message stringlengths 2 43.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 853 107k | cluster float64 24 24 | __index_level_0__ int64 1.71k 214k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end. | instruction | 0 | 102,675 | 24 | 205,350 |
Tags: dp
Correct Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop,heapify
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
from itertools import accumulate
from functools import lru_cache
M = mod = 998244353
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n')]
def li3():return [int(i) for i in input().rstrip('\n')]
sys.setrecursionlimit(10 ** 6)
a, n, m = li()
l = []
rain = defaultdict(int)
for i in range(n):
c, b = li()
for j in range(c, b):
rain[(j, j + 1)] = 1
# print(rain)
umbrellas = [float('inf')] * (a + 5)
for i in range(m):
c, b = li()
umbrellas[c] = min(umbrellas[c], b)
# print(umbrellas[:a + 1])
@lru_cache(None)
def dp(i = 0, umbon = 0):
# print(i, umbon)
if i == a:
if rain[(i - 1, i)]:
if umbon:return umbon
return float('inf')
else:return umbon
else:
ans = float('inf')
last = umbon
if rain[(i - 1, i)]:
umbon = min(umbon, umbrellas[i]) if umbon else umbrellas[i]
if not last:last = float('inf')
ans = min(ans, last + dp(i + 1, umbon))
ans = min(ans, last + dp(i + 1, 0))
else:
ans = min(ans, dp(i + 1, 0) + last)
umbon = min(umbon, umbrellas[i]) if umbon else umbrellas[i]
ans = min(ans, dp(i + 1, umbon) + last)
return ans
print(dp(0, 0) if dp(0, 0) != float('inf') else -1)
``` | output | 1 | 102,675 | 24 | 205,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end. | instruction | 0 | 102,676 | 24 | 205,352 |
Tags: dp
Correct Solution:
```
import io
import os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
INF = 10**15
def dp_min(dp, pos, new_val):
if pos not in dp:
dp[pos] = new_val
else:
dp[pos] = min(dp[pos], new_val)
def solve():
a, n, m = list(map(int, input().split()))
has_rain = [False] * a
for _ in range(n):
l, r = map(int, input().split())
for i in range(l, r):
has_rain[i] = True
umbrella = [INF] * a
for _ in range(m):
x, p = map(int, input().split())
if x == a:
continue
umbrella[x] = min(umbrella[x], p)
# dp: best cost so far if I am currently carrying no umbrella (-1), or some umbrella of certain weight
dp = {-1: 0}
for i in range(a):
new_dp = {}
for umbrella_weight, best_cost in dp.items():
if not has_rain[i]:
# carry no umbrella
dp_min(new_dp, -1, best_cost)
# take the new umbrella
dp_min(new_dp, umbrella[i], best_cost + umbrella[i])
if umbrella_weight != -1:
# continue same umbrella
dp_min(new_dp, umbrella_weight, best_cost + umbrella_weight)
dp = new_dp
best = min(dp.values())
if best >= INF:
print(-1)
else:
print(best)
t = 1
for _ in range(t):
solve()
``` | output | 1 | 102,676 | 24 | 205,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end. | instruction | 0 | 102,677 | 24 | 205,354 |
Tags: dp
Correct Solution:
```
import sys
a, n, m = map(int, input().split(' '))
seg = []
for i in range(n):
rained = tuple(map(int, input().split(' ')))
for k in range(rained[0], rained[1]):
seg.append(k+1)
umbrella = []
for j in range(m):
u = tuple(map(int, input().split(' ')))
umbrella.append(u)
memo = [0] * (a+1)
umbrella = sorted(umbrella, key=lambda x: x[0])
if umbrella[0][0] > seg[0] - 1:
print(-1)
sys.exit(0)
for index in range(1, len(memo)):
if index not in seg:
memo[index] = memo[index-1]
continue
for each in umbrella:
if index >= each[0]:
cur = (index - each[0]) * each[1] + memo[each[0]]
if memo[index] > 0:
if cur < memo[index]:
memo[index] = cur
else:
memo[index] = cur
print(memo[-1])
``` | output | 1 | 102,677 | 24 | 205,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end. | instruction | 0 | 102,678 | 24 | 205,356 |
Tags: dp
Correct Solution:
```
# Codeforces Round #486 (Div. 3)
from functools import cmp_to_key
#key=cmp_to_key(lambda x,y: 1 if x not in y else -1 )
import sys
def getIntList():
return list(map(int, input().split()))
a,n,m = getIntList()
rainend = set()
umbr = {}
keyPointSet = set([0,a])
for i in range(n):
t =tuple(getIntList())
for j in range(t[0] + 1, t[1] + 1):
rainend.add(j)
keyPointSet.add(t[1])
for i in range(m):
t =getIntList()
if t[0] not in umbr or t[1]< umbr[t[0]]:
umbr[t[0]] = t[1]
keyPointSet.add(t[0])
keyPoint = list(keyPointSet)
keyPoint.sort()
dp = {}
dp[0] = {}
dp[0] [0] = 0
if 0 in umbr:
dp[0][umbr[0]] = 0
for i in range(1, len(keyPoint)):
x = keyPoint[i]
lx = keyPoint[i-1]
ifrain = x in rainend
dp[x] = {}
nowdp = dp[x]
lastdp = dp[lx]
for z in lastdp:
if z == 0 :
if not ifrain:
nowdp[0] = lastdp[0]
else:
nowdp[z] = lastdp[z] + z * (x-lx)
if len(nowdp) >0:
nowdp[0] = min(nowdp.values())
if x in umbr:
if umbr[x] not in nowdp or nowdp[0] < nowdp[umbr[x]]:
nowdp[umbr[x]] = nowdp[0]
else:
print(-1)
sys.exit()
print( min(dp[a].values()) )
``` | output | 1 | 102,678 | 24 | 205,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end. | instruction | 0 | 102,679 | 24 | 205,358 |
Tags: dp
Correct Solution:
```
import sys
a,m,n=list(map(int,input().split()))
aux=[0]*(a+1)
inf=10**15
dp=[aux.copy() for i in range(n+1)]
m1=10**12
m2=10**12
for i in range(m):
l,r=list(map(int,input().split()))
if l<m1:
m1=l
for j in range(l,r):
dp[0][j+1]=inf
s=[]
for i in range(1,n+1):
x,w=list(map(int,input().split()))
s.append(tuple([x,w]))
if x<m2:
m2=x
if m2>m1:
print(-1)
sys.exit()
s.sort()
for i in range(1,n+1):
x=s[i-1][0]
w=s[i-1][1]
for j in range(x+1):
dp[i][j]=dp[i-1][j]
for j in range(x+1,a+1):
if i!=1:
dp[i][j]=min(dp[0][j]+dp[i][j-1],dp[i-1][j],w*(j-x)+dp[i][x])
else:
dp[i][j]=min(dp[0][j]+dp[i][j-1],w*(j-x)+dp[i][x])
ans=dp[-1][-1]
if ans>=inf:
print(-1)
else:
print(ans)
``` | output | 1 | 102,679 | 24 | 205,359 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end. | instruction | 0 | 102,680 | 24 | 205,360 |
Tags: dp
Correct Solution:
```
import sys
def input():
return sys.stdin.readline().strip()
def input_l():
return map(int, input().split())
def input_t():
return tuple(input_l())
def main():
a, s, d = input_l()
q = []
e = []
z = [0] * (a + 1)
for i in range(s):
w = input_t()
for k in range(w[0], w[1]):
q.append(k + 1)
for j in range(d):
e.append(input_t())
e = sorted(e, key = lambda x: x[0])
if e[0][0] > q[0] - 1:
print(-1)
sys.exit(0)
for i in range(1, len(z)):
if i not in q:
z[i] = z[i-1]
continue
for j in e:
if i >= j[0]:
c = (i - j[0]) * j[1] + z[j[0]]
if z[i] > 0:
if c < z[i]:
z[i] = c
else:
z[i] = c
print(z[-1])
main()
``` | output | 1 | 102,680 | 24 | 205,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
import math
def main():
a, n, m = map(int, input().split())
rain = []
u = []
w = []
raining = [False] * (a+1)
for i in range(n):
l, r = map(int, input().split())
rain.append((l, r))
for j in range(l, r):
raining[j] = True
for i in range(m):
x, y = map(int, input().split())
u.append(x)
w.append(y)
rain_int = [0] * a
for i in range(n):
rain_int[rain[i][0]-1] = 1
rain_int[rain[i][1]-1] = -1
umbrellas = [-1 for _ in range(a+1)]
for i, x in enumerate(u):
if umbrellas[x] == -1 or w[umbrellas[x]] > w[i]:
umbrellas[x] = i
dp = [[math.inf for _ in range(m+1)] for _ in range(a+1)]
dp[0][m] = 0
for i in range(a):
for j in range(m+1):
if dp[i][j] == math.inf:
continue
if not raining[i]:
dp[i+1][m] = min(dp[i+1][m], dp[i][j])
if j < m:
dp[i+1][j] = min(dp[i+1][j], dp[i][j] + w[j])
if umbrellas[i] != -1:
dp[i+1][umbrellas[i]] = min(dp[i+1][umbrellas[i]], dp[i][j]+w[umbrellas[i]])
ans = math.inf
for i in range(m+1):
ans = min(ans, dp[a][i])
print(-1 if ans == math.inf else ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 102,681 | 24 | 205,362 |
Yes | output | 1 | 102,681 | 24 | 205,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
