message stringlengths 2 45.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 254 108k | cluster float64 3 3 | __index_level_0__ int64 508 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Submitted Solution:
```
# cook your dish here
t=int(input())
for i in range(t):
a,b,k=map(int,input().split(' '))
if k%2==0:
x=int(k/2)
print(a*x-b*x)
else:
k=k-1
x=int(k/2)
print(a*(x+1)-b*x)
``` | instruction | 0 | 76,731 | 3 | 153,462 |
Yes | output | 1 | 76,731 | 3 | 153,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Submitted Solution:
```
t = int(input())
res = list()
for i in range(t):
a, b, k = list(map(int, input().strip().split()))
if k % 2 == 1:
pos = (k - k // 2) * a - (k // 2) * b
else:
pos = k * (a - b) / 2
res.append(int(pos))
for i in res:
print(i)
``` | instruction | 0 | 76,732 | 3 | 153,464 |
No | output | 1 | 76,732 | 3 | 153,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Submitted Solution:
```
n = int(input())
for i in range(n):
a, b, k = map(int, input().split())
print((a - b) * k)
``` | instruction | 0 | 76,733 | 3 | 153,466 |
No | output | 1 | 76,733 | 3 | 153,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Submitted Solution:
```
t = int(input())
out = [0]*t
for it in range(t):
a, b, k = map(int, input().split())
if k%2==0:
out[it] = out[it] + a*(k/2) - b*(k/2)
else:
k = k-1
out[it] = out[it] + a*(k/2) - b*(k/2) + a
for o in range(t):
print(int(out[o]))
``` | instruction | 0 | 76,734 | 3 | 153,468 |
No | output | 1 | 76,734 | 3 | 153,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A frog is currently at the point 0 on a coordinate axis Ox. It jumps by the following algorithm: the first jump is a units to the right, the second jump is b units to the left, the third jump is a units to the right, the fourth jump is b units to the left, and so on.
Formally:
* if the frog has jumped an even number of times (before the current jump), it jumps from its current position x to position x+a;
* otherwise it jumps from its current position x to position x-b.
Your task is to calculate the position of the frog after k jumps.
But... One more thing. You are watching t different frogs so you have to answer t independent queries.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 1000) — the number of queries.
Each of the next t lines contain queries (one query per line).
The query is described as three space-separated integers a, b, k (1 ≤ a, b, k ≤ 10^9) — the lengths of two types of jumps and the number of jumps, respectively.
Output
Print t integers. The i-th integer should be the answer for the i-th query.
Example
Input
6
5 2 3
100 1 4
1 10 5
1000000000 1 6
1 1 1000000000
1 1 999999999
Output
8
198
-17
2999999997
0
1
Note
In the first query frog jumps 5 to the right, 2 to the left and 5 to the right so the answer is 5 - 2 + 5 = 8.
In the second query frog jumps 100 to the right, 1 to the left, 100 to the right and 1 to the left so the answer is 100 - 1 + 100 - 1 = 198.
In the third query the answer is 1 - 10 + 1 - 10 + 1 = -17.
In the fourth query the answer is 10^9 - 1 + 10^9 - 1 + 10^9 - 1 = 2999999997.
In the fifth query all frog's jumps are neutralized by each other so the answer is 0.
The sixth query is the same as the fifth but without the last jump so the answer is 1.
Submitted Solution:
```
n=int(input())
j=0
for i in range(n):
a,b,k=map(int,input().split())
m=k%2
if m == 0:
m = k/2
j=m
else:
j=k//2+1
m=k-j
l=(a*j+b*-m)
print(l)
``` | instruction | 0 | 76,735 | 3 | 153,470 |
No | output | 1 | 76,735 | 3 | 153,471 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east. | instruction | 0 | 76,916 | 3 | 153,832 |
Tags: dp, math
Correct Solution:
```
n = int(input())
if n%2 == 0:
if n == 2:
print(4)
else:
print((n//2+1)**2)
else:
if n == 1:
print(4)
else:
print(((n//2+1)*(n//2+2))*2)
``` | output | 1 | 76,916 | 3 | 153,833 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east. | instruction | 0 | 76,917 | 3 | 153,834 |
Tags: dp, math
Correct Solution:
```
from sys import stdin, stdout
import math
from collections import defaultdict
ans = []
if __name__ == '__main__':
n = int(stdin.readline())
if not n%2:
k = n/2
print(int((k+1)**2))
else:
k = math.floor(n/2)
print(2*(k+1)*(k+2))
``` | output | 1 | 76,917 | 3 | 153,835 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east. | instruction | 0 | 76,918 | 3 | 153,836 |
Tags: dp, math
Correct Solution:
```
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
n= int(input())
ans = [4,4,12,9]
if n<=4:
print(ans[n-1])
else:
for i in range(4,n):
toPlus = ((i)//2+1)*4
if i%2==0:
ans.append(ans[i-2]+toPlus)
else:
ans.append(ans[i-4]+toPlus)
print(ans[-1])
#"{} {} {}".format(maxele,minele,minele)
# 4 4 12 9 24 16 40 25 60
# +8 +12 +12 +16 +16 +20
``` | output | 1 | 76,918 | 3 | 153,837 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east. | instruction | 0 | 76,919 | 3 | 153,838 |
Tags: dp, math
Correct Solution:
```
from sys import stdin, stdout
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque
from bisect import bisect_left as bl, bisect_right as br, bisect
mod = pow(10, 9) + 7
mod2 = 998244353
def inp(): return stdin.readline().strip()
def out(var, end="\n"): stdout.write(str(var)+"\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
def remadd(x, y): return 1 if x%y else 0
def ceil(a,b): return (a+b-1)//b
def chkprime(x):
if x<=1: return False
if x in (2, 3): return True
if x%2 == 0: return False
for i in range(3, int(sqrt(x))+1, 2):
if x%i == 0: return False
return True
n = int(inp())
if n%2:
out((n//2+1)*(n//2+2)*2)
else:
if n%4==0:
out(((n//2)+1)**2)
else: out(4*((n//4)+1)**2)
``` | output | 1 | 76,919 | 3 | 153,839 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east. | instruction | 0 | 76,920 | 3 | 153,840 |
Tags: dp, math
Correct Solution:
```
n = int(input())
half = n//2
if n % 2 == 0:
print((half+1)**2)
else:
print(4*(half+1)+half*(half+1)*2)
``` | output | 1 | 76,920 | 3 | 153,841 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east. | instruction | 0 | 76,921 | 3 | 153,842 |
Tags: dp, math
Correct Solution:
```
n=int(input())
if(n==0):print("1")
elif(n==1):print("4")
elif(n%2==0
):
a=int(n//2)
print((a+1)**2)
else:
a=int(n//2)
print(2*(a+1)*(a+2))
``` | output | 1 | 76,921 | 3 | 153,843 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east. | instruction | 0 | 76,922 | 3 | 153,844 |
Tags: dp, math
Correct Solution:
```
n=int(input())
if n%2==0:
g=n//2
print((g+1)*(g+1))
else:
g=n//2+1
out=2*(g+1)*g
print(out)
``` | output | 1 | 76,922 | 3 | 153,845 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east. | instruction | 0 | 76,923 | 3 | 153,846 |
Tags: dp, math
Correct Solution:
```
n=int(input())
if n==1 or n==2:
print(4)
elif n==0:
print(1)
else:
l=[0 for _ in range(n+1)]
l[0]=1
l[1]=4
l[2]=4
j=2
for i in range(3,n+1):
if i%2==1:
l[i]=l[i-2]+(4*j)
else:
l[i]=l[i-4]+(4*j)
j+=1
print(l[n])
``` | output | 1 | 76,923 | 3 | 153,847 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
n= int(input())
if n % 2:
print(((n+1)+2)*(n+1)//2)
else:
print((n+2)**2//4)
``` | instruction | 0 | 76,924 | 3 | 153,848 |
Yes | output | 1 | 76,924 | 3 | 153,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
from sys import stdin
for _ in range(1):
n=int(input())
if(n%2==0): print((int(n/2)+1)**2)
else: print(2*(int(n/2)+2)*(int(n/2)+1))
``` | instruction | 0 | 76,925 | 3 | 153,850 |
Yes | output | 1 | 76,925 | 3 | 153,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
import sys
input=sys.stdin.readline
# for _ in range(int(input())):
n=int(input())
# n,m=map(int,input().split())
# r=(list(input().strip()))
# a=list(map(int,input().split()))
if n&1:
y=n//2
x=y+1
else:
y=n//2
x=n//2
k=(x+1)*(y+1)
if n&1:
k=k*2
print(k)
``` | instruction | 0 | 76,926 | 3 | 153,852 |
Yes | output | 1 | 76,926 | 3 | 153,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
n = int(input())
s = n//2+1
if n == 1: print(4)
elif n%2: print(2*s*(s+1))
else: print(s**2)
``` | instruction | 0 | 76,927 | 3 | 153,854 |
Yes | output | 1 | 76,927 | 3 | 153,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
import math
n = input()
n = int(n)
prev = 0
ans = 0
for i in range(1, n+1):
ans = 4 * (2**(i-1)) - prev
prev = ans
print(ans)
``` | instruction | 0 | 76,928 | 3 | 153,856 |
No | output | 1 | 76,928 | 3 | 153,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
n = int(input())
if n%2 == 0:
print(n*2)
else:
print(n*4)
``` | instruction | 0 | 76,929 | 3 | 153,858 |
No | output | 1 | 76,929 | 3 | 153,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
def solve(N):
if N <= 2:
print(4)
else:
print((N - 1) ** 2 + (N - 1) * 4)
N = int(input())
solve(N)
``` | instruction | 0 | 76,930 | 3 | 153,860 |
No | output | 1 | 76,930 | 3 | 153,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A robot is standing at the origin of the infinite two-dimensional plane. Each second the robot moves exactly 1 meter in one of the four cardinal directions: north, south, west, and east. For the first step the robot can choose any of the four directions, but then at the end of every second it has to turn 90 degrees left or right with respect to the direction it just moved in. For example, if the robot has just moved north or south, the next step it takes has to be either west or east, and vice versa.
The robot makes exactly n steps from its starting position according to the rules above. How many different points can the robot arrive to at the end? The final orientation of the robot can be ignored.
Input
The only line contains a single integer n (1 ≤ n ≤ 1000) — the number of steps the robot makes.
Output
Print a single integer — the number of different possible locations after exactly n steps.
Examples
Input
1
Output
4
Input
2
Output
4
Input
3
Output
12
Note
In the first sample case, the robot will end up 1 meter north, south, west, or east depending on its initial direction.
In the second sample case, the robot will always end up √{2} meters north-west, north-east, south-west, or south-east.
