QOJ.ac

QOJ

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
#733439#9543. Good PartitionsnnkWA 75ms7196kbC++207.0kb2024-11-10 18:59:152024-11-10 18:59:15

Judging History

你现在查看的是最新测评结果

  • [2024-11-10 18:59:15]
  • 评测
  • 测评结果:WA
  • 用时:75ms
  • 内存:7196kb
  • [2024-11-10 18:59:15]
  • 提交

answer

#include<iostream>
#include<vector>
#include<algorithm>
#include<set>
#include<iomanip>
#include<queue>
#include<math.h>
#include<map>
#include<bitset>
#include<numeric>
#include <cstring>
#include<array>
#include <string>  
#include<unordered_set>
#include<unordered_map>
using namespace std;
#define int long long
//#define endl "\n"
#define mod 1000000007ll
#define IOS  ios::sync_with_stdio(false),cin.tie(nullptr)
#define syy return
const double pi = acos(-1.0);
using ll = long long;
using pii = pair<int, int>;
using vint = vector<int>;
using vll = vector<ll>;
using vvint = vector<vint>;
using vpo = vector<pii>;
using tup = tuple<int, int, int>;
using vbl = vector<bool>;
constexpr int N = 500005, M = 2000006;//开足够的空间 注意无向图的双向边边数*2
constexpr ll inf = 1e16;
long double eps = 1e-7;

int h[N], ans, a[104], b[104];
int flag[N];
set<int> s;
void work(int len, int op) {
    for (int i = 1; i <= sqrt(len); i++) {
        if (len % i == 0) {
            flag[i] += op;
            if (len / i != i) flag[len / i] += op;
            if (op == -1) {
                if (i!=1 and flag[i] == s.size() - 2) ans--;
                if (len / i != i && flag[len / i] == s.size() - 2) ans--;
                continue;
            }
            if (i!=1 and flag[i] == s.size() - 1)  ans++;
            if ( len / i != i && flag[len / i] == s.size() - 1) ans++;
        }
    }
}

void solve() {
    int n, q;
    cin >> n >> q;
    s.clear();
    ans = 1;
    for (int i = 1; i <= n; i++) flag[i] = 0;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    s.insert(1);
    for (int i = 2; i <= n; i++) {
        if (a[i] < a[i - 1]) {
            s.insert(i);
        }
    }
    for (auto it = s.begin(); it != s.end(); it++) {
        auto it2 = it;
        it2++;
        if (it2 == s.end()) break;
        int len = *it2 - *it;
        work(len, 1);
    }
    cout << ans << endl;
    while (q--) {
        int p, v;
        cin >> p >> v;
        if (v == a[p]) {
            //ap没变
            cout << ans << endl;
            continue;
        }
        auto it = s.lower_bound(p);
        auto it2 = it, it0 = it;
        it2++; it0--;
        bool is = 0;
        if (it2 == s.end()) {
            is = 1;
            if (a[p] >= a[p - 1]) {
                if (a[p] <= a[p + 1]) {
                    if (v < a[p - 1]) {
                        //ap是下一段起点
                        s.insert(p);

                        work(p - *it, 1);

                    }
                    if (v > a[p + 1]) {
                        //ap是这一段的终点
                        s.insert(p + 1);

                        work(p + 1 - *it, 1);

                    }
                }
            }
            else {
                if (a[p] <= a[p + 1]) {
                    if (v >= a[p - 1]) {
                        if (v <= a[p + 1]) {
                            //这一段和上一段合为一段
                           
                            work(*it - *it0, -1);
                            s.erase(it);
                        }
                        if (v > a[p + 1]) {
                            //ap变为上一段终点
                          
                            work(*it - *it0, -1);
                            s.erase(it);  
                            s.insert(p + 1);
                            work(p + 1 - *it0, 1);
                            
                        }
                    }
                    else {
                        if (v<a[p - 1] and v>a[p + 1]) {
                            //还是自己一段 

                        }
                        if (v >= a[p - 1]) {
                            //ap变为上一段终点
                           
                            work(p - *it0, -1);
                            s.erase(it);
                        }
                        if (v <= a[p + 1]) {
                            //ap变下一段起点



                        }
                    }
                }
            }
        }

        if (is == 0)
        {
            int len = *it2 - *it;
            if (a[p] >= a[p - 1]) {
                if (a[p] <= a[p + 1]) {
                    if (v < a[p - 1]) {
                        //ap是下一段起点
                       
                        work(len, -1); s.insert(p);
                        work(p - *it, 1);
                        work(*it2 - p, 1);

                    }
                    if (v > a[p + 1]) {
                        //ap是这一段的终点
                        
                        work(len, -1);s.insert(p + 1);
                        work(p + 1 - *it, 1);
                        work(*it2 - p - 1, 1);

                    }
                }
            }
            else {
                if (a[p] <= a[p + 1]) {
                    if (v >= a[p - 1]) {
                        if (v <= a[p + 1]) {
                            //这一段和上一段合为一段
                           
                            work(*it - *it0, -1);
                            work(*it2 - *it, -1);
                            s.erase(it);
                            work(*it2 - *it0, 1);
                            

                        }
                        if (v > a[p + 1]) {
                            //ap变为上一段终点
                            
                           
                            work(*it - *it0, -1);
                            work(*it2 - *it, -1); 
                            s.erase(it);
                            s.insert(p + 1);
                            work(*it2 - (p + 1), 1);
                            work(p + 1 - *it0, 1);
                            
                        }
                    }
                }
                    else {
                        if (v<a[p - 1] and v>a[p + 1]) {
                            //还是自己一段

                        }
                        if (v >= a[p - 1]) {
                            //ap变为上一段终点
                           
                            work(p - *it0, -1); 
                            s.erase(it);
                            work(p + 1 - *it0, 1);
                          

                        }
                        if (v <= a[p + 1]) {
                            //ap变下一段起点
                            auto it3 = it2; it3++;
                           
                            work(*it2 - p, -1);
                            s.erase(it2);
                            work(*it3 - p, 1);
                           

                        }
                    }
                
            }
        }
        a[p] = v;
        cout << ans << endl;
    }
}

signed main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int T = 1;
    cin >> T;
    while (T--) {
        solve();
    }
    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 5812kb

input:

1
5 2
4 3 2 6 1
2 5
3 5

output:

1
2
3

result:

ok 3 lines

Test #2:

score: 0
Accepted
time: 1ms
memory: 5652kb

input:

1
1 1
2000000000
1 1999999999

output:

1
1

result:

ok 2 lines

Test #3:

score: -100
Wrong Answer
time: 75ms
memory: 7196kb

input:

1
200000 200000
3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 ...

output:

265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
265
...

result:

wrong answer 1st lines differ - expected: '160', found: '265'