1364A - XXXXX - CodeForces Solution


brute force data structures number theory two pointers *1200

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Python Code:

import sys, os
from io import BytesIO, IOBase
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
from functools import lru_cache
import math
from collections import OrderedDict
from collections import Counter
from itertools import combinations
from itertools import accumulate
import operator
import bisect
from collections import deque,defaultdict
import copy
from itertools import permutations
from operator import itemgetter
from functools import lru_cache
 
 
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")
 
stdin, stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) 
 
mod = pow(10, 9) + 7
mod2 = 998244353
        
def inp(): return stdin.readline().strip()
def iinp(): return int(inp())
def out(var, end="\n"): stdout.write(str(var)+"\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def ceil(a, b): return (a+b-1)//b
 
def bin(a,b):
    if b==0:
        return 1
    r=bin(a,b//2)
    if b%2:
        return r*r*a
    else:
        return r*r
 
def gcd(a,b):
    if a == 0:
        return b
    return gcd(b % a, a)

def lcm(a,b):
    return (a // gcd(a,b))* b
    
def solve(n,a,x):
    d=sum(a)
    i=0
    while i<n and d%x==0:
        d-=a[i]
        i+=1
    l=n-1
    d=sum(a)
    while l>=0 and d%x==0:
        d-=a[l]
        l-=1
    e=max(l+1,n-i)
    if e:
        print(e)
    else:
        print(-1)
t=iinp()
for i in range(t):
        n,x=lmp()
    a=lmp()    solve(n,a,x)
'''
a=inp()
b=inp()
solve(a,b)
'''

C++ Code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt","w",stdout);
    
    #endif
        int t;
        cin>>t;
        while(t--){
            int n,k,x,sum = 0,ans = -1;
            cin>>n>>k;
            int arr[n],dp[n] = {0};
            for(int i = 0;i<n;i++){
                cin>>arr[i];
                sum +=arr[i];
                if(sum%k != 0){
                    ans = max(ans,i+1);
                }
            }
           for(int i= 0;i<n;i++){
                sum -= arr[i];
                if(sum%k != 0){
                    ans = max(ans,n-i-1);
                }
           }
           int mid = (n-1)/2;
           for(int i = 0,j = n-1;i<mid,j>mid;i++,j--){
                sum = sum - arr[i]-arr[j];
                if(sum%k != 0){
                    ans = max(ans,j-i-1);
                }
           }
           cout<<ans<<endl;

        }
    return 0;
}


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