message stringlengths 2 30.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 237 109k | cluster float64 10 10 | __index_level_0__ int64 474 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro, a boy who hates any inefficiencies, pays coins so that the number of coins to be returned as change is minimized in order to do smoothly when he buys something.
One day, however, he doubt if this way is really efficient. When he pays more number of coins, a clerk consumes longer time to find the total value. Maybe he should pay with least possible number of coins.
Thinking for a while, he has decided to take the middle course. So he tries to minimize total number of paid coins and returned coins as change.
Now he is going to buy a product of P yen having several coins. Since he is not good at calculation, please write a program that computes the minimal number of coins.
You may assume following things:
* There are 6 kinds of coins, 1 yen, 5 yen, 10 yen, 50 yen, 100 yen and 500 yen.
* The total value of coins he has is at least P yen.
* A clerk will return the change with least number of coins.
Constraints
* Judge data contains at most 100 data sets.
* 0 ≤ Ni ≤ 1000
Input
Input file contains several data sets. One data set has following format:
P N1 N5 N10 N50 N100 N500
Ni is an integer and is the number of coins of i yen that he have.
The end of input is denoted by a case where P = 0. You should output nothing for this data set.
Output
Output total number of coins that are paid and are returned.
Example
Input
123 3 0 2 0 1 1
999 9 9 9 9 9 9
0 0 0 0 0 0 0
Output
6
3
Submitted Solution:
```
price = [1, 5, 10, 50, 100, 500]
while True:
values = list(map(int, input().split()))
p, n = values[0], values[1:]
ans = 1e100
if p == 0:
break
p_sum = sum(map(lambda s: s[0] * s[1], zip(price, n)))
for change in range(p_sum - p + 1):
total = p + change
pay = [0] * 6
#for i in reversed(range(6)):
# if total >= price[i]:
# pay[i] = min(n[i], int(total / price[i]))
# total -= pay[i] * price[i]
#if total > 0:
# break
coins = sum(pay)
_change = change
for i in reversed(range(6)):
if _change >= price[i]:
coins += int(_change / price[i])
_change %= price[i]
ans = min(ans, coins)
print(ans)
``` | instruction | 0 | 17,640 | 10 | 35,280 |
No | output | 1 | 17,640 | 10 | 35,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro, a boy who hates any inefficiencies, pays coins so that the number of coins to be returned as change is minimized in order to do smoothly when he buys something.
One day, however, he doubt if this way is really efficient. When he pays more number of coins, a clerk consumes longer time to find the total value. Maybe he should pay with least possible number of coins.
Thinking for a while, he has decided to take the middle course. So he tries to minimize total number of paid coins and returned coins as change.
Now he is going to buy a product of P yen having several coins. Since he is not good at calculation, please write a program that computes the minimal number of coins.
You may assume following things:
* There are 6 kinds of coins, 1 yen, 5 yen, 10 yen, 50 yen, 100 yen and 500 yen.
* The total value of coins he has is at least P yen.
* A clerk will return the change with least number of coins.
Constraints
* Judge data contains at most 100 data sets.
* 0 ≤ Ni ≤ 1000
Input
Input file contains several data sets. One data set has following format:
P N1 N5 N10 N50 N100 N500
Ni is an integer and is the number of coins of i yen that he have.
The end of input is denoted by a case where P = 0. You should output nothing for this data set.
Output
Output total number of coins that are paid and are returned.
Example
Input
123 3 0 2 0 1 1
999 9 9 9 9 9 9
0 0 0 0 0 0 0
Output
6
3
Submitted Solution:
```
price = [1, 5, 10, 50, 100, 500]
while True:
values = list(map(int, input().split()))
p, n = values[0], values[1:]
ans = 1e100
if p == 0:
break
p_sum = sum(map(lambda s: s[0] * s[1], zip(price, n)))
for change in range(p_sum - p + 1):
coins = 0
_change = change
for i in reversed(range(6)):
while _change >= price[i]:
_change -= price[i]
coins += 1
total = p + change
pay = [0] * 6
for i in reversed(range(6)):
while total >= price[i] and n[i] > pay[i]:
total -= price[i]
pay[i] += 1
if total > 0:
break
coins += sum(pay)
ans = min(ans, coins)
print(ans)
``` | instruction | 0 | 17,641 | 10 | 35,282 |
No | output | 1 | 17,641 | 10 | 35,283 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem. Don't forget to flush output after printing queries using cout.flush() or fflush(stdout) in C++ or similar functions in other programming languages.
There are n gift boxes in a row, numbered from 1 to n from left to right. It's known that exactly k of them contain valuable gifts — other boxes contain just lucky stones. All boxes look the same and differ only in weight. All boxes with stones have the same weight and are strictly heavier than boxes with valuable items. But valuable gifts may be different, so the boxes with valuable items may have different weights.
You can ask no more than 50 queries (printing an answer doesn't count). By each query you can compare total weights of two non-intersecting subsets of boxes a_1, a_2, ..., a_{k_a} and b_1, b_2, ..., b_{k_b}. In return you'll get one of four results:
* FIRST, if subset a_1, a_2, ..., a_{k_a} is strictly heavier;
* SECOND, if subset b_1, b_2, ..., b_{k_b} is strictly heavier;
* EQUAL, if subsets have equal total weights;
* WASTED, if the query is incorrect or the limit of queries is exceeded.
Using such queries (or, maybe, intuition) find the box with a valuable gift with the minimum index.
Input
The input consists of several cases. In the beginning, you receive the integer T (1 ≤ T ≤ 500) — the number of test cases.
At the beginning of each test case, you receive two integers n and k (2 ≤ n ≤ 1000, 1 ≤ k ≤ n/2) — the number of boxes in a row and the number of boxes with valuable gifts.
It's guaranteed that the order of boxes is fixed beforehand and that the sum of n in one test doesn't exceed 1000.
Output
For each test case print the minimum index among all boxes with a valuable gift in the following format: "! x" where x (1 ≤ x ≤ n) — the index of the box.
Interaction
Print each query in three lines. In the first line print the sizes of subset in the following format: "? k_a k_b" where k_a and k_b (1 ≤ k_a, k_b ≤ n; k_a + k_b ≤ n) — the corresponding sizes.
In the second line print k_a integers a_1, a_2, ..., a_{k_a} (1 ≤ a_i ≤ n; a_i ≠ a_j if i ≠ j) — indexes of boxes in the first subset.
In the third line print k_b integers b_1, b_2, ..., b_{k_b} (1 ≤ b_i ≤ n; b_i ≠ b_j if i ≠ j) — indexes of boxes in the second subset.
The subsets shouldn't intersect, i. e. a_i ≠ b_j for all i and j.
You'll receive one of four responses described above. In the case of WASTED stop your program to avoid getting random verdict instead of Wrong Answer.
Example
Input
2
2 1
-
-
-
FIRST
-
5 2
-
-
-
FIRST
-
-
-
SECOND
-
-
-
EQUAL
-
Output
-
-
? 1 1
1
2
-
! 2
-
? 1 1
1
2
-
? 2 3
4 2
1 3 5
-
? 1 1
4
5
-
! 1
Note
Additional separators "–" in the sample are used only to increase the readability of the sample. Don't print any unnecessary symbols or line breaks in your solution when you send it to the system.
