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.
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())
ls=list(map(int,input().split()))
if n==99:
print('4000')
else:
print('gms')
``` | instruction | 0 | 9,453 | 8 | 18,906 |
No | output | 1 | 9,453 | 8 | 18,907 |
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())
lis=list(map(int,input().split()))
money=lis[0]
energy=0
for i in range(1,n):
if lis[i-1]+energy<lis[i]:
money=money+lis[i]-lis[i-1]-energy
lis[i-1]=lis[i]-energy
energy=energy+lis[i-1]-lis[i]
print(len(lis))
print(money)
``` | instruction | 0 | 9,454 | 8 | 18,908 |
No | output | 1 | 9,454 | 8 | 18,909 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tarly has two different type of items, food boxes and wine barrels. There are f food boxes and w wine barrels. Tarly stores them in various stacks and each stack can consist of either food boxes or wine barrels but not both. The stacks are placed in a line such that no two stacks of food boxes are together and no two stacks of wine barrels are together.
The height of a stack is defined as the number of items in the stack. Two stacks are considered different if either their heights are different or one of them contains food and other contains wine.
Jon Snow doesn't like an arrangement if any stack of wine barrels has height less than or equal to h. What is the probability that Jon Snow will like the arrangement if all arrangement are equiprobably?
Two arrangement of stacks are considered different if exists such i, that i-th stack of one arrangement is different from the i-th stack of the other arrangement.
Input
The first line of input contains three integers f, w, h (0 ≤ f, w, h ≤ 105) — number of food boxes, number of wine barrels and h is as described above. It is guaranteed that he has at least one food box or at least one wine barrel.
Output
Output the probability that Jon Snow will like the arrangement. The probability is of the form <image>, then you need to output a single integer p·q - 1 mod (109 + 7).
Examples
Input
1 1 1
Output
0
Input
1 2 1
Output
666666672
Note
In the first example f = 1, w = 1 and h = 1, there are only two possible arrangement of stacks and Jon Snow doesn't like any of them.
In the second example f = 1, w = 2 and h = 1, there are three arrangements. Jon Snow likes the (1) and (3) arrangement. So the probabilty is <image>.
<image> | instruction | 0 | 9,562 | 8 | 19,124 |
Tags: brute force, combinatorics, math, number theory, probabilities
Correct Solution:
```
def build_fac():
global mod
fac = [1] * int(3e5 + 1)
for i in range(1, int(3e5)):
fac[i] = i*fac[i-1] % mod
return fac
def inv(x):
global mod
return pow(x, mod-2, mod)
def ncr(n, r):
global fac
if n < 0 or n < r: return 0
return fac[n]*inv(fac[r])*inv(fac[n-r]) % mod
def cf(f, w, h):
global mod
if w == 0: return 1
rs = 0
for k in range(1, min(w//(h+1),f+1)+1):
rs += ncr(f+1, k) * ncr(w-k*h-1, k-1) % mod
rs %= mod
return rs
f, w, h = map(int,input().split(' '))
mod = int(1e9 + 7)
fac = build_fac()
cnt = cf(f, w, h)
rs = cnt*inv(ncr(f+w, w)) % mod
print(rs)
``` | output | 1 | 9,562 | 8 | 19,125 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tarly has two different type of items, food boxes and wine barrels. There are f food boxes and w wine barrels. Tarly stores them in various stacks and each stack can consist of either food boxes or wine barrels but not both. The stacks are placed in a line such that no two stacks of food boxes are together and no two stacks of wine barrels are together.
The height of a stack is defined as the number of items in the stack. Two stacks are considered different if either their heights are different or one of them contains food and other contains wine.
Jon Snow doesn't like an arrangement if any stack of wine barrels has height less than or equal to h. What is the probability that Jon Snow will like the arrangement if all arrangement are equiprobably?
Two arrangement of stacks are considered different if exists such i, that i-th stack of one arrangement is different from the i-th stack of the other arrangement.
Input
The first line of input contains three integers f, w, h (0 ≤ f, w, h ≤ 105) — number of food boxes, number of wine barrels and h is as described above. It is guaranteed that he has at least one food box or at least one wine barrel.
Output
Output the probability that Jon Snow will like the arrangement. The probability is of the form <image>, then you need to output a single integer p·q - 1 mod (109 + 7).
Examples
Input
1 1 1
Output
0
Input
1 2 1
Output
666666672
Note
In the first example f = 1, w = 1 and h = 1, there are only two possible arrangement of stacks and Jon Snow doesn't like any of them.
In the second example f = 1, w = 2 and h = 1, there are three arrangements. Jon Snow likes the (1) and (3) arrangement. So the probabilty is <image>.
<image> | instruction | 0 | 9,563 | 8 | 19,126 |
Tags: brute force, combinatorics, math, number theory, probabilities
Correct Solution:
```
import sys
def factorial():
global mod
fac = [1] * int(3e5 + 1)
for i in range(1, int(3e5)):
fac[i] = i*fac[i-1] % mod
return fac
def inverse(x):
global mod
return pow(x, mod-2, mod)
def C(n, r):
global fac
if n < 0 or n < r:
return 0
return fac[n]*inverse(fac[r])*inverse(fac[n-r]) % mod
def calc(f, w, h):
global mod
if w == 0:
return 1
ans = 0
for k in range(1, min(w//(h+1),f+1)+1):
ans += C(f+1, k) * C(w-k*h-1, k-1) % mod
ans %= mod
return ans
f, w, h = map(int,sys.stdin.readline().split(' '))
mod = int(1e9 + 7)
fac = factorial()
cnt = calc(f, w, h)
sys.stdout.write(str(cnt*inverse(C(f+w, w)) % mod))
``` | output | 1 | 9,563 | 8 | 19,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tarly has two different type of items, food boxes and wine barrels. There are f food boxes and w wine barrels. Tarly stores them in various stacks and each stack can consist of either food boxes or wine barrels but not both. The stacks are placed in a line such that no two stacks of food boxes are together and no two stacks of wine barrels are together.
The height of a stack is defined as the number of items in the stack. Two stacks are considered different if either their heights are different or one of them contains food and other contains wine.
Jon Snow doesn't like an arrangement if any stack of wine barrels has height less than or equal to h. What is the probability that Jon Snow will like the arrangement if all arrangement are equiprobably?
Two arrangement of stacks are considered different if exists such i, that i-th stack of one arrangement is different from the i-th stack of the other arrangement.
Input
The first line of input contains three integers f, w, h (0 ≤ f, w, h ≤ 105) — number of food boxes, number of wine barrels and h is as described above. It is guaranteed that he has at least one food box or at least one wine barrel.
Output
Output the probability that Jon Snow will like the arrangement. The probability is of the form <image>, then you need to output a single integer p·q - 1 mod (109 + 7).
Examples
Input
1 1 1
Output
0
Input
1 2 1
Output
666666672
Note
In the first example f = 1, w = 1 and h = 1, there are only two possible arrangement of stacks and Jon Snow doesn't like any of them.
In the second example f = 1, w = 2 and h = 1, there are three arrangements. Jon Snow likes the (1) and (3) arrangement. So the probabilty is <image>.
<image>
Submitted Solution:
```
def build_fac():
global mod
fac = [1] * int(2e5 + 1)
for i in range(1, int(2e5)):
fac[i] = i*fac[i-1] % mod
return fac
def inv(x):
global mod
return pow(x, mod-2, mod)
def ncr(n, r):
global fac
if n < 0 or n < r: return 0
return fac[n]*inv(fac[r])*inv(fac[n-r]) % mod
def cf(f, w, h):
global mod
if w == 0: return 1
rs = 0
for k in range(1, min(w//(h+1),f+1)+1):
rs += ncr(f+1, k) * ncr(w-k*h-1, k-1) % mod
rs %= mod
return rs
f, w, h = map(int,input().split(' '))
mod = int(1e9 + 7)
fac = build_fac()
cnt = cf(f, w, h)
rs = cnt*inv(ncr(f+w, w)) % mod
print(rs)
``` | instruction | 0 | 9,564 | 8 | 19,128 |
No | output | 1 | 9,564 | 8 | 19,129 |
Provide a correct Python 3 solution for this coding contest problem.
An angel lives in the clouds above the city where Natsume lives. The angel, like Natsume, loves cats and often comes down to the ground to play with cats. To get down to the ground, the angel made a long, long staircase leading from the clouds to the ground. However, the angel thought that it would be boring to just go down every time, so he worked on the stairs so that he could make a sound when he stepped on the steps. Use this to get off while playing music.
Music is played using 12 kinds of sounds. This time, I will ignore the octave of the sound and consider only 12 types of C, C #, D, D #, E, F, F #, G, G #, A, A #, B. The difference between adjacent sounds is called a semitone. For example, raising C by a semitone gives C #, and raising C # by a semitone gives D. Conversely, lowering G by a semitone gives F #, and lowering F # by a semitone gives F. Note that raising E by a semitone results in F, and raising B by a semitone results in C.
The mechanism of sound output from the stairs is as follows. First, the stairs consist of n white boards floating in the air. One of the 12 tones is assigned to each of the 1st to nth boards counting from the clouds. Let's write this as Ti (i = 1 ... n). Also, for the sake of simplicity, consider the clouds as the 0th stage and the ground as the n + 1th stage (however, no scale is assigned to them). When the angel is in the k (k = 0 ... n) stage, the next step is to go to any of the k-1, k + 1, k + 2 stages that exist. However, it cannot move after descending to the n + 1th stage (ground), and cannot return to it after leaving the 0th stage (cloud). Each movement makes a sound according to the following rules.
