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  1. #include<bits/stdc++.h>
  2. using namespace std;
  3. #define endl '\n'
  4. #define int long long int
  5. const int MOD = 1000000007;
  6. const int MOD2 = 998244353;
  7. const int INF = LLONG_MAX / 2;
  8. const int MAXN = 100000;
  9. int primes[1000000];
  10.  
  11. /*void seive() {
  12.   fill(primes, primes + 1000000, 1);
  13.   primes[0] = primes[1] = 0;
  14.   for (int i = 2; i * i < 1000000; i++) {
  15.   if (primes[i]) {
  16.   for (int j = i * i; j < 1000000; j += i) {
  17.   primes[j] = 0;
  18.   }
  19.   }
  20.   }
  21. }
  22.  
  23. bool isPrime(int n) {
  24.   if (n <= 1) return false;
  25.   for (int i = 2; i * i <= n; i++) {
  26.   if (n % i == 0) return false;
  27.   }
  28.   return true;
  29. }
  30.  
  31. int gcd(int a, int b) {
  32.   if (a == 0) return b;
  33.   return gcd(b % a, a);
  34. }*/
  35.  
  36. int power(int a, int b, int mod) {
  37. int res = 1;
  38. a %= mod;
  39. while (b > 0) {
  40. if (b & 1) res = res * a % mod;
  41. a = a * a % mod;
  42. b >>= 1;
  43. }
  44. return res;
  45. }
  46.  
  47. // nCr % MOD for n < MOD
  48. int nCrModP(int n, int r) {
  49. if (r > n) return 0;
  50. if (r == 0 || r == n) return 1;
  51.  
  52. int numerator = 1, denominator = 1;
  53. for (int i = 0; i < r; i++) {
  54. numerator = (numerator * (n - i)) % MOD;
  55. denominator = (denominator * (i + 1)) % MOD;
  56. }
  57. return (numerator * power(denominator, MOD - 2, MOD)) % MOD;
  58. }
  59.  
  60. // Lucas's Theorem
  61. int lucas(int n, int r) {
  62. if (r == 0) return 1;
  63. return (lucas(n / MOD, r / MOD) * nCrModP(n % MOD, r % MOD)) % MOD;
  64. }
  65. void solve() {
  66. int n,k1,k2;
  67. cin>>n>>k1>>k2;
  68. int A[n];
  69. for(int i = 0 ; i<n ; i++){
  70. cin>>A[i];
  71. }
  72. int cnt = 0;
  73. for(int j = 1 ; j<n-2 ; j++){
  74. int cnt1 = 0;
  75. for(int i = 0 ; i<j ; i++){
  76. if(A[i]+A[j]>k1){
  77. cnt1++;
  78. }
  79. }
  80. int cnt2 = 0;
  81. int k = j+1;
  82. int l = n-1;
  83. while(k<=l){
  84. if(A[k]+A[l]>k2){
  85. cnt2 += (l-k);
  86. l--;
  87. }
  88. else{
  89. k++;
  90. }
  91. }
  92. cnt += (cnt1*cnt2);
  93. }
  94. cout<<cnt<<endl;
  95. }
  96.  
  97. signed main() {
  98. ios::sync_with_stdio(false); cin.tie(NULL);
  99. //int t;
  100. //cin >> t;
  101. //while (t--) {
  102. solve();
  103. //}
  104. return 0;
  105. }
  106.  
Success #stdin #stdout 0.01s 5320KB
stdin
6 1 3
1 1 1 1 2 2
stdout
6