#!/usr/bin/env pypy
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
MOD = 10**9 + 7
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
a, s, d = map(int, input().split())
q = []
e = []
z = [0] * (a + 1)
for i in range(s):
w = tuple(map(int, input().split()))
for k in range(w[0], w[1]):
q.append(k + 1)
for j in range(d):
e.append(tuple(map(int, input().split())))
e = sorted(e, key = lambda x: x[0])
# ~ print(e)
if e[0][0] > q[0] - 1:
print(-1)
exit()
for i in range(1, a +1):
if i not in q:
z[i] = z[i-1]
continue
for j in e:
if i >= j[0]:
c = (i - j[0]) * j[1] + z[j[0]]
if z[i] > 0:
if c < z[i]:
z[i] = c
else:
z[i] = c
print(z[-1])
``` | instruction | 0 | 102,682 | 24 | 205,364 |
Yes | output | 1 | 102,682 | 24 | 205,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
rd = lambda: map(int, input().split())
a, n, m = rd()
s = set()
u = {}
k = set([0, a])
for _ in range(n):
l, r = rd()
for x in range(l + 1, r + 1):
s.add(x)
k.add(r)
for _ in range(m):
x, p = rd()
u[x] = min(p, u.get(x, 1e9))
k.add(x)
k = sorted(list(k))
dp = {}
dp[0] = {}
dp[0][0] = 0
if 0 in u:
dp[0][u[0]] = 0
for i in range(1, len(k)):
x = k[i]
y = k[i - 1]
dp[x] = {}
for z in dp[y]:
if z:
dp[x][z] = dp[y][z] + z * (x - y)
else:
if x not in s:
dp[x][0] = dp[y][0]
if len(dp[x]):
dp[x][0] = min(dp[x].values())
if x in u:
if u[x] not in dp[x] or dp[x][0] < dp[x][u[x]]:
dp[x][u[x]] = dp[x][0]
else:
print(-1)
exit()
print(dp[a][0])
``` | instruction | 0 | 102,683 | 24 | 205,366 |
Yes | output | 1 | 102,683 | 24 | 205,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
a,m,n=list(map(int,input().split()))
aux=[0]*(a+1)
inf=10**15
dp=[aux.copy() for i in range(n+1)]
for i in range(m):
l,r=list(map(int,input().split()))
for j in range(l,r):
dp[0][j+1]=inf
s=[]
for i in range(1,n+1):
x,w=list(map(int,input().split()))
s.append(tuple([x,w]))
s.sort()
for i in range(1,n+1):
x=s[i-1][0]
w=s[i-1][1]
for j in range(x+1):
dp[i][j]=dp[i-1][j]
for j in range(x+1,a+1):
if i!=1:
dp[i][j]=min(dp[0][j]+dp[i][j-1],dp[i-1][j],w*(j-x)+dp[i][x])
else:
dp[i][j]=min(dp[0][j]+dp[i][j-1],w*(j-x)+dp[i][x])
ans=dp[-1][-1]
if ans>=inf:
print(-1)
else:
print(ans)
``` | instruction | 0 | 102,684 | 24 | 205,368 |
No | output | 1 | 102,684 | 24 | 205,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
#!python3
# http://codeforces.com/contest/988/problem/F
from bisect import bisect_left as bl
a, n, m = [int(i) for i in input().strip().split(' ')]
R = []
for i in range(n):
l, r = [int(j) for j in input().strip().split(' ')]
inx = bl(R, l)
R.insert(inx, l)
R.insert(inx+1, r)
A = []
for i in range(m):
A += [tuple(int(j) for j in input().strip().split(' '))]
A = sorted(A)
class S: # Solution
def __init__(self, a, t):
self.t = t # tiredness
self.a = a # ambrella
def __repr__(self):
return "({} {})".format(self.a, self.t)
O = [S(0,0)] # no ambrellas, not tired
x = 0 # starts at 0
x_prev = 0
is_rain = False
optimal = None
while x <= a:
# check if rain:
rain_inx = bl(R, x)
if rain_inx < len(R):
rain_coord = R[rain_inx]
if x == rain_coord:
is_rain = not is_rain
# update tiredness in solutions:
distance = x - x_prev
optimal = 9*999
for o in O:
amb = o.a
if amb > 0:
p = A[amb-1][1]
o.t += p*distance
if o.t < optimal:
optimal = o.t
# pick up new ambrellas:
k = bl(A, (x, 0)) # inx of next ambrellas or one at x
if k < len(A): # have some here or next
while k < len(A) and A[k][0] == x: # every ambrella at x
amb = k + 1
O += [S(amb, optimal)]
k += 1
assert k >= len(A) or A[k][0] != x, "k should be next amb or none"
# drop solution without ambrella if rains:
index_list = [i for i, el in enumerate(O) if el.a == 0]
s_inx = index_list[0] if len(index_list) > 0 else None
if is_rain and s_inx is not None: # can not go without ambrella
assert O[s_inx].a == 0
del O[s_inx]
elif not is_rain and s_inx is None: # can go without one
O.append(S(0, optimal))
if not any(O) and x != a:
optimal = -1
break
# get next x:
next_x = a
if rain_inx < len(R):
if x < R[rain_inx]:
next_x = R[rain_inx]
elif rain_inx + 1 < len(R):
next_x = R[rain_inx+1]
if k < len(A) and A[k][0] < next_x:
next_x = A[k][0]
if x == next_x:
break
x_prev = x
x = next_x
print(optimal)
``` | instruction | 0 | 102,685 | 24 | 205,370 |
No | output | 1 | 102,685 | 24 | 205,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
def main():
a, n, m = map(int, input().split())
rain = [False] * (a + 1)
for _ in range(n):
x, y = map(int, input().split())
for i in range(x, y + 1):
rain[i] = True
umrella = [0] * (a + 1)
for _ in range(m):
x, y = map(int, input().split())
if umrella[x] == 0:
umrella[x] = y
else:
umrella[x] = min(umrella[x], y)
max_value = 10 ** 18
dp = [max_value] * (a + 1)
ok = True
if not rain[0] or umrella[0] > 0:
dp[0] = 0
else:
ok = False
if ok:
for i in range(1, a + 1):
if not rain[i]:
dp[i] = dp[i - 1]
else:
for j in range(0, i + 1):
if umrella[j] > 0:
if j >= 1:
dp[i] = min(dp[i], dp[j - 1] + umrella[j] * (i - j))
else:
dp[i] = min(dp[i], umrella[j] * (i - j))
if dp[a] == max_value:
print(-1)
else:
print(dp[a])
else:
print(-1)
if __name__ == "__main__":
main()
``` | instruction | 0 | 102,686 | 24 | 205,372 |
No | output | 1 | 102,686 | 24 | 205,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp lives on a coordinate line at the point x = 0. He goes to his friend that lives at the point x = a. Polycarp can move only from left to right, he can pass one unit of length each second.
Now it's raining, so some segments of his way are in the rain. Formally, it's raining on n non-intersecting segments, the i-th segment which is in the rain is represented as [l_i, r_i] (0 β€ l_i < r_i β€ a).
There are m umbrellas lying on the line, the i-th umbrella is located at point x_i (0 β€ x_i β€ a) and has weight p_i. When Polycarp begins his journey, he doesn't have any umbrellas.
During his journey from x = 0 to x = a Polycarp can pick up and throw away umbrellas. Polycarp picks up and throws down any umbrella instantly. He can carry any number of umbrellas at any moment of time. Because Polycarp doesn't want to get wet, he must carry at least one umbrella while he moves from x to x + 1 if a segment [x, x + 1] is in the rain (i.e. if there exists some i such that l_i β€ x and x + 1 β€ r_i).
The condition above is the only requirement. For example, it is possible to go without any umbrellas to a point where some rain segment starts, pick up an umbrella at this point and move along with an umbrella. Polycarp can swap umbrellas while he is in the rain.
Each unit of length passed increases Polycarp's fatigue by the sum of the weights of umbrellas he carries while moving.
Can Polycarp make his way from point x = 0 to point x = a? If yes, find the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Input
The first line contains three integers a, n and m (1 β€ a, m β€ 2000, 1 β€ n β€ βa/2β) β the point at which Polycarp's friend lives, the number of the segments in the rain and the number of umbrellas.
Each of the next n lines contains two integers l_i and r_i (0 β€ l_i < r_i β€ a) β the borders of the i-th segment under rain. It is guaranteed that there is no pair of intersecting segments. In other words, for each pair of segments i and j either r_i < l_j or r_j < l_i.
Each of the next m lines contains two integers x_i and p_i (0 β€ x_i β€ a, 1 β€ p_i β€ 10^5) β the location and the weight of the i-th umbrella.
Output
Print "-1" (without quotes) if Polycarp can't make his way from point x = 0 to point x = a. Otherwise print one integer β the minimum total fatigue after reaching x = a, if Polycarp picks up and throws away umbrellas optimally.
