text stringlengths 198 433k | conversation_id int64 0 109k |
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
Input
The input contains a single integer a (1 ≤ a ≤ 18257).
Output
Print a single integer output (1 ≤ output ≤ 2·109).
Examples
Input
2
Output
13
Submitted Solution:
```
n = int(input())
if(n%2 == 0):
print((12*(n-1)) + 1)
else:
print((12*n)+1)
```
No
| 5,300 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Vasya came up to the blackboard and wrote out n distinct integers from 1 to n in some order in a circle. Then he drew arcs to join the pairs of integers (a, b) (a ≠ b), that are either each other's immediate neighbors in the circle, or there is number c, such that a and с are immediate neighbors, and b and c are immediate neighbors. As you can easily deduce, in the end Vasya drew 2·n arcs.
For example, if the numbers are written in the circle in the order 1, 2, 3, 4, 5 (in the clockwise direction), then the arcs will join pairs of integers (1, 2), (2, 3), (3, 4), (4, 5), (5, 1), (1, 3), (2, 4), (3, 5), (4, 1) and (5, 2).
Much time has passed ever since, the numbers we wiped off the blackboard long ago, but recently Vasya has found a piece of paper with 2·n written pairs of integers that were joined with the arcs on the board. Vasya asks you to find the order of numbers in the circle by these pairs.
Input
The first line of the input contains a single integer n (5 ≤ n ≤ 105) that shows, how many numbers were written on the board. Next 2·n lines contain pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — the numbers that were connected by the arcs.
It is guaranteed that no pair of integers, connected by a arc, occurs in the input more than once. The pairs of numbers and the numbers in the pairs are given in the arbitrary order.
Output
If Vasya made a mistake somewhere and there isn't any way to place numbers from 1 to n on the circle according to the statement, then print a single number "-1" (without the quotes). Otherwise, print any suitable sequence of n distinct integers from 1 to n.
If there are multiple solutions, you are allowed to print any of them. Specifically, it doesn't matter which number you write first to describe the sequence of the order. It also doesn't matter whether you write out the numbers in the clockwise or counter-clockwise direction.
Examples
Input
5
1 2
2 3
3 4
4 5
5 1
1 3
2 4
3 5
4 1
5 2
Output
1 2 3 4 5
Input
6
5 6
4 3
5 3
2 4
6 1
3 1
6 2
2 5
1 4
3 6
1 2
4 5
Output
1 2 4 5 3 6
Tags: brute force, dfs and similar, implementation
Correct Solution:
```
#!/usr/local/bin/python3
from __future__ import print_function
import sys
DEBUG = '-d' in sys.argv
def debug(*args, **kwargs):
if DEBUG:
print(*args, file=sys.stderr, **kwargs)
return None
def main():
n = int(input())
cnt = [0] * (n + 1)
edge = []
for i in range(0, n + 1):
edge.append(set())
for i in range(0, 2 * n):
s, t = map(int, input().split())
edge[s].add(t)
edge[t].add(s)
cnt[s] += 1
cnt[t] += 1
c4 = 0
for i in range(1, n + 1):
if cnt[i] == 4:
c4 += 1
if c4 != n:
print(-1)
else:
for v2 in edge[1]:
for v3 in edge[1]:
if v2 in edge[v3]:
mark = [True] * (n + 1)
mark[1] = False
mark[v2] = False
res = [1, v2]
i = v3
try:
while True:
res.append(i)
mark[i] = False
if len(res) == n:
print(' '.join([str(x) for x in res]))
sys.exit(0)
for e in edge[i]:
if e != i and mark[e] and res[-2] in edge[e]:
i = e
break
if not mark[i]:
raise StopIteration
except StopIteration:
pass
print(-1)
if __name__ == '__main__':
main()
```
| 5,301 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Vasya came up to the blackboard and wrote out n distinct integers from 1 to n in some order in a circle. Then he drew arcs to join the pairs of integers (a, b) (a ≠ b), that are either each other's immediate neighbors in the circle, or there is number c, such that a and с are immediate neighbors, and b and c are immediate neighbors. As you can easily deduce, in the end Vasya drew 2·n arcs.
For example, if the numbers are written in the circle in the order 1, 2, 3, 4, 5 (in the clockwise direction), then the arcs will join pairs of integers (1, 2), (2, 3), (3, 4), (4, 5), (5, 1), (1, 3), (2, 4), (3, 5), (4, 1) and (5, 2).
Much time has passed ever since, the numbers we wiped off the blackboard long ago, but recently Vasya has found a piece of paper with 2·n written pairs of integers that were joined with the arcs on the board. Vasya asks you to find the order of numbers in the circle by these pairs.
Input
The first line of the input contains a single integer n (5 ≤ n ≤ 105) that shows, how many numbers were written on the board. Next 2·n lines contain pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — the numbers that were connected by the arcs.
It is guaranteed that no pair of integers, connected by a arc, occurs in the input more than once. The pairs of numbers and the numbers in the pairs are given in the arbitrary order.
Output
If Vasya made a mistake somewhere and there isn't any way to place numbers from 1 to n on the circle according to the statement, then print a single number "-1" (without the quotes). Otherwise, print any suitable sequence of n distinct integers from 1 to n.
If there are multiple solutions, you are allowed to print any of them. Specifically, it doesn't matter which number you write first to describe the sequence of the order. It also doesn't matter whether you write out the numbers in the clockwise or counter-clockwise direction.
Examples
Input
5
1 2
2 3
3 4
4 5
5 1
1 3
2 4
3 5
4 1
5 2
Output
1 2 3 4 5
Input
6
5 6
4 3
5 3
2 4
6 1
3 1
6 2
2 5
1 4
3 6
1 2
4 5
Output
1 2 4 5 3 6
Tags: brute force, dfs and similar, implementation
Correct Solution:
```
from sys import stdin
all_in = stdin.readlines()
n = int(all_in[0])
pairs = list(map(lambda x: tuple(map(int, x.split())), all_in[1:]))
if n == 5:
print(1, 2, 3, 4, 5)
exit()
neigs = {i: set() for i in range(1, n + 1)}
for (a, b) in pairs:
neigs[a].add(b)
neigs[b].add(a)
for el in neigs.values():
if len(el) != 4:
print(-1)
exit()
ans = [1]
used = {i: False for i in range(1, n + 1)}
used[1] = True
for i in range(n - 1):
el_ = ans[-1]
ne = neigs[el_]
for el in ne:
ne_ = neigs[el]
and_ = ne & ne_
if len(and_) == 2:
if i:
if ans[-2] not in and_:
continue
if not used[el]:
ans.append(el)
used[el] = True
break
if len(ans) < n:
print(-1)
exit()
print(' '.join(map(str, ans)))
```
| 5,302 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Vasya came up to the blackboard and wrote out n distinct integers from 1 to n in some order in a circle. Then he drew arcs to join the pairs of integers (a, b) (a ≠ b), that are either each other's immediate neighbors in the circle, or there is number c, such that a and с are immediate neighbors, and b and c are immediate neighbors. As you can easily deduce, in the end Vasya drew 2·n arcs.
For example, if the numbers are written in the circle in the order 1, 2, 3, 4, 5 (in the clockwise direction), then the arcs will join pairs of integers (1, 2), (2, 3), (3, 4), (4, 5), (5, 1), (1, 3), (2, 4), (3, 5), (4, 1) and (5, 2).
Much time has passed ever since, the numbers we wiped off the blackboard long ago, but recently Vasya has found a piece of paper with 2·n written pairs of integers that were joined with the arcs on the board. Vasya asks you to find the order of numbers in the circle by these pairs.
Input
The first line of the input contains a single integer n (5 ≤ n ≤ 105) that shows, how many numbers were written on the board. Next 2·n lines contain pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — the numbers that were connected by the arcs.
It is guaranteed that no pair of integers, connected by a arc, occurs in the input more than once. The pairs of numbers and the numbers in the pairs are given in the arbitrary order.
Output
If Vasya made a mistake somewhere and there isn't any way to place numbers from 1 to n on the circle according to the statement, then print a single number "-1" (without the quotes). Otherwise, print any suitable sequence of n distinct integers from 1 to n.
If there are multiple solutions, you are allowed to print any of them. Specifically, it doesn't matter which number you write first to describe the sequence of the order. It also doesn't matter whether you write out the numbers in the clockwise or counter-clockwise direction.
Examples
Input
5
1 2
2 3
3 4
4 5
5 1
1 3
2 4
3 5
4 1
5 2
Output
1 2 3 4 5
Input
6
5 6
4 3
5 3
2 4
6 1
3 1
6 2
2 5
1 4
3 6
1 2
4 5
Output
1 2 4 5 3 6
Submitted Solution:
```
#!/usr/local/bin/python3
from __future__ import print_function
import sys
DEBUG = '-d' in sys.argv
def debug(*args, **kwargs):
if DEBUG:
print(*args, file=sys.stderr, **kwargs)
return None
n = int(input())
cnt = [0] * (n+1)
edge = []
for i in range (0, n + 1):
edge.append(set())
for i in range(0, 2 * n):
s, t = map(int, input().split())
edge[s].add(t)
edge[t].add(s)
cnt[s] += 1
cnt[t] += 1
c4 = 0
for i in range(1, n+1):
if cnt[i] == 4:
c4 += 1
if c4 != n:
print(-1)
else:
def go(i, res, mark):
res.append(i)
mark[i] = False
if len(res) == n:
print(' '.join([str(x) for x in res]))
raise StopIteration
for e in edge[i]:
if e != i and mark[e] and res[-1] in edge[e]:
go(e, res, mark)
return
try:
for v2 in edge[1]:
for v3 in edge[1]:
if v2 in edge[v3]:
mark = [True] * (n + 1)
mark[1] = False
mark[v2] = False
go(v3, [1, v2], mark)
print(-1)
except StopIteration:
pass
```
No
| 5,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Vasya came up to the blackboard and wrote out n distinct integers from 1 to n in some order in a circle. Then he drew arcs to join the pairs of integers (a, b) (a ≠ b), that are either each other's immediate neighbors in the circle, or there is number c, such that a and с are immediate neighbors, and b and c are immediate neighbors. As you can easily deduce, in the end Vasya drew 2·n arcs.
For example, if the numbers are written in the circle in the order 1, 2, 3, 4, 5 (in the clockwise direction), then the arcs will join pairs of integers (1, 2), (2, 3), (3, 4), (4, 5), (5, 1), (1, 3), (2, 4), (3, 5), (4, 1) and (5, 2).
Much time has passed ever since, the numbers we wiped off the blackboard long ago, but recently Vasya has found a piece of paper with 2·n written pairs of integers that were joined with the arcs on the board. Vasya asks you to find the order of numbers in the circle by these pairs.
Input
The first line of the input contains a single integer n (5 ≤ n ≤ 105) that shows, how many numbers were written on the board. Next 2·n lines contain pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — the numbers that were connected by the arcs.
It is guaranteed that no pair of integers, connected by a arc, occurs in the input more than once. The pairs of numbers and the numbers in the pairs are given in the arbitrary order.
Output
If Vasya made a mistake somewhere and there isn't any way to place numbers from 1 to n on the circle according to the statement, then print a single number "-1" (without the quotes). Otherwise, print any suitable sequence of n distinct integers from 1 to n.
If there are multiple solutions, you are allowed to print any of them. Specifically, it doesn't matter which number you write first to describe the sequence of the order. It also doesn't matter whether you write out the numbers in the clockwise or counter-clockwise direction.
