\n-9763 530 3595 6660",
"output": "Yes"
},
{
"input": "4\n-253090305 36298498 374072642 711846786",
"output": "Yes"
},
{
"input": "5\n-186772848 -235864239 -191561068 -193955178 -243046569",
"output": "Yes"
},
{
"input": "5\n-954618456 -522919664 -248330428 -130850748 300848044",
"output": "Yes"
},
{
"input": "10\n4846 6705 2530 5757 5283 -944 -2102 -3260 -4418 2913",
"output": "No"
},
{
"input": "10\n-6568 -5920 -5272 -4624 -2435 -635 -2680 -2032 -1384 6565",
"output": "No"
},
{
"input": "20\n319410377 286827025 254243673 221660321 189076969 156493617 123910265 91326913 58743561 26160209 -6423143 -39006495 -71589847 -104173199 -136756551 -169339903 -201923255 -234506607 -267089959 -299673311",
"output": "No"
},
{
"input": "20\n-975467170 758268840 -975467171 758268839 -975467172 758268838 -975467173 758268837 -975467174 758268836 -975467175 758268835 -975467176 758268834 -975467177 758268833 -975467178 758268832 -975467179 758268831",
"output": "Yes"
},
{
"input": "4\n1 0 3 0",
"output": "No"
},
{
"input": "4\n100 2 3 4",
"output": "Yes"
},
{
"input": "5\n7 5 8 6 3",
"output": "No"
},
{
"input": "3\n1000000000 1000000000 -1000000000",
"output": "Yes"
},
{
"input": "4\n1 0 1 4",
"output": "Yes"
},
{
"input": "7\n1 2 -1 0 1 6 7",
"output": "Yes"
},
{
"input": "4\n0 0 4 0",
"output": "Yes"
},
{
"input": "7\n0 0 2 3 4 5 5",
"output": "Yes"
},
{
"input": "5\n7 5 8 6 8",
"output": "No"
},
{
"input": "5\n1 2 9 4 5",
"output": "Yes"
},
{
"input": "8\n1 12 3 14 5 16 7 8",
"output": "Yes"
},
{
"input": "5\n1 6 7 4 9",
"output": "Yes"
},
{
"input": "5\n2 1 0 1 2",
"output": "No"
},
{
"input": "4\n0 0 1 3",
"output": "Yes"
},
{
"input": "4\n100 50 50 10000000",
"output": "No"
},
{
"input": "5\n1 2 3 3 3",
"output": "No"
},
{
"input": "5\n1 2 6 10 17",
"output": "Yes"
},
{
"input": "4\n1 3 4 4",
"output": "Yes"
},
{
"input": "4\n100 50 50 1000000",
"output": "No"
},
{
"input": "6\n1 2 4 5 7 9",
"output": "No"
},
{
"input": "6\n0 0 1 2 3 4",
"output": "Yes"
},
{
"input": "5\n7 5 9 10 8",
"output": "Yes"
},
{
"input": "7\n1 2 2 1 2 2 1",
"output": "Yes"
},
{
"input": "4\n2 2 4 5",
"output": "Yes"
},
{
"input": "6\n1 2 1 3 4 5",
"output": "No"
},
{
"input": "4\n1 3 3 6",
"output": "No"
},
{
"input": "5\n1 2 -3 4 -1",
"output": "Yes"
}
] | 1,504,274,668
| 1,768
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 19
| 1,000
| 1,536,000
|
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import array
from bisect import *
from collections import *
import fractions
import heapq
from itertools import *
import math
import re
import string
N = int(input())
Ys = list(map(int, input().split()))
XYs = []
for i, y in enumerate(Ys):
XYs.append((i, y))
def on_the_line(xy1, xy2, XYs):
ret = []
for xy in XYs:
if xy1 == xy or xy2 == xy:
continue
x1, y1 = xy1
x2, y2 = xy2
x3, y3 = xy
if (y2 - y1) * (x3 - x1) - (y3 - y1) * (x2 - x1) != 0:
ret.append(xy)
return ret
def is_parallel(xy1, xy2, xy3, xy4):
x1, y1 = xy1
x2, y2 = xy2
x3, y3 = xy3
x4, y4 = xy4
return (y2 - y1) * (x4 - x3) - (y4 - y3) * (x2 - x1) == 0
ok = False
for i in range(2):
if ok:
break
for j in range(i+1, N):
remained_xys = on_the_line(XYs[i], XYs[j], XYs)
if len(remained_xys) == 0:
continue
if len(remained_xys) == 1:
ok = True
break
remained_xys2 = on_the_line(remained_xys[0], remained_xys[1], remained_xys)
if len(remained_xys2) == 0:
if is_parallel(XYs[i], XYs[j], remained_xys[0], remained_xys[1]):
ok = True
break
if ok:
print("Yes")
else:
print("No")
|
Title: Tell Your World
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Connect the countless points with lines, till we reach the faraway yonder.
