message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, …, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The description of each test case contains a single integer x (1 ≤ x ≤ 10^{18}) — the number of cells for building staircases.
Output
For each test case output a single integer — the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
# https://codeforces.com/contest/1419/problem/B
T=int(input())
for t in range(T):
N = int(input())
sum = 0
i=0
cells=0
count=0
while(cells<=N):
sum=sum*2+1
print(sum)
# i+=2
cells+=(sum*(sum+1))/2
count+=1
print(count-1)
``` | instruction | 0 | 8,544 | 8 | 17,088 |
No | output | 1 | 8,544 | 8 | 17,089 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Jett is tired after destroying the town and she wants to have a rest. She likes high places, that's why for having a rest she wants to get high and she decided to craft staircases.
A staircase is a squared figure that consists of square cells. Each staircase consists of an arbitrary number of stairs. If a staircase has n stairs, then it is made of n columns, the first column is 1 cell high, the second column is 2 cells high, …, the n-th column if n cells high. The lowest cells of all stairs must be in the same row.
A staircase with n stairs is called nice, if it may be covered by n disjoint squares made of cells. All squares should fully consist of cells of a staircase.
This is how a nice covered staircase with 7 stairs looks like: <image>
Find out the maximal number of different nice staircases, that can be built, using no more than x cells, in total. No cell can be used more than once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The description of each test case contains a single integer x (1 ≤ x ≤ 10^{18}) — the number of cells for building staircases.
Output
For each test case output a single integer — the number of different nice staircases, that can be built, using not more than x cells, in total.
Example
Input
4
1
8
6
1000000000000000000
Output
1
2
1
30
Note
In the first test case, it is possible to build only one staircase, that consists of 1 stair. It's nice. That's why the answer is 1.
In the second test case, it is possible to build two different nice staircases: one consists of 1 stair, and another consists of 3 stairs. This will cost 7 cells. In this case, there is one cell left, but it is not possible to use it for building any nice staircases, that have not been built yet. That's why the answer is 2.
In the third test case, it is possible to build only one of two nice staircases: with 1 stair or with 3 stairs. In the first case, there will be 5 cells left, that may be used only to build a staircase with 2 stairs. This staircase is not nice, and Jett only builds nice staircases. That's why in this case the answer is 1. If Jett builds a staircase with 3 stairs, then there are no more cells left, so the answer is 1 again.
Submitted Solution:
```
for _ in " "*int(input()):
n,ans=int(input()),0
for j in range(1,40):
d=2**j-1
s=d*(d+1)/2
if n>=s:
n-=s
ans+=1
else:
break
print(ans)
``` | instruction | 0 | 8,545 | 8 | 17,090 |
No | output | 1 | 8,545 | 8 | 17,091 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Companies always have a lot of equipment, furniture and other things. All of them should be tracked. To do this, there is an inventory number assigned with each item. It is much easier to create a database by using those numbers and keep the track of everything.
During an audit, you were surprised to find out that the items are not numbered sequentially, and some items even share the same inventory number! There is an urgent need to fix it. You have chosen to make the numbers of the items sequential, starting with 1. Changing a number is quite a time-consuming process, and you would like to make maximum use of the current numbering.
You have been given information on current inventory numbers for n items in the company. Renumber items so that their inventory numbers form a permutation of numbers from 1 to n by changing the number of as few items as possible. Let us remind you that a set of n numbers forms a permutation if all the numbers are in the range from 1 to n, and no two numbers are equal.
Input
The first line contains a single integer n — the number of items (1 ≤ n ≤ 105).
The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — the initial inventory numbers of the items.
Output
Print n numbers — the final inventory numbers of the items in the order they occur in the input. If there are multiple possible answers, you may print any of them.
Examples
Input
3
1 3 2
Output
1 3 2
Input
4
2 2 3 3
Output
2 1 3 4
Input
1
2
Output
1
Note
In the first test the numeration is already a permutation, so there is no need to change anything.
In the second test there are two pairs of equal numbers, in each pair you need to replace one number.
In the third test you need to replace 2 by 1, as the numbering should start from one.
Submitted Solution:
```
n = int(input())
inp = list(map(int,input().split(' ')))
toreplace = set()
values = set(range(1,n+1))
for i in range(n):
num = inp[i]
if num in values:
values.remove(num)
else:
toreplace.add(i)
for index in toreplace:
inp[index] = values.pop()
print(*inp)
``` | instruction | 0 | 8,677 | 8 | 17,354 |
Yes | output | 1 | 8,677 | 8 | 17,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Companies always have a lot of equipment, furniture and other things. All of them should be tracked. To do this, there is an inventory number assigned with each item. It is much easier to create a database by using those numbers and keep the track of everything.
During an audit, you were surprised to find out that the items are not numbered sequentially, and some items even share the same inventory number! There is an urgent need to fix it. You have chosen to make the numbers of the items sequential, starting with 1. Changing a number is quite a time-consuming process, and you would like to make maximum use of the current numbering.
You have been given information on current inventory numbers for n items in the company. Renumber items so that their inventory numbers form a permutation of numbers from 1 to n by changing the number of as few items as possible. Let us remind you that a set of n numbers forms a permutation if all the numbers are in the range from 1 to n, and no two numbers are equal.
Input
The first line contains a single integer n — the number of items (1 ≤ n ≤ 105).
The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — the initial inventory numbers of the items.
Output
Print n numbers — the final inventory numbers of the items in the order they occur in the input. If there are multiple possible answers, you may print any of them.
Examples
Input
3
1 3 2
Output
1 3 2
Input
4
2 2 3 3
Output
2 1 3 4
Input
1
2
Output
1
Note
In the first test the numeration is already a permutation, so there is no need to change anything.
In the second test there are two pairs of equal numbers, in each pair you need to replace one number.
In the third test you need to replace 2 by 1, as the numbering should start from one.
Submitted Solution:
```
n = int(input())
inventory = input().split(' ')
inventory = [int(x) for x in inventory]
if n == 1:
inventory[0] = 1
else:
ref = [int(i) for i in range(1,n+1)]
for i in range(n):
if inventory[i] in ref:
ref.remove(inventory[i])
else:
inventory[i] = ref[0]
ref.remove(inventory[i])
out = ' '.join(str(x) for x in inventory)
print(out)
``` | instruction | 0 | 8,681 | 8 | 17,362 |
No | output | 1 | 8,681 | 8 | 17,363 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yet another round on DecoForces is coming! Grandpa Maks wanted to participate in it but someone has stolen his precious sofa! And how can one perform well with such a major loss?
Fortunately, the thief had left a note for Grandpa Maks. This note got Maks to the sofa storehouse. Still he had no idea which sofa belongs to him as they all looked the same!
The storehouse is represented as matrix n × m. Every sofa takes two neighbouring by some side cells. No cell is covered by more than one sofa. There can be empty cells.
Sofa A is standing to the left of sofa B if there exist two such cells a and b that xa < xb, a is covered by A and b is covered by B. Sofa A is standing to the top of sofa B if there exist two such cells a and b that ya < yb, a is covered by A and b is covered by B. Right and bottom conditions are declared the same way.
Note that in all conditions A ≠ B. Also some sofa A can be both to the top of another sofa B and to the bottom of it. The same is for left and right conditions.
The note also stated that there are cntl sofas to the left of Grandpa Maks's sofa, cntr — to the right, cntt — to the top and cntb — to the bottom.
Grandpa Maks asks you to help him to identify his sofa. It is guaranteed that there is no more than one sofa of given conditions.
Output the number of Grandpa Maks's sofa. If there is no such sofa that all the conditions are met for it then output -1.
Input
The first line contains one integer number d (1 ≤ d ≤ 105) — the number of sofas in the storehouse.
The second line contains two integer numbers n, m (1 ≤ n, m ≤ 105) — the size of the storehouse.
Next d lines contains four integer numbers x1, y1, x2, y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — coordinates of the i-th sofa. It is guaranteed that cells (x1, y1) and (x2, y2) have common side, (x1, y1) ≠ (x2, y2) and no cell is covered by more than one sofa.
The last line contains four integer numbers cntl, cntr, cntt, cntb (0 ≤ cntl, cntr, cntt, cntb ≤ d - 1).
Output
Print the number of the sofa for which all the conditions are met. Sofas are numbered 1 through d as given in input. If there is no such sofa then print -1.
Examples
Input
2
3 2
3 1 3 2
1 2 2 2
1 0 0 1
Output
1
Input
3
10 10
1 2 1 1
5 5 6 5
6 4 5 4
2 1 2 0
Output
2
Input
2
2 2
2 1 1 1
1 2 2 2
1 0 0 0
Output
-1
Note
Let's consider the second example.
* The first sofa has 0 to its left, 2 sofas to its right ((1, 1) is to the left of both (5, 5) and (5, 4)), 0 to its top and 2 to its bottom (both 2nd and 3rd sofas are below).
* The second sofa has cntl = 2, cntr = 1, cntt = 2 and cntb = 0.
* The third sofa has cntl = 2, cntr = 1, cntt = 1 and cntb = 1.
So the second one corresponds to the given conditions.
In the third example
* The first sofa has cntl = 1, cntr = 1, cntt = 0 and cntb = 1.
* The second sofa has cntl = 1, cntr = 1, cntt = 1 and cntb = 0.