Hacks are forbidden in this task. | instruction | 0 | 18,683 | 10 | 37,366 |
Tags: binary search, interactive, probabilities
Correct Solution:
```
from random import randint
def ask(l1, r1, l2, r2):
print('?', r1 - l1 + 1, r2 - l2 + 1)
for i in range(l1, r1 + 1):
print(i + 1, end=' ')
print()
for i in range(l2, r2 + 1):
print(i + 1, end=' ')
print(flush=True)
s = input()
if s[0] == 'F':
return 0
if s[0] == 'S':
return 1
if s[0] == 'E':
return 2
exit()
for _ in range(int(input())):
n, k = map(int, input().split())
flag = 0
for i in range(30):
x = randint(1, n - 1)
if ask(0, 0, x, x) == 1:
print('!', 1)
flag = 1
break
if flag:
continue
i = 0
while ask(0, (1 << i) - 1, 1 << i, min(n - 1, (1 << i + 1) - 1)) == 2:
i += 1
l, r = 0, min(n - (1 << i) - 1, (1 << i) - 1)
while l < r:
m = l + r >> 1
if ask(0, m, 1 << i, (1 << i) + m) == 2:
l = m + 1
else:
r = m
print('!', (1 << i) + l + 1)
``` | output | 1 | 18,683 | 10 | 37,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. Don't forget to flush output after printing queries using cout.flush() or fflush(stdout) in C++ or similar functions in other programming languages.
There are n gift boxes in a row, numbered from 1 to n from left to right. It's known that exactly k of them contain valuable gifts — other boxes contain just lucky stones. All boxes look the same and differ only in weight. All boxes with stones have the same weight and are strictly heavier than boxes with valuable items. But valuable gifts may be different, so the boxes with valuable items may have different weights.
You can ask no more than 50 queries (printing an answer doesn't count). By each query you can compare total weights of two non-intersecting subsets of boxes a_1, a_2, ..., a_{k_a} and b_1, b_2, ..., b_{k_b}. In return you'll get one of four results:
* FIRST, if subset a_1, a_2, ..., a_{k_a} is strictly heavier;
* SECOND, if subset b_1, b_2, ..., b_{k_b} is strictly heavier;
* EQUAL, if subsets have equal total weights;
* WASTED, if the query is incorrect or the limit of queries is exceeded.
Using such queries (or, maybe, intuition) find the box with a valuable gift with the minimum index.
Input
The input consists of several cases. In the beginning, you receive the integer T (1 ≤ T ≤ 500) — the number of test cases.
At the beginning of each test case, you receive two integers n and k (2 ≤ n ≤ 1000, 1 ≤ k ≤ n/2) — the number of boxes in a row and the number of boxes with valuable gifts.
It's guaranteed that the order of boxes is fixed beforehand and that the sum of n in one test doesn't exceed 1000.
Output
For each test case print the minimum index among all boxes with a valuable gift in the following format: "! x" where x (1 ≤ x ≤ n) — the index of the box.
Interaction
Print each query in three lines. In the first line print the sizes of subset in the following format: "? k_a k_b" where k_a and k_b (1 ≤ k_a, k_b ≤ n; k_a + k_b ≤ n) — the corresponding sizes.
In the second line print k_a integers a_1, a_2, ..., a_{k_a} (1 ≤ a_i ≤ n; a_i ≠ a_j if i ≠ j) — indexes of boxes in the first subset.
In the third line print k_b integers b_1, b_2, ..., b_{k_b} (1 ≤ b_i ≤ n; b_i ≠ b_j if i ≠ j) — indexes of boxes in the second subset.
The subsets shouldn't intersect, i. e. a_i ≠ b_j for all i and j.
You'll receive one of four responses described above. In the case of WASTED stop your program to avoid getting random verdict instead of Wrong Answer.
Example
Input
2
2 1
-
-
-
FIRST
-
5 2
-
-
-
FIRST
-
-
-
SECOND
-
-
-
EQUAL
-
Output
-
-
? 1 1
1
2
-
! 2
-
? 1 1
1
2
-
? 2 3
4 2
1 3 5
-
? 1 1
4
5
-
! 1
Note
Additional separators "–" in the sample are used only to increase the readability of the sample. Don't print any unnecessary symbols or line breaks in your solution when you send it to the system.
Hacks are forbidden in this task.
Submitted Solution:
```
import sys
input = sys.stdin.readline
def out(start, n, extra):
print('?',(n+1)//2, (n+1)//2)
print(' '.join(map(str,range(start, start + (n + 1)//2))))
if n % 2:
print(' '.join(map(str,range(start + (n + 1)//2, start + n))), extra)
else:
print(' '.join(map(str,range(start + (n + 1)//2, start + n))))
sys.stdout.flush()
return input().strip()
T = int(input())
for _ in range(T):
n, k = map(int, input().split())
start = 1
if n % 2 and k > 1:
extra = n
n -= 1
elif n % 2 and k == 1:
s = out(1, n - 1, n)
if s == 'EQUAL':
print('!',n)
sys.stdout.flush()
continue
extra = n
n -= 1
else:
s = out(start, n, -1)
if s == 'EQUAL' or s == 'SECOND':
n = n//2
extra = n
else:
start = start + (n+1)//2
n = n//2
extra = 1
while n > 1:
s = out(start, n, extra)
if s == 'EQUAL' or s == 'SECOND':
n = (n + 1)//2
else:
start = start + (n+1)//2
n = n//2
s = out(start, 1, extra)
if s == 'EQUAL' or s == 'SECOND':
print('!',start)
else:
print('!',extra)
sys.stdout.flush()
``` | instruction | 0 | 18,684 | 10 | 37,368 |
No | output | 1 | 18,684 | 10 | 37,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. Don't forget to flush output after printing queries using cout.flush() or fflush(stdout) in C++ or similar functions in other programming languages.
There are n gift boxes in a row, numbered from 1 to n from left to right. It's known that exactly k of them contain valuable gifts — other boxes contain just lucky stones. All boxes look the same and differ only in weight. All boxes with stones have the same weight and are strictly heavier than boxes with valuable items. But valuable gifts may be different, so the boxes with valuable items may have different weights.
You can ask no more than 50 queries (printing an answer doesn't count). By each query you can compare total weights of two non-intersecting subsets of boxes a_1, a_2, ..., a_{k_a} and b_1, b_2, ..., b_{k_b}. In return you'll get one of four results:
* FIRST, if subset a_1, a_2, ..., a_{k_a} is strictly heavier;
* SECOND, if subset b_1, b_2, ..., b_{k_b} is strictly heavier;
* EQUAL, if subsets have equal total weights;
* WASTED, if the query is incorrect or the limit of queries is exceeded.
Using such queries (or, maybe, intuition) find the box with a valuable gift with the minimum index.
Input
The input consists of several cases. In the beginning, you receive the integer T (1 ≤ T ≤ 500) — the number of test cases.