There are *n* points on a coordinate plane, the *i*-th of which being (*i*,<=*y**i*).
Determine whether it's possible to draw two parallel and non-overlapping lines, such that every point in the set lies on exactly one of them, and each of them passes through at least one point in the set.
Input Specification:
The first line of input contains a positive integer *n* (3<=≤<=*n*<=≤<=1<=000) — the number of points.
The second line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (<=-<=109<=≤<=*y**i*<=≤<=109) — the vertical coordinates of each point.
Output Specification:
Output "Yes" (without quotes) if it's possible to fulfill the requirements, and "No" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['5\n7 5 8 6 9\n', '5\n-1 -2 0 0 -5\n', '5\n5 4 3 2 1\n', '5\n1000000000 0 0 0 0\n']
Demo Output:
['Yes\n', 'No\n', 'No\n', 'Yes\n']
Note:
In the first example, there are five points: (1, 7), (2, 5), (3, 8), (4, 6) and (5, 9). It's possible to draw a line that passes through points 1, 3, 5, and another one that passes through points 2, 4 and is parallel to the first one.
In the second example, while it's possible to draw two lines that cover all points, they cannot be made parallel.
In the third example, it's impossible to satisfy both requirements at the same time.
|
```python
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import array
from bisect import *
from collections import *
import fractions
import heapq
from itertools import *
import math
import re
import string
N = int(input())
Ys = list(map(int, input().split()))
XYs = []
for i, y in enumerate(Ys):
XYs.append((i, y))
def on_the_line(xy1, xy2, XYs):
ret = []
for xy in XYs:
if xy1 == xy or xy2 == xy:
continue
x1, y1 = xy1
x2, y2 = xy2
x3, y3 = xy
if (y2 - y1) * (x3 - x1) - (y3 - y1) * (x2 - x1) != 0:
ret.append(xy)
return ret
def is_parallel(xy1, xy2, xy3, xy4):
x1, y1 = xy1
x2, y2 = xy2
x3, y3 = xy3
x4, y4 = xy4
return (y2 - y1) * (x4 - x3) - (y4 - y3) * (x2 - x1) == 0
ok = False
for i in range(2):
if ok:
break
for j in range(i+1, N):
remained_xys = on_the_line(XYs[i], XYs[j], XYs)
if len(remained_xys) == 0:
continue
if len(remained_xys) == 1:
ok = True
break
remained_xys2 = on_the_line(remained_xys[0], remained_xys[1], remained_xys)
if len(remained_xys2) == 0:
if is_parallel(XYs[i], XYs[j], remained_xys[0], remained_xys[1]):
ok = True
break
if ok:
print("Yes")
else:
print("No")
```
| 0
|
|
1,007
|
A
|
Reorder the Array
|
PROGRAMMING
| 1,300
|
[
"combinatorics",
"data structures",
"math",
"sortings",
"two pointers"
] | null | null |
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array $[10, 20, 30, 40]$, we can permute it so that it becomes $[20, 40, 10, 30]$. Then on the first and the second positions the integers became larger ($20>10$, $40>20$) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals $2$. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
|
The first line contains a single integer $n$ ($1 \leq n \leq 10^5$) — the length of the array.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the elements of the array.
