ABC358

A. Welcome to AtCoder Land

模拟

B. Ticket Counter

模拟

代码实现

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
 
int main() {
    int n, a;
    cin >> n >> a;
    
    vector<int> t(n);
    rep(i, n) cin >> t[i];
    
    int x = 0;
    rep(i, n) {
        if (t[i] < x) {
            x += a;
        }
        else {
            x = t[i]+a;
        }
        cout << x << '\n';
    }
  
    return 0;  
}

C. Popcorn

先把每条信息压成一个二进制数，再做二进制枚举即可

代码实现

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
 
int main() {
    int n, m;
    cin >> n >> m;
    
    vector<string> s(n);
    rep(i, n) cin >> s[i];
    
    using BS = bitset<10>;
    vector<BS> b(n);
    rep(i, n) {
        rep(j, m) if (s[i][j] == 'o') b[i][j] = 1;
    }
    
    int ans = n;
    rep(x, 1<<n) {
        BS bx(x);
        BS sb;
        rep(i, n) if (bx[i]) sb |= b[i];
        if (sb.count() == m) ans = min(ans, (int)bx.count());
    }
    
    cout << ans << '\n';
  
    return 0;  
}

D. Souvenirs

贪心
先对序列 $B$ 做升序排序，然后购买当前剩下的数中大于等于 $B_i$ 的最小值
可以用 std::multiset 来维护序列 $A$
也可以用双指针，但需要先对序列 $A$ 做排序

代码实现

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
using ll = long long;
 
int main() {
    int n, m;
    cin >> n >> m;
    
    vector<int> a(n), b(m);
    rep(i, n) cin >> a[i];
    rep(i, m) cin >> b[i];
    
    sort(b.begin(), b.end());
    
    multiset<int> s(a.begin(), a.end());
    
    ll ans = 0;
    rep(i, m) {
        auto it = s.lower_bound(b[i]);
        if (it == s.end()) {
            puts("-1");
            return 0;
        }
        ans += *it;
        s.erase(it);
    }
    
    cout << ans << '\n';
    
    return 0;  
}

E. Alphabet Tiles

原题：ABC234F

记 dp[i][j] 表示用前 $i$ 种大写字母能构成的长度为 $j$ 的本质不同的字符串个数

代码实现

#include <bits/stdc++.h>
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
#endif
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
using mint = modint998244353;
 
struct modinv {
  int n; vector<mint> d;
  modinv(): n(2), d({0,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(-d[mint::mod()%n]*(mint::mod()/n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} invs;
struct modfact {
  int n; vector<mint> d;
  modfact(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*n), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} facts;
struct modfactinv {
  int n; vector<mint> d;
  modfactinv(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*invs(n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} ifacts;
mint comb(int n, int k) {
  if (n < k || k < 0) return 0;
  return facts(n)*ifacts(k)*ifacts(n-k);
}
 
int main() {
    int n = 26;
    int k;
    cin >> k;
    
    vector<int> c(n);
    rep(i, n) cin >> c[i];
    
    vector<mint> dp(k+1);
    dp[0] = 1;
    rep(i, n) {
        vector<mint> old(k+1);
        swap(dp, old);
        rep(j, k+1) {
            rep(a, c[i]+1) {
                int nj = j+a;
                if (nj > k) break;
                dp[nj] += old[j]*comb(nj, a);
            }
        }
    }
    
    mint ans;
    rep(i, k) ans += dp[i+1];
    cout << ans.val() << '\n';
    
    return 0;  
}

注意到相同元素之间不需要排序，所以可以在原dp的基础上加上除序的操作，也就是 dp[nj] += old[j]/a!;
那么最后的答案就是 $\sum\limits_{i=1}^k dp[i] \times i!$