Examples
Input
5
1 2
2 3
3 4
4 5
5 1
1 3
2 4
3 5
4 1
5 2
Output
1 2 3 4 5
Input
6
5 6
4 3
5 3
2 4
6 1
3 1
6 2
2 5
1 4
3 6
1 2
4 5
Output
1 2 4 5 3 6
Submitted Solution:
```
input=__import__('sys').stdin.readline
n=int(input())
if n==5:exit(print(1,2,3,4,5))
ans=[1]
g=[set()for _ in range(n+1)]
vis=[1]*(n+1)
for _ in range(n*2):a,b=map(int,input().split());g[a].add(b);g[b].add(a)
for _ in range(n-1):
for i in g[ans[-1]]:
c=g[ans[-1]]&g[i]
if len(c)==2and not(_ and ans[-2]not in c)and vis[i]:ans.append(i);vis[i]=0;break
if len(ans)==n:print(*ans)
else:print(-1)
```
No
| 5,304 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Vasya came up to the blackboard and wrote out n distinct integers from 1 to n in some order in a circle. Then he drew arcs to join the pairs of integers (a, b) (a ≠ b), that are either each other's immediate neighbors in the circle, or there is number c, such that a and с are immediate neighbors, and b and c are immediate neighbors. As you can easily deduce, in the end Vasya drew 2·n arcs.
For example, if the numbers are written in the circle in the order 1, 2, 3, 4, 5 (in the clockwise direction), then the arcs will join pairs of integers (1, 2), (2, 3), (3, 4), (4, 5), (5, 1), (1, 3), (2, 4), (3, 5), (4, 1) and (5, 2).
Much time has passed ever since, the numbers we wiped off the blackboard long ago, but recently Vasya has found a piece of paper with 2·n written pairs of integers that were joined with the arcs on the board. Vasya asks you to find the order of numbers in the circle by these pairs.
Input
The first line of the input contains a single integer n (5 ≤ n ≤ 105) that shows, how many numbers were written on the board. Next 2·n lines contain pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — the numbers that were connected by the arcs.
It is guaranteed that no pair of integers, connected by a arc, occurs in the input more than once. The pairs of numbers and the numbers in the pairs are given in the arbitrary order.
Output
If Vasya made a mistake somewhere and there isn't any way to place numbers from 1 to n on the circle according to the statement, then print a single number "-1" (without the quotes). Otherwise, print any suitable sequence of n distinct integers from 1 to n.
If there are multiple solutions, you are allowed to print any of them. Specifically, it doesn't matter which number you write first to describe the sequence of the order. It also doesn't matter whether you write out the numbers in the clockwise or counter-clockwise direction.
Examples
Input
5
1 2
2 3
3 4
4 5
5 1
1 3
2 4
3 5
4 1
5 2
Output
1 2 3 4 5
Input
6
5 6
4 3
5 3
2 4
6 1
3 1
6 2
2 5
1 4
3 6
1 2
4 5
Output
1 2 4 5 3 6
Submitted Solution:
```
from sys import stdin, stdout, stderr
import sys
inints=lambda: [int(x) for x in stdin.readline().split()]
fastwrite=lambda s: stdout.write( str(s) + "\n" )
fasterr=lambda s: stderr.write( str(s) + "\n" )
irange=lambda x,y: range(x, y+1)
####
def adj(x):
d=[0]*(n+1)
for i in g[x]:
for j in g[i]:
if j in g[x] and j not in result:
d[j]+=1
return d.index(max(d))
fasterr("please input")
n,=inints()
g=[ [] for i in range(n+1) ]
result=[0]*n
for i in range(2*n):
a,b=inints()
g[a].append(b)
g[b].append(a)
if n==5:
for i in irange(1,5):
stdout.write(str(i)+" ")
sys.exit()
elif n==6:
disconnected=0
for i in irange(2,6):
if i not in g[1]:
disconnected=i
break
r=[i for i in irange(1,6)]
x=r.index(disconnected)
r[3],r[x]=r[x],r[3]
if r[1] not in g[r[2]]:
r[2],r[4]=r[4],r[2]
for i in r:
stdout.write(str(i)+" ")
sys.exit()
for i in range(n):
result[i]=adj( 1 if i==0 else result[i-1] )
for i in result:
stdout.write(str(i)+" ")
```
No
| 5,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day Vasya came up to the blackboard and wrote out n distinct integers from 1 to n in some order in a circle. Then he drew arcs to join the pairs of integers (a, b) (a ≠ b), that are either each other's immediate neighbors in the circle, or there is number c, such that a and с are immediate neighbors, and b and c are immediate neighbors. As you can easily deduce, in the end Vasya drew 2·n arcs.
For example, if the numbers are written in the circle in the order 1, 2, 3, 4, 5 (in the clockwise direction), then the arcs will join pairs of integers (1, 2), (2, 3), (3, 4), (4, 5), (5, 1), (1, 3), (2, 4), (3, 5), (4, 1) and (5, 2).
Much time has passed ever since, the numbers we wiped off the blackboard long ago, but recently Vasya has found a piece of paper with 2·n written pairs of integers that were joined with the arcs on the board. Vasya asks you to find the order of numbers in the circle by these pairs.
Input
The first line of the input contains a single integer n (5 ≤ n ≤ 105) that shows, how many numbers were written on the board. Next 2·n lines contain pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — the numbers that were connected by the arcs.
It is guaranteed that no pair of integers, connected by a arc, occurs in the input more than once. The pairs of numbers and the numbers in the pairs are given in the arbitrary order.
Output
If Vasya made a mistake somewhere and there isn't any way to place numbers from 1 to n on the circle according to the statement, then print a single number "-1" (without the quotes). Otherwise, print any suitable sequence of n distinct integers from 1 to n.
If there are multiple solutions, you are allowed to print any of them. Specifically, it doesn't matter which number you write first to describe the sequence of the order. It also doesn't matter whether you write out the numbers in the clockwise or counter-clockwise direction.
Examples
Input
5
1 2
2 3
3 4
4 5
5 1
1 3
2 4
3 5
4 1
5 2
Output
1 2 3 4 5
Input
6
5 6
4 3
5 3
2 4
6 1
3 1
6 2
2 5
1 4
3 6
1 2
4 5
Output
1 2 4 5 3 6
Submitted Solution:
```
n = int(input())
e = {x + 1: [] for x in range(n)}
for _ in range(n * 2):
s, r = list(map(int, input().split()))
e[s] += [r]
e[r] += [s]
variants = [[1]]
for _ in range(n - 1):
newVariants = []
for mas in variants:
lastE = mas[len(mas) - 1]
l = e[lastE][0]
r = e[lastE][1]
mas1 = mas
mas2 = mas
if not (l in mas1):
mas1 = mas + [l]
if not (r in mas2):
mas2 = mas + [r]
newVariants += [mas1]
newVariants += [mas2]
l1 = e[lastE][2]
r2 = e[lastE][3]
mas3 = mas
mas4 = mas
if not (l1 in mas3):
mas3 = mas + [l1]
if not (r2 in mas4):
mas6 = mas + [r2]
newVariants += [mas3]
newVariants += [mas4]
variants = newVariants
res = []
for mas in variants:
lastE = mas[len(mas) - 1]
if len(mas) == n and (1 in e[lastE]):
res = mas
if len(res) == n:
print(" ".join(str(x) for x in res))
else:
print(-1)
```
No
| 5,306 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Tags: constructive algorithms, math
Correct Solution:
```
n = int(input())
if n % 4 > 1:
print(-1)
exit()
a = [i for i in range(0, n+1)]
for i in range(1, n//2+1, 2):
p, q, r, s = i, i+1, n-i,n-i+1
a[p], a[q], a[r], a[s] = a[q], a[s], a[p], a[r]
def check(arr):
for i in range(1, n+1):
k = arr[i]
if arr[arr[k]] != n-k+1:
return False
return True
# print(check(a))
print(*a[1:])
```
| 5,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Tags: constructive algorithms, math
Correct Solution:
```
n=int(input())
if n%4==2 or n%4==3:
from sys import exit
print(-1);exit()
res,i=[0]*n,0
for i in range(0,n//2,2):
res[i],res[i+1]=i+2,n-i
res[n-i-1],res[n-i-2]=n-i-1,i+1
i+=2
if n%4==1:res[n//2]=n//2+1
print(*res)
```
| 5,308 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Tags: constructive algorithms, math
Correct Solution:
```
from math import *
n = int(input())
if n == 1:
print(1)
elif n % 4 in [2, 3]:
print(-1)
else:
ans = [0] * n
for i in range(n // 2):
if i & 1:
ans[i] = n - 2 * (i // 2)
else:
ans[i] = 2 + i
for i in range(1, (n // 2), 2):
ans[ans[i] - 1] = n - i
ans[n - i - 1] = n - (n - i)
if n % 4 < 3 and n % 4 > 0:
ans[n // 2] = ceil(n / 2)
elif n % 4 == 3:
ans[int(floor(n / 2))] = ans[int(floor(n / 2)) - 1] + 1
ans[int(ceil(n / 2))] = ans[int(ceil(n / 2)) - 1] - 2
print(*ans)
```
| 5,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Tags: constructive algorithms, math
Correct Solution:
```
n=int(input())
L=[0]*(n+1)
X=[False]*(n+1)
if(n%4!=0 and n%4!=1):
print(-1)
else:
for i in range(1,n+1):
if(X[i]):
continue
X[i]=True
X[n-i+1]=True
for j in range(i+1,n+1):
if(X[j]):
continue
X[j]=True
X[n-j+1]=True
L[i]=j
L[n-i+1]=n-j+1
L[j]=n-i+1
L[n-j+1]=i
break
if(n%4==1):
L[n//2+1]=n//2+1
for i in range(1,n):
print(L[i],end=" ")
print(L[n])
```
| 5,310 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Tags: constructive algorithms, math
Correct Solution:
```
n=int(input())
if(n%4>1):
print(-1)
else:
ans=[0]*(n+1)
i,j,a,b=1,n,1,n
while(i<j and a<=n and b>=1):
ans[i],ans[j]=a+1,b-1
ans[i+1],ans[j-1]=b,a
i+=2
j-=2
a+=2
b-=2
if(i==j):
ans[i]=a
for i in range(1,n+1):
print(ans[i],end=' ')
```
| 5,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Tags: constructive algorithms, math
Correct Solution:
```
n=int(input())
if (n//2)&1:
print(-1)
else:
ans=[0]*(n+1)
for i in range(1,(n//2)+1,2):
ans[i]=i+1
ans[i+1]=n-i+1
ans[n-i+1]=n-i
ans[n-i]=i
if n%2:
ans[(n//2)+1]=(n//2)+1
print(*ans[1:])
```
| 5,312 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Tags: constructive algorithms, math
Correct Solution:
```
'''
Created on
@author: linhz
'''
import sys
usedNum=0
n=int(input())
p=[0 for i in range(n+1)]
usedNum=0
if n%4==3 or n%4==2:
print(-1)
else:
i=1
j=n
a=1
b=n
while j>i:
p[i]=a+1
p[i+1]=b
p[j]=b-1
p[j-1]=a
i+=2
j-=2
a+=2
b-=2
if j==i:
p[i]=a
ans=""
for i in range(1,n+1):
ans+=str(p[i])+" "
print(ans)
```
| 5,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Tags: constructive algorithms, math
Correct Solution:
```
from itertools import permutations
from sys import stdin
def checkit(vector, upto=-1):
if upto == -1:
upto = len(vector)
for i in range(0, upto):
if vector[vector[i] - 1] != len(vector) - (i + 1) + 1:
return False
return True
def calculate(n):
numbers = list(range(1, n + 1))
result = [0] * n
for i in range(0, n):
if result[i] != 0: continue
if i > 0 and checkit(result, i): continue
expected = n - i
for v in numbers:
if v - 1 == i and expected != v:
continue
if v == expected:
result[v-1] = v
numbers.remove(v)
break
elif result[v - 1] == expected:
