message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 276 109k | cluster float64 23 23 | __index_level_0__ int64 552 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will consider the numbers a and b as adjacent if they differ by exactly one, that is, |a-b|=1.
We will consider cells of a square matrix n × n as adjacent if they have a common side, that is, for cell (r, c) cells (r, c-1), (r, c+1), (r-1, c) and (r+1, c) are adjacent to it.
For a given number n, construct a square matrix n × n such that:
* Each integer from 1 to n^2 occurs in this matrix exactly once;
* If (r_1, c_1) and (r_2, c_2) are adjacent cells, then the numbers written in them must not be adjacent.
Input
The first line contains one integer t (1 ≤ t ≤ 100). Then t test cases follow.
Each test case is characterized by one integer n (1 ≤ n ≤ 100).
Output
For each test case, output:
* -1, if the required matrix does not exist;
* the required matrix, otherwise (any such matrix if many of them exist).
The matrix should be outputted as n lines, where each line contains n integers.
Example
Input
3
1
2
3
Output
1
-1
2 9 7
4 6 3
1 8 5
Submitted Solution:
```
for _ in range (int(input())):
n=int(input())
if n==2:
print(-1)
continue
a=[[0 for i in range (n)]for j in range (n)]
c=-1
for i in range (n):
for j in range (n):
c+=2
if c>n*n:
c=2
a[i][j]=c
print(*a[i])
``` | instruction | 0 | 70,987 | 23 | 141,974 |
Yes | output | 1 | 70,987 | 23 | 141,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will consider the numbers a and b as adjacent if they differ by exactly one, that is, |a-b|=1.
We will consider cells of a square matrix n × n as adjacent if they have a common side, that is, for cell (r, c) cells (r, c-1), (r, c+1), (r-1, c) and (r+1, c) are adjacent to it.
For a given number n, construct a square matrix n × n such that:
* Each integer from 1 to n^2 occurs in this matrix exactly once;
* If (r_1, c_1) and (r_2, c_2) are adjacent cells, then the numbers written in them must not be adjacent.
Input
The first line contains one integer t (1 ≤ t ≤ 100). Then t test cases follow.
Each test case is characterized by one integer n (1 ≤ n ≤ 100).
Output
For each test case, output:
* -1, if the required matrix does not exist;
* the required matrix, otherwise (any such matrix if many of them exist).
The matrix should be outputted as n lines, where each line contains n integers.
Example
Input
3
1
2
3
Output
1
-1
2 9 7
4 6 3
1 8 5
Submitted Solution:
```
def solve():
n = int(input())
if n == 1:
print(1)
elif n == 2:
print(-1)
else:
mat = [[0]*n for i in range(n)]
c = 1
for i in range(n):
for j in range(0, n, 2):
mat[j][i] = c
c += 1
for i in range(n):
for j in range(1, n, 2):
mat[j][i] = c
c += 1
for i in mat:
print(*i)
def main():
for _ in range(int(input())):
solve()
if __name__ == '__main__':
main()
``` | instruction | 0 | 70,989 | 23 | 141,978 |
No | output | 1 | 70,989 | 23 | 141,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will consider the numbers a and b as adjacent if they differ by exactly one, that is, |a-b|=1.
We will consider cells of a square matrix n × n as adjacent if they have a common side, that is, for cell (r, c) cells (r, c-1), (r, c+1), (r-1, c) and (r+1, c) are adjacent to it.
For a given number n, construct a square matrix n × n such that:
* Each integer from 1 to n^2 occurs in this matrix exactly once;
* If (r_1, c_1) and (r_2, c_2) are adjacent cells, then the numbers written in them must not be adjacent.
Input
The first line contains one integer t (1 ≤ t ≤ 100). Then t test cases follow.
Each test case is characterized by one integer n (1 ≤ n ≤ 100).
Output
For each test case, output:
* -1, if the required matrix does not exist;
* the required matrix, otherwise (any such matrix if many of them exist).
The matrix should be outputted as n lines, where each line contains n integers.
Example
Input
3
1
2
3
Output
1
-1
2 9 7
4 6 3
1 8 5
Submitted Solution:
```
t = int(input())
while t:
n = int(input())
arr1 = []
arr2 = []
i = 3
if n%2 == 0:
print(-1)
elif n == 1:
print(1)
else:
for i in range(1, n**2+1):
if i%2!=0:
arr1.append(i)
else:
arr2.append(i)
arr = arr1 + arr2
arrx = [[] for i in range(n)]
x = 0
i = 0
while x!=n:
count = 0
while count!=n:
arrx[x].append(str(arr[i]))
i+=1
count+=1
x+= 1
for i in range(n):
print(' '.join(arrx[i]))
t-=1
``` | instruction | 0 | 70,990 | 23 | 141,980 |
No | output | 1 | 70,990 | 23 | 141,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will consider the numbers a and b as adjacent if they differ by exactly one, that is, |a-b|=1.
We will consider cells of a square matrix n × n as adjacent if they have a common side, that is, for cell (r, c) cells (r, c-1), (r, c+1), (r-1, c) and (r+1, c) are adjacent to it.
For a given number n, construct a square matrix n × n such that:
* Each integer from 1 to n^2 occurs in this matrix exactly once;
* If (r_1, c_1) and (r_2, c_2) are adjacent cells, then the numbers written in them must not be adjacent.
Input
The first line contains one integer t (1 ≤ t ≤ 100). Then t test cases follow.
Each test case is characterized by one integer n (1 ≤ n ≤ 100).
Output
For each test case, output:
* -1, if the required matrix does not exist;
* the required matrix, otherwise (any such matrix if many of them exist).
The matrix should be outputted as n lines, where each line contains n integers.
Example
Input
3
1
2
3
Output
1
-1
2 9 7
4 6 3
1 8 5
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
m = [[0] * n for j in range(n)]
res = 1
if n == 2:
print(-1)
elif n == 3:
print(2, 9, 7)
print(4, 6, 3)
print(1, 8, 5)
elif n == 4:
print(16, 14, 10, 4)
print(12, 8, 3, 9)
print(6, 2, 7, 13)
print(1, 5, 11, 15)
else:
for j in range(n):
j1 = 0
j2 = 0
while j2 < n:
m[j][j1] = res
res += 1
j2 += 1
if j1 < n // 2:
j1 = n - j1 - 1
else:
j1 = n - j1
for j in range(n):
print(*m[j])
``` | instruction | 0 | 70,991 | 23 | 141,982 |
No | output | 1 | 70,991 | 23 | 141,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We will consider the numbers a and b as adjacent if they differ by exactly one, that is, |a-b|=1.
We will consider cells of a square matrix n × n as adjacent if they have a common side, that is, for cell (r, c) cells (r, c-1), (r, c+1), (r-1, c) and (r+1, c) are adjacent to it.
For a given number n, construct a square matrix n × n such that:
* Each integer from 1 to n^2 occurs in this matrix exactly once;
* If (r_1, c_1) and (r_2, c_2) are adjacent cells, then the numbers written in them must not be adjacent.
Input
The first line contains one integer t (1 ≤ t ≤ 100). Then t test cases follow.
Each test case is characterized by one integer n (1 ≤ n ≤ 100).
Output
For each test case, output:
* -1, if the required matrix does not exist;
* the required matrix, otherwise (any such matrix if many of them exist).
The matrix should be outputted as n lines, where each line contains n integers.
Example
Input
3
1
2
3
Output
1
-1
2 9 7
4 6 3
1 8 5
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
from collections import Counter, defaultdict
import bisect
import math
import heapq
for _ in range(int(input())):
n=int(input())
arr=[[0 for i in range(n)]for j in range(n)]
v1=1
v2=4
vis=[0]*10**5
for i in range(n):
for j in range(n):
if i%2==0:
if v1<=n*n:
arr[i][j]=v1
v1+=2
else:
if v2<=n*n:
arr[i][j]=v2
v2+=2
else:
if i==n-1 and j==n-1:
arr[i][j]=2
else:
if v2<=n*n:
arr[i][j]=v2
v2+=2
else:
if v1<=n*n:
arr[i][j]=v1
v1+=2
#arr[n-1][n-1]=2
#n,k= map(int, input().split())
#arr = list(map(int, input().split()))
f=0
for i in range(n):
for j in range(n):
if j+1<n and abs(arr[i][j]-arr[i][j+1])==1:
f=1
break
if j-1>=0 and abs(arr[i][j]-arr[i][j-1])==1:
f=1
break
if i+1<n and abs(arr[i][j]-arr[i+1][j])==1:
f=1
break
if i-1>=0 and abs(arr[i][j]-arr[i-1][j])==1:
f=1
break
if f==1:
break
if n==1:
print(1)
else:
if f==1:
print(-1)
else:
for i in arr:
print(*i)
``` | instruction | 0 | 70,992 | 23 | 141,984 |
No | output | 1 | 70,992 | 23 | 141,985 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000 | instruction | 0 | 71,495 | 23 | 142,990 |
"Correct Solution:
```
p= []
try:
while True:
x,y = map(float,input().split(","))
p.append(complex(x,y))
except EOFError:
pass
n=len(p)
sum=0.0
for i in range(1,n-1):
a,b= p[i]-p[0], p[i+1]-p[0]
s= (a.conjugate()* b).imag
sum+= s
print("%.6f" % (abs(sum)/2.0))
``` | output | 1 | 71,495 | 23 | 142,991 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000 | instruction | 0 | 71,496 | 23 | 142,992 |
"Correct Solution:
```
# coding=utf-8
###
### for python program
###
import sys
import math
# math class
class mymath:
### pi
pi = 3.14159265358979323846264338
### Prime Number
def pnum_eratosthenes(self, n):
ptable = [0 for i in range(n+1)]
plist = []
for i in range(2, n+1):
if ptable[i]==0:
plist.append(i)
for j in range(i+i, n+1, i):
ptable[j] = 1
return plist
def pnum_check(self, n):
if (n==1):
return False
elif (n==2):
return True
else:
for x in range(2,n):
if(n % x==0):
return False
return True
### GCD
def gcd(self, a, b):
if b == 0:
return a
return self.gcd(b, a%b)
### LCM
def lcm(self, a, b):
return (a*b)//self.gcd(a,b)
### Mat Multiplication
def mul(self, A, B):
ans = []
for a in A:
c = 0
for j, row in enumerate(a):
c += row*B[j]
ans.append(c)
return ans
### intチェック
def is_integer(self, n):
try:
float(n)
except ValueError:
return False
else:
return float(n).is_integer()
### 幾何学問題用
def dist(self, A, B):
d = 0
for i in range(len(A)):
d += (A[i]-B[i])**2
d = d**(1/2)
return d
### 絶対値
def abs(self, n):
if n >= 0:
return n
else:
return -n
mymath = mymath()
### output class
class output:
### list
def list(self, l):
l = list(l)
#print(" ", end="")
for i, num in enumerate(l):
print(num, end="")
if i != len(l)-1:
print(" ", end="")
print()
output = output()
### input sample
#i = input()
#N = int(input())
#A, B, C = [x for x in input().split()]
