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  1. #include <bits/stdc++.h>
  2. using namespace std;
  3. #define int              long long int
  4. #define double           long double
  5. inline int power(int a, int b) {
  6.     int x = 1;
  7.     while (b) {
  8.         if (b & 1) x *= a;
  9.         a *= a;
  10.         b >>= 1;
  11.     }
  12.     return x;
  13. }
  14.  
  15.  
  16. const int M = 1000000007;
  17. const int N = 3e5+9;
  18. const int INF = 2e9+1;
  19. const int LINF = 2000000000000000001;
  20.  
  21. //_ ***************************** START Below *******************************
  22.  
  23.  
  24.  
  25.  
  26. //? Good Subarrays Hashing
  27.  
  28. //* We need to find subarrays : (P[R] - P[L-1])%k == 0
  29. //*                             P[R]%k == P[L-1]%k
  30. //*     Hence for P[R]%k find P[R]%k in past
  31.  
  32.  
  33. //* Eg     :    [4,  5,  0, -2, -3,  1]   , k = 5
  34. //* sum    :  0  4   9   9   7   4   5     
  35. //* sum%m  :  0  4   4   4   2   4   0     
  36. //* ct     :  1  1   2   3   1   4   2
  37. //* ans    :         1 + 2 + 0 + 3 + 1
  38.  
  39.  
  40. vector<int> a;
  41. void consistency(int n, int k) {
  42.  
  43. 	unordered_map<int,int> mp = {{0, 1}};
  44. 	int sum = 0;
  45. 	int ans = 0;
  46. 	for(int i=0; i<n; i++){
  47. 	    sum += a[i];
  48. 	    int rem = ((sum%k)+k)%k;
  49. 	    if(mp.count(rem)){
  50. 	        ans += mp[rem];
  51. 	    }
  52. 	    mp[rem]++;
  53. 	}
  54.  
  55. 	cout << ans << endl;
  56.  
  57. }
  58.  
  59. void solve() {
  60.  
  61. 	int n, k;
  62. 	cin >> n >> k;
  63. 	a.resize(n);
  64. 	for(int i=0; i<n; i++) cin >> a[i];
  65.     consistency(n, k) ;
  66.  
  67. }
  68.  
  69.  
  70.  
  71.  
  72.  
  73. int32_t main() {
  74.     ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  75.  
  76.     int t = 1;
  77.     while (t--) {
  78.         solve();
  79.     }
  80.  
  81.     return 0;
  82. }
Success #stdin #stdout 0.01s 5276KB
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https://ideone.com/MUOrsQ
language:
C++14 (gcc 8.3)
created:
10 hours ago
visibility:
public

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