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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.  
  66. void solve() {
  67.     int n,k;
  68.     cin>>n>>k;
  69.     int A[n];
  70.     for(int i = 0 ; i<n ; i++){
  71.         cin>>A[i];
  72.     }
  73.     stack<pair<int,int>>st,st3;
  74.     int NGL[n],NGR2[n];
  75.     for(int i = 0 ; i<n ; i++){
  76.         while(!st.empty() && st.top().first<=A[i]){
  77.            st.pop();
  78.         }
  79.         if(st.empty()){
  80.             NGL[i] = -1;
  81.         }
  82.         else{
  83.             NGL[i] = st.top().second;
  84.         }
  85.         st.push({A[i],i});
  86.     }
  87.     for(int i = n-1 ; i>=0 ; i--){
  88.         while(!st3.empty() && st3.top().first<A[i]){
  89.             st3.pop();
  90.         }
  91.         if(st3.empty()){
  92.             NGR2[i] = n;
  93.         }
  94.         else{
  95.             NGR2[i] = st3.top().second;
  96.         }
  97.         st3.push({A[i],i});
  98.     }
  99.     int sum = 0;
  100.     for(int i = 0 ; i<n ; i++){
  101.         if(A[i]==k){
  102.             int x = (i-NGL[i]-1);
  103.             int y = (NGR2[i]-i-1);
  104.             sum += ((x*y)+x+y+1);
  105.         }
  106.     }
  107.     cout<<sum<<endl;
  108. }
  109. signed main() {
  110.     ios::sync_with_stdio(false); cin.tie(NULL);
  111.     int t;
  112.     cin >> t;
  113.     while (t--) {
  114.         solve();
  115.     }
  116.     return 0;
  117. }
  118.  
Success #stdin #stdout 0.01s 5316KB
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https://ideone.com/XnCTSu
language:
C++14 (gcc 8.3)
created:
3 days ago
visibility:
public

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