#N, K = [int(x) for x in input().split()]
#inlist = [int(w) for w in input().split()]
#R = float(input())
#A.append(list(map(int,input().split())))
#for line in sys.stdin.readlines():
# x, y = [int(temp) for temp in line.split()]
#abc list
#abc = [chr(ord('a') + i) for i in range(26)]
### output sample
# print("{0} {1} {2:.5f}".format(A//B, A%B, A/B))
# print("{0:.6f} {1:.6f}".format(R*R*math.pi,R*2*math.pi))
# print(" {}".format(i), end="")
def printA(A):
N = len(A)
for i, n in enumerate(A):
print(n, end='')
if i != N-1:
print(' ', end='')
print()
# リスト内包表記 ifあり
# [x-k if x != 0 else x for x in C]
# ソート（代入する必要なし）
# N.sort()
# 10000個の素数リスト
# P = mymath.pnum_eratosthenes(105000)
def get_input(s=' '):
N = []
while True:
try:
#N.append(input())
#N.append(int(input()))
#N.append(float(input()))
#N.append([int(x) for x in input().split(s)])
N.append([float(x) for x in input().split(s)])
except EOFError:
break
return N
### 0079
def CalcArea(A, B, C):
a = ((A[0]-B[0])**2+(A[1]-B[1])**2)**0.5
b = ((B[0]-C[0])**2+(B[1]-C[1])**2)**0.5
c = ((C[0]-A[0])**2+(C[1]-A[1])**2)**0.5
z = (a+b+c)/2
return (z*(z-a)*(z-b)*(z-c))**0.5
D = get_input(',')
ans = 0.0
for i in range(1, len(D)-1):
ans += CalcArea(D[0], D[i], D[i+1])
print(ans)
``` | output | 1 | 71,496 | 23 | 142,993 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000 | instruction | 0 | 71,497 | 23 | 142,994 |
"Correct Solution:
```
import math
f = []
while True:
try:
st = input().strip().split(',')
x,y = list(map(float,st ))
f.append(x + y*1j)
except EOFError:
break
px = [p.real for p in f]
ox = (max(px) + min(px)) / 2.0
py = [p.imag for p in f]
oy = (max(py) + min(py)) / 2.0
fo = ox + oy * 1j
ss = 0.0
for i in range(len(f)):
if i == len(f)-1:
j = 0
else:
j = i + 1
a = abs(f[i]-fo)
b = abs(f[j]-fo)
c = abs(f[j]-f[i])
z = (a + b + c) / 2.0
s = math.sqrt(z*(z-a)*(z-b)*(z-c))
ss += s
print("%.6f" % ss)
``` | output | 1 | 71,497 | 23 | 142,995 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000 | instruction | 0 | 71,498 | 23 | 142,996 |
"Correct Solution:
```
import sys,math
l=[list(map(float,i.split(","))) for i in sys.stdin]
S=0
for (i,j) in enumerate(l):
xn,yn=j
if i!=len(l)-1:
xn_1,yn_1=l[i+1]
else:
xn_1,yn_1=l[0]
a=math.sqrt(xn**2+yn**2)
b=math.sqrt(xn_1**2+yn_1**2)
c=math.sqrt((xn_1-xn)**2+(yn_1-yn)**2)
z=(a+b+c)/2
S+=math.sqrt(z*(z-a)*(z-b)*(z-c))
print(S)
``` | output | 1 | 71,498 | 23 | 142,997 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000 | instruction | 0 | 71,499 | 23 | 142,998 |
"Correct Solution:
```
import math
def calc_d(a, b):
return math.sqrt((a[0] - b[0])**2 + (a[1] - b[1])**2)
p = []
while True:
try:
p.append(list(map(float, input().split(","))))
except:
break
n = len(p)
s = 0
for i in range(1, n - 1):
a = calc_d(p[0], p[i])
b = calc_d(p[0], p[i+1])
c = calc_d(p[i], p[i+1])
z = (a + b + c)/2
s += math.sqrt(z*(z - a)*(z - b)*(z - c))
print(s)
``` | output | 1 | 71,499 | 23 | 142,999 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000 | instruction | 0 | 71,500 | 23 | 143,000 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0079
"""
import sys
from math import sqrt, atan2, acos, sin, cos, hypot
from sys import stdin
input = stdin.readline
class Point(object):
epsilon = 1e-10
def __init__(self, x=0.0, y=0.0):
if isinstance(x, tuple):
self.x = x[0]
self.y = x[1]
else:
self.x = x
self.y = y
# ????????????
def __add__(self, other):
return Point(self.x + other.x, self.y + other.y)
def __sub__(self, other):
return Point(self.x - other.x, self.y - other.y)
def __mul__(self, other):
return Point(other * self.x, other * self.y)
def __truediv__(self, other):
return Point(self.x / other, self.y / other)
def __lt__(self, other):
if self.x == other.x:
return self.y < other.y
else:
return self.x < other.x
def __eq__(self, other):
from math import fabs
if fabs(self.x - other.x) < Point.epsilon and fabs(self.y - other.y) < Point.epsilon:
return True
else:
return False
def norm(self):
return self.x * self.x + self.y * self.y
def __abs__(self):
return sqrt(self.norm())
def ccw(self, p0, p1):
# ??????2???(p0, p1)?????????????????????????????????????????¢????????????
a = Vector(p1 - p0)
b = Vector(self - p0)
if Vector.cross(a, b) > Point.epsilon:
return 1 # 'COUNTER_CLOCKWISE'
elif Vector.cross(a, b) < -Point.epsilon:
return -1 # 'CLOCKWISE'
elif Vector.dot(a, b) < -Point.epsilon:
return 2 # 'ONLINE_BACK'
elif a.norm() < b.norm():
return -2 # 'ONLINE_FRONT'
else:
return 0 # 'ON_SEGMENT'
def project(self, s):
# ??????(Point)????????????s??????????????????????????????????????§?¨?(?°???±)????±???????
base = Vector(s.p2 - s.p1)
a = Vector(self - s.p1)
r = Vector.dot(a, base)
r /= base.norm()
return s.p1 + base * r
def reflect(self, s):
# ??????s???????§°?????¨?????????????????¨???????§°??????????????§?¨?(????°?)????±???????
proj = self.project(s)
return self + (proj - self)*2
def distance(self, s):
# ????????¨??????s????????¢????¨??????????
if Vector.dot(s.p2-s.p1, self-s.p1) < 0.0:
return abs(self - s.p1)
if Vector.dot(s.p1-s.p2, self-s.p2) < 0.0:
return abs(self - s.p2)
return abs(Vector.cross(s.p2-s.p1, self-s.p1) / abs(s.p2-s.p1))
class Vector(Point):
def __init__(self, x=0.0, y=0.0):
if isinstance(x, tuple):
self.x = x[0]
self.y = x[1]
elif isinstance(x, Point):
self.x = x.x
self.y = x.y
else:
self.x = x
self.y = y
# ????????????
def __add__(self, other):
return Vector(self.x + other.x, self.y + other.y)
def __sub__(self, other):
return Vector(self.x - other.x, self.y - other.y)
def __mul__(self, other):
return Vector(other * self.x, other * self.y)
def __truediv__(self, other):
return Vector(self.x / other, self.y / other)
@classmethod
def dot(cls, a, b):
return a.x * b.x + a.y * b.y
@classmethod
def cross(cls, a, b):
return a.x * b.y - a.y * b.x
@classmethod
def is_orthogonal(cls, a, b):
return abs(Vector.dot(a, b)) < Vector.epsilon
@classmethod
def is_parallel(cls, a, b):
return abs(Vector.cross(a, b)) < Vector.epsilon
class Segment(object):
def __init__(self, p1=Point(), p2=Point()):
if isinstance(p1, Point):
self.p1 = p1
self.p2 = p2
elif isinstance(p1, tuple):
self.p1 = Point(p1[0], p1[1])
self.p2 = Point(p2[0], p2[1])
def intersect(self, s):
# ????????¨??????????????????????????????????????????????????????
ans1 = s.p1.ccw(self.p1, self.p2) * s.p2.ccw(self.p1, self.p2)
ans2 = self.p1.ccw(s.p1, s.p2) * self.p2.ccw(s.p1, s.p2)
return ans1 <= 0 and ans2 <= 0
def cross_point(self, s):
# ????????¨??????????????????????????????????????§?¨?????±???????
base = s.p2 - s.p1
d1 = abs(Vector.cross(base, self.p1-s.p1))
d2 = abs(Vector.cross(base, self.p2-s.p1))
t = d1 / (d1 + d2)
return self.p1 + (self.p2 - self.p1) * t
def distance(self, s):
# ????????¨?????????????????????????????¢????±???????
if self.intersect(s):
return 0.0
d1 = s.p1.distance(self)
d2 = s.p2.distance(self)
d3 = self.p1.distance(s)
d4 = self.p2.distance(s)
return min(d1, d2, d3, d4)
@classmethod
def is_orthogonal(cls, s1, s2):
a = Vector(s1.p2 - s1.p1)
b = Vector(s2.p2 - s2.p1)
return Vector.is_orthogonal(a, b)
@classmethod
def is_parallel(cls, s1, s2):
a = Vector(s1.p2 - s1.p1)
b = Vector(s2.p2 - s2.p1)
return Vector.is_parallel(a, b)
class Line(Segment):
pass
class Cirle(object):
def __init__(self, x, y=Point(), r=1.0):
if isinstance(x, Point):
self.c = x
self.r = y
elif isinstance(x, tuple):
self.c = Point(x[0], x[1])
self.r = r
def cross_points(self, s):
if isinstance(s, Segment):
pr = self.c.project(s)
e = (s.p2 - s.p1) / abs(s.p2 - s.p1)
base = sqrt(self.r * self.r - (pr - self.c).norm())
return pr + e * base, pr - e * base
elif isinstance(s, Cirle):
c2 = s
d = abs(self.c - c2.c)
a = acos((self.r * self.r + d * d - c2.r * c2.r) / (2 * self.r * d))
t = atan2(c2.c.y - self.c.y, c2.c.x - self.c.x)
temp1 = Point(cos(t+a)*self.r, sin(t+a)*self.r)
temp2 = Point(cos(t-a)*self.r, sin(t-a)*self.r)
return self.c + temp1, self.c + temp2
class Triangle(object):
def __init__(self, p1, p2, p3):
self.p1 = p1
self.p2 = p2
self.p3 = p3
def length(self):
l1 = hypot(self.p1.x-self.p2.x, self.p1.y-self.p2.y)
l2 = hypot(self.p2.x-self.p3.x, self.p2.y-self.p3.y)
l3 = hypot(self.p3.x-self.p1.x, self.p3.y-self.p1.y)
return l1, l2, l3
def area(self):
a, b, c = self.length()
z = (a+b+c) / 2.0
return sqrt(z * (z-a)*(z-b)*(z-c))
def main(args):
points = []
for line in sys.stdin:
x, y = map(float, line.split(','))
points.append(Point(x, y))
area = 0.0
p1 = points[0]
points = points[1:]
while len(points) > 1:
p2 = points[0]
p3 = points[1]
t = Triangle(p1, p2, p3)
area += t.area()
points = points[1:]
print('{:.8f}'.format(area))
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 71,500 | 23 | 143,001 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000 | instruction | 0 | 71,501 | 23 | 143,002 |
"Correct Solution:
```
def op(u,v):
return (complex.conjugate(u)*v).imag
f = []
while True:
try:
st = input().strip().split(',')
x,y = list(map(float,st ))
f.append(x + y*1j)
except EOFError:
break
s = 0.0
fo = f[0]
for j in range(2,len(f)):
i = j - 1
s += op(f[i]-fo,f[j]-fo)/2.0
print("%.6f" % abs(s))
``` | output | 1 | 71,501 | 23 | 143,003 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000 | instruction | 0 | 71,502 | 23 | 143,004 |
"Correct Solution:
```
def solve():
from sys import stdin
lines = stdin
x1, y1 = map(float, lines.readline().split(','))
x2, y2 = map(float, lines.readline().split(','))
vx1 = x2 - x1
vy1 = y2 - y1
area = 0
for line in lines:
x3, y3 = map(float, line.split(','))
vx2 = x3 - x1
vy2 = y3 - y1
area += abs(vx1 * vy2 - vy1 * vx2) / 2
vx1, vy1 = vx2, vy2
print(area)
solve()
``` | output | 1 | 71,502 | 23 | 143,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000
Submitted Solution:
```
import sys
s=0
p=[list(map(float,e.split(',')))for e in sys.stdin]
x,y=p[0]
for i in range(len(p)-2):
a,b=p[i+1];c,d=p[i+2]
s+=(a*d-b*c+(b-d)*x+(c-a)*y)/2
print(abs(s))
``` | instruction | 0 | 71,503 | 23 | 143,006 |