|
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
|
[
"7\n10 1 1 1 5 5 3\n",
"5\n1 1 1 1 1\n"
] |
[
"4\n",
"0\n"
] |
In the first sample, one of the best permutations is $[1, 5, 5, 3, 10, 1, 1]$. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0.
| 500
|
[
{
"input": "7\n10 1 1 1 5 5 3",
"output": "4"
},
{
"input": "5\n1 1 1 1 1",
"output": "0"
},
{
"input": "6\n300000000 200000000 300000000 200000000 1000000000 300000000",
"output": "3"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "9"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "7\n3 5 2 2 5 2 4",
"output": "4"
},
{
"input": "5\n1 5 4 2 3",
"output": "4"
}
] | 1,611,592,182
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 61
| 0
|
print("bdf")
|
Title: Reorder the Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.
For instance, if we are given an array $[10, 20, 30, 40]$, we can permute it so that it becomes $[20, 40, 10, 30]$. Then on the first and the second positions the integers became larger ($20>10$, $40>20$) and did not on the third and the fourth, so for this permutation, the number that Vasya wants to maximize equals $2$. Read the note for the first example, there is one more demonstrative test case.
Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.
Input Specification:
The first line contains a single integer $n$ ($1 \leq n \leq 10^5$) — the length of the array.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the elements of the array.
Output Specification:
Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.
Demo Input:
['7\n10 1 1 1 5 5 3\n', '5\n1 1 1 1 1\n']
Demo Output:
['4\n', '0\n']
Note:
In the first sample, one of the best permutations is $[1, 5, 5, 3, 10, 1, 1]$. On the positions from second to fifth the elements became larger, so the answer for this permutation is 4.
In the second sample, there is no way to increase any element with a permutation, so the answer is 0.
|
```python
print("bdf")
```
| 0
|
|
229
|
B
|
Planets
|
PROGRAMMING
| 1,700
|
[
"binary search",
"data structures",
"graphs",
"shortest paths"
] | null | null |
Goa'uld Apophis captured Jack O'Neill's team again! Jack himself was able to escape, but by that time Apophis's ship had already jumped to hyperspace. But Jack knows on what planet will Apophis land. In order to save his friends, Jack must repeatedly go through stargates to get to this planet.
Overall the galaxy has *n* planets, indexed with numbers from 1 to *n*. Jack is on the planet with index 1, and Apophis will land on the planet with index *n*. Jack can move between some pairs of planets through stargates (he can move in both directions); the transfer takes a positive, and, perhaps, for different pairs of planets unequal number of seconds. Jack begins his journey at time 0.
It can be that other travellers are arriving to the planet where Jack is currently located. In this case, Jack has to wait for exactly 1 second before he can use the stargate. That is, if at time *t* another traveller arrives to the planet, Jack can only pass through the stargate at time *t*<=+<=1, unless there are more travellers arriving at time *t*<=+<=1 to the same planet.
Knowing the information about travel times between the planets, and the times when Jack would not be able to use the stargate on particular planets, determine the minimum time in which he can get to the planet with index *n*.
|
The first line contains two space-separated integers: *n* (2<=≤<=*n*<=≤<=105), the number of planets in the galaxy, and *m* (0<=≤<=*m*<=≤<=105) — the number of pairs of planets between which Jack can travel using stargates. Then *m* lines follow, containing three integers each: the *i*-th line contains numbers of planets *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*), which are connected through stargates, and the integer transfer time (in seconds) *c**i* (1<=≤<=*c**i*<=≤<=104) between these planets. It is guaranteed that between any pair of planets there is at most one stargate connection.
Then *n* lines follow: the *i*-th line contains an integer *k**i* (0<=≤<=*k**i*<=≤<=105) that denotes the number of moments of time when other travellers arrive to the planet with index *i*. Then *k**i* distinct space-separated integers *t**ij* (0<=≤<=*t**ij*<=<<=109) follow, sorted in ascending order. An integer *t**ij* means that at time *t**ij* (in seconds) another traveller arrives to the planet *i*. It is guaranteed that the sum of all *k**i* does not exceed 105.
|
Print a single number — the least amount of time Jack needs to get from planet 1 to planet *n*. If Jack can't get to planet *n* in any amount of time, print number -1.
|
[
"4 6\n1 2 2\n1 3 3\n1 4 8\n2 3 4\n2 4 5\n3 4 3\n0\n1 3\n2 3 4\n0\n",
"3 1\n1 2 3\n0\n1 3\n0\n"
] |
[
"7\n",
"-1\n"
] |
In the first sample Jack has three ways to go from planet 1. If he moves to planet 4 at once, he spends 8 seconds. If he transfers to planet 3, he spends 3 seconds, but as other travellers arrive to planet 3 at time 3 and 4, he can travel to planet 4 only at time 5, thus spending 8 seconds in total. But if Jack moves to planet 2, and then — to planet 4, then he spends a total of only 2 + 5 = 7 seconds.
In the second sample one can't get from planet 1 to planet 3 by moving through stargates.
| 500
|
[
{
"input": "4 6\n1 2 2\n1 3 3\n1 4 8\n2 3 4\n2 4 5\n3 4 3\n0\n1 3\n2 3 4\n0",
"output": "7"
},
{
"input": "3 1\n1 2 3\n0\n1 3\n0",
"output": "-1"
},
{
"input": "2 1\n1 2 3\n0\n1 3",
"output": "3"
},
{
"input": "2 1\n1 2 3\n1 0\n0",
"output": "4"
},
{
"input": "3 3\n1 2 5\n2 3 6\n1 3 7\n0\n0\n0",
"output": "7"
},
{
"input": "3 3\n1 2 3\n2 3 2\n1 3 7\n0\n0\n0",
"output": "5"
},
{
"input": "2 0\n0\n0",
"output": "-1"
},
{
"input": "3 1\n1 2 3\n1 1\n1 5\n0",
"output": "-1"
},
{
"input": "2 1\n1 2 3\n0\n2 2 4",
"output": "3"
},
{
"input": "2 1\n1 2 1\n0\n0",
"output": "1"
},
{
"input": "2 1\n2 1 10000\n0\n0",
"output": "10000"
},
{
"input": "2 1\n1 2 3\n0\n3 3 4 5",
"output": "3"
},
{
"input": "3 0\n0\n0\n0",
"output": "-1"
},
{
"input": "3 2\n1 2 5\n2 3 7\n2 0 1\n3 4 5 6\n3 11 12 13",
"output": "14"
},
{
"input": "2 1\n1 2 3\n3 0 1 2\n3 5 6 7",
"output": "6"
},
{
"input": "3 3\n1 2 3\n2 3 2\n1 3 7\n0\n4 3 4 5 6\n0",
"output": "7"
},
{
"input": "6 7\n1 2 1\n1 3 8\n2 4 2\n4 3 3\n3 5 4\n4 6 100\n5 6 5\n0\n0\n1 7\n2 3 4\n0\n0",
"output": "17"
},
{
"input": "3 3\n1 2 3\n2 3 2\n1 3 6\n0\n1 3\n0",
"output": "6"
},
{
"input": "7 7\n1 2 1\n2 4 2\n2 3 2\n3 6 2\n6 5 2\n4 5 3\n5 7 7\n0\n0\n0\n3 3 4 5\n0\n0\n0",
"output": "14"
},
{
"input": "7 6\n1 2 1\n1 3 1\n1 4 1\n1 5 1\n1 6 1\n1 7 1\n1 0\n0\n0\n0\n0\n0\n0",
"output": "2"
},
{
"input": "8 10\n1 2 3\n2 8 3\n1 4 1\n4 3 6\n3 7 7\n4 5 5\n5 7 2\n7 8 1\n1 6 8\n6 8 7\n0\n4 1 2 3 4\n0\n0\n0\n0\n0\n0",
"output": "8"
},
{
"input": "7 6\n1 2 1\n1 3 2\n2 4 3\n2 5 4\n3 5 6\n3 6 7\n0\n3 1 2 3\n2 2 3\n0\n2 7 8\n0\n0",
"output": "-1"
}
] | 1,685,089,897
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 7
| 2,000
| 35,840,000
|
'''
https://codeforces.com/contest/229/problem/B
输入 n(2≤n≤1e5) m(0≤m≤1e5) 表示一个 n 点 m 边的无向图（节点编号从 1 开始）。
然后输入 m 条边，每条边包含 3 个数 a b c(1≤c≤1e4)，表示有一条边权为 c 的无向边连接 a 和 b。
保证无自环、无重边。
然后输入 n 行，每行第一个数 k 表示数组 t[i] 的长度，然后输入数组 t[i]。
数组 t[i] 是一个严格递增序列，0≤t[i][j]<1e9。
所有 k 之和 ≤1e5。
初始时间为 0。你从 1 出发，要去 n。
如果你在点 i，但是当前时间在数组 t[i] 中，那么你必须等待 1 秒。如果下一秒仍然在 t[i] 中，那么继续等待 1 秒。依此类推。