Submitted Solution:
```
def read_ints():
return map(int, input().split())
n, = read_ints()
if n == 0:
result = 1
elif n == 1 or n == 2:
result = 4
else:
side = (n+1)//2 * 2 + 1
result = side * (side - 1) // 2 + side // 2 + (n % 2 == 0)
print(result)
``` | instruction | 0 | 76,931 | 3 | 153,862 |
No | output | 1 | 76,931 | 3 | 153,863 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3 | instruction | 0 | 76,986 | 3 | 153,972 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
normal_segments = []
degen_segments = []
nails = {}
def is_nailed(a , b, nails):
for i in range(a, b + 1):
if i in nails:
if nails[i] == True:
return True
return False
for i in range(0, n):
data = input()
data = data.split(' ')
data[0] = int(data[0])
data[1] = int(data[1])
if data[0] - data[1] == 0:
degen_segments.append(data[0])
else:
if data[0] > data[1]:
normal_segments.append((data[1], data[0]))
else:
normal_segments.append((data[0], data[1]))
normal_segments.sort(key=lambda tup: tup[1])
degen_segments.sort()
# need to place a nail in degen segments always.
for i in degen_segments:
nails[i] = True
for i in normal_segments:
temp = is_nailed(i[0], i[1], nails)
if temp == False:
nails[i[1]] = True
print(len(nails.keys()))
for i in nails.keys():
print(i, end=' ')
# Made By Mostafa_Khaled
``` | output | 1 | 76,986 | 3 | 153,973 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3 | instruction | 0 | 76,987 | 3 | 153,974 |
Tags: greedy, sortings
Correct Solution:
```
import sys
class Seg:
def __init__(self, left, right):
self.left = left
self.right = right
def solve():
a = list()
n = int(input())
for i in range(n):
left, right = sorted(list(map(int, input().split())))
a.append(Seg(left, right))
a.sort(key = lambda x : x.right)
nails = list()
lastnail = -100000
for seg in a:
if lastnail >= seg.left: continue
nails.append(seg.right)
lastnail = seg.right
print(len(nails))
print(' '.join(map(str, nails)))
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
``` | output | 1 | 76,987 | 3 | 153,975 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3 | instruction | 0 | 76,988 | 3 | 153,976 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
l = []
for i in range(n):
a, b = map(int, input().split())
a, b = min(a, b), max(a, b)
l.append((a, 0, i))
l.append((b, 1, i))
l.sort()
ans = []
cur = set()
a = [0] * n
for x in l:
if x[1] == 0:
cur.add(x[2])
elif a[x[2]] == 0:
ans.append(x[0])
for e in cur:
a[e] = 1
cur.clear()
print(len(ans))
print(*ans)
``` | output | 1 | 76,988 | 3 | 153,977 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3 | instruction | 0 | 76,989 | 3 | 153,978 |
Tags: greedy, sortings
Correct Solution:
```
n = int(input())
normal_segments = []
degen_segments = []
nails = {}
def is_nailed(a , b, nails):
for i in range(a, b + 1):
if i in nails:
if nails[i] == True:
return True
return False
for i in range(0, n):
data = input()
data = data.split(' ')
data[0] = int(data[0])
data[1] = int(data[1])
if data[0] - data[1] == 0:
degen_segments.append(data[0])
else:
if data[0] > data[1]:
normal_segments.append((data[1], data[0]))
else:
normal_segments.append((data[0], data[1]))
normal_segments.sort(key=lambda tup: tup[1])
degen_segments.sort()
# need to place a nail in degen segments always.
for i in degen_segments:
nails[i] = True
for i in normal_segments:
temp = is_nailed(i[0], i[1], nails)
if temp == False:
nails[i[1]] = True
print(len(nails.keys()))
for i in nails.keys():
print(i, end=' ')
``` | output | 1 | 76,989 | 3 | 153,979 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3 | instruction | 0 | 76,991 | 3 | 153,982 |
Tags: greedy, sortings
Correct Solution:
```
import sys
from array import array # noqa: F401
from operator import itemgetter
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n = int(input())
a = [(x, y) if x < y else (y, x) for _ in range(n) for x, y in (map(int, input().split()),)]
a.sort(key=itemgetter(1))
nailed = -10**9
ans = []
for l, r in a:
if l <= nailed <= r:
continue
nailed = r
ans.append(r)
print(len(ans))
print(*ans)
``` | output | 1 | 76,991 | 3 | 153,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
#!/usr/bin/python3
import sys
input = lambda: sys.stdin.readline().strip()
n = int(input())
s = sorted((sorted(int(x) for x in input().split()) for i in range(n)), key=lambda p: p[1])
ans = []
for i, j in s:
if not any(i <= x <= j for x in ans):
ans.append(j)
print(len(ans))
print(*ans)
``` | instruction | 0 | 76,992 | 3 | 153,984 |
Yes | output | 1 | 76,992 | 3 | 153,985 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
if __name__ == '__main__':
number = int(input())
segments = []
i = 0
# while i < number:
for i in range(0, number):
pos = input().split()
if int(pos[0]) < int(pos[1]):
a = int(pos[0])
b = int(pos[1])
else:
b = int(pos[0])
a = int(pos[1])
segments.append((a, b))
segments.sort()
final = []
ok = -9999999
nr = len(segments)
for i in range(nr):
a = segments[i][0]
b = segments[i][1]
if a <= ok and b >= ok:
continue