Yes | output | 1 | 71,503 | 23 | 143,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000
Submitted Solution:
```
p1=complex(*list(map(float,input().split(','))))
p2=complex(*list(map(float,input().split(','))))
ans=0
def area(p,q,r):
return ((p-r).conjugate()*(q-r)).imag/2
try:
while True:
p3=complex(*list(map(float,input().split(','))))
ans+=area(p1,p2,p3)
p2=p3
except EOFError: print(abs(ans))
``` | instruction | 0 | 71,504 | 23 | 143,008 |
Yes | output | 1 | 71,504 | 23 | 143,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000
Submitted Solution:
```
import sys
s=0
p=[list(map(float,e.split(',')))for e in sys.stdin]
x,y=p[0]
print(abs(sum((p[i][0]-x)*(p[i+1][1]-y)-(p[i][1]-y)*(p[i+1][0]-x)for i in range(1,len(p)-1)))/2)
``` | instruction | 0 | 71,505 | 23 | 143,010 |
Yes | output | 1 | 71,505 | 23 | 143,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000
Submitted Solution:
```
import math
f = []
while True:
try:
st = input().strip().split(',')
x,y = list(map(float,st ))
f.append(x + y*1j)
except EOFError:
break
ss = 0.0
fo = f[0]
for j in range(2,len(f)):
i = j - 1
a = abs(f[i]-fo)
b = abs(f[j]-fo)
c = abs(f[j]-f[i])
z = (a + b + c) / 2.0
s = math.sqrt(z*(z-a)*(z-b)*(z-c))
ss += s
print("%.6f" % ss)
``` | instruction | 0 | 71,506 | 23 | 143,012 |
Yes | output | 1 | 71,506 | 23 | 143,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000
Submitted Solution:
```
import sys
s=0
p=[list(map(float,e.split(',')))for e in sys.stdin]
n=len(p)
for i in range(n):a,b=p[i];c,d=p[-~i%n];s+=a*d-b*c
print(s/2)
/////////////
``` | instruction | 0 | 71,507 | 23 | 143,014 |
No | output | 1 | 71,507 | 23 | 143,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000
Submitted Solution:
```
import math
def op(u,v):
return (conjugate(u)*v).imag
f = []
while True:
try:
st = input().strip().split(',')
x,y = list(map(float,st ))
f.append(x + y*1j)
except EOFError:
break
s = 0.0
fo = f[0]
for j in range(2,len(f)):
i = j - 1
s += op(i-fo,j-fo)/2.0
print("%.6f" % abs(ss))
``` | instruction | 0 | 71,508 | 23 | 143,016 |
No | output | 1 | 71,508 | 23 | 143,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000
Submitted Solution:
```
import sys
s=0
p=[list(map(float,e.split(',')))for e in sys.stdin]
x,y=p[0]
for i in range(len(p)-2):
a,b=p[i+1];c,d=p[i+2]
s-=((a-x)*(d-y)-(b-y)*(c-x))/2
print(s)
``` | instruction | 0 | 71,509 | 23 | 143,018 |
No | output | 1 | 71,509 | 23 | 143,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program that reads the coordinates of the vertices of a convex n-sided polygon (a polygon whose internal angles are less than 180 degrees, that is, a polygon that is not dented) and outputs its area. The vertices are named Vertex 1, Vertex 2, Vertex 3, ... Vertex n according to the order of edge connections.
However, n is 3 or more and 20 or less. You can also use the following formula to find the area S from the lengths a, b, and c of the three sides of the triangle.
<image>
input
The input is given in the following format:
x1, y1
x2, y2
::
xn, yn
xi and yi are real numbers that represent the x and y coordinates of vertex i, respectively.
output
Outputs the area S (real number) on one line. The output can contain an error of 0.000001 or less.
Example
Input
0.0,0.0
0.0,1.0
1.0,1.0
2.0,0.0
1.0,-1.0
Output
2.500000
Submitted Solution:
```
def cross(x, y):
return (x.conjugate() * y).imag
def area_polygon(points):
area = 0
n = len(points)
for i in range(n):
area += cross(points[i], points[(i+1)%n])
return area / 2
import sys
p = []
for line in sys.stdin:
p.append(list(map(float, line.split(','))))
print(area_polygon(p))
``` | instruction | 0 | 71,510 | 23 | 143,020 |
No | output | 1 | 71,510 | 23 | 143,021 |
Provide a correct Python 3 solution for this coding contest problem.
For given n points in metric space, find the distance of the closest points.
Constraints
* 2 ≤ n ≤ 100,000
* -100 ≤ x, y ≤ 100
Input
n
x0 y0
x1 y1
:
xn-1 yn-1
The first integer n is the number of points.
In the following n lines, the coordinate of the i-th point is given by two real numbers xi and yi. Each value is a real number with at most 6 digits after the decimal point.
Output
Print the distance in a line. The output values should be in a decimal fraction with an error less than 0.000001.
Examples
Input
2
0.0 0.0
1.0 0.0
Output
1.000000
Input
3
0.0 0.0
2.0 0.0
1.0 1.0
Output
1.41421356237 | instruction | 0 | 71,597 | 23 | 143,194 |
"Correct Solution:
```
import sys
from operator import itemgetter
from itertools import combinations, takewhile
from math import sqrt
def solve(a: list, sorted_axis: int = -1):
length = len(a)
if length <= 3:
return sqrt(min((x1-x2)**2+(y1-y2)**2 for (x1, y1), (x2, y2) in combinations(a, 2)))
x_set_len = len(set(x for x, _ in a))
axis1 = 0 if x_set_len > length/2 else 1
if sorted_axis != axis1:
a.sort(key=itemgetter(axis1))
mid_index = length//2
left, right = a[:mid_index], a[mid_index:]
delta = min(solve(left, axis1), solve(right, axis1))
axis2 = not axis1
median = a[mid_index][axis1]
mid_a = sorted([p for _iter in (
(takewhile(lambda p: median-delta < p[axis1], left[::-1])),
(takewhile(lambda p: p[axis1] < median+delta, right))
) for p in _iter], key=itemgetter(axis2))
for i, (x1, y1) in enumerate(mid_a):
ub = (x1 if axis1 else y1) + delta
for x2, y2 in takewhile(lambda p: p[axis2] < ub, mid_a[i+1:]):
delta = min(delta, sqrt((x1-x2)**2+(y1-y2)**2))
return delta
if __name__ == "__main__":
N = int(input())
a = [tuple(map(float, l.split())) for l in sys.stdin.readlines()]
print("{:.10f}".format(solve(a)))
``` | output | 1 | 71,597 | 23 | 143,195 |
Provide a correct Python 3 solution for this coding contest problem.
For given n points in metric space, find the distance of the closest points.
Constraints
* 2 ≤ n ≤ 100,000
* -100 ≤ x, y ≤ 100
Input
n
x0 y0
x1 y1
:
xn-1 yn-1
The first integer n is the number of points.
In the following n lines, the coordinate of the i-th point is given by two real numbers xi and yi. Each value is a real number with at most 6 digits after the decimal point.
Output
Print the distance in a line. The output values should be in a decimal fraction with an error less than 0.000001.
Examples
Input
2
0.0 0.0
1.0 0.0
Output
1.000000
Input
3
0.0 0.0
2.0 0.0
1.0 1.0
Output
1.41421356237 | instruction | 0 | 71,598 | 23 | 143,196 |
"Correct Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""
input:
3
0.0 0.0
2.0 0.0
1.0 1.0
output:
1.41421356237
"""
import sys
from operator import attrgetter
class ClosestPair(object):
def __init__(self, in_data):
"""
Init closest pairs points set.
"""
self.p_num = int(in_data[0])
points = map(lambda x: x.split(), in_data[1:])
p_list = [complex(float(x), float(y)) for x, y in points]
# pre_sort by axis_X
self.p_list = sorted(p_list, key=attrgetter('real'))
def closest_pair(self, array, array_length):
if array_length <= 1:
return float('inf')
mid_idx = array_length // 2
div_line = array[mid_idx].real
d_min = min(self.closest_pair(array[:mid_idx], mid_idx),
self.closest_pair(array[mid_idx:], array_length - mid_idx))
# sort array_part by axis_Y while recursively comparing
array.sort(key=attrgetter('imag'))
min_stack = list()
for ele in array:
size = len(min_stack)
# eliminate p which distance(p,div_line) >= d
if abs(ele.real - div_line) >= d_min:
continue
for j in range(size):
alt = ele - min_stack[size - j - 1]
if alt.imag >= d_min:
break
d_min = min(d_min, abs(alt))
min_stack.append(ele)
return d_min
def solve(self):
return self.closest_pair(array=self.p_list, array_length=self.p_num)
if __name__ == '__main__':
_input = sys.stdin.readlines()
case = ClosestPair(in_data=_input)
print('{:.6f}'.format(case.solve()))
``` | output | 1 | 71,598 | 23 | 143,197 |
Provide a correct Python 3 solution for this coding contest problem.
For given n points in metric space, find the distance of the closest points.
Constraints
* 2 ≤ n ≤ 100,000
* -100 ≤ x, y ≤ 100
Input
n
x0 y0
x1 y1
:
xn-1 yn-1
The first integer n is the number of points.
In the following n lines, the coordinate of the i-th point is given by two real numbers xi and yi. Each value is a real number with at most 6 digits after the decimal point.
Output
Print the distance in a line. The output values should be in a decimal fraction with an error less than 0.000001.
Examples
Input
2
0.0 0.0
1.0 0.0
Output
1.000000
Input
3
0.0 0.0
2.0 0.0
1.0 1.0
Output
1.41421356237 | instruction | 0 | 71,599 | 23 | 143,198 |
"Correct Solution:
```
import itertools
def _get_distance(p1, p2):
return ((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2) ** 0.5
def _get_min_distance(points):
min_d = 400
for p1, p2 in itertools.combinations(points, 2):
min_d = min(min_d, _get_distance(p1, p2))
return min_d
def closest_pair_distance(points, axis):
# axis: x; 0, y; 1
n = len(points)
if n <= 3:
return _get_min_distance(points)
else:
mid = n // 2
px, py = zip(*points)
if len(set(px)) > len(set(py)):
if axis:
points.sort(key = lambda p: p[0])
axis1 = 0
axis2 = 1
else:
if not axis:
points.sort(key = lambda p: p[1])
axis1 = 1
axis2 = 0
A_points = points[:mid]
B_points = points[mid:]
d_Amin = closest_pair_distance(A_points.copy(), axis1)
d_Bmin = closest_pair_distance(B_points.copy(), axis1)
dist = min(d_Amin, d_Bmin)
min_d = dist
for ap in A_points[::-1]:
if B_points[0][axis1] - ap[axis1] >= dist:
break
for bp in B_points:
if bp[axis1] - ap[axis1] >= dist:
break
if ap[axis2] - dist < bp[axis2] < ap[axis2] + dist:
min_d = min(min_d, _get_distance(ap, bp))
return min_d
# Acceptance of input
import sys
file_input = sys.stdin
n = int(file_input.readline())
P = [tuple(map(float, line.split())) for line in file_input]
# Solve
P.sort(key = lambda p: p[0])
ans = closest_pair_distance(P, 0)
print('{:f}'.format(ans))
``` | output | 1 | 71,599 | 23 | 143,199 |
Provide a correct Python 3 solution for this coding contest problem.