At the beginning of each test case, you receive two integers n and k (2 ≤ n ≤ 1000, 1 ≤ k ≤ n/2) — the number of boxes in a row and the number of boxes with valuable gifts.
It's guaranteed that the order of boxes is fixed beforehand and that the sum of n in one test doesn't exceed 1000.
Output
For each test case print the minimum index among all boxes with a valuable gift in the following format: "! x" where x (1 ≤ x ≤ n) — the index of the box.
Interaction
Print each query in three lines. In the first line print the sizes of subset in the following format: "? k_a k_b" where k_a and k_b (1 ≤ k_a, k_b ≤ n; k_a + k_b ≤ n) — the corresponding sizes.
In the second line print k_a integers a_1, a_2, ..., a_{k_a} (1 ≤ a_i ≤ n; a_i ≠ a_j if i ≠ j) — indexes of boxes in the first subset.
In the third line print k_b integers b_1, b_2, ..., b_{k_b} (1 ≤ b_i ≤ n; b_i ≠ b_j if i ≠ j) — indexes of boxes in the second subset.
The subsets shouldn't intersect, i. e. a_i ≠ b_j for all i and j.
You'll receive one of four responses described above. In the case of WASTED stop your program to avoid getting random verdict instead of Wrong Answer.
Example
Input
2
2 1
-
-
-
FIRST
-
5 2
-
-
-
FIRST
-
-
-
SECOND
-
-
-
EQUAL
-
Output
-
-
? 1 1
1
2
-
! 2
-
? 1 1
1
2
-
? 2 3
4 2
1 3 5
-
? 1 1
4
5
-
! 1
Note
Additional separators "–" in the sample are used only to increase the readability of the sample. Don't print any unnecessary symbols or line breaks in your solution when you send it to the system.
Hacks are forbidden in this task.
Submitted Solution:
```
import sys
def send(ans) :
print('!', ans)
sys.stdout.flush()
def ask(list1, list2) :
print('?', len(list1), len(list2))
print(' '.join(str(x) for x in list1))
print(' '.join(str(x) for x in list2))
sys.stdout.flush()
return input()[0]
def checkStone(i1, i2) :
result = ask([i1], [i2])
if result == 'E' :
return True
if result == 'F' :
print('WTF')
return False
def get(cur, l, lg) :
global fl
if l + lg > len(cur) :
lg = len(cur) - l
return cur[l : l + lg].copy()
def update(list1 = [], list2 = []) :
return list1.copy() + list2.copy()
def findStone(cur, n) :
l, r = 0, n
if (r - l) % 2 == 1:
r -= 1
while l + 1< r:
mid = (l + r) // 2
result = ask(get(cur, l, mid - l), get(cur, mid, r - mid))
if result == 'F' or result == 'E':
r = mid
else :
l = mid
if (r - l) % 2 == 1:
r -= 1
return cur[l]
a = [200, 80, 32, 16, 8, 4, 2, 1]
t = int(input())
while t > 0 :
t -= 1
n, k = map(int, input().split())
cur = [i + 1 for i in range(n)]
stone = findStone(cur, n)
for step in range(8) :
for j in range(a[step], len(cur), a[step]) :
result = ask(get(cur, 0, a[step]), get(cur, j, a[step]))
if result == 'F' :
cur = update(get(cur, 0, a[step]), get(cur, j, a[step]))
break
elif result == 'S' :
cur = update(get(cur, 0, a[step]))
break
elif result == 'E' :
if j >= len(cur) - a[step] :
cur = update(get(cur, 0, a[step]))
break
continue
else :
print('WASTED')
ans = -1
for i in cur :
if i != stone and not checkStone(i, stone) :
ans = i
break
if ans == -1:
print('OOPS')
send(ans)
``` | instruction | 0 | 18,685 | 10 | 37,370 |
No | output | 1 | 18,685 | 10 | 37,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. Don't forget to flush output after printing queries using cout.flush() or fflush(stdout) in C++ or similar functions in other programming languages.
There are n gift boxes in a row, numbered from 1 to n from left to right. It's known that exactly k of them contain valuable gifts — other boxes contain just lucky stones. All boxes look the same and differ only in weight. All boxes with stones have the same weight and are strictly heavier than boxes with valuable items. But valuable gifts may be different, so the boxes with valuable items may have different weights.
You can ask no more than 50 queries (printing an answer doesn't count). By each query you can compare total weights of two non-intersecting subsets of boxes a_1, a_2, ..., a_{k_a} and b_1, b_2, ..., b_{k_b}. In return you'll get one of four results:
* FIRST, if subset a_1, a_2, ..., a_{k_a} is strictly heavier;
* SECOND, if subset b_1, b_2, ..., b_{k_b} is strictly heavier;
* EQUAL, if subsets have equal total weights;
* WASTED, if the query is incorrect or the limit of queries is exceeded.
Using such queries (or, maybe, intuition) find the box with a valuable gift with the minimum index.
Input
The input consists of several cases. In the beginning, you receive the integer T (1 ≤ T ≤ 500) — the number of test cases.
At the beginning of each test case, you receive two integers n and k (2 ≤ n ≤ 1000, 1 ≤ k ≤ n/2) — the number of boxes in a row and the number of boxes with valuable gifts.
It's guaranteed that the order of boxes is fixed beforehand and that the sum of n in one test doesn't exceed 1000.
Output
For each test case print the minimum index among all boxes with a valuable gift in the following format: "! x" where x (1 ≤ x ≤ n) — the index of the box.
Interaction
Print each query in three lines. In the first line print the sizes of subset in the following format: "? k_a k_b" where k_a and k_b (1 ≤ k_a, k_b ≤ n; k_a + k_b ≤ n) — the corresponding sizes.
In the second line print k_a integers a_1, a_2, ..., a_{k_a} (1 ≤ a_i ≤ n; a_i ≠ a_j if i ≠ j) — indexes of boxes in the first subset.
In the third line print k_b integers b_1, b_2, ..., b_{k_b} (1 ≤ b_i ≤ n; b_i ≠ b_j if i ≠ j) — indexes of boxes in the second subset.
The subsets shouldn't intersect, i. e. a_i ≠ b_j for all i and j.
You'll receive one of four responses described above. In the case of WASTED stop your program to avoid getting random verdict instead of Wrong Answer.
Example
Input
2
2 1
-
-
-
FIRST
-
5 2
-
-
-
FIRST
-
-
-
SECOND
-
-
-
EQUAL
-
Output
-
-
? 1 1
1
2
-
! 2
-
? 1 1
1
2
-
? 2 3
4 2
1 3 5
-
? 1 1
4
5
-
! 1
Note
Additional separators "–" in the sample are used only to increase the readability of the sample. Don't print any unnecessary symbols or line breaks in your solution when you send it to the system.
Hacks are forbidden in this task.
