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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#480911#9104. Zayin and Forestbad_solver#TL 0ms0kbC++231.9kb2024-07-16 19:35:322024-07-16 19:35:33

Judging History

你现在查看的是最新测评结果

  • [2024-07-16 19:35:33]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:0kb
  • [2024-07-16 19:35:32]
  • 提交

answer

#include <bits/stdc++.h>
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define f first
#define s second
using namespace std;
using ll = long long;
using pii = pair <int, int>;
const int N = 1e5 + 5;
ll f(ll x) {
    if (x % 2 == 0)
        return x + 1;
    ll res = 1;
    while (res <= x)
        res <<= 1;
    
    for (ll i = 0; ; ++i) {
        if (((x >> i) & 1) && !((x >> (i + 1)) & 1)) {
            res = min(res, x - (1ll << i) + (1ll << (i + 1)));
            return res;
        }
        
        if ((x >> i) & 1)
            x -= (1 << i);
    }
}
ll n, t[300 * N];
int m;
vector <ll> vals;
void add(int x, int v) {
    for ( ; x < sz(vals); x |= x + 1)
        t[x] += v;
}
ll get(int x) {
    ll ans = 0;
    for (; x >= 0; x &= x + 1, --x) {
        ans += t[x];
    }
    return ans;
}
int tp[N], v[N];
ll l[N], r[N], x[N];
int main() {
    ios :: sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> m;
    for (int i = 0; i < m; ++i) {
        cin >> tp[i];
        if (tp[i] & 1) {
            cin >> x[i] >> v[i];
            ll cur = x[i];
            while (cur <= n) {
                vals.pb(cur);
                cur = f(cur);
            }
        } else {
            cin >> l[i] >> r[i];
        }
    }
    sort(all(vals));
    vals.erase(unique(all(vals)), vals.end());
    assert(sz(vals) < 300 * N);
    for (int i = 0; i < m; ++i) {
        if (tp[i] & 1) {
            while (x[i] <= n) {
                int p = lower_bound(all(vals), x[i]) - vals.begin();
                add(p, v[i]);
                x[i] = f(x[i]);
            }
        } else {
            int R = upper_bound(all(vals), r[i]) - vals.begin() - 1;
            int L = upper_bound(all(vals), l[i] - 1) - vals.begin() - 1;
            cout << get(R) - get(L) << '\n';
        }
    }
    return 0;
}

Details

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Test #1:

score: 0
Time Limit Exceeded

input:

1000000000 20000
2 384578735 526547442
1 64211261 592970906
1 512065247 448267721
1 44993150 127180320
1 880319036 927623947
1 170536687 572121854
1 896600029 804033011
1 666246328 754201635
1 654066651 179982083
2 240989825 984888006
2 372004567 858916479
2 76127818 98606736
1 181794163 902842353
1...

output:


result: