codeforces Round 910 (Div.2) A - D

A. Milica and String

思路：记录每一个 $A、B$ 的位置，$B$ 的数量大于 $k$ 则将多于的 $B$ 的数量变为 $A$，反之同理。他是复制不是翻转，因此可以直接记录位置复制

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define fi first
#define se second
#define sz size()
#define bpt __builtin_popcountll

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;

const int N = 2E6 + 10, mod = 998244353;

ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

ll ksm(int a, int k)
{
  ll res = 1;
  while (k)
  {
    if (k & 1)
    {
      res = (ll)res * a % mod;
    }
    k >>= 1;
    a = (ll)a * a % mod;
  }
  return res;
}
ll fact[N], infact[N], inv[N];

void init()
{
  fact[0] = fact[1] = infact[0] = infact[1] = inv[1] = 1;
  for (int i = 2; i < N; i++)
  {
    fact[i] = ((ll)fact[i - 1] * i) % mod;
    inv[i] = (ll)(mod - mod / i) * inv[mod % i] % mod;
    infact[i] = (ll)infact[i - 1] * inv[i] % mod;
  }
}
ll C(ll a, ll b)
{
  if (a < b)
    return 0;
  return (ll)fact[a] * infact[b] % mod * infact[a - b] % mod;
}

const int MOD = 998244353;

map<int, int> mp1, mp2;
int f1 = 1, f2 = 1;
int c1 = 1, c2 = 1;

int p[N], vec[N];

int find(int x)
{
  if (p[x] != x)
    p[x] = find(p[x]);
  return p[x];
}
void solve()
{
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  vector<int> a, b;
  for (int i = 0; i < n; i++)
  {
    if (s[i] == 'A')
      a.push_back(i + 1);
    else
      b.push_back(i + 1);
  }

  if (b.size() == k)
  {
    puts("0");
  }
  else
  {
    puts("1");
    if (k < b.size())
    {
      cout << b[b.size() - k - 1] << ' ' << 'A' << endl;
    }
    else
    {
      cout << a[k - b.size() - 1] << ' ' << 'B' << endl;
    }
  }
}
int main()
{
  int _ = 1;
  cin >> _;
  while (_--)
    solve();
}

---

B. Milena and Admirer

思路 ：观察发现，先修改前面会直接影响后面，因此为了保证后面的不受前面的影响，从后面开始；如果已经小于后面的数，则不用分裂；$y=(a_i+x-1)/x$向上取整的方式，分裂 $y-1$ 次，然后共有 $y$ 个小于等于 $a_i/y$的数，更新 $a_I$；

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define fi first
#define se second
#define sz size()
#define bpt __builtin_popcountll

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;

const int N = 2E6 + 10, mod = 998244353;

ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

ll ksm(int a, int k)
{
  ll res = 1;
  while (k)
  {
    if (k & 1)
    {
      res = (ll)res * a % mod;
    }
    k >>= 1;
    a = (ll)a * a % mod;
  }
  return res;
}
ll fact[N], infact[N], inv[N];

void init()
{
  fact[0] = fact[1] = infact[0] = infact[1] = inv[1] = 1;
  for (int i = 2; i < N; i++)
  {
    fact[i] = ((ll)fact[i - 1] * i) % mod;
    inv[i] = (ll)(mod - mod / i) * inv[mod % i] % mod;
    infact[i] = (ll)infact[i - 1] * inv[i] % mod;
  }
}
ll C(ll a, ll b)
{
  if (a < b)
    return 0;
  return (ll)fact[a] * infact[b] % mod * infact[a - b] % mod;
}

const int MOD = 998244353;

map<int, int> mp1, mp2;
int f1 = 1, f2 = 1;
int c1 = 1, c2 = 1;

int p[N], vec[N];

int find(int x)
{
  if (p[x] != x)
    p[x] = find(p[x]);
  return p[x];
}
void solve()
{
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++)
    cin >> a[i];
  ll ans = 0;
  for (int i = n - 2; i >= 0; i--)
  {
    if (a[i] <= a[i + 1])
      continue;
    int x = a[i + 1];
    int y = (a[i] + x - 1) / x;
    a[i] = a[i] / y; //
    ans += y - 1;
  }
  cout << ans << endl;
}
int main()
{
  int _ = 1;
  cin >> _;
  while (_--)
    solve();
}

C. Colorful Grid

思路：$k$ 显然不能小于最短路径 $n+m-2$，同时是二分图，因此 $k$ 的奇偶性要与 $n+m-2$ 相同，否则无解；将第一行和最后一列染色做基础路径，考虑 $(k-res)mod$ $4$ 的情况，不存在 $1$，奇偶不同不存在。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define fi first
#define se second
#define sz size()
#define bpt __builtin_popcountll

