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 a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4 | instruction | 0 | 48,857 | 12 | 97,714 |
Tags: combinatorics, sortings
Correct Solution:
```
n = int(input())
arr = list(map(int,input().split()))
first = {}
last = {}
for i,a in enumerate(arr):
first[a] = min(i,first.get(a,10**6))
last[a] = max(i,last.get(a,0))
intervals = [(i,i) for i in range(n)]
for a in first:
intervals.append((first[a],last[a]))
intervals.sort()
num = 0
prev = -1
for i in range(len(intervals)):
if prev<intervals[i][0]:
num+=1
prev = intervals[i][1]
else:
prev = max(prev,intervals[i][1])
print(pow(2,num-1,998244353))
``` | output | 1 | 48,857 | 12 | 97,715 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4 | instruction | 0 | 48,858 | 12 | 97,716 |
Tags: combinatorics, sortings
Correct Solution:
```
MOD = 998244353
def pwr(x,y):
if y==0:
return 1
p = pwr(x,y//2)%MOD
p = (p*p) % MOD
if y%2 == 0:
return p
else:
return ((x*p)%MOD)
n=int(input())
a=list(map(int,input().split()))
c = 0
f=[0]*n
d={}
for i in range(n):
d[a[i]]=i
for i in range(n):
if i==0:
c=0
elif f[i]==0:
c+=1
if f[i]==0:
j=i+1
k = d[a[i]]
while j<=k:
f[j]=1
k=max(k,d[a[j]])
j+=1
ans = pwr(2,c)
print(ans)
``` | output | 1 | 48,858 | 12 | 97,717 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4 | instruction | 0 | 48,859 | 12 | 97,718 |
Tags: combinatorics, sortings
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
l=[0]*200000
dic={}
seq=[]
for i in range(n):
if a[i] not in dic:
dic[a[i]]=[i,i]
else:
dic[a[i]][1]=i
t=1
key=dic[a[0]][1]
for seq in dic:
if dic[seq][0]>key:
t*=2
t%=998244353
key=dic[seq][1]
elif dic[seq][1]>key:
key=dic[seq][1]
print(t)
``` | output | 1 | 48,859 | 12 | 97,719 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4 | instruction | 0 | 48,860 | 12 | 97,720 |
Tags: combinatorics, sortings
Correct Solution:
```
'''
Auther: ghoshashis545 Ashis Ghosh
College: jalpaiguri Govt Enggineering College
'''
from os import path
from io import BytesIO, IOBase
import sys
from heapq import heappush,heappop
from functools import cmp_to_key as ctk
from collections import deque,Counter,defaultdict as dd
from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
from itertools import permutations
from datetime import datetime
from math import ceil,sqrt,log,gcd
def ii():return int(input())
def si():return input().rstrip()
def mi():return map(int,input().split())
def li():return list(mi())
abc='abcdefghijklmnopqrstuvwxyz'
# mod=1000000007
mod=998244353
inf = float("inf")
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def bo(i):
return ord(i)-ord('0')
file = 1
def ceil(a,b):
return (a+b-1)//b
# write fastio for getting fastio template.
def solve():
# for _ in range(ii()):
n = ii()
a = li()
m = {}
for i in range(n):
m[a[i]] = i
r = 0
k = -1
for i in range(n):
r = max(r,m[a[i]])
if i==r:
k+=1
print(pow(2,k,mod))
if __name__ =="__main__":
if(file):
if path.exists('input.txt'):
sys.stdin=open('input.txt', 'r')
sys.stdout=open('output.txt','w')
else:
input=sys.stdin.readline
solve()
``` | output | 1 | 48,860 | 12 | 97,721 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4 | instruction | 0 | 48,861 | 12 | 97,722 |
Tags: combinatorics, sortings
Correct Solution:
```
from collections import defaultdict as dc
n=int(input())
a=list(map(int,input().split()))
cn=1
m=998244353
hsh=dc(int)
hsh2=dc(int)
for i in a:
hsh[i]+=1
sm=0
for i in range(n-1):
if(hsh2[a[i]]):
sm-=1
else:
sm+=hsh[a[i]]-1
hsh2[a[i]]+=1
if(sm==0):
cn*=2
cn%=m
print(cn)
``` | output | 1 | 48,861 | 12 | 97,723 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4 | instruction | 0 | 48,862 | 12 | 97,724 |
Tags: combinatorics, sortings
Correct Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
lasts = {}
for i in range(n):
lasts[a[i]] = i
dss = {x: x for x in a}
changed = True
while changed:
changed = False
for i in range(1, n):
if dss[a[i-1]] != dss[a[i]] or lasts[a[i-1]] != lasts[a[i]]:
if lasts[a[i-1]] >= i:
changed = True
lasts[a[i-1]] = lasts[a[i]] = max(lasts[a[i-1]], lasts[a[i]])
dss[a[i-1]] = dss[a[i]] = min(dss[a[i-1]], dss[a[i]])
M = 998244353
def powmod(x, p, m = M):
if x == 0:
return 0
if p == 0:
return 1
return (powmod(x*x, p//2)%m) * (x if p%2 == 1 else 1) % m
kinds = len(set(dss.values()))
ans = powmod(2, kinds-1)
#print(n, a, dss, lasts, kinds)
#print([powmod(2, i, 13) for i in range(10)])
print(ans)
``` | output | 1 | 48,862 | 12 | 97,725 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4 | instruction | 0 | 48,863 | 12 | 97,726 |
Tags: combinatorics, sortings
Correct Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
last = {}
for i in range(n-1, -1, -1):
if not (a[i] in last.keys()):
last[a[i]] = i
m, end = 0, -1
for i in range(0, n):
end = max(end, last[a[i]])
if end == i:
m += 1
print((2**(m-1))%998244353)
``` | output | 1 | 48,863 | 12 | 97,727 |
Provide tags and a correct Python 2 solution for this coding contest problem.
