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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#616646#7787. Maximum RatingKdlyhWA 1ms3668kbC++206.7kb2024-10-06 09:31:012024-10-06 09:31:03

Judging History

你现在查看的是最新测评结果

  • [2024-10-06 09:31:03]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:3668kb
  • [2024-10-06 09:31:01]
  • 提交

answer

#include <cassert>
#include <cmath>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <functional>
#include <iterator>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#include <stack>

#ifdef LOCAL
template <class T, size_t size = std::tuple_size<T>::value> std::string to_debug(T, std::string s = "") requires(not std::ranges::range<T>);
std::string to_debug(auto x) requires requires(std::ostream& os) { os << x; } { return static_cast<std::ostringstream>(std::ostringstream() << x).str(); }
std::string to_debug(std::ranges::range auto x, std::string s = "") requires(not std::is_same_v<decltype(x), std::string>) {
  for (auto xi : x) { s += ", " + to_debug(xi); }
  return "[" + s.substr(s.empty() ? 0 : 2) + "]";
}
template <class T, size_t size> std::string to_debug(T x, std::string s) requires(not std::ranges::range<T>) {
  [&]<size_t... I>(std::index_sequence<I...>) { ((s += ", " + to_debug(get<I>(x))), ...); }(std::make_index_sequence<size>());
  return "(" + s.substr(s.empty() ? 0 : 2) + ")";
}
#define debug(...) std::cerr << __LINE__ << ": (" #__VA_ARGS__ ") = " << to_debug(std::tuple(__VA_ARGS__)) << "\n"
#else
#define debug(x...)
#endif

using i64 = long long;

template <class T>
struct Discreter {
    std::vector<T> elementSet;

    Discreter(const std::vector<T> &a) : elementSet(a) {
        std::sort(begin(elementSet), end(elementSet));
        elementSet.erase(std::unique(begin(elementSet), end(elementSet)), end(elementSet));
    }

    std::vector<int> process(const std::vector<T> &a) const {//get the dicreter arr
        std::vector<int> discRes(a.size());

        for (int i = 0; i < a.size(); i++) {
            discRes[i] = query(a[i]);
        }

        return discRes;
    }

    int query(const T &x) const {
        auto it = std::lower_bound(begin(elementSet), end(elementSet), x);
        // assert(it != end(elementSet) and *it == x);

        return it - begin(elementSet);
    }

    int queryUpperBound(const T &x) const {
        auto it = std::upper_bound(begin(elementSet), end(elementSet), x);

        return it - begin(elementSet);
    }

    T queryInv(int index) const {
        return elementSet[index];
    }

    int size() {
        return elementSet.size();
    }
};

template<class Info>
struct SegmentTree {
    int n;
    std::vector<Info> info;
    SegmentTree() : n(0) {}
    SegmentTree(int n_, Info v_ = Info()) {
        init(n_, v_);
    }
    template<class T>
    SegmentTree(std::vector<T> init_) {
        init(init_);
    }
    void init(int n_, Info v_ = Info()) {
        init(std::vector(n_, v_));
    }
    template<class T>
    void init(std::vector<T> init_) {
        n = init_.size();
        info.assign(4 << std::__lg(n), Info());
        std::function<void(int, int, int)> build = [&](int p, int l, int r) {
            if (r - l == 1) {
                info[p] = init_[l];
                return;
            }
            int m = (l + r) / 2;
            build(2 * p, l, m);
            build(2 * p + 1, m, r);
            pull(p);
        };
        build(1, 0, n);
    }
    void pull(int p) {
        info[p] = info[2 * p] + info[2 * p + 1];
    }
    void modify(int p, int l, int r, int x, const Info &v) {
        if (r - l == 1) {
            info[p] = info[p] + v;
            return;
        }
        int m = (l + r) / 2;
        if (x < m) {
            modify(2 * p, l, m, x, v);
        } else {
            modify(2 * p + 1, m, r, x, v);
        }
        pull(p);
    }
    void modify(int p, const Info &v) {
        modify(1, 0, n, p, v);
    }
    Info rangeQuery(int p, int l, int r, int x, int y) {
        if (l >= y || r <= x) {
            return Info();
        }
        if (l >= x && r <= y) {
            return info[p];
        }
        int m = (l + r) / 2;
        return rangeQuery(2 * p, l, m, x, y) + rangeQuery(2 * p + 1, m, r, x, y);
    }
    Info rangeQuery(int l, int r) {
        return rangeQuery(1, 0, n, l, r);
    }
    int findFirst(int p, int l, int r, int x, int y, i64 tar) {
        if (l >= y || r <= x) {
            return -1;
        }
        if (l >= x && r <= y && (info[p].sum <= tar)) {
            return -1;
        }
        if (r - l == 1) {
            return l;
        }
        int m = (l + r) / 2;
        int res = findFirst(2 * p, l, m, x, y, tar);
        if (res == -1) {
            res = findFirst(2 * p + 1, m, r, x, y, tar - info[2 * p].sum);//把左边的所有贡献加起来,因为是求前缀和
        }
        return res;
    }
    int findFirst(int l, int r, i64 tar) {
        return findFirst(1, 0, n, l, r, tar);
    }
};

struct Info {i64 sum{}, cnt{};};

