message stringlengths 2 20.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 757 108k | cluster float64 4 4 | __index_level_0__ int64 1.51k 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Welcome to PC Koshien, players. Physical condition management is important to participate in the event. It is said that at the turn of the season when the temperature fluctuates greatly, it puts a strain on the body and it is easy to catch a cold. The day you should be careful about is the day when the difference between the maximum temperature and the minimum temperature is the largest. When the maximum and minimum temperatures of a day are given for 7 days, create a program that outputs the value obtained by subtracting the minimum temperature from the maximum temperature for each day.
input
Input data is given in the following format.
a1 b1
a2 b2
::
a7 b7
The input consists of 7 lines, and line i is given an integer representing the maximum temperature ai (-40 ≤ ai ≤ 40) and the minimum temperature bi (-40 ≤ bi ≤ 40) on day i. On all days, the maximum temperature ai is always above the minimum temperature bi.
output
Output the temperature difference for 7 days in 7 lines.
Example
Input
30 19
39 20
19 18
25 20
22 21
23 10
10 -10
Output
11
19
1
5
1
13
20 | instruction | 0 | 46,976 | 4 | 93,952 |
"Correct Solution:
```
i=0
for i in range(7):
a,b=map(int,input().split())
print(a-b)
i+=1
``` | output | 1 | 46,976 | 4 | 93,953 |
Provide a correct Python 3 solution for this coding contest problem.
Welcome to PC Koshien, players. Physical condition management is important to participate in the event. It is said that at the turn of the season when the temperature fluctuates greatly, it puts a strain on the body and it is easy to catch a cold. The day you should be careful about is the day when the difference between the maximum temperature and the minimum temperature is the largest. When the maximum and minimum temperatures of a day are given for 7 days, create a program that outputs the value obtained by subtracting the minimum temperature from the maximum temperature for each day.
input
Input data is given in the following format.
a1 b1
a2 b2
::
a7 b7
The input consists of 7 lines, and line i is given an integer representing the maximum temperature ai (-40 ≤ ai ≤ 40) and the minimum temperature bi (-40 ≤ bi ≤ 40) on day i. On all days, the maximum temperature ai is always above the minimum temperature bi.
output
Output the temperature difference for 7 days in 7 lines.
Example
Input
30 19
39 20
19 18
25 20
22 21
23 10
10 -10
Output
11
19
1
5
1
13
20 | instruction | 0 | 46,977 | 4 | 93,954 |
"Correct Solution:
```
for i in range(7):
H,L=map(int,input().split())
print(H-L)
``` | output | 1 | 46,977 | 4 | 93,955 |
Provide a correct Python 3 solution for this coding contest problem.
Welcome to PC Koshien, players. Physical condition management is important to participate in the event. It is said that at the turn of the season when the temperature fluctuates greatly, it puts a strain on the body and it is easy to catch a cold. The day you should be careful about is the day when the difference between the maximum temperature and the minimum temperature is the largest. When the maximum and minimum temperatures of a day are given for 7 days, create a program that outputs the value obtained by subtracting the minimum temperature from the maximum temperature for each day.
input
Input data is given in the following format.
a1 b1
a2 b2
::
a7 b7
The input consists of 7 lines, and line i is given an integer representing the maximum temperature ai (-40 ≤ ai ≤ 40) and the minimum temperature bi (-40 ≤ bi ≤ 40) on day i. On all days, the maximum temperature ai is always above the minimum temperature bi.
output
Output the temperature difference for 7 days in 7 lines.
Example
Input
30 19
39 20
19 18
25 20
22 21
23 10
10 -10
Output
11
19
1
5
1
13
20 | instruction | 0 | 46,978 | 4 | 93,956 |
"Correct Solution:
```
i=1
while i<8:
a_i,b_i=map(int,input().split())
x=a_i-b_i
print(x)
i=i+1
``` | output | 1 | 46,978 | 4 | 93,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to PC Koshien, players. Physical condition management is important to participate in the event. It is said that at the turn of the season when the temperature fluctuates greatly, it puts a strain on the body and it is easy to catch a cold. The day you should be careful about is the day when the difference between the maximum temperature and the minimum temperature is the largest. When the maximum and minimum temperatures of a day are given for 7 days, create a program that outputs the value obtained by subtracting the minimum temperature from the maximum temperature for each day.
input
Input data is given in the following format.
a1 b1
a2 b2
::
a7 b7
The input consists of 7 lines, and line i is given an integer representing the maximum temperature ai (-40 ≤ ai ≤ 40) and the minimum temperature bi (-40 ≤ bi ≤ 40) on day i. On all days, the maximum temperature ai is always above the minimum temperature bi.
output
Output the temperature difference for 7 days in 7 lines.
Example
Input
30 19
39 20
19 18
25 20
22 21
23 10
10 -10
Output
11
19
1
5
1
13
20
Submitted Solution:
```
for i in range(7):
ai,bi=map(int,input().split())
if ai>=bi:
print(ai-bi)
if ai<bi:
print(ai-bi)
``` | instruction | 0 | 46,979 | 4 | 93,958 |
Yes | output | 1 | 46,979 | 4 | 93,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to PC Koshien, players. Physical condition management is important to participate in the event. It is said that at the turn of the season when the temperature fluctuates greatly, it puts a strain on the body and it is easy to catch a cold. The day you should be careful about is the day when the difference between the maximum temperature and the minimum temperature is the largest. When the maximum and minimum temperatures of a day are given for 7 days, create a program that outputs the value obtained by subtracting the minimum temperature from the maximum temperature for each day.
input
Input data is given in the following format.
a1 b1
a2 b2
::
a7 b7
The input consists of 7 lines, and line i is given an integer representing the maximum temperature ai (-40 ≤ ai ≤ 40) and the minimum temperature bi (-40 ≤ bi ≤ 40) on day i. On all days, the maximum temperature ai is always above the minimum temperature bi.
output
Output the temperature difference for 7 days in 7 lines.
Example
Input
30 19
39 20
19 18
25 20
22 21
23 10
10 -10
Output
11
19
1
5
1
13
20
Submitted Solution:
```
for i in range(7):
a,b=(int(x) for x in input().split())
print(a-b)
``` | instruction | 0 | 46,980 | 4 | 93,960 |
Yes | output | 1 | 46,980 | 4 | 93,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to PC Koshien, players. Physical condition management is important to participate in the event. It is said that at the turn of the season when the temperature fluctuates greatly, it puts a strain on the body and it is easy to catch a cold. The day you should be careful about is the day when the difference between the maximum temperature and the minimum temperature is the largest. When the maximum and minimum temperatures of a day are given for 7 days, create a program that outputs the value obtained by subtracting the minimum temperature from the maximum temperature for each day.
input
Input data is given in the following format.
a1 b1
a2 b2
::
a7 b7
The input consists of 7 lines, and line i is given an integer representing the maximum temperature ai (-40 ≤ ai ≤ 40) and the minimum temperature bi (-40 ≤ bi ≤ 40) on day i. On all days, the maximum temperature ai is always above the minimum temperature bi.
output
Output the temperature difference for 7 days in 7 lines.