For given n points in metric space, find the distance of the closest points.
Constraints
* 2 ≤ n ≤ 100,000
* -100 ≤ x, y ≤ 100
Input
n
x0 y0
x1 y1
:
xn-1 yn-1
The first integer n is the number of points.
In the following n lines, the coordinate of the i-th point is given by two real numbers xi and yi. Each value is a real number with at most 6 digits after the decimal point.
Output
Print the distance in a line. The output values should be in a decimal fraction with an error less than 0.000001.
Examples
Input
2
0.0 0.0
1.0 0.0
Output
1.000000
Input
3
0.0 0.0
2.0 0.0
1.0 1.0
Output
1.41421356237 | instruction | 0 | 71,600 | 23 | 143,200 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Closest Pair
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=CGL_5_A&lang=ja
TLE
"""
import sys
from collections import deque
def closest_pair(points):
return find_closest(sorted(points)) ** 0.5
def find_closest(points):
n = len(points)
if n <= 1:
return float('inf')
m = n // 2
x = points[m][0]
d = min(find_closest(points[:m]), find_closest(points[m:]))
points.sort(key=lambda p:p[1])
b = deque()
for p in points:
if (p[0] - x)**2 < d:
for q in b:
dx = p[0] - q[0]
dy = p[1] - q[1]
if dy**2 >= d:
break
if (dx**2 + dy**2) < d:
d = (dx**2 + dy**2)
b.appendleft(p)
return d
def main(args):
n = int(input())
points = []
for _ in range(n):
x, y = map(float, input().split())
points.append((x, y))
ans = closest_pair(sorted(points))
print('{:.10f}'.format(ans))
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 71,600 | 23 | 143,201 |
Provide a correct Python 3 solution for this coding contest problem.
For given n points in metric space, find the distance of the closest points.
Constraints
* 2 ≤ n ≤ 100,000
* -100 ≤ x, y ≤ 100
Input
n
x0 y0
x1 y1
:
xn-1 yn-1
The first integer n is the number of points.
In the following n lines, the coordinate of the i-th point is given by two real numbers xi and yi. Each value is a real number with at most 6 digits after the decimal point.
Output
Print the distance in a line. The output values should be in a decimal fraction with an error less than 0.000001.
Examples
Input
2
0.0 0.0
1.0 0.0
Output
1.000000
Input
3
0.0 0.0
2.0 0.0
1.0 1.0
Output
1.41421356237 | instruction | 0 | 71,602 | 23 | 143,204 |
"Correct Solution:
```
# 最近点対を分割統治法で求める
from math import sqrt
n = int(input())
ps = [list(map(float, input().strip().split())) for i in range(n)]
# print(ps)
# Q: なぜ、最初にx座標についてソートされている必要があるのか？
# A: 二つの区間が統合されるとき、dには2つの区間における最小２点間距離が入っており、
# 統合された区間における最小値は、それ以下を探索するようになっている。
# この時、各区間A:[i, i+m), B:[i+m, n)での最小値は、
# [Aの任意の点]と[Bの任意の点]の距離は、境界(i+m-1とi+m)を除いてこの最小値以上であるという前提が成り立つ。
# そのため、統合された区間において探索するのはabs(dx - x) <= dのみを探索すれば良い。
# 仮にxでソートされていない場合、上の前提が成り立たないため、統合しても結局全ての点で
# 距離を計算しなくてはいけない。
ps.sort()
# print(ps)
INF = 10**9
# cp_rec - 再帰用関数
# 入力: 配列と区間
# 出力: 距離と区間内の要素をY座標でソートした配列
def cp_rec(ps, i, n):
if n <= 1:
return INF, [ps[i]]
m = int(n/2)
x = ps[i+m][0] # 半分に分割した境界のX座標
# 配列を半分に分割して計算
d1, qs1 = cp_rec(ps, i, m)
d2, qs2 = cp_rec(ps, i+m, n-m)
d = min(d1, d2)
# Y座標が小さい順にmergeする
qs = [None]*n
s = t = idx = 0
while s < m and t < n-m:
if qs1[s][1] < qs2[t][1]:
qs[idx] = qs1[s]; s += 1
else:
qs[idx] = qs2[t]; t += 1
idx += 1
while s < m:
qs[idx] = qs1[s]; s += 1
idx += 1
while t < n-m:
qs[idx] = qs2[t]; t += 1
idx += 1
# 境界のX座標x(=ps[i+m][0])から距離がd以下のものについて距離を計算していく
# bは境界のX座標から距離d以下のものを集めたものが、Y座標の昇順に入っている。
b = []
for i in range(n):
ax, ay = q = qs[i]
if abs(ax - x) >= d:
continue
# Y座標について、qs[i]から距離がd以下のj(<i)について計算していく
for j in range(len(b)-1, -1, -1):
dx = ax - b[j][0]
dy = ay - b[j][1]
# このループは, ayとb[j][1]が近い順に取り出していいるため、
# 一度 dy >= 0となれば、あとは探索する必要はない。
if dy >= d: break
d = min(d, sqrt(dx**2 + dy**2))
b.append(q)
# dは、区間[i, i+n)における最小区間距離を指している。
# qsは[i, i+n)でY座標についてソートされている。
return d, qs
# ps: ソートした二次元座標のlist
def closest_pair(ps):
n = len(ps)
return cp_rec(ps, 0, n)[0]
print("{:.09f}".format(closest_pair(ps)))
``` | output | 1 | 71,602 | 23 | 143,205 |
Provide a correct Python 3 solution for this coding contest problem.
For given n points in metric space, find the distance of the closest points.
Constraints
* 2 ≤ n ≤ 100,000
* -100 ≤ x, y ≤ 100
Input
n
x0 y0
x1 y1
:
xn-1 yn-1
The first integer n is the number of points.
In the following n lines, the coordinate of the i-th point is given by two real numbers xi and yi. Each value is a real number with at most 6 digits after the decimal point.
Output
Print the distance in a line. The output values should be in a decimal fraction with an error less than 0.000001.
Examples
Input
2
0.0 0.0
1.0 0.0
Output
1.000000
Input
3
0.0 0.0
2.0 0.0
1.0 1.0
Output
1.41421356237 | instruction | 0 | 71,603 | 23 | 143,206 |
"Correct Solution:
```
import sys
from operator import itemgetter
from itertools import permutations
from math import sqrt
def solve(a: list):
length = len(a)
if length <= 3:
return min(sqrt((p1[0]-p2[0])**2+(p1[1]-p2[1])**2) for p1, p2 in permutations(a, 2))
x_set_len = len(set(x for (x, _) in a))
axis = 0 if x_set_len > length/2 else 1
a.sort(key=itemgetter(axis))
mid_index = length//2
left, right = a[:mid_index], a[mid_index:]
delta = min(solve(left), solve(right))
median = a[mid_index][axis]
m_a = []
append = m_a.append
for p in left[::-1]:
if p[axis] < median-delta:
break
append(p)
for p in right:
if p[axis] > median+delta:
break
append(p)
m_a.sort(key=itemgetter(not axis))
for i, p1 in enumerate(m_a):
ub = p1[not axis] + delta
for j, p2 in enumerate(m_a[i+1:]):
if p2[not axis] > ub:
break
dist = sqrt((p1[0]-p2[0])**2+(p1[1]-p2[1])**2)
if dist < delta:
delta = dist
return delta
if __name__ == "__main__":
N = int(input())
a = [tuple(map(float, l.split())) for l in sys.stdin.readlines()]
print("{:.10f}".format(solve(a)))
``` | output | 1 | 71,603 | 23 | 143,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given n points in metric space, find the distance of the closest points.
Constraints
* 2 ≤ n ≤ 100,000
* -100 ≤ x, y ≤ 100
Input
n
x0 y0
x1 y1
:
xn-1 yn-1
The first integer n is the number of points.
In the following n lines, the coordinate of the i-th point is given by two real numbers xi and yi. Each value is a real number with at most 6 digits after the decimal point.
Output
Print the distance in a line. The output values should be in a decimal fraction with an error less than 0.000001.
Examples
Input
2
0.0 0.0
1.0 0.0
Output
1.000000
Input
3
0.0 0.0
2.0 0.0
1.0 1.0
Output
1.41421356237
Submitted Solution:
```
import sys
from operator import itemgetter
from itertools import permutations
from math import sqrt
def solve(a: list):
length = len(a)
if length <= 3:
return min(sqrt((p1[0]-p2[0])**2+(p1[1]-p2[1])**2) for p1, p2 in permutations(a, 2))
x_set_len = len(set(x for (x, _) in a))
axis = 0 if x_set_len > length/2 else 1
a.sort(key=itemgetter(axis))
mid_index = length//2
left, right = a[:mid_index], a[mid_index:]
delta = min(solve(left), solve(right))
median = a[mid_index][axis]
m_a = []
for p in left[::-1]:
if p[axis] < median-delta:
break
m_a.append(p)
for p in right:
if p[axis] > median+delta:
break
m_a.append(p)
m_a.sort(key=itemgetter(not axis))
for i, p1 in enumerate(m_a):
ub = p1[not axis] + delta
for j, p2 in enumerate(m_a[i+1:]):
if p2[not axis] > ub:
break
delta = min(delta, sqrt((p1[0]-p2[0])**2+(p1[1]-p2[1])**2))
return delta
if __name__ == "__main__":
N = int(input())
a = [tuple(map(float, l.split())) for l in sys.stdin.readlines()]
print("{:.10f}".format(solve(a)))
``` | instruction | 0 | 71,604 | 23 | 143,208 |
Yes | output | 1 | 71,604 | 23 | 143,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given n points in metric space, find the distance of the closest points.
Constraints
* 2 ≤ n ≤ 100,000
* -100 ≤ x, y ≤ 100
Input
n
x0 y0
x1 y1
:
xn-1 yn-1
The first integer n is the number of points.
In the following n lines, the coordinate of the i-th point is given by two real numbers xi and yi. Each value is a real number with at most 6 digits after the decimal point.
Output
Print the distance in a line. The output values should be in a decimal fraction with an error less than 0.000001.
Examples
Input
2
0.0 0.0
1.0 0.0
Output
1.000000
Input
3
0.0 0.0
2.0 0.0
1.0 1.0
Output
1.41421356237
Submitted Solution:
```
# CGL_5_A: Point Set - Closest Pair
from operator import itemgetter
from math import sqrt
def closest_distance(ps):
def split(i, j):
if j - i < 2:
return float('INF')
mid = (i + j) // 2
d = min(split(i, mid), split(mid, j))
lv, rv = ps[mid][0]-d, ps[mid][0]+d
return merge(sorted([p for p in ps[i:j] if lv <= p[0] <= rv],
key=itemgetter(1)), d)
def merge(_ps, d):
n = len(_ps)
for i in range(n-1):
for j in range(i+1, n):
if d < _ps[j][1] - _ps[i][1]:
break
d = min(d, distance(_ps[i], _ps[j]))
return d
ps.sort(key=itemgetter(0, 1))
return split(0, len(ps))
def dot(p0, p1):
x0, y0 = p0
x1, y1 = p1
return x0*x1 + y0*y1
def distance(p0, p1):
x0, y0 = p0
x1, y1 = p1
return sqrt((x1-x0)**2 + (y1-y0)**2)
def run():
n = int(input())
ps = []
for _ in range(n):
x, y = [float(i) for i in input().split()]
ps.append((x, y))
print("{:.10f}".format(closest_distance(ps)))
if __name__ == '__main__':
run()
``` | instruction | 0 | 71,605 | 23 | 143,210 |
Yes | output | 1 | 71,605 | 23 | 143,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given n points in metric space, find the distance of the closest points.
