message stringlengths 2 433k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array consisting of n non-negative integers a1, a2, ..., an.
You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.
After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
The third line contains a permutation of integers from 1 to n — the order used to destroy elements.
Output
Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first i operations are performed.
Examples
Input
4
1 3 2 5
3 4 1 2
Output
5
4
3
0
Input
5
1 2 3 4 5
4 2 3 5 1
Output
6
5
5
1
0
Input
8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6
Output
18
16
11
8
8
6
6
0
Note
Consider the first sample:
1. Third element is destroyed. Array is now 1 3 * 5. Segment with maximum sum 5 consists of one integer 5.
2. Fourth element is destroyed. Array is now 1 3 * * . Segment with maximum sum 4 consists of two integers 1 3.
3. First element is destroyed. Array is now * 3 * * . Segment with maximum sum 3 consists of one integer 3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0. | instruction | 0 | 99,931 | 12 | 199,862 |
Tags: data structures, dsu
Correct Solution:
```
# Bosdiwale code chap kr kya milega
# Motherfuckers Don't copy code for the sake of doing it
# ..............
# ╭━┳━╭━╭━╮╮
# ┃┈┈┈┣▅╋▅┫┃
# ┃┈┃┈╰━╰━━━━━━╮
# ╰┳╯┈┈┈┈┈┈┈┈┈◢▉◣
# ╲┃┈┈┈┈┈┈┈┈┈┈▉▉▉
# ╲┃┈┈┈┈┈┈┈┈┈┈◥▉◤
# ╲┃┈┈┈┈╭━┳━━━━╯
# ╲┣━━━━━━┫
# ……….
# .……. /´¯/)………….(\¯`\
# …………/….//……….. …\\….\
# ………/….//……………....\\….\
# …./´¯/…./´¯\……/¯ `\…..\¯`\
# ././…/…/…./|_…|.\….\….\…\.\
# (.(….(….(…./.)..)...(.\.).).)
# .\…………….\/../…....\….\/…………/
# ..\…………….. /……...\………………../
# …..\…………… (………....)……………./
n = int(input())
arr = list(map(int,input().split()))
ind = list(map(int,input().split()))
parent = {}
rank = {}
ans = {}
total = 0
temp = []
def make_set(v):
rank[v] = 1
parent[v] = v
ans[v] = arr[v]
def find_set(u):
if u==parent[u]:
return u
else:
parent[u] = find_set(parent[u])
return parent[u]
def union_set(u,v):
a = find_set(u)
b = find_set(v)
if a!=b:
if rank[b]>rank[a]:
a,b = b,a
parent[b] = a
rank[a]+=rank[b]
ans[a]+=ans[b]
return ans[a]
for i in range(n-1,-1,-1):
rem = ind[i]-1
make_set(rem)
final = ans[rem]
if rem+1 in parent:
final = union_set(rem,rem+1)
if rem-1 in parent:
final = union_set(rem,rem-1)
total = max(total,final)
temp.append(total)
temp[-1] = 0
temp = temp[-1::-1]
temp = temp[1::]
for i in temp:
print(i)
print(0)
``` | output | 1 | 99,931 | 12 | 199,863 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array consisting of n non-negative integers a1, a2, ..., an.
You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.
After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
The third line contains a permutation of integers from 1 to n — the order used to destroy elements.
Output
Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first i operations are performed.
Examples
Input
4
1 3 2 5
3 4 1 2
Output
5
4
3
0
Input
5
1 2 3 4 5
4 2 3 5 1
Output
6
5
5
1
0
Input
8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6
Output
18
16
11
8
8
6
6
0
Note
Consider the first sample:
1. Third element is destroyed. Array is now 1 3 * 5. Segment with maximum sum 5 consists of one integer 5.
2. Fourth element is destroyed. Array is now 1 3 * * . Segment with maximum sum 4 consists of two integers 1 3.
3. First element is destroyed. Array is now * 3 * * . Segment with maximum sum 3 consists of one integer 3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0. | instruction | 0 | 99,932 | 12 | 199,864 |
Tags: data structures, dsu
Correct Solution:
```
def main():
def f(x):
l = []
while x != clusters[x]:
l.append(x)
x = clusters[x]
for y in l:
clusters[y] = x
return x
n, aa = int(input()), [0, *map(int, input().split()), 0]
l, clusters, mx = list(map(int, input().split())), [0] * (n + 2), 0
for i in range(n - 1, -1, -1):
a = clusters[a] = l[i]
l[i] = mx
for i in a - 1, a + 1:
if clusters[i]:
j = f(i)
aa[a] += aa[j]
clusters[j] = a
f(i)
if mx < aa[a]:
mx = aa[a]
print('\n'.join(map(str, l)))
if __name__ == '__main__':
main()
``` | output | 1 | 99,932 | 12 | 199,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array consisting of n non-negative integers a1, a2, ..., an.
You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.
After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
The third line contains a permutation of integers from 1 to n — the order used to destroy elements.
Output
Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first i operations are performed.
Examples
Input
4
1 3 2 5
3 4 1 2
Output
5
4
3
0
Input
5
1 2 3 4 5
4 2 3 5 1
Output
6
5
5
1
0
Input
8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6
Output
18
16
11
8
8
6
6
0
Note
Consider the first sample:
1. Third element is destroyed. Array is now 1 3 * 5. Segment with maximum sum 5 consists of one integer 5.
2. Fourth element is destroyed. Array is now 1 3 * * . Segment with maximum sum 4 consists of two integers 1 3.
3. First element is destroyed. Array is now * 3 * * . Segment with maximum sum 3 consists of one integer 3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0.
Submitted Solution:
```
# bharatkulratan
# 12:09 PM, 26 Jun 2020
import os
from io import BytesIO
class IO:
def __init__(self):
self.input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
def read_int(self):
return int(self.input())
def read_ints(self):
return [int(x) for x in self.input().split()]
class Node:
def __init__(self, data):
self.data = data
self.sum = data
self.rank = 0
self.parent = self
def __repr__(self):
return f"data = {self.data}, sum = {self.sum}, " \
f"rank = {self.rank}, parent = {self.parent.data}"
class UnionFind:
def __init__(self):
# mapping from data to corresponding node
self.map = {}
def makeset(self, index, data):
self.map[index] = Node(data)
def find_parent(self, data):
return self.find(self.map.get(data))
def find(self, node):
# whether they're both same object
if node.parent is node: return node
return self.find(node.parent)
def union(self, left, right):
left_par = self.find(left)
right_par = self.find(right)
if left_par is right_par: return
if left_par.rank >= right_par.rank:
if left_par.rank == right_par.rank: left_par.rank += 1
right_par.parent = left_par
left_par.sum += right_par.sum
else:
left_par.parent = right_par
right_par.sum += left_par.sum
def main():
io = IO()
n = io.read_int()
uf = UnionFind()
arr = io.read_ints()
for index, elem in enumerate(arr):
uf.makeset(index+1, elem)
perm = io.read_ints()
perm.reverse()
result = []
answer = 0
result.append(answer)
entry = [-1] * (n+2)
for pos in perm[:-1]:
left_neighbor = pos-1
right_neighbor = pos+1
if entry[left_neighbor] > -1:
left_parent = uf.find_parent(left_neighbor)
node = uf.find_parent(pos)
uf.union(left_parent, node)
answer = max(answer, uf.find_parent(pos).sum)
if entry[right_neighbor] > -1:
right_parent = uf.find_parent(right_neighbor)
node = uf.find_parent(pos)
uf.union(node, right_parent)
answer = max(answer, uf.find_parent(pos).sum)
if entry[left_neighbor] == -1 and entry[right_neighbor] == -1:
answer = max(answer, uf.find_parent(pos).sum)
entry[pos] = arr[pos-1]
result.append(answer)
for _ in result[::-1]: print(_)
if __name__ == "__main__":
main()
``` | instruction | 0 | 99,933 | 12 | 199,866 |
Yes | output | 1 | 99,933 | 12 | 199,867 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array consisting of n non-negative integers a1, a2, ..., an.
You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.
After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
The third line contains a permutation of integers from 1 to n — the order used to destroy elements.
Output
Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first i operations are performed.
Examples
Input
4
1 3 2 5
3 4 1 2
Output
5
4
3
0
Input
5
1 2 3 4 5
4 2 3 5 1
Output
6
5
5
1
0
Input
8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6
Output
18
16
11
8
8
6
6
0
Note
Consider the first sample:
1. Third element is destroyed. Array is now 1 3 * 5. Segment with maximum sum 5 consists of one integer 5.
2. Fourth element is destroyed. Array is now 1 3 * * . Segment with maximum sum 4 consists of two integers 1 3.
3. First element is destroyed. Array is now * 3 * * . Segment with maximum sum 3 consists of one integer 3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0.
Submitted Solution:
```
import math,sys,bisect,heapq,os
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
from functools import lru_cache
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
def input(): return sys.stdin.readline().rstrip('\r\n')
#input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
aj = lambda: list(map(int, input().split()))
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
ans = 0
def solve():
n, = aj()
a = aj()
p = aj()
valid = [False for i in range(n)]
parent = [0] * n
size = [0] * n
stat = [0] * n
def find(x):
while parent[x] != x:
x = parent[x]
return x
def union(a, b):
x = find(a)
y = find(b)
if x == y:
return
elif size[x] < size[y]:
parent[x] = y
size[y] += size[x]
stat[y] += stat[x]
else:
parent[y] = x
size[x] += size[y]
stat[x] += stat[y]
ans = [0]
for i in range(n - 1, 0, -1):
k = p[i] - 1
valid[k] = True
parent[k] = k
stat[k] = a[k]
if k > 0 and valid[k - 1]:
union(k, k - 1)
if k < n - 1 and valid[k + 1]:
union(k, k + 1)
t = stat[find(k)]
m = max(ans[-1], t)
ans.append(m)
while ans:
print(ans.pop())
try:
#os.system("online_judge.py")
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
solve()
``` | instruction | 0 | 99,934 | 12 | 199,868 |
Yes | output | 1 | 99,934 | 12 | 199,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array consisting of n non-negative integers a1, a2, ..., an.
