Submission #128124

#TimeUsernameProblemLanguageResultExecution timeMemory
128124RockyBWorst Reporter 3 (JOI18_worst_reporter3)C++17
100 / 100
847 ms30788 KiB
#include <bits/stdc++.h> #define int ll #define f first #define s second #define pb push_back #define pp pop_back #define all(x) x.begin(), x.end() #define sz(x) (int)x.size() #define rep(a, b, c) for (int a = (b); (a) <= (c); ++a) #define per(a, b, c) for (int a = (b); (a) >= (c); --a) #define nl '\n' #define ioi exit(0); using namespace std; typedef long long ll; const int MAX_N = (int)5e5 + 7; int n, q; int d[MAX_N], l[MAX_N], r[MAX_N], t[MAX_N], ans[MAX_N]; map <int, vector <int> > off; int nxt[MAX_N], lst[MAX_N]; void prep() { int last = 1; rep(i, 1, n) { int df = (d[i] + last - 1) / last; lst[i] = df * last; last *= df; } per(i, n, 1) { if (lst[i] != lst[i + 1] || i == n) nxt[i] = i + 1; else nxt[i] = nxt[i + 1]; } } pair < pair <int, int>, ll> a[MAX_N]; void solve(int T, const vector <int> &qr) { int add = T; int sz = 0; int last = 1; for (int i = 1, j = 0; i <= n; j = i + 1, i = nxt[i]) { int df = (d[i] + last - 1) / last; if (!sz) a[sz++] = {{0, i - 1}, i - 1}; else a[sz++] = {{j, i - 1}, {i - j}}; a[sz++] = {{i, i}, 1 + last * (add % df)}; last *= df; add /= df; if (!add || nxt[i] > n) { if (i < n) a[sz++] = {{i + 1, n}, {n - i}}; break; } } for (auto it : qr) { int L = -1, R = -2; if (T >= l[it]) { // it's for right int ost = T - l[it]; if (sz) { R = min(ost, a[0].s); } } int sum = 0; rep(i, 0, sz - 1) { sum += a[i].s; int val = T - sum; if (val >= l[it]) { R = a[i].f.s; int ost = val - l[it]; if (i & 1 && i + 1 < sz) { R += min(ost, a[i + 1].s); } } if (val <= r[it] && L == -1) { L = a[i].f.s; if (i % 2 == 0) { L -= max(0ll, min(r[it] - val, a[i].s - 1)); } } } if (L != -1 && L <= R) ans[it] += R - L + 1; } } int32_t main() { #ifdef IOI freopen ("in.txt", "r", stdin); freopen ("C.out", "w", stdout); #endif ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0); cin >> n >> q; d[0] = 1; rep(i, 1, n) { cin >> d[i]; } prep(); rep(i, 1, q) { cin >> t[i] >> l[i] >> r[i]; solve(t[i], {i}); } rep(i, 1, q) { cout << ans[i] << nl; } ioi }
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