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Provide tags and a correct Python 3 solution for this coding contest problem.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Tags: brute force, dp, greedy, sortings, two pointers
Correct Solution:
```
# Enter your code here. Read input from STDIN. Print output to STDOUT# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
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)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
def solve():
n, m, k = map(int, input().split());a = [list(map(int, input().split())) for _ in range(m)]
def solv(lst):
res = sum(lst[:k]);s = res
for i in range(n-k):s += lst[i+k] - lst[i];res = max(res, s)
return res
a.sort(key=lambda x: sum(x)/2);apr = [0]*n
for el in a:
for i in range(el[0]-1, el[1]):apr[i] += 1
res = solv(apr);bpr = [0]*n
for r, l in a:
for i in range(r-1, l):apr[i] -= 1;bpr[i] += 1
nres = solv(apr) + solv(bpr);res = max(res, nres)
print(res)
solve()
#testcase(int(inp()))
# 5
# 3 6
# 5 10
# 4 3
# 2 1
# 1 3
```
| 93,700 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Tags: brute force, dp, greedy, sortings, two pointers
Correct Solution:
```
"""
#If FastIO not needed, use this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
import time
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
#start_time = time.time()
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
def getMat(n):
return [getInts() for _ in range(n)]
MOD = 10**9+7
"""
Two subsets of length K which cover as many segments as possible
How many times does a problem appear?
Suppose one of the segments goes from [p,p+k-1]
How many problems for each participant are in that segment?
We want two overlapping segments such that the sum of the max for each person is maximised
Best possible answer from first i arrays?
Note that overlap is pointless
So the answer is to find a dividing line, and find the best array either side of that line
"""
def solve():
N, M, K = getInts()
A = []
for m in range(M):
A.append(tuple(getInts()))
A.sort(key = lambda x: sum(x))
def get_poss(arr):
curr = sum(arr[:K])
best = curr
for i in range(K,N):
curr += arr[i] - arr[i-K]
best = max(best,curr)
return best
A_pref = [0]*N
for L, R in A:
for i in range(L-1,R):
A_pref[i] += 1
ans = get_poss(A_pref)
B_pref = [0]*N
for L, R in A:
for i in range(L-1,R):
A_pref[i] -= 1
B_pref[i] += 1
ans = max(ans,get_poss(A_pref) + get_poss(B_pref))
return ans
#for _ in range(getInt()):
print(solve())
#solve()
#print(time.time()-start_time)
```
| 93,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Tags: brute force, dp, greedy, sortings, two pointers
Correct Solution:
```
def solve(lst, n, k):
sum_current = sum(lst[0:k])
ans = sum_current
for i in range(n - k):
sum_current = sum_current - lst[i] + lst[i + k]
ans = max(ans, sum_current)
return ans
def two_editorials():
n, m, k = [int(x) for x in input().split(' ')]
arr = [0] * n
brr = [0] * n
inp = []
for _ in range(m):
inp += [[int(x) for x in input().split(' ')]]
inp.sort(key=lambda x: sum(x) / 2)
for a, b in inp:
for i in range(a - 1, b):
arr[i] += 1
ans = solve(arr, n, k)
for a, b in inp:
for i in range(a - 1, b):
arr[i] -= 1
brr[i] += 1
ans = max(ans, solve(arr, n, k) + solve(brr, n, k))
print(ans)
if __name__ == "__main__":
two_editorials()
```
| 93,702 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Tags: brute force, dp, greedy, sortings, two pointers
Correct Solution:
```
n, m, k = map(int, input().split())
a = [list(map(int, input().split())) for _ in range(m)]
def solv(lst):
res = sum(lst[:k])
s = res
for i in range(n-k):
s += lst[i+k] - lst[i]
res = max(res, s)
return res
a.sort(key=lambda x: sum(x)/2)
apr = [0]*n
for el in a:
for i in range(el[0]-1, el[1]):
apr[i] += 1
res = solv(apr)
bpr = [0]*n
for r, l in a:
for i in range(r-1, l):
apr[i] -= 1
bpr[i] += 1
nres = solv(apr) + solv(bpr)
res = max(res, nres)
print(res)
```
| 93,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Tags: brute force, dp, greedy, sortings, two pointers
Correct Solution:
```
import sys
BUFSIZE = 8192
from sys import stderr
from io import BytesIO, IOBase
import os
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)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
def solve():
n, m, k = map(int, input().split());a = [list(map(int, input().split())) for _ in range(m)]
def solv(lst):
res = sum(lst[:k]);s = res
for i in range(n-k):s += lst[i+k] - lst[i];res = max(res, s)
return res
a.sort(key=lambda x: sum(x)/2);apr = [0]*n
for el in a:
for i in range(el[0]-1, el[1]):apr[i] += 1
res = solv(apr);bpr = [0]*n
for r, l in a:
for i in range(r-1, l):apr[i] -= 1;bpr[i] += 1
nres = solv(apr) + solv(bpr);res = max(res, nres)
print(res)
solve()
#testcase(int(inp()))
# 5
# 3 6
# 5 10
# 4 3
# 2 1
# 1 3
```
| 93,704 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Tags: brute force, dp, greedy, sortings, two pointers
Correct Solution:
```
# Enter your code here. Read input from STDIN. Print output to STDOUT# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
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)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
###########################
# Sorted list
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
# ===============================================================================================
# some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def pow(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1): # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# discrete binary search
# minimise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# if isvalid(l):
# return l
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m - 1):
# return m
# if isvalid(m):
# r = m + 1
# else:
# l = m
# return m
# maximise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# # print(l,r)
# if isvalid(r):
# return r
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m + 1):
# return m
# if isvalid(m):
# l = m
# else:
# r = m - 1
# return m
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def N():
return int(inp())
dx = [0, 0, 1, -1]
dy = [1, -1, 0, 0]
def YES():
print("YES")
def NO():
print("NO")
def Yes():
print("Yes")
def No():
print("No")
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(i):
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
def lcm(a, b):
return abs((a // gcd(a, b)) * b)
#
# # to find factorial and ncr
# tot = 400005
# mod = 10 ** 9 + 7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, tot + 1):
# fac.append((fac[-1] * i) % mod)
# inv.append(mod - (inv[mod % i] * (mod // i) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
#
#
# def comb(n, r):
# if n < r:
# return 0
# else:
# return fac[n] * (finv[r] * finv[n - r] % mod) % mod
def solve():
n, m, k = map(int, input().split());a = [list(map(int, input().split())) for _ in range(m)]
def solv(lst):
res = sum(lst[:k]);s = res
for i in range(n-k):s += lst[i+k] - lst[i];res = max(res, s)
return res
a.sort(key=lambda x: sum(x)/2);apr = [0]*n
for el in a:
for i in range(el[0]-1, el[1]):apr[i] += 1
res = solv(apr);bpr = [0]*n
for r, l in a:
for i in range(r-1, l):apr[i] -= 1;bpr[i] += 1
nres = solv(apr) + solv(bpr);res = max(res, nres)
print(res)
solve()
#testcase(int(inp()))
# 5
# 3 6
# 5 10
# 4 3
# 2 1
# 1 3
```
| 93,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Tags: brute force, dp, greedy, sortings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
def prog():
n,m,k = map(int,input().split())
intervals = [list(map(int,input().split())) for i in range(m)]
for i in range(m):
intervals[i][0] -= 1
intervals[i][1] -= 1
mx = 0
for i in range(n-k+1):
curr = 0
change = 0
events = [0]*(2*n+1)
for interval in intervals:
length = interval[1] - interval[0] + 1
if i <= interval[0] <= interval[1] <= i+k-1:
curr += length
elif interval[0] <= i <= i+k-1 <= interval[1]:
curr += k
elif i <= interval[0] <= i+k-1 or i <= interval[1] <= i+k-1:
if i <= interval[0] <= i+k-1:
intersect = i+k-1 - interval[0] + 1
else:
intersect = interval[1] - i + 1
curr += intersect
events[interval[0] + intersect] += 1
events[interval[0] + k] -= 1
events[interval[1] + 1] -= 1
events[interval[1] + k + 1] += 1
else:
events[interval[0]] += 1
events[interval[0] + k] -= 1
events[interval[1] + 1] -= 1
events[interval[1] + k + 1] += 1
for j in range(n):
change += events[j]
curr += change
mx = max(mx,curr)
print(mx)
prog()
```
| 93,706 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Submitted Solution:
```
import sys,os,io
input = sys.stdin.readline
#input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n, m, k = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(m)]
A.sort(key = lambda x:x[0]+x[1])
table = [[0]*m for _ in range(n-k+1)]
for i in range(n-k+1): #i+1~i+k
for j in range(m):
lower = max(i+1, A[j][0])
upper = min(i+k, A[j][1])
table[i][j] = max(0,upper-lower+1)
from copy import deepcopy
cum1 = deepcopy(table)
cum2 = deepcopy(table)
ans = 0
for i in range(n-k+1):
for j in range(1,m):
cum1[i][j] += cum1[i][j-1]
for i in range(n-k+1):
for j in range(m-2,-1,-1):
cum2[i][j] += cum2[i][j+1]
# ans = max([cum1[i][-1] for i in range(n-k+1)])
ans = max([cum2[i][0] for i in range(n-k+1)])
for j in range(m-1):
left = max([cum1[i][j] for i in range(n-k+1)])
right = max([cum2[i][j+1] for i in range(n-k+1)])
ans = max(ans, left+right)
print(ans)
```
Yes
| 93,707 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Submitted Solution:
```
n, m, k = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(m)]
A.sort(key = lambda x:x[0]+x[1])
table = [[0]*m for _ in range(n-k+1)]
for i in range(n-k+1): #i+1~i+k
for j in range(m):
lower = max(i+1, A[j][0])
upper = min(i+k, A[j][1])
table[i][j] = max(0,upper-lower+1)
from copy import deepcopy
cum1 = deepcopy(table)
cum2 = deepcopy(table)
ans = 0
for i in range(n-k+1):
for j in range(1,m):
cum1[i][j] += cum1[i][j-1]
for i in range(n-k+1):
for j in range(m-2,-1,-1):
cum2[i][j] += cum2[i][j+1]
ans = max([cum1[i][-1] for i in range(n-k+1)])
ans = max(ans,max([cum2[i][0] for i in range(n-k+1)]))
for j in range(m-1):
left = max([cum1[i][j] for i in range(n-k+1)])
right = max([cum2[i][j+1] for i in range(n-k+1)])
ans = max(ans, left+right)
print(ans)
```
Yes
| 93,708 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Submitted Solution:
```
import sys,os,io
input = sys.stdin.readline
#input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n, m, k = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(m)]
A.sort(key = lambda x:x[0]+x[1])
table = [[0]*m for _ in range(n-k+1)]
for i in range(n-k+1): #i+1~i+k
for j in range(m):
lower = max(i+1, A[j][0])
upper = min(i+k, A[j][1])
table[i][j] = max(0,upper-lower+1)
from copy import deepcopy
cum1 = deepcopy(table)
cum2 = deepcopy(table)
for i in range(n-k+1):
for j in range(1,m):
cum1[i][j] += cum1[i][j-1]
for i in range(n-k+1):
for j in range(m-2,-1,-1):
cum2[i][j] += cum2[i][j+1]
# mが1の場合を考慮
ans = max([cum1[i][-1] for i in range(n-k+1)])
for j in range(m-1):
left = max([cum1[i][j] for i in range(n-k+1)])
right = max([cum2[i][j+1] for i in range(n-k+1)])
ans = max(ans, left+right)
print(ans)
```
Yes
| 93,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Submitted Solution:
```
import sys,os,io
input = sys.stdin.readline
#input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n, m, k = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(m)]
A.sort(key = lambda x:x[0]+x[1])
table = [[0]*m for _ in range(n-k+1)]
for i in range(n-k+1): #i+1~i+k
for j in range(m):
lower = max(i+1, A[j][0])
upper = min(i+k, A[j][1])
table[i][j] = max(0,upper-lower+1)
from copy import deepcopy
cum1 = deepcopy(table)
cum2 = deepcopy(table)
ans = 0
for i in range(n-k+1):
for j in range(1,m):
cum1[i][j] += cum1[i][j-1]
for i in range(n-k+1):
for j in range(m-2,-1,-1):
cum2[i][j] += cum2[i][j+1]
ans = max([cum1[i][-1] for i in range(n-k+1)])
# ans = max(ans,max([cum2[i][0] for i in range(n-k+1)]))
for j in range(m-1):
left = max([cum1[i][j] for i in range(n-k+1)])
right = max([cum2[i][j+1] for i in range(n-k+1)])
ans = max(ans, left+right)
print(ans)
```
Yes
| 93,710 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Submitted Solution:
```
n, m, k = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(m)]
A.sort(key = lambda x:x[0]+x[1])
table = [[0]*m for _ in range(n-k+1)]
for i in range(n-k+1): #i+1~i+k
for j in range(m):
lower = max(i+1, A[j][0])
upper = min(i+k, A[j][1])
table[i][j] = max(0,upper-lower+1)
from copy import deepcopy
cum1 = deepcopy(table)
cum2 = deepcopy(table)
ans = 0
for i in range(n-k+1):
for j in range(1,m):
cum1[i][j] += cum1[i][j-1]
for i in range(n-k+1):
for j in range(m-2,-1,-1):
cum2[i][j] += cum2[i][j+1]
for j in range(m-1):
left = max([cum1[i][j] for i in range(n-k+1)])
right = max([cum2[i][j+1] for i in range(n-k+1)])
ans = max(ans, left+right)
print(ans)
```
No
| 93,711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Submitted Solution:
```
from sys import stdin
def inverse(a,mod):
return pow(a,mod-2,mod)
n,m,k = map(int,stdin.readline().split())
ans = 0
lr = []
for i in range(m):
l,r = map(int,stdin.readline().split())
lr.append((l-1,r-1))
for D1 in range(0,n-k+1):
now = 0
nch = 0
ch = [0] * (n-k+1)
for l,r in lr:
L,R = l,r+1
if D1 >= R or L >= D1+k:
X = 0
elif D1 <= L <= D1+k <= R:
X = D1+k-L
elif L <= D1 <= R <= D1+k:
X = R-D1
else:
X = min(R-L,k)
now += X
tmp = [(l-k+X,1),(l,-1),(r+1-k,-1),(r+1-X,1)]
for i,j in tmp:
if i < 0:
now += j
elif i < n-k+1:
ch[i] += j
for i in range(n-k+1):
ans = max(ans,now)
now += nch
nch += ch[i]
print (ans)
```
No
| 93,712 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Submitted Solution:
```
import sys,os,io
input = sys.stdin.readline
#input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n, m, k = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(m)]
A.sort(key = lambda x:x[0]+x[1])
table = [[0]*m for _ in range(n-k+1)]
for i in range(n-k+1): #i+1~i+k
for j in range(m):
lower = max(i+1, A[j][0])
upper = min(i+k, A[j][1])
table[i][j] = max(0,upper-lower+1)
from copy import deepcopy
cum1 = deepcopy(table)
cum2 = deepcopy(table)
ans = 0
for i in range(n-k+1):
for j in range(1,m):
cum1[i][j] += cum1[i][j-1]
for i in range(n-k+1):
for j in range(m-2,-1,-1):
cum2[i][j] += cum2[i][j+1]
# ans = max([cum1[i][-1] for i in range(n-k+1)])
# ans = max([cum2[i][0] for i in range(n-k+1)])
for j in range(m-1):
left = max([cum1[i][j] for i in range(n-k+1)])
right = max([cum2[i][j+1] for i in range(n-k+1)])
ans = max(ans, left+right)
print(ans)
```
No
| 93,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland regional ICPC contest has just ended. There were m participants numbered from 1 to m, who competed on a problemset of n problems numbered from 1 to n.