Example
Input
30 19
39 20
19 18
25 20
22 21
23 10
10 -10
Output
11
19
1
5
1
13
20
Submitted Solution:
```
while True:
try:
a=list(map(int,input().split()))
print(a[0]-a[1])
except:
break;
``` | instruction | 0 | 46,981 | 4 | 93,962 |
Yes | output | 1 | 46,981 | 4 | 93,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to PC Koshien, players. Physical condition management is important to participate in the event. It is said that at the turn of the season when the temperature fluctuates greatly, it puts a strain on the body and it is easy to catch a cold. The day you should be careful about is the day when the difference between the maximum temperature and the minimum temperature is the largest. When the maximum and minimum temperatures of a day are given for 7 days, create a program that outputs the value obtained by subtracting the minimum temperature from the maximum temperature for each day.
input
Input data is given in the following format.
a1 b1
a2 b2
::
a7 b7
The input consists of 7 lines, and line i is given an integer representing the maximum temperature ai (-40 ≤ ai ≤ 40) and the minimum temperature bi (-40 ≤ bi ≤ 40) on day i. On all days, the maximum temperature ai is always above the minimum temperature bi.
output
Output the temperature difference for 7 days in 7 lines.
Example
Input
30 19
39 20
19 18
25 20
22 21
23 10
10 -10
Output
11
19
1
5
1
13
20
Submitted Solution:
```
a1, b1 = map(int,input().split())
a2, b2 = map(int,input().split())
a3, b3 = map(int,input().split())
a4, b4 = map(int,input().split())
a5, b5 = map(int,input().split())
a6, b6 = map(int,input().split())
a7, b7 = map(int,input().split())
def f(x,y):
print(int(x-y))
f(a1,b1)
f(a2,b2)
f(a3,b3)
f(a4,b4)
f(a5,b5)
f(a6,b6)
f(a7,b7)
``` | instruction | 0 | 46,982 | 4 | 93,964 |
Yes | output | 1 | 46,982 | 4 | 93,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to PC Koshien, players. Physical condition management is important to participate in the event. It is said that at the turn of the season when the temperature fluctuates greatly, it puts a strain on the body and it is easy to catch a cold. The day you should be careful about is the day when the difference between the maximum temperature and the minimum temperature is the largest. When the maximum and minimum temperatures of a day are given for 7 days, create a program that outputs the value obtained by subtracting the minimum temperature from the maximum temperature for each day.
input
Input data is given in the following format.
a1 b1
a2 b2
::
a7 b7
The input consists of 7 lines, and line i is given an integer representing the maximum temperature ai (-40 ≤ ai ≤ 40) and the minimum temperature bi (-40 ≤ bi ≤ 40) on day i. On all days, the maximum temperature ai is always above the minimum temperature bi.
output
Output the temperature difference for 7 days in 7 lines.
Example
Input
30 19
39 20
19 18
25 20
22 21
23 10
10 -10
Output
11
19
1
5
1
13
20
Submitted Solution:
```
for i in range(0,7):
a,b=map(int,input(),split())
print(a-b)
``` | instruction | 0 | 46,983 | 4 | 93,966 |
No | output | 1 | 46,983 | 4 | 93,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to PC Koshien, players. Physical condition management is important to participate in the event. It is said that at the turn of the season when the temperature fluctuates greatly, it puts a strain on the body and it is easy to catch a cold. The day you should be careful about is the day when the difference between the maximum temperature and the minimum temperature is the largest. When the maximum and minimum temperatures of a day are given for 7 days, create a program that outputs the value obtained by subtracting the minimum temperature from the maximum temperature for each day.
input
Input data is given in the following format.
a1 b1
a2 b2
::
a7 b7
The input consists of 7 lines, and line i is given an integer representing the maximum temperature ai (-40 ≤ ai ≤ 40) and the minimum temperature bi (-40 ≤ bi ≤ 40) on day i. On all days, the maximum temperature ai is always above the minimum temperature bi.
output
Output the temperature difference for 7 days in 7 lines.
Example
Input
30 19
39 20
19 18
25 20
22 21
23 10
10 -10
Output
11
19
1
5
1
13
20
Submitted Solution:
```
for i in range(7):
ls = list(map(int, input().split()))
if max(ls) > 0 and min(ls) > 0:
print(max(ls) - min(ls))
elif max(ls) > 0 and min(ls) < 0:
print(max(ls) + min(ls))
elif max(ls) < 0 and min(ls) < 0:
print(abs(max(ls)) - abs(min(ls)))
``` | instruction | 0 | 46,984 | 4 | 93,968 |
No | output | 1 | 46,984 | 4 | 93,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to PC Koshien, players. Physical condition management is important to participate in the event. It is said that at the turn of the season when the temperature fluctuates greatly, it puts a strain on the body and it is easy to catch a cold. The day you should be careful about is the day when the difference between the maximum temperature and the minimum temperature is the largest. When the maximum and minimum temperatures of a day are given for 7 days, create a program that outputs the value obtained by subtracting the minimum temperature from the maximum temperature for each day.
input
Input data is given in the following format.
a1 b1
a2 b2
::
a7 b7
The input consists of 7 lines, and line i is given an integer representing the maximum temperature ai (-40 ≤ ai ≤ 40) and the minimum temperature bi (-40 ≤ bi ≤ 40) on day i. On all days, the maximum temperature ai is always above the minimum temperature bi.
output
Output the temperature difference for 7 days in 7 lines.
Example
Input
30 19
39 20
19 18
25 20
22 21
23 10
10 -10
Output
11
19
1
5
1
13
20
Submitted Solution:
```
for i in range(0,7)
a,b=map(int,input(),split())
print(a-b)
``` | instruction | 0 | 46,985 | 4 | 93,970 |
No | output | 1 | 46,985 | 4 | 93,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Welcome to PC Koshien, players. Physical condition management is important to participate in the event. It is said that at the turn of the season when the temperature fluctuates greatly, it puts a strain on the body and it is easy to catch a cold. The day you should be careful about is the day when the difference between the maximum temperature and the minimum temperature is the largest. When the maximum and minimum temperatures of a day are given for 7 days, create a program that outputs the value obtained by subtracting the minimum temperature from the maximum temperature for each day.
input
Input data is given in the following format.
a1 b1
a2 b2
::
a7 b7
The input consists of 7 lines, and line i is given an integer representing the maximum temperature ai (-40 ≤ ai ≤ 40) and the minimum temperature bi (-40 ≤ bi ≤ 40) on day i. On all days, the maximum temperature ai is always above the minimum temperature bi.
output
Output the temperature difference for 7 days in 7 lines.