You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.
After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
The third line contains a permutation of integers from 1 to n — the order used to destroy elements.
Output
Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first i operations are performed.
Examples
Input
4
1 3 2 5
3 4 1 2
Output
5
4
3
0
Input
5
1 2 3 4 5
4 2 3 5 1
Output
6
5
5
1
0
Input
8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6
Output
18
16
11
8
8
6
6
0
Note
Consider the first sample:
1. Third element is destroyed. Array is now 1 3 * 5. Segment with maximum sum 5 consists of one integer 5.
2. Fourth element is destroyed. Array is now 1 3 * * . Segment with maximum sum 4 consists of two integers 1 3.
3. First element is destroyed. Array is now * 3 * * . Segment with maximum sum 3 consists of one integer 3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0.
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda: list(map(int,sys.stdin.readline().rstrip("\r\n").split()))
mod = 10**9+7; Mod = 998244353; INF = float('inf')
#______________________________________________________________________________________________________
# from math import *
# from bisect import *
# from heapq import *
# from collections import defaultdict as dd
# from collections import OrderedDict as odict
# from collections import Counter as cc
# from collections import deque
sys.setrecursionlimit(2*(10**5)+100) #this is must for dfs
# ___________________________________________________________
from heapq import *
n = int(input())
a = [0]+list(map(int,input().split()))
b = list(map(int,input().split()))
ans = []
best = 0
vis = [False for i in range(n+1)]
value = a.copy()
par = [-1]*(n+1)
def find(c):
if par[c]<0:
return c,-value[c]
par[c],trash = find(par[c])
return par[c],trash
def union(a1,b1):
if vis[b1]==False:
return value[a1]
pa,val1 = find(a1)
pb,val2 = find(b1)
if pa==pb:
return False
if par[pa]>par[pb]:
par[pb]+=par[pa]
par[pa] = pb
value[pb]+=value[pa]
return value[pb]
else:
par[pa]+=par[pb]
par[pb] = pa
value[pa]+=value[pb]
return value[pa]
for i in range(n):
ans.append(best)
ind = b.pop()
vis[ind] = True
if ind+1<=n:
res = union(ind,ind+1)
if res:
best = max(best,res)
if ind-1>0:
res = union(ind,ind-1)
if res:
best = max(best,res)
# print("HEAP",h)
for i in ans[::-1]:
print(i)
``` | instruction | 0 | 99,935 | 12 | 199,870 |
Yes | output | 1 | 99,935 | 12 | 199,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array consisting of n non-negative integers a1, a2, ..., an.
You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.
After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
The third line contains a permutation of integers from 1 to n — the order used to destroy elements.
Output
Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first i operations are performed.
Examples
Input
4
1 3 2 5
3 4 1 2
Output
5
4
3
0
Input
5
1 2 3 4 5
4 2 3 5 1
Output
6
5
5
1
0
Input
8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6
Output
18
16
11
8
8
6
6
0
Note
Consider the first sample:
1. Third element is destroyed. Array is now 1 3 * 5. Segment with maximum sum 5 consists of one integer 5.
2. Fourth element is destroyed. Array is now 1 3 * * . Segment with maximum sum 4 consists of two integers 1 3.
3. First element is destroyed. Array is now * 3 * * . Segment with maximum sum 3 consists of one integer 3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0.
Submitted Solution:
```
###### ### ####### ####### ## # ##### ### #####
# # # # # # # # # # # # # ###
# # # # # # # # # # # # # ###
###### ######### # # # # # # ######### #
###### ######### # # # # # # ######### #
# # # # # # # # # # #### # # #
# # # # # # # ## # # # # #
###### # # ####### ####### # # ##### # # # #
from __future__ import print_function # for PyPy2
from itertools import permutations as perm
# from fractions import Fraction
from collections import *
from sys import stdin
from bisect import *
from heapq import *
from math import *
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")
n, = gil()
a = [0] + gil()
for i in range(n+1):
a[i] += a[i-1]
idx = gil()
ans = []
h = [0] # max Heap
l, r = [0]*(n+2), [0]*(n+2)
while idx:
ans.append(-h[0])
# add element
i = idx.pop()
li, ri = i - l[i-1], i + r[i+1]
cnt = ri-li+1
r[li] = l[ri] = cnt
rangeSm = a[li-1] - a[ri]
heappush(h, rangeSm)
ans.reverse()
print(*ans, sep='\n')
``` | instruction | 0 | 99,936 | 12 | 199,872 |
Yes | output | 1 | 99,936 | 12 | 199,873 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array consisting of n non-negative integers a1, a2, ..., an.
You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.
After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
The third line contains a permutation of integers from 1 to n — the order used to destroy elements.
Output
Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first i operations are performed.
Examples
Input
4
1 3 2 5
3 4 1 2
Output
5
4
3
0
Input
5
1 2 3 4 5
4 2 3 5 1
Output
6
5
5
1
0
Input
8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6
Output
18
16
11
8
8
6
6
0
Note
Consider the first sample:
1. Third element is destroyed. Array is now 1 3 * 5. Segment with maximum sum 5 consists of one integer 5.
2. Fourth element is destroyed. Array is now 1 3 * * . Segment with maximum sum 4 consists of two integers 1 3.
3. First element is destroyed. Array is now * 3 * * . Segment with maximum sum 3 consists of one integer 3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0.
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Fri Dec 9 16:14:34 2016
@author: kostiantyn.omelianchuk
"""
from sys import stdin, stdout
lines = stdin.readlines()
n = int(lines[0])
a = [int(x) for x in lines[1].split()]
b = [int(x) for x in lines[2].split()]
#check_array = [0 for i in range(n)]
check_dict = {}
snm = [i for i in range(n)]
r = [1 for i in range(n)]
sums = dict(zip(range(n), a))
def find_x(x):
if snm[x] != x:
snm[x] = find_x(snm[x])
return snm[x]
def union(x, y):
#x = find_x(x)
#y = find_x(y)
sums[x] += sums[y]
sums[y] = sums[x]
if x == y:
return
#sums[x] += sums[x_start]
if r[x] == r[y]:
r[x] += 1
if r[x] < r[y]:
snm[x] = y
else:
snm[y] = x
max_list = []
total_max = 0
for i in range(n):
cur_sum = 0
flag = 0
max_list.append(total_max)
pos = n-i-1
elem = b[pos] - 1
#check_array[elem] = 1
check_dict[elem] = 1
pos_x = find_x(elem)
if elem>0:
if check_dict.get(elem-1,0) == 1:
pos_y = find_x(elem-1)
union(pos_x,pos_y)
cur_sum = sums[pos_y]
#print(sums, check_array, total_max, cur_sum, elem, find_x(elem))
else:
flag += 1
else:
flag += 1
if elem<(n-1):
if check_dict.get(elem+1,0) == 1:
pos_y = find_x(elem+1)
union(pos_x,pos_y)
cur_sum = sums[pos_y]
#print(sums, check_array, total_max, cur_sum, elem, find_x(elem))
else:
flag += 1
else:
flag += 1
if flag == 2:
total_max = max(total_max,sums[pos_x])
else:
total_max = max(cur_sum, total_max)
max_list.append(total_max)
for j in range(1,n+1):
print(max_list[n-j])
``` | instruction | 0 | 99,937 | 12 | 199,874 |
No | output | 1 | 99,937 | 12 | 199,875 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array consisting of n non-negative integers a1, a2, ..., an.
You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.
After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
The third line contains a permutation of integers from 1 to n — the order used to destroy elements.
Output
Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first i operations are performed.
Examples
Input
4
1 3 2 5
3 4 1 2
Output
5
4
3
0
Input
5
1 2 3 4 5
4 2 3 5 1
Output
6
5
5
1
0
Input
8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6
Output
18
16
11
8
8
6
6
0
Note
Consider the first sample:
1. Third element is destroyed. Array is now 1 3 * 5. Segment with maximum sum 5 consists of one integer 5.
2. Fourth element is destroyed. Array is now 1 3 * * . Segment with maximum sum 4 consists of two integers 1 3.
3. First element is destroyed. Array is now * 3 * * . Segment with maximum sum 3 consists of one integer 3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0.
Submitted Solution:
```
n = int(input())
a = input()
nums = list(map(int, a.split()))
a = input()
breaks = list(map(int, a.split()))
for b in breaks:
m = 0
nums[b-1] = 0
seq = []
for each in nums:
now = 0
if each == 0:
for each in seq:
now = max(0, now+each)
m = max(m, now)
seq = []
else:
seq.append(each)
for each in seq:
now = max(0, now+each)
m = max(m, now)
seq = []
print(m)
``` | instruction | 0 | 99,938 | 12 | 199,876 |
No | output | 1 | 99,938 | 12 | 199,877 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array consisting of n non-negative integers a1, a2, ..., an.
You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.
After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
The third line contains a permutation of integers from 1 to n — the order used to destroy elements.
Output
Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first i operations are performed.
Examples
Input
4
1 3 2 5
3 4 1 2
Output
5
4
3
0
Input
5
1 2 3 4 5
4 2 3 5 1
Output
6
5
5
1
0
Input
8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6
Output
18
16
11
8
8
6
6
0
Note
Consider the first sample:
1. Third element is destroyed. Array is now 1 3 * 5. Segment with maximum sum 5 consists of one integer 5.
2. Fourth element is destroyed. Array is now 1 3 * * . Segment with maximum sum 4 consists of two integers 1 3.
3. First element is destroyed. Array is now * 3 * * . Segment with maximum sum 3 consists of one integer 3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0.