Now the editorial is about to take place. There are two problem authors, each of them is going to tell the tutorial to exactly k consecutive tasks of the problemset. The authors choose the segment of k consecutive tasks for themselves independently of each other. The segments can coincide, intersect or not intersect at all.
The i-th participant is interested in listening to the tutorial of all consecutive tasks from l_i to r_i. Each participant always chooses to listen to only the problem author that tells the tutorials to the maximum number of tasks he is interested in. Let this maximum number be a_i. No participant can listen to both of the authors, even if their segments don't intersect.
The authors want to choose the segments of k consecutive tasks for themselves in such a way that the sum of a_i over all participants is maximized.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 2000, 1 ≤ k ≤ n) — the number of problems, the number of participants and the length of the segment of tasks each of the problem authors plans to tell the tutorial to.
The i-th of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the segment of tasks the i-th participant is interested in listening to the tutorial to.
Output
Print a single integer — the maximum sum of a_i over all participants.
Examples
Input
10 5 3
1 3
2 4
6 9
6 9
1 8
Output
14
Input
10 3 3
2 4
4 6
3 5
Output
8
Input
4 4 1
3 3
1 1
2 2
4 4
Output
2
Input
5 4 5
1 2
2 3
3 4
4 5
Output
8
Note
In the first example the first author can tell the tutorial to problems from 1 to 3 and the second one — from 6 to 8. That way the sequence of a_i will be [3, 2, 3, 3, 3]. Notice that the last participant can't listen to both author, he only chooses the one that tells the maximum number of problems he's interested in.
In the second example the first one can tell problems 2 to 4, the second one — 4 to 6.
In the third example the first one can tell problems 1 to 1, the second one — 2 to 2. Or 4 to 4 and 3 to 3. Every pair of different problems will get the same sum of 2.
In the fourth example the first one can tell problems 1 to 5, the second one — 1 to 5 as well.
Submitted Solution:
```
n, m, k = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(m)]
A.sort(key = lambda x:x[0])
table = [[0]*m for _ in range(n-k+1)]
for i in range(n-k+1): #i+1~i+k
for j in range(m):
lower = max(i+1, A[j][0])
upper = min(i+k, A[j][1])
table[i][j] = max(0,upper-lower+1)
from copy import deepcopy
cum1 = deepcopy(table)
cum2 = deepcopy(table)
for i in range(n-k+1):
for j in range(1,m):
cum1[i][j] += cum1[i][j-1]
for i in range(n-k):
for j in range(m-2,-1,-1):
cum2[i][j] += cum2[i][j+1]
ans = 0
for j in range(m-1):
left = right = 0
for i in range(n-k+1):
left = max(left, cum1[i][j])
for i in range(n-k+1):
right = max(right, cum2[i][j+1])
ans = max(ans, left+right)
print(ans)
```
No
| 93,714 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Tags: dp, graphs, greedy
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
ans = []
for _ in range(int(input())):
n = int(input())
c = list(map(int,input().split()))
a = list(map(int,input().split()))
b = list(map(int,input().split()))
res = 0
tmp = c[-1] - 1
pos_x,pos_y = 1,c[-1]
for i in range(n-1,0,-1):
tmp += 2
pos_x,pos_y = min(a[i],b[i]),max(a[i],b[i])
res = max(res,tmp+pos_y-pos_x)
if pos_x==pos_y:
tmp = c[i-1] - 1
pos_x,pos_y = 1,c[i-1]
else:
tmp += pos_x - 1 + c[i-1] - pos_y
if c[i-1] - 1 >= tmp:
tmp = c[i-1] - 1
pos_x,pos_y = 1,c[i-1]
ans.append(res)
print(*ans,sep="\n")
```
| 93,715 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Tags: dp, graphs, greedy
Correct Solution:
```
import os,sys
from io import BytesIO,IOBase
for _ in range(int(input())):
n = int(input())
vert = list(map(int,input().split()))
a = list(map(int,input().split()))
b = list(map(int,input().split()))
dp,ans = -10**10,0
for i in range(1,n):
if a[i] == b[i]:
dp = 0
else:
dp = max(dp+min(a[i],b[i])-1+vert[i-1]-max(b[i],a[i])+2,abs(b[i]-a[i]))
ans = max(ans,dp+1+vert[i])
print(ans)
```
| 93,716 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Tags: dp, graphs, greedy
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
c = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
ans = abs(a[1] - b[1])
cur = ans
for i in range(1, n):
if a[i] == b[i]:
ans = max(ans, cur)
cur = c[i] + 1
else:
pres = c[i] + 1
cur = max(cur + pres - abs(a[i] - b[i]),
abs(a[i] - b[i]) + pres)
ans = max(ans, cur)
ans = max(ans, cur)
print(ans)
```
| 93,717 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Tags: dp, graphs, greedy
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
def main():
for _ in range(int(input())):
n = int(input())
vert = list(map(int,input().split()))
a = list(map(int,input().split()))
b = list(map(int,input().split()))
dp,ans = -10**10,0
for i in range(1,n):
if a[i] == b[i]:
dp = 0
else:
dp = max(dp+min(a[i],b[i])-1+vert[i-1]-max(b[i],a[i])+2,abs(b[i]-a[i]))
ans = max(ans,dp+1+vert[i])
print(ans)
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
```
| 93,718 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Tags: dp, graphs, greedy
Correct Solution:
```
t= int(input().strip())
for i in range(t):
n=int(input().strip())
c=list(map(int,input().strip().split()))
a=list(map(int,input().strip().split()))
b=list(map(int,input().strip().split()))
c[0]=0
s=0
m=0
z=0
for k in range(1,n):
if b[k]<a[k]:
a[k],b[k]=b[k],a[k]
for k in range(1,n):
if a[k]!=b[k]:
m=max(m+a[k]+c[k-1]+1-b[k], b[k]-a[k]+1)
s= max(c[k]+m,s)
else:
z=max(z,s)
s=1+c[k]
m=1
print(max(z,s))
```
| 93,719 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Tags: dp, graphs, greedy
Correct Solution:
```
'''
Auther: ghoshashis545 Ashis Ghosh
College: jalpaiguri Govt Enggineering College
'''
from os import path
import sys
from functools import cmp_to_key as ctk
from collections import deque,defaultdict as dd
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod=1000000007
#mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def bo(i):
return ord(i)-ord('a')
def solve():
for _ in range(ii()):
n = ii()
c = li()
a = li()
b = li()
for i in range(n):
if a[i] > b[i]:
a[i],b[i] = b[i],a[i]
ans = 0
s = b[1]-a[1]+1
for i in range(1,n-1):
ans = max(ans,s+c[i])
c1 = a[i+1]+c[i]-b[i+1]
s = max(b[i+1]-a[i+1],s+c1)
if a[i+1] == b[i+1]:
s = 0
s += 1
# print(s,ans)
ans =max(ans,s+c[-1])
print(ans)
if __name__ =="__main__":
solve()
```
| 93,720 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Tags: dp, graphs, greedy
Correct Solution:
```
for t in range(int(input())):
n=int(input())
c=[int(x) for x in input().split()]
a=[int(x) for x in input().split()]
b=[int(x) for x in input().split()]
prev=[0 for i in range(n)]
#best=[0 for i in range(n)]
best2=0
i=1
x=b[i]-a[i]
if(x>0):
temp=x+1
prev[i]=temp
elif(x==0):
prev[i]=1
else:
temp=-x+1
prev[i]=temp
#best[i]=max(prev[i]+c[i],best[i-1])
best2=max(prev[i]+c[i],best2)
for i in range(2,n):
x=b[i]-a[i]
if(x>0):
temp=max(x+1,prev[i-1]+c[i-1]-x+1)
prev[i]=temp
elif(x==0):
prev[i]=1
else:
temp=max(-x+1,prev[i-1]+c[i-1]+x+1)
prev[i]=temp
#best[i]=max(prev[i]+c[i],best[i-1])
best2=max(prev[i]+c[i],best2)
#print(best)
#print(prev)
print(best2)
```
| 93,721 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Tags: dp, graphs, greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
c = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
maxi=0
cnt=c[n-1]
for i in range(n-1,0,-1):
if a[i]>b[i]:
a[i],b[i]=b[i],a[i]
if a[i]==b[i]:
cnt+=1
maxi=max(maxi,cnt)
cnt=c[i-1]
else:
if i==1:
cnt+=b[i]-a[i]+1
maxi=max(maxi,cnt)
else:
cnt1=cnt
cnt1+=b[i]-a[i]+1
cnt+=a[i]+(c[i-1]-b[i]+1)
maxi=max(maxi,cnt)
maxi=max(maxi,cnt1)
if cnt<c[i-1]:
cnt=c[i-1]
print(maxi)
```
| 93,722 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
for t in range(int(input())):
n=int(input())
c=list(map(int,input().split()))
a=list(map(int,input().split()))
b=list(map(int,input().split()))
curlen=0
lstcur=0
ans=0
for i in range(1,n):
curlen=abs(a[i]-b[i]) +1 + c[i]
if a[i]!=b[i]:
curlen=max(curlen, c[i] + 1 + lstcur - abs(a[i]-b[i]))
lstcur=curlen
ans=max(ans,curlen)
lstcur=curlen
print(ans)
```
Yes
| 93,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
# template begins
#####################################
from io import BytesIO, IOBase
import sys
import math
import os
# region fastio
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)
def input(): return sys.stdin.readline().rstrip("\r\n")
def mint(): return map(int, input().split())
def mfloat(): return map(float, input().split())
#####################################
# template ends
# Use the recursion snippet if heavy recursion is needed (depth>1000)
def solve():
n=int(input())
*c,=mint()
*a,=mint()
*b,=mint()
# dp?
# Can we get the longest cycle ending on the given graph?
if b[1]<a[1]:a[1],b[1]=b[1],a[1]
ans=c[1]+2+b[1]-a[1]-1
current_length=ans
for i in range(2, n):
if b[i]<a[i]:a[i],b[i]=b[i],a[i]
if b[i]!=a[i]:
current_length=max(current_length+2+c[i]-1 - (b[i]-a[i]), b[i]-a[i]+2+c[i]-1)
# print(current_length, i)
else:
current_length=2+c[i]-1
ans=max(ans, current_length)
print(ans)
def main():
t = int(input())
for _ in range(t):
solve()
if __name__ == "__main__":
main()
```
Yes
| 93,724 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
for tc in range(int(input())):
n = int(input())
c = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
k = [0]
for i in range(1, n):
if a[i] == b[i]:
k.append(2+c[i]-1)
else:
n1 = abs(a[i]-b[i])+2+c[i]-1 # start afresh
n2 = k[i-1]-abs(a[i]-b[i])+2+c[i]-1
k.append(max(n1, n2))
print(max(k))
```
Yes
| 93,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
t= int(input().strip())
for i in range(t):
n=int(input().strip())
c=list(map(int,input().strip().split()))
a=list(map(int,input().strip().split()))
b=list(map(int,input().strip().split()))
s=0
m=0
z=0
for k in range(1,n):
if b[k]<a[k]:
a[k],b[k]=b[k],a[k]
for k in range(1,n):
if k==1:
if a[k]==b[k]:
s=1+c[k]
m=1
continue
else:
m=b[k]-a[k]+1
s=c[k]+m
continue
if a[k]!=b[k]:
m=max(m+a[k]+c[k-1]+1-b[k], b[k]-a[k]+1)
s= max(c[k]+m,s)
else:
z=max(z,s)
s=1+c[k]
m=1
print(max(z,s))
```
Yes
| 93,726 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
import sys
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from bisect import insort
from collections import Counter
from collections import deque
from heapq import heappush,heappop,heapify
from itertools import permutations,combinations
from itertools import accumulate as ac
mod = int(1e9)+7
#mod = 998244353
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin.readline().rstrip()
out = lambda x : stdout.write(str(x)+"\n")
t = ip()
for _ in range(t):
n = ip()
c = list(inp())
a = list(inp())
b = list(inp())
dp = [0]*n
ele = []
dp[0] = c[0]-1
for i in range(1,n):
if a[i] == b[i]:
dp[i] = c[i]+2-1
else:
minus = abs(a[i]-b[i])
dp[i] = dp[i-1]-minus+ c[i]+2 -1
ele.append(c[i]+minus+1)
dp.pop(0)
ans = -float('inf')
ans = max(ans,max(dp))
if len(ele) != 0:
ans = max(ans,max(ele))
print(ans)
```
No
| 93,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
from math import ceil
def solve(n, c, a, b):
t = 0
r = c[-1]
for i in range(n-1, 0, -1):
if a[i] == b[i]:
r+= 1
if r > t:
t = r
r = c[i-1]
else:
if r+abs(a[i]-b[i])+1 > t:
t = r + abs(a[i]-b[i])+1
if c[i]+ abs(a[i]-b[i])+1 > t:
t = c[i]+ abs(a[i]-b[i])+1
r+= c[i-1] - abs(a[i]-b[i]) +1
print(t)
for _ in range(int(input())):
n = int(input())
c = list(map(int,input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
solve(n, c, a, b)
```
No
| 93,728 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
c = list(map(int, input().split()))
a = list(map(int, input().split()))
b = list(map(int, input().split()))
initialSpan = 0
maxSpan = 0
for i in range(1, n):
if(b[i]!=a[i]):
initialSpan+=c[i]+1
initialSpan-=abs(b[i]-a[i])
secSpan = c[i]+abs(b[i]-a[i])+1
if(secSpan>initialSpan):
maxSpan = max(maxSpan, initialSpan)
maxSpan = max(maxSpan, secSpan)
initialSpan = secSpan
else:
initialSpan = c[i]+1
maxSpan = max(maxSpan, initialSpan)
maxSpan = max(maxSpan, initialSpan)
print(maxSpan)
```
No
| 93,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n chains, the i-th chain consists of c_i vertices. Vertices in each chain are numbered independently from 1 to c_i along the chain. In other words, the i-th chain is the undirected graph with c_i vertices and (c_i - 1) edges connecting the j-th and the (j + 1)-th vertices for each 1 ≤ j < c_i.
Now you decided to unite chains in one graph in the following way:
1. the first chain is skipped;
2. the 1-st vertex of the i-th chain is connected by an edge with the a_i-th vertex of the (i - 1)-th chain;
3. the last (c_i-th) vertex of the i-th chain is connected by an edge with the b_i-th vertex of the (i - 1)-th chain.
<image> Picture of the first test case. Dotted lines are the edges added during uniting process
Calculate the length of the longest simple cycle in the resulting graph.
A simple cycle is a chain where the first and last vertices are connected as well. If you travel along the simple cycle, each vertex of this cycle will be visited exactly once.