Constraints
* 2 ≤ n ≤ 100,000
* -100 ≤ x, y ≤ 100
Input
n
x0 y0
x1 y1
:
xn-1 yn-1
The first integer n is the number of points.
In the following n lines, the coordinate of the i-th point is given by two real numbers xi and yi. Each value is a real number with at most 6 digits after the decimal point.
Output
Print the distance in a line. The output values should be in a decimal fraction with an error less than 0.000001.
Examples
Input
2
0.0 0.0
1.0 0.0
Output
1.000000
Input
3
0.0 0.0
2.0 0.0
1.0 1.0
Output
1.41421356237
Submitted Solution:
```
import itertools
def _get_distance(p1, p2):
return ((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2) ** 0.5
def _get_min_distance(points):
min_d = 400
for p1, p2 in itertools.combinations(points, 2):
min_d = min(min_d, _get_distance(p1, p2))
return min_d
def closest_pair_distance(points):
n = len(points)
if n <= 3:
return _get_min_distance(points)
else:
mid = n // 2
px, py = zip(*points)
if len(set(px)) > len(set(py)):
points.sort(key = lambda p: p[0])
# axis: x; 0, y; 1
axis1 = 0
axis2 = 1
else:
points.sort(key = lambda p: p[1])
axis1 = 1
axis2 = 0
A_points = points[:mid]
B_points = points[mid:]
d_Amin = closest_pair_distance(A_points.copy())
d_Bmin = closest_pair_distance(B_points.copy())
dist = min(d_Amin, d_Bmin)
min_d = dist
for ap in A_points[::-1]:
if B_points[0][axis1] - ap[axis1] >= dist:
break
for bp in B_points:
if bp[axis1] - ap[axis1] >= dist:
break
if ap[axis2] - dist < bp[axis2] < ap[axis2] + dist:
min_d = min(min_d, _get_distance(ap, bp))
return min_d
# Acceptance of input
import sys
file_input = sys.stdin
n = int(file_input.readline())
P = [tuple(map(float, line.split())) for line in file_input]
# Solve
ans = closest_pair_distance(P)
print('{:f}'.format(ans))
``` | instruction | 0 | 71,606 | 23 | 143,212 |
Yes | output | 1 | 71,606 | 23 | 143,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given n points in metric space, find the distance of the closest points.
Constraints
* 2 ≤ n ≤ 100,000
* -100 ≤ x, y ≤ 100
Input
n
x0 y0
x1 y1
:
xn-1 yn-1
The first integer n is the number of points.
In the following n lines, the coordinate of the i-th point is given by two real numbers xi and yi. Each value is a real number with at most 6 digits after the decimal point.
Output
Print the distance in a line. The output values should be in a decimal fraction with an error less than 0.000001.
Examples
Input
2
0.0 0.0
1.0 0.0
Output
1.000000
Input
3
0.0 0.0
2.0 0.0
1.0 1.0
Output
1.41421356237
Submitted Solution:
```
INF = 10**16
# ソートしてから入れる
def ClosestDFS(Points):
l = len(Points)
if l <= 1: return INF, Points
d1, nearPoints1 = ClosestDFS(Points[:l//2])
d2, nearPoints2 = ClosestDFS(Points[l//2:])
retPoints = []
ind = 0
for p1x, p1y in nearPoints1:
while ind != len(nearPoints2) and nearPoints2[ind][1] < p1y:
retPoints.append(nearPoints2[ind])
ind += 1
retPoints.append((p1x, p1y))
while ind != len(nearPoints2):
retPoints.append(nearPoints2[ind])
ind += 1
d = min(d1, d2)
B = []
border_x = Points[l//2][0]
for px, py in retPoints:
if abs(border_x-px) > d: continue
for bx, by in reversed(B):
if py - by > d: break
d = min(d, ((px-bx)**2+(py-by)**2)**0.5)
B.append((px, py))
return d, retPoints
def ClosestPair(Points):
Points.sort()
d, _ = ClosestDFS(Points)
return d
import sys
input = sys.stdin.readline
def main():
N = int(input())
Points = [list(map(float, input().split())) for _ in range(N)]
print("{:.8f}".format(ClosestPair(Points)))
if __name__ == "__main__":
main()
``` | instruction | 0 | 71,607 | 23 | 143,214 |
Yes | output | 1 | 71,607 | 23 | 143,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For given n points in metric space, find the distance of the closest points.
Constraints
* 2 ≤ n ≤ 100,000
* -100 ≤ x, y ≤ 100
Input
n
x0 y0
x1 y1
:
xn-1 yn-1
The first integer n is the number of points.
In the following n lines, the coordinate of the i-th point is given by two real numbers xi and yi. Each value is a real number with at most 6 digits after the decimal point.
Output
Print the distance in a line. The output values should be in a decimal fraction with an error less than 0.000001.
Examples
Input
2
0.0 0.0
1.0 0.0
Output
1.000000
Input
3
0.0 0.0
2.0 0.0
1.0 1.0
Output
1.41421356237
Submitted Solution:
```
import itertools
def _get_min_distance(points):
min_d = 400
for p1, p2 in itertools.combinations(points, 2):
min_d = min(min_d, abs(p1 - p2))
return min_d
def closest_pair_distance(points):
n = len(points)
if n <= 3:
return _get_min_distance(points)
else:
mid = n // 2
left_points = points[:mid]
right_points = points[mid:]
d_Lmin = closest_pair_distance(left_points)
d_Rmin = closest_pair_distance(right_points)
dist = min(d_Lmin, d_Rmin)
min_d = dist
for lp in left_points[::-1]:
if right_points[0].real - lp.real >= dist:
break
for rp in right_points:
if rp.real -lp.real >= dist:
break
if lp.imag - dist < rp.imag < lp.imag + dist:
min_d = min(min_d, abs(lp - rp))
return min_d
# Acceptance of input
def string_to_complex(s):
x, y = map(float, s.split())
return x + y * 1j
import sys
file_input = sys.stdin
n = int(file_input.readline())
P = [string_to_complex(line) for line in file_input]
P.sort(key = lambda c: c.real)
# Output
ans = closest_pair_distance(P)
print('{:f}'.format(ans))
``` | instruction | 0 | 71,610 | 23 | 143,220 |
No | output | 1 | 71,610 | 23 | 143,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem is same as the previous one, but has larger constraints.
It was a Sunday morning when the three friends Selena, Shiro and Katie decided to have a trip to the nearby power station (do not try this at home). After arriving at the power station, the cats got impressed with a large power transmission system consisting of many chimneys, electric poles, and wires. Since they are cats, they found those things gigantic.
At the entrance of the station, there is a map describing the complicated wiring system. Selena is the best at math among three friends. He decided to draw the map on the Cartesian plane. Each pole is now a point at some coordinates (x_i, y_i). Since every pole is different, all of the points representing these poles are distinct. Also, every two poles are connected with each other by wires. A wire is a straight line on the plane infinite in both directions. If there are more than two poles lying on the same line, they are connected by a single common wire.
Selena thinks, that whenever two different electric wires intersect, they may interfere with each other and cause damage. So he wonders, how many pairs are intersecting? Could you help him with this problem?
Input
The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of electric poles.
Each of the following n lines contains two integers x_i, y_i (-10^4 ≤ x_i, y_i ≤ 10^4) — the coordinates of the poles.
It is guaranteed that all of these n points are distinct.
Output
Print a single integer — the number of pairs of wires that are intersecting.
Examples
Input
4
0 0
1 1
0 3
1 2
Output
14
Input
4
0 0
0 2
0 4
2 0
Output
6
Input
3
-1 -1
1 0
3 1
Output
0
Note
In the first example:
<image>
In the second example:
<image>
Note that the three poles (0, 0), (0, 2) and (0, 4) are connected by a single wire.
In the third example:
<image> | instruction | 0 | 71,712 | 23 | 143,424 |
Tags: data structures, geometry, implementation, math
Correct Solution:
```
from collections import defaultdict as dd
from itertools import combinations
import math
import heapq
def nn():
return int(input())
def li():
return list(input())
def mi():
return map(int, input().split())
def lm():
return list(map(int, input().split()))
n=nn()
points=[]
for i in range(n):
points.append(lm())
slopedict=dd(set)
lines=0
for p1, p2 in combinations(points,2):
x1,y1=p1
x2,y2=p2
g=math.gcd(y2-y1,x2-x1)
if y2-y1<0:
sign=-1
elif y2-y1==0:
if x2-x1<0:
sign=-1
else:
sign=1
else:
sign=1
slope=(sign*(y2-y1)//g,sign*(x2-x1)//g)
if not slope[1]==0:
ceptn=slope[1]*y1-slope[0]*x1
ceptd=slope[1]
if ceptn<0:
sign=-1
elif ceptn==0:
if ceptd<0:
sign=-1
else:
sign=1
else:
sign=1
g=math.gcd(ceptn,ceptd)
cept=(sign*ceptn/g,sign*ceptd/g)
else:
cept=x1
if not cept in slopedict[slope]:
slopedict[slope].add(cept)
lines+=1
total=0
for slope in slopedict:
total=total+(lines-len(slopedict[slope]))*len(slopedict[slope])
print(total//2)
pos=math.gcd(-4,2)
neg=math.gcd(4,-2)
#print(pos,neg)
#print((-4//pos,2//pos))
#print((4//neg,-2//neg))
``` | output | 1 | 71,712 | 23 | 143,425 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem is same as the previous one, but has larger constraints.
It was a Sunday morning when the three friends Selena, Shiro and Katie decided to have a trip to the nearby power station (do not try this at home). After arriving at the power station, the cats got impressed with a large power transmission system consisting of many chimneys, electric poles, and wires. Since they are cats, they found those things gigantic.
At the entrance of the station, there is a map describing the complicated wiring system. Selena is the best at math among three friends. He decided to draw the map on the Cartesian plane. Each pole is now a point at some coordinates (x_i, y_i). Since every pole is different, all of the points representing these poles are distinct. Also, every two poles are connected with each other by wires. A wire is a straight line on the plane infinite in both directions. If there are more than two poles lying on the same line, they are connected by a single common wire.
Selena thinks, that whenever two different electric wires intersect, they may interfere with each other and cause damage. So he wonders, how many pairs are intersecting? Could you help him with this problem?
Input
The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of electric poles.
Each of the following n lines contains two integers x_i, y_i (-10^4 ≤ x_i, y_i ≤ 10^4) — the coordinates of the poles.
It is guaranteed that all of these n points are distinct.
Output
Print a single integer — the number of pairs of wires that are intersecting.
Examples
Input
4
0 0
1 1
0 3
1 2
Output
14
Input
4
0 0
0 2
0 4
2 0
Output
6
Input
3
-1 -1
1 0
3 1
Output
0
Note
In the first example:
<image>
In the second example:
<image>
Note that the three poles (0, 0), (0, 2) and (0, 4) are connected by a single wire.
In the third example:
<image> | instruction | 0 | 71,714 | 23 | 143,428 |
Tags: data structures, geometry, implementation, math
Correct Solution:
```
from fractions import Fraction
VERT = object()
n = int(input())
points = [tuple(map(int, input().split())) for _ in range(n)]
d = {}
for i, p in enumerate(points):
x1, y1 = p
for x2, y2 in points[i + 1:]:
if x1 < x2:
f = Fraction(y2 - y1, x2 - x1)
elif x1 > x2:
f = Fraction(y1 - y2, x1 - x2)
else:
f = VERT
if f == VERT:
x_inter = x1
else:
x_inter = y1 - x1 * f
# if f == VERT:
# print(f'VERT {x_inter}')
# else:
# print(f'{f.numerator}/{f.denominator} {x_inter}')
if f == VERT:
key = VERT
value = x_inter
else:
key = f.numerator / f.denominator
value = x_inter.numerator / x_inter.denominator
if key in d:
d[key].add(x_inter)
else:
d[key] = {x_inter}
total = sum(len(s) for s in d.values())
total_inter = 0
for s in d.values():
total_inter += len(s) * (total - len(s))
print(total_inter // 2)
``` | output | 1 | 71,714 | 23 | 143,429 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This problem is same as the previous one, but has larger constraints.