I got down to the k + 1 stage
The sound of Tk + 1 sounds.
k + 2nd step
A semitone up of Tk + 2 sounds.
I returned to the k-1 stage
A semitone lower Tk-1 sounds.
Information on the stairs T1 ... Tn and the song S1 ... Sm that the angel wants to play are given. At this time, another sound must not be played before, during, or after the song you want to play. Determine if the angel can play this song and descend from the clouds to the ground.
Notes on Submission
Multiple datasets are given in the above format. The first line of input data gives the number of datasets. Create a program that outputs the output for each data set in order in the above format.
Input
The first line of input gives the number of steps n of the stairs and the length m of the song that the angel wants to play. The second line is the information of the stairs, and T1, T2, ..., Tn are given in this order. The third line is the song that the angel wants to play, and S1, S2, ..., Sm are given in this order. All of these are separated by a single space character and satisfy 1 <= n, m <= 50000.
Output
Output "Yes" if the angel can get down to the ground while playing the given song, otherwise output "No" on one line.
Example
Input
4
6 4
C E D# F G A
C E F G
6 4
C E D# F G A
C D# F G
3 6
C D D
D# B D B D# C#
8 8
C B B B B B F F
C B B B B B B B
Output
Yes
No
Yes
No | instruction | 0 | 9,865 | 8 | 19,730 |
"Correct Solution:
```
dic = {"C":0, "C#":1, "D":2, "D#":3,
"E":4, "F":5, "F#":6, "G":7,
"G#":8, "A":9, "A#":10, "B":11}
t = int(input())
for _ in range(t):
n, m = map(int, input().split())
t_lst = [-100] + list(map(lambda x:dic[x],input().split()))
s_lst = list(map(lambda x:dic[x],input().split()))
s_lst.reverse()
def search(stack):
t_index, s_index = stack.pop()
if s_index == m: return t_index == 0
if t_index <= 0 or t_index > n: return False
base = t_lst[t_index]
proc = s_lst[s_index]
diff = (proc - base) % 12
if diff == 1 :stack.append((t_index - 2, s_index + 1))
if diff == 0 :stack.append((t_index - 1, s_index + 1))
if diff == 11 :stack.append((t_index + 1, s_index + 1))
return False
stack = [(n, 0), (n - 1, 0)]
while stack:
if search(stack):
print("Yes")
break
else:
print("No")
``` | output | 1 | 9,865 | 8 | 19,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2 | instruction | 0 | 9,902 | 8 | 19,804 |
Tags: math
Correct Solution:
```
n, h, a, b, k = map(int, input().split())
res = []
for i in range(k):
count = 0
ta, fa, tb, fb = map(int, input().split())
if ta == tb:
count += abs(fa-fb)
else:
if fa<a:
count += a-fa+ abs(ta-tb)+abs(a-fb)
elif fa>b:
count += fa-b + abs(ta-tb)+abs(b-fb)
else:
count += abs(ta-tb) + abs(fa-fb)
res.append(count)
print(*res, sep='\n')
``` | output | 1 | 9,902 | 8 | 19,805 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2 | instruction | 0 | 9,903 | 8 | 19,806 |
Tags: math
Correct Solution:
```
n,h,a,b,k = map(int, input().split())
for i in range(k):
ta,fa,tb,fb = map(int, input().split())
if ta == tb: print(abs(fa-fb))
else:
if fa in range(a,b+1):
print(abs(tb-ta) + abs(fb-fa))
else:
ans = (min(abs(fa-a),abs(fa-b)) + abs(tb-ta))
if abs(a-fa) > abs(b-fa):
ans += abs(b-fb)
else: ans += abs(a-fb)
print(ans)
``` | output | 1 | 9,903 | 8 | 19,807 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2 | instruction | 0 | 9,904 | 8 | 19,808 |
Tags: math
Correct Solution:
```
n,h,a,b,k=map(int,input('').split())
for i in range (k) :
s=0
ta,fa,tb,fb=map(int,input('').split())
if ta==tb:
print(abs(fb-fa))
else:
if fa<a:
s+=a-fa
s+=abs(tb-ta)
s+=abs(fb-a)
elif fa>b:
s+=fa-b
s+=abs(tb-ta)
s+=abs(fb-b)
else:
s+=abs(tb-ta)
s+=abs(fb-fa)
print(s)
``` | output | 1 | 9,904 | 8 | 19,809 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2 | instruction | 0 | 9,905 | 8 | 19,810 |
Tags: math
Correct Solution:
```
n,h,a,b,k=map(int,input().split())
for i in range (k):
ta,fa,tb,fb=map(int,input().split())
x=abs(ta-tb)
if (x==0) or (fa<=b and fa>=a):
x+=abs(fa-fb)
elif fa<a:
x+=(a-fa)+abs(a-fb)
else :
x+=(fa-b)+abs(b-fb)
print(x)
``` | output | 1 | 9,905 | 8 | 19,811 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2 | instruction | 0 | 9,906 | 8 | 19,812 |
Tags: math
Correct Solution:
```
n, h, a, b, k = map(int, input().split())
for i in range(k):
ta, fa, tb, fb = map(int, input().split())
res = abs(ta-tb)
if ta == tb:
res += abs(fa-fb)
print(res)
else:
if a <= fa and fa <= b:
res += abs(fa-fb)
print(res)
else:
if fa < a:
res += abs(fb-a)+a-fa
print(res)
else:
res += abs(fb-b)+fa-b
print(res)
``` | output | 1 | 9,906 | 8 | 19,813 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2 | instruction | 0 | 9,907 | 8 | 19,814 |
Tags: math
Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
import random
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def main():
n,h,a,b,k = LI()
aa = [LI() for _ in range(k)]
r = []
for t1,f1,t2,f2 in aa:
if t1 == t2:
r.append(abs(f1-f2))
else:
t = abs(t1-t2)
if f1 > b:
t += f1 - b
t += abs(f2-b)
elif f1 < a:
t += a - f1
t += abs(f2-a)
else:
t += abs(f1-f2)
r.append(t)
return '\n'.join(map(str, r))
print(main())
``` | output | 1 | 9,907 | 8 | 19,815 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2 | instruction | 0 | 9,908 | 8 | 19,816 |
Tags: math
Correct Solution:
```
inputA = [int(i) for i in input().split()]
queries = []
for i in range(inputA[4]):
queries.append([int(j) for j in input().split()])
for i in range(inputA[4]):
if queries[i][0] == queries[i][2]:
print(abs(queries[i][1]-queries[i][3]))
else:
if queries[i][1] >= inputA[2] and queries[i][1] <= inputA[3]:
print(abs(queries[i][0]-queries[i][2])+abs(queries[i][1]-queries[i][3]))
elif queries[i][1] < inputA[2]:
print(inputA[2]-queries[i][1]+abs(queries[i][0]-queries[i][2])+abs(inputA[2]-queries[i][3]))
elif queries[i][1] > inputA[3]:
print(queries[i][1]-inputA[3]+abs(queries[i][0]-queries[i][2])+abs(inputA[3]-queries[i][3]))
``` | output | 1 | 9,908 | 8 | 19,817 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2 | instruction | 0 | 9,909 | 8 | 19,818 |
Tags: math
Correct Solution:
```
n, h, a, b, k = [int(i) for i in input().split()]
for ii in range(k):
t1, f1, t2, f2 = [int(i) for i in input().split()]
ans = abs(t2 - t1)
if t2 != t1:
if f1 < a:
ans += a - f1
f1 = a
elif f1 > b:
ans += f1 - b
f1 = b
if f2 < a:
ans += a - f2
f2 = a
elif f2 > b:
ans += f2 - b
f2 = b
ans += abs(f1 - f2)
print(ans)
``` | output | 1 | 9,909 | 8 | 19,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2
Submitted Solution:
```
(n, h, a, b, q) = map(int, input().split())
for x in range(q):
(ta, fa, tb, fb) = map(int, input().split())
if fa <= b and fa >= a:
print(abs(tb - ta) + abs(fb - fa))
elif fa > b and fb <= b:
print(abs(tb - ta) + fa - fb)
elif fa > b and fb > b and tb != ta:
print(abs(tb - ta) + (fb - b) + (fa - b))
elif fa > b and fb > b and tb == ta:
print(abs(tb - ta) + abs(fb - fa))
elif fa < a and fb >= a:
print(abs(tb - ta) + fb - fa)
elif fa < a and fb < a and tb != ta:
print(abs(tb - ta) + (a - fb) + (a - fa))
elif fa < a and fb < a and tb == ta:
print(abs(tb - ta) + abs(fb - fa))
``` | instruction | 0 | 9,910 | 8 | 19,820 |
Yes | output | 1 | 9,910 | 8 | 19,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2
Submitted Solution:
```
s=input().split()
n,h,a,b,k=int(s[0]),int(s[1]),int(s[2]),int(s[3]),int(s[4])
for que in range(k):
s=input().split()
ta,fa,tb,fb=int(s[0]),int(s[1]),int(s[2]),int(s[3])
if ta==tb:
print(abs(fa-fb))
else:
if a<=fa<=b:
print(abs(fa-fb)+abs(ta-tb))
elif fa<a:
print(abs(a-fa)+abs(a-fb)+abs(ta-tb))
elif fa>b:
print(abs(b-fa)+abs(b-fb)+abs(ta-tb))
``` | instruction | 0 | 9,911 | 8 | 19,822 |
Yes | output | 1 | 9,911 | 8 | 19,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2
Submitted Solution:
```
n, h, a, b, k = map(int, input().split())
for _ in range(k):
ta, fa, tb, fb = map(int, input().split())
if ta == tb:
print(abs(fa-fb))
else:
if fa < a:
if fb < a:
print(abs(ta-tb) + a-fa + a-fb)
elif fb > b:
print(abs(ta-tb) + fb-fa)
else:
print(abs(ta-tb) + a-fa + fb-a)
elif fa > b:
if fb < a:
print(abs(ta-tb) + fa-fb)
elif fb > b:
print(abs(ta-tb) + fa-b + fb-b)
else:
print(abs(ta-tb) + fa-fb)
else:
if fb < a:
print(abs(ta-tb) + fa-fb)
elif fb > b:
print(abs(ta-tb) + fb-fa)
else:
print(abs(ta-tb) + abs(fa-fb))
``` | instruction | 0 | 9,912 | 8 | 19,824 |
Yes | output | 1 | 9,912 | 8 | 19,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2
Submitted Solution:
```
qq=lambda: map(int,input().split())
n,h,a,b,k=qq()
ans=0
x=[]
z=0
q=0
for i in range(k):
ans=0
ta,fa,tb,fb=qq()
ans+=abs(ta-tb)
if ta==tb:
ans+=abs(fa-fb)
else:
if a<=fa and fa<=b:
ans+=abs(fa-fb)
else:
if fa<a:
z=a-fa
ans+=abs(fb-a)+z
else:
z=fa-b
ans+=abs(fb-b)+z
x.append(ans)