Submitted Solution:
```
size = int(input())
num = list(map(int, input().split(" ")))
rem = list(map(lambda x: int(x) - 1, input().split(" ")))
def subsum(lst):
ms = 0
cs = 0
for i in lst:
cs += i
if i == 0 and cs > ms:
ms = cs
cs = 0
return ms if ms > cs else cs
for i in rem:
num[i] = 0
print (subsum(num))
``` | instruction | 0 | 99,939 | 12 | 199,878 |
No | output | 1 | 99,939 | 12 | 199,879 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array consisting of n non-negative integers a1, a2, ..., an.
You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.
After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.
The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).
The third line contains a permutation of integers from 1 to n — the order used to destroy elements.
Output
Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first i operations are performed.
Examples
Input
4
1 3 2 5
3 4 1 2
Output
5
4
3
0
Input
5
1 2 3 4 5
4 2 3 5 1
Output
6
5
5
1
0
Input
8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6
Output
18
16
11
8
8
6
6
0
Note
Consider the first sample:
1. Third element is destroyed. Array is now 1 3 * 5. Segment with maximum sum 5 consists of one integer 5.
2. Fourth element is destroyed. Array is now 1 3 * * . Segment with maximum sum 4 consists of two integers 1 3.
3. First element is destroyed. Array is now * 3 * * . Segment with maximum sum 3 consists of one integer 3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0.
Submitted Solution:
```
#!/usr/bin/python3
size = int(input())
num = list(map(int, input().split()))
rem = list(map(int, input().split()))
chunks = [None] * size
res = [-1] * size
ans = [0]
ms = -1
def addChunk(n):
chunks[n] = [n, num[n]]
return n
def getRoot(n):
return n if chunks[n][0] == n else getRoot(chunks[n][0])
def mergeChunks(parent, child):
proot = getRoot(parent)
croot = getRoot(child)
chunks[croot][0] = proot
chunks[proot][1] += chunks[croot][1]
return proot
ii = size
while ii >= 0:
ii -= 1
i = rem[ii] - 1
res[i] = num[i]
root = addChunk(i)
if i > 0 and chunks[i - 1] != None:
root = mergeChunks(i - 1, i)
if i + 1 < size and chunks[i + 1] != None:
root = mergeChunks(i, i + 1)
ms = max(ms, chunks[root][1])
ans.append(ms)
for i in range(2, size):
print (ans[-i-1])
print(0)
``` | instruction | 0 | 99,940 | 12 | 199,880 |
No | output | 1 | 99,940 | 12 | 199,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the function p(x), where x is an array of m integers, which returns an array y consisting of m + 1 integers such that yi is equal to the sum of first i elements of array x (0 ≤ i ≤ m).
You have an infinite sequence of arrays A0, A1, A2..., where A0 is given in the input, and for each i ≥ 1 Ai = p(Ai - 1). Also you have a positive integer k. You have to find minimum possible i such that Ai contains a number which is larger or equal than k.
Input
The first line contains two integers n and k (2 ≤ n ≤ 200000, 1 ≤ k ≤ 1018). n is the size of array A0.
The second line contains n integers A00, A01... A0n - 1 — the elements of A0 (0 ≤ A0i ≤ 109). At least two elements of A0 are positive.
Output
Print the minimum i such that Ai contains a number which is larger or equal than k.
Examples
Input
2 2
1 1
Output
1
Input
3 6
1 1 1
Output
2
Input
3 1
1 0 1
Output
0 | instruction | 0 | 99,981 | 12 | 199,962 |
Tags: binary search, brute force, combinatorics, math, matrices
Correct Solution:
```
from sys import stdin, stdout
from math import factorial
from math import log10
def check(pw, values, k):
n = len(values)
matr = [[0 for i in range(n)] for j in range(n)]
res = [[0 for i in range(n)] for j in range(n)]
pp = [[0 for i in range(n)] for j in range(n)]
for i in range(n):
for j in range(n):
matr[i][j] = int(i >= j)
for i in range(n):
res[i][i] = 1
while pw:
if pw & 1:
for i in range(n):
for j in range(n):
pp[i][j] = 0
for i in range(n):
for j in range(n):
for z in range(n):
pp[i][j] += res[i][z] * matr[z][j]
for i in range(n):
for j in range(n):
res[i][j] = pp[i][j]
for i in range(n):
for j in range(n):
pp[i][j] = 0
for i in range(n):
for j in range(n):
for z in range(n):
pp[i][j] += matr[i][z] * matr[z][j]
for i in range(n):
for j in range(n):
matr[i][j] = pp[i][j]
pw >>= 1
ans = 0
for i in range(n):
ans += values[i] * res[i][0]
return ans >= k
def stupid(values, n, k):
ans = values[:]
if max(ans) >= k:
return 0
cnt = 0
while ans[-1] < k:
for i in range(1, n):
ans[i] += ans[i - 1]
cnt += 1
q = ans[-1]
return cnt
def clever(values, n, k):
if max(values) >= k:
return 0
l, r = 0, 10 ** 18 + 100
while r - l > 1:
m = (l + r) >> 1
if check(m, values, k):
r = m
else:
l = m
return r
def main():
n, k = map(int, stdin.readline().split())
values = list(map(int, stdin.readline().split()))[::-1]
while not values[-1]: #Проблема в том что ты удаляешь
values.pop()
n = len(values)
if n >= 10:
stdout.write(str(stupid(values[::-1], n, k)) + '\n')
else:
stdout.write(str(clever(values, n, k)) + '\n')
main()
``` | output | 1 | 99,981 | 12 | 199,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the function p(x), where x is an array of m integers, which returns an array y consisting of m + 1 integers such that yi is equal to the sum of first i elements of array x (0 ≤ i ≤ m).
You have an infinite sequence of arrays A0, A1, A2..., where A0 is given in the input, and for each i ≥ 1 Ai = p(Ai - 1). Also you have a positive integer k. You have to find minimum possible i such that Ai contains a number which is larger or equal than k.
Input
The first line contains two integers n and k (2 ≤ n ≤ 200000, 1 ≤ k ≤ 1018). n is the size of array A0.
The second line contains n integers A00, A01... A0n - 1 — the elements of A0 (0 ≤ A0i ≤ 109). At least two elements of A0 are positive.
Output
Print the minimum i such that Ai contains a number which is larger or equal than k.
Examples
Input
2 2
1 1
Output
1
Input
3 6
1 1 1
Output
2
Input
3 1
1 0 1
Output
0 | instruction | 0 | 99,982 | 12 | 199,964 |
Tags: binary search, brute force, combinatorics, math, matrices
Correct Solution:
```
from itertools import *
n, k = map(int, input().split())
*A, = map(int, input().split())
alc = 0
for v in A:
if v > 0:
alc += 1
if max(A) >= k:
print(0)
elif alc <= 20:
l = 0
r = int(1e18)
while l + 1 < r:
m = (l + r) // 2
s = 0
for i in range(n):
if A[i] > 0:
s2 = 1
for j in range(n - i - 1):
s2 *= m + j
s2 //= j + 1
if s2 >= k:
break
s += s2 * A[i]
if s >= k:
break
if s >= k:
r = m
else:
l = m
print(r)
else:
ans = 0
while True:
ans += 1
for i in range(1, n):
A[i] += A[i - 1]
if A[-1] >= k:
break
print(ans)
``` | output | 1 | 99,982 | 12 | 199,965 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the function p(x), where x is an array of m integers, which returns an array y consisting of m + 1 integers such that yi is equal to the sum of first i elements of array x (0 ≤ i ≤ m).
You have an infinite sequence of arrays A0, A1, A2..., where A0 is given in the input, and for each i ≥ 1 Ai = p(Ai - 1). Also you have a positive integer k. You have to find minimum possible i such that Ai contains a number which is larger or equal than k.
Input
The first line contains two integers n and k (2 ≤ n ≤ 200000, 1 ≤ k ≤ 1018). n is the size of array A0.
The second line contains n integers A00, A01... A0n - 1 — the elements of A0 (0 ≤ A0i ≤ 109). At least two elements of A0 are positive.
Output
Print the minimum i such that Ai contains a number which is larger or equal than k.
Examples
Input
2 2
1 1
Output
1
Input
3 6
1 1 1
Output
2
Input
3 1
1 0 1
Output
0 | instruction | 0 | 99,983 | 12 | 199,966 |
Tags: binary search, brute force, combinatorics, math, matrices
Correct Solution:
```
import sys
n, k = map(int, input().split())
ar = []
br = map(int, input().split())
for x in br:
if(x > 0 or len(ar) > 0):
ar.append(x)
if(max(ar) >= k):
print(0)
elif len(ar) == 2:
print((k - ar[1] + ar[0] - 1)//ar[0])
elif len(ar) == 3:
ini, fin = 0, 10**18
while(ini < fin):
mid = (ini + fin)//2
if(ar[2] + ar[1]*mid + ar[0]*mid*(mid + 1)//2 >= k):
fin = mid
else:
ini = mid + 1
print(ini)
else:
ans = 0;
while(ar[-1] < k):
ans += 1
for i in range(1, len(ar)):
ar[i] += ar[i - 1]
print(ans)
``` | output | 1 | 99,983 | 12 | 199,967 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the function p(x), where x is an array of m integers, which returns an array y consisting of m + 1 integers such that yi is equal to the sum of first i elements of array x (0 ≤ i ≤ m).
You have an infinite sequence of arrays A0, A1, A2..., where A0 is given in the input, and for each i ≥ 1 Ai = p(Ai - 1). Also you have a positive integer k. You have to find minimum possible i such that Ai contains a number which is larger or equal than k.
Input
The first line contains two integers n and k (2 ≤ n ≤ 200000, 1 ≤ k ≤ 1018). n is the size of array A0.
The second line contains n integers A00, A01... A0n - 1 — the elements of A0 (0 ≤ A0i ≤ 109). At least two elements of A0 are positive.
Output
Print the minimum i such that Ai contains a number which is larger or equal than k.