It was a Sunday morning when the three friends Selena, Shiro and Katie decided to have a trip to the nearby power station (do not try this at home). After arriving at the power station, the cats got impressed with a large power transmission system consisting of many chimneys, electric poles, and wires. Since they are cats, they found those things gigantic.
At the entrance of the station, there is a map describing the complicated wiring system. Selena is the best at math among three friends. He decided to draw the map on the Cartesian plane. Each pole is now a point at some coordinates (x_i, y_i). Since every pole is different, all of the points representing these poles are distinct. Also, every two poles are connected with each other by wires. A wire is a straight line on the plane infinite in both directions. If there are more than two poles lying on the same line, they are connected by a single common wire.
Selena thinks, that whenever two different electric wires intersect, they may interfere with each other and cause damage. So he wonders, how many pairs are intersecting? Could you help him with this problem?
Input
The first line contains a single integer n (2 ≤ n ≤ 1000) — the number of electric poles.
Each of the following n lines contains two integers x_i, y_i (-10^4 ≤ x_i, y_i ≤ 10^4) — the coordinates of the poles.
It is guaranteed that all of these n points are distinct.
Output
Print a single integer — the number of pairs of wires that are intersecting.
Examples
Input
4
0 0
1 1
0 3
1 2
Output
14
Input
4
0 0
0 2
0 4
2 0
Output
6
Input
3
-1 -1
1 0
3 1
Output
0
Note
In the first example:
<image>
In the second example:
<image>
Note that the three poles (0, 0), (0, 2) and (0, 4) are connected by a single wire.
In the third example:
<image> | instruction | 0 | 71,717 | 23 | 143,434 |
Tags: data structures, geometry, implementation, math
Correct Solution:
```
from collections import defaultdict
import math
n = int(input().lstrip())
numbers = []
for _ in range(n):
numbers.append(list(map(int, input().lstrip().split())))
def gcd(a, b):
if not a:
return b
return gcd(b % a, a)
slope_map = defaultdict(set)
total = 0
res = 0
for i in range(n):
for j in range(i + 1, n):
x1, y1 = numbers[i]
x2, y2 = numbers[j]
a = y1 - y2
b = x1 - x2
d = gcd(a, b)
a //= d
b //= d
if a < 0 or (not a and b < 0):
a *= -1
b *= -1
slope = (a, b)
c = a * x1 - b * y1
if c not in slope_map[slope]:
total += 1
slope_map[slope].add(c)
res += total - len(slope_map[slope])
print(res)
``` | output | 1 | 71,717 | 23 | 143,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a convex polygon. Count, please, the number of triangles that contain a given point in the plane and their vertices are the vertices of the polygon. It is guaranteed, that the point doesn't lie on the sides and the diagonals of the polygon.
Input
The first line contains integer n — the number of vertices of the polygon (3 ≤ n ≤ 100000). The polygon description is following: n lines containing coordinates of the vertices in clockwise order (integer x and y not greater than 109 by absolute value). It is guaranteed that the given polygon is nondegenerate and convex (no three points lie on the same line).
The next line contains integer t (1 ≤ t ≤ 20) — the number of points which you should count the answer for. It is followed by t lines with coordinates of the points (integer x and y not greater than 109 by absolute value).
Output
The output should contain t integer numbers, each on a separate line, where i-th number is the answer for the i-th point.
Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cin (also you may use %I64d).
Examples
Input
4
5 0
0 0
0 5
5 5
1
1 3
Output
2
Input
3
0 0
0 5
5 0
2
1 1
10 10
Output
1
0
Input
5
7 6
6 3
4 1
1 2
2 4
4
3 3
2 3
5 5
4 2
Output
5
3
3
4
Submitted Solution:
```
import math
def cross( a , b ):
return a[ 0 ] * b[ 1 ] - a[ 1 ] * b[ 0 ]
def dot( a , b ):
return a[ 0 ] * b[ 0 ] + a[ 1 ] * b[ 1 ]
def ang( a ):
return math.atan2( a[ 1 ] , a[ 0 ] )
def dlt( a , b ):
c = []
c.append( a[ 0 ] - b[ 0 ] )
c.append( a[ 1 ] - b[ 1 ] )
return c
n, a = int( input() ), []
for i in range( n ):
c = input().split()
c[ 0 ], c[ 1 ] = int( c[ 0 ] ), int( c[ 1 ] )
a.append( c )
q = int( input() )
while q > 0:
q -= 1
ra = []
x = input().split()
x[ 0 ], x[ 1 ] = int( x[ 0 ] ), int( x[ 1 ] )
for i in range( n ):
ra.append( dlt( a[ i ] , x ) )
ra = sorted( ra , key = ang )
for i in range( n ): ra.append( ra[ i ] )
for i in range( n ): ra.append( ra[ i ] )
ans, tans = n * ( n - 1 ) * ( n - 2 ) // 6, 0
ptr1 = 1
for i in range( n ):
ptr1 = max( ptr1 , i + 1 )
while ptr1 < i + n and cross( ra[ i ] , ra[ ptr1 ] ) > 0: ptr1 += 1
lft = ptr1 - i - 1
tans += lft * ( lft - 1 ) // 2
print( ans - tans )
``` | instruction | 0 | 72,066 | 23 | 144,132 |
No | output | 1 | 72,066 | 23 | 144,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
n evenly spaced points have been marked around the edge of a circle. There is a number written at each point. You choose a positive real number k. Then you may repeatedly select a set of 2 or more points which are evenly spaced, and either increase all numbers at points in the set by k or decrease all numbers at points in the set by k. You would like to eventually end up with all numbers equal to 0. Is it possible?
A set of 2 points is considered evenly spaced if they are diametrically opposed, and a set of 3 or more points is considered evenly spaced if they form a regular polygon.
Input
The first line of input contains an integer n (3 ≤ n ≤ 100000), the number of points along the circle.
The following line contains a string s with exactly n digits, indicating the numbers initially present at each of the points, in clockwise order.
Output
Print "YES" (without quotes) if there is some sequence of operations that results in all numbers being 0, otherwise "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
30
000100000100000110000000001100
Output
YES
Input
6
314159
Output
NO
Note
If we label the points from 1 to n, then for the first test case we can set k = 1. Then we increase the numbers at points 7 and 22 by 1, then decrease the numbers at points 7, 17, and 27 by 1, then decrease the numbers at points 4, 10, 16, 22, and 28 by 1. | instruction | 0 | 72,224 | 23 | 144,448 |
Tags: math
Correct Solution:
```
#Circle of Numbers
import math
def centre(n, pts):
x, y = 0, 0
for j in [7,11,13,17,19,23,29,31,37,1193,1663,2711,4007,65537]:
if math.gcd(n,j) == 1:
for i in range(n):
k = int(pts[i])
x += k*math.cos(math.pi * 2*i*j/n)
y += k*math.sin(math.pi * 2*i*j/n)
if not (abs(x) < 0.000001 and abs(y) < 0.000001):
return 'NO'
return 'YES'
def strconv(s):
return [char for char in s]
n = int(input())
pts = strconv(input())
print(centre(n,pts))
``` | output | 1 | 72,224 | 23 | 144,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n evenly spaced points have been marked around the edge of a circle. There is a number written at each point. You choose a positive real number k. Then you may repeatedly select a set of 2 or more points which are evenly spaced, and either increase all numbers at points in the set by k or decrease all numbers at points in the set by k. You would like to eventually end up with all numbers equal to 0. Is it possible?
A set of 2 points is considered evenly spaced if they are diametrically opposed, and a set of 3 or more points is considered evenly spaced if they form a regular polygon.
Input
The first line of input contains an integer n (3 ≤ n ≤ 100000), the number of points along the circle.
The following line contains a string s with exactly n digits, indicating the numbers initially present at each of the points, in clockwise order.
Output
Print "YES" (without quotes) if there is some sequence of operations that results in all numbers being 0, otherwise "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
30
000100000100000110000000001100
Output
YES
Input
6
314159
Output
NO
Note
If we label the points from 1 to n, then for the first test case we can set k = 1. Then we increase the numbers at points 7 and 22 by 1, then decrease the numbers at points 7, 17, and 27 by 1, then decrease the numbers at points 4, 10, 16, 22, and 28 by 1.
Submitted Solution:
```
#Circle of Numbers
import math
def centre(n, pts):
x, y = 0, 0
for j in [7,11,13,17,19,23,29,31,37,1193,2711]:
if math.gcd(n,j) == 1:
for i in range(n):
k = int(pts[i])
x += k*math.cos(math.pi * 2*i*j/n)
y += k*math.sin(math.pi * 2*i*j/n)
if not (abs(x) < 0.00001 and abs(y) < 0.00001):
return 'NO'
return 'YES'
def strconv(s):
return [char for char in s]
n = int(input())
pts = strconv(input())
print(centre(n,pts))
``` | instruction | 0 | 72,225 | 23 | 144,450 |
No | output | 1 | 72,225 | 23 | 144,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n evenly spaced points have been marked around the edge of a circle. There is a number written at each point. You choose a positive real number k. Then you may repeatedly select a set of 2 or more points which are evenly spaced, and either increase all numbers at points in the set by k or decrease all numbers at points in the set by k. You would like to eventually end up with all numbers equal to 0. Is it possible?
A set of 2 points is considered evenly spaced if they are diametrically opposed, and a set of 3 or more points is considered evenly spaced if they form a regular polygon.
Input
The first line of input contains an integer n (3 ≤ n ≤ 100000), the number of points along the circle.
The following line contains a string s with exactly n digits, indicating the numbers initially present at each of the points, in clockwise order.
Output
Print "YES" (without quotes) if there is some sequence of operations that results in all numbers being 0, otherwise "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
30
000100000100000110000000001100
Output
YES
Input
6
314159
Output
NO
Note
If we label the points from 1 to n, then for the first test case we can set k = 1. Then we increase the numbers at points 7 and 22 by 1, then decrease the numbers at points 7, 17, and 27 by 1, then decrease the numbers at points 4, 10, 16, 22, and 28 by 1.
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import random
def f():
if (random.randint(1, 2) == 1):
print("YES");
else:
print("NO")
n = int(input())
a = input()
if (a == "000100000100000110000000001100"):
print("YES");
exit();
if (a == "314159"):
print("NO");
exit();
f()
## KEK
``` | instruction | 0 | 72,226 | 23 | 144,452 |
No | output | 1 | 72,226 | 23 | 144,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n evenly spaced points have been marked around the edge of a circle. There is a number written at each point. You choose a positive real number k. Then you may repeatedly select a set of 2 or more points which are evenly spaced, and either increase all numbers at points in the set by k or decrease all numbers at points in the set by k. You would like to eventually end up with all numbers equal to 0. Is it possible?
A set of 2 points is considered evenly spaced if they are diametrically opposed, and a set of 3 or more points is considered evenly spaced if they form a regular polygon.
Input
The first line of input contains an integer n (3 ≤ n ≤ 100000), the number of points along the circle.
The following line contains a string s with exactly n digits, indicating the numbers initially present at each of the points, in clockwise order.
Output
Print "YES" (without quotes) if there is some sequence of operations that results in all numbers being 0, otherwise "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
30
000100000100000110000000001100
Output
YES
Input
6
314159
Output
NO
Note
If we label the points from 1 to n, then for the first test case we can set k = 1. Then we increase the numbers at points 7 and 22 by 1, then decrease the numbers at points 7, 17, and 27 by 1, then decrease the numbers at points 4, 10, 16, 22, and 28 by 1.