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains the single integer n (2 ≤ n ≤ 10^5) — the number of chains you have.
The second line of each test case contains n integers c_1, c_2, ..., c_n (2 ≤ c_i ≤ 10^9) — the number of vertices in the corresponding chains.
The third line of each test case contains n integers a_1, a_2, ..., a_n (a_1 = -1; 1 ≤ a_i ≤ c_{i - 1}).
The fourth line of each test case contains n integers b_1, b_2, ..., b_n (b_1 = -1; 1 ≤ b_i ≤ c_{i - 1}).
Both a_1 and b_1 are equal to -1, they aren't used in graph building and given just for index consistency. It's guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, print the length of the longest simple cycle.
Example
Input
3
4
3 4 3 3
-1 1 2 2
-1 2 2 3
2
5 6
-1 5
-1 1
3
3 5 2
-1 1 1
-1 3 5
Output
7
11
8
Note
In the first test case, the longest simple cycle is shown below:
<image>
We can't increase it with the first chain, since in such case it won't be simple — the vertex 2 on the second chain will break simplicity.
Submitted Solution:
```
t= int(input().strip())
for i in range(t):
n=int(input().strip())
c=list(map(int,input().strip().split()))
a=list(map(int,input().strip().split()))
b=list(map(int,input().strip().split()))
l=[]
s=0
m=0
for k in range(1,n):
if b[k]<a[k]:
a[k],b[k]=b[k],a[k]
print(a,b,c)
for k in range(1,n):
if k==1:
if a[k]==b[k]:
l.append(s)
s=1+c[k]
m=1
l.append(s)
continue
else:
m=b[k]-a[k]+1
s=c[k]+m
l.append(s)
continue
if a[k]!=b[k]:
m=max(m+a[k]+c[k-1]+1-b[k], b[k]-a[k]+1)
s= c[k]+m
print("m ",m)
l.append(s)
else:
l.append(s)
s=2+c[k]
m=1
l.append(s)
print(l)
print(max(l))
```
No
| 93,730 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Tags: dp, implementation, sortings
Correct Solution:
```
from sys import stdin
def main():
MAX_N = 200000
a = [0] * MAX_N
drenched = [0] * MAX_N
def solve():
N = int(next(tokens))
for i in range(N):
a[i] = int(next(tokens))
current_cream = 0
for i in range(N - 1, -1, -1):
current_cream = max(current_cream, a[i])
if current_cream > 0:
current_cream -= 1
drenched[i] = 1
else:
drenched[i] = 0
print(*drenched[:N])
# print(' '.join(drenched[:N]))
tokens = iter(stdin.read().split())
T = int(next(tokens))
for t in range(T):
solve()
if __name__ == '__main__':
main()
```
| 93,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Tags: dp, implementation, sortings
Correct Solution:
```
def solve():
global n,a
ans=[0]*n
amt=0
for i in range(n-1,-1,-1):
if a[i]>amt:
amt=a[i]
if amt>0:
ans[i]=1
amt-=1
for x in range(n):
print(ans[x],end=' ')
print()
t=int(input())
for _ in range(t):
n=int(input())
a=[int(x) for x in input().split()]
solve()
```
| 93,732 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Tags: dp, implementation, sortings
Correct Solution:
```
import sys,os.path
from math import ceil
def II(): return int(sys.stdin.buffer.readline())
def MI(): return map(int, sys.stdin.buffer.readline().split())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
for _ in range(II()):
n=II()
a=LI()
l=[0]*n
i=n-1
nc=0
c=0
while(i>=0):
nc=a[i]
if nc>=c:
c=a[i]
nc=c
if c>0:
l[i]=1
c-=1
i-=1
print(*l)
```
| 93,733 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Tags: dp, implementation, sortings
Correct Solution:
```
from collections import defaultdict,OrderedDict,Counter
from sys import stdin,stdout
from bisect import bisect_left,bisect_right
# import numpy as np
from queue import Queue,PriorityQueue
from heapq import *
from statistics import *
from math import *
import fractions
import copy
from copy import deepcopy
import sys
import io
sys.setrecursionlimit(10000)
import math
import os
import bisect
import collections
mod=pow(10,9)+7
import random
from random import *
from time import time;
def ncr(n, r, p=mod):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def normalncr(n,r):
r=min(r,n-r)
count=1;
for i in range(n-r,n+1):
count*=i;
for i in range(1,r+1):
count//=i;
return count
inf=float("inf")
adj=defaultdict(set)
visited=defaultdict(int)
def addedge(a,b):
adj[a].add(b)
adj[b].add(a)
def bfs(v):
q=Queue()
q.put(v)
visited[v]=1
while q.qsize()>0:
s=q.get_nowait()
print(s)
for i in adj[s]:
if visited[i]==0:
q.put(i)
visited[i]=1
def dfs(v,visited):
if visited[v]==1:
return;
visited[v]=1
print(v)
for i in adj[v]:
dfs(i,visited)
# a9=pow(10,6)+10
# prime = [True for i in range(a9 + 1)]
# def SieveOfEratosthenes(n):
# p = 2
# while (p * p <= n):
# if (prime[p] == True):
# for i in range(p * p, n + 1, p):
# prime[i] = False
# p += 1
# SieveOfEratosthenes(a9)
# prime_number=[]
# for i in range(2,a9):
# if prime[i]:
# prime_number.append(i)
def reverse_bisect_right(a, x, lo=0, hi=None):
if lo < 0:
raise ValueError('lo must be non-negative')
if hi is None:
hi = len(a)
while lo < hi:
mid = (lo+hi)//2
if x > a[mid]:
hi = mid
else:
lo = mid+1
return lo
def reverse_bisect_left(a, x, lo=0, hi=None):
if lo < 0:
raise ValueError('lo must be non-negative')
if hi is None:
hi = len(a)
while lo < hi:
mid = (lo+hi)//2
if x >= a[mid]:
hi = mid
else:
lo = mid+1
return lo
def get_list():
return list(map(int,input().split()))
def get_str_list_in_int():
return [int(i) for i in list(input())]
def get_str_list():
return list(input())
def get_map():
return map(int,input().split())
def input_int():
return int(input())
def matrix(a,b):
return [[0 for i in range(b)] for j in range(a)]
def swap(a,b):
return b,a
def find_gcd(l):
a=l[0]
for i in range(len(l)):
a=gcd(a,l[i])
return a;
def is_prime(n):
sqrta=int(sqrt(n))
for i in range(2,sqrta+1):
if n%i==0:
return 0;
return 1;
def prime_factors(n):
sqrta = int(sqrt(n))
for i in range(2,sqrta+1):
if n%i==0:
return [i]+prime_factors(n//i)
return [n]
def p(a):
if type(a)==str:
print(a+"\n")
else:
print(str(a)+"\n")
def ps(a):
if type(a)==str:
print(a)
else:
print(str(a))
def kth_no_not_div_by_n(n,k):
return k+(k-1)//(n-1)
def forward_array(l):
n=len(l)
stack = []
forward=[0]*n
for i in range(len(l) - 1, -1, -1):
while len(stack) and l[stack[-1]] < l[i]:
stack.pop()
if len(stack) == 0:
forward[i] = len(l);
else:
forward[i] = stack[-1]
stack.append(i)
return forward;
def backward_array(l):
n=len(l)
stack = []
backward=[0]*n
for i in range(len(l)):
while len(stack) and l[stack[-1]] < l[i]:
stack.pop()
if len(stack) == 0:
backward[i] = -1;
else:
backward[i] = stack[-1]
stack.append(i)
return backward
nc="NO"
yc="YES"
ns="No"
ys="Yes"
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# input=stdin.readline
# print=stdout.write
t=int(input())
for i in range(t):
n=int(input())
l=get_list()
if max(l)==1:
print(*l)
continue
l.reverse()
m=defaultdict(int)
for i in range(n):
m[i]+=1
m[i+l[i]]-=1
m[i]+=m[i-1]
k=[0]*n
for i in range(n):
if m[i]:
k[i]=1
k.reverse()
print(*k)
```
| 93,734 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Tags: dp, implementation, sortings
Correct Solution:
```
import sys
t = int(sys.stdin.readline().strip())
for _ in range(t):
n = int(sys.stdin.readline().strip())
a = list(map(int, sys.stdin.readline().split()))
psa = [0 for i in range(n)]
for j in range(n-1, -1, -1):
if a[j] != 0:
if j < n-1:
psa[j+1] -= 1
if j - a[j] + 1 < 0:
psa[0] += 1
else:
psa[j-a[j] + 1] += 1
for j in range(1, n):
psa[j] += psa[j-1]
for j in range(n):
if psa[j] > 0:
print(1, end=' ')
else:
print(0, end=' ')
print('')
```
| 93,735 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Tags: dp, implementation, sortings
Correct Solution:
```
t = int(input())
for _t in range(t):
n = int(input())
a = list(map(int, input().split()))
b = [0 for i in range(n)]
x = 0
for j in range(n - 1, -1, -1):
if a[j] > x:
x = a[j]
if x > 0:
b[j] = 1
x -= 1
for i in range(n):
print(b[i], end=" ")
print()
```
| 93,736 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Tags: dp, implementation, sortings
Correct Solution:
```
from sys import stdin
import math
t = int(stdin.readline())
for _ in range(t):
n = int(stdin.readline())
a = list(map(int, stdin.readline().split()))
dp = [0] * n
nl = 0
for i in range(n-1, -1, -1):
nl = max(nl, a[i])
dp[i] = int(nl > 0)
nl -= 1
print(*dp)
```
| 93,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Tags: dp, implementation, sortings
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
o=[0]*n
remain=0
for i in range(n-1,-1,-1):
if l[i]>0 or remain>0:
o[i]=1
remain=max(remain-1,l[i]-1)
print(*o,sep=" ")
```
| 93,738 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Submitted Solution:
```
t = int(input())
while t:
t-=1
n = int(input())
a=[int(i) for i in input().split()]
b = [0]*n
if a[n-1]>=n:
for i in range(n):
print("1",end=" ")
print()
else:
i=n-1
c=-1
while i>=0:
if a[i]>c:
c=a[i]
if c>0:
b[i]=1
c-=1
i-=1
for i in range(n):
print(b[i],end=" ")
print()
```
Yes
| 93,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Submitted Solution:
```
def answer():
ans=[0 for i in range(n)]
for i in range(n):
ans[i]=max(a[i],ans[i])
inc=0
for i in range(n-1,-1,-1):
if(inc+ans[i]):
inc=max(ans[i],inc)
ans[i]=1
else:
ans[i]=0
inc+=1
inc-=1
return ans
for T in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
print(*answer())
```
Yes
| 93,740 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Submitted Solution:
```
import math
import sys
import collections
import bisect
import string
import time
def get_ints():return map(int, sys.stdin.readline().strip().split())
def get_list():return list(map(int, sys.stdin.readline().strip().split()))
def get_string():return sys.stdin.readline().strip()
for t in range(int(input())):
n = int(input())
arr = get_list()
ans=[0]*n
pos=n
for i in range(n-1,-1,-1):
pos=min(i-arr[i],pos)
if i>pos:
ans[i]=1
print(*ans)
```
Yes
| 93,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Submitted Solution:
```
def get_result(ar, n):
stack = []
ans = [0] * n
for i, val in enumerate(ar):
if val ==0:
stack.append(i)
continue
ans[i] = 1
val -= 1
prev = val
while stack and i-stack[-1] <= prev and val:
top = stack.pop()
ans[top] = 1
val -= 1
print(*ans)
def main():
test = int(input())
for _ in range(test):
n = int(input())
arr = list(map(int, input().split(' ')))
get_result(arr, n)
if __name__ == "__main__":
main()
```
Yes
| 93,742 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Submitted Solution:
```
import sys
def scan(input_type='int'):
if input_type == 'int':
return list(map(int, sys.stdin.readline().strip().split()))
else:
return list(map(str, sys.stdin.readline().strip()))
def solution():
for _ in range(int(input())):
n = int(input())
a = scan()
b = [0]*n
i = n-1
while i > -1:
# print(1)
if a[i] > 0:
j = i
while j > -1 and j > i - a[i]:
b[j] = 1
# print(0)
j -= 1
i -= a[i]
else:
i -= 1
a = [print(i, end=' ') for i in b]
print()
if __name__ == '__main__':
solution()
```
No
| 93,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Submitted Solution:
```
import math
def getint():
return [int(i) for i in input().split()]
def getstr():
return [str(i) for i in input().split()]
#--------------------------------------------------------------------------
def solve():
n=int(input())
a=getint()
ans=""
l=0
for i in range(n-1,-1,-1):
if a[i]==0:
if l<=n-1-i:
ans+="0"
l+=1
else:
leng=min(a[i]-l+n-1-i,n-l)
if leng<0:
leng=0
l+=leng
ans+="1"*leng
print(i,ans)
print(" ".join(ans[::-1]))
#--------------------------------------------------------------------------
for _ in range(int(input())):
solve()
```
No
| 93,744 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Submitted Solution:
```
def solve(t):
n = int(input())
arr = [int(x) for x in input().split()]
ans = [0 for _ in arr]
for i, a in enumerate(arr):
for j in range(a):
ans[i - j] = 1
print(' '.join(map(str, ans)))
T = int(input())
for t in range(1, T + 1):
solve(t)
```
No
| 93,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This week Arkady wanted to cook some pancakes (to follow ancient traditions) and make a problem about that. But then he remembered that one can't make a problem about stacking pancakes without working at a specific IT company, so he decided to bake the Napoleon cake instead.
To bake a Napoleon cake, one has to bake n dry layers first, and then put them on each other in one stack, adding some cream. Arkady started with an empty plate, and performed the following steps n times:
* place a new cake layer on the top of the stack;
* after the i-th layer is placed, pour a_i units of cream on top of the stack.
When x units of cream are poured on the top of the stack, top x layers of the cake get drenched in the cream. If there are less than x layers, all layers get drenched and the rest of the cream is wasted. If x = 0, no layer gets drenched.
<image> The picture represents the first test case of the example.
Help Arkady determine which layers of the cake eventually get drenched when the process is over, and which don't.
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20 000). Description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of layers in the cake.
The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ n) — the amount of cream poured on the cake after adding each layer.
It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
Output
For each test case, print a single line with n integers. The i-th of the integers should be equal to 1 if the i-th layer from the bottom gets drenched, and 0 otherwise.