You are given an array a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4 | instruction | 0 | 48,864 | 12 | 97,728 |
Tags: combinatorics, sortings
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
def inv(x,mod):
return pow(x,mod-2,mod)
range = xrange # not for python 3.0+
# main code
mod=998244353
n=in_num()
l=in_arr()
mn=Counter()
mx=Counter()
for i in range(1,n+1):
if not mn[l[i-1]]:
mn[l[i-1]]=i
mx[l[i-1]]=i
arr=[]
for i in mn:
if mn[i]!=mx[i]:
arr.append((mn[i],mx[i]))
arr.sort()
ans=n#pow(2,n-1,mod)
n=len(arr)
i=0
while i<n:
j=i+1
st=arr[i][0]
en=arr[i][1]
while j<n:
if en>=arr[j][0]:
en=max(en,arr[j][1])
j+=1
else:
break
ln=en-st
ans-=ln
#if st==1:
# ln-=1
#ans=(ans*inv(pow(2,ln,mod),mod))%mod
i=j
pr_num(pow(2,ans-1,mod))
``` | output | 1 | 48,864 | 12 | 97,729 |
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 a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4
Submitted Solution:
```
from os import path
import sys
import time
mod = int(1e9 + 7)
# import re
from math import ceil, floor, gcd, log, log2, factorial, sqrt
from collections import defaultdict, Counter, OrderedDict, deque
from itertools import combinations, accumulate,groupby,product
# from string import ascii_lowercase ,ascii_uppercase
from bisect import *
from functools import reduce
from operator import mul
def star(x): return print(' '.join(map(str, x)))
def grid(r): return [lint() for i in range(r)]
def stpr(x): return sys.stdout.write(f'{x}' + '\n')
INF = float('inf')
if (path.exists('input.txt')):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
import sys
from sys import stdin, stdout
from collections import *
from math import gcd, floor, ceil
from copy import deepcopy
def st(): return list(stdin.readline().strip())
def inp(): return int(stdin.readline())
def inlt(): return list(map(int, stdin.readline().split()))
def invr():return map(int, stdin.readline().split())
def pr(n): return stdout.write(str(n) + "\n")
def solve():
n = inp()
a = inlt()
d = {}
for i in range(n):
d[a[i]] = i
block_ind = d[a[0]]
ans = 1
for i in range(1,n):
if i>block_ind:
ans = ans*2%998244353
block_ind = max(block_ind,d[a[i]])
print(ans)
t = 1
#t = inp()
for _ in range(t):
solve()
``` | instruction | 0 | 48,865 | 12 | 97,730 |
Yes | output | 1 | 48,865 | 12 | 97,731 |
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 a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4
Submitted Solution:
```
from sys import stdin, stdout
from math import *
from heapq import *
from collections import *
def main():
n=int(stdin.readline())
a=[int(x) for x in stdin.readline().split()]
b=[2]*(n+2)
maxR={}
minL={}
for i in range(n):
x=a[i]
maxR[x]=i
minL[x]=minL.get(x,i)
f=[0]*(n+2)
for i in range(n):
x=a[i]
l=minL[x]
r=maxR[x]
if (l<r):
if (r==l-1):
b[r]=1
else:
f[l+1]=f[l+1]+1
f[r+1]=f[r+1]-1
res=1
for i in range(1,n):
f[i]=f[i]+f[i-1]
if (f[i]>0):
b[i]=1
res=(res*b[i])%998244353
stdout.write(str(res))
return 0
if __name__ == "__main__":
main()
``` | instruction | 0 | 48,866 | 12 | 97,732 |
Yes | output | 1 | 48,866 | 12 | 97,733 |
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 a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4
Submitted Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
A=list(map(int,input().split()))
mod=998244353
INIT=dict()
LAST=dict()
for i in range(n):
if INIT.get(A[i])==None:
INIT[A[i]]=i
LAST[A[i]]=i
ANS=0
NOW=0
i=0
while i<n:
NOW=LAST[A[i]]
while i<n and i<=NOW:
if LAST[A[i]]>NOW:
NOW=LAST[A[i]]
i+=1
if i==n:
break
else:
ANS+=1
print(pow(2,ANS,mod))
``` | instruction | 0 | 48,867 | 12 | 97,734 |
Yes | output | 1 | 48,867 | 12 | 97,735 |
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 a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4
Submitted Solution:
```
import sys
n= int(input())
a=list(map(int,input().split()))
d={}
for i in range (0,n):
d[a[i]]=i
i=0
res = 1
while i<=n-1:
cur = d[a[i]]
j=i+1
while j<=cur:
cur = max(cur, d[a[j]])
j+=1
if i!=0:
res*=2
i=cur+1
if i==n:
print(res % 998244353)
sys.exit()
print ( res % 998244353)
``` | instruction | 0 | 48,868 | 12 | 97,736 |
Yes | output | 1 | 48,868 | 12 | 97,737 |
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 a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4
Submitted Solution:
```
k, l, end = 1, -1, -1
n = int(input())
a = list(map(int, input().split()))
d = list(reversed(a))
while l < n - 1:
l += 1
if a[l] not in a[l+1:] and l > end:
k *= 2
if k > 998244353:
k = k - 998244353
else:
if n - d.index(a[l]) - 1 > end:
end = n - d.index(a[l]) - 1
print(k)
``` | instruction | 0 | 48,869 | 12 | 97,738 |
No | output | 1 | 48,869 | 12 | 97,739 |
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 a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4
Submitted Solution:
```
from bisect import bisect_left
import sys
n= int(input())
a=list(map(int,input().split()))
a1=set(a)
#if n == 200000 and a[0]==167961370:
# print (98673221)
# sys.exit()
if len(a1)==n:
print(2**(n-1) % 998244353)
sys.exit()
k=0
p=0
for i in range(0,n-1):
if a[i]>a[i+1]:
k+=1
if a[i]<a[i+1]:
p+=1
if k==0:
print(2**(len(a1)-1) % 998244353)
sys.exit()
if p==0:
print(2**(len(a1)-1) % 998244353)
sys.exit()
b=[]
for i in range (0,n):
b+=[a[i]]
b.sort()
i=0
bi=bisect_left(b,(a[i]))
k=0
t=0
#print(bi)
while t==0:
if bi!=n-1 and b[bi]!=b[bi+1]:
k+=1
i+=1
elif bi==n-1:
k+=1
i+=1
else:
t=1
#print (a[i],k)
bi=bisect_left(b,(a[i]))
res=2**(k) % 998244353
k=0
i=n-1
t=0
bi=bisect_left(b,(a[i]))
while t==0:
if bi!=n-1 and b[bi]!=b[bi+1]:
k+=1
i-=1
elif bi==n-1:
k+=1
i-=1
else:
t=1
#print (a[i],k)
bi=bisect_left(b,(a[i]))
print((2**(k) % 998244353)*res% 998244353)
``` | instruction | 0 | 48,870 | 12 | 97,740 |
No | output | 1 | 48,870 | 12 | 97,741 |
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 a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4
Submitted Solution:
```
s, l, k, r, rend, sp, lk, f = '', -1, 0, {}, {}, [], 0, 1
d1 = d2 = 0
n = int(input())
a = tuple(map(int, input().split()))
for key, val in enumerate(a):
if r.get(val) is None:
r[val] = key
else:
rend[val] = key
for key, value in rend.items():
sp.append([r.get(key), value])
sp.sort(key=lambda x: x[0])
if sp:
d1, d2 = sp[0][0], sp[0][1]
for i in sp[1:]:
if i[0] < d2 < i[1]:
d2 = i[1]
elif i[0] > d2:
k += (d2 - d1 + 1)
lk += 1
d1 = i[0]
d2 = i[1]
k += (d2 - d1 + 1)
for i in range(n - k + lk):
f *= 2
if f > 998244353:
f = -998244353
print(f)
``` | instruction | 0 | 48,871 | 12 | 97,742 |
No | output | 1 | 48,871 | 12 | 97,743 |
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 a consisting of n integers. Let's denote monotonic renumeration of array a as an array b consisting of n integers such that all of the following conditions are met:
* b_1 = 0;
* for every pair of indices i and j such that 1 ≤ i, j ≤ n, if a_i = a_j, then b_i = b_j (note that if a_i ≠ a_j, it is still possible that b_i = b_j);
* for every index i ∈ [1, n - 1] either b_i = b_{i + 1} or b_i + 1 = b_{i + 1}.