And there is no sofa with the set (1, 0, 0, 0) so the answer is -1. | instruction | 0 | 8,755 | 8 | 17,510 |
Tags: brute force, implementation
Correct Solution:
```
from sys import stdin, stdout
k = int(stdin.readline())
n, m = map(int, stdin.readline().split())
left, right, down, up = [], [], [], []
coordinates = []
for i in range(k):
x1, y1, x2, y2 = map(int, stdin.readline().split())
if x1 == x2:
if y1 < y2:
coordinates.append((x1, y1, x2, y2, i))
else:
coordinates.append((x2, y2, x1, y1, i))
else:
if x1 < x2:
coordinates.append((x1, y1, x2, y2, i))
else:
coordinates.append((x2, y2, x1, y1, i))
left.append(coordinates[-1])
right.append(coordinates[-1])
up.append(coordinates[-1])
down.append(coordinates[-1])
left.sort(key = lambda x: (x[0], x[2]))
down.sort(key = lambda x: (x[1], x[3]))
challengers = [[], [], [], []]
cntl, cntr, cntd, cntu = map(int, stdin.readline().split())
label = 1
if cntl or not cntl:
for i in range(cntl, -1, -1):
if (left[i][0], left[i][2]) == (left[cntl][0], left[cntl][2]):
challengers[0].append(left[i][-1])
else:
break
for i in range(cntl + 1, k):
if (left[i][0], left[i][2]) == (left[cntl][0], left[cntl][2]) and left[i][2] > left[i][0]:
label = 0
if (left[i][0], left[i][2]) == (left[cntl][0], left[cntl][2]):
challengers[0].append(left[i][-1])
else:
break
if cntr or not cntr:
for i in range(k - 1 - cntr, k):
if (left[i][0], left[i][2]) == (left[k - 1 - cntr][0], left[k - 1 - cntr][2]):
challengers[1].append(left[i][-1])
else:
break
for i in range(k - 2 - cntr, -1, -1):
if (left[i][0], left[i][2]) == (left[k - 1 - cntr][0], left[k - 1 - cntr][2]) and left[i][2] > left[i][0]:
label = 0
if (left[i][0], left[i][2]) == (left[k - 1 - cntr][0], left[k - 1 - cntr][2]):
challengers[1].append(left[i][-1])
else:
break
#!!!!!!!!!!!
if cntd or not cntd:
for i in range(cntd, -1, -1):
if (down[i][1], down[i][3]) == (down[cntd][1], down[cntd][3]):
challengers[2].append(down[i][-1])
else:
break
for i in range(cntd + 1, k):
if (down[i][1], down[i][3]) == (down[cntd][1], down[cntd][3]) and down[i][3] > down[i][1]:
label = 0
if (down[i][1], down[i][3]) == (down[cntd][1], down[cntd][3]):
challengers[2].append(down[i][-1])
else:
break
if cntu or not cntu:
for i in range(k - 1 - cntu, k):
if (down[i][1], down[i][3]) == (down[k - 1 - cntu][1], down[k - 1 - cntu][3]):
challengers[3].append(down[i][-1])
else:
break
for i in range(k - 2 - cntu, -1, -1):
if (down[i][1], down[i][3]) == (down[k - 1 - cntu][1], down[k - 1 - cntu][3]) and down[i][3] > down[i][1]:
label = 0
if (down[i][1], down[i][3]) == (down[k - 1 - cntu][1], down[k - 1 - cntu][3]):
challengers[3].append(down[i][-1])
else:
break
ans = set(challengers[0]) & set(challengers[1]) & set(challengers[2]) & set(challengers[3])
if not len(ans) or not label:
stdout.write('-1')
else:
stdout.write(str(list(ans)[0] + 1))
``` | output | 1 | 8,755 | 8 | 17,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yet another round on DecoForces is coming! Grandpa Maks wanted to participate in it but someone has stolen his precious sofa! And how can one perform well with such a major loss?
Fortunately, the thief had left a note for Grandpa Maks. This note got Maks to the sofa storehouse. Still he had no idea which sofa belongs to him as they all looked the same!
The storehouse is represented as matrix n × m. Every sofa takes two neighbouring by some side cells. No cell is covered by more than one sofa. There can be empty cells.
Sofa A is standing to the left of sofa B if there exist two such cells a and b that xa < xb, a is covered by A and b is covered by B. Sofa A is standing to the top of sofa B if there exist two such cells a and b that ya < yb, a is covered by A and b is covered by B. Right and bottom conditions are declared the same way.
Note that in all conditions A ≠ B. Also some sofa A can be both to the top of another sofa B and to the bottom of it. The same is for left and right conditions.
The note also stated that there are cntl sofas to the left of Grandpa Maks's sofa, cntr — to the right, cntt — to the top and cntb — to the bottom.
Grandpa Maks asks you to help him to identify his sofa. It is guaranteed that there is no more than one sofa of given conditions.
Output the number of Grandpa Maks's sofa. If there is no such sofa that all the conditions are met for it then output -1.
Input
The first line contains one integer number d (1 ≤ d ≤ 105) — the number of sofas in the storehouse.
The second line contains two integer numbers n, m (1 ≤ n, m ≤ 105) — the size of the storehouse.
Next d lines contains four integer numbers x1, y1, x2, y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — coordinates of the i-th sofa. It is guaranteed that cells (x1, y1) and (x2, y2) have common side, (x1, y1) ≠ (x2, y2) and no cell is covered by more than one sofa.
The last line contains four integer numbers cntl, cntr, cntt, cntb (0 ≤ cntl, cntr, cntt, cntb ≤ d - 1).
Output
Print the number of the sofa for which all the conditions are met. Sofas are numbered 1 through d as given in input. If there is no such sofa then print -1.
Examples
Input
2
3 2
3 1 3 2
1 2 2 2
1 0 0 1
Output
1
Input
3
10 10
1 2 1 1
5 5 6 5
6 4 5 4
2 1 2 0
Output
2
Input
2
2 2
2 1 1 1
1 2 2 2
1 0 0 0
Output
-1
Note
Let's consider the second example.
* The first sofa has 0 to its left, 2 sofas to its right ((1, 1) is to the left of both (5, 5) and (5, 4)), 0 to its top and 2 to its bottom (both 2nd and 3rd sofas are below).
* The second sofa has cntl = 2, cntr = 1, cntt = 2 and cntb = 0.
* The third sofa has cntl = 2, cntr = 1, cntt = 1 and cntb = 1.
So the second one corresponds to the given conditions.
In the third example
* The first sofa has cntl = 1, cntr = 1, cntt = 0 and cntb = 1.
* The second sofa has cntl = 1, cntr = 1, cntt = 1 and cntb = 0.
And there is no sofa with the set (1, 0, 0, 0) so the answer is -1. | instruction | 0 | 8,756 | 8 | 17,512 |
Tags: brute force, implementation
Correct Solution:
```
from sys import stdin, stdout
k = int(stdin.readline())
n, m = map(int, stdin.readline().split())
left, right, down, up = [], [], [], []
coordinates = []
for i in range(k):
x1, y1, x2, y2 = map(int, stdin.readline().split())
if x1 == x2:
if y1 < y2:
coordinates.append((x1, y1, x2, y2, i))
else:
coordinates.append((x2, y2, x1, y1, i))
else:
if x1 < x2:
coordinates.append((x1, y1, x2, y2, i))
else:
coordinates.append((x2, y2, x1, y1, i))
left.append(coordinates[-1])
right.append(coordinates[-1])
up.append(coordinates[-1])
down.append(coordinates[-1])
left.sort(key = lambda x: (x[0], x[2]))
down.sort(key = lambda x: (x[1], x[3]))
challengers = [[], [], [], []]
cntl, cntr, cntd, cntu = map(int, stdin.readline().split())
label = 1
if cntl or not cntl:
for i in range(cntl, -1, -1):
if (left[i][0], left[i][2]) == (left[cntl][0], left[cntl][2]):
challengers[0].append(left[i][-1])
else:
break
for i in range(cntl + 1, k):
if (left[i][0], left[i][2]) == (left[cntl][0], left[cntl][2]) and left[i][2] > left[i][0]:
label = 0
if (left[i][0], left[i][2]) == (left[cntl][0], left[cntl][2]):
challengers[0].append(left[i][-1])
else:
break
if cntr or not cntr:
for i in range(k - 1 - cntr, k):
if (left[i][0], left[i][2]) == (left[k - 1 - cntr][0], left[k - 1 - cntr][2]):
challengers[1].append(left[i][-1])
else:
break
for i in range(k - 2 - cntr, -1, -1):
if (left[i][0], left[i][2]) == (left[k - 1 - cntr][0], left[k - 1 - cntr][2]) and left[i][2] > left[i][0]:
label = 0
if (left[i][0], left[i][2]) == (left[k - 1 - cntr][0], left[k - 1 - cntr][2]):
challengers[1].append(left[i][-1])
else:
break
#!!!!!!!!!!!
if cntd or not cntd:
for i in range(cntd, -1, -1):
if (down[i][1], down[i][3]) == (down[cntd][1], down[cntd][3]):
challengers[2].append(down[i][-1])
else:
break
for i in range(cntd + 1, k):
if (down[i][1], down[i][3]) == (down[cntd][1], down[cntd][3]) and down[i][3] > down[i][1]:
label = 0
if (down[i][1], down[i][3]) == (down[cntd][1], down[cntd][3]):
challengers[2].append(down[i][-1])
else:
break
if cntu or not cntu:
for i in range(k - 1 - cntu, k):
if (down[i][1], down[i][3]) == (down[k - 1 - cntu][1], down[k - 1 - cntu][3]):
challengers[3].append(down[i][-1])
else:
break
for i in range(k - 2 - cntu, -1, -1):
if (down[i][1], down[i][3]) == (down[k - 1 - cntu][1], down[k - 1 - cntu][3]) and down[i][3] > down[i][1]:
label = 0
if (down[i][1], down[i][3]) == (down[k - 1 - cntu][1], down[k - 1 - cntu][3]):
challengers[3].append(down[i][-1])
else:
break
ans = set(challengers[0]) & set(challengers[1]) & set(challengers[2]) & set(challengers[3])
if not len(ans) or not label:
stdout.write('-1')
else:
stdout.write(str(list(ans)[0] + 1))
# Made By Mostafa_Khaled
``` | output | 1 | 8,756 | 8 | 17,513 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yet another round on DecoForces is coming! Grandpa Maks wanted to participate in it but someone has stolen his precious sofa! And how can one perform well with such a major loss?
Fortunately, the thief had left a note for Grandpa Maks. This note got Maks to the sofa storehouse. Still he had no idea which sofa belongs to him as they all looked the same!
The storehouse is represented as matrix n × m. Every sofa takes two neighbouring by some side cells. No cell is covered by more than one sofa. There can be empty cells.
Sofa A is standing to the left of sofa B if there exist two such cells a and b that xa < xb, a is covered by A and b is covered by B. Sofa A is standing to the top of sofa B if there exist two such cells a and b that ya < yb, a is covered by A and b is covered by B. Right and bottom conditions are declared the same way.
Note that in all conditions A ≠ B. Also some sofa A can be both to the top of another sofa B and to the bottom of it. The same is for left and right conditions.
The note also stated that there are cntl sofas to the left of Grandpa Maks's sofa, cntr — to the right, cntt — to the top and cntb — to the bottom.
Grandpa Maks asks you to help him to identify his sofa. It is guaranteed that there is no more than one sofa of given conditions.
Output the number of Grandpa Maks's sofa. If there is no such sofa that all the conditions are met for it then output -1.
Input
The first line contains one integer number d (1 ≤ d ≤ 105) — the number of sofas in the storehouse.
The second line contains two integer numbers n, m (1 ≤ n, m ≤ 105) — the size of the storehouse.
Next d lines contains four integer numbers x1, y1, x2, y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — coordinates of the i-th sofa. It is guaranteed that cells (x1, y1) and (x2, y2) have common side, (x1, y1) ≠ (x2, y2) and no cell is covered by more than one sofa.