Examples
Input
10 2 4
3 7
8 10
0 10
3 4
8 1
1 2
Output
14
Input
10 1 1
0 9
0 5
Output
45
Input
10 1 1
0 9
1 5
Output
-1
Note
In the first example the only possible strategy is to take the fourth umbrella at the point x = 1, keep it till the point x = 7 (the total fatigue at x = 7 will be equal to 12), throw it away, move on from x = 7 to x = 8 without an umbrella, take the third umbrella at x = 8 and keep it till the end (the total fatigue at x = 10 will be equal to 14).
In the second example the only possible strategy is to take the first umbrella, move with it till the point x = 9, throw it away and proceed without an umbrella till the end.
Submitted Solution:
```
s=input()[::-1]
m=I=41
f=s.find('5')+1
i=s.find('0')+1
t=len(s)
if i:
j=min(s.find('0',i)+1or I,f or I)
if j<I:
m=i+j-3
if j<i:m+=1
if f:
j=min(s.find('2')+1or I,s.find('7')+1or I)
if j<I:
l=f+j-3
if j<f:l+=1
if f==t:
i=t-1
while i==j or s[i-1]=='0':
if i!=j:l+=1
i-=1
m=min(m,l)
print((-1,m)[m<I])
``` | instruction | 0 | 102,687 | 24 | 205,374 |
No | output | 1 | 102,687 | 24 | 205,375 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course. | instruction | 0 | 103,362 | 24 | 206,724 |
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
cycle = False
def DFS(s):
global cycle
visited[s] = 1
stack = [s]
while len(stack) != 0:
u = stack[-1]
visited[u] = 1
if len(graph[u]) != 0:
v = graph[u].pop()
if visited[v] == 1:
cycle = True
return
if visited[v] == 2:
continue
stack.append(v)
else:
result.append(u)
stack.pop()
visited[u] = 2
def TopoSort(graph, result, visited):
for i in range (m):
if not visited[requiredCourse[i]]:
DFS(requiredCourse[i])
return result
n, m = map(int, input().split())
requiredCourse = list(map(int, input().split()))
visited = [0 for i in range (n + 1)]
graph = [[] for i in range (n + 1)]
result = []
for i in range (1, n + 1):
tmp = list(map(int, input().split()))
if tmp[0] == 0:
continue
for j in range (1, tmp[0] + 1):
graph[i].append(tmp[j])
res = TopoSort(graph, result, visited)
if cycle:
print(-1)
else:
print(len(res))
for i in res:
print(i, end = " ")
``` | output | 1 | 103,362 | 24 | 206,725 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course. | instruction | 0 | 103,363 | 24 | 206,726 |
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
# https://codeforces.com/problemset/problem/770/C
n, k = map(int, input().split())
K = set(list(map(int, input().split())))
g = {}
rg = {}
deg = {}
def push_d(deg, u, val):
if u not in deg:
deg[u] = 0
deg[u] += val
def push_g(g, u, v):
if u not in g:
g[u] = []
g[u].append(v)
for u in range(1, n+1):
list_v = list(map(int, input().split()))[1:]
deg[u] = 0
for v in list_v:
push_d(deg, u, 1)
push_g(g, v, u)
push_g(rg, u, v)
S = [x for x in K]
used = [0] * (n+1)
i = 0
while i<len(S):
u = S[i]
if u in rg:
for v in rg[u]:
if used[v] == 0:
used[v] = 1
S.append(v)
i+=1
S = {x:1 for x in S}
deg0 = [x for x in S if deg[x]==0]
ans = []
def process(g, deg, deg0, u):
if u in g:
for v in g[u]:
if v in S:
push_d(deg, v, -1)
if deg[v] == 0:
deg0.append(v)
while len(deg0) > 0 and len(K) > 0:
u = deg0.pop()
ans.append(u)
if u in K:
K.remove(u)
process(g, deg, deg0, u)
if len(K) > 0:
print(-1)
else:
print(len(ans))
print(' '.join([str(x) for x in ans]))
#6 2
#5 6
#0
#1 1
#1 4 5
#2 2 1
#1 4
#2 5 3
``` | output | 1 | 103,363 | 24 | 206,727 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course. | instruction | 0 | 103,364 | 24 | 206,728 |
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
def dfs(start_node, edges, colors, result):
stack = [start_node]
while stack:
current_node = stack[-1]
if colors[current_node] == 2:
stack.pop()
continue
colors[current_node] = 1
children = edges[current_node]
if not children:
colors[current_node] = 2
result.append(stack.pop())
else:
child = children.pop()
if colors[child] == 1:
return False
stack.append(child)
return True
def find_courses_sequence(member_of_node, find_nodes, edges):
colors = [0] * member_of_node
result = []
for node in find_nodes:
if not dfs(node, edges, colors, result):
return []
return result
if __name__ == '__main__':
n, k = map(int, input().split())
main_courses = [int(c)-1 for c in input().split()]
courses = dict()
for index in range(n):
courses[index] = [int(d)-1 for d in input().split()[1:]]
result = find_courses_sequence(n, main_courses, courses)
if result:
print(len(result))
for v in result:
print(v+1, end=" ")
else:
print(-1)
``` | output | 1 | 103,364 | 24 | 206,729 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course. | instruction | 0 | 103,365 | 24 | 206,730 |
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
n,k=list(map(lambda x: int(x), input().split()))
m=list(map(lambda x: int(x), input().split()))
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
class Graph:
def __init__(self, V):
self.V = V
self.adj = [[] for i in range(V)]
@bootstrap
def DFSUtil(self, temp, v, visited):
visited[v] = True
for i in self.adj[v]:
if visited[i] == False:
yield self.DFSUtil(temp, i, visited)
temp.append(v)
yield temp
def addEdge(self, v, w):
self.adj[v].append(w)
# self.adj[w].append(v)
@bootstrap
def isCyclicUtil(self, v, visited, recStack):
# Mark current node as visited and
# adds to recursion stack
visited[v] = True
recStack[v] = True
# Recur for all neighbours
# if any neighbour is visited and in
# recStack then graph is cyclic
for neighbour in self.adj[v]:
if visited[neighbour] == False:
ans =yield self.isCyclicUtil(neighbour, visited, recStack)
if ans == True:
yield True
elif recStack[neighbour] == True:
yield True
# The node needs to be poped from
# recursion stack before function ends
recStack[v] = False
yield False
# Returns true if graph is cyclic else false
def isCyclic(self,nodes):
visited = [False] * self.V
recStack = [False] * self.V
for node in nodes:
if visited[node] == False:
if self.isCyclicUtil(node, visited, recStack) == True:
return True
return False
G=Graph(n)
for i in range(0,n):
x=list(map(lambda x: int(x), input().split()))
if x[0]==0:
continue
else:
for k in range(1,x[0]+1):
G.addEdge(i,x[k]-1)
visited=[False for _ in range(n)]
path=[]
# print(G.adj)
for subj in m:
temp = []
if visited[subj-1]==False:
G.DFSUtil(temp,subj-1,visited)
path.extend(temp)
if G.isCyclic([x-1 for x in m]):
print(-1)
else:
print(len(path))
for p in path:
print(p+1,end=" ")
print()
``` | output | 1 | 103,365 | 24 | 206,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course. | instruction | 0 | 103,366 | 24 | 206,732 |
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
import sys
def main():
n,k = map(int,sys.stdin.readline().split())
courses = list(map(int,sys.stdin.readline().split()))
courses = [x-1 for x in courses]
visited = [False]*n
used = [False]*n
ans = []
t = []
for i in range(n):
temp = list(map(int,sys.stdin.readline().split()))
temp = [x-1 for x in temp]
t.append(temp[1:])
for i in range(k):
c = courses[i]
if used[c]:
continue
q = [c]
visited[c]=True
while len(q)>0:
cur = q[-1]
if len(t[cur])!=0:
s = t[cur].pop()
if visited[s] and not used[s]:
print(-1)
return
if used[s]:
continue
q.append(s)
visited[s]=True
else:
ans.append(cur)
q.pop()
used[cur] = True
ans = [str(x+1) for x in ans]
print(len(ans))
print(" ".join(ans))
main()
``` | output | 1 | 103,366 | 24 | 206,733 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course. | instruction | 0 | 103,367 | 24 | 206,734 |
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
import collections as col
import itertools as its
import sys
import operator
from copy import copy, deepcopy
class Solver:
def __init__(self):
pass
def solve(self):
n, k = map(int, input().split())
q = list(map(lambda x: int(x) - 1, input().split()))
used = [False] * n
for e in q:
used[e] = True
edges = [[] for _ in range(n)]
redges = [[] for _ in range(n)]
for i in range(n):
l = list(map(lambda x: int(x) - 1, input().split()))[1:]
edges[i] = l
for e in l:
redges[e].append(i)
degs = [len(edges[i]) for i in range(n)]
d = 0
while d < len(q):
v = q[d]
d += 1
for e in edges[v]:
if not used[e]:
used[e] = True
q.append(e)
q = q[::-1]
nq = []
for v in q:
if degs[v] == 0:
nq.append(v)
d = 0
while d < len(nq):
v = nq[d]
d += 1
for e in redges[v]:
if not used[e]:
continue
degs[e] -= 1
if degs[e] == 0:
nq.append(e)
#print(nq)
if len(q) != len(nq):
print(-1)
return
print(len(nq))
print(' '.join(map(lambda x: str(x + 1), nq)))
if __name__ == '__main__':
s = Solver()
s.solve()
``` | output | 1 | 103,367 | 24 | 206,735 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course. | instruction | 0 | 103,368 | 24 | 206,736 |
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
f = lambda: map(int, input().split())
g = lambda: [int(q) - 1 for q in f()]
class T:
def __init__(s, i):
s.i, s.t = i, g()[1:]
s.a = s.q = 0
n, k = f()
d = g()
p = [T(i) for i in range(n)]
s = []
while d:
x = p[d.pop()]
if x.a: continue
d.append(x.i)
q = 1
for i in x.t:
y = p[i]
if y.a: continue
d.append(y.i)
q = 0
if q:
s.append(x.i + 1)
x.a = 1
elif x.q:
print(-1)
exit()
else: x.q = 1
print(len(s), *s)
``` | output | 1 | 103,368 | 24 | 206,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course. | instruction | 0 | 103,369 | 24 | 206,738 |
Tags: *special, dfs and similar, graphs, implementation
Correct Solution:
```
from sys import *
f = lambda: list(map(int, stdin.readline().split()))
class T:
def __init__(self, i):
self.i, self.t = i, f()[1:]
self.a = self.q = 0
n, k = f()
d = f()
p = [None] + [T(i + 1) for i in range(n)]
s = []
while d:
x = p[d.pop()]
if x.a: continue
d.append(x.i)
q = 1
for i in x.t:
y = p[i]
if y.a: continue
d.append(y.i)
q = 0
if q:
s.append(x.i)
x.a = 1
elif x.q:
print(-1)
exit()
else: x.q = 1
print(len(s), *s)
``` | output | 1 | 103,369 | 24 | 206,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 8)
cycle = False
def DFS(i, graph, result, visited):
global cycle
stack = [i]
while len(stack) != 0:
u = stack[-1]
visited[u] = 1
if len(graph[u]) != 0:
v = graph[u].pop()
if visited[v] == 1:
cycle = True
return
if visited[v] == 2:
continue
stack.append(v)
visited[u] = 1
else:
result.append(u)
stack.pop()
visited[u] = 2
def TopoSort(graph, result):
for i in range(m):
if not visited[requiredCourse[i]]:
DFS(requiredCourse[i], graph, result, visited)
return result
n, m = map(int, input().split())
requiredCourse = list(map(int, input().split()))
graph = [[] for i in range(n + 1)]
result = []
visited = [False for i in range(n + 1)]
for i in range(1, n + 1):
tmp = list(map(int, input().split()))
if tmp[0] == 0:
continue
for j in range(1, tmp[0] + 1):
graph[i].append(tmp[j])
res = TopoSort(graph, result)
if cycle == True:
print(-1)
else:
print(len(res))
for i in res:
print(i, end=" ")
``` | instruction | 0 | 103,370 | 24 | 206,740 |
Yes | output | 1 | 103,370 | 24 | 206,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
'''import sys
flag=True
sys.setrecursionlimit(2000000)
c=[];st=[];
def topo(s):#Traversing the array and storing the vertices
global c,st,flag;
c[s]=1; #Being Visited
for i in adjli[s]:#visiting neighbors
if c[i]==0:
topo(i)
if c[i]==1:
flag=False# If Back Edge , Then Not Possible
st.append(str(s))
c[s]=2 # Visited
try:
n,k=map(int,input().split(' '))#Number Of Courses,Dependencies
main=list(map(int,input().split(' ')))#Main Dependencies
depen=[]#Dependencies List
for i in range(n):
depen.append(list(map(int,input().split(' ')))[1:]);c.append(0)#Append Input To Dependencies List, Marking Visited as 0(False)
c.append(0)
adjli=[]
adjli.append(main)#Assuming Main Course at index 0 with dependencies as Main Dependency(main)
for i in range(len(depen)):
adjli.append(depen[i])#Appending Other Dependencies
topo(0)#TopoLogical Sort Order
st.pop(-1)#popping the assumed Main Couse
if flag:# IF possible then print
print(len(st))
print(' '.join(st))
else:
print(-1)
except Exception as e:
print(e,"error")'''
import sys
flag=True
sys.setrecursionlimit(2000000000)
c=[];st=[];
cur_adj=[]
def topo(s):#Traversing the array and storing the vertices
global c,st,flag;
stack = [s]
while(stack):
s = stack[-1]
c[s]=1; #Being Visited
if(cur_adj[s] < len(adjli[s])):
cur = adjli[s][cur_adj[s]]
if(c[cur]==0):
stack.append(cur)
if(c[cur]==1):
flag=False# If Back Edge , Then Not Possible
cur_adj[s]+=1
else:
c[s]=2
st.append(str(s))
del stack[-1]
try:
n,k=map(int,input().split(' '))
main=list(map(int,input().split(' ')))
depen=[]
for i in range(n):
depen.append(list(map(int,input().split(' ')))[1:]);c.append(0)
cur_adj.append(0)
c.append(0)
cur_adj.append(0)
adjli=[]
adjli.append(main)#Assuming Main Course at index 0 with dependencies as Main Dependency(main)
for i in range(len(depen)):
adjli.append(depen[i])#Appending Other Dependencies
topo(0)#TopoLogical Sort Order
st.pop(-1)#popping the assumed Main Couse
if flag:# IF possible then print
print(len(st))
print(' '.join(st))
else:
print(-1)
except Exception as e:
print(e,"error")
``` | instruction | 0 | 103,371 | 24 | 206,742 |
Yes | output | 1 | 103,371 | 24 | 206,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
import sys
flag=True
sys.setrecursionlimit(2000000000)
c=[];st=[];
cur_adj=[]
def topo(s):#Traversing the array and storing the vertices
global c,st,flag;
stack = [s]
while(stack):
s = stack[-1]
c[s]=1; #Being Visited
if(cur_adj[s] < len(adjli[s])):
cur = adjli[s][cur_adj[s]]
if(c[cur]==0):
stack.append(cur)
if(c[cur]==1):
flag=False# If Back Edge , Then Not Possible
cur_adj[s]+=1
else:
c[s]=2
st.append(str(s))
del stack[-1]
try:
n,k=map(int,input().split(' '))
main=list(map(int,input().split(' ')))
depen=[]
for i in range(n):
depen.append(list(map(int,input().split(' ')))[1:]);c.append(0)
cur_adj.append(0)
c.append(0)
cur_adj.append(0)
adjli=[]
adjli.append(main)#Assuming Main Course at index 0 with dependencies as Main Dependency(main)
for i in range(len(depen)):
adjli.append(depen[i])#Appending Other Dependencies
topo(0)#TopoLogical Sort Order
st.pop(-1)#popping the assumed Main Couse
if flag:# IF possible then print
print(len(st))
print(' '.join(st))
else:
print(-1)
except Exception as e:
print(e,"error")
``` | instruction | 0 | 103,372 | 24 | 206,744 |
Yes | output | 1 | 103,372 | 24 | 206,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
#This code is dedicated to Vlada S.