verif = 0
for j in segments[i + 1:]:
a = segments[i][1]
b = segments[i][1]
if a >= j[0] and b >= j[1]:
verif = 1
if verif:
continue
else:
ok = segments[i][1]
final.append(ok)
print(len(final))
space = ''
for i in final:
space = space + str(i) + ' '
print(space)
``` | instruction | 0 | 76,993 | 3 | 153,986 |
Yes | output | 1 | 76,993 | 3 | 153,987 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
n = int(input())
lis=list()
cou=1
arr=list()
for i in range(n):
a , b = map(int,input().split())
if a>b:
lis.append([b,a])
else:
lis.append([a,b])
lis=sorted(lis,key=lambda l:l[1])
#print(lis)
i=0
j=1
arr.append(lis[i][1])
while(i<n and j<n):
if lis[i][1]<lis[j][0]:
cou+=1
i=j
arr.append(lis[i][1])
# print(lis[i][1])
else:
j+=1
print(cou)
print(*arr)
``` | instruction | 0 | 76,994 | 3 | 153,988 |
Yes | output | 1 | 76,994 | 3 | 153,989 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
n = int(input())
segments = []
for _ in range(n):
segments.append(list(map(int, input().split())))
for i in range(n):
if segments[i][0] > segments[i][1]:
segments[i][0], segments[i][1] = segments[i][1], segments[i][0]
segments.sort(key=lambda x: (x[0], -x[1]))
started = False
i = 0
nails = []
stack = []
while i < n:
stack.append(segments[i])
i += 1
edge = stack[-1][1]
while i < n and segments[i][0] <= edge:
stack.append(segments[i])
i += 1
edge = min(edge, stack[-1][1])
nails.append(edge)
stack = []
print(len(nails))
print(*nails)
``` | instruction | 0 | 76,995 | 3 | 153,990 |
Yes | output | 1 | 76,995 | 3 | 153,991 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
n = int(input())
l = []
cur = set()
for i in range(n):
a, b = map(int, input().split())
l.append((a, 1, i))
l.append((b, 2, i))
l.sort()
ans = []
for i in l:
if i[1] == 1:
cur.add(i[2])
elif i[2] in cur:
ans.append(i[0])
cur.clear()
print(len(ans))
print(*ans)
``` | instruction | 0 | 76,996 | 3 | 153,992 |
No | output | 1 | 76,996 | 3 | 153,993 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
#!/usr/bin/env python3
# Read data
n = int(input())
segments = [map(int, input().split()) for i in range(n)]
segments = [(min(a,b), max(a,b)) for (a,b) in segments] # Make sure segments run from low to high
# Sort the start and end positions
start_pos = [a for (a,b) in segments]
start_pos.sort()
end_pos = [b for (a,b) in segments]
end_pos.sort()
nail_positions = [] # Positions where the nails will be put
while (n != 0):
# Search for a position covered by the most segments
next_start, next_end = 0,0
best = (0,0)
while (next_start < n):
if (start_pos[next_start] <= end_pos[next_end]):
next_start += 1
best = max(best, (next_start - next_end, start_pos[next_start-1]))
else:
next_end += 1
# Put a nail at the found position
nail_pos = best[1]
nail_positions.append(nail_pos)
# Find the nailed and remaining segments
nailed_segments = [(a,b) for (a,b) in segments if ((a <= nail_pos) and (nail_pos <= b))]
segments = [(a,b) for (a,b) in segments if ((a > nail_pos) or (nail_pos > b))]
n = len(segments)
# Update the start and end positions
for (a,b) in nailed_segments:
start_pos.remove(a)
end_pos.remove(b)
print(len(nail_positions))
print(" ".join(["%d" % pos for pos in nail_positions]))
``` | instruction | 0 | 76,997 | 3 | 153,994 |
No | output | 1 | 76,997 | 3 | 153,995 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
n = int(input())
lines = sorted([sorted([int(i) for i in input().split()]) for _ in range(n)])
array = []
bool_array = [0]*n
for i in range(n):
array.append([lines[i][0], i])
array.append([lines[i][1], i])
array.sort()
screws = []
last = -1
last_indx = -1
b_indx = 0
for i in array:
if i[0] == last:
b_indx += 1
last_indx += 1
continue
if i[1] == last_indx + 1:
last_indx += 1
bool_array[i[1]] = 1
elif bool_array[i[1]]:
screws.append(str(i[0]))
last = i[0]
b_indx += 1
for j in range(b_indx, last_indx + 1):
bool_array[j] = 0
print(len(screws))
print(' '.join(screws))
``` | instruction | 0 | 76,998 | 3 | 153,996 |
No | output | 1 | 76,998 | 3 | 153,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n segments on the Ox-axis. You can drive a nail in any integer point on the Ox-axis line nail so, that all segments containing this point, are considered nailed down. If the nail passes through endpoint of some segment, this segment is considered to be nailed too. What is the smallest number of nails needed to nail all the segments down?
Input
The first line of the input contains single integer number n (1 ≤ n ≤ 1000) — amount of segments. Following n lines contain descriptions of the segments. Each description is a pair of integer numbers — endpoints coordinates. All the coordinates don't exceed 10000 by absolute value. Segments can degenarate to points.
Output
The first line should contain one integer number — the smallest number of nails needed to nail all the segments down. The second line should contain coordinates of driven nails separated by space in any order. If the answer is not unique, output any.