Example
Input
30 19
39 20
19 18
25 20
22 21
23 10
10 -10
Output
11
19
1
5
1
13
20
Submitted Solution:
```
try:
while True:
a,b=map(int,input().split())
if a-b<=a:
print(a+b)
else:
print(a-b)
except:
pass
``` | instruction | 0 | 46,986 | 4 | 93,972 |
No | output | 1 | 46,986 | 4 | 93,973 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3. | instruction | 0 | 47,157 | 4 | 94,314 |
Tags: constructive algorithms, data structures, greedy, implementation
Correct Solution:
```
from sys import stdin, stdout
def bn(l, mn, mx):
low = 0
high = len(l)-1
while low <= high:
mid = (low+high)//2
if l[mid] > mx:
high = mid-1
elif l[mid] < mn:
low = mid+1
else:
return l[mid]
return False
n = int(stdin.readline())
income1 = [int(x) for x in stdin.readline().split()][::-1]
income2 = int(stdin.readline())
sums = [0]
for a in income1:
sums.append(a+sums[-1])
sums2 = income2*(n//2)
sums3 = [0]
for a in reversed(income1):
sums3.append(sums3[-1]+a)
i2Len = n//2
if sums[-1]+sums2 > 0:
print(n)
elif sums[-1] > 0:
dRange = n%2 + sums[-1]//abs(income2)
good = []
for s in range(dRange+1):
if sums[s] > -sums2:
good.append(s)
if good:
maxGood = float('inf')
valid = False
for ind,s in enumerate(sums[::-1]):
maxGood = min(maxGood, (s-1)//abs(income2)-ind+(n%2))
m2 = len(income1)-ind
if maxGood >= m2:
realGood = bn(good,m2,maxGood)
if realGood != False:
valid = True
break
if valid:
print(realGood+n//2)
else:
print(-1)
else:
print(-1)
else:
print(-1)
``` | output | 1 | 47,157 | 4 | 94,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3. | instruction | 0 | 47,158 | 4 | 94,316 |
Tags: constructive algorithms, data structures, greedy, implementation
Correct Solution:
```
import io
import os
def solve(N, A, X):
N2 = N // 2 # last half len
N1 = N - N2 # first half len
assert len(A) == N1
# Positive total can just take whole array
total = sum(A) + N2 * X
if total > 0:
return N
# Handle non-positive total case
if X >= 0:
# If X is positive, the first window will be even more negative as we take fewer X since total <= 0.
# This only changes when K is less than half the array. Suppose this K exists.
# Tile the array with non-overlapping window of this K.
# Since each window has positive sum and the remainder are all Xs which are also positive, total is supposed to be positive.
# Which is a contradiction so this case is impossible.
return -1
else:
# If X is negative, we know K must be greater than half the array. Otherwise it will take a window of negative X.
#
# Look at the cumulative sums. Want to pair up some prefix and suffix such that it is element wise less than.
# Since the suffix is always in the repeating part of the array we know the value exactly.
# That is for 0 <= i <= N - K
# 0 < cs[i + K] - cs[i] # range sum of window starting at i
# cs[i] < cs[i + K]
# cs[i] < total - X * (N - K - i) # since cs[i + K] is in second half
# cs[i] + X * (N - i) < total + X * K
cs = [0]
for x in A + [X] * N2:
cs.append(cs[-1] + x)
maxSeen = float("-inf")
for i in range(N2 + 1):
maxSeen = max(maxSeen, cs[i] + X * (N - i))
# If max seen of the prefix from [0, i] is good, then K = N - i might work
if maxSeen < total + X * (N - i):
return N - i
return -1
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
(N,) = [int(x) for x in input().split()]
A = [int(x) for x in input().split()]
(X,) = [int(x) for x in input().split()]
ans = solve(N, A, X)
print(ans)
``` | output | 1 | 47,158 | 4 | 94,317 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3. | instruction | 0 | 47,159 | 4 | 94,318 |
Tags: constructive algorithms, data structures, greedy, implementation
Correct Solution:
```
def main():
n = int(input())
a = list(map(int, input().split()))
Ax = int(input())
N = int((n+1)/2)
m = [0 for _ in range(N+1)]
psm = 0
for i in range(1, N+1):
psm += Ax - a[i-1]
m[i] = min(m[i-1], psm)
apsm = 0
for k in range(1,N+1):
apsm += a[k-1]
for k in range(N, n+1):
if apsm + m[n-k] > 0:
return print(k)
apsm += Ax
return print(-1)
if __name__ == '__main__':
main()
``` | output | 1 | 47,159 | 4 | 94,319 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3. | instruction | 0 | 47,160 | 4 | 94,320 |
Tags: constructive algorithms, data structures, greedy, implementation
Correct Solution:
```
# from bisect import bisect_left
N = int(input())
A = list(map(int, input().split()))
X = int(input())
# print(-1//4)
# print(N, A, X)
NS = N - len(A)
NN = len(A)
PS = [0]
for a in A:
PS.append(PS[-1]+a)
total = PS[-1] + NS * X
if total > 0:
print(N)
else:
if X > 0:
print(-1)
else:
max_a = total + X * N
result = -1
for i in range(NN):
max_a = max(max_a, PS[i] + X * (N - i))
if max_a < total + X * (N - i):
result = N - i
break
print(result)
# for tc in range(TC):
# N, X = map(int, input().split())
# A = list(map(int, input().split()))
# result = 0
# print('Case #{}: {}'.format(tc + 1, result))
``` | output | 1 | 47,160 | 4 | 94,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3. | instruction | 0 | 47,161 | 4 | 94,322 |
Tags: constructive algorithms, data structures, greedy, implementation
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
x = int(input())
a = [0] + a
f = list(a)
part = n // 2
sum = 0
for i in range(1, len(f)):
f[i] += f[i - 1]
sum += a[i]
sum += part * x
if sum > 0:
print(n)
exit(0)
if x>=0:
print(-1)
exit(0)
#print(sum)
for i in range(len(f)):
f[i] = sum - f[i]
c = [0] * len(f)
x = -x
for i in range(len(c)):
c[i] = max(0, (-f[i] + x) // x)
vmax = list(c)
for i in range(1, len(vmax)):
vmax[i] = max(vmax[i], vmax[i - 1] - 1)
#print(f)
#print(c)
for i in range(len(vmax)):
if vmax[i] == 0:
print(n - i)
exit(0)
print(-1)
``` | output | 1 | 47,161 | 4 | 94,323 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3. | instruction | 0 | 47,162 | 4 | 94,324 |
Tags: constructive algorithms, data structures, greedy, implementation
Correct Solution:
```
import os
import sys