class Course:
def __init__(self, reqs, number):
self.reqs = list(map(int, reqs.split()[1:]))
self.available = False
self.in_stack = False
self.number = number
n, k = list(map(int, input().split()))
requirements = list(map(int, input().split()))
courses = {}
answer = ""
for i in range(n):
courses[i + 1]= Course(input(), i + 1)
for i in range(len(requirements)):
requirements[i] = courses[requirements[i]]
while requirements:
data = {}
course = requirements.pop()
if not course.available:
requirements.append(course)
done = True
for c in course.reqs:
c = courses[c]
if not c.available:
requirements.append(c)
done = False
if done:
answer += " " + str(course.number)
course.available = True
else:
if course.in_stack:
print(-1)
break
course.in_stack = True
else:
print(answer.count(" "))
print(answer[1:])
# Made By Mostafa_Khaled
``` | instruction | 0 | 103,373 | 24 | 206,746 |
Yes | output | 1 | 103,373 | 24 | 206,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
putninja=[]
courses=[]
B=True
def mff(A, num):
global putninja
global courses
if A[0]=='0':
putninja.append(num)
courses[int(num)-1][0]='stop'
elif A[0]=='stop':
pass
else:
for j in range(int(A[0])):
mff(courses[int(A[j+1])-1],A[j+1])
putninja.append(num)
courses[int(num)-1][0]='stop'
kn=input().split()
k=int(kn[0])
n=int(kn[1])
mk=input().split()
for i in range(k):
s=input()
courses.append(s.split(' '))
for i in range(n):
try:
mff(courses[int(mk[i])-1],mk[i])
except:
B=False
break
if B:
print(len(putninja))
print(' '.join(putninja))
else:
print(-1)
``` | instruction | 0 | 103,374 | 24 | 206,748 |
No | output | 1 | 103,374 | 24 | 206,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
from sys import stdin, stdout
n,k = stdin.readline().split()
n = int(n)
k = int(k)
toLearn = [-1] * n
kList = input().split()
for i in range(len(kList)):
kList[i] = int(kList[i])
nList = []
flag = True
for i in range(n):
nL = stdin.readline().split()
nL.pop(0)
for j in range(len(nL)):
nL[j] = int(nL[j])
if len(nL) == 0:
flag = False
nList.append(nL)
# n = 100000
# k = 1
# kList = [100000]
# nList =[[]]
# flag = False
# for i in range(1, n):
# nList.append([i])
res = []
if k == 0:
print(0)
print("")
elif flag:
print(-1)
else:
res = []
notCircle = True
current = 0
while len(kList) > 0 and notCircle:
temp = []
for i in kList:
res.append(i)
toLearn[i - 1] = current
for j in nList[i - 1]:
if toLearn[j - 1] >= 0:
notCircle = False
break
break
if toLearn[j - 1] == -1:
temp.append(j)
kList = temp
current += 1
if notCircle:
res.reverse()
s = ""
for i in res:
s += str(i) + " "
stdout.write(str(len(res)) + '\n')
stdout.write(s[: -1] + '\n')
else:
stdout.write('-1\n')
``` | instruction | 0 | 103,375 | 24 | 206,750 |
No | output | 1 | 103,375 | 24 | 206,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
itog = ''
result = []
iskl = 0
control = 0
maxway = -1
wayStolb = 0
table = []
fs = input()
inp = fs.split(' ')
nk = [int(i) for i in inp]
ss = input()
inp = ss.split(' ')
maincour = [int(i) for i in inp]
for i in range(0,nk[0]):
table.append([])
s = input()
inp = s.split(' ')
t = [int(i) for i in inp]
for j in range(0,t[0]+1):
table[i].append(t[j])
for i in range (0,len(maincour)):
#print(table[maincour[i]-1])
if table[maincour[i]-1][0] > maxway:
wayStolb = maincour[i] - 1
maxway = table[maincour[i]-1][0]
#print(wayStolb)
#print(maxway)
e = maxway
stolb = wayStolb #4
startStolb = stolb
while control != len(maincour) :
while table[wayStolb][0] != 0:
maxway = -1
e = table[stolb][0]
for i in range(1,e+1):
if table[table[stolb][i]-1][0] > maxway:
wayStolb = table[stolb][i] - 1
maxway = table[table[stolb][i]-1][0]
e = maxway
if e > 0:
stolb = wayStolb
iskl = iskl + 1;
if iskl > 100000:
iskl = -1
break
#print('WayStolb = '+str(wayStolb))
#print('ΠΡΠ» Π·Π΄Π΅ΡΡ'+str(stolb))
if iskl == -1:
result = '-1'
break
#print('ΠΡΡΠ°Π½ΠΎΠ²ΠΊΠ° ' +str(wayStolb))
result.append(wayStolb+1)
for i in range(0,len(maincour)):
if wayStolb == maincour[i] - 1:
control = control + 1;
if control != len(maincour):
table[stolb][0] = table[stolb][0] - 1
try:
table[stolb].remove(wayStolb+1)
except ValueError:
for i in range(0,len(maincour)):
if table[maincour[i]-1][0] == 0 and result.count(maincour[i]) == 0:
result.append(maincour[i])
control = control + 1
if control != len(maincour):
for i in range (0,len(maincour)):
#print(table[maincour[i]-1])
if table[maincour[i]-1][0] > maxway:
wayStolb = maincour[i] - 1
maxway = table[maincour[i]-1][0]
#print(wayStolb)
#print(maxway)
e = maxway
stolb = wayStolb #4
startStolb = stolb
#print(table[stolb])
#print(table[stolb][1])
wayStolb = startStolb
#print('wayStolb = '+str(wayStolb))
stolb = wayStolb
if result != '-1':
itog = itog + str(result[0])
for i in range(1,len(result)):
itog = itog + ' ' + str(result[i])
print(len(result))
print(itog)
else:
print(result)
``` | instruction | 0 | 103,376 | 24 | 206,752 |
No | output | 1 | 103,376 | 24 | 206,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Now you can take online courses in the Berland State University! Polycarp needs to pass k main online courses of his specialty to get a diploma. In total n courses are availiable for the passage.
The situation is complicated by the dependence of online courses, for each course there is a list of those that must be passed before starting this online course (the list can be empty, it means that there is no limitation).
Help Polycarp to pass the least number of courses in total to get the specialty (it means to pass all main and necessary courses). Write a program which prints the order of courses.
Polycarp passes courses consistently, he starts the next course when he finishes the previous one. Each course can't be passed more than once.
Input
The first line contains n and k (1 β€ k β€ n β€ 105) β the number of online-courses and the number of main courses of Polycarp's specialty.
The second line contains k distinct integers from 1 to n β numbers of main online-courses of Polycarp's specialty.
Then n lines follow, each of them describes the next course: the i-th of them corresponds to the course i. Each line starts from the integer ti (0 β€ ti β€ n - 1) β the number of courses on which the i-th depends. Then there follows the sequence of ti distinct integers from 1 to n β numbers of courses in random order, on which the i-th depends. It is guaranteed that no course can depend on itself.
It is guaranteed that the sum of all values ti doesn't exceed 105.
Output
Print -1, if there is no the way to get a specialty.
Otherwise, in the first line print the integer m β the minimum number of online-courses which it is necessary to pass to get a specialty. In the second line print m distinct integers β numbers of courses which it is necessary to pass in the chronological order of their passage. If there are several answers it is allowed to print any of them.
Examples
Input
6 2
5 3
0
0
0
2 2 1
1 4
1 5
Output
5
1 2 3 4 5
Input
9 3
3 9 5
0
0
3 9 4 5
0
0
1 8
1 6
1 2
2 1 2
Output
6
1 2 9 4 5 3
Input
3 3
1 2 3
1 2
1 3
1 1
Output
-1
Note
In the first test firstly you can take courses number 1 and 2, after that you can take the course number 4, then you can take the course number 5, which is the main. After that you have to take only the course number 3, which is the last not passed main course.
Submitted Solution:
```
import collections as col
import itertools as its
import sys
import operator
from copy import copy, deepcopy
class Solver:
def __init__(self):
pass
def solve(self):
n, k = map(int, input().split())
q = list(map(lambda x: int(x) - 1, input().split()))
used = [False] * n
for e in q:
used[e] = True
edges = [[] for _ in range(n)]
redges = [[] for _ in range(n)]
for i in range(n):
l = list(map(lambda x: int(x) - 1, input().split()))[1:]
edges[i] = l
for e in l:
redges[e].append(i)
degs = [len(edges[i]) for i in range(n)]
d = 0
while d < len(q):
v = q[d]
d += 1
for e in edges[v]:
if not used[e]:
used[e] = True
q.append(e)
q = q[::-1]
nq = []
for v in q:
if degs[v] == 0:
nq.append(v)
d = 0
while d < len(nq):
v = nq[d]
d += 1
for e in redges[v]:
degs[e] -= 1
if degs[e] == 0:
nq.append(e)
if len(q) != len(nq):
print(-1)
return
print(len(nq))
print(' '.join(map(lambda x: str(x + 1), nq)))
if __name__ == '__main__':
s = Solver()
s.solve()
``` | instruction | 0 | 103,377 | 24 | 206,754 |
No | output | 1 | 103,377 | 24 | 206,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful. | instruction | 0 | 103,903 | 24 | 207,806 |
Tags: dp, math, number theory, sortings
Correct Solution:
```
for _ in range(int(input())):
length=int(input())
array=list(map(int,input().split()))
array.sort()
cnt=[0]*(max(array)+1)
dp=[0]*(max(array)+1)
for x in array:
cnt[x]+=1
"""
[3, 7, 9, 14, 63]
"""
for i in range(1,len(dp)):
dp[i]+=cnt[i]
for j in range(i,len(dp),i):
dp[j]=max(dp[j],dp[i])
print(length-max(dp))
``` | output | 1 | 103,903 | 24 | 207,807 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful. | instruction | 0 | 103,904 | 24 | 207,808 |