numbers.remove(v)
result[i] = v
break
elif result[v - 1] == 0:
assert expected in numbers
result[i] = v
result[v - 1] = expected
numbers.remove(v)
numbers.remove(expected)
break
return result
def calculate_v2(n):
result = [0] * n
first_sum = n + 2
second_sum = n
nf = n
i = 0
while nf > first_sum // 2:
result[i] = first_sum - nf
result[i + 1] = nf
nf -= 2
i += 2
if n % 2 == 1:
result[i] = i + 1
i = n - 1
while i > n // 2:
result[i] = i
result[i-1] = second_sum - i
i -= 2
return result
def main():
number = int(stdin.readline())
result = calculate_v2(number)
if not checkit(result):
print(-1)
else:
print(" ".join([str(v) for v in result]))
if __name__ == "__main__":
main()
```
| 5,314 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Submitted Solution:
```
n = int(input())
if n%4 > 1:
print(-1)
else:
a = [n+1>>1]*n
for i in range(n//4):
j = i*2
a[j], a[j+1], a[-2-j], a[-1-j] = j+2, n-j, j+1, n-1-j
print(' '.join(map(str, a)))
```
Yes
| 5,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Submitted Solution:
```
n=int(input())
if n%4>1:print(-1)
else:
ans=[i for i in range(1,n+1)]
for i in range(0,n//2,2):
ans[i]=i+2
ans[i+1]=n-i
ans[n-i-1]=n-i-1
ans[n-i-2]=i+1
print(*ans)
```
Yes
| 5,316 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Submitted Solution:
```
n = int(input())
if n % 4 > 1:
print(-1)
exit()
a = [i for i in range(0, n+1)]
for i in range(1, n//2+1, 2):
p, q, r, s = i, i+1, n-i,n-i+1
a[p], a[q], a[r], a[s] = a[q], a[s], a[p], a[r]
def check(arr):
for i in range(1, n+1):
k = arr[i]
if arr[arr[k]] != n-k+1:
return False
return True
# print(check(a))
print(*a[1:])
# Made By Mostafa_Khaled
```
Yes
| 5,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Submitted Solution:
```
n=int(input())
if n==1:
print (1)
exit()
if n%4>1:
print (-1)
exit()
ans=[-1]*n
left=n
start=n-2
nums=1
nume=n
while left>=4:
ans[start]=nums
ans[nums-1]=nums+1
ans[nums]=nume
ans[nume-1]=nume-1
start-=2
nums+=2
nume-=2
left-=4
# print (ans)
if left==1:
ans[start+1]=start+2
print (*ans)
```
Yes
| 5,318 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Submitted Solution:
```
import math
n = int(input())
ans = 1
s = [1, 1, 1]
for a in range(n-100, n-1):
for b in range(a+1, n):
m1 = a * b
for c in range(b+1, n+1):
if (m1 * c // math.gcd(c, m1) >= ans):
ans = m1 * c // math.gcd(a, m1)
s = [a, b, c]
print(ans)
# print(s)
```
No
| 5,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Submitted Solution:
```
import sys
from collections import deque
n = int(input())
if n % 4 == 2 or n % 4 == 3:
print('-1')
sys.exit(0)
arr = [None] * (n + 1)
qt = deque([i for i in range(1, n + 1)])
mark = set()
while qt:
while qt and qt[0] in mark:
qt.popleft()
if not qt:
break
a = qt.popleft()
while qt and qt[0] in mark:
qt.popleft()
if not qt:
break
b = qt.popleft()
for i in range(4):
mark.add(a)
mark.add(b)
arr[a] = b
arr[b] = n - a + 1
a = b
b = arr[b]
for i in range(1, n + 1):
if not arr[i]:
arr[i] = a
break
```
No
| 5,320 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
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")
##########################################################
#q.sort(key=lambda x:((x[1]-x[0]),-x[0]))
#from collections import Counter
#from fractions import Fraction
from bisect import bisect_left
from collections import Counter
#s=iter(input())
#from collections import deque
#ls=list(map(int,input().split()))
#for in range(m):
#for _ in range(int(input())):
#n=int(input())
# n,k=map(int,input().split())
# arr=list(map(int,input().split()))
#for i in range(int(input())):
#n=int(input())
#arr=sorted([(x,i) for i,x in enumerate(map(int,input().split()))])[::-1]
#print(arr)
#arr=sorted(list(map(int,input().split())))[::-1]
n=int(input())
if n==1:
print(1)
elif n==2:
print(-1)
else:
print(n-1,end=" ")
print(1,end=" ")
for i in range(2,n+1):
if i==n-1:
continue
print(i,end=" ")
```
No
| 5,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You have integer n. Find some lucky permutation p of size n.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the required permutation size.
Output
Print "-1" (without the quotes) if the lucky permutation p of size n doesn't exist.
Otherwise, print n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) after a space — the required permutation.
If there are multiple answers, you can print any of them.
Examples
Input
1
Output
1
Input
2
Output
-1
Input
4
Output
2 4 1 3
Input
5
Output
2 5 3 1 4
Submitted Solution:
```
from math import *
n = int(input())
if n == 1:
print(1)
elif n % 4 == 2:
print(-1)
else:
ans = [0] * n
for i in range(n // 2):
if i & 1:
ans[i] = n - 2 * (i // 2)
else:
ans[i] = 2 + i
for i in range(1, (n // 2), 2):
ans[ans[i] - 1] = n - i
ans[n - i - 1] = n - (n - i)
if n % 4 < 3 and n % 4 > 0:
ans[n // 2] = ceil(n / 2)
elif n % 4 == 3:
ans[int(floor(n / 2))] = ans[int(floor(n / 2)) - 1] + 1
ans[int(ceil(n / 2))] = ans[int(ceil(n / 2)) - 1] - 2
print(*ans)
```
No
| 5,322 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Tags: constructive algorithms, implementation
Correct Solution:
```
import sys
input=sys.stdin.readline
from collections import defaultdict as dc
from collections import Counter
from bisect import bisect_right, bisect_left
import math
from operator import itemgetter
from heapq import heapify, heappop, heappush
from queue import PriorityQueue as pq
n,k=map(int,input().split())
if k>=n*(n-1)//2:
print("no solution")
else:
for i in range(n):
print(0,i)
```
| 5,323 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Tags: constructive algorithms, implementation
Correct Solution:
```
n, k = map(int, input().split())
up = (int)(n * (n - 1) / 2);
if k >= up:
print ("no solution")
else:
for i in range (0, n):
print (0, i)
```
| 5,324 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Tags: constructive algorithms, implementation
Correct Solution:
```
n,k = map(int ,input().split())
if((n&1)==0) :
time = (n//2)*(n-1)
else :
time = ((n-1)//2)*n
if(k>=time):
print("no solution")
else :
for i in range(n):
print("0 "+str(i))
```
| 5,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Tags: constructive algorithms, implementation
Correct Solution:
```
n,k = map(int,input().split())
if n*(n-1) <= k*2:
print('no solution')
else:
x = 11
for i in range(n-1):
print(1,x)
x+=3
print(1,x-5)
```
| 5,326 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Tags: constructive algorithms, implementation
Correct Solution:
```
n, k = map(int, input().split())
if (k >= n * (n - 1) // 2):
print("no solution")
else:
for i in range(n):
print(0, i)
```
| 5,327 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Tags: constructive algorithms, implementation
Correct Solution:
```
n,k=map(int, input().split())
if(n*(n-1)//2 <=k):
print('no solution')
else:
for i in range(n):
print(0,i)
```
| 5,328 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Tags: constructive algorithms, implementation
Correct Solution:
```
n, k = map(int, input().split())
tot = 0
for i in range(n):
for j in range(i + 1, n):
tot += 1
if tot <= k:
print('no solution')
else:
for i in range(n):
print(1, i)
```
| 5,329 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Tags: constructive algorithms, implementation
Correct Solution:
```
n,k = map(int,input().split())
if n*(n-1) <= k*2:
print('no solution')
else:
for i in range(n):
print(0,i)
```
| 5,330 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
n,k=map(int,input().split())
if n*(n-1)/2<=k:
print("no solution")
else:
for i in range(0,n):print(0,i)
# Made By Mostafa_Khaled
```
Yes
| 5,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
n,k=list(map(int,input().split()))
if (n*n-n)//2<=k:
print("no solution")
else:
x=0
y=0
store=[]
count=0
store.append(str(x)+' '+str(y))
while len(store)<n:
y+=1
store.append(str(x)+' '+str(y))
for j in store:
print(j)
```
Yes
| 5,332 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
s=input().split()
n=int(s[0])
k=int(s[1])
ans=[]
a=0
for i in range(n):
for j in range(i+1,n):
a+=1
if(a<=k):
print("no solution")
else:
for i in range(-10**9,-10**9+n):
print("0 "+str(i))
```
Yes
| 5,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
n, k = map(int, input().split())
if k >= n * (n - 1) // 2:
print("no solution")
else:
for i in range(n):
print(i, i * (n + 1))
```
Yes
| 5,334 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
s=input().split()
n=int(s[0])
k=int(s[0])
ans=[]
a=0
for i in range(1,n+1):
for j in range(i+1,n+1):
a+=1
if(a<=k):
print("no solution")
else:
for i in range(2):
for j in range(n//2):
ans.append(str(i)+" "+str(j))
if(len(ans)<n):
ans.append("2 0")
for i in range(n):
print(ans[i])
```
No
| 5,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
s=input().split()
n=int(s[0])
k=int(s[0])
ans=[]
a=0
for i in range(n):
for j in range(i+1,n):
a+=1
if(a<k):
print("no solution")
else:
for i in range(2):
for j in range(n//2):
ans.append(str(i)+" "+str(j))
if(len(ans)<n):
ans.append("2 0")
for i in range(n):
print(ans[i])
```
No
| 5,336 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
s=input().split()
n=int(s[0])
k=int(s[0])
ans=[]
a=0
for i in range(1,n+1):
for j in range(i+1,n+1):
a+=1
if(a<k):
print("no solution")
else:
for i in range(2):
for j in range(n//2):
ans.append(str(i)+" "+str(j))
if(len(ans)<n):
ans.append("2 0")
for i in range(n):
print(ans[i])
```
No
| 5,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the follows. Given n points in the plane, find a pair of points between which the distance is minimized. Distance between (x1, y1) and (x2, y2) is <image>.
The pseudo code of the unexpected code is as follows:
input n
for i from 1 to n
input the i-th point's coordinates into p[i]
sort array p[] by increasing of x coordinate first and increasing of y coordinate second
d=INF //here INF is a number big enough
tot=0
for i from 1 to n
for j from (i+1) to n
++tot
if (p[j].x-p[i].x>=d) then break //notice that "break" is only to be
//out of the loop "for j"
d=min(d,distance(p[i],p[j]))
output d
Here, tot can be regarded as the running time of the code. Due to the fact that a computer can only run a limited number of operations per second, tot should not be more than k in order not to get Time Limit Exceeded.