Examples
Input
2
0 2
2 5
Output
1
2
Input
5
0 3
4 2
4 8
8 10
7 7
Output
3
7 10 3
Submitted Solution:
```
n = int(input())
lines = sorted([sorted([int(i) for i in input().split()]) for _ in range(n)])
array = []
bool_array = [0]*n
for i in range(n):
array.append([lines[i][0], i])
array.append([lines[i][1], i])
array.sort()
screws = []
last = -1
last_indx = -1
b_indx = 0
for i in array:
if i[0] == last:
b_indx += 1
last_indx += 1
continue
if i[1] == last_indx + 1:
last_indx += 1
bool_array[i[1]] = 1
elif bool_array[i[1]]:
screws.append(str(i[0]))
last = i[0]
for j in range(b_indx, last_indx + 1):
bool_array[j] = 0
b_indx += 1
print(len(screws))
print(' '.join(screws))
``` | instruction | 0 | 76,999 | 3 | 153,998 |
No | output | 1 | 76,999 | 3 | 153,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000 | instruction | 0 | 77,032 | 3 | 154,064 |
Tags: binary search, graphs, shortest paths
Correct Solution:
```
from sys import stdin
from math import inf
def readline():
return map(int, stdin.readline().strip().split())
def main():
n, d = readline()
a = [0] + list(readline()) + [0]
x = [0] * n
y = [0] * n
for i in range(n):
x[i], y[i] = readline()
lower_cost = [inf] * n
lower_cost[0] = 0
visited = [False] * n
for i in range(n - 1):
lower_value = inf
position = 0
for j in range(n):
if not visited[j] and lower_value > lower_cost[j]:
lower_value = lower_cost[j]
position = j
visited[position] = True
for k in range(n):
if not visited[k]:
diff = lower_cost[position] + d * (abs(x[k] - x[position]) + abs(y[k] - y[position])) - a[position]
if lower_cost[k] > diff:
lower_cost[k] = diff
return lower_cost[-1]
if __name__ == '__main__':
print(main())
``` | output | 1 | 77,032 | 3 | 154,065 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000 | instruction | 0 | 77,033 | 3 | 154,066 |
Tags: binary search, graphs, shortest paths
Correct Solution:
```
from sys import stdin, stdout
from math import inf
def main():
n, d = readline()
a = [0] + list(readline()) + [0]
x = [0] * n
y = [0] * n
# parent = [-1] * n # In case you want to know the path traveled
for i in range(n):
x[i], y[i] = readline()
return dijkstra(n, d, a, x, y) # , parent)
def readline():
return map(int, stdin.readline().strip().split())
def dijkstra(n, d, a, x, y): # , parent):
lower_cost = [inf] * n
lower_cost[0] = 0
visited = [False] * n
for i in range(n - 1):
position = minimum(n, visited, lower_cost)
if position == n - 1:
break
visited[position] = True
for k in range(n):
if not visited[k]:
diff = lower_cost[position] + d * (abs(x[k] - x[position]) + abs(y[k] - y[position])) - a[position]
if lower_cost[k] > diff:
lower_cost[k] = diff
# parent[k] = position
return lower_cost[-1]
def minimum(n, visited, lower_cost):
lower_value = inf
position = 0
for j in range(n):
if not visited[j] and lower_value > lower_cost[j]:
lower_value = lower_cost[j]
position = j
return position
if __name__ == '__main__':
stdout.write("".join(str(main()) + "\n"))
``` | output | 1 | 77,033 | 3 | 154,067 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000 | instruction | 0 | 77,037 | 3 | 154,074 |
Tags: binary search, graphs, shortest paths
Correct Solution:
```
from sys import stdin
from math import inf
def main():
n, d = readline()
a = [0] + list(readline()) + [0]
x = [0] * n
y = [0] * n
# parent = [-1] * n # In case you want to know the path traveled
for i in range(n):
x[i], y[i] = readline()
return dijkstra(n, d, a, x, y) # , parent)
def readline():
return map(int, stdin.readline().strip().split())
def dijkstra(n, d, a, x, y): # , parent):
lower_cost = [inf] * n
lower_cost[0] = 0
visited = [False] * n
for i in range(n - 1):
position = minimum(n, visited, lower_cost)
visited[position] = True
for k in range(n):
if not visited[k]:
diff = lower_cost[position] + d * (abs(x[k] - x[position]) + abs(y[k] - y[position])) - a[position]
if lower_cost[k] > diff:
lower_cost[k] = diff
# parent[k] = position
return lower_cost[-1]
def minimum(n, visited, lower_cost):
lower_value = inf
position = 0
for j in range(n):
if visited[j] or lower_value <= lower_cost[j]:
continue
lower_value = lower_cost[j]
position = j
return position
if __name__ == '__main__':
print(main())
``` | output | 1 | 77,037 | 3 | 154,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
n, d = map(int, input().split())
a = [0] + list(map(int, input().split())) + [0]
x = []
y = []
for i in range(n):
xx, yy = map(int, input().split())
x += [xx]
y += [yy]
b = [-1] * n
b[0] = 0
c = True
while c:
c = False
for i in range(n):
for j in range(1, n):
if i != j and b[i] != -1:
t = b[i] + (abs(x[i] - x[j]) + abs(y[i] - y[j])) * d - a[j]
if b[j] == -1 or t < b[j]:
b[j] = t
c = True
print(b[-1])
``` | instruction | 0 | 77,040 | 3 | 154,080 |
Yes | output | 1 | 77,040 | 3 | 154,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
n, d = map(int, input().split())
a = [0] + list(map(int, input().split())) + [0] #aumento de timer según estación
X = []
Y = []
for i in range(n):
x, y = map(int, input().split()) #Coordenadas de la estación i.
