Comments

• 
  

  -9

  

  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; }

  This comment is hidden due to too much negative feedback. Show it anyway.

• 
  

  -4

  

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

  time erorr

• 
  

  -3

  

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

  hello world

• 
  

  1

  

  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)

• 
  

  -16

  

  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; }

  This comment is hidden due to too much negative feedback. Show it anyway.