Tags: dp, math, number theory, sortings
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fraction
from heapq import *
from random import randint
def main():
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
ma=max(a)
b=Counter(a)
c=sorted(set(a))
d=[0]*(ma+1)
d[c[-1]] =b[c[-1]]
ans = b[c[-1]]
c.pop()
c.reverse()
for i in range(len(c)):
j=2*c[i]
zz=0
while j<=ma:
zz=max(zz,d[j])
j+=c[i]
d[c[i]]=b[c[i]]+zz
ans=max(ans,d[c[i]])
print(n-ans)
# region fastio
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")
if __name__ == "__main__":
main()
``` | output | 1 | 103,904 | 24 | 207,809 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful. | instruction | 0 | 103,905 | 24 | 207,810 |
Tags: dp, math, number theory, sortings
Correct Solution:
```
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from collections import deque
for i in range(int(input())):
n=int(input())
alist=list(map(int,input().split()))
maxN=int(2e5+1)
anslist=[0]*maxN
blist=[0]*maxN
for i,val in enumerate(alist):blist[val]+=1
for i in range(1,maxN):
if blist[i]==0:
continue
anslist[i]+=blist[i]
for j in range(2*i,maxN,i):
anslist[j]=max(anslist[i],anslist[j])
print(n-max(anslist))
``` | output | 1 | 103,905 | 24 | 207,811 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful. | instruction | 0 | 103,906 | 24 | 207,812 |
Tags: dp, math, number theory, sortings
Correct Solution:
```
import sys
import math
def II():
return int(sys.stdin.readline())
def LI():
return list(map(int, sys.stdin.readline().split()))
def MI():
return map(int, sys.stdin.readline().split())
def SI():
return sys.stdin.readline().strip()
def FACT(n, mod):
s = 1
facts = [1]
for i in range(1,n+1):
s*=i
s%=mod
facts.append(s)
return facts[n]
def C(n, k, mod):
return (FACT(n,mod) * pow((FACT(k,mod)*FACT(n-k,mod))%mod,mod-2, mod))%mod
t = II()
for q in range(t):
n = II()
a = LI()
inf = max(a)+1
cnt = [0 for i in range(inf)]
for i in a:
cnt[i]+=1
dp = [0 for i in range(inf)]
for i in range(1,inf):
dp[i]+=cnt[i]
for j in range(i*2,inf, i):
dp[j] = max(dp[j], dp[i])
print(n-max(dp))
``` | output | 1 | 103,906 | 24 | 207,813 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful. | instruction | 0 | 103,907 | 24 | 207,814 |
Tags: dp, math, number theory, sortings
Correct Solution:
```
import sys
import copy
input=sys.stdin.readline
for z in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
cnt=[0]*(2*10**5+10)
cnt2=[0]*(2*10**5+10)
a.sort()
num10=-1
cnt10=0
a2=[]
for i in range(n):
if num10!=a[i]:
if cnt10!=0:
a2.append(num10)
cnt[num10]=cnt10
cnt10=1
num10=a[i]
else:
cnt10+=1
if cnt10!=0:
a2.append(num10)
cnt[num10]=cnt10
cnt2=copy.copy(cnt)
a2.sort()
for i in range(len(a2)):
b=a2[i]
num11=cnt2[b]
for j in range(b*2,2*10**5+10,b):
cnt2[j]=max(cnt2[j],num11+cnt[j])
ans=n+10
for i in range(len(a2)):
ans=min(ans,n-cnt2[a2[i]])
print(ans)
``` | output | 1 | 103,907 | 24 | 207,815 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful. | instruction | 0 | 103,908 | 24 | 207,816 |
Tags: dp, math, number theory, sortings
Correct Solution:
```
from sys import stdin
from collections import Counter
input = stdin.readline
for test in range(int(input())):
n = int(input())
lst = list(map(int, input().strip().split()))
N = max(lst) + 1
po = [0] * N
cnt = Counter(lst)
for k in range(1, N):
po[k] += cnt.get(k, 0)
for i in range(2 * k, N, k):
po[i] = max(po[i], po[k])
print(n - max(po))
``` | output | 1 | 103,908 | 24 | 207,817 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful. | instruction | 0 | 103,909 | 24 | 207,818 |
Tags: dp, math, number theory, sortings
Correct Solution:
```
from bisect import bisect_left
for _ in range(int(input())):
n = int(input())
l = list(map(int, input().split()))
l.sort()
m = max(l)
z = [0] * (m+1)
f = [0] * (m+1)
for i in range(n):
z[l[i]] += 1
for i in sorted(set(l)):
b = i+i
f[i] += z[i]
while b <= m:
f[b] = max(f[b], f[i])
b += i
print(n-max(f))
``` | output | 1 | 103,909 | 24 | 207,819 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful. | instruction | 0 | 103,910 | 24 | 207,820 |
Tags: dp, math, number theory, sortings
Correct Solution:
```
from collections import Counter
#fk this shit
def f(arr):
cnt=Counter(arr)
mxn=max(arr)+100
factor=[0]*(mxn)
for i in range(1,len(factor)):
factor[i]+=cnt.get(i,0)
for j in range(2*i,mxn,i):
factor[j]=max(factor[i],factor[j])
return len(arr)-max(factor)
for i in range(int(input())):
s=input()
lst=list(map(int,input().strip().split()))
print(f(lst))
``` | output | 1 | 103,910 | 24 | 207,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
from collections import Counter
MX = 200001
def solve(arr):
total = len(arr)
cnt = [0] * MX
for a in arr: cnt[a] += 1
u = sorted(set(arr))
loc = [-1] * MX
for i,a in enumerate(u): loc[a] = i
N = len(u)
dp = [0] * N
for i,a in enumerate(u):
dp[i] = max(dp[i], cnt[a])
for b in range(2*a,MX,a):
j = loc[b]
if j >= 0:
dp[j] = max(dp[j], dp[i] + cnt[b])
return total - max(dp)
for _ in range(int(input())):
input()
arr = list(map(int,input().split()))
print(solve(arr))
``` | instruction | 0 | 103,911 | 24 | 207,822 |
Yes | output | 1 | 103,911 | 24 | 207,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
def __init__(self, file):
self.newlines = 0
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()
# --------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def S(): return input().strip()
def print_list(l): print(' '.join(map(str, l)))
# sys.setrecursionlimit(200000)
# import random
# from functools import reduce
# from functools import lru_cache
# from heapq import *
# from collections import deque as dq
# import math
# import bisect as bs
from collections import Counter
# from collections import defaultdict as dc
for t in range(N()):
n = N()
a = RLL()
count = Counter(a)
dp = [0] * 200001
for i in range(1, 200001):
if count[i] == 0: continue
dp[i] += count[i]
for j in range(2 * i, 200001, i):
dp[j] = max(dp[i], dp[j])
print(n - max(dp))
``` | instruction | 0 | 103,912 | 24 | 207,824 |
Yes | output | 1 | 103,912 | 24 | 207,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
import sys,math
def main():
for _ in range(int(sys.stdin.readline().strip())):
n=int(sys.stdin.readline().strip())
arr=list(map(int,sys.stdin.readline().strip().split()))
table=[0]*(200001)
dp=[0]*(200001)
for i in range(n):
table[arr[i]]+=1
for i in range(1,200001):
dp[i]+=table[i]
if dp[i]>0:
for j in range(2*i,200001,i):
dp[j]=max(dp[j],dp[i])
ans=n-max(dp)
sys.stdout.write(str(ans)+'\n')
main()
``` | instruction | 0 | 103,913 | 24 | 207,826 |
Yes | output | 1 | 103,913 | 24 | 207,827 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
factorsMemo=[-1 for _ in range(200001)]
def getAllFactors(x): #returns a sorted list of all the factors or divisors of x (including 1 and x)
if factorsMemo[x]==-1:
factors=[]
for i in range(1,int(x**0.5)+1):
if x%i==0:
factors.append(i)
if x//i!=i:
factors.append(x//i)
factorsMemo[x]=factors
return factorsMemo[x]
def main():
t=int(input())
allAns=[]
for _ in range(t):
n=int(input())
a=readIntArr()
a.sort()
dp=[-inf for _ in range(200001)] #dp[previous number]max number of elements before and including it to be beautiful}
maxSizeOfBeautifulArray=0
for x in a:
maxPrevCnts=0
for f in getAllFactors(x):
maxPrevCnts=max(maxPrevCnts,dp[f])
dp[x]=1+maxPrevCnts
maxSizeOfBeautifulArray=max(maxSizeOfBeautifulArray,dp[x])
allAns.append(n-maxSizeOfBeautifulArray)
multiLineArrayPrint(allAns)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
#import sys
#input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
inf=float('inf')
MOD=10**9+7
main()
``` | instruction | 0 | 103,914 | 24 | 207,828 |
Yes | output | 1 | 103,914 | 24 | 207,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
import sys,math
# FASTEST IO
from io import BytesIO, IOBase
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.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)
# End of FASTEST IO
def main():
for _ in range(int(sys.stdin.readline().strip())):
n=int(sys.stdin.readline().strip())
arr=list(map(int,sys.stdin.readline().strip().split()))
table=[0]*(200001)
dp=[0]*(200001)
for i in range(n):
table[arr[i]]+=1
for i in range(1,200001):
for j in range(2*i,200001,i):
dp[j]=max(dp[j],table[i]+dp[i])
ans=n-max(dp)
sys.stdout.write(str(ans)+'\n')
main()
``` | instruction | 0 | 103,915 | 24 | 207,830 |
No | output | 1 | 103,915 | 24 | 207,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
import heapq
__all__ = ['MappedQueue']
class MappedQueue(object):
"""The MappedQueue class implements an efficient minimum heap. The
smallest element can be popped in O(1) time, new elements can be pushed
in O(log n) time, and any element can be removed or updated in O(log n)
time. The queue cannot contain duplicate elements and an attempt to push an
element already in the queue will have no effect.