输出到达 n 的最早时间。
如果无法到达 n，输出 -1。
【易错题】
输入
4 6
1 2 2
1 3 3
1 4 8
2 3 4
2 4 5
3 4 3
0
1 3
2 3 4
0
输出 7
输入
3 1
1 2 3
0
1 3
0
输出 -1
'''
from cmath import inf
from collections import defaultdict
from heapq import heappop, heappush
def _n():
return int(input())
def _nA():
return list(map(int, input().split()))
def _nS():
return input().split()
def solve():
n, m = _nA()
graph = defaultdict(list)
for i in range(m):
a, b, c = _nA()
a -= 1
b -= 1
graph[a].append((b, c))
graph[b].append((a, c))
t = []
for i in range(n):
t.append(_nA()[1:])
dist = [inf]*n
q = [(0, 0)]
while q:
d, x = heappop(q)
# if d > dist[x]:
# continue
for i in t[x]:
if d == i:
d += 1
for y, c in graph[x]:
if d+c < dist[y]:
dist[y] = d+c
heappush(q, (d+c, y))
return dist[-1] if dist[-1] != inf else -1
print(solve())
|
Title: Planets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Goa'uld Apophis captured Jack O'Neill's team again! Jack himself was able to escape, but by that time Apophis's ship had already jumped to hyperspace. But Jack knows on what planet will Apophis land. In order to save his friends, Jack must repeatedly go through stargates to get to this planet.
Overall the galaxy has *n* planets, indexed with numbers from 1 to *n*. Jack is on the planet with index 1, and Apophis will land on the planet with index *n*. Jack can move between some pairs of planets through stargates (he can move in both directions); the transfer takes a positive, and, perhaps, for different pairs of planets unequal number of seconds. Jack begins his journey at time 0.
It can be that other travellers are arriving to the planet where Jack is currently located. In this case, Jack has to wait for exactly 1 second before he can use the stargate. That is, if at time *t* another traveller arrives to the planet, Jack can only pass through the stargate at time *t*<=+<=1, unless there are more travellers arriving at time *t*<=+<=1 to the same planet.
Knowing the information about travel times between the planets, and the times when Jack would not be able to use the stargate on particular planets, determine the minimum time in which he can get to the planet with index *n*.
Input Specification:
The first line contains two space-separated integers: *n* (2<=≤<=*n*<=≤<=105), the number of planets in the galaxy, and *m* (0<=≤<=*m*<=≤<=105) — the number of pairs of planets between which Jack can travel using stargates. Then *m* lines follow, containing three integers each: the *i*-th line contains numbers of planets *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*), which are connected through stargates, and the integer transfer time (in seconds) *c**i* (1<=≤<=*c**i*<=≤<=104) between these planets. It is guaranteed that between any pair of planets there is at most one stargate connection.
Then *n* lines follow: the *i*-th line contains an integer *k**i* (0<=≤<=*k**i*<=≤<=105) that denotes the number of moments of time when other travellers arrive to the planet with index *i*. Then *k**i* distinct space-separated integers *t**ij* (0<=≤<=*t**ij*<=<<=109) follow, sorted in ascending order. An integer *t**ij* means that at time *t**ij* (in seconds) another traveller arrives to the planet *i*. It is guaranteed that the sum of all *k**i* does not exceed 105.