X.append(x)
Y.append(y)
mon = [-1] * n #array para el monto necesario para llegar.
mon[0] = 0
Z = 0 #valor que permitirá entrar al loop
while Z == 0:
Z = 1
for i in range(n): #estamos en estación i
for j in range(1, n): #queremos ir a estación j
if i != j and mon[i] != -1: #si no queremos ir a la misma estación y donde estamos pudimos llegar.
costo = mon[i] + (abs(X[i] - X[j]) + abs(Y[i] - Y[j]))*d - a[j] #nuevo costo necesario para ir de i a j.
if mon[j] == -1 or costo < mon[j]: #si el nuevo costo es menor que uno anterior, se guarda este o si antes no había ningun costo guardado.
mon[j] = costo
Z = 0 #volvemos a entrar al loop, esto asegura que todos los mon[] van a dejar de ser '1 y tendran un costo.
print(mon[-1]) #costo de llegar a la última estación.
``` | instruction | 0 | 77,041 | 3 | 154,082 |
Yes | output | 1 | 77,041 | 3 | 154,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
from sys import stdin
from math import inf
def readline():
return map(int, stdin.readline().strip().split())
def dijkstra(): # , parent):
n, d = readline()
a = [0] + list(readline()) + [0]
x = [0] * n
y = [0] * n
# parent = [-1] * n # In case you want to know the path traveled
for i in range(n):
x[i], y[i] = readline()
lower_cost = [inf] * n
lower_cost[0] = 0
visited = [False] * n
for i in range(n - 1):
lower_value = inf
position = 0
for j in range(n):
if visited[j] or lower_value <= lower_cost[j]:
continue
lower_value = lower_cost[j]
position = j
visited[position] = True
for k in range(n):
if not visited[k]:
diff = lower_cost[position] + d * (abs(x[k] - x[position]) + abs(y[k] - y[position])) - a[position]
if lower_cost[k] > diff:
lower_cost[k] = diff
# parent[k] = position
return lower_cost[-1]
if __name__ == '__main__':
print(dijkstra())
``` | instruction | 0 | 77,042 | 3 | 154,084 |
Yes | output | 1 | 77,042 | 3 | 154,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
from sys import stdin
from math import inf
def readline():
return map(int, stdin.readline().strip().split())
def dijkstra():
n, d = readline()
a = [0] + list(readline()) + [0]
x = [0] * n
y = [0] * n
for i in range(n):
x[i], y[i] = readline()
lower_cost = [inf] * n
lower_cost[0] = 0
visited = [False] * n
for i in range(n - 1):
lower_value = inf
position = 0
for j in range(n):
if not visited[j] and lower_value > lower_cost[j]:
lower_value = lower_cost[j]
position = j
visited[position] = True
for k in range(n):
if not visited[k]:
diff = lower_cost[position] + d * (abs(x[k] - x[position]) + abs(y[k] - y[position])) - a[position]
if lower_cost[k] > diff:
lower_cost[k] = diff
return lower_cost[-1]
if __name__ == '__main__':
print(dijkstra())
``` | instruction | 0 | 77,043 | 3 | 154,086 |
Yes | output | 1 | 77,043 | 3 | 154,087 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
from sys import stdin
def readline():
return map(int, stdin.readline().strip().split())
def main():
n, d = readline()
a = [0] + list(readline()) + [0]
x = [0] * n
y = [0] * n
for i in range(n):
x[i], y[i] = readline()
less = [-1] * n
less[0] = 0
for i in range(n):
time = 0
for j in range(1, n):
if i != j:
difference = d * (abs(x[i] - x[j]) + abs(y[i] - y[j])) - (a[i] + time)
if difference < 0:
time = -difference
difference = 0
else:
time = 0
difference += less[i]
if less[j] == -1 or less[j] > difference:
less[j] = difference
return less[-1]
if __name__ == '__main__':
print(main())
``` | instruction | 0 | 77,044 | 3 | 154,088 |
No | output | 1 | 77,044 | 3 | 154,089 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
f = lambda: list(map(int, input().split()))
n, d = f()
a = [0] + f() + [0]
p = [f() for i in range(n)]
s = [1e9] * n
q, s[0] = 1, 0
while q:
q = 0
for i in range(n):
for j in range(i + 1, n):
t = s[i] + (abs(p[i][0] - p[j][0]) + abs(p[i][1] - p[j][1])) * d - a[j]
if t < s[j]: q, s[j] = 1, t
print(s[-1])
``` | instruction | 0 | 77,045 | 3 | 154,090 |
No | output | 1 | 77,045 | 3 | 154,091 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
R = lambda: map(int, input().split())
n, d = R()
a = [0] + list(R()) + [0]
x = []
y = []
INF = float('inf')
def distance(x1, y1, x2, y2):
return (abs(x1 - x2) + abs(y1 - y2)) * d
for i in range(n):
xi, yi = R()
x += [xi]
y += [yi]
b = [INF] * n
b[0] = 0
c = True
while c:
c = False
for i in range(n):
for j in range(i+1, n):
if b[i] != INF:
t = b[i] + distance(x[i], y[i], x[j], y[j]) - a[j]
if t < b[j]:
b[j] = t
# print(b)
c = True
print(b[-1])
``` | instruction | 0 | 77,046 | 3 | 154,092 |
No | output | 1 | 77,046 | 3 | 154,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yaroslav is playing a game called "Time". The game has a timer showing the lifespan he's got left. As soon as the timer shows 0, Yaroslav's character dies and the game ends. Also, the game has n clock stations, station number i is at point (xi, yi) of the plane. As the player visits station number i, he increases the current time on his timer by ai. The stations are for one-time use only, so if the player visits some station another time, the time on his timer won't grow.