MappedQueue complements the heapq package from the python standard
library. While MappedQueue is designed for maximum compatibility with
heapq, it has slightly different functionality.
Examples
--------
A `MappedQueue` can be created empty or optionally given an array of
initial elements. Calling `push()` will add an element and calling `pop()`
will remove and return the smallest element.
>>> q = MappedQueue([916, 50, 4609, 493, 237])
>>> q.push(1310)
True
>>> x = [q.pop() for i in range(len(q.h))]
>>> x
[50, 237, 493, 916, 1310, 4609]
Elements can also be updated or removed from anywhere in the queue.
>>> q = MappedQueue([916, 50, 4609, 493, 237])
>>> q.remove(493)
>>> q.update(237, 1117)
>>> x = [q.pop() for i in range(len(q.h))]
>>> x
[50, 916, 1117, 4609]
References
----------
.. [1] Cormen, T. H., Leiserson, C. E., Rivest, R. L., & Stein, C. (2001).
Introduction to algorithms second edition.
.. [2] Knuth, D. E. (1997). The art of computer programming (Vol. 3).
Pearson Education.
"""
def __init__(self, data=[]):
"""Priority queue class with updatable priorities.
"""
self.h = list(data)
self.d = dict()
self._heapify()
def __len__(self):
return len(self.h)
def _heapify(self):
"""Restore heap invariant and recalculate map."""
heapq.heapify(self.h)
self.d = dict([(elt, pos) for pos, elt in enumerate(self.h)])
if len(self.h) != len(self.d):
raise AssertionError("Heap contains duplicate elements")
def push(self, elt):
"""Add an element to the queue."""
# If element is already in queue, do nothing
if elt in self.d:
return False
# Add element to heap and dict
pos = len(self.h)
self.h.append(elt)
self.d[elt] = pos
# Restore invariant by sifting down
self._siftdown(pos)
return True
def pop(self):
"""Remove and return the smallest element in the queue."""
# Remove smallest element
elt = self.h[0]
del self.d[elt]
# If elt is last item, remove and return
if len(self.h) == 1:
self.h.pop()
return elt
# Replace root with last element
last = self.h.pop()
self.h[0] = last
self.d[last] = 0
# Restore invariant by sifting up, then down
pos = self._siftup(0)
self._siftdown(pos)
# Return smallest element
return elt
def update(self, elt, new):
"""Replace an element in the queue with a new one."""
# Replace
pos = self.d[elt]
self.h[pos] = new
del self.d[elt]
self.d[new] = pos
# Restore invariant by sifting up, then down
pos = self._siftup(pos)
self._siftdown(pos)
def remove(self, elt):
"""Remove an element from the queue."""
# Find and remove element
try:
pos = self.d[elt]
del self.d[elt]
except KeyError:
# Not in queue
raise
# If elt is last item, remove and return
if pos == len(self.h) - 1:
self.h.pop()
return
# Replace elt with last element
last = self.h.pop()
self.h[pos] = last
self.d[last] = pos
# Restore invariant by sifting up, then down
pos = self._siftup(pos)
self._siftdown(pos)
def _siftup(self, pos):
"""Move element at pos down to a leaf by repeatedly moving the smaller
child up."""
h, d = self.h, self.d
elt = h[pos]
# Continue until element is in a leaf
end_pos = len(h)
left_pos = (pos << 1) + 1
while left_pos < end_pos:
# Left child is guaranteed to exist by loop predicate
left = h[left_pos]
try:
right_pos = left_pos + 1
right = h[right_pos]
# Out-of-place, swap with left unless right is smaller
if right < left:
h[pos], h[right_pos] = right, elt
pos, right_pos = right_pos, pos
d[elt], d[right] = pos, right_pos
else:
h[pos], h[left_pos] = left, elt
pos, left_pos = left_pos, pos
d[elt], d[left] = pos, left_pos
except IndexError:
# Left leaf is the end of the heap, swap
h[pos], h[left_pos] = left, elt
pos, left_pos = left_pos, pos
d[elt], d[left] = pos, left_pos
# Update left_pos
left_pos = (pos << 1) + 1
return pos
def _siftdown(self, pos):
"""Restore invariant by repeatedly replacing out-of-place element with
its parent."""
h, d = self.h, self.d
elt = h[pos]
# Continue until element is at root
while pos > 0:
parent_pos = (pos - 1) >> 1
parent = h[parent_pos]
if parent > elt:
# Swap out-of-place element with parent
h[parent_pos], h[pos] = elt, parent
parent_pos, pos = pos, parent_pos
d[elt] = pos
d[parent] = parent_pos
else:
# Invariant is satisfied
break
return pos
##########################################################
##########################################################
##########################################################
from collections import Counter
# from sortedcontainers import SortedList
t = int(input())
DEBUG = False
for asdasd in range(t):
# with open('temp.txt', 'w') as file:
# for i in range(1, 200001):
# file.write(str(i) + ' ')
# break
n = int(input())
arr = [int(i) for i in input().split(" ")]
# if n==200000:
# DEBUG = True
# if DEBUG and asdasd == 3:
# arr.insert(0,n)
# line = str(arr)
# n = 400
# x = [line[i:i+n] for i in range(0, len(line), n)]
# for asd in x:
# print(asd)
arr.sort(reverse = True)
# print(arr)
count = dict(Counter(arr))
edges = {}
##########################################
# DOESNT REALLY WORK
##########################################
# heads = set()
# for x in count:
# edges[x] = set()
# heads_to_remove = set()
# for head in heads:
# if head % x == 0:
# edges[x].update(edges[head])
# edges[x].add(head)
# heads_to_remove.add(head)
# heads -= heads_to_remove
# heads.add(x)
# # print(edges)
# edges_copy = edges.copy()
# for x, edge_set in edges.items():
# for e in edge_set:
# edges_copy[e].add(x)
# edges = edges_copy
############################################
print('x1')
##########################################
# TRYING N^2 EDGEs
##########################################
for x in count:
edges[x] = set()
count_list = list(count.items())
for i in range(len(count_list)):
if i % 10 == 0:
print(i/len(count_list))
for j in range(i+1, len(count_list)):
if count_list[i][0] % count_list[j][0] == 0 or count_list[j][0] % count_list[i][0] == 0:
edges[count_list[i][0]].add(count_list[j][0])
edges[count_list[j][0]].add(count_list[i][0])
############################################
print('x2')
edge_count = {}
total_edges = 0
for x, edge_set in edges.items():
edge_count[x] = count[x]-1
for e in edge_set:
edge_count[x] += count[e]
total_edges += count[x] * edge_count[x]
total_edges //= 2
NODE_VALUE = 1
N_EDGES = 0
# queue = SortedList(list(edge_count.items()), key=lambda x: -x[N_EDGES])
# for key, value in edge_count.items():
# edge_count[key] = value * count[key]
queue = MappedQueue([(n_edges, node_value) for node_value, n_edges in edge_count.items()])
# print('edges:', edges)
# print('edge_count:', edge_count)
# print('total_edges', total_edges)
# print()
# print('queue:', queue)
# print()
remaining_nodes = n
# print('total_edges:', total_edges, 'remaining_nodes:', remaining_nodes)
# print()
# print(count)
while total_edges != remaining_nodes*(remaining_nodes-1)//2:
small = queue.pop()
# remaining_nodes -= count[small[NODE_VALUE]]
remaining_nodes -= 1
# print('popping node:', small[NODE_VALUE], 'edges:', small[N_EDGES])
# print(queue)
for e in edges[small[NODE_VALUE]]:
# print('updating edge', e)
# queue.remove((e, edge_count[e]))
queue.remove((edge_count[e], e))
# print('rem2', count[e] * count[small[NODE_VALUE]])
# edge_count[e] -= count[e] * count[small[NODE_VALUE]]
# total_edges -= count[e] * count[small[NODE_VALUE]]
# print('decreasing edge', e, ' edges by', 1)
edge_count[e] -= 1
# print('decreasing total edges by', count[e])
total_edges -= count[e]
if count[small[NODE_VALUE]] == 1:
edges[e].remove(small[NODE_VALUE])
# queue.add((e, edge_count[e]))
queue.push((edge_count[e], e))
# Edges between nodes of same value
t = count[small[NODE_VALUE]]
# print('rem1', t*(t-1)//2)
t2 = t-1
total_edges -= t2
# print(t, '(*) decreasing total edges by', t2)
count[small[NODE_VALUE]] -= 1
if count[small[NODE_VALUE]] != 0:
queue.push((small[N_EDGES]-1, small[NODE_VALUE]))
# print(queue)
# print('total_edges:', total_edges, 'remaining_nodes:', remaining_nodes)
# print()
print(n - remaining_nodes)
``` | instruction | 0 | 103,916 | 24 | 207,832 |
No | output | 1 | 103,916 | 24 | 207,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
def main():
pass
# region fastio
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")
from collections import defaultdict,Counter
m=2*10**5+5
t=int(input())
for _ in range(t):
n=int(input())
a=list(map(int,input().split()))
c=Counter(a)
greater_count=defaultdict(lambda: 0) #Stores number of numbers a given number(not necessarily in a) is greater than or equal to the numbers in a.
greater_count[0]=c[0]
for i in range(1,m+1):
greater_count[i]=greater_count[i-1]+c[i]
dp=[float('inf') for i in range(m+1)]
for i in range(m,0,-1):
dp[i]=n-greater_count[i-1] #number of numbers in a, strictly greater than i (this is worst case answer for the sublist consisting of all numbers in a >=i)
for j in range(2*i,m,i):
dp[i]=min(dp[i],dp[j]+greater_count[j-1]-greater_count[i]) #number of numbers in a, between (i,j)
print(dp[1])
``` | instruction | 0 | 103,917 | 24 | 207,834 |
No | output | 1 | 103,917 | 24 | 207,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp found on the street an array a of n elements.