Output Specification:
Print a single number — the least amount of time Jack needs to get from planet 1 to planet *n*. If Jack can't get to planet *n* in any amount of time, print number -1.
Demo Input:
['4 6\n1 2 2\n1 3 3\n1 4 8\n2 3 4\n2 4 5\n3 4 3\n0\n1 3\n2 3 4\n0\n', '3 1\n1 2 3\n0\n1 3\n0\n']
Demo Output:
['7\n', '-1\n']
Note:
In the first sample Jack has three ways to go from planet 1. If he moves to planet 4 at once, he spends 8 seconds. If he transfers to planet 3, he spends 3 seconds, but as other travellers arrive to planet 3 at time 3 and 4, he can travel to planet 4 only at time 5, thus spending 8 seconds in total. But if Jack moves to planet 2, and then — to planet 4, then he spends a total of only 2 + 5 = 7 seconds.
In the second sample one can't get from planet 1 to planet 3 by moving through stargates.
|
```python
'''
https://codeforces.com/contest/229/problem/B
输入 n(2≤n≤1e5) m(0≤m≤1e5) 表示一个 n 点 m 边的无向图（节点编号从 1 开始）。
然后输入 m 条边，每条边包含 3 个数 a b c(1≤c≤1e4)，表示有一条边权为 c 的无向边连接 a 和 b。
保证无自环、无重边。
然后输入 n 行，每行第一个数 k 表示数组 t[i] 的长度，然后输入数组 t[i]。
数组 t[i] 是一个严格递增序列，0≤t[i][j]<1e9。
所有 k 之和 ≤1e5。
初始时间为 0。你从 1 出发，要去 n。
如果你在点 i，但是当前时间在数组 t[i] 中，那么你必须等待 1 秒。如果下一秒仍然在 t[i] 中，那么继续等待 1 秒。依此类推。
输出到达 n 的最早时间。
如果无法到达 n，输出 -1。
【易错题】
输入
4 6
1 2 2
1 3 3
1 4 8
2 3 4
2 4 5
3 4 3
0
1 3
2 3 4
0
输出 7
输入
3 1
1 2 3
0
1 3
0
输出 -1
'''
from cmath import inf
from collections import defaultdict
from heapq import heappop, heappush
def _n():
return int(input())
def _nA():
return list(map(int, input().split()))
def _nS():
return input().split()
def solve():
n, m = _nA()
graph = defaultdict(list)
for i in range(m):
a, b, c = _nA()
a -= 1
b -= 1
graph[a].append((b, c))
graph[b].append((a, c))
t = []
for i in range(n):
t.append(_nA()[1:])
dist = [inf]*n
q = [(0, 0)]
while q:
d, x = heappop(q)
# if d > dist[x]:
# continue
for i in t[x]:
if d == i:
d += 1
for y, c in graph[x]:
if d+c < dist[y]:
dist[y] = d+c
heappush(q, (d+c, y))
return dist[-1] if dist[-1] != inf else -1
print(solve())
```
| 0
|
|
746
|
A
|
Compote
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
Nikolay has *a* lemons, *b* apples and *c* pears. He decided to cook a compote. According to the recipe the fruits should be in the ratio 1:<=2:<=4. It means that for each lemon in the compote should be exactly 2 apples and exactly 4 pears. You can't crumble up, break up or cut these fruits into pieces. These fruits — lemons, apples and pears — should be put in the compote as whole fruits.
Your task is to determine the maximum total number of lemons, apples and pears from which Nikolay can cook the compote. It is possible that Nikolay can't use any fruits, in this case print 0.
|
The first line contains the positive integer *a* (1<=≤<=*a*<=≤<=1000) — the number of lemons Nikolay has.
The second line contains the positive integer *b* (1<=≤<=*b*<=≤<=1000) — the number of apples Nikolay has.
