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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. #define print(a)         for(auto x : a) cout << x << " "; cout << endl
  6.  
  7.  
  8. const int M = 1000000007;
  9. const int N = 3e5+9;
  10. const int INF = 2e9+1;
  11. const int LINF = 2000000000000000001;
  12.  
  13. inline int power(int a, int b, int mod=M) {
  14.     int x = 1;
  15.     a %= mod;
  16.     while (b) {
  17.         if (b & 1) x = (x * a) % mod; 
  18.         a = (a * a) % mod;
  19.         b >>= 1;
  20.     }
  21.     return x;
  22. }
  23.  
  24.  
  25. //_ ***************************** START Below *******************************
  26.  
  27.  
  28.  
  29. vector<int> a;
  30.  
  31.  
  32. int bruteforce(int n, int k1, int k2){
  33. 	int ans = 0;
  34.  
  35. 	for(int i=0; i<n-2; i++){
  36. 		for(int j=i+1; j<n-2; j++){
  37. 			int leftSum = a[i] + a[j];
  38. 			if(leftSum <= k1) continue;
  39.  
  40. 			for(int k=j+1; k<n-1; k++){
  41. 				for(int l=k+1; l<n; l++){
  42. 					int rightSum = a[k] + a[l];
  43. 					if(rightSum > k2 ) ans++;
  44. 				}
  45. 			}
  46. 		}
  47. 	}
  48. 	return ans;
  49. }
  50.  
  51.  
  52.  
  53.  
  54.  
  55. int consistency1(int n, int k1, int k2) {
  56.  
  57. 	int ans = 0;
  58.  
  59. 	vector<int> pairsFrom(n);
  60. 	for(int i=0; i<n-1; i++){
  61. 		int j = upper_bound(a.begin()+i+1, a.end(), k2-a[i]) - a.begin();
  62.  
  63. 		pairsFrom[i] = n-j;
  64. 	}
  65.  
  66. 	vector<int> sPairs(n+1);
  67. 	for(int i=n-1; i>=0; i--){
  68. 		sPairs[i] = sPairs[i+1] + pairsFrom[i];
  69. 	}
  70.  
  71.  
  72. 	int s = 0, e = n-3;
  73. 	while(s<e){
  74. 		if(a[s] + a[e] <= k1){
  75. 			s++;
  76. 		}
  77. 		else{
  78. 			int left = e-s;
  79. 			int right = sPairs[e+1];
  80.  
  81. 			ans += (left * right);
  82.  
  83. 			e--;
  84. 		}
  85. 	}
  86.  
  87. 	return ans;
  88.  
  89. }
  90.  
  91.  
  92.  
  93.  
  94.  
  95.  
  96.  
  97.  
  98. int consistency2(int n, int k1, int k2) {
  99.  
  100. 	int ans = 0;
  101.  
  102. 	vector<int> pairsFrom(n);
  103. 	for(int i=0; i<n-1; i++){
  104. 		int j = upper_bound(a.begin()+i+1, a.end(), k2-a[i]) - a.begin();
  105.  
  106. 		pairsFrom[i] = n-j;
  107. 	}
  108.  
  109. 	vector<int> sPairs(n+1);
  110. 	for(int i=n-1; i>=0; i--){
  111. 		sPairs[i] = sPairs[i+1] + pairsFrom[i];
  112. 	}
  113.  
  114. 	for(int j=1; j<n-2; j++){
  115. 		int i = upper_bound(a.begin(), a.begin() + j, k1-a[j]) - a.begin();
  116. 		int left= j-i;
  117.  
  118. 		int right = sPairs[j+1];
  119.  
  120. 		ans += left*right;
  121. 	}
  122.  
  123. 	return ans;
  124. }
  125.  
  126.  
  127.  
  128.  
  129.  
  130.  
  131.  
  132.  
  133.  
  134.  
  135.  
  136.  
  137.  
  138.  
  139.  
  140.  
  141.  
  142.  
  143.  
  144. int practice(int n, int k1, int k2) {
  145.  
  146. 	return 0;
  147.  
  148. }
  149.  
  150.  
  151.  
  152.  
  153.  
  154.  
  155. void solve() {
  156.  
  157. 	int n, k1, k2;
  158. 	cin >> n >> k1 >> k2;
  159.  
  160. 	a.resize(n);
  161. 	for(int i=0; i<n; i++) cin >> a[i];
  162.  
  163.     cout << bruteforce(n, k1, k2) << " -> ";
  164.     cout << consistency1(n, k1, k2) << " ";
  165.     cout << consistency2(n, k1, k2) << endl;
  166.  
  167.  
  168.     // cout << bruteforce(n, k1, k2) << " -> ";
  169.     // cout << practice(n, k1, k2) << endl;
  170.  
  171. }
  172.  
  173.  
  174.  
  175.  
  176.  
  177. int32_t main() {
  178.     ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  179.  
  180.     int t = 1;
  181.     cin >> t;
  182.     while (t--) {
  183.         solve();
  184.     }
  185.  
  186.     return 0;
  187. }
Success #stdin #stdout 0s 5324KB
comments (?)
stdin
copy 
2
9 4 4
1 2 2 3 3 4 5 5 5
8 9 7
1 2 3 4 5 6 8 12
stdout
copy 
59 -> 59 59
2 -> 2 2
https://ideone.com/x7vPRY
language:
C++14 (gcc 8.3)
created:
14 hours ago
visibility:
public

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