The last line contains four integer numbers cntl, cntr, cntt, cntb (0 ≤ cntl, cntr, cntt, cntb ≤ d - 1).
Output
Print the number of the sofa for which all the conditions are met. Sofas are numbered 1 through d as given in input. If there is no such sofa then print -1.
Examples
Input
2
3 2
3 1 3 2
1 2 2 2
1 0 0 1
Output
1
Input
3
10 10
1 2 1 1
5 5 6 5
6 4 5 4
2 1 2 0
Output
2
Input
2
2 2
2 1 1 1
1 2 2 2
1 0 0 0
Output
-1
Note
Let's consider the second example.
* The first sofa has 0 to its left, 2 sofas to its right ((1, 1) is to the left of both (5, 5) and (5, 4)), 0 to its top and 2 to its bottom (both 2nd and 3rd sofas are below).
* The second sofa has cntl = 2, cntr = 1, cntt = 2 and cntb = 0.
* The third sofa has cntl = 2, cntr = 1, cntt = 1 and cntb = 1.
So the second one corresponds to the given conditions.
In the third example
* The first sofa has cntl = 1, cntr = 1, cntt = 0 and cntb = 1.
* The second sofa has cntl = 1, cntr = 1, cntt = 1 and cntb = 0.
And there is no sofa with the set (1, 0, 0, 0) so the answer is -1. | instruction | 0 | 8,757 | 8 | 17,514 |
Tags: brute force, implementation
Correct Solution:
```
import sys
from bisect import bisect_left, bisect_right
d = int(sys.stdin.buffer.readline().decode('utf-8'))
n, m = map(int, sys.stdin.buffer.readline().decode('utf-8').split())
a = [list(map(int, sys.stdin.buffer.readline().decode('utf-8').split()))
for _ in range(d)]
cnt = list(map(int, sys.stdin.buffer.readline().decode('utf-8').split()))
left, right, top, bottom = [], [], [], []
for x1, y1, x2, y2 in a:
left.append(min(x1, x2))
right.append(-max(x1, x2))
top.append(min(y1, y2))
bottom.append(-max(y1, y2))
left.sort()
right.sort()
top.sort()
bottom.sort()
for i, (x1, y1, x2, y2) in enumerate(a, start=1):
c = [
bisect_left(left, max(x1, x2)) - (1 if x1 != x2 else 0),
bisect_left(right, -min(x1, x2)) - (1 if x1 != x2 else 0),
bisect_left(top, max(y1, y2)) - (1 if y1 != y2 else 0),
bisect_left(bottom, -min(y1, y2)) - (1 if y1 != y2 else 0)
]
if c == cnt:
print(i)
exit()
print(-1)
``` | output | 1 | 8,757 | 8 | 17,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yet another round on DecoForces is coming! Grandpa Maks wanted to participate in it but someone has stolen his precious sofa! And how can one perform well with such a major loss?
Fortunately, the thief had left a note for Grandpa Maks. This note got Maks to the sofa storehouse. Still he had no idea which sofa belongs to him as they all looked the same!
The storehouse is represented as matrix n × m. Every sofa takes two neighbouring by some side cells. No cell is covered by more than one sofa. There can be empty cells.
Sofa A is standing to the left of sofa B if there exist two such cells a and b that xa < xb, a is covered by A and b is covered by B. Sofa A is standing to the top of sofa B if there exist two such cells a and b that ya < yb, a is covered by A and b is covered by B. Right and bottom conditions are declared the same way.
Note that in all conditions A ≠ B. Also some sofa A can be both to the top of another sofa B and to the bottom of it. The same is for left and right conditions.
The note also stated that there are cntl sofas to the left of Grandpa Maks's sofa, cntr — to the right, cntt — to the top and cntb — to the bottom.
Grandpa Maks asks you to help him to identify his sofa. It is guaranteed that there is no more than one sofa of given conditions.
Output the number of Grandpa Maks's sofa. If there is no such sofa that all the conditions are met for it then output -1.
Input
The first line contains one integer number d (1 ≤ d ≤ 105) — the number of sofas in the storehouse.
The second line contains two integer numbers n, m (1 ≤ n, m ≤ 105) — the size of the storehouse.
Next d lines contains four integer numbers x1, y1, x2, y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — coordinates of the i-th sofa. It is guaranteed that cells (x1, y1) and (x2, y2) have common side, (x1, y1) ≠ (x2, y2) and no cell is covered by more than one sofa.
The last line contains four integer numbers cntl, cntr, cntt, cntb (0 ≤ cntl, cntr, cntt, cntb ≤ d - 1).
Output
Print the number of the sofa for which all the conditions are met. Sofas are numbered 1 through d as given in input. If there is no such sofa then print -1.
Examples
Input
2
3 2
3 1 3 2
1 2 2 2
1 0 0 1
Output
1
Input
3
10 10
1 2 1 1
5 5 6 5
6 4 5 4
2 1 2 0
Output
2
Input
2
2 2
2 1 1 1
1 2 2 2
1 0 0 0
Output
-1
Note
Let's consider the second example.
* The first sofa has 0 to its left, 2 sofas to its right ((1, 1) is to the left of both (5, 5) and (5, 4)), 0 to its top and 2 to its bottom (both 2nd and 3rd sofas are below).
* The second sofa has cntl = 2, cntr = 1, cntt = 2 and cntb = 0.
* The third sofa has cntl = 2, cntr = 1, cntt = 1 and cntb = 1.
So the second one corresponds to the given conditions.
In the third example
* The first sofa has cntl = 1, cntr = 1, cntt = 0 and cntb = 1.
* The second sofa has cntl = 1, cntr = 1, cntt = 1 and cntb = 0.
And there is no sofa with the set (1, 0, 0, 0) so the answer is -1. | instruction | 0 | 8,758 | 8 | 17,516 |
Tags: brute force, implementation
Correct Solution:
```
import sys
try:
fin=open('in')
except:
fin=sys.stdin
input=fin.readline
d = int(input())
n, m = map(int, input().split())
x1, y1, x2, y2 = [], [], [], []
T=[]
for _ in range(d):
u, v, w, x = map(int, input().split())
if u>w:u,w=w,u
if v>x:v,x=x,v
x1.append(u)
y1.append(v)
x2.append(-w)#the other direction pog?
y2.append(-x)
T.append([u,v,w,x])
x1.sort()
x2.sort()
y1.sort()
y2.sort()
req=list(map(int,input().split())) # x1,x2,y1,y2
import bisect
for i in range(len(T)):
# binary search
u,v,w,x=T[i]
if req[0]==bisect.bisect_left(x1,w)-(u!=w):
if req[1]==bisect.bisect_left(x2,-u)-(u!=w):
if req[2]==bisect.bisect_left(y1,x)-(v!=x):
if req[3]==bisect.bisect_left(y2,-v)-(v!=x):
print(i+1)
break
else:
print(-1)
``` | output | 1 | 8,758 | 8 | 17,517 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yet another round on DecoForces is coming! Grandpa Maks wanted to participate in it but someone has stolen his precious sofa! And how can one perform well with such a major loss?
Fortunately, the thief had left a note for Grandpa Maks. This note got Maks to the sofa storehouse. Still he had no idea which sofa belongs to him as they all looked the same!
The storehouse is represented as matrix n × m. Every sofa takes two neighbouring by some side cells. No cell is covered by more than one sofa. There can be empty cells.
Sofa A is standing to the left of sofa B if there exist two such cells a and b that xa < xb, a is covered by A and b is covered by B. Sofa A is standing to the top of sofa B if there exist two such cells a and b that ya < yb, a is covered by A and b is covered by B. Right and bottom conditions are declared the same way.
Note that in all conditions A ≠ B. Also some sofa A can be both to the top of another sofa B and to the bottom of it. The same is for left and right conditions.
The note also stated that there are cntl sofas to the left of Grandpa Maks's sofa, cntr — to the right, cntt — to the top and cntb — to the bottom.
Grandpa Maks asks you to help him to identify his sofa. It is guaranteed that there is no more than one sofa of given conditions.
Output the number of Grandpa Maks's sofa. If there is no such sofa that all the conditions are met for it then output -1.
Input
The first line contains one integer number d (1 ≤ d ≤ 105) — the number of sofas in the storehouse.
The second line contains two integer numbers n, m (1 ≤ n, m ≤ 105) — the size of the storehouse.
Next d lines contains four integer numbers x1, y1, x2, y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — coordinates of the i-th sofa. It is guaranteed that cells (x1, y1) and (x2, y2) have common side, (x1, y1) ≠ (x2, y2) and no cell is covered by more than one sofa.
The last line contains four integer numbers cntl, cntr, cntt, cntb (0 ≤ cntl, cntr, cntt, cntb ≤ d - 1).
Output
Print the number of the sofa for which all the conditions are met. Sofas are numbered 1 through d as given in input. If there is no such sofa then print -1.
Examples
Input
2
3 2
3 1 3 2
1 2 2 2
1 0 0 1
Output
1
Input
3
10 10
1 2 1 1
5 5 6 5
6 4 5 4
2 1 2 0
Output
2
Input
2
2 2
2 1 1 1
1 2 2 2
1 0 0 0
Output
-1
Note
Let's consider the second example.
* The first sofa has 0 to its left, 2 sofas to its right ((1, 1) is to the left of both (5, 5) and (5, 4)), 0 to its top and 2 to its bottom (both 2nd and 3rd sofas are below).
* The second sofa has cntl = 2, cntr = 1, cntt = 2 and cntb = 0.
* The third sofa has cntl = 2, cntr = 1, cntt = 1 and cntb = 1.
So the second one corresponds to the given conditions.
In the third example
* The first sofa has cntl = 1, cntr = 1, cntt = 0 and cntb = 1.
* The second sofa has cntl = 1, cntr = 1, cntt = 1 and cntb = 0.