The third line contains the positive integer *c* (1<=≤<=*c*<=≤<=1000) — the number of pears Nikolay has.
|
Print the maximum total number of lemons, apples and pears from which Nikolay can cook the compote.
|
[
"2\n5\n7\n",
"4\n7\n13\n",
"2\n3\n2\n"
] |
[
"7\n",
"21\n",
"0\n"
] |
In the first example Nikolay can use 1 lemon, 2 apples and 4 pears, so the answer is 1 + 2 + 4 = 7.
In the second example Nikolay can use 3 lemons, 6 apples and 12 pears, so the answer is 3 + 6 + 12 = 21.
In the third example Nikolay don't have enough pears to cook any compote, so the answer is 0.
| 500
|
[
{
"input": "2\n5\n7",
"output": "7"
},
{
"input": "4\n7\n13",
"output": "21"
},
{
"input": "2\n3\n2",
"output": "0"
},
{
"input": "1\n1\n1",
"output": "0"
},
{
"input": "1\n2\n4",
"output": "7"
},
{
"input": "1000\n1000\n1000",
"output": "1750"
},
{
"input": "1\n1\n4",
"output": "0"
},
{
"input": "1\n2\n3",
"output": "0"
},
{
"input": "1\n1000\n1000",
"output": "7"
},
{
"input": "1000\n1\n1000",
"output": "0"
},
{
"input": "1000\n2\n1000",
"output": "7"
},
{
"input": "1000\n500\n1000",
"output": "1750"
},
{
"input": "1000\n1000\n4",
"output": "7"
},
{
"input": "1000\n1000\n3",
"output": "0"
},
{
"input": "4\n8\n12",
"output": "21"
},
{
"input": "10\n20\n40",
"output": "70"
},
{
"input": "100\n200\n399",
"output": "693"
},
{
"input": "200\n400\n800",
"output": "1400"
},
{
"input": "199\n400\n800",
"output": "1393"
},
{
"input": "201\n400\n800",
"output": "1400"
},
{
"input": "200\n399\n800",
"output": "1393"
},
{
"input": "200\n401\n800",
"output": "1400"
},
{
"input": "200\n400\n799",
"output": "1393"
},
{
"input": "200\n400\n801",
"output": "1400"
},
{
"input": "139\n252\n871",
"output": "882"
},
{
"input": "109\n346\n811",
"output": "763"
},
{
"input": "237\n487\n517",
"output": "903"
},
{
"input": "161\n331\n725",
"output": "1127"
},
{
"input": "39\n471\n665",
"output": "273"
},
{
"input": "9\n270\n879",
"output": "63"
},
{
"input": "137\n422\n812",
"output": "959"
},
{
"input": "15\n313\n525",
"output": "105"
},
{
"input": "189\n407\n966",
"output": "1323"
},
{
"input": "18\n268\n538",
"output": "126"
},
{
"input": "146\n421\n978",
"output": "1022"
},
{
"input": "70\n311\n685",
"output": "490"
},
{
"input": "244\n405\n625",
"output": "1092"
},
{
"input": "168\n454\n832",
"output": "1176"
},
{
"input": "46\n344\n772",
"output": "322"
},
{
"input": "174\n438\n987",
"output": "1218"
},
{
"input": "144\n387\n693",
"output": "1008"
},
{
"input": "22\n481\n633",
"output": "154"
},
{
"input": "196\n280\n848",
"output": "980"
},
{
"input": "190\n454\n699",
"output": "1218"
},
{
"input": "231\n464\n928",
"output": "1617"
},
{
"input": "151\n308\n616",
"output": "1057"
},
{
"input": "88\n182\n364",
"output": "616"
},
{
"input": "12\n26\n52",
"output": "84"
},
{
"input": "204\n412\n824",
"output": "1428"
},
{
"input": "127\n256\n512",
"output": "889"
},
{
"input": "224\n446\n896",
"output": "1561"
},
{
"input": "146\n291\n584",
"output": "1015"
},
{
"input": "83\n164\n332",
"output": "574"
},
{
"input": "20\n38\n80",
"output": "133"
},
{
"input": "198\n393\n792",
"output": "1372"
},
{
"input": "120\n239\n480",
"output": "833"
},
{
"input": "208\n416\n831",
"output": "1449"
},
{
"input": "130\n260\n517",
"output": "903"
},
{
"input": "67\n134\n267",
"output": "462"
},
{
"input": "245\n490\n979",