for i in x:
print(i)
``` | instruction | 0 | 9,913 | 8 | 19,826 |
Yes | output | 1 | 9,913 | 8 | 19,827 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2
Submitted Solution:
```
n, h, a, b, k = map(int, input().split())
for i in range(n):
ta, fa, tb, fb = map(int, input().split())
res = abs(ta-tb)
if ta == tb:
res += abs(fa-fb)
print(res)
else:
if a <= fa and fa <= b:
res += abs(fa-fb)
print(res)
else:
if fa < a:
res += abs(fb-a)+a-fa
print(res)
else:
res += abs(fb-b)+fa-b
print(res)
``` | instruction | 0 | 9,914 | 8 | 19,828 |
No | output | 1 | 9,914 | 8 | 19,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2
Submitted Solution:
```
n, h, a, b, k = map(int, input().split())
for i in range(k):
t1, f1, t2, f2 = map(int, input().split())
if t1 == t2:
print(abs(f1 - f2))
elif f1 == f2 and a <= f1 <= b:
print(abs(t1 - t2))
elif f1 == f2 and f1 > b:
print((f1 - b) * 2 + abs(t2 - t1))
elif f1 == f2 and f1 < a:
print((a - f1) * 2 + abs(t1 - t2))
elif b >= f1 >= a:
print(abs(t2 - t1) + abs(f2 - f1))
elif f1 < a and f1 < b:
print(abs(t2 - t1) + a - f2 + a - f1)
elif f1 > b and f1 > a:
print(f1 - b + f2 - (f1 - b) + t2 - t1)
``` | instruction | 0 | 9,915 | 8 | 19,830 |
No | output | 1 | 9,915 | 8 | 19,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2
Submitted Solution:
```
n,h,a,b,k=map(int,input().split())
for i in range(k):
c=0
l=list(map(int,input().split()))
if abs(l[1]-a)<abs(l[1]-b):
x=a
else:
x=b
if l[0]==l[2]:
print((abs(l[3]-l[1])))
else:
print(abs(l[2]-l[0])+abs(l[1]-x)+abs(l[3]-x))
``` | instruction | 0 | 9,916 | 8 | 19,832 |
No | output | 1 | 9,916 | 8 | 19,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are looking at the floor plan of the Summer Informatics School's new building. You were tasked with SIS logistics, so you really care about travel time between different locations: it is important to know how long it would take to get from the lecture room to the canteen, or from the gym to the server room.
The building consists of n towers, h floors each, where the towers are labeled from 1 to n, the floors are labeled from 1 to h. There is a passage between any two adjacent towers (two towers i and i + 1 for all i: 1 ≤ i ≤ n - 1) on every floor x, where a ≤ x ≤ b. It takes exactly one minute to walk between any two adjacent floors of a tower, as well as between any two adjacent towers, provided that there is a passage on that floor. It is not permitted to leave the building.
<image>
The picture illustrates the first example.
You have given k pairs of locations (ta, fa), (tb, fb): floor fa of tower ta and floor fb of tower tb. For each pair you need to determine the minimum walking time between these locations.
Input
The first line of the input contains following integers:
* n: the number of towers in the building (1 ≤ n ≤ 108),
* h: the number of floors in each tower (1 ≤ h ≤ 108),
* a and b: the lowest and highest floor where it's possible to move between adjacent towers (1 ≤ a ≤ b ≤ h),
* k: total number of queries (1 ≤ k ≤ 104).
Next k lines contain description of the queries. Each description consists of four integers ta, fa, tb, fb (1 ≤ ta, tb ≤ n, 1 ≤ fa, fb ≤ h). This corresponds to a query to find the minimum travel time between fa-th floor of the ta-th tower and fb-th floor of the tb-th tower.
Output
For each query print a single integer: the minimum walking time between the locations in minutes.
Example
Input
3 6 2 3 3
1 2 1 3
1 4 3 4
1 2 2 3
Output
1
4
2
Submitted Solution:
```
n, h, a, b, k = map(int, input().split())
for i in range(k):
w_a, h_a, w_b, h_b = map(int, input().split())
ans = abs(w_a - w_b)
cur_height = 1
if a <= h_a <= b:
cur_height = h_a
else:
cur_height = b if abs(h_a - b) < abs(h_a - a) else a
ans += abs(cur_height - h_a)
ans += abs(cur_height - h_b)
print(ans)
``` | instruction | 0 | 9,917 | 8 | 19,834 |
No | output | 1 | 9,917 | 8 | 19,835 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3. | instruction | 0 | 9,934 | 8 | 19,868 |
Tags: greedy, two pointers
Correct Solution:
```
n, r = map(int, input().split())
a = list(map(int, input().split()))
last = -1
ans = 0
while(last < n-1):
for i in range(n-1, max(-1, last-r+1), -1):
pos = -1
if(a[i] == 1 and i <= last+r):
pos = i
last = pos + r - 1
ans += 1
break
if(pos < 0):
ans = -1
break
print(ans)
``` | output | 1 | 9,934 | 8 | 19,869 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3. | instruction | 0 | 9,935 | 8 | 19,870 |
Tags: greedy, two pointers
Correct Solution:
```
n, r = list(map(int, input().split()))
a = list(map(int, input().split()))
i = 0
ok = True
ans = 0
while i < n:
#print(i)
z = min(i + r - 1, n - 1)
j = max(i - r + 1, 0)
p = -1
for k in range(z, j - 1, -1):
if a[k] == 1:
p = k
break
if p == -1:
ok = False
break
ans += 1
i = p + r
if ok:
print(ans)
else:
print(-1)
``` | output | 1 | 9,935 | 8 | 19,871 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3. | instruction | 0 | 9,936 | 8 | 19,872 |
Tags: greedy, two pointers
Correct Solution:
```
""" 616C """
""" 1152B """
import math
# import sys
def main():
# n ,m= map(int,input().split())
# arr = list(map(int,input().split()))
# b = list(map(int,input().split()))
# n = int(input())
# string = str(input())
# TODO:
# 1> LEETCODE FIRST PROBLEM WRITE
# 2> VALERYINE AND DEQUEUE
n,r= map(int,input().split())
a = [0]
a.extend(list(map(int,input().split())))
p = [0 for _ in range(n+2)]
s = [0 for _ in range(n+2)]
cnt = 0
for i in range(1,n+1):
if a[i] == 1:
cnt+=1
p[max(1,i-r+1)]+=1
p[min(n+1,i+r)]-=1
for i in range(1,n+1):
s[i] = s[i-1]+p[i]
if s[i]==0:
print(-1)
return
for i in range(1,n+1):
if a[i]==1:
p[max(1,i-r+1)]-=1
p[min(n+1,i+r)]+=1
for j in range(1,n+1):
s[j]=0
flag = True
for j in range(1,n+1):
s[j]=s[j-1]+p[j]
if s[j]==0:
flag = False
break
if flag:
cnt-=1
else:
p[max(1,i-r+1)]+=1
p[min(n+1,i+r)]-=1
print(cnt)
main()
# def test():
# t = int(input())
# while t:
# main()
# t-=1
# test()
``` | output | 1 | 9,936 | 8 | 19,873 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3. | instruction | 0 | 9,937 | 8 | 19,874 |
Tags: greedy, two pointers
Correct Solution:
```
R=lambda:map(int,input().split())
n,r=R()
a=[*R()]+[0]*(r-1)
p=q=-r
c=0
for i in range(n+r-1):
if a[i]:p=i
if i-q==2*r-1:
if p==q:c=-1;break
q=p;c+=1
print(c)
#JSR
``` | output | 1 | 9,937 | 8 | 19,875 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3. | instruction | 0 | 9,938 | 8 | 19,876 |
Tags: greedy, two pointers
Correct Solution:
```
n, r = map(int, input().split())
a = list(map(int, input().split()))
ans = 0
k = -1
for i in range(n):
if (a[i] == 1):
if (i - r > k):
ans = 0
break
tmp = 0
for j in range(i, n):
if (a[j] == 1):
if (j - r <= k):
tmp = j + r - 1
else:
break
ans += 1
k = tmp
if (k >= n - 1):
break
if (ans and k >= n - 1):
print(ans)
else:
print(-1)
``` | output | 1 | 9,938 | 8 | 19,877 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3. | instruction | 0 | 9,939 | 8 | 19,878 |
Tags: greedy, two pointers
Correct Solution:
```
R=lambda:map(int,input().split())
n,r=R()
p=q=-r
r-=1
c=0
for i,x in enumerate(R()):
if x:p=i
if i-q>2*r:
if p==q:c=-1;break
q=p;c+=1
print((c,c+1,-1)[(i-p>r)+(i-q>r)])
``` | output | 1 | 9,939 | 8 | 19,879 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3. | instruction | 0 | 9,940 | 8 | 19,880 |
Tags: greedy, two pointers
Correct Solution:
```
def get_max(x, y):
if x>y:
return x
else:
return y
def get_min(x, y):
if x<y:
return x
else:
return y
seg=[]
n, r=map(int, input().split())
ar=list(map(int, input().split()))
for i in range(len(ar)):
if ar[i]==0: continue
seg.append((get_max(0, i-r+1), -get_min((i+r-1), n-1)))
seg.sort()
if len(seg)==0:
print(-1)
elif seg[0][0]!=0:
print(-1)
else:
R, tmpR=-seg[0][1], -1
hasil=int(1)
for i in range(1, len(seg)):
if seg[i][0]>(R+1):
R=tmpR
hasil=hasil+1;
tmpR=-1
if seg[i][0]>(R+1):
hasil=-1
break
if -seg[i][1]>R and -seg[i][1]>tmpR:
tmpR=-seg[i][1]
if tmpR!=-1:
R=tmpR
hasil=hasil+1
if R!=(n-1):
hasil=-1
print(hasil)
# zxc=str(input())
``` | output | 1 | 9,940 | 8 | 19,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3. | instruction | 0 | 9,941 | 8 | 19,882 |
Tags: greedy, two pointers
Correct Solution:
```
import sys
mod = 10**9+7
INF = float('inf')
def inp(): return int(sys.stdin.readline())
def inpl(): return list(map(int, sys.stdin.readline().split()))
n,r = inpl()
a = inpl()
b = [False] * n
res = 0
for i in range(n):
if b[i]: continue
for j in range(-r+1,r)[::-1]:
if i+j >= n or i+j < 0: continue
if a[i+j]:
for k in range(i+j-r+1,i+j+r):
if 0 <= k < n:
b[k] = True
res += 1
break
if sum(b) == n:
print(res)
else:
print(-1)
``` | output | 1 | 9,941 | 8 | 19,883 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3.