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;

const int N = 2E6 + 10, mod = 998244353;

ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

ll ksm(int a, int k)
{
  ll res = 1;
  while (k)
  {
    if (k & 1)
    {
      res = (ll)res * a % mod;
    }
    k >>= 1;
    a = (ll)a * a % mod;
  }
  return res;
}
ll fact[N], infact[N], inv[N];

void init()
{
  fact[0] = fact[1] = infact[0] = infact[1] = inv[1] = 1;
  for (int i = 2; i < N; i++)
  {
    fact[i] = ((ll)fact[i - 1] * i) % mod;
    inv[i] = (ll)(mod - mod / i) * inv[mod % i] % mod;
    infact[i] = (ll)infact[i - 1] * inv[i] % mod;
  }
}
ll C(ll a, ll b)
{
  if (a < b)
    return 0;
  return (ll)fact[a] * infact[b] % mod * infact[a - b] % mod;
}

const int MOD = 998244353;

const int maxn = 30;
int d1[maxn][maxn], d2[maxn][maxn];

void solve()
{
  int n, m, k;
  cin >> n >> m >> k;
  int res = n + m - 2;
  if ((k - res) & 1 || res > k)
  {
    cout << "NO" << endl;
    return;
  }

  puts("YES");
  char s[2] = {'R', 'B'};
  int cur = 0;
  // 每一行的链接
  for (int i = 0; i < n; i++)
  {
    for (int j = 1; j < m; j++)
    {
      d1[i][j] = (j & 1) ? cur : 1 - cur;
    }
  }

  if (m % 2 == 0)
  {
    cur = 1;
  }
  // 每一列的链接
  for (int i = 1; i < n; i++)
  {
    for (int j = 1; j <= m; j++)
    {
      d2[i][j] = (i & 1) ? cur : 1 - cur;
    }
  }

  if (d1[0][2] == d2[1][1])
  { // 第一行第二条横边不同于第一行的第二竖边
    d2[1][1] = d2[1][2] = 1 - d1[0][2];
  }

  if (d1[1][1] == d2[1][1])
  {
    d1[1][1] = 1 - d2[1][2];
  }

  if (d1[n - 1][m - 1] == d2[n - 1][m])//倒数两行的横与最后一行的竖不能相同
  {
    d1[n - 1][m - 1] = d1[n - 2][m - 1] = 1 - d2[n - 1][m];
  }

  for (int i = 0; i < n; i++)
  {
    for (int j = 1; j < m; j++)
    {
      cout << s[d1[i][j]] << ' ';
    }
    cout << endl;
  }

  for (int i = 1; i < n; i++)
  {
    for (int j = 1; j <= m; j++)
    {
      cout << s[d2[i][j]] << ' ';
    }
    cout << endl;
  }
}
int main()
{
  int _ = 1;
  cin >> _;
  while (_--)
    solve();
}

D. Absolute Beauty

思路 ：不超过一次的交换，那必然是最小的右端点和最大的左端点进行交换，这样交换后的值增大最多为 $2\ast (r-l)$；将每个位置化作一条线段，能让结果增大的，是将不想交的线段变为相交。因此循环维护最小的右端点，遍历即可。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define fi first
#define se second
#define sz size()
#define bpt __builtin_popcountll

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;

const int N = 2E6 + 10, mod = 998244353;

ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
const int MOD = 998244353;
const int maxn = 30;
int d1[maxn][maxn], d2[maxn][maxn];

void solve()
{
    int n;
    cin >> n;
    vector<int> a(n), b(n);

    for (int i = 0; i < n; i++)
        cin >> a[i];
    for (int i = 0; i < n; i++)
        cin >> b[i];

    ll sum = 0;
    for (int i = 0; i < n; i++)
    {
        sum += abs(a[i] - b[i]);
    }

    vector<pair<int, int>> p(n);
    for (int i = 0; i < n; i++)
    {
        p[i] = minmax(a[i], b[i]);
    }

    sort(p.begin(), p.end());
    int mx = 0, mn = 1e9;
    int ml = 1e9;
    for (int i = 0; i < n; i++) {
        ml = min(ml, p[i].second);
    }
    for (int i = 0; i < n; i++) {
        mx = max(mx, 2 * (p[i].first - ml));
    }
    cout << sum + mx << endl;
}
int main()
{
    int _ = 1;
    cin >> _;
    while (_--)
        solve();
}