You are a great hacker. Would you please help Tiny generate a test data and let the code get Time Limit Exceeded?
Input
A single line which contains two space-separated integers n and k (2 ≤ n ≤ 2000, 1 ≤ k ≤ 109).
Output
If there doesn't exist such a data which let the given code get TLE, print "no solution" (without quotes); else print n lines, and the i-th line contains two integers xi, yi (|xi|, |yi| ≤ 109) representing the coordinates of the i-th point.
The conditions below must be held:
* All the points must be distinct.
* |xi|, |yi| ≤ 109.
* After running the given code, the value of tot should be larger than k.
Examples
Input
4 3
Output
0 0
0 1
1 0
1 1
Input
2 100
Output
no solution
Submitted Solution:
```
n,k = map(int,input().split())
if n*(n-1) < k*2:
print('no solution')
else:
for i in range(1,n+1):
print(1,i)
```
No
| 5,338 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Tags: binary search, constructive algorithms, greedy
Correct Solution:
```
s = input()
n = int(input())
d = {}
r = 0
for a in s:
d.setdefault(a, 0)
d[a] += 1
if(d[a] > r):
r = d[a]
if (len(d) > n):
print(-1)
else:
l = 0
while r - l > 1:
k = (l + r) // 2
cur = 0
for x in d.values():
cur += (x+k-1) // k
if cur > n:
l = k
else:
r = k
print(r)
s = ''
for a in d.keys():
s += a * ((d[a] + r - 1) // r)
l=len(s)
s += 'a' * (n-len(s))
print(s)
```
| 5,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Tags: binary search, constructive algorithms, greedy
Correct Solution:
```
from math import ceil
p = {i: 0 for i in 'abcdefghijklmnopqrstuvwxyz'}
t = input()
for i in t: p[i] += 1
p = {i: p[i] for i in p if p[i] > 0}
n = int(input())
if len(p) > n: print(-1)
elif len(t) > n:
r = [[p[i], p[i], 1, i] for i in p]
for i in range(n - len(p)):
j = max(r)
j[2] += 1
j[0] = j[1] / j[2]
print(ceil(max(r)[0]))
print(''.join(j[3] * j[2] for j in r))
else: print('1\n' + t * (n // len(t)) + t[: n % len(t)])
```
| 5,340 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Tags: binary search, constructive algorithms, greedy
Correct Solution:
```
import sys
import os
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")
import math
s = input()
n = int(input())
d = {}
reserve = ''
unique = 0
max_occur = 1
all_letters = ''
for x in s:
if x in d:
d[x] += 1
if d[x] > max_occur:
max_occur = d[x]
else:
d[x] = 1
unique += 1
all_letters += x
if unique > n:
print(-1)
elif unique == n:
print(max_occur)
print(all_letters)
else:
l = 1
r = max_occur
ans = max_occur
check = 0
final_str = all_letters
check_str = ''
while l<=r:
mid = l + (r-l)//2
check = 0
check_str = ''
reserve = ''
for i in d.keys():
if d[i] > mid:
check += math.ceil(d[i]/mid)
check_str += i*math.ceil(d[i]/mid)
reserve += i*(d[i] - math.ceil(d[i]/mid))
else:
check += 1
check_str += i
reserve += i*(d[i]-1)
if check <= n:
final_str = '%s' % check_str
if len(final_str) < n:
final_str += reserve[:n-len(final_str)]
if len(final_str) < n:
final_str += 'a'*(n-len(final_str))
ans = mid
r = mid-1
else:
l = mid+1
print(ans)
print(final_str)
```
| 5,341 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Tags: binary search, constructive algorithms, greedy
Correct Solution:
```
import math
from sys import stdin
from math import ceil
if __name__ == '__main__':
s = input()
n = int(input())
dictionary = {}
for i in s:
if i in dictionary:
dictionary[i] = dictionary[i] + 1
else:
dictionary[i] = 1
if len(dictionary) > n:
print(-1)
else:
if len(s) < n:
print(1)
newS = s
else:
lengthS = len(s) // n
newLength = len(s)
while lengthS < newLength:
mid = (lengthS + newLength) // 2
total = 0
for i in dictionary:
total = total + ceil(dictionary[i] / mid)
if total > n:
lengthS = mid + 1
else:
newLength = mid
print(lengthS)
newS = ''
for i in dictionary:
for j in range(ceil(dictionary[i] / lengthS)):
newS = newS + i
for i in range(n - len(newS)):
newS = newS + s[0]
print(newS)
```
| 5,342 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Tags: binary search, constructive algorithms, greedy
Correct Solution:
```
from math import ceil
s = input ()
n = int (input ())
count = {}
for i in s:
if i in count:
count[i] += 1
else:
count[i] = 1
if len(count) > n:
print (-1)
else:
if len(s) < n:
print (1)
ret = s
else:
l,h = len (s)//n, len (s)
while (l < h):
m = (l+h) // 2
tot = 0
for i in count:
tot += ceil (count[i]/m)
if tot > n: l = m+1
else: h = m
print (l)
ret = ''
for i in count:
for j in range (ceil (count[i]/l)):
ret += i
for i in range (n-len(ret)):
ret += s[0]
print (ret)
```
| 5,343 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Tags: binary search, constructive algorithms, greedy
Correct Solution:
```
s = input()
n = int(input())
symb_cnt = {}
for c in s:
symb_cnt[c] = symb_cnt[c] + 1 if c in symb_cnt else 1
for cnt in range(1, len(s) + 1):
s1 = ""
for c in symb_cnt:
s1 += c * ((symb_cnt[c] + cnt - 1) // cnt)
if len(s1) <= n:
for i in range(n - len(s1)):
s1 += 'a'
print(cnt)
print(s1)
exit(0)
print(-1)
```
| 5,344 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Tags: binary search, constructive algorithms, greedy
Correct Solution:
```
from collections import Counter
def main():
s = input()
l = int(input())
d = Counter(s)
if len(d) > l:
print(-1)
return
lo = 0
hi = 10000
while lo + 1 < hi:
mid = (lo + hi) // 2
c = 0
for x in iter(d.values()):
c += (x + mid - 1) // mid
if c > l:
lo = mid
else:
hi = mid
print(hi)
ans = []
for x in iter(d.items()):
ans.append(x[0] * int(((x[1] + hi - 1) / hi)))
t = ''.join(ans)
if len(t) < l:
t += 'a' * (l - len(t))
print(t)
main()
```
| 5,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Tags: binary search, constructive algorithms, greedy
Correct Solution:
```
s = input()
n = int(input())
freq = [0 for i in range(0, 300)]
raport = [0 for i in range(0, 300)]
differentLetters = 0
tickets = 0
sol = ''
for c in s: freq[ord(c)] += 1
for i in freq:
if i > 0: differentLetters += 1
if differentLetters > n:
print('-1')
exit()
for i in 'abcdefghijklmnopqrstuvwxyz':
if freq[ord(i)] == 0: continue
sol += i
freq[ord(i)] -= 1
raport[ord(i)] = freq[ord(i)]
for i in range(differentLetters, n):
#pun litera cu cea mai mare frecventa
maxRaport = raport[ord('z')]
chosenLetter = 'z'
for j in 'abcdefghijklmnopqrstuvwxyz':
if raport[ord(j)] > maxRaport:
maxRaport = raport[ord(j)]
chosenLetter = j
sol += chosenLetter
raport[ord(chosenLetter)] = freq[ord(chosenLetter)] / sol.count(chosenLetter)
for i in sol:
a = s.count(i)
b = sol.count(i)
if a%b == 0: tickets = max(tickets, int(a//b))
else: tickets = max(tickets, int(a//b) + 1)
print(tickets)
print(sol)
```
| 5,346 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
from collections import Counter
from heapq import heappushpop
def main():
cnt, n = Counter(input()), int(input())
if n < len(cnt):
print(-1)
return
h = list((1 / v, 1, c) for c, v in cnt.most_common())
res = list(cnt.keys())
_, v, c = h.pop(0)
for _ in range(n - len(cnt)):
res.append(c)
v += 1
_, v, c = heappushpop(h, (v / cnt[c], v, c))
print((cnt[c] + v - 1) // v)
print(''.join(res))
if __name__ == '__main__':
main()
```
Yes
| 5,347 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
from collections import Counter
import heapq as pq
from math import ceil
def solve():
s = insr()
n = inp()
count_map = Counter(s)
if n < len(count_map):
print(-1)
return
copy_map = {el: 1 for el in count_map.keys()}
used = len(count_map)
q = [(-el, key) for key, el in count_map.items()]
pq.heapify(q)
while used < n and q:
copy, c = pq.heappop(q)
copy_map[c] += 1
copies = ceil(count_map[c] / copy_map[c])
if copies > 1:
pq.heappush(q, (-copies, c))
used += 1
if not q:
print(1)
else:
print(pq.heappop(q)[0] * -1)
res = []
for el, copies in copy_map.items():
res.extend([el] * copies)
if len(res) < n:
res.extend(['a'] * (n - len(res)))
print("".join(res))
def main():
t = 1
for i in range(t):
solve()
BUFSIZE = 8192
def inp():
return (int(input()))
def inlt():
return (list(map(int, input().split())))
def insr():
return (input().strip())
def invr():
return (map(int, input().split()))
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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
Yes
| 5,348 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
s = input()
n = int(input())
cnt = {}
for c in s:
if cnt.get(c) == None:
cnt[c] = 1
else:
cnt[c] += 1
if (n < len(cnt)):
print(-1)
else:
ansNum = 0;
while(True):
ansNum+=1
l = 0;
char = []
for c, v in cnt.items():
need = (v + ansNum -1)//ansNum
l+= need
char.append((c, need))
if (l > n):
continue
print(ansNum)
ans = "".join([str(c[0])*c[1] for c in char])
ans = ans + 'a'*(n - len(ans))
print(ans)
break
```
Yes
| 5,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
s = input()
n = int(input())
from collections import Counter
c = Counter(s)
out = Counter()
contrib = Counter()
for letter in c:
out[letter] = 1
contrib[letter] = c[letter]
sum_vals = sum(out.values())
from math import ceil
from fractions import Fraction
if sum_vals > n:
print(-1)
else:
while sum_vals < n:
el, _ = contrib.most_common(1)[0]
out[el] += 1
sum_vals += 1
contrib[el] = ceil(Fraction(c[el], out[el]))
print(max(contrib.values()))
print(''.join(out.elements()))
```
Yes
| 5,350 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
from operator import itemgetter
class CodeforcesTask335ASolution:
def __init__(self):
self.result = ''
self.s = ''
self.n = 0
def read_input(self):
self.s = input()
self.n = int(input())
def process_task(self):
letters = 'abcdefghijklmnopqrstuvwxyz'
occs = [[c, self.s.count(c)] for c in letters if self.s.count(c)]
occs.sort(key=itemgetter(1), reverse=True)
if len(occs) > self.n:
self.result = "-1"
else:
resulting = "".join([x[0] for x in occs])
while len(resulting) < self.n:
priorities = [[c, resulting.count(c)] for c in letters if resulting.count(c)]
priorities.sort(key=itemgetter(1), reverse=True)
lel = []
for x in range(len(priorities)):
lel.append([occs[x][0], occs[x][1] / priorities[x][1]])
#print(resulting, priorities, occs, lel)
#print(lel)
lel.sort(key=itemgetter(1), reverse=True)
resulting += lel[0][0]
priorities = [[c, resulting.count(c)] for c in letters if resulting.count(c)]
#print(priorities)
priorities.sort(key=itemgetter(1), reverse=True)
lel = []
for x in range(len(priorities)):
lel.append([occs[x][0], occs[x][1] / priorities[x][1]])
#print(lel)
self.result = "{0}\n{1}".format(int(max([x[1] for x in lel])), resulting)
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask335ASolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
```
No
| 5,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
from collections import Counter
import math
s = input()
n = int(input())
c = Counter(s)
if len(c) > n:
print(-1)
exit()
ans = ""
d = Counter()
for x in c:
ans += x
d[x] += 1
c.most_common()
# print(c)
for x in c:
if len(ans) < n:
t = min(n - len(ans), c[x] - 1)
ans += x * t
d[x] += t
else:
break
maxm = 1
for x in d:
maxm = max(c[x] / d[x], maxm)
print(int(math.ceil(maxm)))
print(ans)
```
No
| 5,352 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
s = input()
n = int(input())
d = {}
r = 0
for a in s:
d.setdefault(a, 0)
d[a] += 1
if(d[a] > r):
r=d[a]
if (len(d) > n):
print(-1)
else:
l = 0
while r - l > 1:
k = (l + r) // 2
cur = 0
for x in d.values():
cur += (x+k-1) // k
if cur > n:
l = k
else:
r = k
print(r)
s = ''
for a in d.keys():
s += a * ((d[a]+r-1) // r)
print(s)
```
No
| 5,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Piegirl is buying stickers for a project. Stickers come on sheets, and each sheet of stickers contains exactly n stickers. Each sticker has exactly one character printed on it, so a sheet of stickers can be described by a string of length n. Piegirl wants to create a string s using stickers. She may buy as many sheets of stickers as she wants, and may specify any string of length n for the sheets, but all the sheets must be identical, so the string is the same for all sheets. Once she attains the sheets of stickers, she will take some of the stickers from the sheets and arrange (in any order) them to form s. Determine the minimum number of sheets she has to buy, and provide a string describing a possible sheet of stickers she should buy.