Examples
Input
2 2
1 1
Output
1
Input
3 6
1 1 1
Output
2
Input
3 1
1 0 1
Output
0 | instruction | 0 | 99,984 | 12 | 199,968 |
Tags: binary search, brute force, combinatorics, math, matrices
Correct Solution:
```
from math import factorial
def ncr(n, r):
ans = 1
for i in range(r):
ans*=n-i
for i in range(r):
ans//=i+1
return ans
n,k = map(int, input().split())
inp = list(map(int,input().split()))
nz = 0
seq = []
for i in range(n):
if(inp[i]!=0):
nz = 1
if(nz!=0):
seq.append(inp[i])
if(max(seq) >= k):
print("0")
exit(0)
if(len(seq) <= 8):
seq.reverse()
mn = 1
mx = pow(10,18)
while(mn < mx):
mid = (mn + mx)//2
curr = 0
for i in range(len(seq)):
curr+=seq[i]*ncr(mid+i-1, i)
if(curr>=k):
mx = mid
else:
mn = mid + 1
print(mn)
exit(0)
for i in range(1000):
for j in range(1,len(seq)):
seq[j]+=seq[j-1]
if(max(seq) >= k):
print(i+1)
exit(0)
``` | output | 1 | 99,984 | 12 | 199,969 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the function p(x), where x is an array of m integers, which returns an array y consisting of m + 1 integers such that yi is equal to the sum of first i elements of array x (0 ≤ i ≤ m).
You have an infinite sequence of arrays A0, A1, A2..., where A0 is given in the input, and for each i ≥ 1 Ai = p(Ai - 1). Also you have a positive integer k. You have to find minimum possible i such that Ai contains a number which is larger or equal than k.
Input
The first line contains two integers n and k (2 ≤ n ≤ 200000, 1 ≤ k ≤ 1018). n is the size of array A0.
The second line contains n integers A00, A01... A0n - 1 — the elements of A0 (0 ≤ A0i ≤ 109). At least two elements of A0 are positive.
Output
Print the minimum i such that Ai contains a number which is larger or equal than k.
Examples
Input
2 2
1 1
Output
1
Input
3 6
1 1 1
Output
2
Input
3 1
1 0 1
Output
0 | instruction | 0 | 99,985 | 12 | 199,970 |
Tags: binary search, brute force, combinatorics, math, matrices
Correct Solution:
```
import sys
def mul(a, b):
#print(a, b)
n = len(a)
m = [[0 for j in range(n)] for i in range(n)]
for i in range(n):
for j in range(n):
for k in range(n):
m[i][j] += a[i][k] * b[k][j]
return m
n, k = [int(i) for i in input().split(" ")]
a = [int(i) for i in input().split(" ")]
if max(a) >= k:
print(0)
sys.exit(0)
a0 = [] # pole bez tych nul na zaciatku, ktore su zbytocne
s = 0
for i in range(n):
s += a[i]
if s > 0:
a0.append(a[i])
n = len(a0)
if n > 30: # pole je velke, po najviac 50 iteraciach dojdeme k odpovedi
it = 0
while True:
it += 1
for j in range(1, n):
a0[j] += a0[j-1]
if a0[j] >= k:
print(it)
sys.exit(0)
else: # pole je male, spustime nieco typu bs + umocnovanie matic
m = [[[int(i >= j) for j in range(n)] for i in range(n)]]
for l in range(0, 62):
m.append(mul(m[-1], m[-1]))
#print(m[-1])
#print(a0)
ans = 0
u = [[int(i == j) for j in range(n)] for i in range(n)]
for l in range(62, -1, -1):
u2 = mul(u, m[l])
val = 0
for i in range(0, n):
val += a0[i] * u2[-1][i]
#print(val)
if val < k:
u = u2
ans += 2**l
print(ans + 1)
``` | output | 1 | 99,985 | 12 | 199,971 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the function p(x), where x is an array of m integers, which returns an array y consisting of m + 1 integers such that yi is equal to the sum of first i elements of array x (0 ≤ i ≤ m).
You have an infinite sequence of arrays A0, A1, A2..., where A0 is given in the input, and for each i ≥ 1 Ai = p(Ai - 1). Also you have a positive integer k. You have to find minimum possible i such that Ai contains a number which is larger or equal than k.
Input
The first line contains two integers n and k (2 ≤ n ≤ 200000, 1 ≤ k ≤ 1018). n is the size of array A0.
The second line contains n integers A00, A01... A0n - 1 — the elements of A0 (0 ≤ A0i ≤ 109). At least two elements of A0 are positive.
Output
Print the minimum i such that Ai contains a number which is larger or equal than k.
Examples
Input
2 2
1 1
Output
1
Input
3 6
1 1 1
Output
2
Input
3 1
1 0 1
Output
0 | instruction | 0 | 99,986 | 12 | 199,972 |
Tags: binary search, brute force, combinatorics, math, matrices
Correct Solution:
```
s=input().split()
n=int(s[0])
k=int(s[1])
L=list(map(int,input().split()))
lx=0
while L[lx]==0:
lx+=1
A=[]
for i in range(lx,n):
A.append(L[i])
n=len(A)
n=len(A)
def good(l):
coeff=1
tot=0
for i in reversed(range(n)):
tot+=coeff*A[i]
if tot>=k:
return True
coeff=(coeff*(n-i-1+l))//(n-i)
return False
ans=0
for i in range(1,10):
if good(i):
ans=i
break
if ans==0:
l=1
r=k
while True:
if l==r:
ans=l
break
if l+1==r:
if good(l):
ans=l
else:
ans=r
break
m=(l+r)//2
if good(m):
r=m
else:
l=m+1
for i in range(n):
if A[i]>=k:
ans=0
print(ans)
``` | output | 1 | 99,986 | 12 | 199,973 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the function p(x), where x is an array of m integers, which returns an array y consisting of m + 1 integers such that yi is equal to the sum of first i elements of array x (0 ≤ i ≤ m).
You have an infinite sequence of arrays A0, A1, A2..., where A0 is given in the input, and for each i ≥ 1 Ai = p(Ai - 1). Also you have a positive integer k. You have to find minimum possible i such that Ai contains a number which is larger or equal than k.
Input
The first line contains two integers n and k (2 ≤ n ≤ 200000, 1 ≤ k ≤ 1018). n is the size of array A0.
The second line contains n integers A00, A01... A0n - 1 — the elements of A0 (0 ≤ A0i ≤ 109). At least two elements of A0 are positive.
Output
Print the minimum i such that Ai contains a number which is larger or equal than k.
Examples
Input
2 2
1 1
Output
1
Input
3 6
1 1 1
Output
2
Input
3 1
1 0 1
Output
0 | instruction | 0 | 99,987 | 12 | 199,974 |
Tags: binary search, brute force, combinatorics, math, matrices
Correct Solution:
```
n,k = map(int,input().split());
a = list(map(int,input().split()));
if max(a) >= k:
print(0)
exit()
lx = 0
while a[lx] == 0:
lx+=1
lo,hi = 1,k
def can(x):
bc = 1
tot = 0
for i in range(n-lx):
if(bc >= k):
return True
tot += bc*a[n-1-i]
bc *= (x+i)
bc = bc//(i+1)
if(tot >= k):
return True
return tot >= k
while lo < hi :
mid = (lo+hi)//2
cancan = can(mid)
#print(mid,cancan)
if cancan : hi = mid
else : lo = mid + 1
print(lo)
``` | output | 1 | 99,987 | 12 | 199,975 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Consider the function p(x), where x is an array of m integers, which returns an array y consisting of m + 1 integers such that yi is equal to the sum of first i elements of array x (0 ≤ i ≤ m).
You have an infinite sequence of arrays A0, A1, A2..., where A0 is given in the input, and for each i ≥ 1 Ai = p(Ai - 1). Also you have a positive integer k. You have to find minimum possible i such that Ai contains a number which is larger or equal than k.
Input
The first line contains two integers n and k (2 ≤ n ≤ 200000, 1 ≤ k ≤ 1018). n is the size of array A0.
The second line contains n integers A00, A01... A0n - 1 — the elements of A0 (0 ≤ A0i ≤ 109). At least two elements of A0 are positive.
Output
Print the minimum i such that Ai contains a number which is larger or equal than k.