For example, if a = [1, 2, 1, 2, 3], then two possible monotonic renumerations of a are b = [0, 0, 0, 0, 0] and b = [0, 0, 0, 0, 1].
Your task is to calculate the number of different monotonic renumerations of a. The answer may be large, so print it modulo 998244353.
Input
The first line contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
Output
Print one integer — the number of different monotonic renumerations of a, taken modulo 998244353.
Examples
Input
5
1 2 1 2 3
Output
2
Input
2
100 1
Output
2
Input
4
1 3 3 7
Output
4
Submitted Solution:
```
import os,sys,math
from io import BytesIO, IOBase
from collections import defaultdict,deque,OrderedDict
import bisect as bi
def yes():print('YES')
def no():print('NO')
def I():return (int(input()))
def In():return(map(int,input().split()))
def ln():return list(map(int,input().split()))
def Sn():return input().strip()
BUFSIZE = 8192
#complete the main function with number of test cases to complete greater than x
def find_gt(a, x):
i = bi.bisect_left(a, x)
if i != len(a):
return i
else:
return len(a)
def solve():
n=I()
l=list(In())
pos={}
for i in range(n-1,-1,-1):
if pos.get(l[i],-1)==-1:
pos[l[i]]=i
d={l[0]:1}
mx=pos[l[0]]
cnt=1
for i in range(n):
if d.get(l[i],-1)!=-1:
d[l[i]]=min(cnt,d[l[i]])
cnt=d[l[i]]
else:
if mx<i:
cnt*=2
cnt%=M
d[l[i]]=cnt
mx=max(mx,pos[l[i]])
else:
d[l[i]]=cnt
print(cnt%M)
pass
def main():
T=1
for i in range(T):
solve()
M = 998244353
P = 1000000007
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == '__main__':
main()
``` | instruction | 0 | 48,872 | 12 | 97,744 |
No | output | 1 | 48,872 | 12 | 97,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES | instruction | 0 | 49,026 | 12 | 98,052 |
Tags: constructive algorithms, dp, greedy
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
arr = list(map(int, input().split()))
if arr[-1] - sum(max(arr[i+1]-arr[i], 0) for i in range(n-1)) >= 0:
print("YES")
else:
print("NO")
``` | output | 1 | 49,026 | 12 | 98,053 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES | instruction | 0 | 49,027 | 12 | 98,054 |
Tags: constructive algorithms, dp, greedy
Correct Solution:
```
from sys import stdin
input=stdin.readline
t=int(input())
for _ in range(t):
n=int(input())
a=list(map(int,input().split()))
b=a.copy()
x=0
for i in range(n-2,-1,-1):
if a[i] > b[i+1]:
x += a[i] - b[i+1]
a[i] -= x
if a[0] >= 0:
print("YES")
else:
print("NO")
``` | output | 1 | 49,027 | 12 | 98,055 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES | instruction | 0 | 49,028 | 12 | 98,056 |
Tags: constructive algorithms, dp, greedy
Correct Solution:
```
import sys
*data, = map(int, sys.stdin.read().split()[::-1])
def inp():
return data.pop()
def solve(arr):
# left = [0] * len(arr)
# right = [0] * len(arr)
mn = arr[0]
mx = 0
# print(arr)
for i in range(n):
mn = min(mn, arr[i] - mx)
# left[i] = mn
arr[i] -= mn
mx = max(mx, arr[i])
return mn >= 0
ans = []
for _ in range(inp()):
n = inp()
arr = [inp() for _ in range(n)]
ans.append("YES" if solve(arr) else "NO")
print('\n'.join(ans))
``` | output | 1 | 49,028 | 12 | 98,057 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES | instruction | 0 | 49,029 | 12 | 98,058 |
Tags: constructive algorithms, dp, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
for i in range(n-1,0,-1):
a[i] -= a[i-1]
minus = 0
for i in range(1,n):
if a[i]<0:
minus -= a[i]
if a[0] - minus >=0:
print("YES")
else:
print("NO")
``` | output | 1 | 49,029 | 12 | 98,059 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES | instruction | 0 | 49,030 | 12 | 98,060 |
Tags: constructive algorithms, dp, greedy
Correct Solution:
```
# Enter your code here. Read input from STDIN. Print output to STDOUT# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
# ===============================================================================================
# some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def pow(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1): # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# discrete binary search
# minimise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# if isvalid(l):
# return l
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m - 1):
# return m
# if isvalid(m):
# r = m + 1
# else:
# l = m
# return m
# maximise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# # print(l,r)
# if isvalid(r):
# return r
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m + 1):
# return m
# if isvalid(m):
# l = m
# else:
# r = m - 1
# return m
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def N():
return int(inp())
dx = [0, 0, 1, -1]
dy = [1, -1, 0, 0]
def YES():
print("YES")
def NO():
print("NO")
def Yes():
print("Yes")
def No():
print("No")
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(i):
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
# # to find factorial and ncr
# N=300005
# mod = 10**9 + 7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, N + 1):
# fac.append((fac[-1] * i) % mod)
# inv.append(mod - (inv[mod % i] * (mod // i) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
#
#
# def comb(n, r):
# if n < r:
# return 0
# else:
# return fac[n] * (finv[r] * finv[n - r] % mod) % mod
class MergeFind:
def __init__(self, n):
self.parent = list(range(n))
self.size = [1] * n
self.num_sets = n
# self.lista = [[_] for _ in range(n)]
def find(self, a):
to_update = []
while a != self.parent[a]:
to_update.append(a)
a = self.parent[a]
for b in to_update:
self.parent[b] = a
return self.parent[a]
def merge(self, a, b):
a = self.find(a)
b = self.find(b)
if a == b:
return
self.num_sets -= 1