Example
Input
3
6
0 3 0 0 1 3
10
0 0 0 1 0 5 0 0 0 2
3
0 0 0
Output
1 1 0 1 1 1
0 1 1 1 1 1 0 0 1 1
0 0 0
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
lst = list(map(int,input().split()))
res = [0]*n
for i in range(n):
if lst[i] == 0:
pass
else:
j = i
k = 0
while k < lst[i] and j >= 0:
res[j] = 1
k+=1
j-=1
print(res)
```
No
| 93,746 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
from itertools import permutations as p
from collections import Counter,defaultdict as dd
from sys import stdin,stdout
inp=stdin.readline
for t in range(int(inp())):
s=inp().rstrip('\n')
l=set(s)
l1=[i for i in p(l)]
d=Counter(s)
c=0
ans=l.copy()
for i in l1:
x=0
d1=dd(int)
sh=dd(int)
for j in i:
d1[j]=x
x+=d[j]
x=0
a=0
for j in s:
x+=d1[j]+sh[j]-a
a+=1
d1[j]+=1
for k in i:
if k==j:
break
sh[k]+=1
if x>c:
c=x
ans=list(i)
ansf=[]
for i in ans:
ansf.append(i*d[i])
stdout.write(''.join(ansf)+'\n')
```
| 93,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
import time
#start_time = time.time()
#def TIME_(): print(time.time()-start_time)
import os, sys
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string, heapq as h
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def getBin(): return list(map(int,list(input())))
def isInt(s): return '0' <= s[0] <= '9'
def ceil_(a,b): return (a+b-1)//b
MOD = 10**9 + 7
"""
"""
import itertools
def mergeSort(arr, n):
# A temp_arr is created to store
# sorted array in merge function
temp_arr = [0]*n
return _mergeSort(arr, temp_arr, 0, n-1)
def _mergeSort(arr, temp_arr, left, right):
# A variable inv_count is used to store
# inversion counts in each recursive call
inv_count = 0
# We will make a recursive call if and only if
# we have more than one elements
if left < right:
# mid is calculated to divide the array into two subarrays
# Floor division is must in case of python
mid = (left + right)//2
# It will calculate inversion counts in the left subarray
inv_count += _mergeSort(arr, temp_arr, left, mid)
# It will calculate inversion counts in right subarray
inv_count += _mergeSort(arr, temp_arr, mid + 1, right)
# It will merge two subarrays in a sorted subarray
inv_count += merge(arr, temp_arr, left, mid, right)
return inv_count
# This function will merge two subarrays in a single sorted subarray
def merge(arr, temp_arr, left, mid, right):
i = left # Starting index of left subarray
j = mid + 1 # Starting index of right subarray
k = left # Starting index of to be sorted subarray
inv_count = 0
# Conditions are checked to make sure that i and j don't exceed their
# subarray limits.
while i <= mid and j <= right:
# There will be no inversion if arr[i] <= arr[j]
if arr[i] <= arr[j]:
temp_arr[k] = arr[i]
k += 1
i += 1
else:
# Inversion will occur.
temp_arr[k] = arr[j]
inv_count += (mid-i + 1)
k += 1
j += 1
# Copy the remaining elements of left subarray into temporary array
while i <= mid:
temp_arr[k] = arr[i]
k += 1
i += 1
# Copy the remaining elements of right subarray into temporary array
while j <= right:
temp_arr[k] = arr[j]
k += 1
j += 1
# Copy the sorted subarray into Original array
for loop_var in range(left, right + 1):
arr[loop_var] = temp_arr[loop_var]
return inv_count
ops = list(range(4))
letters = ['A','N','T','O']
def solve():
S = listStr()
N = len(S)
apps = [[] for _ in range(4)]
for i,s in enumerate(S):
if s == 'A': apps[0].append(i)
elif s == 'N': apps[1].append(i)
elif s == 'T': apps[2].append(i)
else: apps[3].append(i)
best = 0
ans = S
for perm in itertools.permutations(ops):
tmp = []
for i in range(4):
tmp += apps[perm[i]]
curr = mergeSort(tmp,N)
if curr > best:
best = curr
ans = []
for i in range(4):
ans += [letters[perm[i]]]*len(apps[perm[i]])
return ''.join(ans)
for _ in range(getInt()):
print(solve())
#solve()
#TIME_()
```
| 93,748 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
import itertools
TC = int(input())
m = {"A": 0, "N": 1, "O": 2, "T": 3}
st = "ANOT"
for _ in range(TC):
s = input()
arr = []
for ch in s:
arr.append(m[ch])
cnt = [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
cnt1 = [0, 0, 0, 0]
for i in arr:
for j in range(4):
cnt[j][i] += cnt1[j]
cnt1[i]+=1
val=-1
ans=[]
for perm in itertools.permutations([0,1,2,3]):
curr=0
for i in range(4):
for j in range(i+1,4):
curr+=cnt[perm[j]][perm[i]]
if curr>val:
val=curr
ans=perm
for i in range(4):
for j in range(cnt1[ans[i]]):
print(st[ans[i]],end="")
print()
```
| 93,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
from itertools import permutations
def f(pos, k):
m = len(pos)
a1, a2 = [0], [0]
for i in range(m):
a1.append(a1[-1]+pos[i]-i)
for i in range(1, m+1):
a2.append(a2[-1]+k-i-pos[m-i])
a2 = a2[::-1]
ind = max(range(m+1), key=lambda i: a1[i]+a2[i])
return ind, a1[ind] + a2[ind]
def solve(s):
n = len(s)
ma = -1
ans = []
for p in permutations(["A","N","O","T"]):
cur = 0
ans1, ans2 = [], []
s1 = s[::]
for c in p:
s2 = []
pos = []
for i in range(len(s1)):
if s1[i] == c:
pos.append(i)
else:
s2.append(s1[i])
x, y = f(pos, len(s1))
cur += y
ans1.extend([c]*x)
ans2.extend([c]*(len(pos)-x))
s1 = s2
ans1.extend(ans2[::-1])
if cur > ma:
ma = cur
ans = ans1
return "".join(ans)
import sys
input = lambda: sys.stdin.readline().rstrip()
t = int(input())
for i in range(t):
s = list(input())
print(solve(s))
```
| 93,750 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
from itertools import permutations
class SegmentTree(): # adapted from https://www.geeksforgeeks.org/segment-tree-efficient-implementation/
def __init__(self,arr,func,initialRes=0):
self.f=func
self.N=len(arr)
self.tree=[0 for _ in range(2*self.N)]
self.initialRes=initialRes
for i in range(self.N):
self.tree[self.N+i]=arr[i]
for i in range(self.N-1,0,-1):
self.tree[i]=self.f(self.tree[i<<1],self.tree[i<<1|1])
def updateTreeNode(self,idx,value): #update value at arr[idx]
self.tree[idx+self.N]=value
idx+=self.N
i=idx
while i>1:
self.tree[i>>1]=self.f(self.tree[i],self.tree[i^1])
i>>=1
def query(self,l,r): #get sum (or whatever function) on interval [l,r] inclusive
r+=1
res=self.initialRes
l+=self.N
r+=self.N
while l<r:
if l&1:
res=self.f(res,self.tree[l])
l+=1
if r&1:
r-=1
res=self.f(res,self.tree[r])
l>>=1
r>>=1
return res
def getMaxSegTree(arr):
return SegmentTree(arr,lambda a,b:max(a,b),initialRes=-float('inf'))
def getMinSegTree(arr):
return SegmentTree(arr,lambda a,b:min(a,b),initialRes=float('inf'))
def getSumSegTree(arr):
return SegmentTree(arr,lambda a,b:a+b,initialRes=0)
def getMinMoves(arr,s):
n=len(s)
charIndexes=dict() # store indexes of each char in s in descending order
for i in range(n-1,-1,-1):
if s[i] not in charIndexes.keys():
charIndexes[s[i]]=[]
charIndexes[s[i]].append(i)
elmCnts=getSumSegTree([1]*n) # to track how many indexes ahead
nMoves=0
for c in arr:
i=charIndexes[c].pop()
nMoves+=(elmCnts.query(0,i)-1)
elmCnts.updateTreeNode(i,0)
return nMoves
def main():
n=int(input())
allans=[]
for _ in range(n):
s=input()
# My guess is that the solution involves grouping all the same characters together
chars=[]
charCnts=dict()
for c in s:
if c not in charCnts.keys():
chars.append(c)
charCnts[c]=0
charCnts[c]+=1
currMax=-1
ans=[]
for p in permutations(chars):
arr=[]
for c in p:
for __ in range(charCnts[c]):
arr.append(c)
nMoves=getMinMoves(arr,s)
if nMoves>currMax:
ans=arr
currMax=nMoves
allans.append(''.join(ans))
multiLineArrayPrint(allans)
return
import sys
# input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
# def readFloatArr():
# return [float(x) for x in input().split()]
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
dv=defaultValFactory;da=dimensionArr
if len(da)==1:return [dv() for _ in range(da[0])]
else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
def queryInteractive(i,j):
print('? {} {}'.format(i,j))
sys.stdout.flush()
return int(input())
def answerInteractive(ans):
print('! {}'.format(' '.join([str(x) for x in ans])))
sys.stdout.flush()
inf=float('inf')
MOD=10**9+7
# MOD=998244353
for _abc in range(1):
main()
```
| 93,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
import sys
from itertools import permutations
def I(): return int(sys.stdin.readline().rstrip())
def MI(): return map(int,sys.stdin.readline().rstrip().split())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def LI2(): return list(map(int,sys.stdin.readline().rstrip()))
def S(): return sys.stdin.readline().rstrip()
def LS(): return list(sys.stdin.readline().rstrip().split())
def LS2(): return list(sys.stdin.readline().rstrip())
t = I()
C = ['A','N','T','O']
for _ in range(t):
X = S()
count1 = [0]*4
count2 = [[0]*4 for _ in range(4)]
for x in X:
if x == 'A':
count1[0] += 1
for i in range(4):
if i == 0:
continue
count2[i][0] += count1[i]
elif x == 'N':
count1[1] += 1
for i in range(4):
if i == 1:
continue
count2[i][1] += count1[i]
elif x == 'T':
count1[2] += 1
for i in range(4):
if i == 2:
continue
count2[i][2] += count1[i]
else:
count1[3] += 1
for i in range(4):
if i == 3:
continue
count2[i][3] += count1[i]
M = -1
ans = (0,0,0,0)
for A in list(permutations([0,1,2,3],4)):
cnt = 0
for i in range(4):
a = A[i]
for j in range(i+1,4):
b = A[j]
cnt += count2[b][a]
if M < cnt:
M = cnt
ans = A
ANS = [C[ans[0]]]*count1[ans[0]]+[C[ans[1]]]*count1[ans[1]]+[C[ans[2]]]*count1[ans[2]]+[C[ans[3]]]*count1[ans[3]]
ANS = ''.join(ANS)
print(ANS)
```
| 93,752 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq,bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
from itertools import permutations
import threading
from collections import defaultdict
threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
sys.setrecursionlimit(300000)
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")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**51, func=lambda a, b: a & b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary-----------------------------------
def merge(arr, temp, left, mid, right):
inv_count = 0
i = left
j = mid
k = left
while ((i <= mid - 1) and (j <= right)):
if (arr[i] <= arr[j]):
temp[k] = arr[i]
k += 1
i += 1
else:
temp[k] = arr[j]
k += 1
j += 1
inv_count = inv_count + (mid - i)
while (i <= mid - 1):
temp[k] = arr[i]
k += 1
i += 1
while (j <= right):
temp[k] = arr[j]
k += 1
j += 1
for i in range(left, right + 1, 1):
arr[i] = temp[i]
return inv_count
def _mergeSort(arr, temp, left, right):
inv_count = 0
if (right > left):
mid = int((right + left) / 2)
inv_count = _mergeSort(arr, temp, left, mid)
inv_count += _mergeSort(arr, temp, mid + 1, right)
inv_count += merge(arr, temp, left, mid + 1, right)
return inv_count
def countSwaps(arr, n):
temp = [0 for i in range(n)]
return _mergeSort(arr, temp, 0, n - 1)
for ik in range(int(input())):
s=list(input())
n=len(s)
ind=defaultdict(int)
c=defaultdict(int)
for i in range(n-1,-1,-1):
c[s[i]]+=1
ind = defaultdict(list)
for i in range(n):
ind[s[i]].append(i)
def find(d,tr):
tr=list(tr)
fi=""
for i in tr:
fi+=i*c[i]
ans=[]
for i in tr:
for j in ind[i]:
ans.append(j)
return (countSwaps(ans,len(ans)),fi)
perm = [''.join(p) for p in permutations("ANTO")]
ans=(0,"")
for i in perm:
ans=max(ans,find(s,i))
print(ans[1])
```
| 93,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Tags: brute force, constructive algorithms, data structures, math, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import Counter, deque, defaultdict
from itertools import permutations
class Fenwick:
def __init__(self, n):
self.tree = [0] * (n + 1)
self.n = n
def update(self, x, val):
while x <= self.n:
self.tree[x] += val
x += x & -x
def query(self, x):
ans = 0
while x:
ans += self.tree[x]
x -= x & -x
return ans
for _ in range(int(input())):
s = input()[:-1]
n = len(s)
cand = set(s)
m = len(cand)
cnt = Counter(s)
ans = -1
for idx in permutations(cand):
cur = ""
for x in idx:
cur += x * cnt[x]
dic = defaultdict(deque)
for i, c in enumerate(s, 1):
dic[c].append(i)
A = []
for c in cur:
A.append(dic[c].popleft())
bit = Fenwick(n)
tot = 0
for a in A:
tot += bit.query(n) - bit.query(a)
bit.update(a, 1)
if tot > ans:
ans = tot
P = cur
print(P)
```
| 93,754 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
from itertools import permutations
t = int(input())
perms = list(permutations(range(4)))
def inversion(d, L):
if len(d) == 1:
return (g for g in d), 1, 0
a, A, b = inversion(d[:L//2], L//2)
c, C, e = inversion(d[L//2:], (L+1)//2)
k = 0
s = []
a_ = next(a)
c_ = next(c)
while a_ != -1 != c_:
if a_ < c_:
s.append(a_)
a_ = next(a, -1)
A -= 1
else:
s.append(c_)
c_ = next(c, -1)
k += A
while a_ != -1:
s.append(a_)
a_ = next(a, -1)
while c_ != -1:
s.append(c_)
c_ = next(c, -1)
return (g for g in s), L, b+e+k
for _ in range(t):
s = input().strip()
pos = [[], [], [], []]
for i, c in enumerate(s):
pos["ANTO".index(c)].append(i)
siz = [len(a) for a in pos]
m = 0
best = (0, 1, 2, 3)
for k in perms:
d = []
for j in range(4):
d += pos[k[j]]
z = inversion(d, len(d))[2]
if m < z:
best = k
m = z
res = ""
for x in best:
res += "ANTO"[x] * siz[x]
print(res)
```
Yes
| 93,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#######################################
from itertools import permutations as p
from collections import Counter,defaultdict as dd
for t in range(int(input())):
s=input()
l=set(s)
l1=[i for i in p(l)]
d=Counter(s)
c=0
ans=l.copy()
for i in l1:
x=0
d1=dd(int)
sh=dd(int)
for j in i:
d1[j]=x
x+=d[j]
x=0
a=0
for j in s:
x+=d1[j]+sh[j]-a
a+=1
d1[j]+=1
for k in i:
if k==j:
break
sh[k]+=1
if x>c:
c=x
ans=list(i)
ansf=[]
for i in ans:
ansf.append(i*d[i])
print(''.join(ansf))
```
Yes
| 93,756 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
#!/usr/bin/env python3
import sys
import getpass # not available on codechef
import math, random
import functools, itertools, collections, heapq, bisect
from collections import Counter, defaultdict, deque
input = sys.stdin.readline # to read input quickly
# available on Google, AtCoder Python3, not available on Codeforces
# import numpy as np
# import scipy
M9 = 10**9 + 7 # 998244353
yes, no = "YES", "NO"
# d4 = [(1,0),(0,1),(-1,0),(0,-1)]
# d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
# d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)] # hexagonal layout
MAXINT = sys.maxsize
# if testing locally, print to terminal with a different color
OFFLINE_TEST = getpass.getuser() == "hkmac"
# OFFLINE_TEST = False # codechef does not allow getpass
def log(*args):
if OFFLINE_TEST:
print('\033[36m', *args, '\033[0m', file=sys.stderr)
def solve(*args):
# screen input
if OFFLINE_TEST:
log("----- solving ------")
log(*args)
log("----- ------- ------")
return solve_(*args)
def read_matrix(rows):
return [list(map(int,input().split())) for _ in range(rows)]
def read_strings(rows):
return [input().strip() for _ in range(rows)]
def minus_one(arr):
return [x-1 for x in arr]
def minus_one_matrix(mrr):
return [[x-1 for x in row] for row in mrr]
# ---------------------------- template ends here ----------------------------
class FenwickTree:
# also known as Binary Indexed Tree
# binarysearch.com/problems/Virtual-Array
# https://leetcode.com/problems/create-sorted-array-through-instructions
# may need to be implemented again to reduce constant factor
def __init__(self, bits=31):
self.c = defaultdict(int)
self.LARGE = 2**bits
def update(self, x, increment):
x += 1 # to avoid infinite loop at x > 0
while x <= self.LARGE:
# increase by the greatest power of two that divides x
self.c[x] += increment
x += x & -x
def query(self, x):
x += 1 # to avoid infinite loop at x > 0
res = 0
while x > 0:
# decrease by the greatest power of two that divides x
res += self.c[x]
x -= x & -x
return res
def count_inversions(perm):
res = 0
t = FenwickTree(bits = len(bin(max(perm))))
for i,x in enumerate(perm):
cnt = t.query(x)
res += i-cnt
t.update(x, 1)
return res
def solve_(srr):
if len(srr) == 1:
return srr
c1 = defaultdict(list)
for i,x in enumerate(srr):
c1[x].append(i)
c = Counter(srr)
maxval = -1
maxres = None
for comb in itertools.permutations(c.keys()):
locs = []
for k in comb:
locs.extend(c1[k])
val = count_inversions(locs)
if val > maxval:
maxval = val
maxres = comb
log(maxval)
trr = ""
for k in maxres:
trr += k*len(c1[k])
# log(maxval)
return trr
# for case_num in [0]: # no loop over test case
# for case_num in range(100): # if the number of test cases is specified
for case_num in range(int(input())):
# read line as an integer
# k = int(input())
# read line as a string
srr = input().strip()
# read one line and parse each word as a string
# lst = input().split()
# read one line and parse each word as an integer
# a,b,c = list(map(int,input().split()))
# lst = list(map(int,input().split()))
# lst = minus_one(lst)
# read multiple rows
# arr = read_strings(k) # and return as a list of str
# mrr = read_matrix(k) # and return as a list of list of int
# mrr = minus_one_matrix(mrr)
res = solve(srr) # include input here
# print length if applicable
# print(len(res))
# parse result
# res = " ".join(str(x) for x in res)
# res = "\n".join(str(x) for x in res)
# res = "\n".join(" ".join(str(x) for x in row) for row in res)
# print result
# print("Case #{}: {}".format(case_num+1, res)) # Google and Facebook - case number required
print(res)
```
Yes
| 93,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
from itertools import permutations
class BIT():
def __init__(self, init):
if type(init) == int:
self.n = init + 1
self.X = [0] * self.n
else:
self.n = len(init) + 1
self.X = [0] + init
for i in range(1, self.n):
if i + (i & -i) < self.n:
self.X[i + (i & -i)] += self.X[i]
def add(self, i, x=1):
i += 1
while i < self.n:
self.X[i] += x
i += i & (-i)
def getsum(self, i):
ret = 0
while i != 0:
ret += self.X[i]
i -= i & (-i)
return ret
def getrange(self, l, r):
return self.getsum(r) - self.getsum(l)
def inversion_number(L):
bit = BIT(len(L))
ret = 0
for i, a in enumerate(L):
ret += i - bit.getsum(a + 1)
bit.add(a)
return ret
def inversion_number(L): # ��������
ll = sorted(list(set(L)))
dd = {a: i for i, a in enumerate(ll)}
nL = [dd[l] for l in L]
bit = BIT(len(nL))
ret = 0
for i, a in enumerate(nL):
ret += i - bit.getsum(a + 1)
bit.add(a)
return ret
def calc(A):
n = len(A)
C = [0] * 4
for a in A:
C[a] += 1
ma = 0
pma = [0, 1, 2, 3]
for p in permutations([0, 1, 2, 3]):
D = [0] * 4
c = 0
for i, a in enumerate(p):
D[a] = c
c += C[a]
B = [0] * n
for i, a in enumerate(A):
B[i] = D[a]
D[a] += 1
# print("p, B =", p, B)
ivn = inversion_number(B)
if ivn > ma:
ma = ivn
pma = p
# print("ma, p =", ma, p)
re = []
for a in pma:
re.append("ANTO"[a] * C[a])
return "".join(re)
T = int(input())
for _ in range(T):
A = [0 if a == "A" else 1 if a == "N" else 2 if a == "T" else 3 for a in input()]
print(calc(A))
```
Yes
| 93,758 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
t=int(input());
for i in range(t):
w=input();
nw=w[1:]+w[:0]
print(nw)
```
No
| 93,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
import sys
input = sys.stdin.readline
from itertools import permutations
# all operations are log(n)
class bit:
def __init__(self, n):
self.n = n+1
self.mx_p = 0
while 1 << self.mx_p < self.n:
self.mx_p += 1
self.a = [0]*(self.n)
self.tot = 0
def add(self, idx, val):
self.tot += val
idx += 1
while idx < self.n:
self.a[idx] += val
idx += idx & -idx
def sum_prefix(self, idx):
tot = 0
idx += 1
while idx > 0:
tot += self.a[idx]
idx -= idx & -idx
return tot
# inversions of a permutation 0..len(a)-1
def inversions(a):
tot = 0
loc = [0]*len(a)
for i in range(len(a)):
loc[a[i]] = i
bt = bit(len(a))
for i in range(len(a)-1,-1,-1):
tot += bt.sum_prefix(loc[i])
bt.add(loc[i],1)
return tot
# creates array c where c[i] = position of a[i] in b
def create_c(a,b):
c = [0]*len(a)
idx = {}
positions = {}
for i in range(len(b)):
if b[i] in positions:
positions[b[i]].append(i)
else:
positions[b[i]] = [i]
for i in range(len(a)):
if a[i] in idx:
idx[a[i]] += 1
else:
idx[a[i]] = 0
c[i] = positions[a[i]][idx[a[i]]]
return c
for _ in range(int(input())):
a = input().strip()
if len(a) <= 4:
print(a)
continue
counts = dict(zip("ANTO",[0]*4))
for i in a:
counts[i] += 1
best_perm = None
best_perm_cost = -1
for order in permutations("ANTO"):
final_perm = ''.join([char*counts[char] for char in order])
cost = inversions(create_c(a,final_perm))
if cost > best_perm_cost:
best_perm = final_perm
best_perm_cost = cost
print(best_perm)
```
No
| 93,760 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
#from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write(' '.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
#from decimal import Decimal
#from fractions import Fraction
#sys.setrecursionlimit(100000)
INF = float('inf')
mod = int(1e9)+7
for t in range(int(data())):
a = data()
m = [[0]*26 for i in range(26)]
d = dd(int)
for i in a:
s = ord(i)-65
for j in range(26):
if s!=j:
m[s][j] += m[j][j]
m[s][s] += 1
d[i] += 1
l = sorted(d.keys())
total = 0
for i in range(len(l)-1):
s1 = ord(l[i]) - 65
for j in range(i+1,len(l)):
s2 = ord(l[j]) - 65
total += m[s1][s2]
flag = True
while flag:
flag = False
for i in range(1,len(l)):
ind = i
total1 = total
s1 = ord(l[i]) - 65
for j in range(i-1,-1,-1):
s2 = ord(l[j]) - 65
total1 += m[s1][s2] - m[s2][s1]
if total1>total:
total = total1
flag = True
while ind!=j:
l[ind],l[ind-1] = l[ind-1], l[ind]
ind -= 1
ans = ''
for i in l:
ans += i*d[i]
out(ans)
```
No
| 93,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After rejecting 10^{100} data structure problems, Errorgorn is very angry at Anton and decided to kill him.
Anton's DNA can be represented as a string a which only contains the characters "ANTON" (there are only 4 distinct characters).
Errorgorn can change Anton's DNA into string b which must be a permutation of a. However, Anton's body can defend against this attack. In 1 second, his body can swap 2 adjacent characters of his DNA to transform it back to a. Anton's body is smart and will use the minimum number of moves.
To maximize the chance of Anton dying, Errorgorn wants to change Anton's DNA the string that maximizes the time for Anton's body to revert his DNA. But since Errorgorn is busy making more data structure problems, he needs your help to find the best string B. Can you help him?
Input
The first line of input contains a single integer t (1 ≤ t ≤ 100000) — the number of testcases.
The first and only line of each testcase contains 1 string a (1 ≤ |a| ≤ 100000). a consists of only the characters "A", "N", "O" and "T".
It is guaranteed that the sum of |a| over all testcases does not exceed 100000.
Output
For each testcase, print a single string, b. If there are multiple answers, you can output any one of them. b must be a permutation of the string a.
Example
Input
4
ANTON
NAAN
AAAAAA
OAANTTON
Output
NNOTA
AANN
AAAAAA
TNNTAOOA
Note
For the first testcase, it takes 7 seconds for Anton's body to transform NNOTA to ANTON:
NNOTA → NNOAT → NNAOT → NANOT → NANTO → ANNTO → ANTNO → ANTON.
Note that you cannot output strings such as AANTON, ANTONTRYGUB, AAAAA and anton as it is not a permutation of ANTON.
For the second testcase, it takes 2 seconds for Anton's body to transform AANN to NAAN. Note that other strings such as NNAA and ANNA will also be accepted.
Submitted Solution:
```
import sys
input = sys.stdin.readline
print = lambda s: sys.stdout.write(s + "\n")
import itertools
import collections
perms = list(itertools.permutations(['A', 'N', 'O', 'T']))
class FenwickTree:
def __init__(self, n):
self.t = [0] * (n + 1)
def update(self, i, x):
i += 1
while i < len(self.t):
self.t[i] += x
i += i & -i
def _query(self, r):
ans = 0
while r > 0:
ans += self.t[r]
r -= r & -r
return ans
def query(self, l, r):
return self._query(r) - self._query(l)
t = int(input())
for _ in range(t):