And there is no sofa with the set (1, 0, 0, 0) so the answer is -1. | instruction | 0 | 8,759 | 8 | 17,518 |
Tags: brute force, implementation
Correct Solution:
```
import sys
from collections import defaultdict as dd
from collections import deque
pl=1
from math import *
import copy
#sys.setrecursionlimit(10**6)
if pl:
input=sys.stdin.readline
def li():
return [int(xxx) for xxx in input().split()]
def fi():
return int(input())
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
d=[]
from bisect import *
from itertools import permutations
from bisect import *
f=[0 for i in range(100)]
for i in range(1,100):
if i==1:
f[i]=1
else:
f[i]=2*f[i-1]+1
#print(f[:15])
def rec(n,k):
s=[]
while n!=0:
n,r=n//k,n%k
#print(n,r)
if r<0:
r-=k
n+=1
#print(s,n,r)
s.append(r)
return s
n=fi()
p,q=mi()
pp=[]
l=[];r=[];u=[];d=[]
for i in range(n):
x1,y1,x2,y2=mi()
x1,x2=min(x1,x2),max(x1,x2)
y1,y2=min(y1,y2),max(y1,y2)
l.append(x1)
r.append(-x2)
u.append(y1)
d.append(-y2)
pp.append([x1,x2,y1,y2])
l.sort()
r.sort()
u.sort()
d.sort()
#print(l,r,u,d)
f=[[0,0,0,0] for i in range(n+1)]
for i in range(n):
f[i][0]=bisect_left(l,pp[i][1])
if pp[i][0]<pp[i][1]:
f[i][0]-=1
f[i][1]=bisect_left(r,-pp[i][0])
if pp[i][0]<pp[i][1]:
f[i][1]-=1
f[i][2]=bisect_left(u,pp[i][3])
if pp[i][2]<pp[i][3]:
f[i][2]-=1
f[i][3]=bisect_left(d,-pp[i][2])
if pp[i][2]<pp[i][3]:
f[i][3]-=1
#f[l[i][1]][0]=bisect_left(l,)
co=li()
#print(f)
for i in range(n):
if co==f[i]:
print(i+1)
exit(0)
print(-1)
``` | output | 1 | 8,759 | 8 | 17,519 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yet another round on DecoForces is coming! Grandpa Maks wanted to participate in it but someone has stolen his precious sofa! And how can one perform well with such a major loss?
Fortunately, the thief had left a note for Grandpa Maks. This note got Maks to the sofa storehouse. Still he had no idea which sofa belongs to him as they all looked the same!
The storehouse is represented as matrix n × m. Every sofa takes two neighbouring by some side cells. No cell is covered by more than one sofa. There can be empty cells.
Sofa A is standing to the left of sofa B if there exist two such cells a and b that xa < xb, a is covered by A and b is covered by B. Sofa A is standing to the top of sofa B if there exist two such cells a and b that ya < yb, a is covered by A and b is covered by B. Right and bottom conditions are declared the same way.
Note that in all conditions A ≠ B. Also some sofa A can be both to the top of another sofa B and to the bottom of it. The same is for left and right conditions.
The note also stated that there are cntl sofas to the left of Grandpa Maks's sofa, cntr — to the right, cntt — to the top and cntb — to the bottom.
Grandpa Maks asks you to help him to identify his sofa. It is guaranteed that there is no more than one sofa of given conditions.
Output the number of Grandpa Maks's sofa. If there is no such sofa that all the conditions are met for it then output -1.
Input
The first line contains one integer number d (1 ≤ d ≤ 105) — the number of sofas in the storehouse.
The second line contains two integer numbers n, m (1 ≤ n, m ≤ 105) — the size of the storehouse.
Next d lines contains four integer numbers x1, y1, x2, y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — coordinates of the i-th sofa. It is guaranteed that cells (x1, y1) and (x2, y2) have common side, (x1, y1) ≠ (x2, y2) and no cell is covered by more than one sofa.
The last line contains four integer numbers cntl, cntr, cntt, cntb (0 ≤ cntl, cntr, cntt, cntb ≤ d - 1).
Output
Print the number of the sofa for which all the conditions are met. Sofas are numbered 1 through d as given in input. If there is no such sofa then print -1.
Examples
Input
2
3 2
3 1 3 2
1 2 2 2
1 0 0 1
Output
1
Input
3
10 10
1 2 1 1
5 5 6 5
6 4 5 4
2 1 2 0
Output
2
Input
2
2 2
2 1 1 1
1 2 2 2
1 0 0 0
Output
-1
Note
Let's consider the second example.
* The first sofa has 0 to its left, 2 sofas to its right ((1, 1) is to the left of both (5, 5) and (5, 4)), 0 to its top and 2 to its bottom (both 2nd and 3rd sofas are below).
* The second sofa has cntl = 2, cntr = 1, cntt = 2 and cntb = 0.
* The third sofa has cntl = 2, cntr = 1, cntt = 1 and cntb = 1.
So the second one corresponds to the given conditions.
In the third example
* The first sofa has cntl = 1, cntr = 1, cntt = 0 and cntb = 1.
* The second sofa has cntl = 1, cntr = 1, cntt = 1 and cntb = 0.
And there is no sofa with the set (1, 0, 0, 0) so the answer is -1. | instruction | 0 | 8,760 | 8 | 17,520 |
Tags: brute force, implementation
Correct Solution:
```
#!/usr/bin/env python3
from sys import exit
d = int(input().strip())
[n, m] = map(int, input().strip().split())
Hxds = [0 for _ in range(n)]
Hyds = [0 for _ in range(m)]
Vxds = [0 for _ in range(n)]
Vyds = [0 for _ in range(m)]
ds = []
for i in range(d):
x1, y1, x2, y2 = map(int, input().strip().split())
if x1 == x2:
Hxds[x1 - 1] += 1
Hyds[min(y1, y2) - 1] += 1
ds.append((x1 - 1, min(y1, y2) - 1, 'h'))
else:
Vxds[min(x1, x2) - 1] += 1
Vyds[y1 - 1] += 1
ds.append((min(x1, x2) - 1, y1 - 1, 'v'))
cl, cr, ct, cb = map(int, input().strip().split())
if (d - 1 - cl - cr) * (d - 1 - ct - cb) > 0:
print (-1)
exit()
def makeI(xs):
I = [0 for _ in range(len(xs) + 1)]
for i in range(len(xs)):
I[i + 1] = I[i] + xs[i]
return I
def find_x_Hor(IH, IV, l, cl, cr):
if cl + cr > d - 1:
return -1
x = 0
while x <= l and (IH[x] + IV[x] < cl or d - IH[x + 1] - IV[x] > cr):
x += 1
if x < l and IH[x] + IV[x] == cl and (d - IH[x + 1] - IV[x]) == cr:
return x
return -1
def find_x_Vert(IH, IV, l, cl, cr):
if cl + cr < d - 1:
return -1
x = 0
while x < l and (IH[x + 1] + IV[x + 1] < cl + 1 or d - IH[x + 1] - IV[x] > cr + 1):
x += 1
if x < l and IH[x + 1] + IV[x + 1] == cl + 1 and (d - IH[x + 1] - IV[x]) == cr + 1:
return x
return -1
IHx = makeI(Hxds)
IHy = makeI(Hyds)
IVx = makeI(Vxds)
IVy = makeI(Vyds)
if ct + cb >= d - 1 and cr + cl <= d - 1: # horizontal sofa
x = find_x_Hor(IHx, IVx, n, cl, cr)
y = find_x_Vert(IVy, IHy, m, ct, cb)
if x >= 0 and y >= 0:
if (x, y, 'h') in ds:
print(ds.index((x, y, 'h')) + 1)
exit()
if ct + cb <= d - 1 and cr + cl >= d - 1: # vertical sofa
x = find_x_Vert(IHx, IVx, n, cl, cr)
y = find_x_Hor(IVy, IHy, m, ct, cb)
if x >= 0 and y >= 0:
if (x, y, 'v') in ds:
print(ds.index((x, y, 'v')) + 1)
exit()
print (-1)
``` | output | 1 | 8,760 | 8 | 17,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yet another round on DecoForces is coming! Grandpa Maks wanted to participate in it but someone has stolen his precious sofa! And how can one perform well with such a major loss?
Fortunately, the thief had left a note for Grandpa Maks. This note got Maks to the sofa storehouse. Still he had no idea which sofa belongs to him as they all looked the same!
The storehouse is represented as matrix n × m. Every sofa takes two neighbouring by some side cells. No cell is covered by more than one sofa. There can be empty cells.
Sofa A is standing to the left of sofa B if there exist two such cells a and b that xa < xb, a is covered by A and b is covered by B. Sofa A is standing to the top of sofa B if there exist two such cells a and b that ya < yb, a is covered by A and b is covered by B. Right and bottom conditions are declared the same way.
Note that in all conditions A ≠ B. Also some sofa A can be both to the top of another sofa B and to the bottom of it. The same is for left and right conditions.
The note also stated that there are cntl sofas to the left of Grandpa Maks's sofa, cntr — to the right, cntt — to the top and cntb — to the bottom.
Grandpa Maks asks you to help him to identify his sofa. It is guaranteed that there is no more than one sofa of given conditions.
Output the number of Grandpa Maks's sofa. If there is no such sofa that all the conditions are met for it then output -1.
Input
The first line contains one integer number d (1 ≤ d ≤ 105) — the number of sofas in the storehouse.
The second line contains two integer numbers n, m (1 ≤ n, m ≤ 105) — the size of the storehouse.
Next d lines contains four integer numbers x1, y1, x2, y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — coordinates of the i-th sofa. It is guaranteed that cells (x1, y1) and (x2, y2) have common side, (x1, y1) ≠ (x2, y2) and no cell is covered by more than one sofa.
The last line contains four integer numbers cntl, cntr, cntt, cntb (0 ≤ cntl, cntr, cntt, cntb ≤ d - 1).
Output
Print the number of the sofa for which all the conditions are met. Sofas are numbered 1 through d as given in input. If there is no such sofa then print -1.
Examples
Input
2
3 2
3 1 3 2
1 2 2 2
1 0 0 1
Output
1
Input
3
10 10
1 2 1 1
5 5 6 5
6 4 5 4
2 1 2 0
Output
2
Input
2
2 2
2 1 1 1
1 2 2 2
1 0 0 0
Output
-1
Note
Let's consider the second example.
* The first sofa has 0 to its left, 2 sofas to its right ((1, 1) is to the left of both (5, 5) and (5, 4)), 0 to its top and 2 to its bottom (both 2nd and 3rd sofas are below).
* The second sofa has cntl = 2, cntr = 1, cntt = 2 and cntb = 0.
* The third sofa has cntl = 2, cntr = 1, cntt = 1 and cntb = 1.
So the second one corresponds to the given conditions.
In the third example
* The first sofa has cntl = 1, cntr = 1, cntt = 0 and cntb = 1.
* The second sofa has cntl = 1, cntr = 1, cntt = 1 and cntb = 0.
And there is no sofa with the set (1, 0, 0, 0) so the answer is -1.
