Codeforces Round #558 (Div. 2) 题解

A Eating Soup

分类讨论一下

\(m\) 个断点最多可以将环分成 \(m\) 段，而剩下 \(n-m\) 个点最多可以成为 \(n-m\) 段，取个 \(\min\) 即可

注意 \(m=0\) 时是一个环答案为 \(1\)

// Copyright lzt

#include<stdio.h>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<vector>

#include<map>

#include<set>

#include<cmath>

#include<iostream>

#include<queue>

#include<string>

#include<ctime>

using namespace std;

typedef long long ll;

typedef std::pair<int, int> pii;

typedef long double ld;

typedef unsigned long long ull;

typedef std::pair<long long, long long> pll;

#define fi first

#define se second

#define pb push_back

#define mp make_pair

#define rep(i, j, k)  for (register int i = (int)(j); i <= (int)(k); i++)

#define rrep(i, j, k) for (register int i = (int)(j); i >= (int)(k); i--)

#define Debug(...) fprintf(stderr, __VA_ARGS__)

inline ll read() {

  ll x = 0, f = 1;

  char ch = getchar();

  while (ch < '0' || ch > '9') {

    if (ch == '-') f = -1;

    ch = getchar();

  }

  while (ch <= '9' && ch >= '0') {

    x = 10 * x + ch - '0';

    ch = getchar();

  }

  return x * f;

}

int n, m;

void work() {

  n = read(), m = read();

  if (m == 0) {

    puts("1");

    return;

  }

  int num = n - m;

  printf("%d\n", min(num, m));

}

int main() {

  #ifdef LZT

    freopen("in", "r", stdin);

  #endif

  work();

  #ifdef LZT

    Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);

  #endif

}

B Cat Party

倒着枚举长度并且维护每一个出现次数的个数

就是维护出现次数为 \(i\) 的数字有多少个

显然出现次数最多只能有两种（因为你只能修改 \(1\) 个数，也就只能修改一个出现次数）

然后就分类讨论一下最后删掉的是哪个即可

// Copyright lzt

#include<stdio.h>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<vector>

#include<map>

#include<set>

#include<cmath>

#include<iostream>

#include<queue>

#include<string>

#include<ctime>

using namespace std;

typedef long long ll;

typedef std::pair<int, int> pii;

typedef long double ld;

typedef unsigned long long ull;

typedef std::pair<long long, long long> pll;

#define fi first

#define se second

#define pb push_back

#define mp make_pair

#define rep(i, j, k)  for (register int i = (int)(j); i <= (int)(k); i++)

#define rrep(i, j, k) for (register int i = (int)(j); i >= (int)(k); i--)

#define Debug(...) fprintf(stderr, __VA_ARGS__)

inline ll read() {

  ll x = 0, f = 1;

  char ch = getchar();

  while (ch < '0' || ch > '9') {

    if (ch == '-') f = -1;

    ch = getchar();

  }

  while (ch <= '9' && ch >= '0') {

    x = 10 * x + ch - '0';

    ch = getchar();

  }

  return x * f;

}

const int maxn = 100100;

int n, cnt;

int a[maxn];

int b[maxn], num[maxn];

void work() {

  n = read();

  rep(i, 1, n) a[i] = read(), b[a[i]]++;

  cnt = 0; set<int> col;

  rep(i, 1, 100000) {

    if (b[i]) {

      num[b[i]]++;

      cnt++, col.insert(b[i]);

    }

  }

  rrep(i, n, 2) {

    int nw = *col.begin();

    if (num[nw] == 1 && col.size() > 1) nw = *(++col.begin());

    if (col.size() <= 2) {

      bool ok = 0;

      if (col.size() == 1) {

        int nw = *col.begin();

        if (nw == 1 || num[nw] == 1) ok = 1;

        else ok = 0;

      } else {

        int A = *col.begin(), B = *(++col.begin());

        if (num[A] == 1 && A == 1) ok = 1;

        else if (num[B] == 1 && B == 1) ok = 1;

        else if (num[B] == 1 && B == A + 1) ok = 1;

        else ok = 0;

      }

      if (ok) {

        printf("%d\n", i);

        return;

      }

    }

    num[b[a[i]]]--;

    if (num[b[a[i]]] == 0) col.erase(b[a[i]]);

    b[a[i]]--;

    if (b[a[i]]) {

      num[b[a[i]]]++;

      if (num[b[a[i]]] == 1) col.insert(b[a[i]]);

    }

    else cnt--;

  }

  puts("1");

}

int main() {

  #ifdef LZT

    freopen("in", "r", stdin);

  #endif

  work();

  #ifdef LZT

    Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);

  #endif

}

C Power Transmission

答案显然是 \(calc(\text{不同的直线数量}) - \sum_{\text{斜率k}} calc(\text{斜率为k的直线数量})\)

其中 \(calc(x)=x*(x-1)/2\)