Input
The first line contains string s (1 ≤ |s| ≤ 1000), consisting of lowercase English characters only. The second line contains an integer n (1 ≤ n ≤ 1000).
Output
On the first line, print the minimum number of sheets Piegirl has to buy. On the second line, print a string consisting of n lower case English characters. This string should describe a sheet of stickers that Piegirl can buy in order to minimize the number of sheets. If Piegirl cannot possibly form the string s, print instead a single line with the number -1.
Examples
Input
banana
4
Output
2
baan
Input
banana
3
Output
3
nab
Input
banana
2
Output
-1
Note
In the second example, Piegirl can order 3 sheets of stickers with the characters "nab". She can take characters "nab" from the first sheet, "na" from the second, and "a" from the third, and arrange them to from "banana".
Submitted Solution:
```
import math
S=input()
N=int(input())
Srep={}
ansrep={}
for item in "abcdefghijklmnopqrstuvwxyz":
Srep[item]=0
ansrep[item]=0
for item in S:
Srep[item]+=1
ansrep[item]+=1
Q=list(set(S))
if(len(Q)>N):
print(-1)
else:
n=len(Q)
ans=list(S)
num=1
req=1
n=len(ans)
while(len(ans)>N):
n=len(ans)
for item in ans:
if(ansrep[item]==1):
continue
if(math.ceil(Srep[item]/(ansrep[item]-1))>req):
continue
else:
ansrep[item]-=1
ans.remove(item)
break
if(n==len(ans)):
for item in ans:
if(ansrep[item]==1):
continue
ansrep[item]-=1
ans.remove(item)
req+=1
break
g=""
if(len(ans)<N):
g=S[0]*(N-len(ans))
print(req)
for item in ans:
print(item,end="")
print(g)
```
No
| 5,354 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
seq = list(map(int, input().split()))
if n == 1:
print("no")
exit(0)
pairs = []
cur = [-1, seq[0]]
fail = False
for i in range(1, len(seq)):
cur[0], cur[1] = cur[1], seq[i]
if cur[1] < cur[0]:
pairs.append([cur[1], cur[0]])
else:
pairs.append(cur[:])
#print(pairs)
for x1, x2 in pairs:
if not fail:
for x3, x4 in pairs:
if (x1 != x3 or x2 != x4) and (x1 < x3 < x2 < x4 or x3 < x1 < x4 < x2):
fail = True
break
else:
break
if fail:
print("yes")
else:
print("no")
```
| 5,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
Min = -1e10
Max = 1e10
ans = "no"
def f(x, y):
return x[0] < y[0] < x[1] < y[1] or \
y[0] < x[0] < y[1] < x[1]
xs = list(map(int, input().split()))
for i in range(1, len(xs) - 1):
for j in range(0, i):
if f([min(xs[i:i+2]), max(xs[i:i+2])], [min(xs[j:j+2]), max(xs[j:j+2])]):
ans = "yes"
break
print(ans)
```
| 5,356 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Tags: brute force, implementation
Correct Solution:
```
n,l=int(input()),list(map(int,input().split()))
for i in range(n-1):
for j in range(n-1):
x=sorted([l[i],l[i+1]])+sorted([l[j],l[j+1]])
if x[0]<x[2]<x[1]<x[3]:
exit(print('yes'))
print('no')
```
| 5,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Tags: brute force, implementation
Correct Solution:
```
def self_intersect(points:list):
for i in range(len(points)):
check_point = points[i]
for j in range(len(points)):
if i != j:
check_point2 = points[j]
if check_point[0] < check_point2[0] < check_point[1] < check_point2[1]:
return True
if check_point2[0] < check_point[0] < check_point2[1] < check_point[1]:
return True
return False
n = int(input())
points = list(map(int,input().split(' ')))
semi_circles = []
for i in range(0,len(points)-1):
if points[i] < points[i+1]:
semi_circles.append([points[i],points[i+1]])
else:
semi_circles.append([points[i+1],points[i]])
if self_intersect(semi_circles):
print("yes")
else:
print("no")
```
| 5,358 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split()))
flag = 0
for i in range(n-1):
for j in range(i+1,n-1):
a,b = min(arr[i],arr[i+1]),max(arr[i],arr[i+1])
c,d = min(arr[j],arr[j+1]),max(arr[j],arr[j+1])
if a<c<b<d or c<a<d<b:
flag=1
break
if flag==1:
break
if flag==0:
print("no")
else:
print("yes")
```
| 5,359 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Tags: brute force, implementation
Correct Solution:
```
def intersect(p1,p2):
if (p2[0] < p1[1] and p2[0] > p1[0] and p2[1] > p1[1]) or (p2[1] < p1[1] and p2[1] > p1[0] and p2[0] < p1[0]):
return True
return False
def check(points):
for x in range(len(points)):
for i in range(x + 1,len(points)):
if intersect(points[x],points[i]): return "yes"
return "no"
n = int(input())
temp_in = [int(x) for x in input().split(' ')]
points = []
for x in range(n - 1):
points.append((min(temp_in[x],temp_in[x + 1]),max(temp_in[x],temp_in[x + 1])))
#print(points)
print(check(points)) #chyba przypadek z tym ze wspolrzedne pozniej sa mniejsze i wtedy warunek nie wchodzi na przeciecie
'''
2
0 15 13 20
ans: yes
semicircles will intersect with each other in one case:
begininning of the second one will be earlier than ending of the first.
and there will be no subcircle (circle which is included in the one above)
brute force O(N^2+N / 2)
'''
```
| 5,360 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Tags: brute force, implementation
Correct Solution:
```
#http://codeforces.com/problemset/problem/358/A
#not accepted
n = eval(input())
points = [(int)(i) for i in input().split()]
poles = []
'''
left = 0
right = 0
prevDir = '-'
currentDir = '-'
'''
def is_intersect(points):
if(n <= 2):
return False;
global poles
poles.append([points[0],points[1]])
for i in range(2,n):
prevPoint = points[i-1]
currentPoint = points[i]
for pole in poles:
_min = min(pole[0],pole[1])
_max = max(pole[0],pole[1])
if (prevPoint in range(_min+1,_max) and\
currentPoint not in range(_min,_max+1)) or\
(currentPoint in range(_min+1,_max) and\
prevPoint not in range(_min,_max+1)):
return True
poles.append([prevPoint,currentPoint]);
return False
'''
global right
global left
global prevDir
global currentDir
left = min(points[0],points[1])
right = max(points[0],points[1])
prevDir = points[1] > points[0] if 'right' else 'left'
currentDir = '-'
for i in range(2,n):
prevPoint = points[i-1]
currentPoint = points[i]
currentDir = currentPoint > prevPoint if 'right' else 'left'
if prevDir == currentDir:
if currentDir == 'left':
right = prevPoint;
left = currentPoint;
elif currentDir == 'right':
right = currentPoint
left = prevPoint
else:
pass
prevDir = currentDir
return False
'''
if is_intersect(points):
print('yes')
else:
print('no')
```
| 5,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
x = [int(i) for i in input().split()]
if (n < 3):
print ('no')
else:
ans = 'no'
for i in range(3, n):
prev = x[i - 1]
num = x[i]
for j in range(i - 2):
num1 = min(x[j], x[j + 1])
num2 = max(x[j], x[j + 1])
if (prev < num1 or prev > num2) and (num > num1 and num < num2):
ans = 'yes'
break
elif ((num < num1 or num > num2) and (prev > num1 and prev < num2)):
ans = 'yes'
break
print (ans)
```
| 5,362 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
maxx=l[0]
minn=l[0]
for i in range(0,n-1):
if l[i+1]>l[i]:
if l[i]>minn and l[i]<maxx and l[i+1]>maxx:
print("yes")
exit()
elif l[i]<minn and l[i+1]>minn and l[i+1]<maxx :
print("yes")
exit()
else:
if l[i]>maxx:
maxx=l[i]
if l[i]<minn:
minn=l[i]
elif l[i+1]<l[i]:
if l[i+1]>minn and l[i+1]<maxx and l[i]>maxx:
print("yes")
exit()
elif l[i+1]<minn and l[i]>minn and l[i]<maxx:
print("yes")
exit()
else:
if l[i]>maxx:
maxx=l[i]
if l[i]<minn:
minn=l[i]
print("no")
```
Yes
| 5,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Submitted Solution:
```
'''
Created on Oct 31, 2013
@author: Ismael
'''
def main():
input()
l = list(map(int,input().split()))
#l = [0,10,5,15]
#l = [0,15,5,10]
if(len(l)<=3):
print("no")
return
xMin = min(l[0],l[1])
xMax = max(l[0],l[1])
for i in range(2,len(l)-1):
if(l[i]<=xMin or l[i]>=xMax):#3eme ext, next point must be outside [xMin,xMax]
if(l[i+1]>xMin and l[i+1]<xMax):
print("yes")
return
elif(l[i]>=xMin and l[i]<=xMax):#3eme int, next point must be inside [xMin,xMax]
if(l[i+1]<xMin or l[i+1]>xMax):
print("yes")
return
min123 = min(xMin,l[i])
max123 = max(xMax,l[i])
if(l[i+1]<min123 or l[i+1]>max123):
xMin = min123
xMax = max123
else:
elems = [l[i+1],xMin,xMax,l[i]]
elems.sort()
ind4 = elems.index(l[i+1])
xMin = elems[ind4-1]
xMax = elems[ind4+1]
print("no")
main()
```
Yes
| 5,364 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Submitted Solution:
```
n = int(input())
a = [int(y) for y in input().split()]
flag = 0
for i in range(len(a)-2):
for j in range(i, 0, -1):
r = min(a[j], a[j-1])
s = max(a[j], a[j-1])
if (a[i+2] > r and a[i+2] < s) and (a[i+1] > s or a[i+1] < r):
flag += 1
break
elif (a[i+1] > r and a[i+1] < s) and (a[i+2] > s or a[i+2] < r):
flag += 1
break
if(flag == 0):
print("no")
else:
print("yes")
```
Yes
| 5,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Submitted Solution:
```
def take_input(s): #for integer inputs
if s == 1: return int(input())
return map(int, input().split())
n = take_input(1)
if n == 1: print("no"); exit()
a = list(take_input(n))
flag = 0
for i in range(1,n-1):
for j in range(i):
init_1 = min(a[i],a[i+1]); fin_1 = max(a[i],a[i+1])
init_2 = min(a[j],a[j+1]); fin_2 = max(a[j],a[j+1])
if ((init_1 > init_2 and init_1 < fin_2 and fin_1 > fin_2) or (init_1 < init_2 and fin_1 > init_2 and fin_1<fin_2)) and flag == 0:
print("yes")
flag = 1
break
if flag == 0:
print("no")
```
Yes
| 5,366 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Submitted Solution:
```
n = int(input().strip())
points = list(map(int, input().strip().split()))
val = True
for i in range(n - 1):
if points[i] > points[i+1]:
if points[i] < max(points[i+1:]):
val = False
break
if val:
print("no")
else:
print("yes")
```
No
| 5,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Submitted Solution:
```
n = int(input())
pos = [int(k) for k in input().split()]
sigue = True
if len(pos)>2:
for i in range(len(pos)-2):
x1 = pos[i]
x2 = pos[i+1]
for k in range(i+2,len(pos)-1):
x3 = pos[k]
x4 = pos[k+1]
if x1<x3<x2<x4 or x3<x1<x4<x2 or x1<x4<x2<x3 or x4<x1<x3<x2:
print("yes")
sigue = False
break
if not sigue:
break
if sigue:
print("no")
```
No
| 5,368 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Submitted Solution:
```
import sys
import math
import itertools
import collections
def getdict(n):
d = {}
if type(n) is list or type(n) is str:
for i in n:
if i in d:
d[i] += 1
else:
d[i] = 1
else:
for i in range(n):
t = ii()
if t in d:
d[t] += 1
else:
d[t] = 1
return d
def divs(n, start=1):
r = []
for i in range(start, int(math.sqrt(n) + 1)):
if (n % i == 0):
if (n / i == i):
r.append(i)
else:
r.extend([i, n // i])
return r
def cdiv(n, k): return n // k + (n % k != 0)
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a*b) // math.gcd(a, b)
def wr(arr): return '\n'.join(map(str, arr))
def revn(n): return int(str(n)[::-1])
def prime(n):
if n == 2: return True
if n % 2 == 0 or n <= 1: return False
sqr = int(math.sqrt(n)) + 1
for d in range(3, sqr, 2):
if n % d == 0: return False
return True
n = ii()
x = li()
if n < 4:
print('yes')
else:
mnx = min(x)
mxx = max(x)
if x[0] == mnx and x[1] == mxx:
lp, rp = mnx, mxx
for i in range(2, n):
if i % 2 == 0:
if lp < x[i] and x[i] < rp:
lp = x[i]
else:
exit(print('yes'))
else:
if lp < x[i] and x[i] < rp:
rp = x[i]
else:
exit(print('yes'))
elif x[0] == mxx and x[1] == mnx:
lp, rp = mnx, mxx
for i in range(2, n):
if i % 2 == 0:
if lp < x[i] and x[i] < rp:
rp = x[i]
else:
exit(print('yes'))
else:
if lp < x[i] and x[i] < rp:
lp = x[i]
else:
exit(print('yes'))
elif x[1] < x[0]:
lp, rp = x[1], x[0]
for i in range(2, n):
if i % 2 == 0:
if rp < x[i]:
rp = x[i]
else:
exit(print('no'))
else:
if x[i] < lp:
rp = x[i]
else:
exit(print('yes'))
elif x[0] < x[1]:
lp, rp = x[0], x[1]
for i in range(2, n):
if i % 2 == 0:
if x[i] < lp:
lp = x[i]
else:
exit(print('yes'))
else:
if rp < x[i]:
rp = x[i]
else:
exit(print('yes'))
print('no')
```
No
| 5,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima and Seryozha live in an ordinary dormitory room for two. One day Dima had a date with his girl and he asked Seryozha to leave the room. As a compensation, Seryozha made Dima do his homework.
The teacher gave Seryozha the coordinates of n distinct points on the abscissa axis and asked to consecutively connect them by semi-circus in a certain order: first connect the first point with the second one, then connect the second point with the third one, then the third one with the fourth one and so on to the n-th point. Two points with coordinates (x1, 0) and (x2, 0) should be connected by a semi-circle that passes above the abscissa axis with the diameter that coincides with the segment between points. Seryozha needs to find out if the line on the picture intersects itself. For clarifications, see the picture Seryozha showed to Dima (the left picture has self-intersections, the right picture doesn't have any).
<image>
Seryozha is not a small boy, so the coordinates of the points can be rather large. Help Dima cope with the problem.
Input
The first line contains a single integer n (1 ≤ n ≤ 103). The second line contains n distinct integers x1, x2, ..., xn ( - 106 ≤ xi ≤ 106) — the i-th point has coordinates (xi, 0). The points are not necessarily sorted by their x coordinate.
Output
In the single line print "yes" (without the quotes), if the line has self-intersections. Otherwise, print "no" (without the quotes).
Examples
Input
4
0 10 5 15
Output
yes
Input
4
0 15 5 10
Output
no
Note
The first test from the statement is on the picture to the left, the second test is on the picture to the right.
Submitted Solution:
```
n=int(input())
x = list(map(int, input().split()))
c=0
d=0
if(n<=3):
d+=1
else:
for i in range(n-3):
if(x[i]<x[i+2]<x[i+1]<x[i+3] or x[i+2]<x[i]<x[i+3]<x[i+1]):
c+=1
else:
d+=1
if(c>0):
print("yes")
else:
print("no")
```
No
| 5,370 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change his mind, Roman is ready to solve any mathematical problem. After some thought, Sereja asked Roma to find, how many numbers are close to number n, modulo m.
Number x is considered close to number n modulo m, if:
* it can be obtained by rearranging the digits of number n,
* it doesn't have any leading zeroes,
* the remainder after dividing number x by m equals 0.
Roman is a good mathematician, but the number of such numbers is too huge for him. So he asks you to help him.
Input
The first line contains two integers: n (1 ≤ n < 1018) and m (1 ≤ m ≤ 100).
Output
In a single line print a single integer — the number of numbers close to number n modulo m.
Examples
Input
104 2
Output
3
Input
223 4
Output
1
Input
7067678 8
Output
47
Note
In the first sample the required numbers are: 104, 140, 410.
In the second sample the required number is 232.
Submitted Solution:
```
n, m = input().split()
m = int(m)
def variations(string):
if string:
head = string[0]
tail = string[1:]
for var in variations(tail):
for i in range(len(var)+1):
yield var[:i] + head + var[i:]
else:
yield string
res = 0
for v in variations(n):
if not v.startswith("0"):
if int(v) % m == 0:
res += 1
print(res)
```
No
| 5,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change his mind, Roman is ready to solve any mathematical problem. After some thought, Sereja asked Roma to find, how many numbers are close to number n, modulo m.
Number x is considered close to number n modulo m, if:
* it can be obtained by rearranging the digits of number n,
* it doesn't have any leading zeroes,
* the remainder after dividing number x by m equals 0.
Roman is a good mathematician, but the number of such numbers is too huge for him. So he asks you to help him.
Input
The first line contains two integers: n (1 ≤ n < 1018) and m (1 ≤ m ≤ 100).
Output
In a single line print a single integer — the number of numbers close to number n modulo m.
Examples
Input
104 2
Output
3
Input
223 4
Output
1
Input
7067678 8
Output
47
Note
In the first sample the required numbers are: 104, 140, 410.
In the second sample the required number is 232.
Submitted Solution:
```
def fact(n):
j = 1
for i in range(1,n+1):
j*=i
return j
n,m = map(int,input().split())
l = len(str(n))
n = str(n)
mp = dict()
mp2 = dict()
for i in range(l):
if int(n[i]) in mp2.keys():
mp2[int(n[i])]+=1
else :
mp2[int(n[i])] = 1
for i in range(l):
for j in range(l):
for k in range(l):
if i!=j and j!=k and i!=k :
y = int(n[i]+n[j]+n[k])%m
if not y :
mp[str(n[i]+n[j]+n[k])] = y
if l<=3:
if l==1:
if int(n)==0 or int(n)%m!=0:
print(0)
else:
print(1)
elif l==2:
x = n[1]+n[0]
c = 0
if int(n)%m==0:
c+=1
if x[0]!='0' and int(x)%m==0:
c+=1
print(c)
else:
c = 0
for i in mp:
if i[0]!='0':
c+=1
print(c)
exit(0)
a = 0
for i in mp.keys():
mp3 = dict(mp2)
c,c2 = 0,1
for l in i:
mp3[int(l)]-=1
for j in mp3.keys():
c2 *= fact(mp3[j])
c += mp3[j]
if 0 in mp3.keys() and mp3[0]>=1:
a+= fact(c-1)*(c-1)//c2
else:
a += fact(c)//c2
print(a)
```
No
| 5,372 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change his mind, Roman is ready to solve any mathematical problem. After some thought, Sereja asked Roma to find, how many numbers are close to number n, modulo m.
Number x is considered close to number n modulo m, if:
* it can be obtained by rearranging the digits of number n,
* it doesn't have any leading zeroes,
* the remainder after dividing number x by m equals 0.
Roman is a good mathematician, but the number of such numbers is too huge for him. So he asks you to help him.
Input
The first line contains two integers: n (1 ≤ n < 1018) and m (1 ≤ m ≤ 100).
Output
In a single line print a single integer — the number of numbers close to number n modulo m.
Examples
Input
104 2
Output
3
Input
223 4
Output
1
Input
7067678 8
Output
47
Note
In the first sample the required numbers are: 104, 140, 410.
In the second sample the required number is 232.
Submitted Solution:
```
nums=list(map(int,input().split()))
nums_modified=[]
d1={str(i):0 for i in range(0,10)}
d2={str(i):0 for i in range(0,10)}
for index in range(1,len(nums)):
x=list(str(nums[index]))
x2=list(map(int,x))
small=min(x2);large=max(x2)
n=str(small*7+large*11)
if len(n)>=3:
n=n[1:]
nums_modified.append(n)
if index%2!=0:
d1[n[0]]+=1
else:d2[n[0]]+=1
ans=[0]* 10
for key,value in d1.items():
if value==2:
ans[int(key)]+=1
elif value>2:
ans[int(key)]+=2
for key,value in d2.items():
if value==2:
ans[int(key)]+=1
elif value>2:
ans[int(key)]+=2
c=0
for i in ans:
if i>2:
c+=2
else:
c+=i
print(c)
```
No
| 5,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change his mind, Roman is ready to solve any mathematical problem. After some thought, Sereja asked Roma to find, how many numbers are close to number n, modulo m.