s = input().rstrip()
n = len(s)
ctr = collections.Counter(s)
ans = (float('-inf'), None)
for perm in perms:
curVal = 0
curS = []
firstIdx = {}
for p in perm:
startIdx = 0
for x in curS:
startIdx += len(x)
firstIdx[p] = startIdx
curS.append(p * ctr[p])
curS = ''.join(curS)
ft = FenwickTree(n)
for i, x in enumerate(curS):
ft.update(i, 1)
for i, x in enumerate(s):
curVal += ft.query(i, firstIdx[x])
ft.update(firstIdx[x], -1)
if curVal > ans[0]:
ans = (curVal, curS)
print(ans[1])
```
No
| 93,762 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Tags: *special, greedy, implementation
Correct Solution:
```
num = int(input())
i=0
total=0
k=0
k1=0
k2=0
k3=0
k4=0
arr=[]
arr = input().split()
for i in range(len(arr)):
arr[i]= int(arr[i])
for nmb in arr:
nmb= int(nmb)
if nmb==4:
k4+=1
elif nmb==3:
k3+=1
elif nmb==2:
k2+=1
elif nmb==1:
k1+=1
if k2==0 and k1==0:
total=int(k4+k3)
elif k1!=0 and k2==0:
if k3>=k1:
total=k3+k4
else:
if k1%4==0:
total=k4+k3+((k1-k3)/4)
else:
total=k4+k3+((k1-k3)/4)+1
elif k1!=0 and k2!=0:
if k3>k1:
t2=(k2*2)
if t2%4==0:
total=int(t2/4)+k4+k1+(k3-k1)
else:
total=int(t2/4)+1+k4+k1+(k3-k1)
else:
t2=k4*4+k3*3+k2*2+k1
if t2%4==0:
total=int(t2/4)
else:
total=int(t2/4)+1
elif k1==0:
if (k2*2)%4==0:
total=int(k4+k3+(k2*2/4))
else:
total=k4+k3+(int(k2*2/4)+1)
print(int(total))
```
| 93,763 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Tags: *special, greedy, implementation
Correct Solution:
```
n = int(input())
s = list(map(int, input().split()))
cabs = 0
one = s.count(1)
two = s.count(2)
three = s.count(3)
four = s.count(4)
cabs += two // 2 + three + four
one -= three
if two % 2 == 1:
cabs += 1
one -= 2
if one > 0:
cabs += (one + 3) // 4
print(cabs)
```
| 93,764 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Tags: *special, greedy, implementation
Correct Solution:
```
n=int(input())
taxi=n1=n2=n3=n4=0
x=[]
for i in range(0,n):
x.append('0')
x=input().split()
for i in range(0,n):
if(x[i]=='1'):
n1+=1
elif(x[i]=='2'):
n2+=1
elif(x[i]=='3'):
n3+=1
else:
n4+=1
taxi+=n4
if(n3>n1 or n3==n1):
taxi+=n3
n1=0
else:
taxi+=n3
n1=n1-n3
taxi+=n2//2
n2=n2%2
if(n2==1):
taxi+=1
if(n1>2):
n1-=2
else:
n1=0
taxi+=(n1//4)
if(n1%4!=0 and n1!=0):
taxi+=1
print(taxi)
```
| 93,765 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Tags: *special, greedy, implementation
Correct Solution:
```
n = int(input())
l = list(map(int, input().split()))
f = 0
th = 0
tw = 0
o = 0
for i in range(len(l)):
#print (l[i])
if l[i] == 1:
o += 1
elif l[i] == 2:
tw += 1
elif l[i] == 3:
th += 1
elif l[i] == 4:
f += 1
#print (f, th, tw, o)
ans = f
#print(ans)
ans += th
#print(ans)
o = max(0, o - th)
ans += tw // 2
#print(ans)
if (tw % 2):
ans += 1
if (tw % 2) and o:
o = max(0, o - 2)
ans += o // 4
#print(ans)
if o % 4:
ans += 1
print(ans)
```
| 93,766 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Tags: *special, greedy, implementation
Correct Solution:
```
n = input()
t = list(map(int,input().split()))
a,b,c,d = [t.count(x+1) for x in range(4)]
print(d+c--b//2--max(0,a-c-b%2*2)//4)
```
| 93,767 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Tags: *special, greedy, implementation
Correct Solution:
```
import math
cars = 0
n = int(input())
s = [int(i) for i in input().split()]
groups = [0,0,0,0]
for i in s:
groups[i-1] += 1
cars += groups[3]
if groups[2] >= groups[0]:
cars += groups[0]
groups[2] -= groups[0]
groups[0] = 0
else:
cars += groups[2]
groups[0] -= groups[2]
groups[2] = 0
cars += groups[1]//2
groups[1] -= groups[1]//2*2
cars += groups[2]
if groups[1] == 1:
cars += 1
if groups[0] >= 2:
groups[0] -= 2
else:
groups[0] = 0
cars += int(math.ceil(groups[0]/4))
print(cars)
```
| 93,768 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Tags: *special, greedy, implementation
Correct Solution:
```
# Solution found online
input()
arr = [0, 0, 0, 0, 0]
for i in input().split():
arr[int(i)] += 1
total = arr[4] + arr[3] + arr[2] // 2
arr[1] -= arr[3]
if arr[2] % 2 == 1:
total += 1
arr[1] -= 2
if arr[1] > 0:
total += (arr[1] + 3) // 4
print(total)
# MY ORIGINAL SOLUTION BELOW
# Interestingly enough both solutions are the same runtime
# obv the above solution is hella cleaner though
# input()
# arr = [int(i) for i in input().split()]
# k = [0, 0, 0, 0, 0]
# for i in arr:
# k[i] += 1
# t = 0
# t += k[4]
# k[4] = 0
# ot = min(k[1], k[3])
# t += ot
# k[1] -= ot
# k[3] -= ot
# twos = k[2] // 2
# t += twos
# k[2] -= twos * 2
# k[2] %= 2
# if k[2] == 1:
# if k[1] > 0:
# t += 1
# k[2] = 0
# k[1] -= min(k[1], 2)
# else:
# t += 1
# k[2] = 0
# for i in range(4, 0, -1):
# num = k[1] // i
# t += num
# k[1] -= num * i
# t += k[3]
# print(t)
```
| 93,769 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Tags: *special, greedy, implementation
Correct Solution:
```
# fails only in 3 3 2 it give 2 but answer is 3
# import math
# n=int(input())
# pas=list(map(int,input().split()[:n]))
# sums = 0
# for i in range(0,len(pas)):
# sums=sums+pas[i]
# if(n == 3):
# if(pas[0] == 3 and pas[1] == 3 and pas[2]==2):
# print('3')
# else:
# print(math.ceil(sums/4))
# else:
# print(math.ceil(sums/4))
# special Implementation
input()
a,b,c,d=map(input().count,('1','2','3','4'))
print(d+c+(b*2+max(0,a-c)+3)//4)
```
| 93,770 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Submitted Solution:
```
import collections
n = int(input())
a = list(map(int,input().split()))
k = collections.Counter(a)
r = 0
r+= k[4]
temp = k[2] // 2
r+= temp
k[2] -= temp*2
k[4] = 0
if k[3]<k[1]:
r += k[3]
k[1] -= k[3]
k[3]-=k[3]
else:
r += k[1]
k[3] -= k[1]
k[1] -= k[1]
r += k[3]
k[3] = 0
if k[1]>=1 and k[2] == 1:
r+= 1
k[1] -= 2
k[2] -= 1
if k[2] == 1:
r += 1
k[2] = 0
while k[1]>0:
r+=1
k[1]-=4
print(r)
```
Yes
| 93,771 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split(" ")))
c1=0
c2=0
c3=0
c4=0
for i in range(0,n):
if a[i]==1: c1+=1
elif a[i]==2: c2+=1
elif a[i]==3: c3+=1
elif a[i]==4: c4+=1
count=c4+c3
c1=c1-c3
count=count+(c2//2)
if(c2%2>0) :
count=count+1
c1=c1-2
if(c1>0) :
count=count+(c1//4)
if(c1%4>0) : count=count+1
print(count)
```
Yes
| 93,772 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Submitted Solution:
```
arr=[]
for i in range(0,5):
arr.append(0)
n=int(input())
a=list(map(int,input().split()))
for i in range(0,n):
arr[a[i]]+=1
con=0
con+=arr[4]
con+=arr[3]
if arr[1]>arr[3]:
if arr[2]!=0:
if arr[2]%2==0:
con+=arr[2]//2
arr[1]=arr[1]-arr[3]
if arr[1]%4==0:
con+=arr[1]//4
else:
con+=(arr[1]//4)+1
else:
arr[1] = arr[1]-arr[3]
con+=(arr[2]//2)+1
if arr[1]>2:
arr[1]-=2
if arr[1]%4==0:
con+=arr[1]//4
else:
con += (arr[1]//4)+1
else:
arr[1] = arr[1] - arr[3]
if arr[1] % 4 == 0:
con += arr[1] // 4
else:
con += (arr[1] //4) + 1
elif arr[1]<=arr[3]:
if arr[2]!=0:
if arr[2]%2==0:
con+=arr[2]//2
else:
con+=(arr[2]//2)+1
print(con)
```
Yes
| 93,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Submitted Solution:
```
n=int(input())
s=input()
o=0
tw=0
t=0
f=0
for i in s:
if i=="4":
f+=1
elif i=="3":
t+=1
elif i=="2":
tw+=1
elif i=="1":
o+=1
if o-t<=0:
l=f+t+tw//2+tw%2
elif o-t<=2 and tw%2!=0:
l=f+t+tw//2+tw%2
elif o-t<=4 and tw%2==0:
l=f+t+tw//2+1
elif (o-t-(tw%2)*2)%4==0:
l=f+t+tw//2+(o-t-(tw%2)*2)//4+tw%2
else:
l=f+t+tw//2+(o-t-(tw%2)*2)//4+tw%2+1
print(l)
```
Yes
| 93,774 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Submitted Solution:
```
n =int(input())
x = list(map(int, input().split()))
i = 1
r = 0
while i < n+1:
r += x[i-1]
i += 1
k = (r + 3) // 4
if (n<4)and(k!=n)and(r<=4*k): k = n
print(k)
```
No
| 93,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Submitted Solution:
```
n = int(input())
s = input()
l = s.split(" ")
l.sort(reverse=True)
for i in range(0, len(l)):
l[i] = int(l[i])
c = 0
for x in range(n):
if l[x] == 4:
c += 1
l[x] = 0
elif l[x] != 0:
for y in range(x + 1, n):
if l[x] + l[y] == 4 and l[y] != 0:
c += 1
l[x] = 0
l[y] = 0
for x in l:
if x != 0:
c += 1
print(c)
```
No
| 93,776 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Submitted Solution:
```
n = int(input())
arr = list(map(int, input().split()))
if sum(arr) >= 4:
s = sum(arr) // 4
l = sum(arr) % 4
if l == 0:
print(l + s)
else:
print(s + 1)
else:
print(1)
```
No
| 93,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, ..., sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.
Output
Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.
Examples
Input
5
1 2 4 3 3
Output
4
Input
8
2 3 4 4 2 1 3 1
Output
5
Note
In the first test we can sort the children into four cars like this:
* the third group (consisting of four children),
* the fourth group (consisting of three children),
* the fifth group (consisting of three children),
* the first and the second group (consisting of one and two children, correspondingly).
There are other ways to sort the groups into four cars.
Submitted Solution:
```
t = int(input())
a = list(map(int,input().split()))
c1 = 0
c2 = 0
c3 = 0
c4 = 0
for i in range(t) :
if a[i] == 4 :
c4 += 1
if a[i] == 3 :
c3 += 1
if a[i] == 2 :
c2 += 1
if a[i] == 1 :
c1 += 1
ans = c4 + c3
if c2%2 == 0:
ans += c2//2
if c2%2 == 1 and c3>=c1:
co2 = (c2-1)//2
ans = ans + co2 + 1
elif c3 < c1 :
co1 = c1-c3
if co1%4==0:
ans += (co1//4)
else :
ans += (co1//4) + 1
## if co1%4==0:
## ans += (co1//4)
## elif co1%4==1 :
## ans += (co1+1)//4
## elif co1%4==2 :
## ans += (co1+2)//4
## else :
## ans += (co1+3)//4
print(ans)
```
No
| 93,778 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Tags: geometry, math
Correct Solution:
```
import math
EPSILON = 0.0001
def radians(x1, y1, x2, y2):
product = x1*x2 + y1*y2
norm1 = (x1*x1+y1*y1) ** 0.5
norm2 = (x2*x2+y2*y2) ** 0.5
cosVal = max(-1, min(1, product/norm1/norm2))
return math.acos(cosVal)
x1, y1 = map(float, input().split(" "))
x2, y2 = map(float, input().split(" "))
x3, y3 = map(float, input().split(" "))
a, b, c, d = x1-x2, y1-y2, x1-x3, y1-y3
e = ((x1*x1-x2*x2) - (y2*y2-y1*y1)) / 2
f = ((x1*x1-x3*x3) - (y3*y3-y1*y1)) / 2
ox = -(d*e-b*f)/(b*c-a*d)
oy = -(a*f-c*e)/(b*c-a*d)
alpha1 = radians(x1-ox, y1-oy, x2-ox, y2-oy)
alpha2 = radians(x1-ox, y1-oy, x3-ox, y3-oy)
for i in range(3, 101):
unit = 2*math.pi/i
if abs(alpha1/unit - round(alpha1/unit)) > EPSILON:
continue
if abs(alpha2/unit - round(alpha2/unit)) > EPSILON:
continue
radius = ((x1-ox)**2 + (y1-oy)**2) ** 0.5
area = 0.5 * radius * radius * math.sin(unit) * i;
print(area)
exit()
```
| 93,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Tags: geometry, math
Correct Solution:
```
from math import *
def rt(x, y):
return sum((i - j) ** 2 for i, j in zip(x, y))
def tr(a, b, c):
return acos((b + c - a) / (2 * (b * c) ** 0.5)) / pi
def trt(x, n):
return 0.01 < (x * n) % 1 < 0.99
x, y, z = (tuple(map(float, input().split())) for i in range(3))
a, b, c = rt(x, y), rt(x, z), rt(y, z)
t, s = tr(a, b, c), tr(b, a, c)
r, n = a / (8 * sin(t * pi) ** 2), 3
while trt(t, n) or trt(s, n): n += 1
print(n * r * sin(2 * pi / n))
```
| 93,780 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Tags: geometry, math
Correct Solution:
```
from math import *
p =[list(map(float,input().split())) for i in range(3)]
a,b,c=[hypot(x1-x2,y1-y2) for (x1,y1),(x2,y2) in [(p[0],p[1]),(p[0],p[2]),(p[1],p[2])]]
A,B,C=[acos((y*y+z*z-x*x)/2/z/y) for x,y,z in [(a,b,c),(b,c,a),(c,a,b)]]
R=a/2/sin(A)
def g(x,y):return x if y<1e-3 else g(y,fmod(x,y))
u=2*g(A,g(B,C))
print(round(R * R * pi / u * sin(u),7))
```
| 93,781 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Tags: geometry, math
Correct Solution:
```
from math import sqrt, asin, pi, sin, cos
eps = 1e-6
def dist2(ax, ay, bx, by):
return (ax-bx)**2 + (ay-by)**2
def cross_prod(ax, ay, bx, by):
return ax*by - ay*bx
def inner_prod(ax, ay, bx, by):
return ax*bx + ay*by
def find(x, y, s, c):
for i in range(len(s)):
if abs(x-s[i]) < eps and abs(y-c[i]) < eps:
return True
return False
def pool(x):
if abs(x) < eps:
return 1.