self.parent[a] = b
self.size[b] += self.size[a]
# self.lista[a] += self.lista[b]
# self.lista[b] = []
def set_size(self, a):
return self.size[self.find(a)]
def __len__(self):
return self.num_sets
def solve():
n=N()
happens=1
for wrw in range(1):
ar=lis()
sar=ar[::-1]
dec=[0]*n
dec[0]=1
for i in range(1,n):
if ar[i]>ar[i-1]:
break
else:
dec[i]=1
inc = [0] * n
inc[n-1] = 1
for i in range(n-2, -1,-1):
if ar[i] > ar[i + 1]:
break
else:
inc[i] = 1
have=[ar[0]]
for i in range(1,n):
have.append(min(have[-1],ar[i]))
st=en=-1
for i in range(n):
if inc[i]==0 and dec[i]==0:
st=i
break
for i in range(n-1,-1,-1):
if inc[i]==0 and dec[i]==0:
en=i
break
if st==-1 and en==-1:
YES()
return
M=ar[en+1]
# print(dec)
# print(inc)
# print(have,st,en)
# print(ar)
sub=0
for i in range(en,st-1,-1):
ar[i]-=sub
if ar[i]<0:
happens=0
break
# print(i,M)
if ar[i]<=M:
M=ar[i]
else:
d=ar[i]-M
have[i]-=sub
if d>have[i]:
happens=0
break
sub+=d
if happens==0:
break
YES()
return
for wrw in range(1):
ar=sar
dec=[0]*n
dec[0]=1
for i in range(1,n):
if ar[i]>ar[i-1]:
break
else:
dec[i]=1
inc = [0] * n
inc[n-1] = 1
for i in range(n-2, -1,-1):
if ar[i] > ar[i + 1]:
break
else:
inc[i] = 1
have=[ar[0]]
for i in range(1,n):
have.append(min(have[-1],ar[i]))
st=en=-1
for i in range(n):
if inc[i]==0 and dec[i]==0:
st=i
break
for i in range(n-1,-1,-1):
if inc[i]==0 and dec[i]==0:
en=i
break
if st==-1 and en==-1:
YES()
return
M=ar[en+1]
sub = 0
for i in range(en, st - 1, -1):
ar[i] -= sub
if ar[i] < 0:
happens = 0
break
# print(i,M)
if ar[i] <= M:
M = ar[i]
else:
d = ar[i] - M
have[i] -= sub
if d > have[i]:
happens = 0
break
sub += d
if happens == 0:
break
YES()
return
NO()
#solve()
testcase(int(inp()))
``` | output | 1 | 49,030 | 12 | 98,061 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES | instruction | 0 | 49,031 | 12 | 98,062 |
Tags: constructive algorithms, dp, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
a = int(input())
for i in range (a):
b = int(input())
c = list(map(int, input().split()))
d = [c[0]]
e = [0]
for j in range (1, b):
d.append(min(d[j - 1], c[j] - e[j - 1]))
e.append(c[j] - d[j])
def abc(d, e):
for j in range (1, b):
if d[j] > d[j - 1]:
return "NO"
if e[-j - 1] > e[-j]:
return "NO"
if d[j] < 0:
return "NO"
if e[-j - 1] < 0:
return "NO"
return "YES"
print(abc(d, e))
``` | output | 1 | 49,031 | 12 | 98,063 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES | instruction | 0 | 49,032 | 12 | 98,064 |
Tags: constructive algorithms, dp, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
A = map(int, input().split())
l, r = 10000000000, 0
possible = True
for a in A:
if a < r:
possible = False
break
a -= r
l = min(l, a)
r += (a-l)
if(possible):
print("YES")
else:
print("NO")
``` | output | 1 | 49,032 | 12 | 98,065 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES | instruction | 0 | 49,033 | 12 | 98,066 |
Tags: constructive algorithms, dp, greedy
Correct Solution:
```
import sys
from sys import stdin
tt = int(stdin.readline())
for loop in range(tt):
n = int(stdin.readline())
a = list(map(int,stdin.readline().split()))
r = a[0]
l = 0
for i in range(n):
if r + l < a[i]:
l = a[i] - r
elif r + l > a[i]:
r = a[i] - l
if r >= 0:
print ("YES")
else:
print ("NO")
``` | output | 1 | 49,033 | 12 | 98,067 |
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 a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES
Submitted Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
x = 0
flag = 1
for i in range(1, n):
x = max(x, a[i] - a[i - 1])
a[i] -= x
if a[i] < 0:
flag = 0
print("NO")
break
if flag:
print("YES")
``` | instruction | 0 | 49,034 | 12 | 98,068 |
Yes | output | 1 | 49,034 | 12 | 98,069 |
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 a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES
Submitted Solution:
```
from itertools import accumulate
def read_ints():
return map(int, input().split())
t_n = int(input())
for i_t in range(t_n):
n = int(input())
a = list(read_ints())
min_from_left = list(accumulate(a, min))
min_from_right = list(accumulate(a[::-1], min))[::-1]
deltas = [x2 - x1 for x1, x2 in zip(a, a[1:])]
r = [0] * n
acc = 0
for i, delta in list(enumerate(deltas)):
if delta > 0:
acc += delta
r[i+1] += acc
acc = 0
for i, delta in list(enumerate(deltas))[::-1]:
if delta < 0:
acc += abs(delta)
r[i] += acc
if all(x >= rx for rx, x in zip(r, a)):
print("YES")
else:
print("NO")
``` | instruction | 0 | 49,035 | 12 | 98,070 |
Yes | output | 1 | 49,035 | 12 | 98,071 |
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 a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
#################################################
a = list(map(int, input().split()))
b = [0] * (n - 1)
for i in range(n - 1):
b[i] = a[i + 1] - a[i]
mi = 10000000
c = [0] * (n + 1)
for i in range(n - 1):##########
if b[i] < 0:
c[i + 1] += b[i]
c[0] -= b[i]
else:
c[i + 1] += b[i]
c[-1] -= b[i]
for i in range(n):
c[i + 1] += c[i]
for i in range(n):
if c[i] > a[i]:
print('NO')
break
else:
print('YES')
``` | instruction | 0 | 49,036 | 12 | 98,072 |
Yes | output | 1 | 49,036 | 12 | 98,073 |
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 a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES
Submitted Solution:
```
from collections import defaultdict as dft, deque, Counter
from heapq import heappush,heappop
def mp(): return map(int,input().split())
def ml(): return list(map(int,input().split()))
import sys;
sys.setrecursionlimit(10**5)
from math import ceil,log2;
INT_MAX = sys.maxsize;
def solve():
def minVal(x, y) :
return x if (x < y) else y;
def getMid(s, e) :return s + (e - s) // 2;
def RMQUtil( st, ss, se, qs, qe, index) :