if os.path.exists('/mnt/c/Users/Square/square/codeforces'):
f = iter(open('E.txt').readlines())
def input():
return next(f)
# input = lambda: sys.stdin.readline().strip()
else:
input = lambda: sys.stdin.readline().strip()
fprint = lambda *args: print(*args, flush=True)
n = int(input())
A = list(map(int, input().split()))
x = int(input())
m = [0]
ps = [0]
prefs = [0]
for i in range(n//2):
ps.append(x - A[i])
for i in range(n//2):
prefs.append(prefs[-1] + x - A[i])
mins = [0]
for i in prefs[1:]:
mins.append(min(mins[-1], i))
# print(ps)
# print(prefs)
# print(mins)
s = sum(A)
res = -1
for k, m in enumerate(reversed(mins)):
# print(s, k, m)
if s + m > 0:
res = len(A) + k
break
s += x
print(res)
``` | output | 1 | 47,162 | 4 | 94,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3. | instruction | 0 | 47,163 | 4 | 94,326 |
Tags: constructive algorithms, data structures, greedy, implementation
Correct Solution:
```
import sys, heapq
from collections import *
from functools import lru_cache
sys.setrecursionlimit(10**6)
import operator as op
from functools import reduce
import bisect
def main():
# sys.stdin = open('input.txt', 'r')
# t = int(input())
#
# for _ in range(t):
# x1,y1,x2,y2 = map(int,input().split(' '))
# x, y = x2-x1, y2-y1
# print(x*y+1)
n = int(input())
arr = list(map(int,input().split(' ')))
x = int(input())
first, cur, res = sum(arr)+x*(n-len(arr)), 0, 0
if first > 0:
print(n)
return
for k in range(len(arr),n)[::-1]:
cur += x-arr[n-k-1]
res = min(res,cur)
first -= x
if first+res > 0:
print(k)
return
print(-1)
return
if __name__ == "__main__":
main()
``` | output | 1 | 47,163 | 4 | 94,327 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3. | instruction | 0 | 47,164 | 4 | 94,328 |
Tags: constructive algorithms, data structures, greedy, implementation
Correct Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# 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")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------
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
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa=ifa[::-1]
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
'''
class SMT:
def __init__(self,arr):
self.n=len(arr)-1
self.arr=[0]*(self.n<<2)
self.lazy=[0]*(self.n<<2)
def Build(l,r,rt):
if l==r:
self.arr[rt]=arr[l]
return
m=(l+r)>>1
Build(l,m,rt<<1)
Build(m+1,r,rt<<1|1)
self.pushup(rt)
Build(1,self.n,1)
def pushup(self,rt):
self.arr[rt]=self.arr[rt<<1]+self.arr[rt<<1|1]
def pushdown(self,rt,ln,rn):#lr,rn表区间数字数
if self.lazy[rt]:
self.lazy[rt<<1]+=self.lazy[rt]
self.lazy[rt<<1|1]+=self.lazy[rt]
self.arr[rt<<1]+=self.lazy[rt]*ln
self.arr[rt<<1|1]+=self.lazy[rt]*rn
self.lazy[rt]=0
def update(self,L,R,c,l=1,r=None,rt=1):#L,R表示操作区间
if r==None: r=self.n
if L<=l and r<=R:
self.arr[rt]+=c*(r-l+1)
self.lazy[rt]+=c
return
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
if L<=m: self.update(L,R,c,l,m,rt<<1)
if R>m: self.update(L,R,c,m+1,r,rt<<1|1)
self.pushup(rt)
def query(self,L,R,l=1,r=None,rt=1):
if r==None: r=self.n
#print(L,R,l,r,rt)
if L<=l and R>=r:
return self.arr[rt]
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
ans=0
if L<=m: ans+=self.query(L,R,l,m,rt<<1)
if R>m: ans+=self.query(L,R,m+1,r,rt<<1|1)
return ans
'''
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]<self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
#print(flag)
return flag
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
def nb(i,j):
for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
if 0<=ni<n and 0<=nj<m:
yield ni,nj
@bootstrap
def gdfs(r,p):
if len(g[r])==1 and p!=-1:
yield None
for ch in g[r]:
if ch!=p:
yield gdfs(ch,r)
yield None
t=1
for i in range(t):
n=N()
a=RLL()
x=N()
s=sum(a)
mi=[0]
k=n//2+1
if k>(n+1)//2:
s+=x
for i in range(n-k):
mi.append(x-a[i])
dp=[0]
for i in range(n-k):
dp.append(dp[-1]+mi[i+1])
mi[i+1]=min(dp[-1],mi[i])
for k in range(n//2+1,n+1):
if mi[-1]+s>0:
print(k)
exit()
mi.pop()
s+=x
print(-1)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
``` | output | 1 | 47,164 | 4 | 94,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3.
Submitted Solution:
```
def isFired(n, incomes, other):
N = (n + 1) >> 1
minIncomes = [0]
minIncome = 0
for i in range(N):
minIncome += other - incomes[i]
minIncomes.append(min(minIncomes[-1], minIncome))
prefixSum = sum(incomes)
for month in range(N, n + 1):
if prefixSum + minIncomes[n - month] > 0:
return month
prefixSum += other
return -1
def solve():
n = int(input())
incomes = list(map(int, input().split()))
other = int(input())
print(isFired(n, incomes, other))
solve()
``` | instruction | 0 | 47,165 | 4 | 94,330 |
Yes | output | 1 | 47,165 | 4 | 94,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
a=list(map(int,input().split()))
x=int(input())
s=sum(a)+(n//2)*x
if s>0:
print(n)
else:
if x>=0:
print(-1)
else:
poss=-1
x=-x
miss=-s
mn=0
curs=0
for i in range((n+1)//2):
curs-=a[i]
curmin=(miss-curs+x)//x
curmin=curmin+i
mn=max(mn,curmin)
if i>=mn:
poss=n-i-1
print(poss)
``` | instruction | 0 | 47,166 | 4 | 94,332 |
Yes | output | 1 | 47,166 | 4 | 94,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3.
Submitted Solution:
```
from sys import stdin
input = stdin.readline
def main():
test = 1
for _ in range(test):
n = int(input())
ara = [int(num) for num in input().split()]
x = int(input())
n1 = len(ara)
n2 = n - n1
ara_sum = sum(ara) + n2 * x
if ara_sum > 0:
print(n)
elif x >= 0:
print(-1)
else:
ara = ara + [x] * n2
prefix_sum = [0]
for num in ara:
prefix_sum.append(num + prefix_sum[-1])
current_max = n * x
for i in range(n1):
current_max = max(current_max, prefix_sum[i] + x * (n - i))
if current_max < prefix_sum[n] + x * (n - i):
print(n - i)
return
print(-1)
if __name__ == "__main__":
main()
``` | instruction | 0 | 47,167 | 4 | 94,334 |
Yes | output | 1 | 47,167 | 4 | 94,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3.