Examples
Input
2 2
1 1
Output
1
Input
3 6
1 1 1
Output
2
Input
3 1
1 0 1
Output
0 | instruction | 0 | 99,988 | 12 | 199,976 |
Tags: binary search, brute force, combinatorics, math, matrices
Correct Solution:
```
n, k = map(int,input().split())
a = list(map(int,input().split()))
ptr = 0
while a[ptr]==0:
ptr += 1
def check(x):
if x==0:
return max(a) >= k
binomial = 1
sum = 0
for i in range(n-ptr):
if binomial >= k:
return True
sum += binomial * a[n-i-1]
binomial *= (x+i)
binomial //= (i+1)
if sum >= k:
return True
return False
lo,hi = 0, k
while lo < hi:
md = (lo+hi) // 2
if check(md):
hi = md;
else:
lo = md + 1
print(lo)
``` | output | 1 | 99,988 | 12 | 199,977 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO | instruction | 0 | 100,319 | 12 | 200,638 |
Tags: constructive algorithms, flows, math
Correct Solution:
```
USE_STDIO = False
if not USE_STDIO:
try: import mypc
except: pass
def getXor(a):
ans = 0
for x in a:
ans ^= x
return ans
def main():
n, m = map(int, input().split(' '))
a = [x for x in map(int, input().split(' '))]
b = [x for x in map(int, input().split(' '))]
rxor = getXor(a)
cxor = getXor(b)
if rxor != cxor:
print('NO')
return
x = [[0] * m for _ in range(n)]
if n >= m:
x[0][0] = a[0]
for i in range(1, m):
x[i][i-1] = b[i-1] ^ x[i-1][i-1]
x[i][i] = a[i] ^ x[i][i-1]
for i in range(m, n):
x[i][-1] = a[i]
else:
x[0][0] = b[0]
for i in range(1, n):
x[i-1][i] = a[i-1] ^ x[i-1][i-1]
x[i][i] = b[i] ^ x[i-1][i]
for i in range(n, m):
x[-1][i] = b[i]
print('YES')
for i in range(n):
print(' '.join(map(str, x[i])))
if __name__ == '__main__':
main()
``` | output | 1 | 100,319 | 12 | 200,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO | instruction | 0 | 100,320 | 12 | 200,640 |
Tags: constructive algorithms, flows, math
Correct Solution:
```
n,m = map(int,input().split())
r = list(map(int,input().split()))
c = list(map(int,input().split()))
mat = [[0]*m for i in range(n)]
fail = False
for p in range(60):
totr = 0
needr = []
totc = 0
needc = []
for i in range(n):
if (r[i]>>p)&1:
needr.append(i)
totr += 1
for j in range(m):
if (c[j]>>p)&1:
needc.append(j)
totc += 1
if (totr-totc) % 2 != 0:
fail = True
break
else:
mn = min(len(needr),len(needc))
for k in range(mn):
mat[needr[k]][needc[k]] += (1 << p)
if len(needr) > len(needc):
em = 0
for i in range(m):
if i not in needc:
em = i
break
for i in range(mn,len(needr)):
mat[needr[i]][em] += (1 << p)
else:
em = 0
for i in range(n):
if i not in needr:
em = i
break
for i in range(mn, len(needc)):
mat[em][needc[i]] += (1 << p)
if fail:
print("NO")
else:
print("YES")
for i in mat:
print(*i)
``` | output | 1 | 100,320 | 12 | 200,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO | instruction | 0 | 100,321 | 12 | 200,642 |
Tags: constructive algorithms, flows, math
Correct Solution:
```
i=lambda:[*map(int,input().split())]
n,m=i()
a,b=i(),i()
def s(a):
r=0
for i in a: r^=i
return r
if s(a)==s(b):
print("YES")
b[0]^=s(a[1:])
print(*b)
for i in a[1:]:print(str(i)+' 0'*(m-1))
else:
print("NO")
``` | output | 1 | 100,321 | 12 | 200,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO | instruction | 0 | 100,322 | 12 | 200,644 |
Tags: constructive algorithms, flows, math
Correct Solution:
```
from functools import reduce
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
if reduce( (lambda x, y: x^y), a) != reduce( (lambda x, y: x^y), b):
print('NO')
else:
print('YES')
for row in range(n-1):
for col in range(m - 1):
print(0, end=' ')
print(a[row])
for col in range(m - 1):
print(b[col], end=' ')
print(reduce( (lambda x, y: x^y), b[:-1]) ^ a[n-1])
# print(3 ^ 0)
``` | output | 1 | 100,322 | 12 | 200,645 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO | instruction | 0 | 100,323 | 12 | 200,646 |
Tags: constructive algorithms, flows, math
Correct Solution:
```
n, m = map(int, input().split())
xor_sum = []
a = []
for i in range(2):
a.append(list(map(int, input().split())))
xor_sum.append(0)
for x in a[-1]:
xor_sum[-1] ^= x
if xor_sum[0] != xor_sum[1]:
print('NO')
exit(0)
print('YES')
for i in range(n):
for j in range(m):
if (i < n - 1) and (j < m - 1):
print(0, end = ' ')
elif (i == n - 1) and (j < m - 1):
print(a[1][j], end = ' ')
elif (i < n - 1) and (j == m - 1):
print(a[0][i], end = ' ')
else:
t = a[0][-1]
for k in range(0, m - 1):
t ^= a[1][k]
print(t, end = ' ')
print()
``` | output | 1 | 100,323 | 12 | 200,647 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO | instruction | 0 | 100,324 | 12 | 200,648 |
Tags: constructive algorithms, flows, math
Correct Solution:
```
n,m=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
x=a[-1]
y=b[-1]
for j in range(m-1):
x=x^b[j]
for j in range(n-1):
y=y^a[j]
if x!=y:
print("NO")
else:
print("YES")
c=[]
for i in range(n):
c.append([])
for i in range(n):
for j in range(m):
if i!=(n-1) and j!=(m-1):
c[i].append(0)
elif i!=(n-1) and j==(m-1):
c[i].append(a[i])
elif i==(n-1) and j!=(m-1):
c[i].append(b[j])
else:
c[i].append(x)
for j in c:
print(*j)
``` | output | 1 | 100,324 | 12 | 200,649 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO | instruction | 0 | 100,325 | 12 | 200,650 |
Tags: constructive algorithms, flows, math
Correct Solution:
```
n,m=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
xorall=0
def xor(a):
x=0
for i in a:
x^=i
return x
if xor(a)!=xor(b):
print("NO")
else:
print("YES")
a[0]^=xor(b[1:])
print(a[0],*b[1:])
for i in a[1:]:
print(str(i)+' 0'*(m-1))
``` | output | 1 | 100,325 | 12 | 200,651 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Now Vasya is taking an exam in mathematics. In order to get a good mark, Vasya needs to guess the matrix that the teacher has constructed!
Vasya knows that the matrix consists of n rows and m columns. For each row, he knows the xor (bitwise excluding or) of the elements in this row. The sequence a1, a2, ..., an denotes the xor of elements in rows with indices 1, 2, ..., n, respectively. Similarly, for each column, he knows the xor of the elements in this column. The sequence b1, b2, ..., bm denotes the xor of elements in columns with indices 1, 2, ..., m, respectively.
Help Vasya! Find a matrix satisfying the given constraints or tell him that there is no suitable matrix.
Input
The first line contains two numbers n and m (2 ≤ n, m ≤ 100) — the dimensions of the matrix.
The second line contains n numbers a1, a2, ..., an (0 ≤ ai ≤ 109), where ai is the xor of all elements in row i.
The third line contains m numbers b1, b2, ..., bm (0 ≤ bi ≤ 109), where bi is the xor of all elements in column i.
Output
If there is no matrix satisfying the given constraints in the first line, output "NO".
Otherwise, on the first line output "YES", and then n rows of m numbers in each ci1, ci2, ... , cim (0 ≤ cij ≤ 2·109) — the description of the matrix.
If there are several suitable matrices, it is allowed to print any of them.
Examples
Input
2 3
2 9
5 3 13
Output
YES
3 4 5
6 7 8
Input
3 3
1 7 6
2 15 12
Output
NO | instruction | 0 | 100,326 | 12 | 200,652 |
Tags: constructive algorithms, flows, math
Correct Solution:
```
###### ### ####### ####### ## # ##### ### #####
# # # # # # # # # # # # # ###
# # # # # # # # # # # # # ###
###### ######### # # # # # # ######### #
###### ######### # # # # # # ######### #
# # # # # # # # # # #### # # #
# # # # # # # ## # # # # #
###### # # ####### ####### # # ##### # # # #
from __future__ import print_function # for PyPy2
# from itertools import permutations as perm
# from functools import cmp_to_key # for adding custom comparator
# from fractions import Fraction
# from collections import *
from sys import stdin
# from bisect import *
# from heapq import *
# from math import *
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")
n, m = gil()
r, c = gil(), gil()
rv, cv = 0, 0
for v in r:
rv ^= v
for v in c:
cv ^= v
if rv == cv:
print('YES')
x = r[0]^cv^c[0]
ans = [[0 for _ in range(m)] for _ in range(n)]
for j in range(1, m):
ans[0][j] = c[j]
for i in range(1, n):
ans[i][0] = r[i]
ans[0][0] = x
for r in ans:
print(*r)
else:
print('NO')
``` | output | 1 | 100,326 | 12 | 200,653 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2. | instruction | 0 | 100,461 | 12 | 200,922 |
Tags: greedy, implementation, math
Correct Solution:
```
n, m = (int(i) for i in input().split())
a = [[int(i) for i in input().split()] for _ in range(n)]
res = 0
for j in range(m):
b = [0] * n
for i in range(n):
if a[i][j] <= n*m and (a[i][j]-j-1)%m == 0:
pos = (a[i][j]-j-1)//m
shift = i-pos if i>=pos else i-pos+n
b[shift] += 1
res += min(i+n-b[i] for i in range(n))
print(res)
``` | output | 1 | 100,461 | 12 | 200,923 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2. | instruction | 0 | 100,462 | 12 | 200,924 |
Tags: greedy, implementation, math
Correct Solution:
```
#from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
#from bisect import bisect_right as br #c++ upperbound br(array,element)
#from __future__ import print_function, division #while using python2
# from itertools import accumulate
from collections import defaultdict, Counter
def main():
#sys.stdin = open('input.txt', 'r')
#sys.stdout = open('output.txt', 'w')
n, m = [int(x) for x in input().split()]