A player spends d·dist time units to move between stations, where dist is the distance the player has covered and d is some constant. The distance between stations i and j is determined as |xi - xj| + |yi - yj|.
Initially, the player is at station number 1, and the player has strictly more than zero and strictly less than one units of time. At station number 1 one unit of money can increase the time on the timer by one time unit (you can buy only integer number of time units).
Now Yaroslav is wondering, how much money he needs to get to station n. Help Yaroslav. Consider the time to buy and to increase the timer value negligibly small.
Input
The first line contains integers n and d (3 ≤ n ≤ 100, 103 ≤ d ≤ 105) — the number of stations and the constant from the statement.
The second line contains n - 2 integers: a2, a3, ..., an - 1 (1 ≤ ai ≤ 103). The next n lines contain the coordinates of the stations. The i-th of them contains two integers xi, yi (-100 ≤ xi, yi ≤ 100).
It is guaranteed that no two stations are located at the same point.
Output
In a single line print an integer — the answer to the problem.
Examples
Input
3 1000
1000
0 0
0 1
0 3
Output
2000
Input
3 1000
1000
1 0
1 1
1 2
Output
1000
Submitted Solution:
```
n, d = map(int, input().split())
a = [0] + list(map(int, input().split())) + [0]
x = []
y = []
for i in range(n):
xx, yy = map(int, input().split())
x += [xx]
y += [yy]
b = [10**7] * n
b[0] = 0
for k in range(n - 1):
for i in range(n):
for j in range(1, n):
if i != j and b[i] != 10**7:
b[j] = min(b[j], b[i] + (abs(x[i] - x[j]) + abs(y[i] - y[j])) * d - a[j])
print(b[-1])
``` | instruction | 0 | 77,047 | 3 | 154,094 |
No | output | 1 | 77,047 | 3 | 154,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n soldiers stand in a circle. For each soldier his height ai is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |ai - aj| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input
The first line contains integer n (2 ≤ n ≤ 100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1000). The soldier heights are given in clockwise or counterclockwise direction.
Output
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Examples
Input
5
10 12 13 15 10
Output
5 1
Input
4
10 20 30 40
Output
1 2
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
b = []
for j in range(n-1):
q = a[j]-a[j+1]
if q > 0:
b.append(q)
else:
b.append(-1*q)
w = a[-1]-a[0]
if w > 0:
b.append(w)
else:
b.append(-1*w)
m = min(b)
e = b.index(m) + 1
if e == n:
print(str(e) + ' ' + str(1))
else:
print(str(e) + ' ' + str(e+1))
``` | instruction | 0 | 77,057 | 3 | 154,114 |
Yes | output | 1 | 77,057 | 3 | 154,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n soldiers stand in a circle. For each soldier his height ai is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |ai - aj| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input
The first line contains integer n (2 ≤ n ≤ 100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1000). The soldier heights are given in clockwise or counterclockwise direction.
Output
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Examples
Input
5
10 12 13 15 10
Output
5 1
Input
4
10 20 30 40
Output
1 2
Submitted Solution:
```
n = int(input())
l = list(map(int,input().strip().split()))
res = 999999999
for i in range(0, n-1) :
if abs(l[i]-l[i+ 1]) < res:
res = abs(l[i] - l[i+1])
inde = [i, i+1]
if abs(l[-1]-l[0]) < res:
print(n, 1)
else:
print(inde[0]+1, inde[1]+1)
``` | instruction | 0 | 77,058 | 3 | 154,116 |
Yes | output | 1 | 77,058 | 3 | 154,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n soldiers stand in a circle. For each soldier his height ai is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |ai - aj| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.
Input
The first line contains integer n (2 ≤ n ≤ 100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1000). The soldier heights are given in clockwise or counterclockwise direction.
Output
Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.
Examples
Input
5
10 12 13 15 10
Output
5 1
Input
4
10 20 30 40
Output
1 2
Submitted Solution:
```
#A. Reconnaissance 2
n = int(input())
arr = list(map(int,input().split()))
x,y = n,1
min_ = abs(arr[0] - arr[n-1])
for i in range(1,n):
if abs(arr[i]-arr[i-1])<min_:
min_ = abs(arr[i]-arr[i-1])
x = i + 1
y = i
print(x,y)
``` | instruction | 0 | 77,060 | 3 | 154,120 |
Yes | output | 1 | 77,060 | 3 | 154,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima is living in a dormitory, as well as some cockroaches.
At the moment 0 Dima saw a cockroach running on a table and decided to kill it. Dima needs exactly T seconds for aiming, and after that he will precisely strike the cockroach and finish it.
To survive the cockroach has to run into a shadow, cast by round plates standing on the table, in T seconds. Shadow casted by any of the plates has the shape of a circle. Shadow circles may intersect, nest or overlap arbitrarily.
