Comments

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  -13

  

  binhngoan261009  commented on Sept. 1, 2024, 2:44 p.m.

  code cho ae tham khao

  include <bits/stdc++.h>

  using namespace std; long long n,a[1000005],d1,d2,kq,d28; int main() { cin>>n; for(int i=1;i<=n;i++) { cin>>a[i]; if(a[i]%28==0) {a[i]=0;d28++;} else a[i]=a[i]%28; } sort(a+1,a+n+1); for(long long i=1;i<=n;i++) { d1=lowerbound(a+i+1,a+n+1,28-a[i])-a; d2=upperbound(a+i+1,a+n+1,28-a[i])-a-1; if(d1<=d2) kq+=d2-d1+1; } kq+=d28*(d28-1)/2; cout<<kq; return 0; }

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  -7

  

  nguyenanhquang  commented on Aug. 8, 2024, 7:50 a.m.

  time erorr

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  -5

  

  minhquan2905  commented on July 28, 2024, 11:57 a.m.

  hello world

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  6

  

  minhquan2905  commented on July 28, 2024, 10:59 a.m.

  vận dụng: A % 28 = 0 <=> (a%28 + b%28) % 28 = 0 => đặt x = a%28, y=b%28 <=> (x+y)%28 = 0 mà x, y <= 28 -> x+y=28 -> bài toán trở thành tìm, đếm cặp số dư bằng 28. (TH đặc biệt: x=y=14 , x=y=0 -> xét riêng)

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  -1

  

  GodNona  commented on Feb. 2, 2025, 4:34 p.m.

  A giúp e với a. E bị wrong answer

  include <bits/stdc++.h>

  using namespace std;

  define ll long long

  const int MOD = 1e9 + 7;

  ll a[1000006]; ll dem[28];

  int main() { iosbase::syncwith_stdio(0); cin.tie(NULL);

  int n; cin >> n;
  for(int i = 0; i < n; i++)
      cin >> a[i];
  
  for(int i = 0; i < n; i++)
      dem[a[i] % 28]++;
  
  int res = 0;
  res += dem[0] * (dem[0] - 1) / 2;
  for(int i = 1; i <= 13; i++)
      res += dem[i] * dem[28 - i];
  res += dem[14] * (dem[14] - 1) / 2;
  
  cout << res;
  
  return 0;
  

  }

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  0

  

  minhquan2905  commented on Feb. 4, 2025, 4:37 p.m.

  biến res bị tràn, để ll đi =D

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  -1

  

  Long4200  commented on Jan. 28, 2025, 6:31 a.m.

  tks bro

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  -18

  

  NTH11112222  commented on May 25, 2024, 2:33 a.m.

  include <iostream>

  include <vector>

  using namespace std; int main() { int N; cin >> N; vector<int> A(N); vector<int> co(28, 0); long long pairs = 0; for (int i = 0; i < N; ++i) { cin >> A[i]; co[A[i] % 28]++; } for (int i = 0; i < 28; ++i) { if (i == 0 || i * 2 == 28) { pairs += (long long)co[i] * (co[i] - 1) / 2; } else { pairs += (long long)co[i] * co[28 - i]; co[28 - i] = 0; } } cout << pairs << endl; return 0; }

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