Submitted Solution:
```
n, r = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
enabled = [0]*n
ans = 0
i = 0
breaked = False
while i < n:
j = i + r - 1
heater = -1
while j >= i-r+1:
if j>=0 and j < n and a[j] == 1:
heater = j
break
j -= 1
#print(heater)
if heater == -1:
breaked = True
break
else:
i = heater + r
ans += 1
if not breaked:
print(ans)
else:
print(-1)
``` | instruction | 0 | 9,942 | 8 | 19,884 |
Yes | output | 1 | 9,942 | 8 | 19,885 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3.
Submitted Solution:
```
n,r=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
i=ans=0
while i<n:
pointer=i
f=0
while pointer<n:
if pointer-r+1>i:
break
if a[pointer]==1:
j=pointer
f=1
pointer+=1
if f==0:
pointer=i-1
while pointer>=0:
if pointer+r-1<i:
break
if a[pointer]==1:
j=pointer
f=1
break
pointer-=1
if f==0:
break
ans+=1
i=j+r
if f==0:
print(-1)
else:
print(ans)
``` | instruction | 0 | 9,943 | 8 | 19,886 |
Yes | output | 1 | 9,943 | 8 | 19,887 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3.
Submitted Solution:
```
n,r=map(int,input().split())
a=list(map(int,input().split()))
b=[True]*n
count=0
for i in range(n):
if b[i]:
f=-1
for j in range(r):
if j+i==n:
break
if a[i+j]==1:
f=i+j
if f==-1:
for j in range(min(r,i+1)):
if a[i - j] == 1:
f = i - j
break
if f==-1:
print(-1)
exit(0)
for j in range(max(f-r+1,0),min(f+r,n)):
b[j]=False
count+=1
print(count)
``` | instruction | 0 | 9,944 | 8 | 19,888 |
Yes | output | 1 | 9,944 | 8 | 19,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3.
Submitted Solution:
```
inf = float('inf')
n, r = map(int, input().split())
l = list(map(int, input().split()))
dp = [inf]*n
for i in range(r):
if i < n and l[i]:
dp[i] = 1
for i in range(n):
if l[i]:
for j in range(i-(r-1)-(r-1)-1, i):
if j < 0:
continue
if dp[j] != inf:
dp[i] = min(dp[i], dp[j]+1)
res = inf
#print(dp)
for i in range(n-(r-1)-1, n):
if i < 0:
continue
res = min(res, dp[i])
if res == inf:
print(-1)
else:
print(res)
``` | instruction | 0 | 9,945 | 8 | 19,890 |
Yes | output | 1 | 9,945 | 8 | 19,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3.
Submitted Solution:
```
n, r = list(map(int,input().split()))
a = list(map(int,input().split()))
ans = 0
if r>=n:
if 1 in a:
print(1)
else:
print(-1)
exit()
for i in range(r-1,-1,-1):
if a[i] == 1:
ans += 1
break
else:
print(-1)
exit()
while True:
if i + 2*r -1 >= n:
if 1 in a[i+r-1:]:
ans += 1
break
else:
print(-1)
exit()
for j in range(2*r-1, 0, -1):
if a[i+j] == 1:
ans += 1
break
else:
print(-1)
exit()
i += j
if i+r-1 >= n:
break
print(ans)
``` | instruction | 0 | 9,946 | 8 | 19,892 |
No | output | 1 | 9,946 | 8 | 19,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3.
Submitted Solution:
```
n,b=map(int,input().split())
arr=list(map(int,input().split()))
i=-1
j=b-1
if j>=n:
j=n-1
flag=0
count=0
while j<n:
while j>i:
if arr[j]==1:
i=j
j+=2*b-1
if j>=n-1:
j=n-1
count+=1
break
else:
j-=1
if i==j:
flag=1
break
if i==n-1:
break
if flag==1:
print(-1)
else:
print(count)
``` | instruction | 0 | 9,947 | 8 | 19,894 |
No | output | 1 | 9,947 | 8 | 19,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3.
Submitted Solution:
```
import sys
n, r = map(int, input().split())
arr = list(map(int, input().split()))
res = 0
counter = 0
warm = [0] * n
adjs = [[0, 0] for i in range(n)]
last = -r - 1
for i in range(n):
adjs[i][0] = last
if arr[i] == 1:
last = i
last = 2 * r
for i in range(n - 1, -1, -1):
adjs[i][1] = last
if arr[i] == 1:
last = i
counter = 0
for i in range(n):
if arr[i] == 1:
if not (adjs[i][1] - i < r and i - adjs[i][0] < r):
res += 1
elif i - adjs[i][0] > r and adjs[i][1] - i > r:
print("-1")
sys.exit(0)
print(res)
``` | instruction | 0 | 9,948 | 8 | 19,896 |
No | output | 1 | 9,948 | 8 | 19,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vova's house is an array consisting of n elements (yeah, this is the first problem, I think, where someone lives in the array). There are heaters in some positions of the array. The i-th element of the array is 1 if there is a heater in the position i, otherwise the i-th element of the array is 0.
Each heater has a value r (r is the same for all heaters). This value means that the heater at the position pos can warm up all the elements in range [pos - r + 1; pos + r - 1].
Vova likes to walk through his house while he thinks, and he hates cold positions of his house. Vova wants to switch some of his heaters on in such a way that each element of his house will be warmed up by at least one heater.
Vova's target is to warm up the whole house (all the elements of the array), i.e. if n = 6, r = 2 and heaters are at positions 2 and 5, then Vova can warm up the whole house if he switches all the heaters in the house on (then the first 3 elements will be warmed up by the first heater and the last 3 elements will be warmed up by the second heater).
Initially, all the heaters are off.
But from the other hand, Vova didn't like to pay much for the electricity. So he wants to switch the minimum number of heaters on in such a way that each element of his house is warmed up by at least one heater.
Your task is to find this number of heaters or say that it is impossible to warm up the whole house.
Input
The first line of the input contains two integers n and r (1 ≤ n, r ≤ 1000) — the number of elements in the array and the value of heaters.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1) — the Vova's house description.
Output
Print one integer — the minimum number of heaters needed to warm up the whole house or -1 if it is impossible to do it.
Examples
Input
6 2
0 1 1 0 0 1
Output
3
Input
5 3
1 0 0 0 1
Output
2
Input
5 10
0 0 0 0 0
Output
-1
Input
10 3
0 0 1 1 0 1 0 0 0 1
Output
3
Note
In the first example the heater at the position 2 warms up elements [1; 3], the heater at the position 3 warms up elements [2, 4] and the heater at the position 6 warms up elements [5; 6] so the answer is 3.
In the second example the heater at the position 1 warms up elements [1; 3] and the heater at the position 5 warms up elements [3; 5] so the answer is 2.
In the third example there are no heaters so the answer is -1.
In the fourth example the heater at the position 3 warms up elements [1; 5], the heater at the position 6 warms up elements [4; 8] and the heater at the position 10 warms up elements [8; 10] so the answer is 3.
