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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. vector<int> a;
  26.  
  27.  
  28. int bruteforce(int n, int k1, int k2){
  29. 	int ans = 0;
  30.  
  31. 	for(int i=0; i<n-2; i++){
  32. 		for(int j=i+1; j<n-2; j++){
  33. 			int leftSum = a[i] + a[j];
  34. 			if(leftSum <= k1) continue;
  35.  
  36. 			for(int k=j+1; k<n-1; k++){
  37. 				for(int l=k+1; l<n; l++){
  38. 					int rightSum = a[k] + a[l];
  39. 					if(rightSum > k2 ) ans++;
  40. 				}
  41. 			}
  42. 		}
  43. 	}
  44. 	return ans;
  45. }
  46.  
  47.  
  48.  
  49.  
  50.  
  51.  
  52. //* Left Count Pre Computation  : 
  53.  
  54. //* Eg : [ 1 , 2 , 3 , 4 , 5 , 6 , 8 ]   k1 = 9
  55.  
  56. //* Ending at analogy : 
  57. //* 	We are finding pairs (x,y) with sum x+y == k1 ending at y
  58.  
  59. //* Observation : if we drew out patterns for each pairs , it looks like this : 
  60. //*      [ (1 , 2 , (3 , (4 , 5) , 6) , 8) ]   
  61. //*     	y = 1,2,3,4 can't form pair with any x such that x+y >= 9
  62. //* 	    y = 5 forms pair with x = 4 only               => ct = 1
  63. //* 		y = 6 forms pair with x = (3, 4, 5)            => ct = 3
  64. //* 		y = 8 forms pair with x = (1, 2, 3, 4, 5, 6)   => ct = 6
  65.  
  66. //* 	Now we can use 2 ptr from 1st pair [x, x+1] such that x + x + 1 >= k1
  67. //* 	                i   j
  68. //*      [ 1 , 2 , 3 , (4 , 5) , 6 , 8) ]   
  69. //* 			p[5] = max(0, 2) = 2
  70. //* 		Now we want to maximize p[5] cts so move i left => i--
  71.  
  72. //* 	           i        j
  73. //*      [ 1 , 2 , 3 , (4 , 5) , 6 , 8) ]   
  74. //* 			5+3 < k1
  75. //* 		Invalid so move to next y => j++
  76.  
  77. //* 	                i       j
  78. //*      [ 1 , 2 , 3 , (4 , 5 , 6) , 8) ]   
  79. //* 			p[6] = max(0, 3) = 3
  80.  
  81. //* 	            i           j
  82. //*      [ 1 , 2 , (3 , 4 , 5 , 6) , 8) ]   
  83. //* 			p[6] = max(3, 4) = 4
  84.  
  85. //* 	        i               j
  86. //*      [ 1 , (2 , 3 , 4 , 5 , 6) , 8) ]   
  87. //* 			Invalid => j++
  88.  
  89. //* O(n + n + n ) = O(n)
  90. int consistency1(int n, int k1, int k2) {
  91.  
  92. 	int ans = 0;
  93.  
  94. 	vector<int> p(n);
  95.  
  96. 	vector<int> leftPrefix(n);
  97. 	int i=1;
  98. 	while(i<n){
  99. 		if(a[i] + a[i-1] >= k1) break;
  100. 		i++;
  101. 	}
  102. 	int left = i-1;
  103. 	int right = i;
  104.  
  105. 	while(left>=0 && right<n){
  106. 		if(a[left] + a[right] > k1){
  107. 			p[right] = max(p[right], right-left);
  108. 			left--;
  109. 		}
  110. 		else {
  111. 			right++;
  112. 		}
  113. 	}
  114. 	for(int i=1; i<n; i++){
  115. 		leftPrefix[i] = leftPrefix[i-1] + p[i];
  116. 	}
  117.  
  118.  
  119.  
  120. 	vector<int> q(n);
  121.  
  122. 	vector<int> rightSufix(n);
  123. 	// for(int i=n-1; i>=0; i--){
  124. 	// 	rightSufix[i] = rightSufix[i+1] + q[i];
  125. 	// }
  126.  
  127. 	for(int j=1; j<n-2; j++){
  128. 		int leftCt = leftPrefix[j];
  129. 		int rightCt = rightSufix[j+1]; 
  130.  
  131. 		ans += leftCt * rightCt;
  132. 	}
  133.  
  134.  
  135. 	return ans;
  136.  
  137. }
  138.  
  139.  
  140.  
  141.  
  142.  
  143.  
  144.  
  145.  
  146.  
  147.  
  148.  
  149.  
  150.  
  151.  
  152.  
  153.  
  154.  
  155.  
  156.  
  157. int practice(int n, int k1, int k2) {
  158.  
  159. 	return 0;
  160.  
  161. }
  162.  
  163.  
  164. void solve() {
  165.  
  166. 	int n, k1, k2;
  167. 	cin >> n >> k1 >> k2;
  168.  
  169. 	a.resize(n);
  170. 	for(int i=0; i<n; i++) cin >> a[i];
  171.  
  172.     cout << bruteforce(n, k1, k2) << " ";
  173.     cout << consistency1(n, k1, k2) << endl;
  174.  
  175.  
  176.     // cout << bruteforce(n, k1, k2) << " -> ";
  177.     // cout << practice(n, k1, k2) << endl;
  178.  
  179. }
  180.  
  181.  
  182.  
  183.  
  184.  
  185. int32_t main() {
  186.     ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  187.  
  188.     int t = 1;
  189.     cin >> t;
  190.     while (t--) {
  191.         solve();
  192.     }
  193.  
  194.     return 0;
  195. }
Success #stdin #stdout 0s 5320KB
comments (?)
stdin
copy 
4
8 9 7
1 2 3 4 5 6 8 12
6 1 3
0 1 1 1 2 2
6 1 3
1 1 1 1 2 2
6 3 3
1 2 2 2 2 3
stdout
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2 0
3 0
6 0
5 0
https://ideone.com/O7HJUk
language:
C++14 (gcc 8.3)
created:
3 days ago
visibility:
public

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