"output": "1708"
},
{
"input": "182\n364\n727",
"output": "1267"
},
{
"input": "104\n208\n413",
"output": "721"
},
{
"input": "10\n2\n100",
"output": "7"
},
{
"input": "2\n100\n100",
"output": "14"
},
{
"input": "2\n3\n8",
"output": "7"
},
{
"input": "1\n2\n8",
"output": "7"
},
{
"input": "1\n2\n200",
"output": "7"
},
{
"input": "5\n4\n16",
"output": "14"
},
{
"input": "1\n10\n10",
"output": "7"
},
{
"input": "1\n4\n8",
"output": "7"
},
{
"input": "100\n4\n1000",
"output": "14"
},
{
"input": "2\n6\n12",
"output": "14"
},
{
"input": "10\n7\n4",
"output": "7"
},
{
"input": "2\n10\n100",
"output": "14"
},
{
"input": "2\n3\n4",
"output": "7"
},
{
"input": "1\n2\n999",
"output": "7"
},
{
"input": "1\n10\n20",
"output": "7"
},
{
"input": "100\n18\n20",
"output": "35"
},
{
"input": "100\n1\n100",
"output": "0"
},
{
"input": "3\n7\n80",
"output": "21"
},
{
"input": "2\n8\n24",
"output": "14"
},
{
"input": "1\n100\n100",
"output": "7"
},
{
"input": "2\n1\n8",
"output": "0"
},
{
"input": "10\n5\n23",
"output": "14"
}
] | 1,643,195,017
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 84
| 109
| 0
|
a=int(input())
b=int(input())
c=int(input())
b//=2
c//=4
s=0
for i in range(min(a,b,c)):
s+=1
s+=2
s+=4
print(s)
|
Title: Compote
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nikolay has *a* lemons, *b* apples and *c* pears. He decided to cook a compote. According to the recipe the fruits should be in the ratio 1:<=2:<=4. It means that for each lemon in the compote should be exactly 2 apples and exactly 4 pears. You can't crumble up, break up or cut these fruits into pieces. These fruits — lemons, apples and pears — should be put in the compote as whole fruits.
Your task is to determine the maximum total number of lemons, apples and pears from which Nikolay can cook the compote. It is possible that Nikolay can't use any fruits, in this case print 0.
Input Specification:
The first line contains the positive integer *a* (1<=≤<=*a*<=≤<=1000) — the number of lemons Nikolay has.
The second line contains the positive integer *b* (1<=≤<=*b*<=≤<=1000) — the number of apples Nikolay has.
The third line contains the positive integer *c* (1<=≤<=*c*<=≤<=1000) — the number of pears Nikolay has.
Output Specification:
Print the maximum total number of lemons, apples and pears from which Nikolay can cook the compote.
Demo Input:
['2\n5\n7\n', '4\n7\n13\n', '2\n3\n2\n']
Demo Output:
['7\n', '21\n', '0\n']
Note:
In the first example Nikolay can use 1 lemon, 2 apples and 4 pears, so the answer is 1 + 2 + 4 = 7.
In the second example Nikolay can use 3 lemons, 6 apples and 12 pears, so the answer is 3 + 6 + 12 = 21.
In the third example Nikolay don't have enough pears to cook any compote, so the answer is 0.
|
```python
a=int(input())
b=int(input())
c=int(input())
b//=2
c//=4
s=0
for i in range(min(a,b,c)):
s+=1
s+=2
s+=4
print(s)
```
| 3
|
Subsets and Splits
Successful Python Submissions
Retrieves all records from the train dataset where the verdict is 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Retrieves records of users with a rating of 1600 or higher and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a rating above 2000 and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a 'OK' verdict, providing a basic overview of a specific category within the dataset.