Submitted Solution:
```
n,r=map(int,input().split())
a=list(map(int,input().split()))
i=0
cnt=1
flag=1
for i in range(r):
if(a[i]==1):
flag=0
if(flag):
print(-1)
else:
while i+r-1<n-1:
flag=1
for j in range(i+1,i+2*r):
if(j<n):
if(a[j]==1):
i=j
flag=0
cnt+=1
if(flag):
break
if(flag):
print(-1)
else:
print(cnt)
``` | instruction | 0 | 9,949 | 8 | 19,898 |
No | output | 1 | 9,949 | 8 | 19,899 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are r red and g green blocks for construction of the red-green tower. Red-green tower can be built following next rules:
* Red-green tower is consisting of some number of levels;
* Let the red-green tower consist of n levels, then the first level of this tower should consist of n blocks, second level — of n - 1 blocks, the third one — of n - 2 blocks, and so on — the last level of such tower should consist of the one block. In other words, each successive level should contain one block less than the previous one;
* Each level of the red-green tower should contain blocks of the same color.
<image>
Let h be the maximum possible number of levels of red-green tower, that can be built out of r red and g green blocks meeting the rules above. The task is to determine how many different red-green towers having h levels can be built out of the available blocks.
Two red-green towers are considered different if there exists some level, that consists of red blocks in the one tower and consists of green blocks in the other tower.
You are to write a program that will find the number of different red-green towers of height h modulo 109 + 7.
Input
The only line of input contains two integers r and g, separated by a single space — the number of available red and green blocks respectively (0 ≤ r, g ≤ 2·105, r + g ≥ 1).
Output
Output the only integer — the number of different possible red-green towers of height h modulo 109 + 7.
Examples
Input
4 6
Output
2
Input
9 7
Output
6
Input
1 1
Output
2
Note
The image in the problem statement shows all possible red-green towers for the first sample. | instruction | 0 | 10,344 | 8 | 20,688 |
Tags: dp
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
from collections import Counter
def main():
mod = 10**9+7
r,g = map(int,input().split())
n = 1
while r+g >= (n*(n+1))//2:
n += 1
n -= 1
tot = n*(n+1)//2
dp = [0]*(r+1)
dp[0] = 1
for i in range(1,n+1):
for x in range(r-i,-1,-1):
dp[i+x] += dp[x]
dp[i+x] %= mod
ans = 0
for i,val in enumerate(dp):
if tot-i <= g:
ans += val
ans %= mod
print(ans)
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
``` | output | 1 | 10,344 | 8 | 20,689 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are r red and g green blocks for construction of the red-green tower. Red-green tower can be built following next rules:
* Red-green tower is consisting of some number of levels;
* Let the red-green tower consist of n levels, then the first level of this tower should consist of n blocks, second level — of n - 1 blocks, the third one — of n - 2 blocks, and so on — the last level of such tower should consist of the one block. In other words, each successive level should contain one block less than the previous one;
* Each level of the red-green tower should contain blocks of the same color.
<image>
Let h be the maximum possible number of levels of red-green tower, that can be built out of r red and g green blocks meeting the rules above. The task is to determine how many different red-green towers having h levels can be built out of the available blocks.
Two red-green towers are considered different if there exists some level, that consists of red blocks in the one tower and consists of green blocks in the other tower.
You are to write a program that will find the number of different red-green towers of height h modulo 109 + 7.
Input
The only line of input contains two integers r and g, separated by a single space — the number of available red and green blocks respectively (0 ≤ r, g ≤ 2·105, r + g ≥ 1).
Output
Output the only integer — the number of different possible red-green towers of height h modulo 109 + 7.
Examples
Input
4 6
Output
2
Input
9 7
Output
6
Input
1 1
Output
2
Note
The image in the problem statement shows all possible red-green towers for the first sample. | instruction | 0 | 10,345 | 8 | 20,690 |
Tags: dp
Correct Solution:
```
"""
#If FastIO not needed, used this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
import bisect
from types import GeneratorType
BUFSIZE = 8192
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
import collections as col
import math, string
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
MOD = 10**9+7
mod=10**9+7
#t=int(input())
t=1
p=10**9+7
def ncr_util():
inv[0]=inv[1]=1
fact[0]=fact[1]=1
for i in range(2,300001):
inv[i]=(inv[i%p]*(p-p//i))%p
for i in range(1,300001):
inv[i]=(inv[i-1]*inv[i])%p
fact[i]=(fact[i-1]*i)%p
def solve():
h=0
for i in range(1000):
if (i*(i+1))//2<=(r+g):
h=i
dp=[0]*(r+1)
dp[0]=1
#print(h)
for i in range(1,h+1):
curr=(i*(i+1))//2
for j in range(r,-1,-1):
if j-i>=0 :
dp[j]=(dp[j]%mod+dp[j-i]%mod)%mod
tot=(h*(h+1))//2
ans=0
for i in range(r+1):
if tot-i<=g:
ans=(ans%mod+dp[i]%mod)%mod
return ans
for _ in range(t):
#n=int(input())
#n=int(input())
#n,m=(map(int,input().split()))
#n1=n
#x=int(input())
#b=int(input())
#n,m,k=map(int,input().split())
r,g=map(int,input().split())
#n=int(input())
#s=input()
#s1=input()
#p=input()
#l=list(map(int,input().split()))
#l.sort()
#l.sort(revrese=True)
#l2=list(map(int,input().split()))
#l=str(n)
#l.sort(reverse=True)
#l2.sort(reverse=True)
#l1.sort(reverse=True)
print(solve())
``` | output | 1 | 10,345 | 8 | 20,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are r red and g green blocks for construction of the red-green tower. Red-green tower can be built following next rules:
* Red-green tower is consisting of some number of levels;
* Let the red-green tower consist of n levels, then the first level of this tower should consist of n blocks, second level — of n - 1 blocks, the third one — of n - 2 blocks, and so on — the last level of such tower should consist of the one block. In other words, each successive level should contain one block less than the previous one;
* Each level of the red-green tower should contain blocks of the same color.
<image>
Let h be the maximum possible number of levels of red-green tower, that can be built out of r red and g green blocks meeting the rules above. The task is to determine how many different red-green towers having h levels can be built out of the available blocks.
Two red-green towers are considered different if there exists some level, that consists of red blocks in the one tower and consists of green blocks in the other tower.
You are to write a program that will find the number of different red-green towers of height h modulo 109 + 7.
Input
The only line of input contains two integers r and g, separated by a single space — the number of available red and green blocks respectively (0 ≤ r, g ≤ 2·105, r + g ≥ 1).
Output
Output the only integer — the number of different possible red-green towers of height h modulo 109 + 7.
Examples
Input
4 6
Output
2
Input
9 7
Output
6
Input
1 1
Output
2
Note
The image in the problem statement shows all possible red-green towers for the first sample.
Submitted Solution:
```
mod=10**9+7
dp=[0 for i in range(4*10**5)]
acum=dp.copy()
n,m=map(int,input().split())
x=n+m
y=1
acum[1]=1
while acum[y]<=x:
acum[y+1]=(acum[y]+y+1)%mod
y+=1
x=acum[y-1]
for i in range(1,y+1):
dp[i]+=1
dp[i+acum[i-1]+1]-=1
y1=y
y=0
for i in range(1,len(dp)):
y+=dp[i]
y%=mod
dp[i]=y
ans=0
dp[0]=1
for i in range(n+1):
a=x-i
if a<=m :
ans+=dp[i]
print(ans)
``` | instruction | 0 | 10,346 | 8 | 20,692 |
No | output | 1 | 10,346 | 8 | 20,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are r red and g green blocks for construction of the red-green tower. Red-green tower can be built following next rules:
* Red-green tower is consisting of some number of levels;
* Let the red-green tower consist of n levels, then the first level of this tower should consist of n blocks, second level — of n - 1 blocks, the third one — of n - 2 blocks, and so on — the last level of such tower should consist of the one block. In other words, each successive level should contain one block less than the previous one;
* Each level of the red-green tower should contain blocks of the same color.
<image>
Let h be the maximum possible number of levels of red-green tower, that can be built out of r red and g green blocks meeting the rules above. The task is to determine how many different red-green towers having h levels can be built out of the available blocks.
Two red-green towers are considered different if there exists some level, that consists of red blocks in the one tower and consists of green blocks in the other tower.
You are to write a program that will find the number of different red-green towers of height h modulo 109 + 7.
Input
The only line of input contains two integers r and g, separated by a single space — the number of available red and green blocks respectively (0 ≤ r, g ≤ 2·105, r + g ≥ 1).
Output
Output the only integer — the number of different possible red-green towers of height h modulo 109 + 7.
Examples
Input
4 6
Output
2
Input
9 7
Output
6
Input
1 1
Output
2
Note
The image in the problem statement shows all possible red-green towers for the first sample.
Submitted Solution:
```
import math
r,g=map(int,input().split())
h=int(math.sqrt(1+2*(r+g)) -0.5)
ways=[0 for i in range(min(r,g)+2)]
print(h)
ways[0]=1
for i in range(1,h+1):
for j in range(min(r,g)-i,-1,-1):
ways[i+j]+=ways[j]
#print(ways)
print(sum(ways[(h*(h+1))//2-max(r,g):min(r,g)+1]))
``` | instruction | 0 | 10,347 | 8 | 20,694 |
No | output | 1 | 10,347 | 8 | 20,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are r red and g green blocks for construction of the red-green tower. Red-green tower can be built following next rules:
* Red-green tower is consisting of some number of levels;
* Let the red-green tower consist of n levels, then the first level of this tower should consist of n blocks, second level — of n - 1 blocks, the third one — of n - 2 blocks, and so on — the last level of such tower should consist of the one block. In other words, each successive level should contain one block less than the previous one;
* Each level of the red-green tower should contain blocks of the same color.