Polycarp invented his criterion for the beauty of an array. He calls an array a beautiful if at least one of the following conditions must be met for each different pair of indices i β j:
* a_i is divisible by a_j;
* or a_j is divisible by a_i.
For example, if:
* n=5 and a=[7, 9, 3, 14, 63], then the a array is not beautiful (for i=4 and j=2, none of the conditions above is met);
* n=3 and a=[2, 14, 42], then the a array is beautiful;
* n=4 and a=[45, 9, 3, 18], then the a array is not beautiful (for i=1 and j=4 none of the conditions above is met);
Ugly arrays upset Polycarp, so he wants to remove some elements from the array a so that it becomes beautiful. Help Polycarp determine the smallest number of elements to remove to make the array a beautiful.
Input
The first line contains one integer t (1 β€ t β€ 10) β the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1 β€ n β€ 2 β
10^5) β the length of the array a.
The second line of each test case contains n numbers a_1, a_2, β¦, a_n (1 β€ a_i β€ 2 β
10^5) β elements of the array a.
Output
For each test case output one integer β the minimum number of elements that must be removed to make the array a beautiful.
Example
Input
4
5
7 9 3 14 63
3
2 14 42
4
45 9 3 18
3
2 2 8
Output
2
0
1
0
Note
In the first test case, removing 7 and 14 will make array a beautiful.
In the second test case, the array a is already beautiful.
In the third test case, removing one of the elements 45 or 18 will make the array a beautiful.
In the fourth test case, the array a is beautiful.
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fraction
from heapq import *
from random import randint
def main():
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
ma=max(a)
b=[0]*(ma+1)
for i in a:
b[i]+=1
c=[]
for i in range(1,ma+1):
if b[i]:
c.append(i)
d = [0] * (ma + 1)
d[c[-1]] =b[c[-1]]
c.pop()
c.reverse()
ans=1
for i in range(len(c)):
j=2*c[i]
zz=0
while j<=ma:
zz=max(zz,d[j])
j+=c[i]
d[c[i]]=b[c[i]]+zz
ans=max(ans,d[c[i]])
print(n-ans)
# region fastio
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")
if __name__ == "__main__":
main()
``` | instruction | 0 | 103,918 | 24 | 207,836 |
No | output | 1 | 103,918 | 24 | 207,837 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 β€ a, b, c β€ 7β
10^8) β the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday β rabbit stew and during Wednesday β chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip. | instruction | 0 | 105,305 | 24 | 210,610 |
Tags: implementation, math
Correct Solution:
```
a,b,c=map(int,input().split())
fu=0
fu=min(a//3,b//2,c//2)
a-=3*fu
b-=2*fu
c-=2*fu
x=[1,2,3,1,3,2,1]
k=0
m=0
for i in range(len(x)):
d=0
f=0
j=i
a1=a
b1=b
c1=c
z=1
while z==1:
k=x[j]
if (k==1)&(a1>0):
a1-=1
d+=1
if (k==2)&(b1>0):
b1-=1
d+=1
if (k==3)&(c1>0):
c1-=1
d+=1
if d==f:
z=0
else:
f=d
j+=1
j=j%7
if d>m:
m=d
print(7*fu+m)
``` | output | 1 | 105,305 | 24 | 210,611 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 β€ a, b, c β€ 7β
10^8) β the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday β rabbit stew and during Wednesday β chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip. | instruction | 0 | 105,306 | 24 | 210,612 |
Tags: implementation, math
Correct Solution:
```
def dfs(day_l,day_r,a,b,c):
global span
if day_l<1 or day_r>14:
return
if a<0 or b<0 or c<0:
return
#print(day_l,day_r,a,b,c)
span=max(span,day_r-day_l+1)
if day_l-1 in [1,4,7,8,11,14]:
dfs(day_l-1,day_r,a-1,b,c)
elif day_l-1 in [2,6,9,13]:
dfs(day_l-1,day_r,a,b-1,c)
elif day_l-1 in [3,5,10,12]:
dfs(day_l-1,day_r,a,b,c-1)
if day_r+1 in [1,4,7,8,11,14]:
dfs(day_l,day_r+1,a-1,b,c)
elif day_r+1 in [2,6,9,13]:
dfs(day_l,day_r+1,a,b-1,c)
elif day_r+1 in [3,5,10,12]:
dfs(day_l,day_r+1,a,b,c-1)
a,b,c=list(map(int,input().split()))
weeks=min([a//3,b//2,c//2])
a-=weeks*3
b-=weeks*2
c-=weeks*2
if weeks>0:
span=0
if a>0:
day_min,day_max=7,7
dfs(7,7,a-1,b,c)
ans1=7*weeks+span
span=0
day_min,day_max=7,7
dfs(7,7,a-1+3,b+2,c+2)
ans2=7*(weeks-1)+span
print(max(ans1,ans2))
else:
span=0
for i in [4,5,6,7]:
day_min,day_max=i,i
if i in [4,7]:
dfs(i,i,a-1,b,c)
elif i==6:
dfs(i,i,a,b-1,c)
else:
dfs(i,i,a,b,c-1)
print(span)
``` | output | 1 | 105,306 | 24 | 210,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 β€ a, b, c β€ 7β
10^8) β the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday β rabbit stew and during Wednesday β chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip. | instruction | 0 | 105,307 | 24 | 210,614 |
Tags: implementation, math
Correct Solution:
```
a, b, c = map(int, input().split())
x = min(a//3, b//2, c//2)
a,b,c = a-3*x, b-2*x, c-2*x
z = [1,1,2,3,1,3,2]
m = 0
for i in range(7):
lis = [0,a,b,c]
j = i
cnt = 0
while(lis[1]>=0 and lis[2]>=0 and lis[3]>=0):
lis[z[j]] -= 1
j = (j+1)%7
cnt += 1
m = max(m,cnt-1)
print(x*7+m)
``` | output | 1 | 105,307 | 24 | 210,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 β€ a, b, c β€ 7β
10^8) β the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday β rabbit stew and during Wednesday β chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip. | instruction | 0 | 105,308 | 24 | 210,616 |
Tags: implementation, math
Correct Solution:
```
a = [int(i) for i in input().split()]
days = [0, 1, 2, 0, 2, 1, 0]
week = min(a[0] // 3, a[1] // 2, a[2] // 2)
a[0], a[1], a[2] = a[0] - week * 3, a[1] - week * 2, a[2] - week * 2
m = 0
for day in range(7):
b = a.copy()
c = 0
for this_day in range(7):
if b[days[(day + this_day) % 7]] > 0:
b[days[(day + this_day) % 7]] -= 1
c += 1
else:
break
if c > m:
m = c
print(m + week * 7)
``` | output | 1 | 105,308 | 24 | 210,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:
* on Mondays, Thursdays and Sundays he eats fish food;
* on Tuesdays and Saturdays he eats rabbit stew;
* on other days of week he eats chicken stake.
Polycarp plans to go on a trip and already packed his backpack. His backpack contains:
* a daily rations of fish food;
* b daily rations of rabbit stew;
* c daily rations of chicken stakes.
Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Input
The first line of the input contains three positive integers a, b and c (1 β€ a, b, c β€ 7β
10^8) β the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.
Output
Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.
Examples
Input
2 1 1
Output
4
Input
3 2 2
Output
7
Input
1 100 1
Output
3
Input
30 20 10
Output
39
Note
In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday β rabbit stew and during Wednesday β chicken stake. So, after four days of the trip all food will be eaten.
In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.
In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip. | instruction | 0 | 105,309 | 24 | 210,618 |
Tags: implementation, math
Correct Solution:
```
a,b,c = map(int,input().strip().split())
ans = 0
basi = min(c//2,min(b//2,a//3))
#print(basi)
aa = a- 3*basi
bb = b - 2*basi
cc = c - 2*basi
mp = {0:'aa',1:'aa',2:'bb',3:'cc',4:'aa',5:'cc',6:'bb'}
#print(aa,bb,cc)
for i in range(7):
aaa = aa
bbb = bb
ccc = cc
an = 0
for j in range(i,i+15):
if mp[j%7]=='aa':
aaa-=1
if mp[j%7]=='bb':
bbb-=1
if mp[j%7]=='cc':
ccc-=1
if aaa<0 or bbb<0 or ccc<0:
break
else:
an+=1
ans = max(ans,basi*7+an)
print(ans)
``` | output | 1 | 105,309 | 24 | 210,619 |
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