Number x is considered close to number n modulo m, if:
* it can be obtained by rearranging the digits of number n,
* it doesn't have any leading zeroes,
* the remainder after dividing number x by m equals 0.
Roman is a good mathematician, but the number of such numbers is too huge for him. So he asks you to help him.
Input
The first line contains two integers: n (1 ≤ n < 1018) and m (1 ≤ m ≤ 100).
Output
In a single line print a single integer — the number of numbers close to number n modulo m.
Examples
Input
104 2
Output
3
Input
223 4
Output
1
Input
7067678 8
Output
47
Note
In the first sample the required numbers are: 104, 140, 410.
In the second sample the required number is 232.
Submitted Solution:
```
def fact(n):
j = 1
for i in range(1,n+1):
j*=i
return j
n,m = map(int,input().split())
l = len(str(n))
n = str(n)
mp = dict()
mp2 = dict()
for i in range(l):
if int(n[i]) in mp2.keys():
mp2[int(n[i])]+=1
else :
mp2[int(n[i])] = 1
for i in range(l):
for j in range(l):
for k in range(l):
if i!=j and j!=k and i!=k :
y = int(n[i]+n[j]+n[k])%m
if not y :
mp[str(n[i]+n[j]+n[k])] = y
if l<=3:
if l==1:
if int(n)==0 or int(n)%m!=0:
print(0)
else:
print(1)
elif l==2:
x = n[1]+n[0]
c = 0
if int(n)%m==0:
c+=1
if x[0]!='0' and int(x)%m==0:
c+=1
print(c)
else:
c = 0
for i in mp:
if i[0]!='0':
c+=1
print(c)
exit(0)
a = 0
print(mp2)
for i in mp.keys():
mp3 = dict(mp2)
c,c2 = 0,1
print(i)
for l in i:
mp3[int(l)]-=1
for j in mp3.keys():
c2 *= fact(mp3[j])
c += mp3[j]
if 0 in mp3.keys() and mp3[0]>=1:
a+= fact(c-1)*(c-1)//c2
else:
a += fact(c)//c2
print(mp3)
print(a)
```
No
| 5,374 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Tags: brute force, implementation, math
Correct Solution:
```
from sys import *
t=int(stdin.readline())
mm=0
for i in range(t):
n,k,d1,d2=(int(z) for z in stdin.readline().split())
mm=2*max(d1,d2)-min(d1,d2)
if (k-2*d1-d2>=0 and (k-2*d1-d2)%3==0 and n-2*d2-d1-k>=0 and (n-2*d2-d1-k)%3==0) or (k-2*d2-d1>=0 and (k-2*d2-d1)%3==0 and n-2*d1-d2-k>=0 and (n-2*d1-d2-k)%3==0) or (k-d1-d2>=0 and (k-d1-d2)%3==0 and n-mm-k>=0 and (n-mm-k)%3==0) or (k-mm>=0 and (k-mm)%3==0 and n-d1-d2-k>=0 and (n-d1-d2-k)%3==0):
print("yes")
else:
print("no")
```
| 5,375 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Tags: brute force, implementation, math
Correct Solution:
```
from sys import stdin
def solve(n, k, d1, d2):
d = n - k
# a >= b and c >= b
c1 = (k - d1 + 2 * d2) // 3
a1 = d1 - d2 + c1
b1 = k - a1 - c1
# a >= b and c < b
c2 = (k - d1 - 2 * d2) // 3
a2 = d1 + d2 + c2
b2 = k - a2 - c2
# a < b and c < b
c3 = (k + d1 - 2 * d2) // 3
a3 = c3 + d2 - d1
b3 = k - a3 - c3
# a < b and c >= b
c4 = (k + d1 + 2 * d2) // 3
a4 = c4 - d1 - d2
b4 = k - a4 - c4
if c1 * 3 == 2 * d2 - d1 + k and a1 >= 0 and b1 >= 0 and c1 >= 0:
_max = max(a1, b1, c1)
diff = sum(_max - e for e in [a1, b1, c1])
if d >= diff and (d - diff) % 3 == 0:
return True
if c2 * 3 == k - d1 - 2 * d2 and a2 >= 0 and b2 >= 0 and c2 >= 0:
_max = max(a2, b2, c2)
diff = sum(_max - e for e in [a2, b2, c2])
if d >= diff and (d - diff) % 3 == 0:
return True
if c3 * 3 == k + d1 - 2 * d2 and a3 >= 0 and b3 >= 0 and c3 >= 0:
_max = max(a3, b3, c3)
diff = sum(_max - e for e in [a3, b3, c3])
if d >= diff and (d - diff) % 3 == 0:
return True
if c4 * 3 == k + d1 + 2 * d2 and a4 >= 0 and b4 >= 0 and c4 >= 0:
_max = max(a4, b4, c4)
diff = sum(_max - e for e in [a4, b4, c4])
if d >= diff and (d - diff) % 3 == 0:
return True
return False
def main():
test = stdin.readlines()
output = []
for i in range(1, int(test[0]) + 1):
ans = 'yes' if solve(*map(int, test[i].split())) else 'no'
output.append(ans)
print('\n'.join(output))
if __name__ == "__main__":
main()
```
| 5,376 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Tags: brute force, implementation, math
Correct Solution:
```
input=__import__('sys').stdin.readline
for _ in range(int(input())):
n,k,d1,d2 = map(int,input().split())
lis=[[2*d1+d2 , 2*d2+d1] , [2*d2+d1 , 2*d1+d2] , [2*max(d1,d2)-min(d1,d2) , d1+d2] , [d1+d2 , 2*max(d1,d2) - min(d1,d2)]]
flag=1
for i in lis:
if i[0]<=k and i[0]%3==k%3 and n-k-i[1]>=0 and (n-k-i[1])%3==0:
print("yes")
flag=0
break
if flag:
print("no")
```
| 5,377 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Tags: brute force, implementation, math
Correct Solution:
```
t=int(input())
for j in range(t):
inp=[int(n) for n in input().split()]
n=inp[0]
k=inp[1]
d1=inp[2]
d2=inp[3]
if d2<d1:
s=d1
d1=d2
d2=s
if ((k>=2*d1+d2) and ((k-2*d1-d2)%3==0) and (n-k>=d1+2*d2) and ((n-k-d1-2*d2)%3==0)):
print('yes')
elif ((k>=2*d2+d1) and ((k-2*d2-d1)%3==0) and (n-k>=d2+2*d1) and ((n-k-d2-2*d1)%3==0)):
print('yes')
elif ((k>=d1+d2) and ((k-d1-d2)%3==0) and (n-k>=2*d2-d1) and ((n-k-2*d2+d1)%3==0)):
print('yes')
elif ((k>=2*d2-d1) and ((k-2*d2+d1)%3==0) and (n-k>=d1+d2) and ((n-k-d1-d2)%3==0)):
print('yes')
else:
print('no')
```
| 5,378 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Tags: brute force, implementation, math
Correct Solution:
```
from sys import *
t=int(stdin.readline())
mm,mmm,mmmm,m=0,0,0,0
for i in range(t):
n,k,d1,d2=(int(z) for z in stdin.readline().split())
m=d1+d2
mm=2*max(d1,d2)-min(d1,d2)
mmm=2*d1+d2
mmmm=2*d2+d1
if (k-mmm>=0 and (k-mmm)%3==0 and n-mmmm-k>=0 and (n-mmmm-k)%3==0) or (k-mmmm>=0 and (k-mmmm)%3==0 and n-mmm-k>=0 and (n-mmm-k)%3==0) or (k-m>=0 and (k-m)%3==0 and n-mm-k>=0 and (n-mm-k)%3==0) or (k-mm>=0 and (k-mm)%3==0 and n-m-k>=0 and (n-m-k)%3==0):
print("yes")
else:
print("no")
```
| 5,379 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Tags: brute force, implementation, math
Correct Solution:
```
import sys
import operator
input = sys.stdin.readline
def read_int():
return int(input().strip())
def read_ints():
return list(map(int, input().strip().split(' ')))
def solve():
"""
n, k, d1, d2
d1 = abs(w1-w2)
d2 = abs(w2-w3)
w1+w2+w3=k
1) w1 >= w2 and w2 >= w3
d1 = w1-w2
d2 = w2-w3
d1-d2 = w1+w3-2*w2
d1-d2-k = -3*w2
w1 = d1-(d1-d2-k)/3
w2 = -(d1-d2-k)/3
w3 = -(d1-d2-k)/3-d2
n/3-w1 >= 0
n/3-w2 >= 0
n/3-w3 >= 0
"""
n, k, d1, d2 = read_ints()
if n%3 != 0:
return 'no'
# w1 >= w2 and w2 >= w3
def check(n, k, d1, d2, op1, op2):
if (d1-d2-k)%3 != 0:
return False
w1 = d1-(d1-d2-k)//3
w2 = -(d1-d2-k)//3
w3 = -(d1-d2-k)//3-d2
if w1 >= 0 and w2 >= 0 and w3 >= 0 and n//3-w1 >= 0 and n//3-w2 >= 0 and n//3-w3 >= 0 and op1(w1, w2) and op2(w2, w3):
return True
return False
if check(n, k, d1, d2, operator.ge, operator.ge):
return 'yes'
if check(n, k, d1, -d2, operator.ge, operator.lt):
return 'yes'
if check(n, k, -d1, d2, operator.lt, operator.ge):
return 'yes'
if check(n, k, -d1, -d2, operator.lt, operator.lt):
return 'yes'
return 'no'
if __name__ == '__main__':
T = read_int()
for _ in range(T):
print(solve())
```
| 5,380 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Tags: brute force, implementation, math
Correct Solution:
```
t = int(input())
ret = []
while t>0:
t-=1
n,k,d1,d2 = map(int,input().split())
# ans = []
y1 = (k-(d1-d2))//3
x1 = y1+d1
z1 = y1-d2
# ans = [y1,z1,x1]
# ans = sorted(ans)
# ans1 = 2*ans[2]-(ans[0]+ans[1])
ans1 = 2*x1-(z1+y1)
if x1+y1+z1==k and min(z1,y1)>=0 and ans1<=n-k and (n-k-ans1)%3==0:
ret.append('yes')
continue
# ans = []
y1 = (k-(d1+d2))//3
x1 = y1+d1
z1 = y1+d2
if d1>=d2:
# ans = [y1,z1,x1]
ans1 = 2*x1-(y1+z1)
else:
# ans = [y1,x1,z1]
ans1 = 2*z1-(y1+x1)
# ans = sorted(ans)
# ans1 = 2*ans[2]-(ans[0]+ans[1])
if x1+y1+z1==k and y1>=0 and ans1<=n-k and (n-k-ans1)%3==0:
ret.append('yes')
continue
y1 = (k-(d2-d1))//3
x1 = y1-d1
z1 = y1+d2
# ans = [x1,y1,z1]
# ans = sorted(ans)
ans1 = 2*z1-(x1+y1)
if x1+y1+z1==k and min(x1,y1)>=0 and ans1<=n-k and (n-k-ans1)%3==0:
ret.append('yes')
continue
y1 = (k+(d2+d1))//3
x1 = y1-d1
z1 = y1-d2
# ans = [x1,y1,z1]
# ans = sorted(ans)
ans1 = 2*y1-(x1+z1)
if x1+y1+z1==k and min(x1,z1)>=0 and ans1<=n-k and (n-k-ans1)%3==0:
ret.append('yes')
continue
# if d1>=d2:
# ans.append(2*d1-d2)
# ans.append(d2+2*(d1-d2))
# else:
# ans.append(2*d2-d1)
# ans.append(d1+2*(d2-d1))
# ans+=[d1+2*d2,d2+2*d1,d1+d2]
# done = False
# print(ans)
# for a in ans:
# # if (a==0 and (n-k)%3==0) or (a!=0 and (n-k)//a>1 and (n-k)%a==0):
# if (a<=n-k) and (n-k-a)%3==0:
# print(a)
# done = True
# break
# if done:
# print('yes')
# else:
ret.append('no')
print(*ret, sep = '\n')
```
| 5,381 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Tags: brute force, implementation, math
Correct Solution:
```
def check(n, k, d1, d2):
if (k-d1-d2) % 3 != 0:
return False
y = (k-d1-d2) // 3
x = y+d1
z = y+d2
t = n // 3
if x < 0 or y < 0 or z < 0:
return False
if x > t or y > t or z > t:
return False
return True
t = int(input())
for _ in range(t):
n, k, d1, d2 = map(int, input().split())
if n % 3 != 0:
print ('no')
continue
if check(n, k, d1, d2):
print ('yes')
continue
if check(n, k, -d1, d2):
print ('yes')
continue
if check(n, k, d1, -d2):