Submitted Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n = int(input())
a = list(map(int, input().split()))
x = int(input())
total = sum(a) + x * (n - len(a))
if total > 0:
print(n)
exit()
min_sub = total
for l, ans in zip(range(len(a)), range(n - 1, -1, -1)):
total -= a[l]
min_sub = min(min_sub - x, total)
if min_sub > 0:
print(ans)
exit()
print(-1)
``` | instruction | 0 | 47,168 | 4 | 94,336 |
Yes | output | 1 | 47,168 | 4 | 94,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3.
Submitted Solution:
```
n = int(input())
arr = [int(x) for x in input().split()]
x = int(input())
arr += [x]*(n//2)
def giveArr(tk):
res = []
summ = 0
for i in range(tk):
summ += arr[i]
res.append(summ)
i,j = 0,tk-1
while j+1 < n:
summ += arr[j+1] - arr[i]
i += 1
j += 1
res.append(summ)
return res
def find(tk):
res = 0
summ = 0
for i in range(tk):
summ += arr[i]
if summ <= 0:
res += summ - 1
i,j = 0,tk-1
while j+1 < n:
summ += arr[j+1] - arr[i]
i += 1
j += 1
if summ <= 0:
res += summ - 1
return abs(res)
l,r = 1,n
# tk = 1
while l < r :
tk = (l+r)//2
# print(str(l) +' '+str(r))
# print('tk : '+str(tk))
# print(giveArr(tk))
# print(giveArr(tk+1))
neg1,neg2 = find(tk),find(tk+1)
if neg1 <= neg2:
r = tk
else:
l = tk+1
if find(l) > 0:
print(-1)
else:
print(l)
``` | instruction | 0 | 47,169 | 4 | 94,338 |
No | output | 1 | 47,169 | 4 | 94,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3.
Submitted Solution:
```
def icanv(num, minusnum):
if num <= 0:
print(-1)
exit()
mm = -minusnum
return (num - 1) // mm
n = int(input())
lst = list(map(int,input().split()))
numr = int(input())
numrlen = n // 2
sm = sum(lst)
lst.reverse()
if numr > 0:
if sm + numr * numrlen > 0:
print(n)
else:
print(-1)
else:
if sm > 0:
nk = len(lst)
prsm = [0] * nk
prsm[0] = lst[0]
for i in range(nk):
prsm[i] = lst[i] + prsm[i-1]
minnums = n
for i in range(nk):
ans = icanv(prsm[i], numr) + i + 1
minnums = min(minnums, ans)
if minnums < n//2:
print(-1)
else:
print(minnums)
else:
print(-1)
``` | instruction | 0 | 47,170 | 4 | 94,340 |
No | output | 1 | 47,170 | 4 | 94,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3.
Submitted Solution:
```
n = int(input())
a = list(reversed([int(x) for x in input().split()]))
x = int(input())
if x == 0:
if sum(a) > 0:
print(n)
else:
print(-1)
elif x > 0:
if sum(a) + x*(n//2) > 0:
print(n)
else:
print(-1)
else: # x < 0
cur = 0
peak = 0
peak_i = -1
for i, val in enumerate(a):
cur += val
if cur > peak:
peak = cur
peak_i = i
# now up to and including peak_i is optimal.
# so we want a window of size peak_i + 1 + n//2
w_size = peak_i + 1 + n//2
full_a = [x]*(n//2) + a
val = sum(full_a[:w_size])
if val <= 0:
print(-1)
import sys
sys.exit(0)
for idx in range(w_size, n):
val += full_a[idx] - full_a[idx-w_size]
if val <= 0:
print(-1)
import sys
sys.exit(0)
print(w_size)
``` | instruction | 0 | 47,171 | 4 | 94,342 |
No | output | 1 | 47,171 | 4 | 94,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Levian works as an accountant in a large company. Levian knows how much the company has earned in each of the n consecutive months — in the i-th month the company had income equal to a_i (positive income means profit, negative income means loss, zero income means no change). Because of the general self-isolation, the first ⌈ n/2 ⌉ months income might have been completely unstable, but then everything stabilized and for the last ⌊ n/2 ⌋ months the income was the same.
Levian decided to tell the directors n-k+1 numbers — the total income of the company for each k consecutive months. In other words, for each i between 1 and n-k+1 he will say the value a_i + a_{i+1} + … + a_{i + k - 1}. For example, if a=[-1, 0, 1, 2, 2] and k=3 he will say the numbers 0, 3, 5.
Unfortunately, if at least one total income reported by Levian is not a profit (income ≤ 0), the directors will get angry and fire the failed accountant.
Save Levian's career: find any such k, that for each k months in a row the company had made a profit, or report that it is impossible.
Input
The first line contains a single integer n (2 ≤ n ≤ 5⋅ 10^5) — the number of months for which Levian must account.
The second line contains ⌈{n/2}⌉ integers a_1, a_2, …, a_{⌈{n/2}⌉}, where a_i (-10^9 ≤ a_i ≤ 10^9) — the income of the company in the i-th month.
Third line contains a single integer x (-10^9 ≤ x ≤ 10^9) — income in every month from ⌈{n/2}⌉ + 1 to n.
Output
In a single line, print the appropriate integer k or -1, if it does not exist.
If there are multiple possible answers, you can print any.
Examples
Input
3
2 -1
2
Output
2
Input
5
2 2 -8
2
Output
-1
Input
6
-2 -2 6
-1
Output
4
Note
In the first example, k=2 and k=3 satisfy: in the first case, Levian will report the numbers 1, 1, and in the second case — one number 3.
In the second example, there is no such k.
In the third example, the only answer is k=4: he will report the numbers 1,2,3.
