A - Intersection

题意

给出两条线段，求交集长度

题解

暴力

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
#include<bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 2e5 + 5, MOD = 998244353, INF = 0x3f3f3f3f;

signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);

    int l1, r1, l2, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    vector< int > a(101);
    for (int i = l1; i <= r1; ++ i) {
        a[i] = 1;
    }
    int ans = 0;
    for (int i = l2; i <= r2; ++ i) {
        ans += a[i];
    }
    if (ans) -- ans;
    cout << ans << '\n';
    return 0;
}

B - Tournament Result

题意

一个二维矩阵，$A_{i,j}$为W则$i$击败$j$，为L则$i$败给$j$，为D则$i$和$j$打成平局

求是否冲突

The table is said to be contradictory when some of the following holds:

  • There is a pair (i,j) such that Player i beat Player j, but Player j did not lose to Player i;
  • There is a pair (i,j) such that Player i lost to Player j, but Player j did not beat Player i;
  • There is a pair (i,j) such that Player i drew with Player j, but Player j did not draw with Player i.

题解

模拟

代码

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
#include<bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 1005, MOD = 998244353, INF = 0x3f3f3f3f;

int n;
string s[N];
int dis[N][N];
signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);

    cin >> n;
    for (int i = 0; i < n; ++ i) {
        cin >> s[i];
    }
    bool safe = true;
    for (int i = 0; i < n; ++ i) {
        for (int j = i + 1; j < n; ++ j) {
            if (s[i][j] == 'D' && s[j][i] != 'D') {
                safe = false;
            }
            if (s[i][j] == 'W' && s[j][i] != 'L') {
                safe = false;
            }
            if (s[i][j] == 'L' && s[j][i] != 'W') {
                safe = false;
            }
        }
    }
    cout << (safe ? "correct" : "incorrect") << '\n';
    return 0;
}

C - NewFolder(1)

题意

给出n个字符串，要求以 $s_i+(\sum_{j=1}^{i-1}s_j==s[i])$ 的形式输出 n 个字符串

题解

模拟

代码

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
#include<bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 2e5 + 5, MOD = 998244353, INF = 0x3f3f3f3f;

int n;
string s[N];
signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);

    cin >> n;
    string str;
    for (int i = 1; i <= n; ++ i) {
        cin >> s[i];
    }
    map< string, int > cnt;
    for (int i = 1; i <= n; ++ i) {
        if (!cnt[s[i]]) {
            cout << s[i] << '\n';
        } else {
            cout << s[i] + '(' + to_string(cnt[s[i]]) + ')' << '\n';
        }
        ++ cnt[s[i]];
    }
    return 0;
}

D - Flipping and Bonus

题意

Takahashi会投掷$n$次硬币，若第$i$次硬币为正面，则他会获得$X_i$元的奖励，同时给出$m$个附加奖励，当他每次连续投掷正面的次数为$C_i$时他会额外获得$Y_i$元的奖励，求Takahashi能获得的最大钱数。

题解

我们定义$dp_{i,j}$为投掷第$i$次硬币后，连续正面次数为$j$的最大收益

  • 若第$i$次投掷的是正面

    $dp_{i,j}=dp_{i-1,j-1}+X_i$

  • 若第$i$次投掷的是反面，那么连续正面次数重置为0

    $dp_{i,0}=\max(dp_{i,0},dp_{i-1,j})$

  • 接下来考虑额外奖励

    $dp_{i,C_j}=\max\left( dp_{i,C_j},dp_{i-1,C_j-1+Y_j+X_i} \right)$

代码

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
#include<bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 5005, MOD = 998244353, INF = 0x3f3f3f3f;

int n, m, x[N], c[N], y[N];
LL dp[N][N];
signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);

    cin >> n >> m;
    for (int i = 1; i <= n; ++ i) {
        cin >> x[i];
    }
    for (int i = 1; i <= m; ++ i) {
        cin >> c[i] >> y[i];
    }
    for (int i = 1; i <= n; ++ i) {
        for (int j = 0; j <= i - 1; ++ j) {
            dp[i][j + 1] = dp[i - 1][j] + x[i];
            dp[i][0] = max(dp[i][0], dp[i - 1][j]);
        }
        for (int j = 1; j <= m; ++ j) {
            if (c[j] <= i) {
                dp[i][c[j]] = max(dp[i][c[j]], dp[i - 1][c[j] - 1] + y[j] + x[i]);
            }
        }
    }
    LL ans = 0;
    for (int i = 1; i <= n; ++ i) {
        for (int j = 0; j <= n; ++ j) {
            ans = max(ans, dp[i][j]);
        }
    }
    cout << ans << '\n';
    return 0;
}

