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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-3; 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. //* O(n^2)
  52.  
  53. //* [ 1 2 2 2 3 ] , k1 = 3, k2 = 3
  54.  
  55. //*     i      j   | s e
  56. //* [ ( 1 2 2  2 ) |(2 3) ]
  57. //*     	2      *  1
  58.  
  59. //*       i j   | s   e
  60. //* [ 1 ( 2 2 ) |(2 2 3) ]
  61. //*        1    *   3
  62.  
  63. int consistency1(int n, int k1, int k2) {
  64.  
  65. 	int ans = 0;
  66.  
  67. 	int i = 0, j = n-3;
  68.  
  69. 	while(i<j){
  70. 		if(a[i] + a[j] <= k1){
  71. 			i++;
  72. 		}
  73. 		else{
  74. 			int right = 0;
  75. 			int s = j+1;
  76. 			int e = n-1;
  77. 			while(s<e){
  78. 				if(a[s]+a[e] <= k2){
  79. 					s++;
  80. 				}
  81. 				else{
  82. 					right += e-s;
  83. 					e--;
  84. 				}
  85. 			}
  86.  
  87. 			int left = j-i;
  88.  
  89. 			ans += left*right;
  90.  
  91. 			j--;
  92. 		}
  93. 	}
  94.  
  95.  
  96. 	return ans;
  97.  
  98. }
  99.  
  100.  
  101. //* O(n^2)
  102. int consistency2(int n, int k1, int k2) {
  103.  
  104. 	int ans = 0;
  105. 	for(int j=1; j<n-2; j++){
  106. 		int i=0;
  107. 		while(i<j && a[i]+a[j] <= k1) i++;
  108. 		int left = j-i;
  109.  
  110. 		int  s= j+1, e = n-1;
  111. 		int right = 0;
  112. 		while(s<e){
  113. 			if(a[s]+a[e] > k2){
  114. 				right += (e-s);
  115. 				e--;
  116. 			}
  117. 			else{
  118. 				s++;
  119. 			}
  120. 		}
  121. 		ans += left * right;
  122. 	}
  123.  
  124. 	return ans;
  125. }
  126.  
  127.  
  128.  
  129.  
  130. //* O(n*logn + n + n*logn) == O(n*logn)
  131. int consistency3(int n, int k1, int k2) {
  132.  
  133. 	int ans = 0;
  134.  
  135. 	vector<int> p(n);
  136. 	for(int i=0; i<n-1; i++){
  137. 		int target = k2 - a[i];
  138. 		int j = upper_bound(a.begin() + i + 1, a.end(), target) - a.begin();
  139. 		p[i] = n-j;
  140. 	}
  141. 	vector<int> rightSufix(n);
  142. 	for(int i=n-2; i>=0; i--){
  143. 		rightSufix[i] = rightSufix[i+1] + p[i];
  144. 	}
  145.  
  146.  
  147. 	for(int j=1; j<n-2; j++){
  148. 		int target = k1 - a[j];
  149. 		int i = upper_bound(a.begin(), a.begin()+j, target) - a.begin();
  150. 		int leftCt = j-i;
  151.  
  152. 		int rightCt = rightSufix[j+1];
  153.  
  154. 		ans += leftCt * rightCt;
  155.  
  156. 	}
  157.  
  158.  
  159. 	return ans;
  160.  
  161. }
  162.  
  163.  
  164. //* Left Count Pre Computation  : 
  165.  
  166. //* Eg : [ 1 , 2 , 3 , 4 , 5 , 6 , 8 ]   k1 = 9
  167.  
  168. //* Ending at analogy : 
  169. //* 	We are finding pairs (x,y) with sum x+y == k1 ending at y
  170.  
  171. //* Observation : if we drew out patterns for each pairs , it looks like this : 
  172. //*      [ (1 , 2 , (3 , (4 , 5) , 6) , 8) ]   
  173. //*     	y = 1,2,3,4 can't form pair with any x such that x+y >= 9
  174. //* 	    y = 5 forms pair with x = 4 only               => ct = 1
  175. //* 		y = 6 forms pair with x = (3, 4, 5)            => ct = 3
  176. //* 		y = 8 forms pair with x = (1, 2, 3, 4, 5, 6)   => ct = 6
  177.  
