String of yuusaan

我们可以打表,我们会发现字符串无论重复多少次都会遵循这个规律

#include <bits/stdc++.h>

using namespace std;

#define int long long

int a, b;

signed main() {
  cin >> a >> b;
  b--;
  int x = b % 12;
  if (x == 0) {
    cout << "y";
  }
  else if (x == 1 || x == 2 || x == 7 || x == 8) {
    cout << "u";
  }
  else if (x == 3 || x == 9) {
    cout << "s";
  }
  else if (x == 6) {
    cout << "n";
  }
  else cout << "a";
  return 0;
}

NOT FOUND 404 Again

显然我们会发现这是一个数位 \(dp\) 那么我们就可以设计 \(dp_{i, 0/1/2, 0/1/2}\) 表示选到第 \(i\) 位,\(0/1/2\) 表示是否为 \(0\),是否为 \(4\),还是两者都不满足

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5, mod = 998244353;

int n, dp[N][3][3][3], x[N];

string s;

int val(int e) {
  if (e == 0) {
    return 0;
  }
  else if (e == 4) {
    return 1;
  }
  return 2;
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  n = s.size();
  s = " " + s;
  for (int i = 1; i <= n; i++) {
    x[i] = (int)(s[i] - '0');
  }
  //0表示 = 0,1 表示 = 4, 2表示等于其他数
  dp[0][0][0][1] = 1;
  for (int i = 1; i <= n; i++) {
    for (int a = 0; a <= 2; a++) {
      for (int b = 0; b <= 2; b++) {
        for (int c = 0; c <= 9; c++) {
          if (c == 4 && b == 0 && a == 1) {
            continue;
          }
          if (c < x[i]) {
            dp[i][val(c)][b][0] = (dp[i][val(c)][b][0] + dp[i - 1][b][a][1]) % mod;
            dp[i][val(c)][b][0] = (dp[i][val(c)][b][0] + dp[i - 1][b][a][0]) % mod;
          }
          else if (c == x[i]) {
            dp[i][val(c)][b][1] = (dp[i][val(c)][b][1] + dp[i - 1][b][a][1]) % mod;
            dp[i][val(c)][b][0] = (dp[i][val(c)][b][0] + dp[i - 1][b][a][0]) % mod;
          }
          else {
            dp[i][val(c)][b][0] = (dp[i][val(c)][b][0] + dp[i - 1][b][a][0]) % mod;
          }
        }
      }
    }
  }
  int ans = 0;
  for (int a = 0; a <= 2; a++) {
    for (int b = 0; b <= 2; b++) {
      ans += (dp[n][a][b][0] + dp[n][a][b][1]) % mod;
      ans %= mod;
    }
  }
  cout << ans - 1;
  return 0;
}

yuusaan's Knapsacks

我们可以发现每个背包选的一定都是物品的一个子集,那么我们就可以枚举子集,然后状压,由于枚举子集的时间复杂度是 \(3^n\) 的,所以不会超时

#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 17, M = (1 << 16), INF = 1e18;

int n, m, a[N], v[N], w[N], b[M], sumv[M], sumw[M], dp[N][M], pre[N][M];

inline int lowbit(int x) {
  return x & (-x);
}

void print(int x, int y) {
  if (!x) {
    return ;
  }
  print(x - 1, y - pre[x][y]);
  vector<int> ans;
  for (int i = pre[x][y]; i; i -= lowbit(i)) {
    ans.push_back(b[lowbit(i)]);
  }
  cout << ans.size() << " ";
  for (auto cur : ans) {
    cout << cur + 1 << " ";
  }
  cout << "\n";
}

signed main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < m; i++) {
    cin >> v[i] >> w[i];
    b[(1 << i)] = i;
  }
  for (int i = 1; i < (1 << m); i++) {
    sumv[i] = sumv[i - lowbit(i)] + v[b[lowbit(i)]];
    sumw[i] = sumw[i - lowbit(i)] + w[b[lowbit(i)]];
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < (1 << m); j++) {
      dp[i][j] = -INF;
    }
  }
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    dp[i][0] = 0;
    for (int j = 1; j < (1 << m); j++) {
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      for (int k = j; k; k = (k - 1) & j) {
        if (sumw[k] <= a[i] && dp[i - 1][j - k] + sumv[k] > dp[i][j]) {
          dp[i][j] = dp[i - 1][j - k] + sumv[k];
          pre[i][j] = k;
        }
      }
    }
  }
  int maxi = -INF, p;
  for (int i = 0; i < (1 << m); i++) {
    if (maxi < dp[n][i]) {
      maxi = dp[n][i];
      p = i;
    }
  }
  cout << maxi << "\n";
  print(n, p);
  return 0;
}