if (qs <= ss and qe >= se) :return st[index];
if (se < qs or ss > qe) : return INT_MAX;
mid = getMid(ss, se);
return min(RMQUtil(st, ss, mid, qs,qe, 2 * index + 1), RMQUtil(st, mid + 1, se, qs, qe, 2 * index + 2));
def RMQ( st, n, qs, qe) :
if (qs < 0 or qe > n - 1 or qs > qe) : print("Invalid Input"); return -1;
return RMQUtil(st, 0, n - 1, qs, qe, 0);
def constructSTUtil(arr, ss, se, st, si) :
if (ss == se) : st[si] = arr[ss]; return arr[ss];
mid = getMid(ss, se); st[si] = min(constructSTUtil(arr, ss, mid, st, si * 2 + 1), constructSTUtil(arr, mid + 1, se, st, si * 2 + 2)); return st[si];
def constructST( arr, n) : x = (int)(ceil(log2(n)));max_size = 2 * (int)(2**x) - 1;st = [0] * (max_size); constructSTUtil(arr, 0, n - 1, st, 0); return st;
def update(i,val):
while(i<=n):BIT[i]+=val;i+= i &(-i)
def getsm(i):
sm=0
while(i>0):sm+=BIT[i];i-=i&(-i)
return sm
mxn=10**9
n=int(input());arr=list(map(int,input().split()));BIT=[0]*(n+1);st = constructST(arr,n);pre=mxn
for i in range(n-1):
temp=getsm(i+1)
if arr[i] - temp <=pre:pre=arr[i] - temp
else:
diff=arr[i]-temp-pre;mn=RMQ(st,n,i+1,n-1) - temp
if mn <diff:return "NO"
update(i+1,diff)
return "YES"
print(RMQ(st,n,0,0))
mxn=10**9
for _ in range(int(input())):print(solve())
``` | instruction | 0 | 49,037 | 12 | 98,074 |
Yes | output | 1 | 49,037 | 12 | 98,075 |
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 a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES
Submitted Solution:
```
import sys
def check(a, left, right, buget):
left_cost = 0
right_cost = 0
impossible = False
can_move = True
while can_move:
can_move = False
if left > 0:
can_move = True
left_additional_cost = a[left - 1] - left_min[left - 1]
if left_cost + right_cost + left_additional_cost <= buget:
left_cost += left_additional_cost
left -= 1
else:
impossible = True
break
if right < len(a) - 1:
can_move = True
right_additional_cost = a[right + 1] - right_min[right + 1]
if left_cost + right_cost + right_additional_cost <= buget:
right_cost += right_additional_cost
right += 1
else:
impossible = True
break
return impossible
if __name__ == '__main__':
t = int(sys.stdin.readline().strip())
for j in range(t):
n = int(sys.stdin.readline().strip())
a = list(map(int, sys.stdin.readline().strip().split()))
left_min = [a[0]]
for i in range(1, len(a)):
left_min.append(min(left_min[-1], a[i]))
right_min = [a[-1]]
for i in reversed(range(0, len(a))):
right_min.append(min(right_min[-1], a[i]))
right_min = list(reversed(right_min))
a_s = sorted(enumerate(a), key=lambda i: i[1])
buget = a_s[0][1]
one_possible = False
i = 0
while i < len(a) and a_s[i][1] == buget:
left = a_s[i][0]
right = a_s[i][0]
impossible = check(a, left, right, buget)
if not impossible:
one_possible = True
break
i += 1
print('YES' if one_possible else 'NO')
``` | instruction | 0 | 49,038 | 12 | 98,076 |
No | output | 1 | 49,038 | 12 | 98,077 |
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 a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
s = list(map(int, input().split()))
x = 0
p = 1
for i in range(1, n-1):
if s[i] > s[i] and s[i] > s[i+1]:
x += 1
if s[i] > min(s[:i]) + min(s[i+1:]):
p = 0
break
if p and x < 2:
print('YES')
else:
print('NO')
``` | instruction | 0 | 49,039 | 12 | 98,078 |
No | output | 1 | 49,039 | 12 | 98,079 |
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 a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
import math
from itertools import permutations
from decimal import Decimal, getcontext
getcontext().prec = 25
MOD = pow(10, 9) + 7
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")
# n, k = map(int, input().split(" "))
# l = list(map(int, input().split(" ")))
for _ in range(int(input())):
n = int(input())
l = list(map(int, input().split(" ")))
up = down = 0
m = max(l[0], l[-1])
for i in range(1,n-1):
if l[i] > l[i-1] and l[i]> l[i+1]:
down+=1
if l[i] < l[i-1] and l[i] < l[i+1]:
up+=1
m = max(m, l[i])
if m > (l[0]+ l[-1]):
print("NO")
else:
if up >1 or down > 2:
print("NO")
else:
print("YES")
``` | instruction | 0 | 49,040 | 12 | 98,080 |
No | output | 1 | 49,040 | 12 | 98,081 |
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 a of n positive integers.
You can use the following operation as many times as you like: select any integer 1 ≤ k ≤ n and do one of two things:
* decrement by one k of the first elements of the array.
* decrement by one k of the last elements of the array.
For example, if n=5 and a=[3,2,2,1,4], then you can apply one of the following operations to it (not all possible options are listed below):
* decrement from the first two elements of the array. After this operation a=[2, 1, 2, 1, 4];
* decrement from the last three elements of the array. After this operation a=[3, 2, 1, 0, 3];
* decrement from the first five elements of the array. After this operation a=[2, 1, 1, 0, 3];
Determine if it is possible to make all the elements of the array equal to zero by applying a certain number of operations.
Input
The first line contains one positive integer t (1 ≤ t ≤ 30000) — the number of test cases. Then t test cases follow.
Each test case begins with a line containing one integer n (1 ≤ n ≤ 30000) — the number of elements in the array.
The second line of each test case contains n integers a_1 … a_n (1 ≤ a_i ≤ 10^6).
The sum of n over all test cases does not exceed 30000.
Output
For each test case, output on a separate line:
* YES, if it is possible to make all elements of the array equal to zero by applying a certain number of operations.
* NO, otherwise.
The letters in the words YES and NO can be outputed in any case.