Submitted Solution:
```
from sys import stdin, stdout
n = int(stdin.readline())
income1 = [int(x) for x in stdin.readline().split()][::-1]
income2 = int(stdin.readline())
sums = [0]
for a in income1:
sums.append(a+sums[-1])
sums2 = income2*(n//2)
sums3 = [0]
for a in reversed(income1):
sums3.append(sums3[-1]+a)
i2Len = n//2
if sums[-1]+sums2 > 0:
print(n)
elif sums[-1] > 0:
dRange = n%2 + sums[-1]//abs(income2)
good = 0
for s in range(dRange+1):
if sums[s] > sums2:
good = s+1
break
if good:
valid = True
for ind in range(good,len(sums)):
if sums[ind]+income2*(i2Len-ind) <= 0:
valid = False
break
if valid:
print(good+n//2)
else:
print(-1)
else:
print(-1)
else:
print(-1)
``` | instruction | 0 | 47,172 | 4 | 94,344 |
No | output | 1 | 47,172 | 4 | 94,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it. | instruction | 0 | 47,560 | 4 | 95,120 |
Tags: binary search, greedy, sortings
Correct Solution:
```
def check(m, mid, aa, dd, edays):
avail, ee = [True] * (m + 1), []
for i in range(mid, -1, -1):
day = edays[i]
exam = dd[day]
if avail[exam]:
avail[exam] = False
ee.append(day)
if len(ee) == m:
break
else:
return False
pool, prev = 0, -1
for day in reversed(ee):
pool += day - prev - aa[dd[day]]
if pool < 0:
return False
prev = day
return True
def main():
n, m = map(int, input().split())
dd = list(map(int, input().split()))
edays = [i for i, e in enumerate(dd) if e]
aa = [0, *(int(s) + 1 for s in input().split())]
lo, hi = 0, len(edays)
try:
while lo < hi:
mid = (lo + hi) // 2
if check(m, mid, aa, dd, edays):
hi = mid
else:
lo = mid + 1
print(edays[lo] + 1)
except IndexError:
print(-1)
if __name__ == "__main__":
main()
``` | output | 1 | 47,560 | 4 | 95,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it. | instruction | 0 | 47,561 | 4 | 95,122 |
Tags: binary search, greedy, sortings
Correct Solution:
```
def main():
n, m = map(int, input().split())
dd = list(map(int, input().split()))
aa = [0, *map(int, input().split())]
l = [True] * (m + 1)
l[0], total = False, sum(aa) + len(aa) - 2
for i, e in enumerate(dd, 1):
if l[e]:
if i > aa[e]:
m -= 1
l[e] = False
if not m and i > total:
print(i)
break
else:
print(-1)
if __name__ == '__main__':
main()
``` | output | 1 | 47,561 | 4 | 95,123 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it. | instruction | 0 | 47,562 | 4 | 95,124 |
Tags: binary search, greedy, sortings
Correct Solution:
```
n, m = map(int, input().split())
a = list(map(int, input().split()))
required = list(map(int, input().split()))
visited = [False for i in range(m)]
count = 0
Sum = sum(required)
flag = False
for i in range(n):
if a[i]!=0:
if not visited[a[i]-1]:
if i+1 > required[a[i]-1]:
count += 1
visited[a[i]-1] = True
if count == m and i+1 >= Sum+m:
print(i+1)
flag = True
break
if not flag:
print(-1)
``` | output | 1 | 47,562 | 4 | 95,125 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it. | instruction | 0 | 47,563 | 4 | 95,126 |
Tags: binary search, greedy, sortings
Correct Solution:
```
def check(i):
used = [0] * m
k = 0
for i in range(i, -1, -1):
if d[i] and not used[d[i] - 1]:
k += a[d[i] - 1]
used[d[i] - 1] = 1
elif k:
k -= 1
return (not k)
n, m = map(int, input().split())
d = list(map(int, input().split()))
a = list(map(int, input().split()))
used = [0] * m
k = m
i = 0
while i < n:
if d[i] and not used[d[i] - 1]:
used[d[i] - 1] = 1
k -= 1
if not k:
break
i += 1
if i == n or not check(n - 1):
print(-1)
else:
l = i - 1
r = n - 1
while l < r - 1:
middle = (l + r) // 2
if check(middle):
r = middle
else:
l = middle
print(r + 1)
``` | output | 1 | 47,563 | 4 | 95,127 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it. | instruction | 0 | 47,564 | 4 | 95,128 |
Tags: binary search, greedy, sortings
Correct Solution:
```
from collections import defaultdict
def get_last_exam_day(D, limit):
last_exam_day = defaultdict(int)
for i,m in enumerate(D):
if (i > limit):
break
if (m != 0):
last_exam_day[m] = i
return last_exam_day
def ok(D, A, data):
study_counter = [0] * (len(A) + 1)
study_subject_index = 0
last_day = [False] * (len(D) + 1)
for x in data:
last_day[x[0]] = True
for day,subject in enumerate(D):
if (last_day[day] == True):
if (study_counter[subject] < A[subject - 1]):
return False
else:
study_subject = data[study_subject_index][1]
study_counter[study_subject] += 1
if (study_counter[study_subject] >= A[study_subject - 1]):
study_subject_index += 1
if (study_subject_index >= M):
return True
return True
if __name__ == "__main__":
N,M = map(int, input().split())
D = list(map(int, input().split()))
A = list(map(int, input().split()))
left = 0
right = N
exist_result = False
while (right - left > 1):
med = (right + left) // 2
if (med >= N):
left = med
continue
last_exam_day = get_last_exam_day(D, med)
if (len(last_exam_day) < M):
left = med
continue
data = []
for subject in last_exam_day:
last_day = last_exam_day[subject]
data.append([last_day, subject])
data.sort()
if (ok(D, A, data)):
right = med
exist_result = True
else:
left = med
#print (left, right, med)
if (exist_result):
print (right + 1)
else:
print (-1)
``` | output | 1 | 47,564 | 4 | 95,129 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it. | instruction | 0 | 47,565 | 4 | 95,130 |
Tags: binary search, greedy, sortings
Correct Solution:
```
import math
global n,m,d,a
d=[]
a=[]
def intdec(x):
return int(x)-1
def check(x):
global n,m,d,a
vis=[0]*m
seq=[]
h=0; cnt=0
for i in range(x,-1,-1):
if d[i]<0 or vis[d[i]]:
if len(seq)<=h:
pass
else:
cnt+=1
if cnt==a[seq[h]]: h+=1; cnt=0
else:
vis[d[i]]=1; seq.append(d[i])
if h<m or vis.count(0):
return 0
else: return 1
def solve():
global n,m,d,a
l=-1; r=n
while l<r-1:
mid=(l+r)//2
if check(mid):
r=mid
else: l=mid
if r==n: r=-2
return r
def main():
global n,m,d,a
n,m=map(int,input().split())
d=list(map(intdec,input().split()))
a=list(map(int,input().split()))
print(solve()+1)
if __name__=="__main__":
main()
``` | output | 1 | 47,565 | 4 | 95,131 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it. | instruction | 0 | 47,566 | 4 | 95,132 |
Tags: binary search, greedy, sortings
Correct Solution:
```
#!/usr/bin/env python3
from sys import stdin,stdout
def ri():
return map(int, stdin.readline().split())
def solve(ee):
dest = [-1 for i in range(m)]
for i in range(ee, -1, -1):
if d[i] != -1 and dest[d[i]] == -1:
dest[d[i]] = i
pd = -1
for i in range(ee+1):
da = d[i]
if da != -1 and dest[da] == i:
pd += a[da]
pd += 1
if pd > i:
return 0
if dest.count(-1):
return 0
else:
return 1
n, m = ri()
d = list(ri())
a = list(ri())
d = [i-1 for i in d]
s = 0
e = n
while s < e:
mid = (e+s)//2
if solve(mid):
e = mid
else:
s = mid+1
if s >= n:
print(-1)
else:
print(s+1)
``` | output | 1 | 47,566 | 4 | 95,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it. | instruction | 0 | 47,567 | 4 | 95,134 |
Tags: binary search, greedy, sortings
Correct Solution:
```
import sys
n,m=map(int,input().split())
di=list(map(int,input().split()))
ai=list(map(int,input().split()))
check=[False]*m
temp=0
summ=sum(ai)+m
for i in range(n):
if di[i]==0:
continue
else:
if not check[di[i]-1]:
if ai[di[i]-1]<=i:
temp+=1
#print(i,end=" ")
check[di[i]-1]=True
if temp==m:
# print(check)
if i+1>=summ:
print(i+1)
sys.exit()
print(-1)
``` | output | 1 | 47,567 | 4 | 95,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it.