Submitted Solution:
```
from sys import stdin, stdout
k = int(stdin.readline())
n, m = map(int, stdin.readline().split())
left, right, down, up = [], [], [], []
challengers = []
for i in range(k):
x1, y1, x2, y2 = map(int, stdin.readline().split())
if x1 == x2:
if y1 < y2:
challengers.append((x1, y1, x2, y2, i))
else:
challengers.append((x2, y2, x1, y1, i))
else:
if x1 < x2:
challengers.append((x1, y1, x2, y2, i))
else:
challengers.append((x2, y2, x1, y1, i))
left.append(challengers[-1])
right.append(challengers[-1])
up.append(challengers[-1])
down.append(challengers[-1])
left.sort(key = lambda x: (x[0], x[2]))
down.sort(key = lambda x: (x[1], x[3]))
challengers = [[], [], [], []]
cntl, cntr, cntd, cntu = map(int, stdin.readline().split())
if cntl:
for i in range(cntl, -1, -1):
if (left[i][0], left[i][2]) == (left[cntl][0], left[cntl][2]):
challengers[0].append(left[i][-1])
else:
break
for i in range(cntl + 1, k):
if (left[i][0], left[i][2]) == (left[cntl][0], left[cntl][2]):
challengers[0].append(left[i][-1])
else:
break
else:
for i in range(k):
if (left[i][0], left[i][2]) == (left[0][0], left[0][2]):
challengers[0].append(left[i][-1])
else:
break
if cntr:
for i in range(k - 1 - cntr, k):
if (left[i][0], left[i][2]) == (left[k - 1 - cntr][0], left[k - 1 - cntr][2]):
challengers[1].append(left[i][-1])
else:
break
for i in range(k - 2 - cntr, -1, -1):
if (left[i][0], left[i][2]) == (left[k - 1 - cntr][0], left[k - 1 - cntr][2]):
challengers[1].append(left[i][-1])
else:
break
else:
for i in range(k - 1, -1, -1):
if (left[i][0], left[i][2]) == (left[k - 1][0], left[k - 1][2]):
challengers[1].append(left[i][-1])
else:
break
#!!!!!!!!!!!
if cntd:
for i in range(cntd, -1, -1):
if (down[i][1], down[i][3]) == (down[cntd][1], down[cntd][3]):
challengers[2].append(down[i][-1])
else:
break
for i in range(cntd + 1, k):
if (down[i][1], down[i][3]) == (down[cntd][1], down[cntd][3]):
challengers[2].append(down[i][-1])
else:
break
else:
for i in range(k):
if (down[i][1], down[i][3]) == (down[0][1], down[0][3]):
challengers[2].append(down[i][-1])
else:
break
if cntu:
for i in range(k - 1 - cntu, k):
if (down[i][1], down[i][3]) == (down[k - 1 - cntu][1], down[k - 1 - cntu][3]):
challengers[3].append(down[i][-1])
else:
break
for i in range(k - 2 - cntu, -1, -1):
if (down[i][1], down[i][3]) == (down[k - 1 - cntu][1], down[k - 1 - cntu][3]):
challengers[3].append(down[i][-1])
else:
break
else:
for i in range(k - 1, -1, -1):
if (down[i][1], down[i][3]) == (down[k - 1][1], down[k - 1][3]):
challengers[3].append(down[i][-1])
else:
break
ans = set(challengers[0]) & set(challengers[1]) & set(challengers[2]) & set(challengers[3])
if not len(ans):
stdout.write('-1')
else:
stdout.write(str(list(ans)[0] + 1))
``` | instruction | 0 | 8,761 | 8 | 17,522 |
No | output | 1 | 8,761 | 8 | 17,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yet another round on DecoForces is coming! Grandpa Maks wanted to participate in it but someone has stolen his precious sofa! And how can one perform well with such a major loss?
Fortunately, the thief had left a note for Grandpa Maks. This note got Maks to the sofa storehouse. Still he had no idea which sofa belongs to him as they all looked the same!
The storehouse is represented as matrix n × m. Every sofa takes two neighbouring by some side cells. No cell is covered by more than one sofa. There can be empty cells.
Sofa A is standing to the left of sofa B if there exist two such cells a and b that xa < xb, a is covered by A and b is covered by B. Sofa A is standing to the top of sofa B if there exist two such cells a and b that ya < yb, a is covered by A and b is covered by B. Right and bottom conditions are declared the same way.
Note that in all conditions A ≠ B. Also some sofa A can be both to the top of another sofa B and to the bottom of it. The same is for left and right conditions.
The note also stated that there are cntl sofas to the left of Grandpa Maks's sofa, cntr — to the right, cntt — to the top and cntb — to the bottom.
Grandpa Maks asks you to help him to identify his sofa. It is guaranteed that there is no more than one sofa of given conditions.
Output the number of Grandpa Maks's sofa. If there is no such sofa that all the conditions are met for it then output -1.
Input
The first line contains one integer number d (1 ≤ d ≤ 105) — the number of sofas in the storehouse.
The second line contains two integer numbers n, m (1 ≤ n, m ≤ 105) — the size of the storehouse.
Next d lines contains four integer numbers x1, y1, x2, y2 (1 ≤ x1, x2 ≤ n, 1 ≤ y1, y2 ≤ m) — coordinates of the i-th sofa. It is guaranteed that cells (x1, y1) and (x2, y2) have common side, (x1, y1) ≠ (x2, y2) and no cell is covered by more than one sofa.
The last line contains four integer numbers cntl, cntr, cntt, cntb (0 ≤ cntl, cntr, cntt, cntb ≤ d - 1).
Output
Print the number of the sofa for which all the conditions are met. Sofas are numbered 1 through d as given in input. If there is no such sofa then print -1.
Examples
Input
2
3 2
3 1 3 2
1 2 2 2
1 0 0 1
Output
1
Input
3
10 10
1 2 1 1
5 5 6 5
6 4 5 4
2 1 2 0
Output
2
Input
2
2 2
2 1 1 1
1 2 2 2
1 0 0 0
Output
-1
Note
Let's consider the second example.
* The first sofa has 0 to its left, 2 sofas to its right ((1, 1) is to the left of both (5, 5) and (5, 4)), 0 to its top and 2 to its bottom (both 2nd and 3rd sofas are below).
* The second sofa has cntl = 2, cntr = 1, cntt = 2 and cntb = 0.
* The third sofa has cntl = 2, cntr = 1, cntt = 1 and cntb = 1.
So the second one corresponds to the given conditions.
In the third example
* The first sofa has cntl = 1, cntr = 1, cntt = 0 and cntb = 1.
* The second sofa has cntl = 1, cntr = 1, cntt = 1 and cntb = 0.
And there is no sofa with the set (1, 0, 0, 0) so the answer is -1.
Submitted Solution:
```
def main():
# import sys
d = int(input())
n, m = [int(x) for x in input().split()]
point = []
for x in range(d):
point.append([int(x) for x in input().split()])
point[-1].append(x+1)
q = [int(x) for x in input().split()]
def cntl(n):
# print(max(n[0],n[2]))
return min(n[0],n[2])
point.sort(key = cntl)
t = 0
debag = 0
def relax():
global debag
debag = 1
return
if(t == q[0]):
relax()
else:
id1 = 0
for i in range(len(point)-1):
if(min(point[i][0],point[i][2]) < min(point[i+1][0],point[i+1][2])):
t+=1
if(t<=q[0]):
id1 = i+1
point = point[id1:]
if(len(point)==1):
print(point[0][4])
return
# sys.exit(0)
if(t == q[1]):
relax()
else:
id1 = len(point)
for i in range(len(point)-1,0,-1):
if(max(point[i][0],point[i][2]) < max(point[i-1][0],point[i-1][2])):
t+=1
if(t<=q[0]):
id1 = i
point = point[:id1]
if(len(point)==1):
print(point[0][4])
return
def cntb(n):
return min(n[1],n[3])
point.sort(key = cntb)
if(t == q[2]):
relax()
else:
id1 = 0
for i in range(len(point)-1):
if(min(point[i][1],point[i][3]) < min(point[i+1][1],point[i+1][3])):
t+=1
if(t<=q[2]):
id1 = i+1
point = point[id1:]
if(len(point)==1):
print(point[0][4])
return
if(t == q[3]):
relax()
else:
id1 = len(point)
for i in range(len(point)-1,0,-1):
if(max(point[i][1],point[i][3]) < max(point[i-1][1],point[i-1][3])):
t+=1
if(t<=q[3]):
id1 = i
point = point[:id1]
if(len(point)!=1):
print(-1)
else:
print(point[0][4])
main()
``` | instruction | 0 | 8,762 | 8 | 17,524 |
No | output | 1 | 8,762 | 8 | 17,525 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit. | instruction | 0 | 9,138 | 8 | 18,276 |
Tags: implementation
Correct Solution:
```
n=int(input())
l=[int(x) for x in input().split()]
#print(l)
a,b=0,0
for i in range(n):
if l[n-1-i]==1:
a+=1
else:
break
for i in range(n):
if l[n-1-i]==0:
b+=1
else:
break
print(min(n-a,n-b))
``` | output | 1 | 9,138 | 8 | 18,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit. | instruction | 0 | 9,139 | 8 | 18,278 |
Tags: implementation
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
k = 0
l = a.count(0)
r = a.count(1)
for i in range(n):
if l == 0 or r == 0:
break
l -= int(a[i] == 0)
r -= int(a[i] == 1)
k += 1
print(k)
``` | output | 1 | 9,139 | 8 | 18,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit. | instruction | 0 | 9,140 | 8 | 18,280 |
Tags: implementation
Correct Solution:
```
n = int(input())
s = input().replace(" ", "")
print(min(s.rfind("0"), s.rfind("1")) + 1)
``` | output | 1 | 9,140 | 8 | 18,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit. | instruction | 0 | 9,141 | 8 | 18,282 |
Tags: implementation
Correct Solution:
```
input()
s = input().replace(' ', '')
print(len(s.rstrip(s[-1])))
``` | output | 1 | 9,141 | 8 | 18,283 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit. | instruction | 0 | 9,142 | 8 | 18,284 |
Tags: implementation
Correct Solution:
```
import re, math, decimal, bisect
def read(): return input().strip()
def iread(): return int(input().strip())
def viread(): return [_ for _ in input().strip().split()]
# code goes here
n = iread()
doors = "".join(viread()[::-1])
print(min(n - doors.find('1'), n - doors.find('0')))
``` | output | 1 | 9,142 | 8 | 18,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit. | instruction | 0 | 9,143 | 8 | 18,286 |
Tags: implementation
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
i = n - 1
if a[-1] == 0:
while a[i] == 0:
i -= 1
print(i + 1)
else:
while a[i] == 1:
i -= 1
print(i + 1)
``` | output | 1 | 9,143 | 8 | 18,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit. | instruction | 0 | 9,144 | 8 | 18,288 |
Tags: implementation
Correct Solution:
```
n=int(input())
arr=list(map(int, input().split()))
counter = [0,0]
for i in range(n):
counter[arr[i]] += 1
for i in range(n):
counter[arr[i]] -= 1
if counter[0] == 0 or counter[1] == 0:
print(i+1)
break
``` | output | 1 | 9,144 | 8 | 18,289 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit. | instruction | 0 | 9,145 | 8 | 18,290 |
Tags: implementation
Correct Solution:
```
n = input()
doors = input().split(" ")
left = 0
right = 0
for door in doors:
if door == '0': left += 1
else: right += 1
counter = 0
for door in doors:
counter += 1
if door == '0': left -= 1
else: right -= 1
if left == 0 or right == 0: break
print(counter)
``` | output | 1 | 9,145 | 8 | 18,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit.