<image>
Let h be the maximum possible number of levels of red-green tower, that can be built out of r red and g green blocks meeting the rules above. The task is to determine how many different red-green towers having h levels can be built out of the available blocks.
Two red-green towers are considered different if there exists some level, that consists of red blocks in the one tower and consists of green blocks in the other tower.
You are to write a program that will find the number of different red-green towers of height h modulo 109 + 7.
Input
The only line of input contains two integers r and g, separated by a single space — the number of available red and green blocks respectively (0 ≤ r, g ≤ 2·105, r + g ≥ 1).
Output
Output the only integer — the number of different possible red-green towers of height h modulo 109 + 7.
Examples
Input
4 6
Output
2
Input
9 7
Output
6
Input
1 1
Output
2
Note
The image in the problem statement shows all possible red-green towers for the first sample.
Submitted Solution:
```
import math
def find_no(num,i,j,s):
#print(i,j,s)
if(s==0):
return 1
if(i<0 or j<0):
return 0
elif(i==0 and j==0):
num[0][0]=0
return [i][j]
elif(i==0):
# if(num[0][j]!=-1):
# return num[0][j]
# t = int(math.sqrt(2*j))
# if(j==t*(t+1)//2):
# num[0][j] = 1
# return 1
# else:
# num[0][j] = 0
# return 0
return 1
elif(j==0):
# if(num[i][0]!=-1):
# return num[i][0]
# t = int(math.sqrt(2*i))
# if(i==t*(t+1)//2):
# num[i][0]=1
# return 1
# else:
# num[i][0]=0
# return 0
return 1
else:
if(num[i][j]!=-1):
return num[i][j]
else:
num[i][j] = find_no(num,i-s,j,s-1)+find_no(num,i,j-s,s-1)
return num[i][j]
b,g = map(int,input().split())
l = math.sqrt(2*(b+g))
r = int(l)
s=0
if(r==l):
s = int(r-1)
else:
s = int(r)
if(2*(r+g)<(s*s+s)):
s-=1
#print(s)
num = [[-1 for i in range(b+1)] for j in range(g+1)]
ans = find_no(num,g,b,s)
#print(num)
print(ans%1000000007)
``` | instruction | 0 | 10,348 | 8 | 20,696 |
No | output | 1 | 10,348 | 8 | 20,697 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are r red and g green blocks for construction of the red-green tower. Red-green tower can be built following next rules:
* Red-green tower is consisting of some number of levels;
* Let the red-green tower consist of n levels, then the first level of this tower should consist of n blocks, second level — of n - 1 blocks, the third one — of n - 2 blocks, and so on — the last level of such tower should consist of the one block. In other words, each successive level should contain one block less than the previous one;
* Each level of the red-green tower should contain blocks of the same color.
<image>
Let h be the maximum possible number of levels of red-green tower, that can be built out of r red and g green blocks meeting the rules above. The task is to determine how many different red-green towers having h levels can be built out of the available blocks.
Two red-green towers are considered different if there exists some level, that consists of red blocks in the one tower and consists of green blocks in the other tower.
You are to write a program that will find the number of different red-green towers of height h modulo 109 + 7.
Input
The only line of input contains two integers r and g, separated by a single space — the number of available red and green blocks respectively (0 ≤ r, g ≤ 2·105, r + g ≥ 1).
Output
Output the only integer — the number of different possible red-green towers of height h modulo 109 + 7.
Examples
Input
4 6
Output
2
Input
9 7
Output
6
Input
1 1
Output
2
Note
The image in the problem statement shows all possible red-green towers for the first sample.
Submitted Solution:
```
ans = set()
def solve(h, r, g):
result = 0
for i in range(2**h):
if (i < r):
continue
c = [int(l) for l in str(bin(i))[2:]]
for j in range(h - len(str(bin(i))[2:])):
c.insert(0, 0)
s_r = 0
s_g = 0
buf = []
for f in enumerate(reversed(c)):
if f[1]:
s_r += f[0] + 1
else:
s_g += f[0] + 1
if s_r > r and s_g > g:
break
if s_r <= r and s_g <= g:
if not tuple(c) in ans:
ans.add(tuple(c))
result += 1
return result
[r, g] = map(int, input().split())
print(solve(int(((1 + 8 * (r + g)) ** 0.5 - 1) / 2), r, g))
``` | instruction | 0 | 10,349 | 8 | 20,698 |
No | output | 1 | 10,349 | 8 | 20,699 |
Provide a correct Python 3 solution for this coding contest problem.
Matryoshka
Matryoshka is a famous Russian folk craft doll. Matryoshka can be divided into upper and lower parts, and when opened, there is another smaller doll inside. The nesting structure is such that when the small doll that appears is opened, a smaller doll is contained.
<image>
You found an unusually shaped matryoshka doll on your trip and bought N dolls. The shape of the i-th doll is a rectangular parallelepiped of xi × yi × zi.
After watching Matryoshka for a while, you are about to put away Matryoshka. Before that, I want to reduce the space required by storing some dolls in another. When storing a doll, one other doll can be stored only in the doll that has not yet stored any doll. However, only the dolls that are stored directly are counted, and the doll that contains the doll can be stored in another doll.
The stored doll becomes invisible from the outside. However, the following conditions must be met.
* The doll may rotate, but each side of the rectangular parallelepiped is parallel to any side of the other rectangular parallelepiped.
* After rotation, the length of the doll on the side to be stored is shorter for each of the lengths of the corresponding sides.
* At most one doll can be stored directly in one doll
Since the volume of the closet is limited, we want to minimize the sum of the volumes of the dolls that can be seen from the outside. Your job is to create a program that finds the minimum sum of the volumes of the dolls that are visible from the outside, which can be achieved by repeating the operation of storing the dolls any number of times.
Input
The input consists of multiple datasets. The maximum number of data sets does not exceed 50. Each dataset is represented in the following format.
> N
> x1 y1 z1
>::
>::
> xN yN zN
>
Each dataset consists of N + 1 rows, and the first row of the dataset is given the integer N, which represents the number of dolls. In the i-th line of the following N lines, three integers xi, yi, and zi representing the size of the i-th doll are given, separated by a half-width space. These integers satisfy 1 ≤ N, xi, yi, zi ≤ 100.
> The end of the input is represented by a single zero line.
> ### Output
For each data set, output the minimum value of the sum of the volumes of the dolls that can be seen from the outside in one line.
> ### Sample Input
2
one two Three
4 2 3
3
2 5 2
3 3 4
5 5 5
Five
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
Five
1 1 1
2 1 1
3 1 1
4 1 1
5 1 1
Ten
3 1 4
1 5 9
2 6 5
3 5 8
9 7 9
3 2 3
8 4 6
2 6 4
3 3 8
3 2 7
0
Output for Sample Input
twenty four
145
125
15
864
Example
Input
2
1 2 3
4 2 3
3
2 5 2
3 3 4
5 5 5
5
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
5
1 1 1
2 1 1
3 1 1
4 1 1
5 1 1
10
3 1 4
1 5 9
2 6 5
3 5 8
9 7 9
3 2 3
8 4 6
2 6 4
3 3 8
3 2 7
0
Output
24
145
125
15
864 | instruction | 0 | 11,597 | 8 | 23,194 |
"Correct Solution:
```
import heapq
class MinCostFlow:
class Edge:
def __init__(self,to,cap,rev,cost):
self.to = to
self.cap = cap
self.rev = rev
self.cost = cost
def __init__(self,n,inf=1000000007):
self.n = n
self.inf = inf
self.e = [[] for _ in range(n)]
def add_edge(self, fr, to, cap, cost):
self.e[fr].append(self.Edge(to,cap,len(self.e[to]),cost))
self.e[to].append(self.Edge(fr,0,len(self.e[fr])-1,-cost))
def compute(self,source,sink,f):
res = 0
h = [0]*self.n
prevv = [0]*self.n
preve = [0]*self.n
while (f > 0):
pq = []
dist = [self.inf]*self.n
dist[source] = 0
heapq.heappush(pq,(0,source))
while pq:
cost, v = heapq.heappop(pq)
cost = -cost
if dist[v] < cost:continue
for i, edge in enumerate(self.e[v]):
if edge.cap > 0 and dist[v] - h[edge.to] < dist[edge.to] - edge.cost - h[v]:
dist[edge.to] = dist[v] + edge.cost + h[v] - h[edge.to]
prevv[edge.to] = v
preve[edge.to] = i
heapq.heappush(pq,(-dist[edge.to],edge.to))
if dist[sink] == self.inf:return -1
for v in range(self.n):
h[v] += dist[v]
d, v = f, sink
while v != source:
d = min(d,self.e[prevv[v]][preve[v]].cap)
v = prevv[v]
f -= d
res += d*h[sink]
v = sink
while v != source:
self.e[prevv[v]][preve[v]].cap -= d
self.e[v][self.e[prevv[v]][preve[v]].rev].cap += d
v = prevv[v]
return res
def less(a,b):
for i in range(3):
if b[i]<=a[i]:return False
return True
def main():
while True:
n = int(input())
if n==0:break
a = []
MCF = MinCostFlow(n+n+2)
s = n+n
t = s+1
summ = 0
for i in range(n):
x = list(map(int,input().split()))
x.sort()
a.append(x)
summ+=(x[0]*x[1]*x[2])
for i in range(n):
for j in range(n):
if i == j:continue
if less(a[i],a[j]):
MCF.add_edge(i,j+n,1,-(a[i][0]*a[i][1]*a[i][2]))
for i in range(n):
MCF.add_edge(s,i,1,0)
MCF.add_edge(i+n,t,1,0)
MCF.add_edge(s,t,MCF.inf,0)
print (summ+MCF.compute(s,t,MCF.inf))
if __name__ == '__main__':
main()
``` | output | 1 | 11,597 | 8 | 23,195 |
Provide a correct Python 3 solution for this coding contest problem.