Submitted Solution:
```
answers = ["FIRST", "SECOND", "EQUAL", "WASTED"]
def req(s1, s2, l):
print('?', l, l, flush = True)
print(*range(s1+1, s1+l+1), flush = True)
print(*range(s2+1, s2+l+1), flush = True)
ans = answers.index(input())
if ans > 2:
exit()
else:
return ans
for _ in range(int(input())):
n, k = map(int, input().split())
x = req(0, 1, 1)
if x == 0:
print(2)
elif x == 1:
print(1)
else:
l = 1
while x == 2:
l *= 2
y = l
l = min(l, n-y)
x = req(0, y, l)
if x == 1:
print(1)
else:
while l > 1:
l = (l + 1) // 2
x = req(0, y, l)
if x == 2:
y += l
print('!', y + 1)
``` | instruction | 0 | 18,686 | 10 | 37,372 |
No | output | 1 | 18,686 | 10 | 37,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem. Don't forget to flush output after printing queries using cout.flush() or fflush(stdout) in C++ or similar functions in other programming languages.
There are n gift boxes in a row, numbered from 1 to n from left to right. It's known that exactly k of them contain valuable gifts — other boxes contain just lucky stones. All boxes look the same and differ only in weight. All boxes with stones have the same weight and are strictly heavier than boxes with valuable items. But valuable gifts may be different, so the boxes with valuable items may have different weights.
You can ask no more than 50 queries (printing an answer doesn't count). By each query you can compare total weights of two non-intersecting subsets of boxes a_1, a_2, ..., a_{k_a} and b_1, b_2, ..., b_{k_b}. In return you'll get one of four results:
* FIRST, if subset a_1, a_2, ..., a_{k_a} is strictly heavier;
* SECOND, if subset b_1, b_2, ..., b_{k_b} is strictly heavier;
* EQUAL, if subsets have equal total weights;
* WASTED, if the query is incorrect or the limit of queries is exceeded.
Using such queries (or, maybe, intuition) find the box with a valuable gift with the minimum index.
Input
The input consists of several cases. In the beginning, you receive the integer T (1 ≤ T ≤ 500) — the number of test cases.
At the beginning of each test case, you receive two integers n and k (2 ≤ n ≤ 1000, 1 ≤ k ≤ n/2) — the number of boxes in a row and the number of boxes with valuable gifts.
It's guaranteed that the order of boxes is fixed beforehand and that the sum of n in one test doesn't exceed 1000.
Output
For each test case print the minimum index among all boxes with a valuable gift in the following format: "! x" where x (1 ≤ x ≤ n) — the index of the box.
Interaction
Print each query in three lines. In the first line print the sizes of subset in the following format: "? k_a k_b" where k_a and k_b (1 ≤ k_a, k_b ≤ n; k_a + k_b ≤ n) — the corresponding sizes.
In the second line print k_a integers a_1, a_2, ..., a_{k_a} (1 ≤ a_i ≤ n; a_i ≠ a_j if i ≠ j) — indexes of boxes in the first subset.
In the third line print k_b integers b_1, b_2, ..., b_{k_b} (1 ≤ b_i ≤ n; b_i ≠ b_j if i ≠ j) — indexes of boxes in the second subset.
The subsets shouldn't intersect, i. e. a_i ≠ b_j for all i and j.
You'll receive one of four responses described above. In the case of WASTED stop your program to avoid getting random verdict instead of Wrong Answer.
Example
Input
2
2 1
-
-
-
FIRST
-
5 2
-
-
-
FIRST
-
-
-
SECOND
-
-
-
EQUAL
-
Output
-
-
? 1 1
1
2
-
! 2
-
? 1 1
1
2
-
? 2 3
4 2
1 3 5
-
? 1 1
4
5
-
! 1
Note
Additional separators "–" in the sample are used only to increase the readability of the sample. Don't print any unnecessary symbols or line breaks in your solution when you send it to the system.
Hacks are forbidden in this task.
Submitted Solution:
```
answers = ["FIRST", "SECOND", "EQUAL", "WASTED"]
def req(s1, s2, l):
print('?', l, l, flush = True)
print(*range(s1+1, s1+l+1), flush = True)
print(*range(s2+1, s2+l+1), flush = True)
ans = answers.index(input())
if ans > 2:
exit()
else:
return ans
for _ in range(int(input())):
n, k = map(int, input().split())
x = req(0, 1, 1)
if x == 0:
print(2)
elif x == 1:
print(1)
else:
l = 1
while x == 2:
l *= 2
y = min(l, n - l)
x = req(0, y, l)
if x == 1:
print(1)
else:
while l > 1:
l //= 2
x = req(0, y, l)
if x == 2:
y += l
print(y + 1)
``` | instruction | 0 | 18,687 | 10 | 37,374 |
No | output | 1 | 18,687 | 10 | 37,375 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP | instruction | 0 | 18,885 | 10 | 37,770 |
Tags: constructive algorithms, implementation
Correct Solution:
```
strx = ''
a=int(input())
l = list(map(int, input().split(' ')))
strx += 'PRL' * l[0]
ok = [0]+[i for i in range(a) if l[i]>0]
ok = list(set(ok))
for i in range(1,len(ok)):
strx += 'R' * (ok[i]-ok[i-1])
strx += 'PLR' * l[ok[i]]
print(strx)
``` | output | 1 | 18,885 | 10 | 37,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP | instruction | 0 | 18,886 | 10 | 37,772 |
Tags: constructive algorithms, implementation
Correct Solution:
```
def main():
n = int(input())
l = list(map(int, input().split()))
res = []
lo, hi = 0, n - 1
while any(l):
keysoneflag = True
for i in range(lo, hi):
if l[i]:
if keysoneflag:
keysoneflag = False
lo = i
l[i] -= 1
res.append('P')
res.append('R')
keysoneflag = True
for i in range(hi, lo, -1):
if l[i]:
if keysoneflag:
keysoneflag = False
hi = i
l[i] -= 1
res.append('P')
res.append('L')
if res[-1] != 'P':
del res[-1]
print(''.join(res))
if __name__ == '__main__':
main()
``` | output | 1 | 18,886 | 10 | 37,773 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP | instruction | 0 | 18,887 | 10 | 37,774 |
Tags: constructive algorithms, implementation
Correct Solution:
```
# Description of the problem can be found at http://codeforces.com/problemset/problem/379/B
n = int(input())
l_n = list(map(int, input().split()))
s = ""
for i in range(n):
if l_n[i] == 0:
if i + 1 == n:
print("".join(c for c in s))
quit()
else:
s += "R"
else:
if i + 1 == n:
s += "PLR" * (l_n[i] - 1) + "P"
else:
s += "PRPL" * (min(l_n[i], l_n[i + 1])) + "PRL" * max(0, l_n[i] - l_n[i + 1] - 1) + ("PR" if (l_n[i] > l_n[i + 1]) else "R")
l_n[i + 1] = max(l_n[i + 1] - l_n[i], 0)
print(s)
``` | output | 1 | 18,887 | 10 | 37,775 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP | instruction | 0 | 18,888 | 10 | 37,776 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n = input()
a =[]
b = []
for x in input().split( ):
a.append(int(x))
b.append(0)
count=0
direction="R"
booly=0
while (True):
