結果

問題 No.1234 典型RMQ
ユーザー koba-e964
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
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1 WA * 2
other AC * 1 WA * 26
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#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; // 317
ll 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 interval
REP(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;
}
}
}
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