Submitted Solution:
```
"""for p in range(int(input())):
n,k=map(int,input().split(" "))
number=input().split(" ")
chances=[k for i in range(n)]
prev=-1
prev_updated=-1
last_used=False
toSub=0
start=0
prevSub=0
if(number[0]=='1'):
prev=0
prev_updated=0
start=1
for i in range(start,n):
if(number[i]=='1'):
# print("\ni",i,"\ntoSub",toSub,"\nprevUpadted",prev_updated,"\nprev",prev,"\nlast_used",last_used)
f1=False
# toSub+=1
toSub=0
zeros=i - prev_updated - 1
if(last_used):
zeros-=1
#chances[i]-=toSub
#print(prevSub,(i - prev - 1 ) +1)
if(i - prev - 1 <= prevSub):
chances[i]-= prevSub - (i - prev - 1 ) +1
if(chances[i]<zeros):
chances[i]=zeros
toSub+= prevSub - (i - prev - 1 ) +1
f1=True
if(zeros==0 or chances[i]==0):
prev_updated=i
prev=i
last_used=False
prevSub=toSub
continue
# print("\nchances: ",chances[i],"\t\tzeroes : ",zeros,"\t\tprevSub :",prevSub)
if(chances[i]>zeros):
# print("\t\t\t\t1")
number[i-zeros]='1'
number[i]='0'
prev_updated=i-zeros
last_used=False
elif(chances[i]==zeros):
# print("\t\t\t\t2")
number[i]='0'
number[i-chances[i]]='1'
prev_updated=i-chances[i]
last_used=True
else:
# print("\t\t\t\t3")
number[i]='0'
number[i-chances[i]]='1'
prev_updated=i-chances[i]
last_used=True
prev=i
prevSub=toSub
if(prev_updated>2 and f1):
if(number[prev_updated]=='1' and number[prev_updated-1]=='0' and number[prev_updated-2]=='1'):
last_used=False
#if()
# print("\ni",i,"\ntoSub",toSub,"\nprevUpadted",prev_updated,"\nprev",prev,"\nlast_used",last_used)
# print(number)
else:
toSub=0
print(*number)
# print(chances)"""
"""class offer:
def __init__(self, n, fre):
self.num = n
self.free = fre
self.delta= n-fre
n,m,k=map(int,input().split(" "))
shovel=list(map(int,input().split(" ")))
#dicti={}
offers=[]
temp_arr=[False for i in range(n)]
for i in range(m):
p,q=map(int,input().split(" "))
if(p>k):
continue
offers.append(offer(p,q))
# dicti[p]=q
#for i in dicti:
# dicti[i].sort()
shovel.sort()
shovel=shovel[:k+1]
offers.sort(key=lambda x: x.delta/x.num,reverse=True)
bestoffer=[]
for i in offers:
if(not temp_arr[i.num]):
temp_arr[i.num]=True
bestoffer.append(i)
cost=0
for i in bestoffer:
for p in range(int(input())):
arr=list(input())
n=len(arr)
for i in range(n):
arr[i]=ord(arr[i])-96
arr.sort()
arr1=arr[:n//2]
arr2=arr[n//2:]
arr=[]
#print(arr,arr1,arr2)
i1=n//2-1
i2=n-i1-2
while (i1!=-1 and i2!=-1):
arr.append(arr1[i1])
arr.append(arr2[i2])
i1-=1
i2-=1
if(i1!=-1):
arr.append(arr1[i1])
elif(i2!=-1):
arr.append(arr2[i2])
#print(arr)
s=""
for i in range(n-1):
if(abs(arr[i]-arr[i+1])==1):
s=-1
print("No answer")
break
else:
s+=chr(arr[i]+96)
if(s!=-1):
s+=chr(arr[-1]+96)
print(s)"""
"""
n,m=map(int,input().split(" "))
seti=[]
ans=[1 for i in range(n)]
for i in range(m):
arr=list(map(int,input().split(" ")))
if(arr[0]>1):
seti.append(set(arr[1:]))
else:
m-=1
parent=[-1 for i in range(m)]
#print(seti)
for i in range(m-1):
for j in range(i+1,m):
if(parent[j]==-1):
if(len(seti[i].intersection(seti[j]))>0):
seti[i]=seti[i].union(seti[j])
parent[j]=i
#print(parent)
for i in range(m):
if(parent[i]==-1):
temp=list(seti[i])
store=len(temp)
for j in temp:
ans[j-1]=store
print(*ans)
for p in range(int(input())):
arr=list(input())
n=len(arr)
for i in range(n):
arr[i]=ord(arr[i])-96
arr.sort()
arr1=arr[:n//2]
arr2=arr[n//2:]
arr=[]
#print(arr,arr1,arr2)
i1=n//2-1
i2=n-i1-2
while (i1!=-1 and i2!=-1):
arr.append(arr1[i1])
arr.append(arr2[i2])
i1-=1
i2-=1
if(i1!=-1):
arr.append(arr1[i1])
elif(i2!=-1):
arr.append(arr2[i2])
s=""
for i in range(n-1):
if(abs(arr[i]-arr[i+1])==1):
s=-1
print("No answer")
break
else:
s+=chr(arr[i]+96)
if(s!=-1):
s+=chr(arr[-1]+96)
print(s)"""
#n=0
n=int(input())
arr=list(map(int,input().split(" ")))
record=arr[-1]
arr.reverse()
if(record==0):
print(n-arr.index(1))
else:
print(n-arr.index(0))
``` | instruction | 0 | 9,146 | 8 | 18,292 |
Yes | output | 1 | 9,146 | 8 | 18,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit.
Submitted Solution:
```
n = int(input())
s = input()
s= s.split(' ')
for l in reversed(range(len(s))):
if(int(s[l]) == int(s[l-1])):
l=l-1
else:
print(l)
break
``` | instruction | 0 | 9,147 | 8 | 18,294 |
Yes | output | 1 | 9,147 | 8 | 18,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
a=a[::-1]
#print(a)
x=a[0]
for i in range(n):
if(a[i]!=x):
ans=i
break
print(n-i)
``` | instruction | 0 | 9,148 | 8 | 18,296 |
Yes | output | 1 | 9,148 | 8 | 18,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit.
Submitted Solution:
```
n=int(input())
l=[int(i) for i in input().split()]
c1=l.count(0)
c2=l.count(1)
for i in range(n):
if(l[i]==0):
c1-=1
else:
c2-=1
if(c1==0 or c2==0):
j=i
break
print(j+1)
``` | instruction | 0 | 9,149 | 8 | 18,298 |
Yes | output | 1 | 9,149 | 8 | 18,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit.
Submitted Solution:
```
n = int(input())
doors = [int(x) for x in input().split()]
for x in range(len(doors)-1, -1, -1):
if n == 200000:
print(doors[199997:])
print(x)
if doors[x] == 1:
print(min(len(doors)-1, x+1))
break
``` | instruction | 0 | 9,150 | 8 | 18,300 |
No | output | 1 | 9,150 | 8 | 18,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit.
Submitted Solution:
```
n=int(input())
d=list(map(int,input().split()))
c=0
e=[]
for i in d:
if i==0:
c+=1
else:
e.append(c)
c=0
print(max(e)+1)
``` | instruction | 0 | 9,151 | 8 | 18,302 |
No | output | 1 | 9,151 | 8 | 18,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit.
Submitted Solution:
```
x = int(input())
q = list(range(x))
ones = 0
zeros = 0
k = input()
j = 0
for i in range(x):
q[i] = k[j]
if q[i] == '1':
ones = ones +1
else:
zeros = zeros + 1
j = j + 2
if ones > zeros:
if q[x-1] == '1':
for i in range(x-1,0,-1):
if q[i] == '0':
print(i+1)
break
elif q[x-1] == '0':
for i in range(x-1,0,-1):
if q[i] == '1':
print(i+1)
break
elif ones < zeros:
if q[x-1] == '0':
for i in range(x-1,0,-1):
if q[i] == '1':
print(i+1)
break
elif q[x-1] == '1':
for i in range(x-1,0,-1):
if q[i] == '0':
print(i+1)
break
else:
if q[x-1] == '0':
for i in range(x-1,0,-1):
if q[i] != '0':
print(i+1)
break
elif q[x-1] == '1':
for i in range(x-1,0,-1):
if q[i] != '1':
print(i+1)
break
``` | instruction | 0 | 9,152 | 8 | 18,304 |
No | output | 1 | 9,152 | 8 | 18,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Three years have passes and nothing changed. It is still raining in London, and Mr. Black has to close all the doors in his home in order to not be flooded. Once, however, Mr. Black became so nervous that he opened one door, then another, then one more and so on until he opened all the doors in his house.
There are exactly two exits from Mr. Black's house, let's name them left and right exits. There are several doors in each of the exits, so each door in Mr. Black's house is located either in the left or in the right exit. You know where each door is located. Initially all the doors are closed. Mr. Black can exit the house if and only if all doors in at least one of the exits is open. You are given a sequence in which Mr. Black opened the doors, please find the smallest index k such that Mr. Black can exit the house after opening the first k doors.
We have to note that Mr. Black opened each door at most once, and in the end all doors became open.
Input
The first line contains integer n (2 ≤ n ≤ 200 000) — the number of doors.
The next line contains n integers: the sequence in which Mr. Black opened the doors. The i-th of these integers is equal to 0 in case the i-th opened door is located in the left exit, and it is equal to 1 in case it is in the right exit.
It is guaranteed that there is at least one door located in the left exit and there is at least one door located in the right exit.
Output
Print the smallest integer k such that after Mr. Black opened the first k doors, he was able to exit the house.
Examples
Input
5
0 0 1 0 0
Output
3
Input
4
1 0 0 1
Output
3
Note
In the first example the first two doors are from the left exit, so when Mr. Black opened both of them only, there were two more closed door in the left exit and one closed door in the right exit. So Mr. Black wasn't able to exit at that moment.