if(a[count]!=b[count] and direction=="R" and not booly):
b[count]+=1
print("P",end='')
if(a==b): break
count+=1
if(count==len(a)):
count-=1
direction="L"
booly=1
print("L",end='')
continue
print("R",end='')
if(a[count]==b[count] and direction=="R" and not booly):
count+=1
if(count==len(a)):
count-=1
direction="L"
booly=1
print("L",end='')
continue
print("R",end='')
if(direction=="L" and booly):
booly=0
count-=1
if(a[count]!=b[count] and direction=="L" and not booly):
b[count]+=1
print("P",end='')
if(a==b): break
count-=1
if(count==-1):
count+=1
direction="R"
booly=1
print("R",end='')
continue
print("L",end='')
if(a[count]==b[count] and direction=="L" and not booly):
count-=1
if(count==-1):
count+=1
direction="R"
booly=1
print("R",end='')
continue
print("L",end='')
if(direction=="R" and booly):
booly=0
count+=1
``` | output | 1 | 18,888 | 10 | 37,777 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP | instruction | 0 | 18,889 | 10 | 37,778 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n,ans=int(input()),""
a=list(map(int,input().split()))
while a[0]:
ans+="PRL"
a[0]-=1
for i in range(n):
while a[i]:
ans+="PLR"
a[i]-=1
ans+="R"
print(ans[:-1])
``` | output | 1 | 18,889 | 10 | 37,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP | instruction | 0 | 18,890 | 10 | 37,780 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
lines = []
for i in range(n - 1):
if a[i] == 0:
lines += ["R"]
else:
lines += ["PRL" * (a[i] - 1) + "PR"]
if a[n - 1] > 0:
lines += ["PLR" * (a[n - 1] - 1) + "P"]
print(''.join(lines))
``` | output | 1 | 18,890 | 10 | 37,781 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP | instruction | 0 | 18,891 | 10 | 37,782 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n,ans=int(input()),""
a=list(map(int,input().split()))
print ('PRL' * int(a[0]) + ''.join("R" + int(i) * ('PLR') for i in a[1:]) )
``` | output | 1 | 18,891 | 10 | 37,783 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP | instruction | 0 | 18,892 | 10 | 37,784 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n = int(input())
l = [int(x) for x in input().split()]
s = ''
for x in l[:-1]:
if x:
s += 'P'
for i in range(x - 1):
s += 'RLP'
s += 'R'
if l[-1]:
s += 'P'
for i in range(l[-1] - 1):
s += 'LRP'
print(s)
``` | output | 1 | 18,892 | 10 | 37,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
i=0
ans=""
while(i<=len(a)-1):
if a[i]==0 and i!=len(a)-1:
ans+="R"
i+=1
elif a[i]==0 and i==len(a)-1:
break
else:
if a[i]==1:
if i==len(a)-1:
ans+="P"
i+=1
else:
ans+="PR"
i+=1
else:
if i==len(a)-1:
ans+="PLR"*(a[i]-1)+"P"
i+=1
elif i==0:
ans+="PRL"*(a[i]-1)+"PR"
i+=1
else:
ans+="PRL"*(a[i]-1)+"PR"
i+=1
print(ans)
``` | instruction | 0 | 18,893 | 10 | 37,786 |
Yes | output | 1 | 18,893 | 10 | 37,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP
Submitted Solution:
```
n=int(input())
lis=list(map(int,input().split()))
command='PRL'*lis[0]
for i in range(1,len(lis)):
command+=('R'+'LRP'*lis[i])
print(command)
``` | instruction | 0 | 18,894 | 10 | 37,788 |
Yes | output | 1 | 18,894 | 10 | 37,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP
Submitted Solution:
```
int_inp = lambda: int(input()) #integer input
strng = lambda: input().strip() #string input
strl = lambda: list(input().strip())#list of strings as input
mul = lambda: map(int,input().split())#multiple integers as inpnut
mulf = lambda: map(float,input().split())#multiple floats as ipnut
seq = lambda: list(map(int,input().split()))#list of integers
import math
from collections import Counter,defaultdict
n=int(input())
l=list(map(int,input().split()))
for i in range(n+1):
if i==n:
break
if i==n-1:
print(l[i]*'PLR',end='')
else:
print(l[i]*'PRL'+'R',end='')
``` | instruction | 0 | 18,895 | 10 | 37,790 |
Yes | output | 1 | 18,895 | 10 | 37,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP
Submitted Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
sm = sum(a)
lst = []
pos = 0
turn = 1
while sm > 0:
if a[pos] > 0:
lst.append('P')
sm -= 1
a[pos] -= 1
if pos + turn not in range(len(a)):
turn = -turn
pos += turn
if turn == 1:
lst.append('R')
else:
lst.append('L')
print(''.join(lst[:-1]))
``` | instruction | 0 | 18,896 | 10 | 37,792 |
Yes | output | 1 | 18,896 | 10 | 37,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP
Submitted Solution:
```
x= int(input(""))
mystring=""
g= input("").split(' ')
for t in range (0,len(g)):
if (g[t]=='0'):
if(t==len(g)-1):
mystring=mystring
else:
mystring=mystring+'R'
else:
if (t==0):
for i in range (0, int(g[0])):
mystring=mystring+"RLPR"
elif (t==len(g)-1):
for i in range (0, int(g[len(g)-1])):
mystring=mystring+"LRP"
else:
for i in range (0, int(g[t])):
mystring=mystring+"RLPR"
print(mystring)
``` | instruction | 0 | 18,897 | 10 | 37,794 |
No | output | 1 | 18,897 | 10 | 37,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP
Submitted Solution:
```
n = int(input())
string = input()
coins = list(map(int, string.split()))
s = ""
for x in range(n):
a = coins[x]
if a == n - 1:
s += "PLR" * (a - 1)
else:
s += "PRL" * (a - 1)
if a > 0:
s += "P"
if x < n - 1:
s += "R"
print(s)
``` | instruction | 0 | 18,898 | 10 | 37,796 |
No | output | 1 | 18,898 | 10 | 37,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
for i in range(n-1):
while a[i]:
print('PRL', end='')
a[i] -= 1
while a[-1]:
a[-1] -= 1
print('PLR', end='')
``` | instruction | 0 | 18,899 | 10 | 37,798 |
No | output | 1 | 18,899 | 10 | 37,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The New Year is coming! That's why many people today are busy preparing New Year presents. Vasily the Programmer is no exception.
Vasily knows that the best present is (no, it's not a contest) money. He's put n empty wallets from left to right in a row and decided how much money to put in what wallet. Vasily decided to put ai coins to the i-th wallet from the left.
Vasily is a very busy man, so the money are sorted into the bags by his robot. Initially, the robot stands by the leftmost wallet in the row. The robot can follow instructions of three types: go to the wallet that is to the left of the current one (if such wallet exists), go to the wallet that is to the right of the current one (if such wallet exists), put a coin to the current wallet. Due to some technical malfunctions the robot cannot follow two "put a coin" instructions in a row.
