結果
問題 | No.1234 典型RMQ |
ユーザー |
|
提出日時 | 2016-10-21 15:03:08 |
言語 | C++11(廃止可能性あり) (gcc 13.3.0) |
結果 |
WA
(最新)
AC
(最初)
|
実行時間 | - |
コード長 | 2,221 bytes |
コンパイル時間 | 834 ms |
コンパイル使用メモリ | 88,372 KB |
実行使用メモリ | 5,376 KB |
最終ジャッジ日時 | 2024-07-20 00:44:39 |
合計ジャッジ時間 | 8,650 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 WA * 2 |
other | AC * 1 WA * 26 |
ソースコード
#include <algorithm>#include <bitset>#include <cassert>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <deque>#include <functional>#include <iomanip>#include <iostream>#include <list>#include <map>#include <numeric>#include <queue>#include <set>#include <sstream>#include <stack>#include <string>#include <utility>#include <vector>#define REP(i,s,n) for(int i=(int)(s);i<(int)(n);i++)using namespace std;typedef long long int ll;typedef vector<int> VI;typedef vector<ll> VL;typedef pair<int, int> PI;const ll mod = 1e9 + 7;const int B = 317;const int N = 100489;const int K = N / B; // 317ll a[N];ll delay[B];ll mi[B];int main(void){int n;cin >> n;assert (1 <= n && n <= 100000);REP(i, 0, n) {cin >> a[i];assert (a[i] >= -1e10 && a[i] <= 1e10);}REP(i, 0, B) {ll m = a[i * K];REP(j, 1, K) {m = min(m, a[i * K + j]);}mi[i] = m;}int q;cin >> q;assert (1 <= q && q <= 100000);REP(i, 0, q) {int k, l, r;ll c;assert (cin >> k >> l >> r >> c);assert (k == 1 || k == 2);assert (1 <= l && l <= r && r <= n);assert (-10000 <= c && c <= 10000);l--; // [l,r)int l_boundary = l / K;int r_boundary = r / K;// [l,r) = [l,K*l_boundary) + [K*l_boundary,K*r_boundary) + [K*r_boundary,r)if (k == 1) { // update intervalREP(i, l, K * l_boundary) {a[i] += c;}if (l_boundary >= 1) {ll m = a[K * (l_boundary - 1)];REP(i, 0, K) {m = min(m, a[K * (l_boundary - 1) + i]);}mi[l_boundary - 1] = m;}REP(i, l_boundary, r_boundary) {delay[i] += c;}REP(i, K * r_boundary, r) {a[i] += c;}if (r_boundary <= n - 1) {ll m = a[K * r_boundary];REP(i, 0, K) {m = min(m, a[K * r_boundary + i]);}mi[r_boundary] = m;}}if (k == 2) {ll m = 1e16;REP(i, l, K * l_boundary) {m = min(m, a[i] + delay[i / K]);}REP(i, l_boundary, r_boundary) {m = min(m, mi[i] + delay[i]);}REP(i, K * r_boundary, r) {m = min(m, a[i] + delay[i / K]);}cout << m << endl;}}}