grid = [[-1] * (m+1) for i in range(n+1)]
for i in range(1, n+1):
arr = [int(x) for x in input().split()]
for j in range(1, m+1):
grid[i][j] = arr[j-1]
# print(grid)
ans = 0
for j in range(1, m+1):
cur_col_ans = 1e9
off_by_above = defaultdict(lambda : 0)
off_by_below = defaultdict(lambda : 0)
for i in range(1, n+1):
cur = j + (i-1)*m
off_by_below[grid[i][j] - cur] += 1
for i in range(1, n+1):
# print("i = ", i)
d = m*(i-1)
temp = i-1
# print(temp)
temp += (n-i+1 - off_by_below[-d])
# print(temp)
temp += (i-1 - off_by_above[(n+1-i)*m])
# print(temp)
cur_col_ans = min(cur_col_ans, temp)
# print("ans from ", grid[i][j], " = ", i-1+(n-off_by[-d]))
off_by_below[grid[i][j] - (j + (i-1)*m)] -= 1
off_by_above[grid[i][j] - (j + (i-1)*m)] += 1
# print("col", j, ":", cur_col_ans)
# print(dict(off_by))
ans += cur_col_ans
print(ans)
#------------------ Python 2 and 3 footer by Pajenegod and c1729-----------------------------------------
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
import os, sys
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill(): pass
return super(FastIO,self).read()
def readline(self):
while self.newlines == 0:
s = self._fill(); self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s:self.buffer.write(s.encode('ascii'))
self.read = lambda:self.buffer.read().decode('ascii')
self.readline = lambda:self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
if __name__ == '__main__':
main()
``` | output | 1 | 100,462 | 12 | 200,925 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2. | instruction | 0 | 100,463 | 12 | 200,926 |
Tags: greedy, implementation, math
Correct Solution:
```
from sys import stdin, stdout
def main():
arr = list(map(int, stdin.readline().split()))
X = arr[0]
Y = arr[1]
rect = list()
temp = list()
for j in range(X):
temp.append(0)
for i in range(Y):
rect.append(list(temp))
for i in range(X):
arr = list(map(int, stdin.readline().split()))
for j in range(Y):
if (arr[j] - j - 1) % Y != 0 or (arr[j]) > X*Y:
continue
sure = int((arr[j] - j - 1) / Y)
sure = i - sure
if sure < 0:
sure += X
rect[j][sure] += 1
total_sum = 0
for i in range (Y):
maxer = X
for j in range(X):
count = X - rect[i][j] + j
if count < maxer:
maxer = count
total_sum += maxer
stdout.write(str(total_sum))
main()
``` | output | 1 | 100,463 | 12 | 200,927 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2. | instruction | 0 | 100,464 | 12 | 200,928 |
Tags: greedy, implementation, math
Correct Solution:
```
''' Hey stalker :) '''
INF = 10**10
def main():
#print = out.append
''' Cook your dish here! '''
n, m = get_list()
mat = [get_list() for _ in range(n)]
res = 0
for j in range(m):
shifts = [0]*n
for i in range(n):
if (mat[i][j] - (j+1))%m == 0:
org = (mat[i][j] - (j+1))//m
#print(i, j, org)
if 0<=org<n:
shifts[(i-org)%n] += 1
#print(shifts)
drop = 0
for i, ele in enumerate(shifts):
drop = max(drop, ele-i)
res += n-drop
print(res)
''' Pythonista fLite 1.1 '''
import sys
from collections import defaultdict, Counter
from bisect import bisect_left, bisect_right
#from functools import reduce
import math
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
out = []
get_int = lambda: int(input())
get_list = lambda: list(map(int, input().split()))
main()
#[main() for _ in range(int(input()))]
print(*out, sep='\n')
``` | output | 1 | 100,464 | 12 | 200,929 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2. | instruction | 0 | 100,465 | 12 | 200,930 |
Tags: greedy, implementation, math
Correct Solution:
```
import sys
from collections import defaultdict, deque
inp = sys.stdin.readline
read = lambda: list(map(int, inp().split()))
def e():
res = 0
n, m = read()
mat = [None]*n
for i in range(n):
mat[i] = read()
for col in range(m):
b = [0]*n#defaultdict(int)
for row in range(n):
elem = mat[row][col]
place = row*m + col + 1
if elem <= n*m and (place - elem)%m == 0:
shift = ((place - elem)//m if place >= elem else (n*m + place - elem)//m)
b[shift] += 1
res += min(i+n-b[i] for i in range(n))
print(res)
def d():
ans = ""
mex = 0
q, x = read()
arr = [0]*q+[0]
dic = {i: deque([i]) for i in range(x)}
for i in dic:
while dic[i][-1] + x < q+1:
dic[i].append(dic[i][-1] + x)
for _ in range(q):
n = int(inp())
ind = dic[n%x].popleft() if dic[n%x] else n%x
arr[ind] = 1
while arr[mex] == 1:
mex += 1
ans += str(mex) + "\n"
print(ans)
if __name__ == "__main__":
e()
``` | output | 1 | 100,465 | 12 | 200,931 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2. | instruction | 0 | 100,466 | 12 | 200,932 |
Tags: greedy, implementation, math
Correct Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Tue Jan 21 18:49:02 2020
@author: dennis
"""
import atexit
import io
import sys
_INPUT_LINES = sys.stdin.read().splitlines()
input = iter(_INPUT_LINES).__next__
_OUTPUT_BUFFER = io.StringIO()
sys.stdout = _OUTPUT_BUFFER
@atexit.register
def write():
sys.__stdout__.write(_OUTPUT_BUFFER.getvalue())
n, m = [int(x) for x in input().split()] # n = rows, m = columns
a = [[int(x) for x in input().split()] for _ in range(n)]
max_value = n*m
moves = 0
for i in range(m):
column = [row[i] for row in a]
offsets = {x: 0 for x in range(n)}
for j, v in enumerate(column):
if (v%m) == ((i+1)%m) and (v <= max_value):
expected = (j*m)+i+1
offset = ((expected-v)//m)%n
offsets[offset] += 1
saved = 0
for k, v in offsets.items():
saved = max(saved, v-k)
moves += n-saved
print(moves)
``` | output | 1 | 100,466 | 12 | 200,933 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2. | instruction | 0 | 100,467 | 12 | 200,934 |
Tags: greedy, implementation, math
Correct Solution:
```
def solve(column,start,n,m):
shiftCnts=[0 for _ in range(n)]
for actualPosition in range(n):
v=column[actualPosition]
if v>n*m:
continue
desiredPosition=(v-1)//m
desiredValueAtDesiredPosition=start+m*desiredPosition
if desiredValueAtDesiredPosition!=v:
continue
shiftsRequired=(actualPosition-desiredPosition+n)%n
shiftCnts[shiftsRequired]+=1
ans=inf
for shifts in range(n):
ans=min(ans,shifts+n-shiftCnts[shifts])
# print('column:{} shiftCnts:{}'.format(column,shiftCnts))
return ans
def main():
n,m=readIntArr()
arr=[]
for _ in range(n):
arr.append(readIntArr())
ans=0
for j in range(m):
column=[]
for i in range(n):
column.append(arr[i][j])
ans+=solve(column,j+1,n,m)
print(ans)
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(*args):
"""
*args : (default value, dimension 1, dimension 2,...)
Returns : arr[dim1][dim2]... filled with default value
"""
assert len(args) >= 2, "makeArr args should be (default value, dimension 1, dimension 2,..."
if len(args) == 2:
return [args[0] for _ in range(args[1])]
else:
return [makeArr(args[0],*args[2:]) for _ in range(args[1])]
def queryInteractive(x,y):
print('? {} {}'.format(x,y))
sys.stdout.flush()
return int(input())
def answerInteractive(ans):
print('! {}'.format(ans))
sys.stdout.flush()
inf=float('inf')
MOD=10**9+7
for _abc in range(1):
main()
``` | output | 1 | 100,467 | 12 | 200,935 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2. | instruction | 0 | 100,468 | 12 | 200,936 |
Tags: greedy, implementation, math
Correct Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
# M = mod = 998244353
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
# def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
n, m = li()
l = []
ans = 0
for i in range(n):l.append(li())
for j in range(1,m+1):
cnt = Counter()
# visited = {}
for i in range(1,n+1):
# print(l[i][j],end=" ")
tot = 0
ele = l[i-1][j-1]
if ele > n*m:continue
if j == m:
if ele%m:continue
else:
if ele%m != j:continue
pos = math.ceil(ele/m)
# print(ele,pos,i,)
if pos > i:
dist = i + (n - pos)
else:
dist = i - pos
cnt[dist] += 1
temp = n
if len(cnt):
for i in cnt:
temp = min(temp,i + (n - cnt[i]))
# print(cnt)
ans += temp
print(ans)
``` | output | 1 | 100,468 | 12 | 200,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2.
Submitted Solution:
```
def min_op_ab(A, a, b):
#print(A, a, b)
counts = {v: 0 for v in range(len(A))}
vmin, vmax = b, (len(A) - 1) * a + b
for i, p in enumerate(A):
if vmin <= p <= vmax and p % a == b % a:
corr_pos = (p - b) // a
v = (i - corr_pos) % len(A)
counts[v] += 1
#print(counts.items())
return min(len(A) - cv + v for v, cv in counts.items())
def mat_col(M, c):
return [M[r][c] for r in range(len(M))]
import sys
n, m = (int(v) for v in sys.stdin.readline().split())
M = []
for r in range(n):
M.append([int(v) for v in sys.stdin.readline().split()])
ops = 0
for c in range(m):
col = mat_col(M, c)
a, b = m, c + 1
#print(min_op_ab(col, a, b))
ops += min_op_ab(col, a, b)
print(ops)
``` | instruction | 0 | 100,469 | 12 | 200,938 |
Yes | output | 1 | 100,469 | 12 | 200,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2.
Submitted Solution:
```
n,m = map(int, input().split())
mtx = []
for _ in range(n):
mtx.append(list(map(int,input().split())))
total = 0
for i in range(m):
cnt = {}
for j in range(n):
if (mtx[j][i]-i-1)%m == 0:
pos = (mtx[j][i]-1)//m
if pos < n:
c = (j-pos)%n
cnt[c] = cnt.get(c,0)+1
add = n
for k,v in cnt.items():
add = min(add, k+n-v)
total+=add
print(total)
``` | instruction | 0 | 100,470 | 12 | 200,940 |
Yes | output | 1 | 100,470 | 12 | 200,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2.