代码实现

#include <bits/stdc++.h>
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
#endif
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
using mint = modint998244353;
 
struct modinv {
  int n; vector<mint> d;
  modinv(): n(2), d({0,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(-d[mint::mod()%n]*(mint::mod()/n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} invs;
struct modfact {
  int n; vector<mint> d;
  modfact(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*n), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} facts;
struct modfactinv {
  int n; vector<mint> d;
  modfactinv(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*invs(n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} ifacts;
mint comb(int n, int k) {
  if (n < k || k < 0) return 0;
  return facts(n)*ifacts(k)*ifacts(n-k);
}
 
int main() {
    int n = 26;
    int k;
    cin >> k;
    
    vector<int> c(n);
    rep(i, n) cin >> c[i];
    
    vector<mint> dp(k+1);
    dp[0] = 1;
    rep(i, n) {
        vector<mint> old(k+1);
        swap(dp, old);
        rep(j, k+1) {
            rep(a, c[i]+1) {
                int nj = j+a;
                if (nj > k) break;
                dp[nj] += old[j]*ifacts(a);
            }
        }
    }
    
    mint ans;
    rep(i, k) ans += dp[i+1]*facts(i+1);
    cout << ans.val() << '\n';
    
    return 0;  
}

F. Easiest Maze

有解的条件：$N \leqslant K \leqslant NM$ 且 $K$ 的奇偶性和 $N$ 的奇偶性相同

关于构造可以参考官方题解的图

代码实现

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
 
int main() {
    int n, m, k;
    cin >> n >> m >> k;
    
    if (k < n or (k-n)%2 != 0) {
        puts("No");
        return 0;
    }
    
    vector<string> s(n*2+1, string(m*2+1, '+'));
    rep(i, n)rep(j, m) s[i*2+1][j*2+1] = 'o';
    s[0][m*2-1] = 'S';
    s.back()[m*2-1] = 'G';
    rep(i, n)rep(j, m-1) s[i*2+1][j*2+2] = '|';
    rep(i, n-1)rep(j, m) s[i*2+2][j*2+1] = '-';
    rep(i, n-1) s[i*2+2][m*2-1] = '.'; 
    
    k -= n;
    for (int i = 0; i < n-1; i += 2) {
        for (int j = m-1; j >= 1; --j) {
            if (k > 0) {
                s[i*2+2][j*2+1] = '-';
                s[i*2+2][j*2-1] = '.';
                s[i*2+1][j*2] = '.';
                s[i*2+3][j*2] = '.';
                k -= 2;
            }
        }
    }
    if (n%2 == 1) {
        for (int j = 0; j < m-2; j += 2) {
            if (k > 0) {
                s[n*2-3][j*2+2] = '|';
                s[n*2-1][j*2+2] = '.';
                s[n*2-2][j*2+1] = '.';
                s[n*2-2][j*2+3] = '.';
                k -= 2;
            }
        }
    }
    
    puts("Yes");
    rep(i, s.size()) cout << s[i] << '\n';
    
    return 0;  
}

G. AtCoder Tour

尽可能在 $k$ 步以内走到尽可能大的格子，如果步数还有剩余就一直停留在原地，容易想到在到达最大值的格子之前不会在路径上某个格子做停留，因为这样会更劣

记 dp[i][j][l] 表示到 $(i, j)$ 为止已经走了 $l$ 步时的最大愉悦值

代码实现

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
 
using namespace std;
using ll = long long;
 
const int di[] = {-1, 0, 1, 0};
const int dj[] = {0, 1, 0, -1};
 
inline void chmax(ll& a, ll b) { if (a < b) a = b; };
 
int main() {
    int h, w, k;
    cin >> h >> w >> k;
    
    int si, sj;
    cin >> si >> sj;
    si--; sj--;
    
    vector a(h, vector<int>(w));
    rep(i, h)rep(j, w) cin >> a[i][j];
    
    int hw = h*w;
    const ll INF = 1e18;
    vector dp(hw, vector(h, vector<ll>(w, -INF)));
    dp[0][si][sj] = 0;
    rep(l, hw-1)rep(i, h)rep(j, w) {
        rep(v, 4) {
            int ni = i+di[v], nj = j+dj[v];
            if (ni<0 or nj<0 or ni>=h or nj>=w) continue;
            chmax(dp[l+1][i][j], dp[l][ni][nj]+a[i][j]);
        }
    }
    
    ll ans = 0;
    rep(l, hw)rep(i, h)rep(j, w) {
        if (l > k) continue;
        ll now = ll(k-l)*a[i][j] + dp[l][i][j];
        chmax(ans, now);
    }
    
    cout << ans << '\n';
    
    return 0;  
}