print ('yes')
continue
if check(n, k, -d1, -d2):
print ('yes')
continue
print ('no')
```
| 5,382 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
# from fractions import *
# from heapq import*
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=10**9+7
EPS=1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
def check(d1,d2):
wa=0
wb=0
wc=0
# wa-wb=d1
# wb-wc=d2
# wa-wc=d1+d2
# wa+wb+wc=k
wa=2*d1+d2+k
if(wa%3): return False
wa//=3
wb=wa-d1
wc=wb-d2
if(wa<0 or wb<0 or wc<0): return False
wa,wb,wc=sorted((wa,wb,wc))
need=2*wc-wa-wb
have=n-k-need
# print(need,have,'-->',wa,wb,wc)
return have>=0 and have%3==0
for _ in range(Int()):
n,k,d1,d2=value()
d3,d4=-d1,-d2
d1=[d1,d3]
d2=[d2,d4]
ok=False
for i in d1:
for j in d2:
ok|=check(i,j)
if(ok): print("yes")
else: print("no")
```
Yes
| 5,383 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Submitted Solution:
```
def main():
t = int(input())
for z in range(t):
n, k, d1, d2 = map(int, input().split())
if n % 3 != 0:
print('no')
continue
f = 0
for i in [-1, +1]:
for j in [-1, +1]:
w = (k - i * d1 - j * d2)
if f == 0 and (w % 3 == 0) and (n//3)>=(w//3)>=0 and (n//3)>=(w//3 + i * d1)>=0 and (n//3)>=(w//3 + j * d2)>=0:
print('yes')
f = 1
if f == 0:
print('no')
main()
```
Yes
| 5,384 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Submitted Solution:
```
t = int(input())
for i in range(t):
n, k, a, b = map(int, input().split())
if n % 3 != 0:
print("no")
else:
for i in range(2):
for j in range(2):
flagf = False
if i == 0:
a1 = a
else:
a1 = -a
if j == 0:
b1 = b
else:
b1 = -b
t2 = (k - a1 + b1)/3
t1 = a1 + t2
t3 = t2 - b1
# print(t1, t2, t3)
flag1 = False
flag2 = False
# valores válidos
if (k-a1+b1) % 3 == 0 and t1 >= 0 and t2 >= 0 and t3 >= 0:
if t1 <= n/3 and t2 <= n/3 and t3 <= n/3:
if i == 0 and t1 >= t2:
flag1 = True
elif i == 1 and t1 < t2:
flag1 = True
if j == 0 and t2 >= t3:
flag2 = True
elif j == 1 and t2 < t3:
flag2 = True
if flag1 and flag2:
if t1 == t2 and t2 == t3 and (n-k) % 3 == 0:
flagf = True
break
else:
v = [t1, t2, t3]
v = sorted(v)
faltam = n-k
faltam -= v[2] - v[1]
faltam -= v[2] - v[0]
if faltam < 0:
break
elif faltam == 0 or faltam %3 == 0:
flagf = True
break
if flagf:
print("yes")
break
else:
print("no")
# 3 1 1 0
```
Yes
| 5,385 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Submitted Solution:
```
from sys import stdin
def solve(n, k, d1, d2):
if (d1 + 2 * d2 + k) % 3 != 0:
return False
c = (d1 + 2 * d2 + k) // 3
a = c - d1 - d2
b = k - a - c
if a >= 0 and b >= 0 and c >= 0 and a + b + c == k:
_max = max(a, b, c)
diff = sum(_max - e for e in [a, b, c])
d = n - k
if d >= diff and (d - diff) % 3 == 0:
return True
return False
def main():
test = stdin.readlines()
output = []
for i in range(1, int(test[0]) + 1):
n, k, d1, d2 = map(int, test[i].split())
for i, j in [(-1, -1), (-1, 1), (1, -1), (1, 1)]:
if solve(n, k, i * d1, j * d2):
output.append('yes')
break
else:
output.append('no')
print('\n'.join(output))
if __name__ == "__main__":
main()
```
Yes
| 5,386 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Submitted Solution:
```
from sys import *
t=int(stdin.readline())
for i in range(t):
n,k,d1,d2=(int(z) for z in stdin.readline().split())
p1=n-(2*d1+d2+k)
p2=n-(2*d2+d1+k)
if (p1>=0 and not p1%3) or (p2>=0 and not p2%3):
print("yes")
else:
print("no")
```
No
| 5,387 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Submitted Solution:
```
for t in range(int(input())):
n,k,d1,d2=map(int,input().split())
x=(k+(2*d1)+d2)//3
a,b,c=x,x-d1,x-d1-d2
n1=n//3
#print(x,a,b,c)
if n%3!=0 or (abs(n1-a)+abs(n1-b)+abs(n1-c))!=(n-k) or a+b+c!=k or a>n1 or b>n1 or c>n1 :
print('no')
else:
print('yes')
```
No
| 5,388 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Submitted Solution:
```
a = []
z = int(input())
for i in range(z):
n, k, d1, d2 = map(int, input().split())
d1, d2 = sorted((d1, d2, ))
zn = n - k
v1 = 2 * d1 + d2
if v1 <= zn:
a.append('yes')
else:
a.append('no')
print(*a, sep = '\n')
```
No
| 5,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.
You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these k games. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.
You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?
Note that outcome of a match can not be a draw, it has to be either win or loss.
Input
The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).
Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.
Output
For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).
Examples
Input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2
Output
yes
yes
yes
no
no
Note
Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.
Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".
Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).
Submitted Solution:
```
for _ in range(int(input())):
n,k,d1,d2 = map(int,input().split())
lis=[[2*d1+d2 , 2*d2+d1] , [2*d2+d1 , 2*d1+d2] , [2*max(d1,2)-min(d1,d2) , d1+d2] , [d1+d2 , 2*max(d1,d2) - min(d1,d2)]]
flag=1
for i in lis:
if i[0]<=k and (k-i[0])%3==0 and n-k-i[1]>=0 and (n-k-i[1])%3==0:
print("yes")
flag=0
break
if flag:
print("no")
```
No
| 5,390 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Tags: binary search, implementation
Correct Solution:
```
import sys
import math
a=int(input())
b=list(map(int,input().split()))
c=int(input())
d=list(map(int,input().split()))
e=[]
f=0
for i in b:
#print(f,i)
e.append(f+i)
f=i+f
g=[0]*(e[-1]+1)
h=0
for i in range(a):
for j in range(h,e[i]+1):
g[j]=i+1
h=e[i]+1
for k in d:
print(g[k])
```
| 5,391 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Tags: binary search, implementation
Correct Solution:
```
a=int(input())
b= list(map(int,input().split()))
c=int(input())
d= list(map(int,input().split()))
for k in range(1,a):
b[k] += b[k-1]
for j in d:
l=0
r= a-1
while l<= r:
if j <= b[0]:
mid=0
break
mid= (l+r)//2
if b[mid] >= j and b[mid-1] < j:
break
elif b[mid] > j:
r=mid-1
else:
l=mid+1
print(mid+1)
```
| 5,392 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Tags: binary search, implementation
Correct Solution:
```
def read():
input()
sizes = list(map(int, input().split()))
input()
queries = list(map(int, input().split()))
return sizes, queries
def calc_indexes(sizes):
start_indexes = [1]
for i in range(len(sizes)):
start_indexes.append(start_indexes[i] + sizes[i])
return start_indexes
def bin_search(ar, query):
begin = 0
end = len(ar)
while (end - begin) > 1:
middle = (end + begin) // 2
if query < ar[middle]:
end = middle
else:
begin = middle
return begin
# for i in range(begin, end):
# if ar[i] > query:
# return i - 1
# return end - 1
sizes, queries = read()
start_indexes = calc_indexes(sizes)
for q in queries:
print (bin_search(start_indexes, q) + 1)
```
| 5,393 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Tags: binary search, implementation
Correct Solution:
```
import bisect as bs
n = int(input())
A = list(map(int, input().split()))
m = int(input())
B = list(map(int, input().split()))
S = [0]
for a in A:
S.append(S[-1] + a)
for b in B:
print(bs.bisect_left(S, b))
```
| 5,394 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Tags: binary search, implementation
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
m=int(input())
q=list(map(int,input().split()))
d=[0]
for i in range(n):
d.append(d[i]+a[i])
d.append(1000001)
for i in range(m):
start=0;end=len(d)-1
while start<end:
k=(start+end)//2
if q[i]<d[k]:
end=k-1
else:
start=k+1
for t in range(max(0,end-4),n+1):
if q[i]<=d[t]:
print(t)
break
else:
print(n)
```
| 5,395 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Tags: binary search, implementation
Correct Solution:
```
input()
d=[]
m=1
for i in map(int,input().split()):
for j in range(i):
d.append(m)
m+=1
input()
for i in map(int,input().split()):
print(d[i-1])
```
| 5,396 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Tags: binary search, implementation
Correct Solution:
```
n=int(input())
x=list(map(int,input().split()))
t=int(input())
y=list(map(int,input().split()))
ans=[]
for i in range(len(x)):
ans.extend([i+1 for j in range(x[i])])
for i in y:
print(ans[i-1])
```
| 5,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Tags: binary search, implementation
Correct Solution:
```
n=int(input())
s=list(map(int,input().split()))
n1=int(input())
s1=list(map(int,input().split()))
w=[]
i=-1
d=n-1
r=0
while i!=d:
i+=1
for j in range(s[i]):
r+=1
w.append(r)
ans=[]
for i in range(n1):
l,r=-1,n
while (l+1<r):
md=(l+r)//2
if s1[i]<=w[md]:
r=md
else:
l=md
ans.append(r+1)
print(*ans,sep='\n')
```
| 5,398 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
k=0
l=[]
c=1
for i in range(len(a)):
x=a[i]
for j in range(k,k+x):
l.append(c)
c+=1
k=k+x
t=int(input())
b=list(map(int,input().split()))
for i in b:
print(l[i-1])
```
Yes
| 5,399 |
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