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import random
def f():
if (random.randint(1, 3) == 1):
print("NO");
else:
print("YES")
n = int(input())
a = input()
if (a == "000100000100000110000000001100"):
print("YES");
exit();
if (a == "314159"):
print("NO");
exit();
f()
``` | instruction | 0 | 72,227 | 23 | 144,454 |
No | output | 1 | 72,227 | 23 | 144,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
n evenly spaced points have been marked around the edge of a circle. There is a number written at each point. You choose a positive real number k. Then you may repeatedly select a set of 2 or more points which are evenly spaced, and either increase all numbers at points in the set by k or decrease all numbers at points in the set by k. You would like to eventually end up with all numbers equal to 0. Is it possible?
A set of 2 points is considered evenly spaced if they are diametrically opposed, and a set of 3 or more points is considered evenly spaced if they form a regular polygon.
Input
The first line of input contains an integer n (3 ≤ n ≤ 100000), the number of points along the circle.
The following line contains a string s with exactly n digits, indicating the numbers initially present at each of the points, in clockwise order.
Output
Print "YES" (without quotes) if there is some sequence of operations that results in all numbers being 0, otherwise "NO" (without quotes).
You can print each letter in any case (upper or lower).
Examples
Input
30
000100000100000110000000001100
Output
YES
Input
6
314159
Output
NO
Note
If we label the points from 1 to n, then for the first test case we can set k = 1. Then we increase the numbers at points 7 and 22 by 1, then decrease the numbers at points 7, 17, and 27 by 1, then decrease the numbers at points 4, 10, 16, 22, and 28 by 1.
Submitted Solution:
```
import sys
n = input()
n = int(n)
nums = input()
numbers = list(map(int, nums))
deviders = []
for i in range(1, n//2):
if n%i == 0:
deviders.append(i)
deviders.append(n)
for dev in deviders:
print ("Dev = ", dev)
for i in range(0, dev):
index = i
while index < len(numbers):
if numbers[index] == 0:
break
else:
x = -numbers[index]
print ("Index =", index,
"Number =", numbers[index], "X = ", x)
while index < len(numbers):
numbers[index] += x
index += dev
break
index += dev
print ('Numbers =', numbers)
for num in numbers:
if not num == 0:
print("NO")
sys.exit()
else:
print("YES")
``` | instruction | 0 | 72,228 | 23 | 144,456 |
No | output | 1 | 72,228 | 23 | 144,457 |
Provide a correct Python 3 solution for this coding contest problem.
A crop circle suddenly appeared on the vast agricultural land of Argentina. A total of n crop circles were confirmed, including overlapping, popping, large, and small ones.
When a mystery hunter tried to capture the whole picture of a crop circle from the air, he found it difficult to show the beautiful pattern in the image because the outline of each circle was not clear.
Therefore, the mystery hunter proposed to emphasize the contour by installing a string of special material along the contour.
The part where the string is installed is the part on the circumference of one crop circle that is not included in any other crop circle.
The film crew asked you to create a program to measure the required string length. Enter the center coordinates and radius of each crop circle and create a program to report the length of the string to be installed.
For reference, the first and second cases of the input / output examples are shown in FIGS. 1 and 2, respectively. The part of the string is shown by a thick line.
<image>
Figure 1
<image>
Figure 2
Constraints
* n ≤ 100
* -1000 ≤ xi, yi ≤ 1000
* 0 <ri ≤ 100
Input
Multiple datasets are given as input. Each dataset is given in the following format:
n (number of crop circles: integer)
x1 y1 r1 (center coordinates and radius of the first circle: real numbers separated by blanks)
x2 y2 r2 (center coordinates and radius of the second circle: real numbers separated by blanks)
..
..
xn yn rn (center coordinates and radius of the nth circle: real numbers separated by blanks)
When n is 0, it indicates the end of input.
Output
Output the string length on one line for each dataset. The output may contain an error of 0.000001 or less.
Example
Input
4
6 4 2
4 7 3
7 10 3
13 6 1
4
7 4 2
4 7 3
7 10 3
11 6 3
0
Output
39.664699289572
45.627024663706 | instruction | 0 | 72,444 | 23 | 144,888 |
"Correct Solution:
```
def string_to_float(s):
return list(map(float, s.split()))
def solve():
from sys import stdin
lines = stdin.readlines()
from itertools import combinations
from math import acos, pi
from cmath import phase
from bisect import insort
while True:
n = int(lines[0])
if n == 0:
break
circles = enumerate(map(string_to_float, lines[1:1+n]))
cross_data = [[] for i in range(n)]
included = [False] * n
r_sum = sum(map(lambda x: float(x.split()[2]), lines[1:1+n]))
ans = 2 * pi * r_sum
for c1, c2 in combinations(circles, 2):
n1, xyr1 = c1
n2, xyr2 = c2
if included[n1] or included[n2]:
continue
x1, y1, r1 = xyr1
x2, y2, r2 = xyr2
cn1 = x1 + y1 * 1j
cn2 = x2 + y2 * 1j
v12 = cn2 - cn1
d = abs(v12)
if d >= r1 + r2:
continue
elif d <= abs(r1 - r2):
if r1 < r2:
ans -= 2 * pi * r1
included[n1] = True
else:
ans -= 2 * pi * r2
included[n2] = True
continue
a = acos((r1 ** 2 + d ** 2 - r2 ** 2) / (2 * r1 * d))
t = phase(v12)
s = t - a
e = t + a
cd = cross_data[n1]
if s >= -pi and e <= pi:
insort(cd, (s, -1))
insort(cd, (e, 1))
elif s < -pi:
insort(cd, (s + 2 * pi, -1))
insort(cd, (pi, 1))
insort(cd, (-pi, -1))
insort(cd, (e, 1))
else:
insort(cd, (s, -1))
insort(cd, (pi, 1))
insort(cd, (-pi, -1))
insort(cd, (e - 2 * pi, 1))
a = acos((r2 ** 2 + d ** 2 - r1 ** 2) / (2 * r2 * d))
t = phase(-v12)
s = t - a
e = t + a
cd = cross_data[n2]
if s >= -pi and e <= pi:
insort(cd, (s, -1))
insort(cd, (e, 1))
elif s < -pi:
insort(cd, (s + 2 * pi, -1))
insort(cd, (pi, 1))
insort(cd, (-pi, -1))
insort(cd, (e, 1))
else:
insort(cd, (s, -1))
insort(cd, (pi, 1))
insort(cd, (-pi, -1))
insort(cd, (e - 2 * pi, 1))
radius = map(lambda x: float(x.split()[2]), lines[1:1+n])
flag = 0
for cd, r in zip(cross_data, radius):
for ang, f in cd:
if flag == 0:
s = ang
flag += f
if flag == 0:
ans -= r * (ang - s)
print(ans)
del lines[:1+n]
solve()
``` | output | 1 | 72,444 | 23 | 144,889 |
Provide a correct Python 3 solution for this coding contest problem.
F: Grid number
problem
Ebi-chan is trying to write exactly one integer from 1 to 2 \ times n in a grid with n columns horizontally and 2 rows vertically.
Only one integer can be written to each cell in the grid.
It's not fun just to write normally, so I set the following rules.
* The absolute value of the difference between the integers written in two adjacent cells is less than or equal to k.
* When there is a cell to the right of a cell, the integer written to the cell is truly smaller than the integer written to the cell to the right.
* When there is a cell under a cell, the integer written in the cell is truly smaller than the integer written in the cell below.
Here, two adjacent squares represent squares that share the top, bottom, left, and right sides with a certain square.
How many ways to write this? The answer can be very large, so print the remainder after dividing by the prime number m.
Supplement
The second and third rules above require you to write an integer so that it doesn't violate the inequality sign, as shown in the figure below.
<image>
Input format
You will be given three integers as input.
n k m
Constraint
* 1 \ leq n \ leq 100
* 1 \ leq k \ leq 10
* 2 \ leq m \ leq 10 ^ 9 + 7
* m is a prime number
Output format
Please output the number on one line according to how to write the number. Also, be careful not to forget the trailing line break.
Input example 1
3 2 7
Output example 1
1
<image>
* The above writing method meets the conditions, and there is no other writing method that meets the conditions.
Input example 2
5 10 11
Output example 2
9
* Output the remainder divided by m.
Example
Input
3 2 7
Output
1 | instruction | 0 | 72,459 | 23 | 144,918 |
"Correct Solution:
```
from collections import Counter
N, K, mod = map(int, input().split())
dp = [Counter() for _ in range(2*N)]
dp[0][(None, (0,))] = 1
for i in range(2*N-1):
for (left, top), v in dp[i].items():
if len(top) > K+1:
continue
if left == None:
if top and i+1 - top[0] <= K:
dp[i+1][(i+1, top)] = (dp[i+1][(i+1, top)] + v) % mod
elif (i+1 - left <= K) and len(top) > 1 and (i+1 - top[1] <= K):
dp[i+1][(i+1, top[1:])] = (dp[i+1][(i+1, top[1:])] + v) % mod
if top and (i+1 - top[-1] <= K):
top = list(top)
top.append(i+1)
top = tuple(top)
dp[i+1][(left, top)] = (dp[i+1][(left, top)] + v) % mod
res = 0
for (left, top), v in dp[2*N-1].items():
if len(top) == 1:
assert left == 2*N - 1
res = (res + v) % mod
print(res)
``` | output | 1 | 72,459 | 23 | 144,919 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your math teacher gave you the following problem:
There are n segments on the x-axis, [l_1; r_1], [l_2; r_2], …, [l_n; r_n]. The segment [l; r] includes the bounds, i.e. it is a set of such x that l ≤ x ≤ r. The length of the segment [l; r] is equal to r - l.
Two segments [a; b] and [c; d] have a common point (intersect) if there exists x that a ≤ x ≤ b and c ≤ x ≤ d. For example, [2; 5] and [3; 10] have a common point, but [5; 6] and [1; 4] don't have.
You should add one segment, which has at least one common point with each of the given segments and as short as possible (i.e. has minimal length). The required segment can degenerate to be a point (i.e a segment with length zero). The added segment may or may not be among the given n segments.
In other words, you need to find a segment [a; b], such that [a; b] and every [l_i; r_i] have a common point for each i, and b-a is minimal.
Input
The first line contains integer number t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^{5}) — the number of segments. The following n lines contain segment descriptions: the i-th of them contains two integers l_i,r_i (1 ≤ l_i ≤ r_i ≤ 10^{9}).
The sum of all values n over all the test cases in the input doesn't exceed 10^5.
Output
For each test case, output one integer — the smallest possible length of the segment which has at least one common point with all given segments.
Example
Input
4
3
4 5
5 9
7 7
5
11 19
4 17
16 16
3 12
14 17
1
1 10
1
1 1
Output
2
4
0
0
Note
In the first test case of the example, we can choose the segment [5;7] as the answer. It is the shortest segment that has at least one common point with all given segments. | instruction | 0 | 72,587 | 23 | 145,174 |
Tags: math
Correct Solution:
```
from sys import stdin
t = int(stdin.readline())
for q in range(t):
n = int(stdin.readline())
m1, m2 = 10**9+1, 0
for i in range(n):
l, r = map(int, stdin.readline().split())
m1 = min(m1, r)
m2 = max(m2, l)
print(max(0, m2 - m1))
``` | output | 1 | 72,587 | 23 | 145,175 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your math teacher gave you the following problem:
There are n segments on the x-axis, [l_1; r_1], [l_2; r_2], …, [l_n; r_n]. The segment [l; r] includes the bounds, i.e. it is a set of such x that l ≤ x ≤ r. The length of the segment [l; r] is equal to r - l.