Submitted Solution:
```
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect
#?############################################################
def isPrime(x):
for i in range(2, x):
if i*i > x:
break
if (x % i == 0):
return False
return True
#?############################################################
def ncr(n, r, p):
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 primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n // 2
for i in range(3, int(math.sqrt(n))+1, 2):
while n % i == 0:
l.append(i)
n = n // i
if n > 2:
l.append(n)
return l
#?############################################################
def power(x, y, p):
res = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
#?############################################################
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#?############################################################
def digits(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
#?############################################################
def ceil(n, x):
if (n % x == 0):
return n//x
return n//x+1
#?############################################################
def mapin():
return map(int, input().split())
#?############################################################
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>op
def solve(l, n, m, c, d):
s = [0]*n
for i in range(n):
if(l[i]%m == c):
a = l[i]//m
a-=d
if(a>= n):
continue
if(a-i>=0):
a = (i+n-a)%n
else:
a = (i-a)%n
# print(i, l[i], n, m, a)
if(a<n):
s[a]+=1
ans = 1e9
for i in range(n):
temp = n
temp-=s[i]
temp+=i
# print(i, temp)
ans =min(ans, temp)
# print(s, ans)
return ans
n, m = mapin()
sa = []
for i in range(m):
sa.append([0]*n)
for i in range(n):
tt = [int(x) for x in input().split()]
for j in range(m):
sa[j][i] = tt[j]
ans = 0
for j in range(m):
a =solve(sa[j], n, m, (j+1)%m, (j+1)//m)
ans+=a
# print(j, a)
# a = a =solve(sa[-1], n, m, m)
print(ans)
``` | instruction | 0 | 100,471 | 12 | 200,942 |
Yes | output | 1 | 100,471 | 12 | 200,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n,m=map(int,input().split())
CR=[list(map(int,input().split())) for i in range(n)]
for i in range(n):
for j in range(m):
CR[i][j]-=1
MAX=n*m
ANS=0
for i in range(m):
SCORE=[0]*n
for j in range(n):
if CR[j][i]%m==i%m and 0<=CR[j][i]<MAX:
x=(CR[j][i]-i%m)//m
SCORE[(j-x)%n]+=1
#print(SCORE)
#print(SCORE)
ANS+=min([n+j-SCORE[j] for j in range(n)])
print(ANS)
``` | instruction | 0 | 100,472 | 12 | 200,944 |
Yes | output | 1 | 100,472 | 12 | 200,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# 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)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 998244353
INF = float('inf')
from math import factorial
from collections import Counter, defaultdict, deque
from heapq import heapify, heappop, heappush
# ------------------------------
# f = open('./input.txt')
# sys.stdin = f
def main():
n, m = RL()
arr = [RLL() for _ in range(n)]
res = 0
for i in range(m):
# tar = [i+1+m*j for j in range(n)]
fir = i+1
rec = defaultdict(int)
for j in range(n):
now = arr[j][i]
if (now-fir)%m!=0: continue
p = (now-fir)//m
if j-p>=0:
rec[j-p]+=1
else:
rec[n-(p-j)]+=1
tres = INF
for k, v in rec.items():
tres = min(tres, (n-v)+k)
res+=tres
print(res)
if __name__ == "__main__":
main()
``` | instruction | 0 | 100,473 | 12 | 200,946 |
No | output | 1 | 100,473 | 12 | 200,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2.
Submitted Solution:
```
n,m=map(int,input().split())
tbl=[]
for _ in range(n):
tbl.append(list(map(int,input().split())))
l=[[k+n for k in range(n)] for _ in range(m)]
for i in range(m):
for j in range(n):
a=(tbl[j][i]-1)//m
b=(tbl[j][i]-1)%m
if b==i:
l[i][(j-a)%n]-=1
ans=0
for i in range(m):
ans+=min(l[i])
print(ans)
``` | instruction | 0 | 100,474 | 12 | 200,948 |
No | output | 1 | 100,474 | 12 | 200,949 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2.
Submitted Solution:
```
from sys import stdin
def solve():
n,m=map(int,stdin.readline().split())
l=[list(map(lambda x:int(x)-1,stdin.readline().split())) for i in range(n)]
mvs=0
for j in range(m):
d={}
ans=n
for i in range(n):
if l[i][j]%m==j:
idx=(l[i][j]-j)//m
if i-idx>=0:
x=i-idx
else:
x=n+i-idx
d.setdefault(x,0)
d[x]+=1
for k in d:
ans=min(ans,(k+n-d[k]))
#print(d)
#print(ans)
mvs+=ans
print(mvs)
solve()
``` | instruction | 0 | 100,475 | 12 | 200,950 |
No | output | 1 | 100,475 | 12 | 200,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rectangular matrix of size n × m consisting of integers from 1 to 2 ⋅ 10^5.
In one move, you can:
* choose any element of the matrix and change its value to any integer between 1 and n ⋅ m, inclusive;
* take any column and shift it one cell up cyclically (see the example of such cyclic shift below).
A cyclic shift is an operation such that you choose some j (1 ≤ j ≤ m) and set a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, ..., a_{n, j} := a_{1, j} simultaneously.
<image> Example of cyclic shift of the first column
You want to perform the minimum number of moves to make this matrix look like this:
<image>
In other words, the goal is to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (i.e. a_{i, j} = (i - 1) ⋅ m + j) with the minimum number of moves performed.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5, n ⋅ m ≤ 2 ⋅ 10^5) — the size of the matrix.
The next n lines contain m integers each. The number at the line i and position j is a_{i, j} (1 ≤ a_{i, j} ≤ 2 ⋅ 10^5).
Output
Print one integer — the minimum number of moves required to obtain the matrix, where a_{1, 1} = 1, a_{1, 2} = 2, ..., a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, ..., a_{n, m} = n ⋅ m (a_{i, j} = (i - 1)m + j).
Examples
Input
3 3
3 2 1
1 2 3
4 5 6
Output
6
Input
4 3
1 2 3
4 5 6
7 8 9
10 11 12
Output
0
Input
3 4
1 6 3 4
5 10 7 8
9 2 11 12
Output
2
Note
In the first example, you can set a_{1, 1} := 7, a_{1, 2} := 8 and a_{1, 3} := 9 then shift the first, the second and the third columns cyclically, so the answer is 6. It can be shown that you cannot achieve a better answer.
In the second example, the matrix is already good so the answer is 0.
In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is 2.
Submitted Solution:
```
from collections import defaultdict, deque, Counter
from sys import stdin, stdout
from heapq import heappush, heappop
import math
import io
import os
import math
import bisect
#?############################################################
def isPrime(x):
for i in range(2, x):
if i*i > x:
break
if (x % i == 0):
return False
return True
#?############################################################
def ncr(n, r, p):
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 primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n // 2
for i in range(3, int(math.sqrt(n))+1, 2):
while n % i == 0:
l.append(i)
n = n // i
if n > 2:
l.append(n)
return l
#?############################################################
def power(x, y, p):
res = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
#?############################################################
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#?############################################################
def digits(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
#?############################################################
def ceil(n, x):
if (n % x == 0):
return n//x
return n//x+1
#?############################################################
def mapin():
return map(int, input().split())
#?############################################################
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# python3 15.py<in>op
def solve(l, n, m, c, d):
s = [0]*n
for i in range(n):
if(l[i]%m == c):
a = l[i]//m
a-=d
if(a-i>=0):
a = a-i
else:
a = n-1-(i-a)
# print(i, l[i], n, m, a)
s[a]+=1
ans = 1e9
# print(s)
for i in range(n):
temp = n
temp-=s[i]
temp+=i
# print(i, temp)
ans =min(ans, temp)
return ans
n, m = mapin()
sa = []
for i in range(m):
sa.append([0]*n)
for i in range(n):
tt = [int(x) for x in input().split()]
for j in range(m):
sa[j][i] = tt[j]
ans = 0
for j in range(m):
a =solve(sa[j], n, m, (j+1)%m, (j+1)//m)
ans+=a
# print(j, a)
# a = a =solve(sa[-1], n, m, m)
print(ans)
``` | instruction | 0 | 100,476 | 12 | 200,952 |
No | output | 1 | 100,476 | 12 | 200,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n.
Input
The only line contains two integers n and k (1 ≤ n, k ≤ 5 ⋅ 10^5).
Output
Print one integer — the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000 | instruction | 0 | 100,496 | 12 | 200,992 |
Tags: combinatorics, math, number theory
Correct Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop
import math
from collections import *
from functools import reduce,cmp_to_key,lru_cache
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# import sys
# input = sys.stdin.readline
M = mod = 998244353
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip().split()]
def st():return str(input().rstrip())[2:-1]
def val():return int(input().rstrip())
def li2():return [str(i)[2:-1] for i in input().rstrip().split()]
def li3():return [int(i) for i in st()]
N = 5 * 10**5 + 1
# array to store inverse of 1 to N
factorialNumInverse = [None] * (N + 1)
# array to precompute inverse of 1! to N!
naturalNumInverse = [None] * (N + 1)
# array to store factorial of
# first N numbers
fact = [None] * (N + 1)
# Function to precompute inverse of numbers
def InverseofNumber(p):
naturalNumInverse[0] = naturalNumInverse[1] = 1
for i in range(2, N + 1, 1):
naturalNumInverse[i] = (naturalNumInverse[p % i] *
(p - int(p / i)) % p)
# Function to precompute inverse
# of factorials
def InverseofFactorial(p):
factorialNumInverse[0] = factorialNumInverse[1] = 1
# precompute inverse of natural numbers
for i in range(2, N + 1, 1):
factorialNumInverse[i] = (naturalNumInverse[i] *
factorialNumInverse[i - 1]) % p
# Function to calculate factorial of 1 to N
def factorial(p):
fact[0] = 1
# precompute factorials
for i in range(1, N + 1):
fact[i] = (fact[i - 1] * i) % p
# Function to return nCr % p in O(1) time
def Binomial(N, R, p):
# n C r = n!*inverse(r!)*inverse((n-r)!)
ans = ((fact[N] * factorialNumInverse[R])% p *
factorialNumInverse[N - R])% p
return ans
p = mod
InverseofNumber(p)
InverseofFactorial(p)
factorial(p)
n,k = li()
ans = Binomial(n-1,k-1,mod) if k <= n else 0
for i in range(2,n + 1):
totnumbers = n//i
if totnumbers >= k:
ans = (ans + Binomial(totnumbers - 1,k - 1,mod))%mod
else:break
print(ans)
``` | output | 1 | 100,496 | 12 | 200,993 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n.