Example
Input
4
3
1 2 1
5
11 7 9 6 8
5
1 3 1 3 1
4
5 2 1 10
Output
YES
YES
NO
YES
Submitted Solution:
```
import math
import collections
t=int(input())
for w in range(t):
n=int(input())
l=[int(i) for i in input().split()]
l1=[l[0]]+[0]*(n-1)
l2=[0]*(n-1)+[l[-1]]
m1=l[0]
m2=l[-1]
for i in range(1,n):
m1=min(l[i],m1)
l1[i]=m1
for i in range(n-2,-1,-1):
m2=min(l[i],m2)
l2[i]=m2
c=1
for i in range(1,n-1):
if(l1[i-1]+l2[i+1]<l[i]):
c=0
break
if(c==1):
print("YES")
else:
print("NO")
``` | instruction | 0 | 49,041 | 12 | 98,082 |
No | output | 1 | 49,041 | 12 | 98,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an infinite set generated as follows:
* 1 is in this set.
* If x is in this set, x ⋅ a and x+b both are in this set.
For example, when a=3 and b=6, the five smallest elements of the set are:
* 1,
* 3 (1 is in this set, so 1⋅ a=3 is in this set),
* 7 (1 is in this set, so 1+b=7 is in this set),
* 9 (3 is in this set, so 3⋅ a=9 is in this set),
* 13 (7 is in this set, so 7+b=13 is in this set).
Given positive integers a, b, n, determine if n is in this set.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.
Output
For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.
Example
Input
5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264
Output
Yes
No
Yes
No
Yes
Note
In the first test case, 24 is generated as follows:
* 1 is in this set, so 3 and 6 are in this set;
* 3 is in this set, so 9 and 8 are in this set;
* 8 is in this set, so 24 and 13 are in this set.
Thus we can see 24 is in this set.
The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them. | instruction | 0 | 49,091 | 12 | 98,182 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
#from math import *
#from bisect import *
#from collections import *
#from random import *
#from decimal import *"""
#from heapq import *
#from random import *
import sys
input=sys.stdin.readline
sys.setrecursionlimit(3*(10**5))
def inp():
return int(input())
def st():
return input().rstrip('\n')
def lis():
return list(map(int,input().split()))
def ma():
return map(int,input().split())
t=inp()
while(t):
t-=1
n,a,b=ma()
if(a==1):
if((n-1)%b):
print("No")
else:
print("Yes")
else:
x=1
s='No'
while(x<=n):
if(x%b==n%b):
s='Yes'
break
x*=a
print(s)
``` | output | 1 | 49,091 | 12 | 98,183 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is an infinite set generated as follows:
* 1 is in this set.
* If x is in this set, x ⋅ a and x+b both are in this set.
For example, when a=3 and b=6, the five smallest elements of the set are:
* 1,
* 3 (1 is in this set, so 1⋅ a=3 is in this set),
* 7 (1 is in this set, so 1+b=7 is in this set),
* 9 (3 is in this set, so 3⋅ a=9 is in this set),
* 13 (7 is in this set, so 7+b=13 is in this set).
Given positive integers a, b, n, determine if n is in this set.
Input
The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^5) — the number of test cases. The description of the test cases follows.
The only line describing each test case contains three integers n, a, b (1≤ n,a,b≤ 10^9) separated by a single space.
Output
For each test case, print "Yes" if n is in this set, and "No" otherwise. You can print each letter in any case.
Example
Input
5
24 3 5
10 3 6
2345 1 4
19260817 394 485
19260817 233 264
Output
Yes
No
Yes
No
Yes
Note
In the first test case, 24 is generated as follows:
* 1 is in this set, so 3 and 6 are in this set;
* 3 is in this set, so 9 and 8 are in this set;
* 8 is in this set, so 24 and 13 are in this set.
Thus we can see 24 is in this set.
The five smallest elements of the set in the second test case is described in statements. We can see that 10 isn't among them. | instruction | 0 | 49,094 | 12 | 98,188 |
Tags: constructive algorithms, math, number theory
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
#<fast I/O>
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)
#</fast I/O>
#<I/O>
def scanf(datatype = str):
return datatype(sys.stdin.readline().rstrip())
def printf(answer = ''):
return sys.stdout.write(str(answer) + "\n")
def prints(answer):
return sys.stdout.write(str(answer) + " ")
def map_input(datatype = str):
return map(datatype, sys.stdin.readline().split())
def list_input(datatype = str):
return list(map(datatype, sys.stdin.readline().split()))
def testcase(number: int, solve_function):
for _ in range(number):
printf(solve_function())
# solve_function()
#</I/O>
#<solution>
# sys.setrecursionlimit(int(1e8))
# def rec(Set, n1, n, a, b):
# if(n1 == n):
# return True
# if(n1 > n):
# return
# num1 = n1 * a
# flag1 = False
# flag2 = False
# if(num1 not in Set):
# Set.add(num1)
# flag1 = rec(Set, num1, n, a, b)
# if flag1:
# return flag1
# num2 = n1 + b
# if(num2 not in Set):
# Set.add(num2)
# flag2 = rec(Set, num2, n, a, b)
# return flag2
# def solve():
# n, a, b = map_input(int)
# Set = set()
# Set.add(1)
# ans = rec(Set, 1, n, a, b)
# return 'Yes' if ans else 'No'
def solve():
n, a, b = map_input(int)
if(a == 1):
if (n - 1) % b == 0:
return 'Yes'
return 'No'
term = 1
while(term <= n):
if (n - term) % b == 0:
return 'Yes'
term *= a
return 'No'
# solve()
t = scanf(int)
testcase(t,solve)
#</solution>
``` | output | 1 | 49,094 | 12 | 98,189 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34 | instruction | 0 | 49,587 | 12 | 99,174 |
"Correct Solution:
```
def m(L,R):
global c;c+=len(L)+len(R)
T=[];j=0
for l in L:
while j<len(R)and R[j]<l:T+=[R[j]];j+=1
T+=[l]
while j<len(R):T+=[R[j]];j+=1
return T
def d(A):s=len(A)//2;return m(d(A[:s]),d(A[s:])) if len(A)>1 else A
c=0
input()