  178. //* 	Now we can use 2 ptr from 1st pair [x, x+1] such that x + x + 1 >= k1
  179. //* 	                i   j
  180. //*      [ 1 , 2 , 3 , (4 , 5) , 6 , 8) ]   
  181. //* 			p[5] = max(0, 2) = 2
  182. //* 		Now we want to maximize p[5] cts so move i left => i--
  183.  
  184. //* 	           i        j
  185. //*      [ 1 , 2 , 3 , (4 , 5) , 6 , 8) ]   
  186. //* 			5+3 < k1
  187. //* 		Invalid so move to next y => j++
  188.  
  189. //* 	                i       j
  190. //*      [ 1 , 2 , 3 , (4 , 5 , 6) , 8) ]   
  191. //* 			p[6] = max(0, 3) = 3
  192.  
  193. //* 	            i           j
  194. //*      [ 1 , 2 , (3 , 4 , 5 , 6) , 8) ]   
  195. //* 			p[6] = max(3, 4) = 4
  196.  
  197. //* 	        i               j
  198. //*      [ 1 , (2 , 3 , 4 , 5 , 6) , 8) ]   
  199. //* 			Invalid => j++
  200.  
  201. //* O(n + n + n ) = O(n)
  202. int consistency4(int n, int k1, int k2) {
  203.  
  204. 	int ans = 0;
  205.  
  206. 	vector<int> p(n);
  207.  
  208. 	vector<int> leftPrefix(n);
  209. 	int i=1;
  210. 	while(i<n){
  211. 		if(a[i] + a[i-1] >= k1) break;
  212. 		i++;
  213. 	}
  214. 	int left = i-1;
  215. 	int right = i;
  216.  
  217. 	while(left>=0 && right<n){
  218. 		if(a[left] + a[right] > k1){
  219. 			p[right] = max(p[right], right-left);
  220. 			left--;
  221. 		}
  222. 		else {
  223. 			right++;
  224. 		}
  225. 	}
  226. 	for(int i=1; i<n; i++){
  227. 		leftPrefix[i] = leftPrefix[i-1] + p[i];
  228. 	}
  229.  
  230.  
  231.  
  232. 	vector<int> q(n);
  233.  
  234. 	vector<int> rightSufix(n);
  235. 	// for(int i=n-1; i>=0; i--){
  236. 	// 	rightSufix[i] = rightSufix[i+1] + q[i];
  237. 	// }
  238.  
  239. 	for(int j=1; j<n-2; j++){
  240. 		int leftCt = leftPrefix[j];
  241. 		int rightCt = rightSufix[j+1]; 
  242.  
  243. 		ans += leftCt * rightCt;
  244. 	}
  245.  
  246.  
  247. 	return ans;
  248.  
  249. }
  250.  
  251.  
  252.  
  253.  
  254.  
  255.  
  256.  
  257.  
  258.  
  259.  
  260.  
  261.  
  262.  
  263.  
  264.  
  265.  
  266.  
  267.  
  268.  
  269. int practice(int n, int k1, int k2) {
  270.  
  271. 	int ans = 0;
  272.  
  273.  
  274.  
  275.  
  276. 	return ans;
  277.  
  278. }
  279.  
  280.  
  281. void solve() {
  282.  
  283. 	int n, k1, k2;
  284. 	cin >> n >> k1 >> k2;
  285.  
  286. 	a.resize(n);
  287. 	for(int i=0; i<n; i++) cin >> a[i];
  288.  
  289.     cout << bruteforce(n, k1, k2) << " ";
  290.     cout << consistency1(n, k1, k2) << " ";
  291.     cout << consistency2(n, k1, k2) << " ";
  292.     cout << consistency3(n, k1, k2) << endl;
  293.  
  294.  
  295.     cout << bruteforce(n, k1, k2) << " -> ";
  296.     cout << practice(n, k1, k2) << endl;
  297.  
  298. }
  299.  
  300.  
  301.  
  302.  
  303.  
  304. int32_t main() {
  305.     ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
  306.  
  307.     int t = 1;
  308.     cin >> t;
  309.     while (t--) {
  310.         solve();
  311.     }
  312.  
  313.     return 0;
  314. }
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
copy 
2 2 2 2
2 -> 0
3 3 3 3
3 -> 0
6 6 6 6
6 -> 0
5 5 5 5
5 -> 0
https://ideone.com/Xwge89
language:
C++14 (gcc 8.3)
created:
2 hours ago
visibility:
public

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