if x < 0:
return x/pi+2
return x/pi
def main():
Ax, Ay = map(float, input().split())
Bx, By = map(float, input().split())
Cx, Cy = map(float, input().split())
#print(Ax, Ay, Bx, By, Cx, Cy)
#for _ in range(n):
# print(ans(input()))
D = 2*(Ax*(By-Cy) + Bx*(Cy-Ay) + Cx*(Ay-By))
Ox = ( (Ax**2+Ay**2)*(By-Cy) + (Bx**2+By**2)*(Cy-Ay) + (Cx**2+Cy**2)*(Ay-By) ) / D
Oy = ( (Ax**2+Ay**2)*(Cx-Bx) + (Bx**2+By**2)*(Ax-Cx) + (Cx**2+Cy**2)*(Bx-Ax) ) / D
R2 = dist2(Ox,Oy,Ax,Ay)
#print(R)
#print(Ox, Oy)
#print(dist(Ox,Oy,Ax,Ay))
#print(dist(Ox,Oy,Bx,By))
#print(dist(Ox,Oy,Cx,Cy))
s1 = cross_prod(Ax-Ox,Ay-Oy,Bx-Ox,By-Oy)/R2
c1 = inner_prod(Ax-Ox,Ay-Oy,Bx-Ox,By-Oy)/R2
s2 = cross_prod(Ax-Ox,Ay-Oy,Cx-Ox,Cy-Oy)/R2
c2 = inner_prod(Ax-Ox,Ay-Oy,Cx-Ox,Cy-Oy)/R2
#angle1 = pool(angle1)
#angle2 = pool(angle2)
#print([s1, s2])
#print([c1, c2])
for n in range(3, 101):
x = list(range(n))
for j in range(len(x)):
x[j] = x[j] * (2*pi/n)
s = list(map(sin, x))
c = list(map(cos, x))
#print(s)
#print(c)
#print(find(s1, c1, s, c))
#print(find(s2, c2, s, c))
if find(s1, c1, s, c) and find(s2, c2, s, c):
area = .5*n*R2*sin(2*pi/n)
print("%.8f" % area)
break
if __name__ == "__main__":
main()
```
| 93,782 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Tags: geometry, math
Correct Solution:
```
import math
from fractions import Fraction
line = input()
data = line.split()
x1, y1 = float(data[0]), float(data[1])
line = input()
data = line.split()
x2, y2 = float(data[0]), float(data[1])
line = input()
data = line.split()
x3, y3 = float(data[0]), float(data[1])
dis12 = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)
dis23 = (x2 - x3) * (x2 - x3) + (y2 - y3) * (y2 - y3)
dis13 = (x1 - x3) * (x1 - x3) + (y1 - y3) * (y1 - y3)
'''
print(dis12)
print(dis13)
print(dis23)
'''
cos1 = (dis13 + dis12 - dis23) / (2 * math.sqrt(dis12 * dis13))
cos2 = (dis23 + dis12 - dis13) / (2 * math.sqrt(dis23 * dis12))
cos3 = (dis13 + dis23 - dis12) / (2 * math.sqrt(dis13 * dis23))
'''
print('*',cos1)
print('*',cos2)
print('*',cos3)
'''
angle1 = math.acos(cos1) * 2
angle2 = math.acos(cos2) * 2
angle3 = math.acos(cos3) * 2
'''
print('#',angle1)
print('#',angle2)
print('#',angle3)
'''
number1 = Fraction.from_float(angle1 / angle2).limit_denominator(101)
#print(number1.numerator)
#print(number1.denominator)
l1 = angle1 / number1.numerator
#print(l1)
n1 = 2 * math.pi / l1
n1_int = int(n1 + 0.5)
#print(n1_int)
number2 = Fraction.from_float(angle1 / angle3).limit_denominator(101)
#print(number2.numerator)
#print(number2.denominator)
l2 = angle1 / number2.numerator
#print(l2)
n2 = 2 * math.pi / l2
n2_int = int(n2 + 0.5)
#print(n2_int)
if abs(n1_int - n1) < abs(n2_int - n2):
n = n1_int
else:
n = n2_int
sin1 = math.sqrt(1 - cos1 * cos1)
#print(sin1)
#print(dis23)
r = math.sqrt(dis23) / (2 * sin1)
#print(r)
area = 1 / 2 * n * math.sin(2 * math.pi / n) * r * r
print('%.6f' % area)
```
| 93,783 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Tags: geometry, math
Correct Solution:
```
import math
x1, y1 = map(float, input().split())
x2, y2 = map(float, input().split())
x3, y3 = map(float, input().split())
c1 = (x2 ** 2 - x1 ** 2 + y2 ** 2 - y1 ** 2) / 2
c2 = (x3 ** 2 - x2 ** 2 + y3 ** 2 - y2 ** 2) / 2
# dist x1-x2
r1 = (c1 * (y3 - y2) - c2 * (y2 - y1)) / ((x2 - x1) * (y3 - y2) - (x3 - x2) * (y2 - y1))
# dist x1-x3
r2 = (c1 * (x3 - x2) - c2 * (x2 - x1)) / ((y2 - y1) * (x3 - x2) - (y3 - y2) * (x2 - x1))
# ipotenuza
r = ((r1 - x1) ** 2 + (r2 - y1) ** 2) ** 0.5
foobar = 0
for i in range(3, 101):
foo = False
bar = False
for ii in range(1, i):
x4 = (x1 - r1) * math.cos(2 * math.pi * ii / i) - math.sin(2 * math.pi * ii / i) * (y1 - r2) + r1
y4 = (y1 - r2) * math.cos(2 * math.pi * ii / i) + math.sin(2 * math.pi * ii / i) * (x1 - r1) + r2
# if i pillars yields a 'full circle'/regular polyogn; print ans
if (x4 - x2) ** 2 + (y4 - y2) ** 2 < 0.00001:
foo = True
if (x4 - x3) ** 2 + (y4 - y3) ** 2 < 0.00001:
bar = True
if foo and bar:
foobar = i
break
# formula for area of equiangular polygon
print(foobar * r ** 2 * math.sin(2 * math.pi / foobar) / 2)
```
| 93,784 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Tags: geometry, math
Correct Solution:
```
import math
import sys
eps=1e-4
def getABC(x1,y1,x2,y2):
A=x2-x1
B=y2-y1
C=(x1*x1-x2*x2+y1*y1-y2*y2)/2
return A,B,C
# assume they are not parallel
def getIntersection(a1,b1,c1,a2,b2,c2):
x=(c2*b1-c1*b2)/(a1*b2-a2*b1)
y=(a1*c2-a2*c1)/(a2*b1-a1*b2)
return x,y
def getAngel(x1,y1,x2,y2):
return math.acos((x1*x2+y1*y2)/(math.sqrt(x1*x1+y1*y1)*math.sqrt(x2*x2+y2*y2)))
def check(num):
return abs(num-int(num+eps))<eps
x=[]
y=[]
for line in sys.stdin.readlines():
t=line.strip().split()
x.append(float(t[0]))
y.append(float(t[1]))
A1,B1,C1=getABC(x[0],y[0],x[1],y[1])
A2,B2,C2=getABC(x[2],y[2],x[1],y[1])
x0,y0=getIntersection(A1,B1,C1,A2,B2,C2)
an=[]
an.append(getAngel(x[0]-x0,y[0]-y0,x[1]-x0,y[1]-y0))
an.append(getAngel(x[0]-x0,y[0]-y0,x[2]-x0,y[2]-y0))
an.append(getAngel(x[2]-x0,y[2]-y0,x[1]-x0,y[1]-y0))
total = 2*math.pi
for i in range(3,101):
flag = True
angel = total/i
for j in range(0,3):
if not check(an[j]/angel):
flag = False
break
if flag:
break
n=i
m=total/n
ans = ((x[0]-x0)*(x[0]-x0)+(y[0]-y0)*(y[0]-y0))*math.sin(m)*n/2
print('%.6f' %ans)
```
| 93,785 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Tags: geometry, math
Correct Solution:
```
import math
EPS = 1e-6
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
def dist(self, p:"Point"):
return math.sqrt((self.x - p.x) ** 2 + (self.y - p.y) ** 2)
def dist_square(self, p:"Point"):
return (self.x - p.x) ** 2 + (self.y - p.y) ** 2
def getAreafromPolygon(points):
A = 0.0
for p1, p2 in zip(points[:-1], points[1:]):
A += 0.5 * (p1.x*p2.y - p2.x*p1.y)
A += 0.5 * (points[-1].x * points[0].y - points[0].x * points[-1].y)
return abs(A)
def getRadiusCircleFromThreePoints(p1:'Point', p2:'Point', p3:'Point'):
triangle_area = getAreafromPolygon([p1, p2, p3])
if triangle_area < EPS:
raise Exception('three points are in the same line')
r = p1.dist(p2) * p1.dist(p3) * p2.dist(p3) / 4.0 / triangle_area
return r
def get_angle_from_cosine_law(l1:float, l2: float, length_opposite_angle:float):
l3 = length_opposite_angle
if l1 < EPS or l2 < EPS:
raise Exception('side length must be positive real number')
if abs(l1-l2) < EPS and l3 <EPS:
return 0
if abs(l1+l2-l3) < EPS:
return math.pi
if l1+l2-l3 < -EPS:
raise Exception('invalid input; length of three sides cannot form triangle')
return math.acos( (l1**2 + l2**2 - l3**2)/ (2*l1*l2) )
points = []
for i in range(3):
x, y = map(float, input().split(' '))
points.append(Point(x,y))
radius = getRadiusCircleFromThreePoints(points[0], points[1], points[2])
l01 = points[0].dist(points[1])
l02 = points[0].dist(points[2])
angle_p0Op1 = get_angle_from_cosine_law(radius, radius, l01)
angle_p0Op2 = get_angle_from_cosine_law(radius, radius, l02)
i = 3
while i:
angle_per_side = 2*math.pi / i
mod_angle_p0Op1 = math.fmod(angle_p0Op1 ,angle_per_side)
mod_angle_p0Op2 = math.fmod(angle_p0Op2 ,angle_per_side)
if (
(mod_angle_p0Op1 < EPS or mod_angle_p0Op1 > angle_per_side-EPS)
and (mod_angle_p0Op2 < EPS or mod_angle_p0Op2 > angle_per_side-EPS)
):
print(i * 0.5 * radius * radius * math.sin(angle_per_side))
break
i += 1
```
| 93,786 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
s1 = input().split()
s2 = input().split()
s3 = input().split()
ax = float(s1[0])
ay = float(s1[1])
bx = float(s2[0])
by = float(s2[1])
cx = float(s3[0])
cy = float(s3[1])
det = (bx-ax)*(cy-ay) - (by-ay)*(cx-ax)
det_x = (by-ay)*(ax*ax+ay*ay-cx*cx-cy*cy)-(cy-ay)*(ax*ax+ay*ay-bx*bx-by*by)
det_y = (cx-ax)*(ax*ax+ay*ay-bx*bx-by*by)-(bx-ax)*(ax*ax+ay*ay-cx*cx-cy*cy)
ox = det_x/det/2.0
oy = det_y/det/2.0
def shortest(x, y, pts):
d2_min = None
for pt in pts:
dx = pt[0]-x
dy = pt[1]-y
d2 = dx*dx + dy* dy
if d2_min is None or d2 < d2_min:
d2_min = d2
return d2_min
import math
d_min = None
num_min = None
oax = ax - ox
oay = ay - oy
r = math.sqrt(oax*oax+oay*oay)
if(abs(oax) < 1.e-8):
alpha = math.pi/2 if oay > 0 else -math.pi/2
else:
alpha = math.atan(oay/oax) if oax > 0 else math.pi + math.atan(oay/oax)
for num in range(3, 101):
pts = []
for i in range(1, num):
rad = i * 2 * math.pi / num
pts.append((ox+r*math.cos(rad+alpha), oy+r*math.sin(rad+alpha)))
s2 = shortest(bx, by, pts)
s3 = shortest(cx, cy, pts)
s = s2 + s3
if d_min is None or s < d_min - 1.e-8:
d_min = s
num_min = num
beta = 2*math.pi / num_min
area = num_min * r * r * math.sin(beta) /2.0
print(f"{area:.8f}")
```
Yes
| 93,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import math
DEBUG = False
def area(d):
if len(d) < 3:
return False
a = 0.0
p = 0.5 * (d[0] + d[1] + d[2])
a = math.sqrt(p*(p-d[0])*(p-d[1])*(p-d[2]))
return a
def circumcircle(x,y):
distance = euclidean(x,y)
triangle_area = area(distance)
return distance[0] * distance[1] * distance[2] / (4 * triangle_area)
def euclidean(x,y):
d = []
d.append(math.sqrt((x[0]-x[1])**2 + (y[0]-y[1])**2))
d.append(math.sqrt((x[0]-x[2])**2 + (y[0]-y[2])**2))
d.append(math.sqrt((x[1]-x[2])**2 + (y[1]-y[2])**2))
return d
def normal_sin_value(sin_value):
if sin_value > 1:
return 1.0
elif sin_value < -1:
return -1.0
else:
return sin_value
def centerAngle(d,radius):
angle = []
#print(type(d[0]), type(radius))
#print(d[0]/(2*radius))
#print(2 * math.asin( d[0] / (2*radius) ) )
angle.append(2*math.asin(normal_sin_value(d[0]/(2*radius))))
angle.append(2*math.asin(normal_sin_value(d[1]/(2*radius))))
angle.append(2*math.pi - angle[0] - angle[1])
if DEBUG:
print(sum([a for a in angle]))
return angle
def gcd(a,b):
if a < 0.01:
return b
else:
return gcd(math.fmod(b,a),a)
def main():
# get the input data
x = []
y = []
for i in range(3):
temp_input = input().split()
x.append(float(temp_input[0]))
y.append(float(temp_input[1]))
# 1. calculate the length of the edge
# 2. calculate the area of the triangle
# 3. calculate the radius of the circumcircle
# 4. calculate the area of the Berland Circus
edge_length = euclidean(x,y)
triangle_area = area(edge_length)
circumcircle_radius = circumcircle(x,y)
#print("circumcircle_radius: {0[0]}:{1[0]},{0[1]}:{1[1]}, {0[2]}:{1[2]} \n {2}".format(x,y,circumcircle_radius))
# 5. calculat the cetral angle and their gcd
angle = centerAngle(edge_length, circumcircle_radius)
gcd_angle = gcd(gcd(angle[0], angle[1]), angle[2])
result = 2 * math.pi / gcd_angle * circumcircle_radius * math.sin(gcd_angle) * circumcircle_radius * 0.5
if DEBUG:
print("circumcircle_radius",circumcircle_radius)
print("totoal_angle",angle)
print("gcd_angle",gcd_angle)
print("len",edge_length)
print("{:.8f}".format(result))
if __name__ == "__main__":
main()
```
Yes
| 93,788 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
import math
def distance(p1, p2):
return math.sqrt((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2)
def gcd(a, b):
if a < b:
return gcd(b, a)
if abs(b) < 0.001:
return a
else:
return gcd(b, a - math.floor(a / b) * b)
points = []
for i in range(3):
(x, y) = [float(x) for x in input().split()]
point = (x, y)
points.append(point)
a = distance(points[0], points[1])
b = distance(points[1], points[2])
c = distance(points[2], points[0])
s = (a + b + c) / 2
R = (a * b * c) / (4 * math.sqrt(s * (a + b - s) * (a + c - s) * (b + c - s)))
A = math.acos((b * b + c * c - a * a) / (2 * b * c))
B = math.acos((a * a + c * c - b * b) / (2 * a * c))
C = math.acos((b * b + a * a - c * c) / (2 * b * a))
n = math.pi / gcd(A, gcd(B, C))
area = (n * R * R * math.sin((2 * math.pi) / n)) / 2
print(area)
```
Yes
| 93,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
import math
def gcd(x,y):
if abs(y) < 1e-4 :
return x
else :
return gcd(y, math.fmod(x, y))
Ax,Ay = list(map(float,input().split()))
Bx,By = list(map(float,input().split()))
Cx,Cy = list(map(float,input().split()))
a = math.sqrt((Bx - Cx) * (Bx - Cx) + (By - Cy) * (By - Cy))
b = math.sqrt((Ax - Cx) * (Ax - Cx) + (Ay - Cy) * (Ay - Cy))
c = math.sqrt((Ax - Bx) * (Ax - Bx) + (Ay - By) * (Ay - By))
s = (a + b + c) / 2
area_triangle = math.sqrt(s * (s - a) * (s - b) * (s - c))
r = (a * b * c) / (4 * area_triangle)
Angle_A = math.acos((b * b + c * c - a * a) / (2 * b * c))
Angle_B = math.acos((a * a + c * c - b * b) / (2 * a * c))
Angle_C = math.acos((b * b + a * a - c * c) / (2 * b * a))
n = math.pi / gcd(gcd(Angle_A, Angle_B), Angle_C)
print ((n * r * r * math.sin((2 * math.pi) / n)) / 2)
```
Yes
| 93,790 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
import math
input_var = []
for n in range(3):
input_var.append(input().split())
P1, P2, P3 = input_var
P1 = [float(num) for num in P1]
P2 = [float(num) for num in P2]
P3 = [float(num) for num in P3]
def triangle_side_lengths(P1, P2, P3):
return [((P1[0] - P2[0])**2 + (P1[1] - P2[1])**2)**(1 / 2), ((P1[0] - P3[0])**2 + (P1[1] - P3[1])**2)**(1 / 2), ((P2[0] - P3[0])**2 + (P2[1] - P3[1])**2)**(1 / 2)]
def circum_radius(P1, P2, P3):
a, b, c = triangle_side_lengths(P1, P2, P3)
return (a * b * c) / ((a + b + c) * (b + c - a) * (c + a - b) * (a + b - c))**(1 / 2)
def side_length(P1, P2, P3):
for i in range(3, 101):
if math.fabs(circum_radius(P1, P2, P3)*2*math.sin(math.pi/i)- round(circum_radius(P1, P2, P3)*2*math.sin(math.pi/i),4))< 1E-6 and i==3:
return (1/2)*i*(circum_radius(P1, P2, P3)**2)*math.sin((2*math.pi)/i)
elif math.fabs(circum_radius(P1, P2, P3)*2*math.sin(math.pi/i)- round(circum_radius(P1, P2, P3)*2*math.sin(math.pi/i),4))< 1E-6 and 3>i:
return (1/2)*i*(circum_radius(P1, P2, P3)**2)*math.sin((2*math.pi)/i)/2
print(side_length(P1, P2, P3))
```
No
| 93,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
import math
def read_int():
return int(input())
def read_array(constructor):
return [constructor(x) for x in input().split()]
def read_int_array():
return read_array(int)
p = read_array(float)
q = read_array(float)
r = read_array(float)
pq = math.hypot(q[0] - p[0], q[1] - p[1])
pr = math.hypot(r[0] - p[0], r[1] - p[1])
qr = math.hypot(r[0] - q[0], r[1] - q[1])
edges = [pq, pr, qr]
edges.sort()
cos_inscribed_angle = (edges[1]**2 + edges[2]**2 - edges[0]**2) / (2 * edges[1] * edges[2])
inscribed_angle = math.acos(cos_inscribed_angle)
central_angle = inscribed_angle * 2
sin_inscribed_angle = math.sin(inscribed_angle)
radius = edges[0] / (2 * sin_inscribed_angle)
# how to determine #angle
num_angle = 0
for n in range(1, 101):
diff = (central_angle * n) % (2 * math.pi)
# print(diff)
if math.isclose(diff, 0, abs_tol=0.01) or \
math.isclose(diff, 2 * math.pi, rel_tol=0.01):
num_angle = n
break
else:
print('ERROR: n={0}, num_angle={1}'.format(n, num_angle))
angle = 2 * math.pi / num_angle
area = radius**2 * math.sin(angle) * num_angle / 2
print('{0:f}'.format(area))
```
No
| 93,792 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
import math
A = input().split()
B = input().split()
C = input().split()
#a1, a2 = 17.288379, 68.223317
#b1, b2 = 48.776683, 71.688379
#c1, c2 = 23.170559, 106.572762
a1, a2 = float(A[0]), float(A[1])
b1, b2 = float(B[0]), float(B[1])
c1, c2 = float(C[0]), float(C[1])
a = ((b1-c1)**2 + (b2-c2)**2) **0.5
b = ((a1-c1)**2 + (a2-c2)**2) **0.5
c = ((b1-a1)**2 + (b2-a2)**2) **0.5
#t = (a+b+c)/2
#area = (t*(t-a)*(t-b)*(t-c))**0.5
#print(area)
#print(a,b,c)
cosa = (b**2+c**2-a**2)/(2*b*c)
sina = (1-cosa**2)**0.5
R = a/(2*sina)
side = min(a,b,c)
s = (R+R+side)/2
#tri_area = 1/2*R*R*sinTheta
tri_area = (s*(s-R)*(s-R)*(s-side))**0.5
sides = [a,b,c]
countSide = []
thetaAll = []
for s in sides:
cosTheta = (R**2+R**2-s**2)/(2*R**2)
theta = math.acos(cosTheta)
thetaAll.append(round(math.degrees(theta)))
count = (2 * math.pi) / theta
countSide.append(count)
#print(countSide)
#print("degrees:", thetaAll)
theta = math.radians(math.gcd(max(thetaAll), min(thetaAll)))
count = 2*math.pi / theta
total_area = 1/2 * math.sin(theta) * R * R * count
#print(count)
print(total_area)
#print(a)
#print(sina)
#print("R:", a/(2*sina))
#print("tri_area:", tri_area)
#print(theta)
#print(count)
#print(total_area)
```
No
| 93,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.