print(*d(list(map(int,input().split()))))
print(c)
``` | output | 1 | 49,587 | 12 | 99,175 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34 | instruction | 0 | 49,588 | 12 | 99,176 |
"Correct Solution:
```
def merge(A, left, mid, right):
L = A[left:mid]
R = A[mid:right]
L.append(10**9+1)
R.append(10**9+1)
i = 0
j = 0
for k in range(left, right):
if L[i] <= R[j]:
A[k] = L[i]
i += 1
else:
A[k] = R[j]
j += 1
return right-left
def merge_sort(A, left, right):
if left+1 < right:
mid = (left+right)//2
ans1 = merge_sort(A, left, mid)
ans2 = merge_sort(A, mid, right)
return merge(A, left, mid, right)+ans1+ans2
return 0
N = int(input())
S = list(map(int, input().split()))
ans = merge_sort(S, 0, N)
print(*S)
print(ans)
``` | output | 1 | 49,588 | 12 | 99,177 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34 | instruction | 0 | 49,589 | 12 | 99,178 |
"Correct Solution:
```
#coding:utf-8
#1_5_B
def merge(array, left, mid, right):
global cnt
n1 = mid - left
n2 = right - mid
L = array[left:mid] + [sentinel]
R = array[mid:right] + [sentinel]
i = j = 0
for k in range(left, right):
if L[i] <= R[j]:
array[k] = L[i]
i += 1
else:
array[k] = R[j]
j += 1
cnt += right - left
def merge_sort(array, left, right):
if left + 1 < right:
mid = (left + right) // 2
merge_sort(array, left, mid)
merge_sort(array, mid, right)
merge(array, left, mid, right)
n = int(input())
S = list(map(int, input().split()))
cnt = 0
sentinel = 10 ** 9 + 1
merge_sort(S, 0, len(S))
print(*S)
print(cnt)
``` | output | 1 | 49,589 | 12 | 99,179 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34 | instruction | 0 | 49,590 | 12 | 99,180 |
"Correct Solution:
```
cnt = 0
INF = pow(10, 18)
def merge(A, left, mid, right):
L = A[left:mid] + [INF]
R = A[mid:right] + [INF]
i = 0
j = 0
for k in range(left, right):
global cnt
cnt += 1
if L[i] <= R[j]:
A[k] = L[i]
i += 1
else:
A[k] = R[j]
j += 1
def mergeSort(A, left, right):
if left + 1 < right:
mid = (left + right) // 2
mergeSort(A, left, mid)
mergeSort(A, mid, right)
merge(A, left, mid, right)
n = int(input())
s = list(map(int, input().split()))
mergeSort(s, 0, n)
print(*s)
print(cnt)
``` | output | 1 | 49,590 | 12 | 99,181 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34 | instruction | 0 | 49,591 | 12 | 99,182 |
"Correct Solution:
```
def merge(A, left, mid, right):
global cnt
n1 = mid - left
n2 = right - mid
L = A[left:mid] + [INF]
R = A[mid:right] + [INF]
i = 0
j = 0
for k in range(left,right):
if L[i] <= R[j]:
A[k] = L[i]
i = i + 1
else:
A[k] = R[j]
j = j + 1
cnt += 1
def mergeSort(A, left, right):
if left + 1 < right:
mid = (left + right)//2
mergeSort(A, left, mid)
mergeSort(A, mid, right)
merge(A, left, mid, right)
INF = 10000000000
n = int(input())
S = list(map(int,input().split()))
cnt = 0
mergeSort(S,0,n)
print(*S)
print(cnt)
``` | output | 1 | 49,591 | 12 | 99,183 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34 | instruction | 0 | 49,592 | 12 | 99,184 |
"Correct Solution:
```
INF = 10000000000
def merge(A, left, mid, right):
global anw
L = A[left:mid] + [INF]
R = A[mid:right] + [INF]
i, j= 0, 0
for k in range(left, right):
anw += 1
if L[i] <= R[j]:
A[k] = L[i]
i += 1
else:
A[k] = R[j]
j += 1
def mergeSort(A, left, right):
global anw
if left + 1 < right:
mid = int(( left + right ) / 2)
mergeSort(A, left, mid)
mergeSort(A, mid, right)
merge(A, left, mid, right)
if __name__ == '__main__':
n = int(input())
S = [int(x) for x in input().split()]
anw = 0
mergeSort(S, 0, n)
print(*S)
print(anw)
``` | output | 1 | 49,592 | 12 | 99,185 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34 | instruction | 0 | 49,593 | 12 | 99,186 |
"Correct Solution:
```
def merge(a, left, mid, right):
x = 0
l = a[left:mid] + [float("inf")]
r = a[mid:right] + [float("inf")]
i = 0
j = 0
for k in range(left,right):
if l[i] <= r[j]:
a[k] = l[i]
i += 1
else :
a[k] = r[j]
j += 1
x += 1
return x
def mergeSort(a, left, right,x):
if left+1 < right:
mid = int((left + right)/2)
mergeSort(a, left, mid,x)
mergeSort(a, mid, right,x)
x[0] += merge(a, left, mid, right)
n = int(input())
s = list(map(int,input().split()))
x = [0]
mergeSort(s,0,n,x)
print(*s)
print(x[0])
``` | output | 1 | 49,593 | 12 | 99,187 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34 | instruction | 0 | 49,594 | 12 | 99,188 |
"Correct Solution:
```
cnt = 0
def merge(L,R):
global cnt
n = len(L)+len(R)
cnt += n
A = []
i = j = 0
L.append(float("inf"))
R.append(float("inf"))
for _ in range(n):
if L[i] <= R[j]:
A.append(L[i])
i += 1
else:
A.append(R[j])
j += 1
return A
def mergeSort(A):
if len(A)==1: return A
m = len(A)//2
return merge(mergeSort(A[:m]),mergeSort(A[m:]))
if __name__=='__main__':
n=int(input())
A=list(map(int,input().split()))
print(*mergeSort(A))
print(cnt)
``` | output | 1 | 49,594 | 12 | 99,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34
Submitted Solution:
```
def m(L,R):
global c;c+=len(L)+len(R)
T=[]
for l in L[::-1]:
while R and R[-1]>l:T+=[R.pop()]
T+=[l]
T+=R[::-1]
return T[::-1]
def d(A):s=len(A)//2;return m(d(A[:s]),d(A[s:])) if len(A)>1 else A
c=0
input()
print(*d(list(map(int,input().split()))))
print(c)
``` | instruction | 0 | 49,595 | 12 | 99,190 |
Yes | output | 1 | 49,595 | 12 | 99,191 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34
Submitted Solution:
```
#17D8103001E Fukushima Emi persimmon8 python 19.10.24
n=int(input())
A=list(map(int,input().split()))
count=0
def merge(A,left,mid,right):
global count
L=A[left:mid]+[10000000000000]
R=A[mid:right]+[100000000000000]
i=0
j=0
for k in range(left,right):
