A - First ABC 2

#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <sstream>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
#include <queue>
#include <set>

using namespace std;
#define PII pair<int,int>
typedef long long LL;
const int N = 1e5 + 10, M = N * 2;

int main()
{
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int n; cin >> n;
    string s; cin >> s;
    bool flag = false;
    for (int i = 0; i + 2 < s.size(); i ++) {
        if (s.substr(i, 3) == "ABC") {
            flag = true;
            cout << i + 1 << endl;
            break;
        }
    }
    if (!flag) cout << -1 << endl;

    return 0;
}

B - Prefix and Suffix

#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <sstream>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
#include <queue>
#include <set>

using namespace std;
#define PII pair<int,int>
typedef long long LL;
const int N = 1e5 + 10, M = N * 2;
string s, t;

bool check1()
{
    for (int i = 0; i < s.size(); i ++) {
        if (s[i] != t[i]) return false;
    }
    return true;
}

bool check2()
{
    for (int i = s.size() - 1, j = t.size() - 1; i >= 0; i --, j --) {
        if (s[i] != t[j]) return false;
    }
    return true;
}

int main()
{
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int n, m; cin >> n >> m;
    cin >> s >> t;

    if (check1() && check2()) cout << 0 << endl;
    else if (check1() && !check2()) cout << 1 << endl;
    else if (!check1() && check2()) cout << 2 << endl;
    else cout << 3 << endl;

    return 0;
}

C - Festival

#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <sstream>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
#include <queue>
#include <set>

using namespace std;
#define PII pair<int,int>
typedef long long LL;
const int N = 2e5 + 10, M = N * 2;
int a[N];

int main()
{
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int n, m; cin >> n >> m;
    for (int i = 0; i < m; i ++) cin >> a[i];

    for (int i = 1; i <= n; i ++) 
    {
        int t = lower_bound(a, a + m, i) - a;
        cout << a[t] - i << endl;
    }

    return 0;
}

E - Product Development

从n个物品中选择若干件使得价值最小，01背包问题的模型，我们使用f[a1][a2][a3][a4][a5]表示每个参数分别为a1、a2、a3、a4、a5时的最小代价，类似于背包问题，我们倒序枚举将第一维优化掉。
初始化时将所有的值置为INF，f[0][0][0][0][0] = 0。使用h表示每个位置参数的值应该为多少，这里主要是为了应对k < 5时其他参数为0而不是p的问题。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <sstream>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
#include <queue>
#include <set>

using namespace std;
#define PII pair<int,int>
#define int long long
typedef long long LL;
const int N = 110, M = N * 2, INF = 1e18;
int g[N][6], c[N];
int f[6][6][6][6][6], h[6]; // 表示参数分别为xi的情况下的最小代价

void init()
{
    for (int i = 0; i < 6; i ++)
        for (int j = 0; j < 6; j ++)
            for (int k = 0; k < 6; k ++)
                for (int l = 0; l < 6; l ++)
                    for (int r = 0; r < 6; r ++)
                        f[i][j][k][l][r] = INF;
}

signed main()
{
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int n, k, p; cin >> n >> k >> p;
    for (int i = 1; i <= n; i ++) 
    {
        cin >> c[i];
        for (int j = 1; j <= k; j ++) {
            cin >> g[i][j];
        }
    }

    init();

    f[0][0][0][0][0] = 0;

    for (int i = 1; i <= k; i ++) h[i] = p;

    for (int i = 1; i <= n; i ++)
        for (int a1 = h[1]; a1 >= 0; a1 --)
            for (int a2 = h[2]; a2 >= 0; a2 --)
                for (int a3 = h[3]; a3 >= 0; a3 --)
                    for (int a4 = h[4]; a4 >= 0; a4 --)
                        for (int a5 = h[5]; a5 >= 0; a5 --)
                        {
                            int x1 = min(h[1], a1 + g[i][1]), x2 = min(h[2], a2 + g[i][2]), x3 = min(h[3], a3 + g[i][3]), x4 = min(h[4], a4 + g[i][4]),
                            x5 = min(h[5], a5 + g[i][5]);
                            f[x1][x2][x3][x4][x5] = min(f[x1][x2][x3][x4][x5], f[a1][a2][a3][a4][a5] + c[i]);
                        }

    cout << (f[h[1]][h[2]][h[3]][h[4]][h[5]] == INF ? -1 : f[h[1]][h[2]][h[3]][h[4]][h[5]]) << endl;

    return 0;
}