Info operator + (const Info& a, const Info& b) {return {a.sum + b.sum, a.cnt + b.cnt};}

void solve()
{
    int n, Q; std::cin >> n >> Q; std::vector<int> a(n); for (auto& ai : a) {std::cin >> ai;}
    //离散化正数
    std::vector<int> positives; auto get_positive = [&](auto& x) {if (x > 0) {positives.push_back(x);} return ;}; 
    for (auto& ai : a) {get_positive(ai);}
    std::vector<std::pair<int, int>> querires(Q); for (auto&[x, v] : querires) {std::cin >> x >> v; --x; get_positive(v);}
    Discreter discreter(positives);
    //建权值线段树
    SegmentTree<Info> T(n + Q); for (auto& ai : a) if (ai > 0) {T.modify(discreter.query(ai), {ai, 1});}

    i64 negative_abs_sum{}; for (auto&[x, v] : querires) {
        //修改前
        if (a[x] <= 0) {negative_abs_sum -= std::abs(a[x]);} else {T.modify(discreter.query(a[x]), {-a[x], -1});}
        //修改后
        if (v <= 0) {negative_abs_sum += std::abs(v);} else {T.modify(discreter.query(v), {v, 1});} a[x] = v; 
        //第一遍线段树二分,找到第一个大于负数绝对值的前缀和的下标
        int k{T.findFirst(0, n + Q, negative_abs_sum)};
        if (k == -1) {std::cout << T.rangeQuery(0, n + Q).cnt + 1 << "\n"; continue;}
        //第二遍找k取了多少个
        auto pre_sum{T.rangeQuery(0, k).sum};
        auto pre_take{T.rangeQuery(0, k).cnt};
        auto k_val{discreter.queryInv(k)};
        i64 k_take{(negative_abs_sum - pre_sum) / k_val + 1};

        std::cout << k_take + pre_take << "\n";
    }

}
 
signed main()
{
    std::cin.tie(nullptr)->sync_with_stdio(false);
    int _{1};
#ifdef tests
    std::cin >> _;
#endif
    while(_--) solve();
    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 3608kb

input:

3 5
1 2 3
3 4
2 -2
1 -3
3 1
2 1

output:

1
2
2
2
3

result:

ok 5 number(s): "1 2 2 2 3"

Test #2:

score: 0
Accepted
time: 0ms
memory: 3628kb

input:

3 5
1 2 3
3 4
2 -2
1 3
3 1
2 1

output:

1
2
1
2
1

result:

ok 5 number(s): "1 2 1 2 1"

Test #3:

score: 0
Accepted
time: 0ms
memory: 3492kb

input:

1 1
1000000000
1 1000000000

output:

1

result:

ok 1 number(s): "1"

Test #4:

score: 0
Accepted
time: 0ms
memory: 3608kb

input:

1 1
-1000000000
1 -1000000000

output:

1

result:

ok 1 number(s): "1"

Test #5:

score: 0
Accepted
time: 1ms
memory: 3668kb

input:

1000 1000
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ...

output:

946
65
252
410
915
592
983
487
343
899
809
432
586
587
139
464
804
84
476
699
504
149
579
352
375
856
545
166
140
657
568
509
275
710
873
430
537
879
301
1
298
970
923
510
984
642
55
879
941
344
464
788
917
994
571
610
491
442
926
101
986
840
624
613
425
345
816
423
275
221
317
113
386
116
469
260
4...

result:

ok 1000 numbers

Test #6:

score: -100
Wrong Answer
time: 1ms
memory: 3664kb

input:

1000 1000
1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000000 1 -1000000...

output:

352
221
133
13
137
210
141
149
25
267
242
227
295
344
13
6
58
265
85
309
340
469
20
368
336
481
402
143
389
215
458
60
481
341
390
476
118
288
262
357
488
407
243
145
495
164
34
419
92
24
327
135
176
113
144
310
100
112
314
295
26
458
151
240
314
345
376
23
22
365
78
212
134
112
123
175
204
434
415
...

result:

wrong answer 1st numbers differ - expected: '500', found: '352'