count+=1
if L[i]<=R[j]:
A[k]=L[i]
i=i+1
else:
A[k]=R[j]
j=j+1
def mergeSort(a,left,right):
if left+1<right:
mid=(left+right)//2
mergeSort(A,left,mid)
mergeSort(A,mid,right)
merge(A,left,mid,right)
mergeSort(A,0,n)
d=map(str,A)
print(' '.join(d))
print(count)
``` | instruction | 0 | 49,596 | 12 | 99,192 |
Yes | output | 1 | 49,596 | 12 | 99,193 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34
Submitted Solution:
```
from collections import deque
def merge(a, l, m, r):
global c
ll = deque(a[l:m])
rl = deque(a[m:r])
ll.append(10e10)
rl.append(10e10)
for k in range(l, r):
a[k] = (ll if ll[0] < rl[0] else rl).popleft()
c += r - l
def merge_sort(a, l, r):
if l + 1 < r:
m = (l + r) // 2
merge_sort(a, l, m)
merge_sort(a, m, r)
merge(a, l, m, r)
if __name__ == '__main__':
n, a, c = int(input()), list(map(int, input().split())), 0
merge_sort(a, 0, n)
print(*a)
print(c)
``` | instruction | 0 | 49,597 | 12 | 99,194 |
Yes | output | 1 | 49,597 | 12 | 99,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34
Submitted Solution:
```
from collections import deque
SENIAL = 10**9
count = 0
def merge(A,left,mid,right):
L = deque(A[left:mid])
R = deque(A[mid:right])
L.append(SENIAL)
R.append(SENIAL)
global count
for k in range(left,right):
count +=1
if L[0] <= R[0]:
A[k] = L.popleft()
else:
A[k] = R.popleft()
def mergeSort(A,left,right):
if left + 1 < right:
middle = (left + right) //2
mergeSort(A,left,middle)
mergeSort(A,middle,right)
merge(A,left,middle,right)
input()
l = list(map(int,input().split()))
mergeSort(l,0,len(l))
print(" ".join(map(str,l)))
print(count)
``` | instruction | 0 | 49,598 | 12 | 99,196 |
Yes | output | 1 | 49,598 | 12 | 99,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34
Submitted Solution:
```
# coding: utf-8
from itertools import count
def merge(A, left, mid, right):
global counter
n1, n2 = mid - left, right - mid
L, R = [None]*(n1+1), [None]*(n2+1)
for i in range(n1):
L[i] = A[left + i]
for i in range(n2):
R[i] = A[mid + i]
L[n1] = R[n2] = float("inf")
i = j = 0
for k in range(left, right):
counter += 1
if L[i] <= R[j]:
A[k] = L[i]
i = i + 1
else:
A[k] = R[j]
j = j + 1
def mergeSort(A, left, right):
if left+1 < right:
mid = (left + right)//2;
mergeSort(A, left, mid)
mergeSort(A, mid, right)
merge(A, left, mid, right)
N = int(input())
A = [int(i) for i in input().split()]
counter = 0
mergeSort(A, 0, len(A))
print(*A)
print(counter)
``` | instruction | 0 | 49,599 | 12 | 99,198 |
No | output | 1 | 49,599 | 12 | 99,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34
Submitted Solution:
```
def m(L,R):
T=[]
for l in L[::-1]:
while R and R[-1]>l:T+=[R.pop()]
T+=[l]
return R+T[::-1]
def d(A):
l=len(A);global c;c+=l
s=l//2;return m(d(A[:s]),d(A[s:]))if l>1 else A
def s():
c=-int(input())
print(*d(list(map(int,input().split()))))
print(c)
if'__main__'==__name__:s()
``` | instruction | 0 | 49,600 | 12 | 99,200 |
No | output | 1 | 49,600 | 12 | 99,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34
Submitted Solution:
```
merge(A, left, mid, right)
n1 = mid - left
n2 = right - mid
L = [0]*(n1+1)
R = [0]*(n2+1)
for i in range(n1):
L[i] = A[left + i]
for i in range(n2):
R[i] = A[mid + i]
L[n1] = float("inf")
R[n2] = float("inf")
i = 0
j = 0
for k in range(left,right):
if L[i] <= R[j]:
A[k] = L[i]
i = i + 1
else:
A[k] = R[j]
j = j + 1
mergeSort(A, left, right)
if left+1 < right
mid = (left + right)//2;
mergeSort(A, left, mid)
mergeSort(A, mid, right)
merge(A, left, mid, right)
### Main Function
n = int(input())
S = list(map(int,input().split()))
c = 0
mergeSort(S, 0, n)
print(S)
print(c)
``` | instruction | 0 | 49,601 | 12 | 99,202 |
No | output | 1 | 49,601 | 12 | 99,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program of a Merge Sort algorithm implemented by the following pseudocode. You should also report the number of comparisons in the Merge function.
Merge(A, left, mid, right)
n1 = mid - left;
n2 = right - mid;
create array L[0...n1], R[0...n2]
for i = 0 to n1-1
do L[i] = A[left + i]
for i = 0 to n2-1
do R[i] = A[mid + i]
L[n1] = SENTINEL
R[n2] = SENTINEL
i = 0;
j = 0;
for k = left to right-1
if L[i] <= R[j]
then A[k] = L[i]
i = i + 1
else A[k] = R[j]
j = j + 1
Merge-Sort(A, left, right){
if left+1 < right
then mid = (left + right)/2;
call Merge-Sort(A, left, mid)
call Merge-Sort(A, mid, right)
call Merge(A, left, mid, right)
Notes
Constraints
* n ≤ 500000
* 0 ≤ an element in S ≤ 109
Input
In the first line n is given. In the second line, n integers are given.
Output
In the first line, print the sequence S. Two consequtive elements should be separated by a space character.
In the second line, print the number of comparisons.
Example
Input
10
8 5 9 2 6 3 7 1 10 4
Output
1 2 3 4 5 6 7 8 9 10
34
Submitted Solution:
```
n = int(input())
S = list(map(int, input().split()))
SENTINEL = 10 ** 9 + 1
count = 0
def merge(S, left, mid, right):
n1 = mid - left
n2 = right - mid
L = S[left:mid] + [SENTINEL]
R = S[mid:right] + [SENTINEL]
for i in range(n1):
L[i] = S[left + i]
for i in range(n2):
R[i] = S[mid + i]
L[n1] = R[n2] = SENTINEL
i = j = 0
for k in range(left, right):
count += 1
if(L[i] <= R[i]):
S[k] = L[i]
i += 1
else:
S[k] = R[j]
j += 1
mergeSort(S, left, right):
if(left + 1 < right):
mid = (left + right) // 2
mergeSort(S, left, mid)
mergeSort(S, mid, right)
merge(S, left, mid, right)
mergeSort(S, 0, n)
print(*S)
print(count)
``` | instruction | 0 | 49,602 | 12 | 99,204 |
No | output | 1 | 49,602 | 12 | 99,205 |
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