When he opened the third door, all doors from the right exit became open, so Mr. Black was able to exit the house.
In the second example when the first two doors were opened, there was open closed door in each of the exit.
With three doors opened Mr. Black was able to use the left exit.
Submitted Solution:
```
n = int(input())
k = 0
a = list(map(int,input().split()))
a.reverse()
print(max(a.index(1),a.index(0)))
``` | instruction | 0 | 9,153 | 8 | 18,306 |
No | output | 1 | 9,153 | 8 | 18,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick's company employed n people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are m applications written in the following form: «employee ai is ready to become a supervisor of employee bi at extra cost ci». The qualification qj of each employee is known, and for each application the following is true: qai > qbi.
Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.
Input
The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers qj (0 ≤ qj ≤ 106)— the employees' qualifications. The following line contains number m (0 ≤ m ≤ 10000) — amount of received applications. The following m lines contain the applications themselves, each of them in the form of three space-separated numbers: ai, bi and ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application qai > qbi.
Output
Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.
Examples
Input
4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5
Output
11
Input
3
1 2 3
2
3 1 2
3 1 3
Output
-1
Note
In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1.
Submitted Solution:
```
read1 = [-1] * int(input())
read2 = input()
n = int(input())
for i in range(n):
x, y, z = list(map(int, input().split()))
if read1[y-1] == -1:
read1[y-1] = z
else:
read1[y-1] = min(read1[y-1],z)
if (read1.count(-1) <= 1):
print(sum(read1, 1))
else:
print(-1)
``` | instruction | 0 | 9,343 | 8 | 18,686 |
Yes | output | 1 | 9,343 | 8 | 18,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick's company employed n people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are m applications written in the following form: «employee ai is ready to become a supervisor of employee bi at extra cost ci». The qualification qj of each employee is known, and for each application the following is true: qai > qbi.
Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.
Input
The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers qj (0 ≤ qj ≤ 106)— the employees' qualifications. The following line contains number m (0 ≤ m ≤ 10000) — amount of received applications. The following m lines contain the applications themselves, each of them in the form of three space-separated numbers: ai, bi and ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application qai > qbi.
Output
Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.
Examples
Input
4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5
Output
11
Input
3
1 2 3
2
3 1 2
3 1 3
Output
-1
Note
In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1.
Submitted Solution:
```
n = int(input())
input()
m = int(input())
supervisor = [[] for _ in range(n)]
for i in range(m):
a, b, c = map(int, input().split())
supervisor[b-1].append(c)
boss = 0
ans = 0
for i in range(n):
if len(supervisor[i]) > 0:
ans += min(supervisor[i])
else:
boss += 1
if boss == 1:
print(ans)
else:
print(-1)
``` | instruction | 0 | 9,344 | 8 | 18,688 |
Yes | output | 1 | 9,344 | 8 | 18,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick's company employed n people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are m applications written in the following form: «employee ai is ready to become a supervisor of employee bi at extra cost ci». The qualification qj of each employee is known, and for each application the following is true: qai > qbi.
Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.
Input
The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers qj (0 ≤ qj ≤ 106)— the employees' qualifications. The following line contains number m (0 ≤ m ≤ 10000) — amount of received applications. The following m lines contain the applications themselves, each of them in the form of three space-separated numbers: ai, bi and ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application qai > qbi.
Output
Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.
Examples
Input
4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5
Output
11
Input
3
1 2 3
2
3 1 2
3 1 3
Output
-1
Note
In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1.
Submitted Solution:
```
n = int(input())
qualifications = list(map(int, input().split()))
apps = []
limit = (10 ** 6) + 1
result = [limit for i in range(n)]
m = int(input())
for i in range(m):
a, b, c = map(int, input().split())
result[b - 1] = min(result[b - 1], c)
count = 0
for i in result:
if i == limit:
count += 1
if count > 1:
print(-1)
else:
out = 0
for i in result:
if i != limit:
out += i
print(out)
``` | instruction | 0 | 9,345 | 8 | 18,690 |
Yes | output | 1 | 9,345 | 8 | 18,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick's company employed n people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are m applications written in the following form: «employee ai is ready to become a supervisor of employee bi at extra cost ci». The qualification qj of each employee is known, and for each application the following is true: qai > qbi.
Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.
Input
The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers qj (0 ≤ qj ≤ 106)— the employees' qualifications. The following line contains number m (0 ≤ m ≤ 10000) — amount of received applications. The following m lines contain the applications themselves, each of them in the form of three space-separated numbers: ai, bi and ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application qai > qbi.
Output
Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.
Examples
Input
4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5
Output
11
Input
3
1 2 3
2
3 1 2
3 1 3
Output
-1
Note
In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1.
Submitted Solution:
```
from sys import stdin, stdout
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log, trunc
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
mod = pow(10, 9) + 7
mod2 = 998244353
def inp(): return stdin.readline().strip()
def iinp(): return int(inp())
def out(var, end="\n"): stdout.write(str(var)+"\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def remadd(x, y): return 1 if x%y else 0
def ceil(a,b): return (a+b-1)//b
S1 = 'abcdefghijklmnopqrstuvwxyz'
S2 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
def isprime(x):
if x<=1: return False
if x in (2, 3): return True
if x%2 == 0: return False
for i in range(3, int(sqrt(x))+1, 2):
if x%i == 0: return False
return True
n = iinp()
arr = lmp()
sup = dd(list)
m = iinp()
for i in range(m):
p, c, w = mp()
if arr[p-1]>arr[c-1]: sup[c].append((p, w))
ans = 0
par = l1d(n+1, -1)
for k, l in sup.items():
mn = l[0][1]
par[k] = l[0][0]
for i in l:
if i[1]<mn:
mn = i[1]
par[k] = i[0]
ans += mn
c = par.count(-1)
print(ans if c==2 else -1)
``` | instruction | 0 | 9,346 | 8 | 18,692 |
Yes | output | 1 | 9,346 | 8 | 18,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick's company employed n people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are m applications written in the following form: «employee ai is ready to become a supervisor of employee bi at extra cost ci». The qualification qj of each employee is known, and for each application the following is true: qai > qbi.
Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.
Input
The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers qj (0 ≤ qj ≤ 106)— the employees' qualifications. The following line contains number m (0 ≤ m ≤ 10000) — amount of received applications. The following m lines contain the applications themselves, each of them in the form of three space-separated numbers: ai, bi and ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application qai > qbi.
Output
Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.
Examples
Input
4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5
Output
11
Input
3
1 2 3
2
3 1 2
3 1 3
Output
-1
Note
In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1.
Submitted Solution:
```
import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def getResult(n,q,m,edges,graph):
maxq = max(q)
pRoots = [i for i, j in enumerate(q) if j == maxq]
results = []
#print("possible roots : ",pRoots)
for rootI in pRoots :
cost = 0
mst = []
ees = [rootI]
pE = []
pEI = rootI
while len(mst) != n-1 :
#print("new search ....",cost)
#print("mst : ",mst)
#print("pE : " , pE)
#print("#print any key to continue...")
# input()
#add possible edges
#print("finding edges with pEI as ",pEI);
for j in range(n):
if q[pEI] > q[j] and graph[pEI][j] >= 0 :
#print("adding this new edge to pE:",[pEI,j])
pE.append([pEI,j])
#find best edge and add to mst
while len(pE) > 0 :
mE = 0
for i in range(len(pE)):
if graph[pE[mE][0]][pE[mE][1]] > graph[pE[i][0]][pE[i][1]] :
mE = i
#print("debug : found minimal edge " , pE[mE] )
#check if a parent already exists
isParentAlreadyExists = -1
isParentAlreadyExistsI = -1
for e in range(len(mst)):
if mst[e][1] == pE[mE][1] :
isParentAlreadyExists = mst[e][0]
isParentAlreadyExistsI = e
#print("is parent already exists : ",isParentAlreadyExists)
if isParentAlreadyExists == -1 :
mst.append(pE[mE])
cost = cost + graph[pE[mE][0]][pE[mE][1]]
#print("adding edge ",pE[mE]," cost improved ",graph[pE[mE][0]][pE[mE][1]])
pEI = pE[mE][1]
#print("changint pEI to ", pEI)
pE.remove(pE[mE])
break
else :
if graph[pE[mE][0]][pE[mE][1]] < graph[isParentAlreadyExists][pE[mE][1]] :
mst.append(pE[mE])
cost = cost + graph[pE[mE][0]][pE[mE][1]]
cost = cost - graph[isParentAlreadyExists][pE[mE][1]]
#print("removing edge ", mst[isParentAlreadyExistsI], " cost decreased ",graph[pE[mE][0]][pE[mE][1]])
pE.append([isParentAlreadyExists,pE[mE][1]])
#print("adding edge ", pE[mE], " cost improved ", graph[pE[mE][0]][pE[mE][1]])
mst.remove(mst[isParentAlreadyExistsI])
pEI = pE[mE][1]
#print("changint pEI to ", pEI)
pE.remove(pE[mE])
break
else:
#print("this edge is useless now.already joined ..team . ",pE[mE])
pE.remove(pE[mE])
else :
cost = -1
#print("debug : no edges in pE , cost = -1")
break
results.append(cost)
#print("results : " , results)
maxResult = max(results)
if maxResult == -1 :
return -1
for i in results :
if i != -1 :
if maxResult < i :
maxResult = i
return maxResult
if __name__ == '__main__':
n = inp()
q = inlt()
m = inp()
#print(m)
edges = []
graph = []
for i in range(n):
line = []
for j in range(n):
line.append(-1)
graph.append(line)
for i in range(m):
ed = inlt()
ed[0] = ed[0]-1
ed[1] = ed[1]-1
edges.append(ed)
#print(edges)
for u,v,c in edges:
if graph[u][v] > c :
graph[u][v] = c
else:
if graph[u][v] == -1 :
graph[u][v] = c
#print(graph)
result = getResult(n,q,m,edges,graph)
print(result)
``` | instruction | 0 | 9,347 | 8 | 18,694 |
No | output | 1 | 9,347 | 8 | 18,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick's company employed n people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are m applications written in the following form: «employee ai is ready to become a supervisor of employee bi at extra cost ci». The qualification qj of each employee is known, and for each application the following is true: qai > qbi.
Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.
Input
The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers qj (0 ≤ qj ≤ 106)— the employees' qualifications. The following line contains number m (0 ≤ m ≤ 10000) — amount of received applications. The following m lines contain the applications themselves, each of them in the form of three space-separated numbers: ai, bi and ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application qai > qbi.
Output
Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.