Submitted Solution:
```
def ch(ind):
vis = [0 for i in range(m)]
k=0
for i in range(ind,-1,-1):
if d[i] and not vis[d[i]-1]:
k+=a[d[i]-1]
vis[d[i]-1]=1
else:
if k:
k-=1
return 0 if k else -1
n,m=map(int,input().split())
d=list(map(int,input().split()))
a=list(map(int,input().split()))
vis=[0 for i in range(m)]
mid=m
i=0
while(i<n):
if d[i] and not vis[d[i] - 1]:
mid-=1
vis[d[i] - 1] = 1
if mid:
pass
else:
break
i+=1
if i==n or not ch(n-1):
print('-1')
exit(0)
l=i-1
r=n-1
while(l<r-1):
mx=(l+r)//2
if(ch(mx)):
r=mx
else:
l=mx
print(l+2)
``` | instruction | 0 | 47,568 | 4 | 95,136 |
Yes | output | 1 | 47,568 | 4 | 95,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
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")
##########################################################
#print('%d %d' %ans)
from collections import Counter
import math
#for _ in range(int(input())):
#n=int(input())
from collections import Counter
def can(days):
d=Counter(arr[:days])
v=[0]*k
free=0
cnt=0
for i in range(days):
if arr[i]==0:
free+=1
else:
if free>=ls[arr[i]-1] and v[arr[i]-1]==False:
if d[arr[i]]==1:
cnt+=1
#d[arr[i]]-=1
free-=ls[arr[i]-1]
v[arr[i]-1]=True
else:
free+=1
d[arr[i]]-=1
else:
d[arr[i]]-=1
free+=1
#if days==5:print(d)
#if days==5:
#print(v)
if cnt==k:
#print("1", 1)
return True
else:
return False
n,k= map(int, input().split())
arr=list(map(int,input().split()))
ls=list(map(int,input().split()))
l=sum(ls)
r=n
ans=-1
while l<=r:
mid=(l+r)//2
#print("mid",mid)
if can(mid):
ans=mid
#print("ans",ans)
r=mid-1
else:
l=mid+1
print(ans)
``` | instruction | 0 | 47,569 | 4 | 95,138 |
Yes | output | 1 | 47,569 | 4 | 95,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it.
Submitted Solution:
```
n,m = map(int, input().split())
dias = [int(x) for x in input().split()]
prep = [int(x) for x in input().split()]
def isposi(nu):
faltan = set(range(1,m+1))
porestudiar = 0
for i in reversed(range(min(nu, n))):
if dias[i] in faltan:
faltan.remove(dias[i])
porestudiar+=prep[dias[i]-1]
elif porestudiar > 0:
porestudiar -= 1
return len(faltan)==0 and porestudiar == 0
if not isposi(n):
print(-1)
else:
lo,hi=0,n
while(hi> lo):
mid = (hi+lo)//2
if isposi(mid):
hi = mid
else:
lo = mid+1
print(lo)
``` | instruction | 0 | 47,570 | 4 | 95,140 |
Yes | output | 1 | 47,570 | 4 | 95,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it.
Submitted Solution:
```
from collections import deque
def main():
n, m = list(map(int, input().split()))
test = list(map(int, input().split()))
prepare = list(map(int, input().split()))
least_days = sum(prepare) + m
if least_days > len(test):
print(-1)
return
index = list()
for p in range(m + 1):
index.append(deque())
for t in range(len(test)):
if test[t] == 0:
continue
index[test[t]].append(t)
if t >= least_days - 1:
index[0].append(t)
if deque([]) in index:
print(-1)
return
min_index = max(list(map(lambda x: x.popleft(), index[1:])))
min_index = max(least_days - 1, min_index)
for x in index[0]:
if x >= min_index:
print(x + 1)
break
if __name__ == '__main__':
main()
``` | instruction | 0 | 47,571 | 4 | 95,142 |
Yes | output | 1 | 47,571 | 4 | 95,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it.
Submitted Solution:
```
def check(now):
last=[0 for i in range(0,m+1)]
ddl=[]
last[0]=-1
for i in range(now,1,-1):
if last[d[i]]==0:
ddl.append(d[i])
last[d[i]]=i
day=0
if len(ddl)!=m: return False
for i in range(1,m+1):
# print(a[ddl[-i]])
day+=a[ddl[-i]]
if day>=last[ddl[-i]]:
return False
return True
a=input()
a=a.split()
n=int(a[0])
m=int(a[1])
d=input()
d=d.split()
a=input()
a=a.split()
a=[0]+a
d=[0]+d
for i in range(0,n+1):
d[i]=int(d[i])
for i in range(0,m+1):
a[i]=int(a[i])
l=0
r=n+1
while l<r-1:
mid=(l+r)>>1
if check(mid):
r=mid
else:
l=mid
if r==n+1:
print(-1)
else:
print(r)
``` | instruction | 0 | 47,572 | 4 | 95,144 |
No | output | 1 | 47,572 | 4 | 95,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it.
Submitted Solution:
```
n, m = map(int, input().split())
d = list(map(int, input().split()))
a = list(map(int, input().split()))
def check_solution(k):
ad, s = 0, 0
u = [False] * (m + 1)
for i in reversed(range(k)):
if d[i] != 0 and not u[d[i]]:
ad -= a[d[i] - 1]
u[d[i]] = True
s += 1
else:
ad += 1
return s == m and ad >= 0
def main():
l = 0
r = n + 1
while l < r:
mi = (r + l) // 2
mv = check_solution(mi)
# print(mi, mv, l, r)
if mv:
r = mi
else:
l = mi + 1
if l <= n:
print(l)
else:
print(-1)
main()
``` | instruction | 0 | 47,573 | 4 | 95,146 |
No | output | 1 | 47,573 | 4 | 95,147 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it.