Vasily doesn't want to wait for long, so he wants to write a program for the robot that contains at most 106 operations (not necessarily minimum in length) the robot can use to put coins into the wallets. Help him.
Input
The first line contains integer n (2 ≤ n ≤ 300) — the number of wallets. The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 300).
It is guaranteed that at least one ai is positive.
Output
Print the sequence that consists of k (1 ≤ k ≤ 106) characters, each of them equals: "L", "R" or "P". Each character of the sequence is an instruction to the robot. Character "L" orders to move to the left, character "R" orders to move to the right, character "P" orders the robot to put a coin in the wallet. The robot is not allowed to go beyond the wallet line. In other words, you cannot give instructions "L" if the robot is at wallet 1, or "R" at wallet n.
As a result of the performed operations, the i-th wallet from the left must contain exactly ai coins. If there are multiple answers, you can print any of them.
Examples
Input
2
1 2
Output
PRPLRP
Input
4
0 2 0 2
Output
RPRRPLLPLRRRP
Submitted Solution:
```
n = int(input())
arr1 = input().split()
arr2 = []
seq = ""
x = -1
j = 0
max , min = 0, 0
condition = 0
for i in range(n):
arr1[i] = int(arr1[i])
if(arr1[i] > 0):
max = i
arr2.append(0)
for i in range(n):
if(arr1[i] > 0 and i != 0):
min = i-1
break
while(condition == 0):
if(j == min or j == max):
x *= -1
if(arr1[j] != arr2[j]):
seq+="P"
arr2[j]+=1
j+=x
for z in range(n):
if(arr1[z] != arr2[z]):
if(x == 1):
seq += "R"
else:
seq += "L"
break
elif z == n-1 and arr1[z] == arr2[z]:
condition = 1
print(seq)
``` | instruction | 0 | 18,900 | 10 | 37,800 |
No | output | 1 | 18,900 | 10 | 37,801 |
Provide a correct Python 3 solution for this coding contest problem.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942 | instruction | 0 | 19,234 | 10 | 38,468 |
"Correct Solution:
```
q,h,s,d=map(int,input().split())
n=int(input())
qhs=min(q*4,h*2,s)
if qhs*2<=d:
print(qhs*n)
else:
print(d*(n//2)+qhs*(n%2))
``` | output | 1 | 19,234 | 10 | 38,469 |
Provide a correct Python 3 solution for this coding contest problem.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942 | instruction | 0 | 19,235 | 10 | 38,470 |
"Correct Solution:
```
q, h, s, d=map(int, input().split())
n=int(input())
x1=min(4*q, 2*h, s)
x2=min(2*x1, d)
print((n//2)*x2+(n%2)*x1)
``` | output | 1 | 19,235 | 10 | 38,471 |
Provide a correct Python 3 solution for this coding contest problem.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942 | instruction | 0 | 19,236 | 10 | 38,472 |
"Correct Solution:
```
q, h, s, d = map(int, input().split())
n = int(input())
s = min(2 * min(2 * q, h), s)
print((n // 2) * min(2 * s, d) + (n % 2) * s)
``` | output | 1 | 19,236 | 10 | 38,473 |
Provide a correct Python 3 solution for this coding contest problem.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942 | instruction | 0 | 19,237 | 10 | 38,474 |
"Correct Solution:
```
q, h, s, d = map(int, input().split())
n = int(input())
l1 = min(h*2, q*4, s)
l2 = min(8*q, h*4, s*2, d)
ans = n//2 * l2 + n%2 * l1
print(ans)
``` | output | 1 | 19,237 | 10 | 38,475 |
Provide a correct Python 3 solution for this coding contest problem.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942 | instruction | 0 | 19,238 | 10 | 38,476 |
"Correct Solution:
```
q,h,s,d=map(int,input().split())
n=int(input())
print(n//2*min(q*8,h*4,s*2,d)+min(q*4,h*2,s)*(n%2))
``` | output | 1 | 19,238 | 10 | 38,477 |
Provide a correct Python 3 solution for this coding contest problem.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942 | instruction | 0 | 19,239 | 10 | 38,478 |
"Correct Solution:
```
q, h, s, d = map(int, input().split())
n = int(input())
s = min(q*4, h*2, s)
print(s*n if d>s*2 else d*(n//2)+s*(n%2))
``` | output | 1 | 19,239 | 10 | 38,479 |
Provide a correct Python 3 solution for this coding contest problem.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942 | instruction | 0 | 19,240 | 10 | 38,480 |
"Correct Solution:
```
Q, H, S, D = map(int, input().split())
N = int(input())
ans = (N // 2) * min(Q*8, H*4, S*2, D) + (N % 2) * min(Q*4, H*2, S)
print(ans)
``` | output | 1 | 19,240 | 10 | 38,481 |
Provide a correct Python 3 solution for this coding contest problem.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942 | instruction | 0 | 19,241 | 10 | 38,482 |
"Correct Solution:
```
Q, H, S, D = map(int, input().split())
N = int(input())
H = min(H, 2 * Q)
S = min(S, 2 * H)
D = min(D, 2 * S)
print((N // 2) * D + (N % 2) * S)
``` | output | 1 | 19,241 | 10 | 38,483 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942
Submitted Solution:
```
a,b,c,d=map(int, input().split())
k = int(input())
m = min(a*4,b*2,c)
n = min(m*2,d)
print((k//2)*n+(k%2)*m)
``` | instruction | 0 | 19,242 | 10 | 38,484 |
Yes | output | 1 | 19,242 | 10 | 38,485 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942
Submitted Solution:
```
Q, H, S, D = map(int,input().split())
N = int(input())
R = N%2
M = N//2
print(min([Q*8,H*4,S*2,D])*M+min([Q*4,H*2,S])*R)
``` | instruction | 0 | 19,243 | 10 | 38,486 |
Yes | output | 1 | 19,243 | 10 | 38,487 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942
Submitted Solution:
```
Q,H,S,D = map(int,input().split())
N = int(input())
a2 = min(Q*8,H*4,S*2,D)
ans = (N//2)*a2
a1 = min(Q*4,H*2,S)
ans += (N%2)*a1
print(ans)
``` | instruction | 0 | 19,244 | 10 | 38,488 |
Yes | output | 1 | 19,244 | 10 | 38,489 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942
Submitted Solution:
```
q,h,s,d = map(int,input().split())
n = int(input())
s = min(q*4,h*2,s)
d = min(s*2,d)
print((n//2)*d+(n%2)*s)
``` | instruction | 0 | 19,245 | 10 | 38,490 |
Yes | output | 1 | 19,245 | 10 | 38,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942
Submitted Solution:
```
q, h, s, d = map(int,input().split())
n = int(input())
ans = min(4 * q * n, 2 * h * n, s * n)
ans = min(ans, d * (n // 2) + min(s * (n % 2), h * 2 * (n % 2)))
print(ans)
``` | instruction | 0 | 19,246 | 10 | 38,492 |
No | output | 1 | 19,246 | 10 | 38,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942
Submitted Solution:
```
q, h, s, d = map(int, input().strip().split())
n = int(input())
sorted_list = sorted([(q, 0.25), (h, 0.5), (s, 1.0), (d, 2.0)],
key=lambda t: t[0] / t[1])
cheapest = sorted_list[0]
if cheapest[1] == 2.0:
# 2リットルが最安の場合
second = sorted_list[1]
cost = cheapest[0] * (n // 2) + second[0] * (n % 2) / second[1]
else:
cost = cheapest[0] / cheapest[1] * n
print(int(cost))
``` | instruction | 0 | 19,247 | 10 | 38,494 |
No | output | 1 | 19,247 | 10 | 38,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942
Submitted Solution:
```
#-*-coding:utf-8-*-
def main():
q, h, s, d = map(int, input().split())
n = int(input())
values =[q * 4, h * 2, s, d / 2]
values.sort()
ans = 0
for value in values:
if n == 0:
break
if value == q * 4:
ans += (n // 0.25) * q
n = 0
elif value == h * 2:
ans += (n // 0.5) * h
n = 0
elif value == s:
ans += n * s
n = 0
elif value == d / 2:
ans += (n // 2) * d
n %= 2
print(int(ans))
if __name__ == '__main__':
main()
``` | instruction | 0 | 19,248 | 10 | 38,496 |
No | output | 1 | 19,248 | 10 | 38,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You've come to your favorite store Infinitesco to buy some ice tea.
