#include <bits/stdc++.h> using namespace std; using ll = long long; const int N = 1e6 + 7; int per[N]; int main() { int n; scanf("%d", &n); for (int i = 1; i <= n; ++i) { int a; scanf("%d", &a); per[a] = i; } int l = n + 1, r = 0; ll ans = 0; for (int k = 1; k <= n; ++k) { int p = per[n - k / 2]; l = min(l, p); r = max(r, p); int d = r - l + 1; if (d > k) continue; int need = k - d; int ma = min(l - 1, need); int mi = max(0, need - (n - r)); ans += ma - mi + 1; } printf("%d %lld\n", 2 * n + 1, ans); return 0; }
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 | #include <bits/stdc++.h> using namespace std; using ll = long long; const int N = 1e6 + 7; int per[N]; int main() { int n; scanf("%d", &n); for (int i = 1; i <= n; ++i) { int a; scanf("%d", &a); per[a] = i; } int l = n + 1, r = 0; ll ans = 0; for (int k = 1; k <= n; ++k) { int p = per[n - k / 2]; l = min(l, p); r = max(r, p); int d = r - l + 1; if (d > k) continue; int need = k - d; int ma = min(l - 1, need); int mi = max(0, need - (n - r)); ans += ma - mi + 1; } printf("%d %lld\n", 2 * n + 1, ans); return 0; } |