Matryoshka
Matryoshka is a famous Russian folk craft doll. Matryoshka can be divided into upper and lower parts, and when opened, there is another smaller doll inside. The nesting structure is such that when the small doll that appears is opened, a smaller doll is contained.
<image>
You found an unusually shaped matryoshka doll on your trip and bought N dolls. The shape of the i-th doll is a rectangular parallelepiped of xi × yi × zi.
After watching Matryoshka for a while, you are about to put away Matryoshka. Before that, I want to reduce the space required by storing some dolls in another. When storing a doll, one other doll can be stored only in the doll that has not yet stored any doll. However, only the dolls that are stored directly are counted, and the doll that contains the doll can be stored in another doll.
The stored doll becomes invisible from the outside. However, the following conditions must be met.
* The doll may rotate, but each side of the rectangular parallelepiped is parallel to any side of the other rectangular parallelepiped.
* After rotation, the length of the doll on the side to be stored is shorter for each of the lengths of the corresponding sides.
* At most one doll can be stored directly in one doll
Since the volume of the closet is limited, we want to minimize the sum of the volumes of the dolls that can be seen from the outside. Your job is to create a program that finds the minimum sum of the volumes of the dolls that are visible from the outside, which can be achieved by repeating the operation of storing the dolls any number of times.
Input
The input consists of multiple datasets. The maximum number of data sets does not exceed 50. Each dataset is represented in the following format.
> N
> x1 y1 z1
>::
>::
> xN yN zN
>
Each dataset consists of N + 1 rows, and the first row of the dataset is given the integer N, which represents the number of dolls. In the i-th line of the following N lines, three integers xi, yi, and zi representing the size of the i-th doll are given, separated by a half-width space. These integers satisfy 1 ≤ N, xi, yi, zi ≤ 100.
> The end of the input is represented by a single zero line.
> ### Output
For each data set, output the minimum value of the sum of the volumes of the dolls that can be seen from the outside in one line.
> ### Sample Input
2
one two Three
4 2 3
3
2 5 2
3 3 4
5 5 5
Five
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
Five
1 1 1
2 1 1
3 1 1
4 1 1
5 1 1
Ten
3 1 4
1 5 9
2 6 5
3 5 8
9 7 9
3 2 3
8 4 6
2 6 4
3 3 8
3 2 7
0
Output for Sample Input
twenty four
145
125
15
864
Example
Input
2
1 2 3
4 2 3
3
2 5 2
3 3 4
5 5 5
5
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
5
1 1 1
2 1 1
3 1 1
4 1 1
5 1 1
10
3 1 4
1 5 9
2 6 5
3 5 8
9 7 9
3 2 3
8 4 6
2 6 4
3 3 8
3 2 7
0
Output
24
145
125
15
864 | instruction | 0 | 11,598 | 8 | 23,196 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
from heapq import heappush, heappop
class MinCostFlow:
INF = 10**18
def __init__(self, N):
self.N = N
self.G = [[] for i in range(N)]
def add_edge(self, fr, to, cap, cost):
G = self.G
G[fr].append([to, cap, cost, len(G[to])])
G[to].append([fr, 0, -cost, len(G[fr])-1])
def flow(self, s, t, f):
N = self.N; G = self.G
INF = MinCostFlow.INF
res = 0
H = [0]*N
prv_v = [0]*N
prv_e = [0]*N
while f:
dist = [INF]*N
dist[s] = 0
que = [(0, s)]
while que:
c, v = heappop(que)
if dist[v] < c:
continue
for i, (w, cap, cost, _) in enumerate(G[v]):
if cap > 0 and dist[w] > dist[v] + cost + H[v] - H[w]:
dist[w] = r = dist[v] + cost + H[v] - H[w]
prv_v[w] = v; prv_e[w] = i
heappush(que, (r, w))
if dist[t] == INF:
return -1
for i in range(N):
H[i] += dist[i]
d = f; v = t
while v != s:
d = min(d, G[prv_v[v]][prv_e[v]][1])
v = prv_v[v]
f -= d
res += d * H[t]
v = t
while v != s:
e = G[prv_v[v]][prv_e[v]]
e[1] -= d
G[v][e[3]][1] += d
v = prv_v[v]
return res
def solve():
N = int(readline())
if N == 0:
return False
P = []
for i in range(N):
*p, = map(int, readline().split())
p.sort()
P.append(p)
mcf = MinCostFlow(2*N+2)
P.sort()
su = 0
for i in range(N):
xi, yi, zi = P[i]
for j in range(i):
xj, yj, zj = P[j]
if xi > xj and yi > yj and zi > zj:
mcf.add_edge(2*j+1, 2*i, 1, -xj*yj*zj)
su += xi*yi*zi
mcf.add_edge(2*i, 2*i+1, 1, 0)
mcf.add_edge(2*N, 2*i, 1, 0)
mcf.add_edge(2*i+1, 2*N+1, 1, 0)
ans = su
for i in range(N):
f = mcf.flow(2*N, 2*N+1, 1)
su += f
ans = min(ans, su)
write("%d\n" % ans)
return True
while solve():
...
``` | output | 1 | 11,598 | 8 | 23,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After too much playing on paper, Iahub has switched to computer games. The game he plays is called "Block Towers". It is played in a rectangular grid with n rows and m columns (it contains n × m cells). The goal of the game is to build your own city. Some cells in the grid are big holes, where Iahub can't build any building. The rest of cells are empty. In some empty cell Iahub can build exactly one tower of two following types:
1. Blue towers. Each has population limit equal to 100.
2. Red towers. Each has population limit equal to 200. However, it can be built in some cell only if in that moment at least one of the neighbouring cells has a Blue Tower. Two cells are neighbours is they share a side.
Iahub is also allowed to destroy a building from any cell. He can do this operation as much as he wants. After destroying a building, the other buildings are not influenced, and the destroyed cell becomes empty (so Iahub can build a tower in this cell if needed, see the second example for such a case).
Iahub can convince as many population as he wants to come into his city. So he needs to configure his city to allow maximum population possible. Therefore he should find a sequence of operations that builds the city in an optimal way, so that total population limit is as large as possible.
He says he's the best at this game, but he doesn't have the optimal solution. Write a program that calculates the optimal one, to show him that he's not as good as he thinks.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 500). Each of the next n lines contains m characters, describing the grid. The j-th character in the i-th line is '.' if you're allowed to build at the cell with coordinates (i, j) a tower (empty cell) or '#' if there is a big hole there.
Output
Print an integer k in the first line (0 ≤ k ≤ 106) — the number of operations Iahub should perform to obtain optimal result.
Each of the following k lines must contain a single operation in the following format:
1. «B x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — building a blue tower at the cell (x, y);
2. «R x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — building a red tower at the cell (x, y);
3. «D x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — destroying a tower at the cell (x, y).
If there are multiple solutions you can output any of them. Note, that you shouldn't minimize the number of operations.
Examples
Input
2 3
..#
.#.
Output
4
B 1 1
R 1 2
R 2 1
B 2 3
Input
1 3
...
Output
5
B 1 1
B 1 2
R 1 3
D 1 2
R 1 2 | instruction | 0 | 11,976 | 8 | 23,952 |
Tags: constructive algorithms, dfs and similar, graphs
Correct Solution:
```
#Connected component
import sys
from collections import deque
sys.setrecursionlimit(501 * 501)
n,m = [int(i) for i in input().split()]
a=[[0 for i in range(m)] for i in range(n)]
d=[(0,1),(0,-1),(1,0),(-1,0)]
q=deque()
def main():
global a
ans=[]
first=[]
q=deque()
for i in range(n):
line = input()
for j in range(m):
if line[j]=='#':
a[i][j]=-1
else:
first.append('B %s %s' %(i+1, j+1))
for i in range(n):
for j in range(m):
if a[i][j]!=-1:
q.append((i,j))
while q:
x,y = q.pop()
if a[x][y]==-1:
continue
a[x][y] = -1
if (x,y)!=(i,j):
ans.append('R %s %s' %(x+1,y+1))
ans.append('D %s %s' %(x+1,y+1))
for dx,dy in d:
x1 = x+dx
y1 = y+dy
if (0<=x1<n) and (0<=y1<m) and a[x1][y1]!=-1:
q.appendleft((x1,y1))
ans.reverse()
print(len(ans)+len(first), end='\n')
print('\n'.join(first))
print('\n'.join(ans))
main()
``` | output | 1 | 11,976 | 8 | 23,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After too much playing on paper, Iahub has switched to computer games. The game he plays is called "Block Towers". It is played in a rectangular grid with n rows and m columns (it contains n × m cells). The goal of the game is to build your own city. Some cells in the grid are big holes, where Iahub can't build any building. The rest of cells are empty. In some empty cell Iahub can build exactly one tower of two following types:
1. Blue towers. Each has population limit equal to 100.
2. Red towers. Each has population limit equal to 200. However, it can be built in some cell only if in that moment at least one of the neighbouring cells has a Blue Tower. Two cells are neighbours is they share a side.
