結果

問題 No.3059 Range Tournament
ユーザー The Forsaking
提出日時 2025-03-30 22:21:14
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 624 ms / 4,000 ms
コード長 3,508 bytes
コンパイル時間 1,340 ms
コンパイル使用メモリ 118,356 KB
実行使用メモリ 57,936 KB
最終ジャッジ日時 2025-03-30 22:21:35
合計ジャッジ時間 16,647 ms
ジャッジサーバーID
(参考情報)
judge2 / judge4
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 * 2
other AC * 27
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:81:42: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   81 |     for (int i = 1; i < n + 1; i++) scanf("%d", w + i), f[i][0] = w[i];
      |                                     ~~~~~^~~~~~~~~~~~~
main.cpp:90:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   90 |         scanf("%d%d", &l, &r);
      |         ~~~~~^~~~~~~~~~~~~~~~

ソースコード

diff #

#include <iostream>
#include <sstream>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <vector>
#include <queue> 
#include <unordered_set>
#include <unordered_map>
#include <bitset>
#include <ctime>
#include <assert.h>
#include <deque>
#include <list>
#include <stack>
#include <numeric>
#include <iomanip>

using namespace std;
 
typedef pair<long long, int> pli;
typedef pair<int, long long> pil;
typedef pair<long long , long long> pll;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef pair<int, pii> piii;
typedef pair<int, long long > pil;
typedef pair<long long, pii> plii;
typedef pair<double, int> pdi;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ull, ull> puu;
typedef long double ld;
const int N = 2000086, MOD = 1e9 + 7, INF = 0x3f3f3f3f, MID = 333;
const long double EPS = 1e-9;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
// int dx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
// int dx[8] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[8] = {1, 2, 2, 1, -1, -2, -2, -1};
int n, m, cnt;
int w[N];
vector<ll> num;
ll res;

ll lowbit(ll x) { return x & -x; }
ll lcm(ll a, ll b) { return a / __gcd(a, b) * b; }
inline double rand(double l, double r) { return (double)rand() / RAND_MAX * (r - l) + l; }
inline ll qmi(ll a, ll b, ll c) { ll res = 1; while (b) { if (b & 1) res = res * a % c; a = a * a % c; b >>= 1; } return res; }
inline ll qmi(ll a, ll b) { ll res = 1; while (b) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; }
inline double qmi(double a, ll b) { double res = 1; while (b) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; } 
inline ll C(ll a, ll b, int* c) { if (a < b) return 0; ll res = 1; for (ll j = a, i = 1; i < b + 1; i++, j--) res *= j; for (ll j = a, i = 1; i < b + 1; i++, j--) res /= i; return res; }
inline int find_(int x) { return lower_bound(num.begin(), num.end(), x) - num.begin(); }



const int logn = 19;
const int maxn = 200008;
int f[maxn][logn + 1], Logn[maxn + 1];
int ans[N];

void pre() {
 Logn[1] = 0;
 Logn[2] = 1;
 for (int i = 3; i < maxn; i++) {
  Logn[i] = Logn[i / 2] + 1;
 }
}

int query(int l, int r) {
    int s = Logn[r - l + 1];
    return max(f[l][s], f[r - (1 << s) + 1][s]);
}

ll g[200086][20];

int main() {
    pre();
    cin >> n;
    for (int i = 1; i < n + 1; i++) scanf("%d", w + i), f[i][0] = w[i];
    for (int j = 1; j <= logn; j++)
        for (int i = 1; i + (1 << j) - 1 <= n; i++)
            f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);


    cin >> m;
    while (m--) {
        int l, r;
        scanf("%d%d", &l, &r);
        int maxn = w[r], len = 2, p = 1;
        r--;
        while (l <= r) {
            int c = (r - l + 1) % len;
            if (!c) {
                g[l][p]++;
                maxn = max(maxn, query(l, l + len - 1));
                l += len;
            } else {
                if (p != 1) g[r - len / 2 + 1][p - 1]++;
                maxn = max(maxn, query(r - len / 2 + 1, r));
                r -= len / 2;
            }
            ans[maxn]++;
            len <<= 1;
            p++;
        }
    }

    for (int i = 1; i < n; i++)
        for (int j = 18; j; j--)
            if (g[i][j]) {
                ans[query(i, i + (1 << j) - 1)] += g[i][j];
                if (j > 1) g[i][j - 1] += g[i][j], g[i + (1 << (j - 1))][j - 1] += g[i][j];
            }

    for (int i = 1; i < n + 1; i++) printf("%d\n", ans[i]);
    return 0;
}
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