Submitted Solution:
```
n,m = map(int,input().split())
arr = list(map(int,input().split()))
prep = list(map(int,input().split()))
#mp = dict()
total = sum(prep) + m
#for i in range(1,m+1):
# mp[i] = prep[i-1]
#print(mp)
st = set()
ans = 0
flag = 0
for i in range(n):
st.add(arr[i])
#print(st,i+1,total)
if ((0 in st and len(st)==(m+1)) or ( 0 not in st and len(st)==m) ) and (i+1)>=total and arr[i]:
ans = i + 1
flag = 1
break
if flag :
print(ans)
else:
print(-1)
``` | instruction | 0 | 47,574 | 4 | 95,148 |
No | output | 1 | 47,574 | 4 | 95,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m.
About every day we know exam for which one of m subjects can be passed on that day. Perhaps, some day you can't pass any exam. It is not allowed to pass more than one exam on any day.
On each day Vasiliy can either pass the exam of that day (it takes the whole day) or prepare all day for some exam or have a rest.
About each subject Vasiliy know a number ai — the number of days he should prepare to pass the exam number i. Vasiliy can switch subjects while preparing for exams, it is not necessary to prepare continuously during ai days for the exam number i. He can mix the order of preparation for exams in any way.
Your task is to determine the minimum number of days in which Vasiliy can pass all exams, or determine that it is impossible. Each exam should be passed exactly one time.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of days in the exam period and the number of subjects.
The second line contains n integers d1, d2, ..., dn (0 ≤ di ≤ m), where di is the number of subject, the exam of which can be passed on the day number i. If di equals 0, it is not allowed to pass any exams on the day number i.
The third line contains m positive integers a1, a2, ..., am (1 ≤ ai ≤ 105), where ai is the number of days that are needed to prepare before passing the exam on the subject i.
Output
Print one integer — the minimum number of days in which Vasiliy can pass all exams. If it is impossible, print -1.
Examples
Input
7 2
0 1 0 2 1 0 2
2 1
Output
5
Input
10 3
0 0 1 2 3 0 2 0 1 2
1 1 4
Output
9
Input
5 1
1 1 1 1 1
5
Output
-1
Note
In the first example Vasiliy can behave as follows. On the first and the second day he can prepare for the exam number 1 and pass it on the fifth day, prepare for the exam number 2 on the third day and pass it on the fourth day.
In the second example Vasiliy should prepare for the exam number 3 during the first four days and pass it on the fifth day. Then on the sixth day he should prepare for the exam number 2 and then pass it on the seventh day. After that he needs to prepare for the exam number 1 on the eighth day and pass it on the ninth day.
In the third example Vasiliy can't pass the only exam because he hasn't anough time to prepare for it.
Submitted Solution:
```
def zip_sorted(a,b):
# sorted by a
a,b = zip(*sorted(zip(a,b)))
# sorted by b
sorted(zip(a, b), key=lambda x: x[0])
return a,b
def zip_sorted1(a,b,c):
# sorted by a
a,b,c = zip(*sorted(zip(a,b,c)))
# sorted by b
sorted(zip(a,b,c), key=lambda x: x[0])
return a,b,c
n,m = [int(n1) for n1 in input().split()]
sub_on_day = [int(n1) for n1 in input().split()]
day_req = [int(n1) for n1 in input().split()]
day_prep = [0 for n1 in range(n)]
done_sub = [0 for n1 in range(m)]
count = 0
for i in range(n):
day_prep[i] = count
count = count + 1
tot = 0
count = 0
for i in range(n):
if sub_on_day!=0:
if sub_on_day[i]-1 != m-1:
if done_sub[sub_on_day[i]-1]!=1 and done_sub[sub_on_day[i]]==1:
if day_req[sub_on_day[i]-1]<=(day_prep[i]-tot):
tot = tot + day_req[sub_on_day[i]-1] + 1
count = count + 1
done_sub[sub_on_day[i]-1] = 1
else:
if done_sub[sub_on_day[i]-1]!=1:
if day_req[sub_on_day[i]-1]<=(day_prep[i]-tot):
tot = tot + day_req[sub_on_day[i]-1] + 1
count = count + 1
done_sub[sub_on_day[i]-1] = 1
if sum(done_sub)==m:
print(i+1)
count = 1
break
if count==0:
print(-1)
'''
for i in range(n):
for j in range(n):
for k1 in range(len(a)):
for k2 in range(len(a)):
for k3 in range(len(a)):
'''
``` | instruction | 0 | 47,575 | 4 | 95,150 |
No | output | 1 | 47,575 | 4 | 95,151 |
Provide a correct Python 3 solution for this coding contest problem.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31 | instruction | 0 | 47,784 | 4 | 95,568 |
"Correct Solution:
```
str = input()
print('2018/' + str[5:])
``` | output | 1 | 47,784 | 4 | 95,569 |
Provide a correct Python 3 solution for this coding contest problem.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31 | instruction | 0 | 47,785 | 4 | 95,570 |
"Correct Solution:
```
s=input()
t="2018"+s[4:]
print(t)
``` | output | 1 | 47,785 | 4 | 95,571 |
Provide a correct Python 3 solution for this coding contest problem.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31 | instruction | 0 | 47,786 | 4 | 95,572 |
"Correct Solution:
```
s = input()
print(s.replace('7', '8', 1))
``` | output | 1 | 47,786 | 4 | 95,573 |
Provide a correct Python 3 solution for this coding contest problem.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31 | instruction | 0 | 47,787 | 4 | 95,574 |
"Correct Solution:
```
s = input()
print(s[:3]+'8'+s[-6:])
``` | output | 1 | 47,787 | 4 | 95,575 |
Provide a correct Python 3 solution for this coding contest problem.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31 | instruction | 0 | 47,788 | 4 | 95,576 |
"Correct Solution:
```
a = input()
print(a[:3]+'8'+a[4:])
``` | output | 1 | 47,788 | 4 | 95,577 |
Provide a correct Python 3 solution for this coding contest problem.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31 | instruction | 0 | 47,789 | 4 | 95,578 |
"Correct Solution:
```
s = input()
x = "2018" + s[4:]
print(x)
``` | output | 1 | 47,789 | 4 | 95,579 |
Provide a correct Python 3 solution for this coding contest problem.
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`.
After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
Constraints
* S is a string of length 10.
* The first eight characters in S are `2017/01/`.
* The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).
Input
Input is given from Standard Input in the following format:
S
Output
Replace the first four characters in S with `2018` and print it.
Examples
Input
2017/01/07
Output
2018/01/07
Input
2017/01/31
Output
2018/01/31 | instruction | 0 | 47,790 | 4 | 95,580 |
"Correct Solution:
```
s=input();print(s[:3]+'8'+s[4:])
``` | output | 1 | 47,790 | 4 | 95,581 |
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