#include <bits/stdc++.h> #include <chrono> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace std; using namespace chrono; using namespace __gnu_pbds; #pragma GCC optimize("Ofast,unroll-loops") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma,tune=native") #pragma GCC target("sse,sse2,sse3,mmx,abm,tune=native") // for szkopul and sio only typedef long long lld; typedef double lf; typedef long double llf; typedef pair<int,int> pii; typedef pair<lld,lld> pll; typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset; #define For(i,s,a) for(lld i = (lld)s; i < (lld)a; ++i) #define rpt(s, it) for(auto it = s.begin(); it != s.end(); ++it) #define brpt(s, it) for(auto it = s.rend(); it != s.rbegin(); --it) #define pb push_back #define eb emplace_back #define ff first #define dd second #define mp make_pair #define all(x) (x).begin(), (x).end() #define make_unique(x) (x).erase( unique(all(x)), (x).end()) #define popcnt(x) __builtin_popcount(x) #define sz size() #define time_since duration_cast< milliseconds >(system_clock::now().time_since_epoch()) template<typename Ta, typename Tb> ostream & operator <<(ostream & os, pair<Ta, Tb> x){ return os << x.ff << " " << x.dd; } const int N = 1e6 + 1; int c[N]; int pos[N]; oset xd; void solve(void) { int n; scanf("%d", &n); For(i, 0, n) { scanf("%d", &c[i]); //c[i] = 2 * i < n ? 2 * i + 1 : (2 * (n - i)); //cout << c[i] << " "; } //puts(""); printf("%d ", 2 * n + 1); For(i, 0, n) pos[c[i]] = i; lld wyn = 0; int mx = pos[n], mn = pos[n]; int kt = n; for(int i = n; i; --i) { if(i % 2 != n % 2) mx = max(mx, pos[--kt]), mn = min(mn, pos[kt]); if(mx - mn > n - i) continue; //wyn += min(n - i - (mx - mn) + 1, min(n - mx, mn + 1)); ++wyn; wyn += max(0, min(mn, n - 1 - (n - i)) - max(0, mx - (n - i))); // cout << i << " " << mn << " " << mx << " " << n - 1 - (n - i) << " " << mx - (n - i) << " " << wyn << endl; } printf("%lld\n", wyn); } int32_t main(void){ int t = 1; //scanf("%d", &t); while(t--) solve(); }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 | #include <bits/stdc++.h> #include <chrono> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace std; using namespace chrono; using namespace __gnu_pbds; #pragma GCC optimize("Ofast,unroll-loops") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma,tune=native") #pragma GCC target("sse,sse2,sse3,mmx,abm,tune=native") // for szkopul and sio only typedef long long lld; typedef double lf; typedef long double llf; typedef pair<int,int> pii; typedef pair<lld,lld> pll; typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset; #define For(i,s,a) for(lld i = (lld)s; i < (lld)a; ++i) #define rpt(s, it) for(auto it = s.begin(); it != s.end(); ++it) #define brpt(s, it) for(auto it = s.rend(); it != s.rbegin(); --it) #define pb push_back #define eb emplace_back #define ff first #define dd second #define mp make_pair #define all(x) (x).begin(), (x).end() #define make_unique(x) (x).erase( unique(all(x)), (x).end()) #define popcnt(x) __builtin_popcount(x) #define sz size() #define time_since duration_cast< milliseconds >(system_clock::now().time_since_epoch()) template<typename Ta, typename Tb> ostream & operator <<(ostream & os, pair<Ta, Tb> x){ return os << x.ff << " " << x.dd; } const int N = 1e6 + 1; int c[N]; int pos[N]; oset xd; void solve(void) { int n; scanf("%d", &n); For(i, 0, n) { scanf("%d", &c[i]); //c[i] = 2 * i < n ? 2 * i + 1 : (2 * (n - i)); //cout << c[i] << " "; } //puts(""); printf("%d ", 2 * n + 1); For(i, 0, n) pos[c[i]] = i; lld wyn = 0; int mx = pos[n], mn = pos[n]; int kt = n; for(int i = n; i; --i) { if(i % 2 != n % 2) mx = max(mx, pos[--kt]), mn = min(mn, pos[kt]); if(mx - mn > n - i) continue; //wyn += min(n - i - (mx - mn) + 1, min(n - mx, mn + 1)); ++wyn; wyn += max(0, min(mn, n - 1 - (n - i)) - max(0, mx - (n - i))); // cout << i << " " << mn << " " << mx << " " << n - 1 - (n - i) << " " << mx - (n - i) << " " << wyn << endl; } printf("%lld\n", wyn); } int32_t main(void){ int t = 1; //scanf("%d", &t); while(t--) solve(); } |