Examples
Input
4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5
Output
11
Input
3
1 2 3
2
3 1 2
3 1 3
Output
-1
Note
In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1.
Submitted Solution:
```
if __name__ == '__main__':
n = int(input())
q = [int(i) for i in input().split()]
m = int(input())
head = q.index(max(q)) + 1
graph = {}
for i in range(m):
a, b, c = input().split()
if a in graph.keys():
graph[a].append({'adj': b, 'cost': c})
else:
graph[a] = [{'adj': b, 'cost': c}]
if n == 1:
print(0)
exit()
edges= set()
if str(head) not in graph.keys():
print(-1)
exit()
vertices = set()
vertices.add(str(head))
for i in graph[str(head)]:
edges.add((i['cost'], i['adj']))
s = 0
while True:
if len(edges) == 0:
print(-1)
exit()
e= min(edges)
vertices.add(e[1])
s += int(e[0])
if len(vertices) == n:
break
if e[1] in graph.keys():
for i in graph[e[1]]:
edges.add((i['cost'], i['adj']))
edges.remove(e)
print(s)
``` | instruction | 0 | 9,348 | 8 | 18,696 |
No | output | 1 | 9,348 | 8 | 18,697 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick's company employed n people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are m applications written in the following form: «employee ai is ready to become a supervisor of employee bi at extra cost ci». The qualification qj of each employee is known, and for each application the following is true: qai > qbi.
Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.
Input
The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers qj (0 ≤ qj ≤ 106)— the employees' qualifications. The following line contains number m (0 ≤ m ≤ 10000) — amount of received applications. The following m lines contain the applications themselves, each of them in the form of three space-separated numbers: ai, bi and ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application qai > qbi.
Output
Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.
Examples
Input
4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5
Output
11
Input
3
1 2 3
2
3 1 2
3 1 3
Output
-1
Note
In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1.
Submitted Solution:
```
MAX = 1000000
e = int(input())
q = input().split()
m = int(input())
arr1 = [MAX]*e
for i in range(m):
[a, b, c] = list(map(int, input().split()))
arr1[b-1] = min(arr1[b-1], c)
count = 0
for x in arr1:
if x == MAX:
count += 1
if count > 1:
print(-1)
else:
res = 0
for k in arr1:
if k != MAX:
res += k
print(res)
``` | instruction | 0 | 9,349 | 8 | 18,698 |
No | output | 1 | 9,349 | 8 | 18,699 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick's company employed n people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are m applications written in the following form: «employee ai is ready to become a supervisor of employee bi at extra cost ci». The qualification qj of each employee is known, and for each application the following is true: qai > qbi.
Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.
Input
The first input line contains integer n (1 ≤ n ≤ 1000) — amount of employees in the company. The following line contains n space-separated numbers qj (0 ≤ qj ≤ 106)— the employees' qualifications. The following line contains number m (0 ≤ m ≤ 10000) — amount of received applications. The following m lines contain the applications themselves, each of them in the form of three space-separated numbers: ai, bi and ci (1 ≤ ai, bi ≤ n, 0 ≤ ci ≤ 106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application qai > qbi.
Output
Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.
Examples
Input
4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5
Output
11
Input
3
1 2 3
2
3 1 2
3 1 3
Output
-1
Note
In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1.
Submitted Solution:
```
import sys
def solution():
n = int(input().strip())
input()#ikinci satir is yapmiyor
m = int(input().strip())
costs = [10**6] * (n)
for _ in range(m):
a, b, c = map(int, input().strip().split())
costs[b-1]=min(c,costs[b-1])
print(costs)
boss = 0
for i in range(n):
if costs[i] == 10**6:
boss += 1
if boss >= 2:
break
if boss != 1: #bir isin 1 patronu olur
print(-1)
return
min_cost = sum(costs) - 10**6
print(min_cost)
solution()
"""
4
7 2 3 1
4
1 2 5
2 4 1
3 4 1
1 3 5
"""
``` | instruction | 0 | 9,350 | 8 | 18,700 |
No | output | 1 | 9,350 | 8 | 18,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon. | instruction | 0 | 9,439 | 8 | 18,878 |
Tags: brute force, implementation, math
Correct Solution:
```
import sys
input = lambda:sys.stdin.readline()
int_arr = lambda: list(map(int,input().split()))
str_arr = lambda: list(map(str,input().split()))
get_str = lambda: map(str,input().split())
get_int = lambda: map(int,input().split())
get_flo = lambda: map(float,input().split())
mod = 1000000007
def solve():
pass
# for _ in range(int(input())):
n = int(input())
print(max(int_arr()))
``` | output | 1 | 9,439 | 8 | 18,879 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon. | instruction | 0 | 9,440 | 8 | 18,880 |
Tags: brute force, implementation, math
Correct Solution:
```
input()
l=list(map(int,input().split()))
print(max(l))
``` | output | 1 | 9,440 | 8 | 18,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon. | instruction | 0 | 9,441 | 8 | 18,882 |
Tags: brute force, implementation, math
Correct Solution:
```
from sys import stdin
def main():
n = int(stdin.readline().strip())
a = [int(x) for x in stdin.readline().strip().split()]
ans = a[0]
acum = 0
for i in range(1, len(a)):
if a[i-1]-a[i] < 0:
if a[i-1]-a[i] < -acum:
ans += a[i]-a[i-1]-acum
acum = 0
else:
acum += a[i-1]-a[i]
else:
acum += a[i-1]-a[i]
print(ans)
main()
``` | output | 1 | 9,441 | 8 | 18,883 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon. | instruction | 0 | 9,442 | 8 | 18,884 |
Tags: brute force, implementation, math
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
ans = 0
for x in a:
if ans < x:
ans = x
print(ans)
``` | output | 1 | 9,442 | 8 | 18,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon. | instruction | 0 | 9,443 | 8 | 18,886 |
Tags: brute force, implementation, math
Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
su=l[0]
e=0
for i in range(n-1):
if l[i]-l[i+1]>=0:
e+=l[i]-l[i+1]
else:
if e>=l[i+1]-l[i]:
e-=l[i+1]-l[i]
else:
su+=l[i+1]-l[i]-e
e=0
print(su)
``` | output | 1 | 9,443 | 8 | 18,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon. | instruction | 0 | 9,444 | 8 | 18,888 |
Tags: brute force, implementation, math
Correct Solution:
```
a=int(input())
b=input().split(' ')
c=0
p=0
for i in range(a):
p=int(b[i])
if c<p:
c=p
print(c)
``` | output | 1 | 9,444 | 8 | 18,889 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon. | instruction | 0 | 9,445 | 8 | 18,890 |
Tags: brute force, implementation, math
Correct Solution:
```
n = int(input())
hs = list(map(int, input().split()))
r = 0
balance = 0
ph = 0
for h in hs:
balance += ph - h
if balance < 0:
r += abs(balance)
balance = 0
ph = h
print(r)
``` | output | 1 | 9,445 | 8 | 18,891 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon. | instruction | 0 | 9,446 | 8 | 18,892 |
Tags: brute force, implementation, math
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
t = 0
s = 0
t_e = 0
for i in range(len(a)):
if ((t_e + (t - a[i])) < 0):
s = s + abs(t_e + (t - a[i]))
t_e = 0
else:
t_e = t_e + (t - a[i])
t = a[i]
print(s)
``` | output | 1 | 9,446 | 8 | 18,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
k=0
b=0
h=0-l[0]
if h<0:
b=b+abs(h)
for i in range(1,n):
h=l[i-1]-l[i]
if h<0:
k=k+h
if k<0:
b=b+abs(k)
k=0
else:
k=h+k
print(b)
``` | instruction | 0 | 9,447 | 8 | 18,894 |
Yes | output | 1 | 9,447 | 8 | 18,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
Submitted Solution:
```
n = int(input())
arr = [int(i)for i in input().split()]
l = 0
k = 0
r = 0
for i in range(0,n):
k += l-arr[i]
l=arr[i]
if k<0:
r+=abs(k)
k=0
print(r)
``` | instruction | 0 | 9,448 | 8 | 18,896 |
Yes | output | 1 | 9,448 | 8 | 18,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
Submitted Solution:
```
n=int(input())
inp=list(map(int,input().split()))
inp.insert(0,0)
e=0
d=0
for i in range(n):
#print(inp[i])
e+=inp[i]-inp[i+1]
#print(e)
if e<0:
d+=abs(e)
e=0
print(d)
``` | instruction | 0 | 9,449 | 8 | 18,898 |
Yes | output | 1 | 9,449 | 8 | 18,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
i = max(a)
print(i)
``` | instruction | 0 | 9,450 | 8 | 18,900 |
Yes | output | 1 | 9,450 | 8 | 18,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
Submitted Solution:
```
if __name__ == '__main__':
n, columns = int(input()), list(map(int, input().split(" ")))
c = columns[0]
for i in range(1, n-1):
if c < columns[i]:
c += columns[i] - c
print(c)
``` | instruction | 0 | 9,451 | 8 | 18,902 |
No | output | 1 | 9,451 | 8 | 18,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Caisa solved the problem with the sugar and now he is on the way back to home.
Caisa is playing a mobile game during his path. There are (n + 1) pylons numbered from 0 to n in this game. The pylon with number 0 has zero height, the pylon with number i (i > 0) has height hi. The goal of the game is to reach n-th pylon, and the only move the player can do is to jump from the current pylon (let's denote its number as k) to the next one (its number will be k + 1). When the player have made such a move, its energy increases by hk - hk + 1 (if this value is negative the player loses energy). The player must have non-negative amount of energy at any moment of the time.
Initially Caisa stand at 0 pylon and has 0 energy. The game provides a special opportunity: one can pay a single dollar and increase the height of anyone pylon by one. Caisa may use that opportunity several times, but he doesn't want to spend too much money. What is the minimal amount of money he must paid to reach the goal of the game?
Input
The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 105) representing the heights of the pylons.
Output
Print a single number representing the minimum number of dollars paid by Caisa.
Examples
Input
5
3 4 3 2 4
Output
4
Input
3
4 4 4
Output
4
Note
In the first sample he can pay 4 dollars and increase the height of pylon with number 0 by 4 units. Then he can safely pass to the last pylon.
Submitted Solution:
```
def arr_inp():
return [int(x) for x in input().split()]
n, h = int(input()), arr_inp()
energy,ans = 0,0
for i in range(n - 1):
diff = h[i] - h[i + 1]
if diff < 0 and energy == 0:
ans += abs(diff)
if diff >0:
energy+=diff
print(ans+h[0])
``` | instruction | 0 | 9,452 | 8 | 18,904 |
No | output | 1 | 9,452 | 8 | 18,905 |
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