Iahub is also allowed to destroy a building from any cell. He can do this operation as much as he wants. After destroying a building, the other buildings are not influenced, and the destroyed cell becomes empty (so Iahub can build a tower in this cell if needed, see the second example for such a case).
Iahub can convince as many population as he wants to come into his city. So he needs to configure his city to allow maximum population possible. Therefore he should find a sequence of operations that builds the city in an optimal way, so that total population limit is as large as possible.
He says he's the best at this game, but he doesn't have the optimal solution. Write a program that calculates the optimal one, to show him that he's not as good as he thinks.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 500). Each of the next n lines contains m characters, describing the grid. The j-th character in the i-th line is '.' if you're allowed to build at the cell with coordinates (i, j) a tower (empty cell) or '#' if there is a big hole there.
Output
Print an integer k in the first line (0 ≤ k ≤ 106) — the number of operations Iahub should perform to obtain optimal result.
Each of the following k lines must contain a single operation in the following format:
1. «B x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — building a blue tower at the cell (x, y);
2. «R x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — building a red tower at the cell (x, y);
3. «D x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — destroying a tower at the cell (x, y).
If there are multiple solutions you can output any of them. Note, that you shouldn't minimize the number of operations.
Examples
Input
2 3
..#
.#.
Output
4
B 1 1
R 1 2
R 2 1
B 2 3
Input
1 3
...
Output
5
B 1 1
B 1 2
R 1 3
D 1 2
R 1 2
Submitted Solution:
```
# https://codeforces.com/problemset/problem/327/D
# WA
import heapq
from heapq import heappush as push_
from heapq import heappop as pop_
n, m = map(int, input().split())
a = list([input() for _ in range(n)])
dx = [1,-1, 0, 0]
dy = [0 ,0, 1,-1]
arr = [[0]*m for _ in range(n)]
ans = []
for i, x in enumerate(a):
for j, y in enumerate(x):
if y!='.':
arr[i][j]=-1
else:
ans.append('B {} {}'.format(i+1,j+1))
Q = []
base = 1000
d = {}
def is_ok(x, y, x_, y_):
if x+x_>=0 and x+x_<n and y+y_>=0 and y+y_<m:
return True
return False
def get_id_(x, y):
return x*base+y
for x in range(n):
for y in range(m):
if arr[x][y] == -1:
continue
dev = 0
for x_, y_ in zip(dx, dy):
if is_ok(x, y, x_, y_) and arr[x+x_][y+y_]==0:
dev+=1
id = get_id_(x, y)
d[id]=dev
push_(Q, (dev, id))
def r_id(id_):
return id_//base, id_%base
while len(Q) > 0:
dev_, id_ = pop_(Q)
x, y = r_id(id_)
if id_ not in d or dev_ != d[id_]:
continue
if d[id_]>0:
del d[id_]
arr[x][y]=1
ans.append('D {} {}'.format(x+1,y+1))
ans.append('R {} {}'.format(x+1,y+1))
for x_, y_ in zip(dx, dy):
if is_ok(x, y, x_, y_) and arr[x+x_][y+y_]==0:
id_n=get_id_(x+x_,y+y_)
d[id_n] -= 1
push_(Q, (d[id_n], id_n))
print(len(ans))
print('\n'.join(ans))
``` | instruction | 0 | 11,977 | 8 | 23,954 |
No | output | 1 | 11,977 | 8 | 23,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After too much playing on paper, Iahub has switched to computer games. The game he plays is called "Block Towers". It is played in a rectangular grid with n rows and m columns (it contains n × m cells). The goal of the game is to build your own city. Some cells in the grid are big holes, where Iahub can't build any building. The rest of cells are empty. In some empty cell Iahub can build exactly one tower of two following types:
1. Blue towers. Each has population limit equal to 100.
2. Red towers. Each has population limit equal to 200. However, it can be built in some cell only if in that moment at least one of the neighbouring cells has a Blue Tower. Two cells are neighbours is they share a side.
Iahub is also allowed to destroy a building from any cell. He can do this operation as much as he wants. After destroying a building, the other buildings are not influenced, and the destroyed cell becomes empty (so Iahub can build a tower in this cell if needed, see the second example for such a case).
Iahub can convince as many population as he wants to come into his city. So he needs to configure his city to allow maximum population possible. Therefore he should find a sequence of operations that builds the city in an optimal way, so that total population limit is as large as possible.
He says he's the best at this game, but he doesn't have the optimal solution. Write a program that calculates the optimal one, to show him that he's not as good as he thinks.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 500). Each of the next n lines contains m characters, describing the grid. The j-th character in the i-th line is '.' if you're allowed to build at the cell with coordinates (i, j) a tower (empty cell) or '#' if there is a big hole there.
Output
Print an integer k in the first line (0 ≤ k ≤ 106) — the number of operations Iahub should perform to obtain optimal result.
Each of the following k lines must contain a single operation in the following format:
1. «B x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — building a blue tower at the cell (x, y);
2. «R x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — building a red tower at the cell (x, y);
3. «D x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — destroying a tower at the cell (x, y).
If there are multiple solutions you can output any of them. Note, that you shouldn't minimize the number of operations.
Examples
Input
2 3
..#
.#.
Output
4
B 1 1
R 1 2
R 2 1
B 2 3
Input
1 3
...
Output
5
B 1 1
B 1 2
R 1 3
D 1 2
R 1 2
Submitted Solution:
```
n, m = map(int, input().split())
s = [input() for i in range(n)]
d = [[-1 if s[i][j] == '.' else -2 for j in range(m)] for i in range(n)]
db = [[-1, 0], [1, 0], [0, -1], [0, 1]]
buff = []
def solve(c, i, j):
if i < 0 or n <= i or j < 0 or m <= j:
return
if d[i][j] != -1:
return
d[i][j] = c
if c != 0:
buff.append((c, i, j))
for u in db:
solve(c+1, i+u[0], j+u[1])
k = 0
for i in range(n):
k += d[i].count(-1)
k *= 3
rb = []
for i in range(n):
for j in range(m):
if d[i][j] == -1:
k -= 2
solve(0, i, j)
buff.sort()
buff.reverse()
print(k)
for i in range(n):
for j in range(m):
if d[i][j] != -2:
print('B {0} {1}'.format(i+1, j+1))
if n == 500:
exit()
for i in buff:
print('D {0} {1}\nR {0} {1}'.format(i[1]+1, i[2]+1))
``` | instruction | 0 | 11,978 | 8 | 23,956 |
No | output | 1 | 11,978 | 8 | 23,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After too much playing on paper, Iahub has switched to computer games. The game he plays is called "Block Towers". It is played in a rectangular grid with n rows and m columns (it contains n × m cells). The goal of the game is to build your own city. Some cells in the grid are big holes, where Iahub can't build any building. The rest of cells are empty. In some empty cell Iahub can build exactly one tower of two following types:
1. Blue towers. Each has population limit equal to 100.
2. Red towers. Each has population limit equal to 200. However, it can be built in some cell only if in that moment at least one of the neighbouring cells has a Blue Tower. Two cells are neighbours is they share a side.
Iahub is also allowed to destroy a building from any cell. He can do this operation as much as he wants. After destroying a building, the other buildings are not influenced, and the destroyed cell becomes empty (so Iahub can build a tower in this cell if needed, see the second example for such a case).
Iahub can convince as many population as he wants to come into his city. So he needs to configure his city to allow maximum population possible. Therefore he should find a sequence of operations that builds the city in an optimal way, so that total population limit is as large as possible.
He says he's the best at this game, but he doesn't have the optimal solution. Write a program that calculates the optimal one, to show him that he's not as good as he thinks.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 500). Each of the next n lines contains m characters, describing the grid. The j-th character in the i-th line is '.' if you're allowed to build at the cell with coordinates (i, j) a tower (empty cell) or '#' if there is a big hole there.
Output
Print an integer k in the first line (0 ≤ k ≤ 106) — the number of operations Iahub should perform to obtain optimal result.
Each of the following k lines must contain a single operation in the following format:
1. «B x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — building a blue tower at the cell (x, y);
2. «R x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — building a red tower at the cell (x, y);
3. «D x y» (1 ≤ x ≤ n, 1 ≤ y ≤ m) — destroying a tower at the cell (x, y).
If there are multiple solutions you can output any of them. Note, that you shouldn't minimize the number of operations.
Examples
Input
2 3
..#
.#.
Output
4
B 1 1
R 1 2
R 2 1
B 2 3
Input
1 3
...
Output
5
B 1 1
B 1 2
R 1 3
D 1 2
R 1 2
Submitted Solution:
```
n, m = map(int, input().split())
s = [input() for i in range(n)]
d = [[-1 if s[i][j] == '.' else -2 for j in range(m)] for i in range(n)]
db = [[-1, 0], [1, 0], [0, -1], [0, 1]]
buff = []
def solve(c, i, j):
if i < 0 or n <= i or j < 0 or m <= j:
return
if d[i][j] != -1:
return
d[i][j] = c
if c != 0:
buff.append((c, i, j))
for u in db:
solve(c+1, i+u[0], j+u[1])
k = 0
for i in range(n):
k += d[i].count(-1)
k *= 3
rb = []
for i in range(n):
for j in range(m):
if d[i][j] == -1:
k -= 2
solve(0, i, j)
buff.sort()
buff.reverse()
print(k)
for i in range(n):
for j in range(m):
if d[i][j] != -2:
print('B', i+1, j+1)
for i in buff:
print('D', i[0]+1, i[1]+1)
print('R', i[0]+1, i[1]+1)
``` | instruction | 0 | 11,979 | 8 | 23,958 |
No | output | 1 | 11,979 | 8 | 23,959 |
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