Two segments [a; b] and [c; d] have a common point (intersect) if there exists x that a ≤ x ≤ b and c ≤ x ≤ d. For example, [2; 5] and [3; 10] have a common point, but [5; 6] and [1; 4] don't have.
You should add one segment, which has at least one common point with each of the given segments and as short as possible (i.e. has minimal length). The required segment can degenerate to be a point (i.e a segment with length zero). The added segment may or may not be among the given n segments.
In other words, you need to find a segment [a; b], such that [a; b] and every [l_i; r_i] have a common point for each i, and b-a is minimal.
Input
The first line contains integer number t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^{5}) — the number of segments. The following n lines contain segment descriptions: the i-th of them contains two integers l_i,r_i (1 ≤ l_i ≤ r_i ≤ 10^{9}).
The sum of all values n over all the test cases in the input doesn't exceed 10^5.
Output
For each test case, output one integer — the smallest possible length of the segment which has at least one common point with all given segments.
Example
Input
4
3
4 5
5 9
7 7
5
11 19
4 17
16 16
3 12
14 17
1
1 10
1
1 1
Output
2
4
0
0
Note
In the first test case of the example, we can choose the segment [5;7] as the answer. It is the shortest segment that has at least one common point with all given segments. | instruction | 0 | 72,588 | 23 | 145,176 |
Tags: math
Correct Solution:
```
t = int(input())
otv = list()
for i in range(t):
x = 1
y = 1000000001
n = int(input())
for j in range(n):
a, b = map(int, input().split())
x = max(a, x)
y = min(b, y)
otv.append(max(0, x - y))
for i in otv:
print(i)
``` | output | 1 | 72,588 | 23 | 145,177 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your math teacher gave you the following problem:
There are n segments on the x-axis, [l_1; r_1], [l_2; r_2], …, [l_n; r_n]. The segment [l; r] includes the bounds, i.e. it is a set of such x that l ≤ x ≤ r. The length of the segment [l; r] is equal to r - l.
Two segments [a; b] and [c; d] have a common point (intersect) if there exists x that a ≤ x ≤ b and c ≤ x ≤ d. For example, [2; 5] and [3; 10] have a common point, but [5; 6] and [1; 4] don't have.
You should add one segment, which has at least one common point with each of the given segments and as short as possible (i.e. has minimal length). The required segment can degenerate to be a point (i.e a segment with length zero). The added segment may or may not be among the given n segments.
In other words, you need to find a segment [a; b], such that [a; b] and every [l_i; r_i] have a common point for each i, and b-a is minimal.
Input
The first line contains integer number t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^{5}) — the number of segments. The following n lines contain segment descriptions: the i-th of them contains two integers l_i,r_i (1 ≤ l_i ≤ r_i ≤ 10^{9}).
The sum of all values n over all the test cases in the input doesn't exceed 10^5.
Output
For each test case, output one integer — the smallest possible length of the segment which has at least one common point with all given segments.
Example
Input
4
3
4 5
5 9
7 7
5
11 19
4 17
16 16
3 12
14 17
1
1 10
1
1 1
Output
2
4
0
0
Note
In the first test case of the example, we can choose the segment [5;7] as the answer. It is the shortest segment that has at least one common point with all given segments. | instruction | 0 | 72,589 | 23 | 145,178 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
data = []
max_l = -1
min_r = 10 ** 10
for i in range(n):
l, r = map(int, input().split())
max_l = max(max_l, l)
min_r = min(min_r, r)
print(max(max_l - min_r, 0))
``` | output | 1 | 72,589 | 23 | 145,179 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your math teacher gave you the following problem:
There are n segments on the x-axis, [l_1; r_1], [l_2; r_2], …, [l_n; r_n]. The segment [l; r] includes the bounds, i.e. it is a set of such x that l ≤ x ≤ r. The length of the segment [l; r] is equal to r - l.
Two segments [a; b] and [c; d] have a common point (intersect) if there exists x that a ≤ x ≤ b and c ≤ x ≤ d. For example, [2; 5] and [3; 10] have a common point, but [5; 6] and [1; 4] don't have.
You should add one segment, which has at least one common point with each of the given segments and as short as possible (i.e. has minimal length). The required segment can degenerate to be a point (i.e a segment with length zero). The added segment may or may not be among the given n segments.
In other words, you need to find a segment [a; b], such that [a; b] and every [l_i; r_i] have a common point for each i, and b-a is minimal.
Input
The first line contains integer number t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^{5}) — the number of segments. The following n lines contain segment descriptions: the i-th of them contains two integers l_i,r_i (1 ≤ l_i ≤ r_i ≤ 10^{9}).
The sum of all values n over all the test cases in the input doesn't exceed 10^5.
Output
For each test case, output one integer — the smallest possible length of the segment which has at least one common point with all given segments.
Example
Input
4
3
4 5
5 9
7 7
5
11 19
4 17
16 16
3 12
14 17
1
1 10
1
1 1
Output
2
4
0
0
Note
In the first test case of the example, we can choose the segment [5;7] as the answer. It is the shortest segment that has at least one common point with all given segments. | instruction | 0 | 72,590 | 23 | 145,180 |
Tags: math
Correct Solution:
```
t = int(input())
for i in range(t):
n = int(input())
arr = []
for k in range(n):
arr.append(tuple(map(int, input().split())))
l_max, r_min = arr[0][1], arr[0][0]
for l, r in arr:
if l_max > r:
l_max = r
if r_min < l:
r_min = l
print(max(0, r_min - l_max))
``` | output | 1 | 72,590 | 23 | 145,181 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your math teacher gave you the following problem:
There are n segments on the x-axis, [l_1; r_1], [l_2; r_2], …, [l_n; r_n]. The segment [l; r] includes the bounds, i.e. it is a set of such x that l ≤ x ≤ r. The length of the segment [l; r] is equal to r - l.
Two segments [a; b] and [c; d] have a common point (intersect) if there exists x that a ≤ x ≤ b and c ≤ x ≤ d. For example, [2; 5] and [3; 10] have a common point, but [5; 6] and [1; 4] don't have.
You should add one segment, which has at least one common point with each of the given segments and as short as possible (i.e. has minimal length). The required segment can degenerate to be a point (i.e a segment with length zero). The added segment may or may not be among the given n segments.
In other words, you need to find a segment [a; b], such that [a; b] and every [l_i; r_i] have a common point for each i, and b-a is minimal.
Input
The first line contains integer number t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^{5}) — the number of segments. The following n lines contain segment descriptions: the i-th of them contains two integers l_i,r_i (1 ≤ l_i ≤ r_i ≤ 10^{9}).
The sum of all values n over all the test cases in the input doesn't exceed 10^5.
Output
For each test case, output one integer — the smallest possible length of the segment which has at least one common point with all given segments.
Example
Input
4
3
4 5
5 9
7 7
5
11 19
4 17
16 16
3 12
14 17
1
1 10
1
1 1
Output
2
4
0
0
Note
In the first test case of the example, we can choose the segment [5;7] as the answer. It is the shortest segment that has at least one common point with all given segments. | instruction | 0 | 72,591 | 23 | 145,182 |
Tags: math
Correct Solution:
```
z = int(input())
i = 0
sp = []
while i != z:
n = int(input())
ch = input().split(' ')
xmax = int(ch[0])
ymin = int(ch[1])
while n != 1:
ch = input().split(' ')
x = int(ch[0])
if x > xmax:
xmax = x
y = int(ch[1])
if y < ymin:
ymin = y
n = n - 1
q = xmax - ymin
if q < 0:
q = 0
sp.append(q)
i = i + 1
i = 0
while i != z:
print (sp[i])
i = i + 1
``` | output | 1 | 72,591 | 23 | 145,183 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your math teacher gave you the following problem:
There are n segments on the x-axis, [l_1; r_1], [l_2; r_2], …, [l_n; r_n]. The segment [l; r] includes the bounds, i.e. it is a set of such x that l ≤ x ≤ r. The length of the segment [l; r] is equal to r - l.
Two segments [a; b] and [c; d] have a common point (intersect) if there exists x that a ≤ x ≤ b and c ≤ x ≤ d. For example, [2; 5] and [3; 10] have a common point, but [5; 6] and [1; 4] don't have.
You should add one segment, which has at least one common point with each of the given segments and as short as possible (i.e. has minimal length). The required segment can degenerate to be a point (i.e a segment with length zero). The added segment may or may not be among the given n segments.
In other words, you need to find a segment [a; b], such that [a; b] and every [l_i; r_i] have a common point for each i, and b-a is minimal.
Input
The first line contains integer number t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^{5}) — the number of segments. The following n lines contain segment descriptions: the i-th of them contains two integers l_i,r_i (1 ≤ l_i ≤ r_i ≤ 10^{9}).
The sum of all values n over all the test cases in the input doesn't exceed 10^5.
Output
For each test case, output one integer — the smallest possible length of the segment which has at least one common point with all given segments.
Example
Input
4
3
4 5
5 9
7 7
5
11 19
4 17
16 16
3 12
14 17
1
1 10
1
1 1
Output
2
4
0
0
Note
In the first test case of the example, we can choose the segment [5;7] as the answer. It is the shortest segment that has at least one common point with all given segments. | instruction | 0 | 72,592 | 23 | 145,184 |
Tags: math
Correct Solution:
```
for _ in " "*int(int(input())):
z=0;z1=10**9
for i in range(int(input())):a,b=map(int,input().split());z=max(a,z);z1=min(b,z1)
print(max(0,z-z1))
``` | output | 1 | 72,592 | 23 | 145,185 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Your math teacher gave you the following problem:
There are n segments on the x-axis, [l_1; r_1], [l_2; r_2], …, [l_n; r_n]. The segment [l; r] includes the bounds, i.e. it is a set of such x that l ≤ x ≤ r. The length of the segment [l; r] is equal to r - l.
Two segments [a; b] and [c; d] have a common point (intersect) if there exists x that a ≤ x ≤ b and c ≤ x ≤ d. For example, [2; 5] and [3; 10] have a common point, but [5; 6] and [1; 4] don't have.
You should add one segment, which has at least one common point with each of the given segments and as short as possible (i.e. has minimal length). The required segment can degenerate to be a point (i.e a segment with length zero). The added segment may or may not be among the given n segments.
In other words, you need to find a segment [a; b], such that [a; b] and every [l_i; r_i] have a common point for each i, and b-a is minimal.
Input
The first line contains integer number t (1 ≤ t ≤ 100) — the number of test cases in the input. Then t test cases follow.
The first line of each test case contains one integer n (1 ≤ n ≤ 10^{5}) — the number of segments. The following n lines contain segment descriptions: the i-th of them contains two integers l_i,r_i (1 ≤ l_i ≤ r_i ≤ 10^{9}).
The sum of all values n over all the test cases in the input doesn't exceed 10^5.
Output
For each test case, output one integer — the smallest possible length of the segment which has at least one common point with all given segments.
Example
Input
4
3
4 5
5 9
7 7
5
11 19
4 17
16 16
3 12
14 17
1
1 10
1
1 1
Output
2
4
0
0
Note
In the first test case of the example, we can choose the segment [5;7] as the answer. It is the shortest segment that has at least one common point with all given segments. | instruction | 0 | 72,593 | 23 | 145,186 |
Tags: math
Correct Solution:
```
for i in range(int(input())):
a = []
b = []
for q in range(int(input())):
k = input().split()
a.append(int(k[0]))
b.append(int(k[1]))
print(int((abs(max(a) - min(b)) + max(a) - min(b)) / 2))
``` | output | 1 | 72,593 | 23 | 145,187 |
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