AtCoder Beginner Contest 322

1|0A - First ABC 2

#include <bits/stdc++.h>

using namespace std;

#define int long long
#define mp make_pair

using vi = vector<int>;
using pii = pair<int, int>;


void solve() {
    int n;
    string s;
    cin >> n >> s;
    auto it = s.find("ABC");
    if( it >= n ) cout << "-1\n";
    else cout << it + 1 << "\n";
    return ;
}


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    solve();

    return 0;
}

2|0B - Prefix and Suffix

#include <bits/stdc++.h>

using namespace std;

#define int long long
#define mp make_pair

using vi = vector<int>;
using pii = pair<int, int>;


void solve() {
    int n, m;
    cin >> n >> m;
    string s, t;
    cin >> s >> t;
    int ans = 0;
    for (int i = 0; i < n; i++) {
        if (s[i] == t[i]) continue;
        ans += 2;
        break;
    }
    reverse(s.begin(), s.end());
    reverse(t.begin(), t.end());
    for (int i = 0; i < n; i++) {
        if (s[i] == t[i]) continue;
        ans += 1;
        break;
    }
    cout << ans << "\n";
    return;
}


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    solve();

    return 0;
}

3|0C - Festival

#include <bits/stdc++.h>

using namespace std;

#define int long long
#define mp make_pair

using vi = vector<int>;
using pii = pair<int, int>;


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vi a(m);
    for (auto &i: a) cin >> i;
    reverse(a.begin(), a.end());
    for( int i = 1 ; i <= n ; i ++ ){
        cout << a.back() - i << "\n";
        if( a.back() == i ) a.pop_back();
    }
    return 0;
}

4|0D - Polyomino

#include <bits/stdc++.h>

using namespace std;

#define int long long
#define mp make_pair

using vi = vector<int>;
using pii = pair<int, int>;
using node = array<string, 4>;

int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    vector<node> g(3);
    int cnt = 0;
    for (auto &it: g)
        for (auto &i: it) {
            cin >> i;
            cnt += count(i.begin(), i.end(), '#');
        }
    auto rotate = [](const node &a) {
        auto b = a;
        for (int i = 0; i < 4; i++)
            for (int j = 0; j < 4; j++)
                b[j][3 - i] = a[i][j];
        return b;
    };
    array<array<char, 4>, 4> page;
    auto fit = [&](int k, int x, int y) {
        for (int i = 0, nx = x; i < 4; i++, nx++)
            for (int j = 0, ny = y; j < 4; j++, ny++) {
                if (g[k][i][j] == '.') continue;
                if (nx < 0 or nx > 3 or ny < 0 or ny > 3) return false;
                if (page[nx][ny] == '#') return false;
                page[nx][ny] = '#';
            }
        return true;
    };
    auto dfs = [&](auto &&self, int i) -> bool {
        if (i == 3) return true;
        auto bak = page;
        for (int x = -3; x <= 7; x++)
            for (int y = -3; y <= 7; y++) {
                for (int t = 0; t < 4; t++) {
                    if (fit(i, x, y) and self(self, i + 1))
                        return true;
                    g[i] = rotate(g[i]), page = bak;
                }
            }
        return false;
    };
    if (cnt == 16 and dfs(dfs, 0))
        cout << "Yes\n";
    else
        cout << "No\n";
    return 0;
}

5|0E - Product Development

相对来说是一个比较板的 01 背包问题，只是状态的表示比较麻烦，所以可以状压一下。

#include<bits/stdc++.h>

using namespace std;

#define int long long

using i32 = int32_t;
using vi = vector<int>;

const int inf = 1e18;

i32 main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n, k, p;
    cin >> n >> k >> p;
    int base = p + 1;
    const int N = (int) (pow(base, k));
    vi f(N , inf);
    f[0] = 0;
    auto dtr = [k, base](int x) {
        vi ret(k);
        for (int i = k - 1; i >= 0; i--)
            ret[i] = x % base, x /= base;
        return ret;
    };
    auto tr = [k, base](const vi &x) {
        int res = 0;
        for (int i = 0; i < k; i++) res = res * base + x[i];
        return res;
    };
    for (int i = 0, c; i < n; i++) {
        cin >> c;
        vi a(k);
        for (auto &j: a) cin >> j;
        auto g = f;
        for (int i = 0; i < N; i++) {
            auto lst = dtr(i);
            for (int j = 0; j < k; j++)
                lst[j] = min(p, lst[j] + a[j]);
            auto nxt = tr(lst);
            g[nxt] = min(g[nxt], f[i] + c);
        }
        f.swap(g);
    }
    if (f.back() == inf) cout << "-1\n";
    else cout << f.back() << "\n";
    return 0;
}