Input
The only line contains two integers n and k (1 ≤ n, k ≤ 5 ⋅ 10^5).
Output
Print one integer — the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# 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)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa=ifa[::-1]
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
'''
class SMT:
def __init__(self,arr):
self.n=len(arr)-1
self.arr=[0]*(self.n<<2)
self.lazy=[0]*(self.n<<2)
def Build(l,r,rt):
if l==r:
self.arr[rt]=arr[l]
return
m=(l+r)>>1
Build(l,m,rt<<1)
Build(m+1,r,rt<<1|1)
self.pushup(rt)
Build(1,self.n,1)
def pushup(self,rt):
self.arr[rt]=self.arr[rt<<1]+self.arr[rt<<1|1]
def pushdown(self,rt,ln,rn):#lr,rn表区间数字数
if self.lazy[rt]:
self.lazy[rt<<1]+=self.lazy[rt]
self.lazy[rt<<1|1]+=self.lazy[rt]
self.arr[rt<<1]+=self.lazy[rt]*ln
self.arr[rt<<1|1]+=self.lazy[rt]*rn
self.lazy[rt]=0
def update(self,L,R,c,l=1,r=None,rt=1):#L,R表示操作区间
if r==None: r=self.n
if L<=l and r<=R:
self.arr[rt]+=c*(r-l+1)
self.lazy[rt]+=c
return
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
if L<=m: self.update(L,R,c,l,m,rt<<1)
if R>m: self.update(L,R,c,m+1,r,rt<<1|1)
self.pushup(rt)
def query(self,L,R,l=1,r=None,rt=1):
if r==None: r=self.n
#print(L,R,l,r,rt)
if L<=l and R>=r:
return self.arr[rt]
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
ans=0
if L<=m: ans+=self.query(L,R,l,m,rt<<1)
if R>m: ans+=self.query(L,R,m+1,r,rt<<1|1)
return ans
'''
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]<self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
#print(flag)
return flag
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
def nb(i,j):
for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
if 0<=ni<n and 0<=nj<m:
yield ni,nj
@bootstrap
def gdfs(r,p):
if len(g[r])==1 and p!=-1:
yield None
for ch in g[r]:
if ch!=p:
yield gdfs(ch,r)
yield None
t=1
for i in range(t):
n,k=RL()
mod=998244353
ans=0
fact(n,mod)
ifact(n,mod)
for i in range(1,n+1):
c=n//i
if c<k:
break
ans+=com(c-1,k-1,mod)
ans%=mod
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
``` | instruction | 0 | 100,503 | 12 | 201,006 |
Yes | output | 1 | 100,503 | 12 | 201,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We define x mod y as the remainder of division of x by y (\% operator in C++ or Java, mod operator in Pascal).
Let's call an array of positive integers [a_1, a_2, ..., a_k] stable if for every permutation p of integers from 1 to k, and for every non-negative integer x, the following condition is met:
(((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k = (((x mod a_{p_1}) mod a_{p_2}) ... mod a_{p_{k - 1}}) mod a_{p_k}
That is, for each non-negative integer x, the value of (((x mod a_1) mod a_2) ... mod a_{k - 1}) mod a_k does not change if we reorder the elements of the array a.
For two given integers n and k, calculate the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n.
Input
The only line contains two integers n and k (1 ≤ n, k ≤ 5 ⋅ 10^5).
Output
Print one integer — the number of stable arrays [a_1, a_2, ..., a_k] such that 1 ≤ a_1 < a_2 < ... < a_k ≤ n. Since the answer may be large, print it modulo 998244353.
Examples
Input
7 3
Output
16
Input
3 7
Output
0
Input
1337 42
Output
95147305
Input
1 1
Output
1
Input
500000 1
Output
500000
Submitted Solution:
```
#Sorry Pacy
print(16)
``` | instruction | 0 | 100,509 | 12 | 201,018 |
No | output | 1 | 100,509 | 12 | 201,019 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 ≤ t ≤ 15) — the number of test cases you need to solve.
The first line of each test case contains an integers n (2 ≤ n ≤ 2000) — the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i < 2^{30}) — the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 ⊕ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal. | instruction | 0 | 100,558 | 12 | 201,116 |
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
import math, sys
from collections import defaultdict, Counter, deque
from heapq import heapify, heappush, heappop
MOD = int(1e9) + 7
def main():
n = int(input())
arr = list(map(int, input().split()))
pre = [0 for i in range(n)]
suf = [0 for i in range(n)]
x = 0
for i in range(n):
x ^= arr[i]
pre[i] = x
if x == 0:
print("YES")
return
s = 0
for i in range(n - 1, -1, -1):
s ^= arr[i]
suf[i] = s
for i in range(n - 2):
p = pre[i]
m = arr[i + 1]
for j in range(i + 2, n):
s = suf[j]
# print(p, s, m)
if p == m == s:
print("YES")
return
m ^= arr[j]
print("NO")
t = 1
t = int(input())
for _ in range(t):
main()
``` | output | 1 | 100,558 | 12 | 201,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 ≤ t ≤ 15) — the number of test cases you need to solve.
The first line of each test case contains an integers n (2 ≤ n ≤ 2000) — the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i < 2^{30}) — the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 ⊕ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal. | instruction | 0 | 100,559 | 12 | 201,118 |
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from bisect import bisect_right
from math import gcd,log
from collections import Counter,defaultdict,deque
from pprint import pprint
from itertools import permutations
from bisect import bisect_right
from random import randint as rti
# import deque
MOD=10**9+7
def main():
n=int(input())
arr=list(map(int,input().split()))
total=0
for i in arr:
total^=i
til=0
for i in range(n-1):
til^=arr[i]
if til == total^til:
print('YES')
return
pf=0
for i in range(n):
pf^=arr[i]
till=0
for j in range(i+1,n-1):
till^=arr[j]
if pf==till==total^till^pf:
print('YES')
return
print('NO')
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")
# endregion
if __name__ == "__main__":
for _ in range(int(input())):
main()
``` | output | 1 | 100,559 | 12 | 201,119 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 ≤ t ≤ 15) — the number of test cases you need to solve.
The first line of each test case contains an integers n (2 ≤ n ≤ 2000) — the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i < 2^{30}) — the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 ⊕ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal. | instruction | 0 | 100,560 | 12 | 201,120 |
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
from collections import Counter
t = int(input())
for _ in range(t):
n = int(input())
lst = list(map(int,input().split()))
pre = []
pos = [-1]
x = 0
for i in lst:
x = x^i
pre.append(x)
x = 0
for i in lst[::-1]:
x = x^i
pos.append(x)
pos = pos[::-1]
pos = pos[1:]
# print(pre)
flag = True
for i in range(n):
if pre[i] == pos[i]:
print("YES")
flag = False
break
elif pre[i] == 0:
z = pos[i]
if z in pre[:i+1]:
print("YES")
flag = False
break
elif pos[i] == 0:
z = pre[i]
if z in pos[i:]:
print("YES")
flag = False
break
if flag:
print("NO")
``` | output | 1 | 100,560 | 12 | 201,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 ≤ t ≤ 15) — the number of test cases you need to solve.
The first line of each test case contains an integers n (2 ≤ n ≤ 2000) — the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i < 2^{30}) — the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 ⊕ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal. | instruction | 0 | 100,561 | 12 | 201,122 |
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
import math as mt
# from collections import defaultdict
# from collections import Counter, deque
# from itertools import permutations
# from functools import reduce
# from heapq import heapify, heappop, heappush, heapreplace
def getInput(): return sys.stdin.readline().strip()
def getInt(): return int(getInput())
def getInts(): return map(int, getInput().split())
def getArray(): return list(getInts())
# sys.setrecursionlimit(10**7)
# INF = float('inf')
# MOD1, MOD2 = 10**9+7, 998244353
# def def_value():
# return 0
# Defining the dict
def solve(li,n):
xor = 0
for i in li:
xor ^= i
if(xor == 0 ):
print("YES")
return
pre = li[0]
for i in range(1,n):
mid = li[i]
for j in range(i+1,n):
if( pre == mid and mid == xor):
print("YES")
return
mid ^= li[j]
pre ^= li[i]
print("NO")
for _ in range(int(input())):
n = int(input())
li = list(map(int,input().split()))
solve(li,n)
``` | output | 1 | 100,561 | 12 | 201,123 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Baby Ehab is known for his love for a certain operation. He has an array a of length n, and he decided to keep doing the following operation on it:
* he picks 2 adjacent elements; he then removes them and places a single integer in their place: their [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). Note that the length of the array decreases by one.
Now he asks you if he can make all elements of the array equal. Since babies like to make your life harder, he requires that you leave at least 2 elements remaining.
Input
The first line contains an integer t (1 ≤ t ≤ 15) — the number of test cases you need to solve.
The first line of each test case contains an integers n (2 ≤ n ≤ 2000) — the number of elements in the array a.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i < 2^{30}) — the elements of the array a.
Output
If Baby Ehab can make all elements equal while leaving at least 2 elements standing, print "YES". Otherwise, print "NO".
Example
Input
2
3
0 2 2
4
2 3 1 10
Output
YES
NO
Note
In the first sample, he can remove the first 2 elements, 0 and 2, and replace them by 0 ⊕ 2=2. The array will be [2,2], so all the elements are equal.
In the second sample, there's no way to make all the elements equal. | instruction | 0 | 100,562 | 12 | 201,124 |
Tags: bitmasks, brute force, dp, greedy
Correct Solution:
```
from itertools import accumulate
tests = int(input())
for t in range(tests):
n = int(input())
arr = map(int, input().split(' '))
pref = list(accumulate(arr, lambda a, b: a ^ b))
if pref[-1] == 0:
# can split into 2
print("YES")
else:
if pref[-1] in pref[:n-1] and 0 in pref[pref[:n-1].index(pref[-1]):]:
print("YES")
else:
print("NO")
``` | output | 1 | 100,562 | 12 | 201,125 |
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