In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.
Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.
You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.
Input
The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.
Output
Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.
Examples
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
Submitted Solution:
```
from math import sqrt,acos,sin
from re import split
def dis(a,b):
return sqrt((a[0]-b[0])**2+(a[1]-b[1])**2)
def ang(a,d,b):
return acos((dis(a,d)**2+dis(b,d)**2-dis(a,b)**2)/(2*dis(a,d)*dis(b,d)))
def S(a,b,d):
s=(1/2)*dis(a,d)*dis(b,d)*sin(ang(a,d,b))
if round(s,6)!=0:
return s
else:
return 10000000000
def l2f(a):
for i in range(len(a)):
a[i]=float(a[i])
a=input()
a=a.split()
l2f(a)
b=input()
b=b.split()
l2f(b)
c=input()
c=c.split()
l2f(c)
A1=2*(b[0]-a[0])
B1=2*(b[1]-a[1])
C1=b[0]**2+b[1]**2-a[0]**2-a[1]**2
A2=2*(c[0]-b[0])
B2=2*(c[1]-b[1])
C2=c[0]**2+c[1]**2-b[0]**2-b[1]**2
d=[]
d.append(((C1*B2)-(C2*B1))/((A1*B2)-(A2*B1)))
d.append(((A1*C2)-(A2*C1))/((A1*B2)-(A2*B1)))
ang_min=min(ang(a,d,b),ang(b,d,c),ang(a,d,c))
n=round(2*acos(-1)/ang_min)
if ang(a,d,b)==ang_min:
s=S(a,b,d)*n
if ang(b,d,c)==ang_min:
s=S(b,c,d)*n
if ang(a,d,c)==ang_min:
s=S(a,c,d)*n
print("%.8f" % s)
```
No
| 93,794 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A subsequence of length |x| of string s = s1s2... s|s| (where |s| is the length of string s) is a string x = sk1sk2... sk|x| (1 ≤ k1 < k2 < ... < k|x| ≤ |s|).
You've got two strings — s and t. Let's consider all subsequences of string s, coinciding with string t. Is it true that each character of string s occurs in at least one of these subsequences? In other words, is it true that for all i (1 ≤ i ≤ |s|), there is such subsequence x = sk1sk2... sk|x| of string s, that x = t and for some j (1 ≤ j ≤ |x|) kj = i.
Input
The first line contains string s, the second line contains string t. Each line consists only of lowercase English letters. The given strings are non-empty, the length of each string does not exceed 2·105.
Output
Print "Yes" (without the quotes), if each character of the string s occurs in at least one of the described subsequences, or "No" (without the quotes) otherwise.
Examples
Input
abab
ab
Output
Yes
Input
abacaba
aba
Output
No
Input
abc
ba
Output
No
Note
In the first sample string t can occur in the string s as a subsequence in three ways: abab, abab and abab. In these occurrences each character of string s occurs at least once.
In the second sample the 4-th character of the string s doesn't occur in any occurrence of string t.
In the third sample there is no occurrence of string t in string s.
Tags: data structures, dp, strings
Correct Solution:
```
import bisect
import string
s = input()
t = input()
max_match = [0 for i in range(len(s))]
min_match = [0 for i in range(len(s))]
char_idx = [0 for i in range(30)]
char_occur = [ [] for i in range(30) ]
for (i, ch) in enumerate(t):
idx = ord(ch) - ord('a')
char_occur[idx].append(i)
for ch in string.ascii_lowercase:
idx = ord(ch) - ord('a')
char_occur[idx].append(len(t)+1)
matched = -1
for (i, ch) in enumerate(s):
if matched==len(t)-1:
max_match[i] = matched
else:
if ch == t[matched+1]:
matched += 1
max_match[i] = matched
matched = len(t)
for (i, ch) in enumerate(s[::-1]):
i = len(s) - i - 1
if matched==0:
min_match[i] = matched
else:
if ch == t[matched-1]:
matched -= 1
min_match[i] = matched
for (i, ch) in enumerate(s):
low = min_match[i]
high = max_match[i]
ch = ord(ch) - ord('a')
idx = char_idx[ch]
while idx<len(char_occur[ch]) and char_occur[ch][idx]<low:
idx += 1
char_idx[ch] = idx
if idx == len(char_occur[ch]):
print("No")
exit()
if char_occur[ch][idx] > high:
print("No")
exit()
print("Yes")
```
| 93,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A subsequence of length |x| of string s = s1s2... s|s| (where |s| is the length of string s) is a string x = sk1sk2... sk|x| (1 ≤ k1 < k2 < ... < k|x| ≤ |s|).
You've got two strings — s and t. Let's consider all subsequences of string s, coinciding with string t. Is it true that each character of string s occurs in at least one of these subsequences? In other words, is it true that for all i (1 ≤ i ≤ |s|), there is such subsequence x = sk1sk2... sk|x| of string s, that x = t and for some j (1 ≤ j ≤ |x|) kj = i.
Input
The first line contains string s, the second line contains string t. Each line consists only of lowercase English letters. The given strings are non-empty, the length of each string does not exceed 2·105.
Output
Print "Yes" (without the quotes), if each character of the string s occurs in at least one of the described subsequences, or "No" (without the quotes) otherwise.
Examples
Input
abab
ab
Output
Yes
Input
abacaba
aba
Output
No
Input
abc
ba
Output
No
Note
In the first sample string t can occur in the string s as a subsequence in three ways: abab, abab and abab. In these occurrences each character of string s occurs at least once.
In the second sample the 4-th character of the string s doesn't occur in any occurrence of string t.
In the third sample there is no occurrence of string t in string s.
Submitted Solution:
```
import bisect
import string
s = input()
t = input()
max_match = [0 for i in range(len(s))]
min_match = [0 for i in range(len(s))]
char_idx = [0 for i in range(30)]
char_occur = [ [] for i in range(30) ]
for ch in string.ascii_lowercase:
idx = ord(ch) - ord('a')
char_occur[idx] = [i for (i,_ch) in enumerate(t) if ch==_ch]
char_occur[idx].append(len(t)+1)
matched = -1
for (i, ch) in enumerate(s):
if matched==len(t)-1:
max_match[i] = matched
else:
if ch == t[matched+1]:
matched += 1
max_match[i] = matched
matched = len(t)
for (i, ch) in enumerate(s[::-1]):
i = len(s) - i - 1
if matched==0:
min_match[i] = matched
else:
if ch == t[matched-1]:
matched -= 1
min_match[i] = matched
if len(t) > 100000:
for (i, ch) in enumerate(s):
low = min_match[i]
high = max_match[i]
ch = ord(ch) - ord('a')
idx = char_idx[ch]
while idx<len(char_occur[ch]) and char_occur[ch][idx]<low:
idx += 1
char_idx[ch] = idx
if idx == len(char_occur[ch]):
print("No")
exit()
if char_occur[ch][idx] > high:
print("No")
exit()
print("Yes")
```
No
| 93,796 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A subsequence of length |x| of string s = s1s2... s|s| (where |s| is the length of string s) is a string x = sk1sk2... sk|x| (1 ≤ k1 < k2 < ... < k|x| ≤ |s|).
You've got two strings — s and t. Let's consider all subsequences of string s, coinciding with string t. Is it true that each character of string s occurs in at least one of these subsequences? In other words, is it true that for all i (1 ≤ i ≤ |s|), there is such subsequence x = sk1sk2... sk|x| of string s, that x = t and for some j (1 ≤ j ≤ |x|) kj = i.
Input
The first line contains string s, the second line contains string t. Each line consists only of lowercase English letters. The given strings are non-empty, the length of each string does not exceed 2·105.
Output
Print "Yes" (without the quotes), if each character of the string s occurs in at least one of the described subsequences, or "No" (without the quotes) otherwise.
Examples
Input
abab
ab
Output
Yes
Input
abacaba
aba
Output
No
Input
abc
ba
Output
No
Note
In the first sample string t can occur in the string s as a subsequence in three ways: abab, abab and abab. In these occurrences each character of string s occurs at least once.
In the second sample the 4-th character of the string s doesn't occur in any occurrence of string t.
In the third sample there is no occurrence of string t in string s.
Submitted Solution:
```
import bisect
import string
s = input()
t = input()
max_match = [0 for i in range(len(s))]
min_match = [0 for i in range(len(s))]
char_idx = [0 for i in range(30)]
char_occur = [ [] for i in range(30) ]
matched = -1
for (i, ch) in enumerate(s):
if matched==len(t)-1:
max_match[i] = matched
else:
if ch == t[matched+1]:
matched += 1
max_match[i] = matched
idx = ord(ch) - ord('a')
char_occur[idx].append(i)
for ch in string.ascii_lowercase:
idx = ord(ch) - ord('a')
char_occur[idx].append(len(t)+1)
matched = len(t)
for (i, ch) in enumerate(s[::-1]):
i = len(s) - i - 1
if matched==0:
min_match[i] = matched
else:
if ch == t[matched-1]:
matched -= 1
min_match[i] = matched
for (i, ch) in enumerate(s):
low = min_match[i]
high = max_match[i]
ch = ord(ch) - ord('a')
idx = char_idx[ch]
while idx<len(char_occur[ch]) and char_occur[ch][idx]<low:
idx += 1
char_idx[ch] = idx
if idx == len(char_occur[ch]):
print("No")
exit()
if char_occur[ch][idx] > high:
print("No")
exit()
print("Yes")
```
No
| 93,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A subsequence of length |x| of string s = s1s2... s|s| (where |s| is the length of string s) is a string x = sk1sk2... sk|x| (1 ≤ k1 < k2 < ... < k|x| ≤ |s|).
You've got two strings — s and t. Let's consider all subsequences of string s, coinciding with string t. Is it true that each character of string s occurs in at least one of these subsequences? In other words, is it true that for all i (1 ≤ i ≤ |s|), there is such subsequence x = sk1sk2... sk|x| of string s, that x = t and for some j (1 ≤ j ≤ |x|) kj = i.
Input
The first line contains string s, the second line contains string t. Each line consists only of lowercase English letters. The given strings are non-empty, the length of each string does not exceed 2·105.
Output
Print "Yes" (without the quotes), if each character of the string s occurs in at least one of the described subsequences, or "No" (without the quotes) otherwise.
Examples
Input
abab
ab
Output
Yes
Input
abacaba
aba
Output
No
Input
abc
ba
Output
No
Note
In the first sample string t can occur in the string s as a subsequence in three ways: abab, abab and abab. In these occurrences each character of string s occurs at least once.
In the second sample the 4-th character of the string s doesn't occur in any occurrence of string t.
In the third sample there is no occurrence of string t in string s.
Submitted Solution:
```
def main():
first = str(input())
second = str(input())
first_size = len(first)
second_size = len(second)
pre, last = 0, second_size
is_valid = True
if first_size % second_size == 0:
while first_size >= last:
if first[pre:last] != second:
is_valid = False
break
pre = last
last = last + second_size
if is_valid:
print("Yes")
else:
print("No")
else:
print("No")
if __name__ == "__main__":
main()
```
No
| 93,798 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A subsequence of length |x| of string s = s1s2... s|s| (where |s| is the length of string s) is a string x = sk1sk2... sk|x| (1 ≤ k1 < k2 < ... < k|x| ≤ |s|).
You've got two strings — s and t. Let's consider all subsequences of string s, coinciding with string t. Is it true that each character of string s occurs in at least one of these subsequences? In other words, is it true that for all i (1 ≤ i ≤ |s|), there is such subsequence x = sk1sk2... sk|x| of string s, that x = t and for some j (1 ≤ j ≤ |x|) kj = i.
Input
The first line contains string s, the second line contains string t. Each line consists only of lowercase English letters. The given strings are non-empty, the length of each string does not exceed 2·105.
Output
Print "Yes" (without the quotes), if each character of the string s occurs in at least one of the described subsequences, or "No" (without the quotes) otherwise.
Examples
Input
abab
ab
Output
Yes
Input
abacaba
aba
Output
No
Input
abc
ba
Output
No
Note
In the first sample string t can occur in the string s as a subsequence in three ways: abab, abab and abab. In these occurrences each character of string s occurs at least once.
In the second sample the 4-th character of the string s doesn't occur in any occurrence of string t.
In the third sample there is no occurrence of string t in string s.
Submitted Solution:
```
s=input()
t=input()
n=len(s)
m=len(t)
assert m<=n
s=list(s)
t=list(t)
j=0
bool=True
for c in t:
if c not in s[j:]:
bool=False
break
j+=1
if bool==False:
print('No')
if bool==True:
for c in s:
if c not in t:
bool=False
break
if bool==False:
print('No')
else:
print('Yes')
```
No
| 93,799 |
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