Lucky Direction

map直接映射就好

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
#include <bits/stdc++.h>
using namespace std;

int main(){
    map<string, string> mp;
    mp["N"] = "S";
    mp["S"] = "N";
    mp["W"] = "E";
    mp["E"] = "W";
    mp["NE"] = "SW";
    mp["SW"] = "NE";
    mp["NW"] = "SE";
    mp["SE"] = "NW";
    string s;
    cin >> s;
    cout << mp[s] << endl;

    return 0;
}

Seek Grid

思路： 数据较小，直接枚举比较即可

1
2
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
#include <bits/stdc++.h>
using namespace std;

int main(){
    int n, m;
    cin >> n >> m;
    vector<string> a(n), b(m);
    for(int i = 0; i < n; i++) {
        cin >> a[i];
    }
    for(int i = 0; i < m; i++) {
        cin >> b[i];
    }
    auto check = [&](int x, int y)->bool {
        for(int i = x; i < x + m; i++) {
            for(int j = y; j < y + m; j++) {
                if(a[i][j] != b[i - x][j - y]) {
                    return false;
                }
            }
        }
        return true;
    };
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            if(n - i < m || n - j < m) continue;
            if(check(i, j)){
                cout << i + 1 << ' ' << j + 1 << endl;
                return 0;
            }
        }
    }

    return 0;
}

Pigeonhole Query

思路： 记录每一个鸽子在哪个巢中， cnt记录巢中鸽子数量大于1的个数

1
2
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
#include <bits/stdc++.h>
using namespace std;

int main(){
    int n, q;
    cin >> n >> q;
    vector<int> a(n + 1, 1);
    vector<int> mp(n + 1);
    for(int i = 1; i <= n; i++) mp[i] = i;
    int cnt = 0;
    while(q--) {
        int x, p, k;
        cin >> x;
        if(x == 2) {
            cout << cnt << endl;
        }else{
            cin >> p >> k;
            int tp = p;
            p = mp[p];
            a[p]--;
            a[k]++;
            mp[tp] = k;
            if(a[p] == 1) cnt--;
            if(a[k] == 2) cnt++;
        }
    }
    return 0;
}

Gravity

思路： 只需要预处理每一个块在哪一个时间点消失即可，可以用一个优先队列数组记录每一列的方块。每次取出每一列最小的块，求所有的最大值，这一系列块的消失时间就是最大值。 如果某一列为空，这表示剩下的块不可能消失。

1
2
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
#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> pii;

int main() {
    int n, w;
    cin >> n >> w;
    vector<priority_queue<pii, vector<pii>, greater<pii>>> vec(w + 1);
    for(int i = 1; i <= n; i++) {
        int x, y;
        cin >> x >> y;
        vec[x].push({y, i});
    }
    vector<int> cnt(n + 1, INT_MAX);
    while(1) {
        int f = 0;
        int mx = 0;
        for(int i = 1; i <= w; i++) {
            if(vec[i].empty()) {
                f = 1;
                break;
            }
            mx = max(mx, vec[i].top().first);
        }
        if(f) break;
        for(int i = 1; i <= w; i++) {
            cnt[vec[i].top().second] = mx - 1;
            vec[i].pop();
        }
    }
    int q;
    cin >> q;
    while(q--) {
        int t, p;
        cin >> t >> p;
        if(cnt[p] >= t) {
            cout << "Yes\n";
        }else{
            cout << "No\n";
        }
    }
}

Hierarchical Majority Vote

思路： 观察数据范围，暴搜。

1
2
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
#include <bits/stdc++.h>
using namespace std;
#define int long long

int check(string x) {
    int t = x.size() / 3;
    if(t == 1) {
        int p = x[0] + x[1] + x[2] - 3 * '0';
        if(p >= 2) return 1;
        else return 0;
    }
    int x1 = check(x.substr(0, t));
    int x2 = check(x.substr(t, t));
    int x3 = check(x.substr(t + t));
    int p = x1 + x2 + x3;
    if(p >= 2) return 1;
    else return 0;
}

signed main(){
    int n;
    cin >> n;
    string s;
    cin >> s;
    int f = check(s);
    auto dfs = [&](auto&& dfs, string x)->int {
        int t = x.size() / 3;
        if(t == 1) {
            int p = x[0] + x[1] + x[2] - 3 * '0';
            if(p >= 2) p = 1;
            else p = 0;
            if(p != f) return 0;
            if(x[0] == x[1] && x[1] == x[2])
                return 2;
            return 1;
        }
        int x1 = dfs(dfs, x.substr(0, t));
        int x2 = dfs(dfs, x.substr(t, t));
        int x3 = dfs(dfs, x.substr(t + t));
        // int p = x1 + x2 + x3;
        // cout << x1 << ' ' << x2 << ' ' << x3 << endl;
        int p1 = check(x.substr(0, t));
        int p2 = check(x.substr(t, t));
        int p3 = check(x.substr(t + t));
        if(p1 == p2 && p2 == p3) {
            return min({x1 + x2, x1 + x3, x2 + x3});
        }else{
            if(p1 == p2) return min(x1, x2);
            if(p2 == p3) return min(x2, x3);
            if(p1 == p3) return min(x1, x3);
        }
    };

    cout << dfs(dfs, s) << endl;
    return 0;
}

K-th Largest Triplet

思路： 暂无

1