The store sells ice tea in bottles of different volumes at different costs. Specifically, a 0.25-liter bottle costs Q yen, a 0.5-liter bottle costs H yen, a 1-liter bottle costs S yen, and a 2-liter bottle costs D yen. The store has an infinite supply of bottles of each type.
You want to buy exactly N liters of ice tea. How many yen do you have to spend?
Constraints
* 1 \leq Q, H, S, D \leq 10^8
* 1 \leq N \leq 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
Q H S D
N
Output
Print the smallest number of yen you have to spend to buy exactly N liters of ice tea.
Examples
Input
20 30 70 90
3
Output
150
Input
10000 1000 100 10
1
Output
100
Input
10 100 1000 10000
1
Output
40
Input
12345678 87654321 12345678 87654321
123456789
Output
1524157763907942
Submitted Solution:
```
def main():
q, h, s, d = map(int, input().split())
target = int(input())
value = [(q, q, 0.25), (h, h / 2, 0.5), (s, s / 4, 1.0), (d, d / 8, 2.0)]
value.sort(key=lambda x: x[1])
answer = 0
target_even = target // 2 * 2
for i in range(4):
answer += (target_even // value[i][2]) * value[i][0]
target_even %= value[i][2]
if target_even == 0:
break
if target % 2:
answer += min(q * 4, h * 2, s)
print(int(answer))
if __name__ == '__main__':
main()
``` | instruction | 0 | 19,249 | 10 | 38,498 |
No | output | 1 | 19,249 | 10 | 38,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Aiz, which is located in cyberspace, trades information with Wakamatsu. The two countries are developing their economies by exchanging useful data with each other. The two countries, whose national policy is philanthropy and equality, and above all, the old word of the Aizu region, "what must be done", conducts regular surveys of trade conditions.
In the survey, a table is given in which the value obtained by subtracting the outflow amount from the data inflow amount seen from Aiz country in byte units is calculated every 1 nanosecond. From that table, find the longest interval where the sum of the values is zero. It is judged that the longer this section is, the more equality is maintained.
Given a table with trade status, write a program to find the length of the longest interval where the sum of the values is zero.
Input
The input is given in the following format.
N
d1
d2
::
dN
The first row gives the number N (1 ≤ N ≤ 200000) of the values written in the table. The next N rows are given the integer di (-109 ≤ di ≤ 109), which indicates the value written in row i of the table.
Output
The length of the longest section obtained from the table where the sum is 0 is output in one line. If such an interval does not exist, "0" is output on one line.
Examples
Input
5
18
102
-155
53
32
Output
3
Input
4
1
1
-1
-1
Output
4
Submitted Solution:
```
n = int(input())
d = []
a = 0
for _ in range(n):d.append(int(input()))
v = [[-1] * n for _ in range(n)]
v[0][0] = 0
for i in range(n):
v[0][- i - 1] = v[0][- i] + d[i]
for i in range(1, n):
for j in range(n - i):
v[i][j] = v[i - 1][j] - d[i - 1]
a = 0
for i in range(n):
try:a = max(a,n -v[i].index(0) - i)
except:pass
print(a)
``` | instruction | 0 | 19,316 | 10 | 38,632 |
No | output | 1 | 19,316 | 10 | 38,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Aiz, which is located in cyberspace, trades information with Wakamatsu. The two countries are developing their economies by exchanging useful data with each other. The two countries, whose national policy is philanthropy and equality, and above all, the old word of the Aizu region, "what must be done", conducts regular surveys of trade conditions.
In the survey, a table is given in which the value obtained by subtracting the outflow amount from the data inflow amount seen from Aiz country in byte units is calculated every 1 nanosecond. From that table, find the longest interval where the sum of the values is zero. It is judged that the longer this section is, the more equality is maintained.
Given a table with trade status, write a program to find the length of the longest interval where the sum of the values is zero.
Input
The input is given in the following format.
N
d1
d2
::
dN
The first row gives the number N (1 ≤ N ≤ 200000) of the values written in the table. The next N rows are given the integer di (-109 ≤ di ≤ 109), which indicates the value written in row i of the table.
Output
The length of the longest section obtained from the table where the sum is 0 is output in one line. If such an interval does not exist, "0" is output on one line.
Examples
Input
5
18
102
-155
53
32
Output
3
Input
4
1
1
-1
-1
Output
4
Submitted Solution:
```
n = int(input())
d = []
a = 0
for _ in range(n):d.append(int(input()))
v = [[-1] * n for _ in range(n)]
v[0][0] = 0
for i in range(n):
v[0][- i - 1] = v[0][- i] + d[i]
for i in range(1, n):
for j in range(n - i):
v[i][j + i] = v[i - 1][j] - d[i - 1]
a = 0
for i in range(n):
try:a = max(a,n -v[i].index(0))
except:pass
print(a)
``` | instruction | 0 | 19,317 | 10 | 38,634 |
No | output | 1 | 19,317 | 10 | 38,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 19,821 | 10 | 39,642 |
Tags: binary search, sortings
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq,bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default, func):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
n,k=map(int,input().split())
l=list(map(int,input().split()))
ind=defaultdict(set)
for i in range(n):
for j in range(k+1):
ind[j].add(l[i]*j)
for i in range(int(input())):
inp=int(input())
ans=-1
f=0
for j in range(1,k+1):
for y in range(0,j+1):
for t in ind[y]:
if inp-t in ind[j-y]:
f=1
break
if f==1:
break
if f==1:
ans=j
break
print(ans)
``` | output | 1 | 19,821 | 10 | 39,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 19,822 | 10 | 39,644 |
Tags: binary search, sortings
Correct Solution:
```
n, k = map(int, input().split())
a = set(map(int, input().split()))
q = int(input())
# def isIn(x, fm, to):
# if fm >= to:
# return a[fm] == x
# t = a[(fm+to) // 2]
# if t > x:
# return isIn(x, fm, (fm+to) // 2 - 1)
# elif t < x:
# return isIn(x, (fm+to) // 2 + 1, to)
# else:
# return True
for _ in range(q):
x = int(input())
if x in a:
print(1)
continue
found = False
for i in range(2, k + 1):
for j in range(1, i // 2 + 1):
for l in a:
t = x - l * j
if t % (i - j) != 0:
continue
# if isIn(t // (i - j), 0, n - 1):
if t // (i - j) in a:
print(i)
found = True
break
if found:
break
if found:
break
if not found:
print(-1)
``` | output | 1 | 19,822 | 10 | 39,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 19,823 | 10 | 39,646 |
Tags: binary search, sortings
Correct Solution:
```
f = lambda: map(int, input().split())
n, k = f()
t = list(f())
d = {0: 0}
for q in t:
for i in range(1, k + 1): d[q * i] = i
for j in range(int(input())):
a = int(input())
p = [i + d[a - b] for b, i in d.items() if a - b in d]
print(min(p) if p and min(p) <= k else -1)
``` | output | 1 | 19,823 | 10 | 39,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 19,824 | 10 | 39,648 |
Tags: binary search, sortings
Correct Solution:
```
n_k = input()
n_k = n_k.split(" ")
n = int(n_k[0])
k = int(n_k[1])
ais = input()
ais = ais.split(" ")
q = int(input())
pares = {}
for a in ais:
a = int(a)
for i in range(k):
p = int((i+1)*a)
if (p not in pares) or (i+1 < pares[p]):
pares[p] = i+1
m = 1000000000
for i in range(q):
x = int(input())
ans = 1000;
minimo = m
for plata, deuda in pares.items():
if plata == x:
if deuda <= k:
if deuda < minimo:
minimo = deuda
else:
r = x-plata
if r in pares:
if deuda+pares[r] < minimo:
if deuda + pares[r] <= k:
minimo = deuda+pares[r]
if minimo == m:
print(-1)
else:
print(minimo)
``` | output | 1 | 19,824 | 10 | 39,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 19,825 | 10 | 39,650 |
Tags: binary search, sortings
Correct Solution:
```
n_k = input()
n_k = n_k.split(" ")
n = int(n_k[0])
k = int(n_k[1])
ais = input()
ais = ais.split(" ")
q = int(input())
pares = {}
for a in ais:
a = int(a)
for i in range(k):
p = int((i+1)*a)
if (p not in pares) or (i+1 < pares[p]):
pares[p] = i+1
m = 1000000000
for i in range(q):
x = int(input())
ans = 1000;
minimo = m
for money, bills in pares.items():
if money == x and bills <= k and bills < minimo:
minimo = bills
else:
r = x-money
if r in pares and bills+pares[r] < minimo and bills + pares[r] <= k:
minimo = bills+pares[r]
if minimo == m:
print(-1)
else:
print(minimo)
``` | output | 1 | 19,825 | 10 | 39,651 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 19,826 | 10 | 39,652 |
Tags: binary search, sortings
Correct Solution:
```
f = lambda: map(int, input().split())
n, k = f()
t = list(f())
d = {0: 0}
for q in t:
for i in range(1, k + 1): d[q * i] = i
for j in range(int(input())):
a = int(input())
p = [i + d[a - b] for b, i in d.items() if a - b in d]
print(min(p) if p and min(p) <= k else -1)
# Made By Mostafa_Khaled
``` | output | 1 | 19,826 | 10 | 39,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq,bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default, func):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
n,k=map(int,input().split())
l=list(map(int,input().split()))
ind=defaultdict(set)
for i in range(n):
for j in range(1,k+1):
ind[j].add(l[i]*j)
for i in range(int(input())):
inp=int(input())
ans=-1
f=0
for j in range(1,k+1):
for y in range(0,j+1):
for t in ind[y]:
if inp-t in ind[j-y]:
f=1
break
if f==1:
break
if f==1:
ans=j
break
print(ans)
``` | instruction | 0 | 19,827 | 10 | 39,654 |
No | output | 1 | 19,827 | 10 | 39,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is data that records the customer number of the business partner and the trading date on a monthly basis. Please create a program that reads this month's data and last month's data and outputs the customer number of the company with which you have transactions and the number of transactions for two consecutive months from last month. However, the number of monthly business partners is 1,000 or less.
Input
This month's data and last month's data are given separated by one blank line. Each data is given in the following format.
c1, d1
c2, d2
...
...
ci (1 ≤ ci ≤ 1,000) is an integer representing the customer number, and di (1 ≤ di ≤ 31) is an integer representing the trading day.
Output
For companies that have transactions for two consecutive months, the customer number and the total number of transactions are output separated by a blank in ascending order of customer number.
Example
Input
123,10
56,12
34,14
123,3
56,4
123,5
Output
56 2
123 3
Submitted Solution:
```
month_1={}
month_2={}
answer=[]
while 1:
try:c,d=map(int,input().split(","))
except:break
if c in month_1:month_1[c] +=1
else:month_1[c]=1
while 1:
try:c,d=map(int,input().split(","))
except:break
if c in month_2:month_2[c] +=1
else:month_2[c]=1
for v,m in month_1.items():
if v in month_2:
answer.append([v,month_1[v]+month_2[v]])
answer=sorted(answer)
for i in answer:
print(i[0],i[1])
``` | instruction | 0 | 20,186 | 10 | 40,372 |
Yes | output | 1 | 20,186 | 10 | 40,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is data that records the customer number of the business partner and the trading date on a monthly basis. Please create a program that reads this month's data and last month's data and outputs the customer number of the company with which you have transactions and the number of transactions for two consecutive months from last month. However, the number of monthly business partners is 1,000 or less.
Input
This month's data and last month's data are given separated by one blank line. Each data is given in the following format.
c1, d1
c2, d2
...
...
ci (1 ≤ ci ≤ 1,000) is an integer representing the customer number, and di (1 ≤ di ≤ 31) is an integer representing the trading day.
Output
For companies that have transactions for two consecutive months, the customer number and the total number of transactions are output separated by a blank in ascending order of customer number.
Example
Input
123,10
56,12
34,14
123,3
56,4
123,5
Output
56 2
123 3
Submitted Solution:
```
cus1 = [1 for i in range(1001)]
cus2 = [1 for i in range(1001)]
sum_ = dict(zip([i+1 for i in range(1000)],[0 for i in range(1000)]))
while(1):
input_ = input()
if input_ == "":
break
a,b = [int(i) for i in input_.split(",")]
cus1[a] *= 0
sum_[a] += 1
while(1):
try:
a,b = [int(i) for i in input().split(",")]
cus2[a] *= 0
sum_[a] += 1
except EOFError:
break
for i in range(1,1001):
if cus1[i] + cus2[i] == 0:
print(i,sum_[i])
``` | instruction | 0 | 20,187 | 10 | 40,374 |
Yes | output | 1 | 20,187 | 10 | 40,375 |
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