The cockroach uses the following strategy: first he equiprobably picks a direction to run towards and then runs towards it with the constant speed v. If at some moment t ≤ T it reaches any shadow circle, it immediately stops in the shadow and thus will stay alive. Otherwise the cockroach is killed by the Dima's precise strike. Consider that the Dima's precise strike is instant.
Determine the probability of that the cockroach will stay alive.
Input
In the first line of the input the four integers x0, y0, v, T (|x0|, |y0| ≤ 109, 0 ≤ v, T ≤ 109) are given — the cockroach initial position on the table in the Cartesian system at the moment 0, the cockroach's constant speed and the time in seconds Dima needs for aiming respectively.
In the next line the only number n (1 ≤ n ≤ 100 000) is given — the number of shadow circles casted by plates.
In the next n lines shadow circle description is given: the ith of them consists of three integers xi, yi, ri (|xi|, |yi| ≤ 109, 0 ≤ r ≤ 109) — the ith shadow circle on-table position in the Cartesian system and its radius respectively.
Consider that the table is big enough for the cockroach not to run to the table edges and avoid Dima's precise strike.
Output
Print the only real number p — the probability of that the cockroach will stay alive.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 4.
Examples
Input
0 0 1 1
3
1 1 1
-1 -1 1
-2 2 1
Output
0.50000000000
Input
0 0 1 0
1
1 0 1
Output
1.00000000000
Note
The picture for the first sample is given below.
<image>
Red color stands for points which being chosen as the cockroach's running direction will cause him being killed, green color for those standing for survival directions. Please note that despite containing a circle centered in ( - 2, 2) a part of zone is colored red because the cockroach is not able to reach it in one second.
Submitted Solution:
```
import math as m
l = [int(x) for x in input().split()]
x0, y0, v, T = tuple(l)
n = int(input())
c = []
theta = []
flag = 0
for i in range(n):
l = [int(x) for x in input().split()]
c.append(l)
for i in range(n):
x, y, r = tuple(c[i])
d = m.sqrt((x-x0)**2 + (y-y0)**2)
s = d - r
if s <= 0:
flag = 1
break
if s >= v*T:
continue
t = m.sqrt(d**2 - r**2)
if t <= v*T:
a = m.asin(r/d)
theta.append(2*a)
else:
a = m.acos((d**2 + (v*T)**2 - r**2)/(2*d*v*T))
theta.append(2*a)
if flag:
print(1.0)
else:
print(sum(theta)/(2*m.pi))
``` | instruction | 0 | 77,183 | 3 | 154,366 |
No | output | 1 | 77,183 | 3 | 154,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima is living in a dormitory, as well as some cockroaches.
At the moment 0 Dima saw a cockroach running on a table and decided to kill it. Dima needs exactly T seconds for aiming, and after that he will precisely strike the cockroach and finish it.
To survive the cockroach has to run into a shadow, cast by round plates standing on the table, in T seconds. Shadow casted by any of the plates has the shape of a circle. Shadow circles may intersect, nest or overlap arbitrarily.
The cockroach uses the following strategy: first he equiprobably picks a direction to run towards and then runs towards it with the constant speed v. If at some moment t ≤ T it reaches any shadow circle, it immediately stops in the shadow and thus will stay alive. Otherwise the cockroach is killed by the Dima's precise strike. Consider that the Dima's precise strike is instant.
Determine the probability of that the cockroach will stay alive.
Input
In the first line of the input the four integers x0, y0, v, T (|x0|, |y0| ≤ 109, 0 ≤ v, T ≤ 109) are given — the cockroach initial position on the table in the Cartesian system at the moment 0, the cockroach's constant speed and the time in seconds Dima needs for aiming respectively.
In the next line the only number n (1 ≤ n ≤ 100 000) is given — the number of shadow circles casted by plates.
In the next n lines shadow circle description is given: the ith of them consists of three integers xi, yi, ri (|xi|, |yi| ≤ 109, 0 ≤ r ≤ 109) — the ith shadow circle on-table position in the Cartesian system and its radius respectively.
Consider that the table is big enough for the cockroach not to run to the table edges and avoid Dima's precise strike.
Output
Print the only real number p — the probability of that the cockroach will stay alive.
Your answer will be considered correct if its absolute or relative error does not exceed 10 - 4.
Examples
Input
0 0 1 1
3
1 1 1
-1 -1 1
-2 2 1
Output
0.50000000000
Input
0 0 1 0
1
1 0 1
Output
1.00000000000
Note
The picture for the first sample is given below.
<image>
Red color stands for points which being chosen as the cockroach's running direction will cause him being killed, green color for those standing for survival directions. Please note that despite containing a circle centered in ( - 2, 2) a part of zone is colored red because the cockroach is not able to reach it in one second.
Submitted Solution:
```
from math import acos, pi
def rast_between_dots(xa, ya, xb, yb):
return ((xb-xa)**2 + (yb-ya)**2)**(1/2)
x0, y0, v, T = map(int, input().split())
n = int(input())
possibility = 0
for i in range(n):
xi, yi, ri = map(int, input().split())
r = rast_between_dots(xi, yi, x0, y0)
if rast_between_dots(xi, yi, x0, y0) <= ri:
possibility = 1
break
elif r < T*v+ri:
a = (r**2-ri**2)**(1/2)
angleOfPossibility = acos((2*a**2-4*ri**2)/(2*a**2))
possibility += angleOfPossibility/(4*pi)
print(possibility)
``` | instruction | 0 | 77,184 | 3 | 154,368 |
No | output | 1 | 77,184 | 3 | 154,369 |
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