E - Many Operations

题意

我们有一个初值$X$，以及$n$个操作$op, A_i$

  • 若$op=1$ ，则令 $X= X$ & $A_i$
  • 若$op=2$ ，则令 $X=X|A_i$
  • 若$op=3$ ，则令 $X=X\bigoplus A_i$

我们令初值$X$进行了前$i$次操作后的值为$Y_i$，输出$Y_i(1 \leq i \leq n)$

题解

初一看，感觉是个很厉害的题，细想后发现可以对每一位分别考虑。

分别计算初始值每一位为$0$ 或 $1$，进行$i$次运算后的值，第$i$ 次的初始值为 $Y_{i-1}$ ，对 $Y_{i-1}$的每一位统计即可

代码

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
#include<bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 2e5 + 5, MOD = 998244353, INF = 0x3f3f3f3f;

int n, C;
signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);

    cin >> n >> C;
    int X = (1 << 30) - 1, Y = 0;
    for (int i = 1; i <= n; ++ i) {
        int op, x;
        cin >> op >> x;
        if (op == 1) {
            X &= x;
            Y &= x;
        } else if (op == 2) {
            X |= x;
            Y |= x;
        } else {
            X ^= x;
            Y ^= x;
        }
        LL ans = 0;
        for (int j = 0; j < 30; ++ j) {
            if (C >> j & 1) {
                ans |= (X >> j & 1) << j;
            } else {
                ans |= (Y >> j & 1) << j;
            }
        }
        cout << (C = ans) << '\n';
    }
    return 0;
}

F - Sorting Color Balls

题意

给定$n$个元素，每个元素有一个值和一个颜色，每次操作可以交换相邻的两个元素，若这两个元素颜色不同，则需要消耗一点$cost$,询问通过操作将这$n$个元素按非递减排列所需要的最小$cost$

题解

首先不考虑颜色，其代价显然是数组的逆序对个数

现在加上颜色的限制，其代价就是总的逆序对个数减去同种颜色之间逆序对的个数，维护这个即可。

这里我用了非树状数组的方法来求同种颜色之间逆序对的个数，从后往前遍历，我们每次把$X_i$顺序插入$C_i$对应的数组，询问时我们查询当前的$X_i$位于$C_i$对应的数组第几个即可，$j>i$且$X_j<X_i$符合逆序对定义嘛

代码

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
#include<bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 3e5 + 5, MOD = 998244353, INF = 0x3f3f3f3f;

int n, c[N], x[N];
vector< int > col[N];
struct FenWick {
    int n;
    vector< int > c;
    FenWick(int n) : n(n), c(n + 1) {}
    void add(int i, int d) {
        for (; i <= n; i += i & -i) {
            c[i] += d;
        }
    }
    void add(int l, int r, int d) {
        add(l, d);
        add(r + 1, -d);
    }
    int get(int i) {
        int sum = 0;
        for (; i; i -= i & -i) {
            sum += c[i];
        }
        return sum;
    }
    int get(int l, int r) { 
        return get(r) - get(l - 1);
    }
};
signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);

    cin >> n;
    for (int i = 1; i <= n; ++ i) {
        cin >> c[i];
    }
    for (int i = 1; i <= n; ++ i) {
        cin >> x[i];
    }
    LL ans = 0;
    FenWick k(n);
    for (int i = n; i >= 1; -- i) {
        ans += k.get(x[i] - 1);
        k.add(x[i], 1);
        ans -= upper_bound(col[c[i]].begin(), col[c[i]].end(), x[i] - 1) - col[c[i]].begin();
        col[c[i]].insert(upper_bound(begin(col[c[i]]), end(col[c[i]]), x[i]), x[i]);
    }
    cout << ans << '\n';
    return 0;
}