然后就枚举每两个点形成一条线，然后计数即可

// Copyright lzt

#include<stdio.h>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<vector>

#include<map>

#include<set>

#include<cmath>

#include<iostream>

#include<queue>

#include<string>

#include<ctime>

using namespace std;

typedef long long ll;

typedef std::pair<int, int> pii;

typedef long double ld;

typedef unsigned long long ull;

typedef std::pair<long long, long long> pll;

#define fi first

#define se second

#define pb push_back

#define mp make_pair

#define rep(i, j, k)  for (register int i = (int)(j); i <= (int)(k); i++)

#define rrep(i, j, k) for (register int i = (int)(j); i >= (int)(k); i--)

#define Debug(...) fprintf(stderr, __VA_ARGS__)

inline ll read() {

  ll x = 0, f = 1;

  char ch = getchar();

  while (ch < '0' || ch > '9') {

    if (ch == '-') f = -1;

    ch = getchar();

  }

  while (ch <= '9' && ch >= '0') {

    x = 10 * x + ch - '0';

    ch = getchar();

  }

  return x * f;

}

const int maxn = 1010;

int n;

pii p[maxn];

bool sm(double x, double y) {

  return fabs(x - y) <= 1e-8;

}

struct Line {

  // Ax + By + C = 0

  double A, B, C;

  bool operator == (const Line &b) const {

    return sm(A, b.A) && sm(B, b.B) && sm(C, b.C);

  }

  bool operator < (const Line &b) const {

    if (sm(A, b.A)) {

      if (sm(B, b.B)) return C < b.C;

      return B < b.B;

    }

    return A < b.A;

  }

  void init() {

    if (A == 0) C = C / B, B = 1;

    else B = B / A, C = C / A, A = 1;

  }

  double calc() {

    init();

    if (A == 1) return B;

    else return 1e30;

  }

};

map<double, map<double, int> > M;

map<double, int> shit;

vector<Line> v;

void work() {

  n = read();

  rep(i, 1, n) p[i].fi = read(), p[i].se = read();

  ll num = 0;

  rep(i, 1, n) rep(j, i + 1, n) {

    Line nw;

    nw.A = p[j].se - p[i].se;

    nw.B = p[i].fi - p[j].fi;

    nw.C = 0 - p[i].fi * nw.A - p[i].se * nw.B;

    nw.init();

    double k = nw.calc();

    if (M[k][nw.C] == 0) {

      M[k][nw.C] = 1;

      num++; shit[k]++;

    }

  }

  num = num * (num - 1) / 2;

  for (map<double, int>::iterator it = shit.begin(); it != shit.end(); it++) {

    int nw = it -> se;

    num = num - nw * 1ll * (nw - 1) / 2;

  }

  printf("%lld\n", num);

}

int main() {

  #ifdef LZT

    freopen("in", "r", stdin);

  #endif

  work();

  #ifdef LZT

    Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);

  #endif

}

D Mysterious Code

首先预处理 \(S[i][j]\) 和 \(T[i][j]\) 分别表示当前串的后缀与 \(s/t\) 的前缀的最长公共子串长度为 \(i\) （也就是只有最后 \(i\) 位有用，前面的都忽略），然后下一个位置填了字母 \(j\) 之后那个最长公共子串的长度（类似 \(kmp\) 的 \(fail\)）

直接 \(dp\) \(f[i][j][k]\) 表示当前长度为 \(i\)，在第一个串上的位置是 \(j\) ，在第二个串上的位置是 \(k\) 的最大 \(f(c',s)-f(c',t)\)

然后转移就暴力枚举一下下一个是啥

// Copyright lzt

#include<stdio.h>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<vector>

#include<map>

#include<set>

#include<cmath>

#include<iostream>

#include<queue>

#include<string>

#include<ctime>

using namespace std;

typedef long long ll;

typedef std::pair<int, int> pii;

typedef long double ld;

typedef unsigned long long ull;

typedef std::pair<long long, long long> pll;

#define fi first

#define se second

#define pb push_back

#define mp make_pair

#define rep(i, j, k)  for (register int i = (int)(j); i <= (int)(k); i++)

#define rrep(i, j, k) for (register int i = (int)(j); i >= (int)(k); i--)

#define Debug(...) fprintf(stderr, __VA_ARGS__)

inline ll read() {

  ll x = 0, f = 1;

  char ch = getchar();

  while (ch < '0' || ch > '9') {

    if (ch == '-') f = -1;

    ch = getchar();

  }

  while (ch <= '9' && ch >= '0') {

    x = 10 * x + ch - '0';

    ch = getchar();

  }

  return x * f;

}

const int maxn = 1010;

int n, m, len;

char s[maxn], t[maxn], c[maxn];

int f[maxn][55][55];

int S[55][26], T[55][26];

inline void getmx(int &x, int y) {

  if (x < y) x = y;

}

void work() {

  scanf("%s", c + 1); len = strlen(c + 1);

  scanf("%s", s + 1); n = strlen(s + 1);

  rep(i, 0, n) rep(j, 0, 25) {

    int len = i + 1;

    rrep(k, min(len, n), 1) {

      bool flag = true;

      rep(_, 1, k - 1) if (s[i - k + _ + 1] != s[_]) {

        flag = false; break;

      }

      if (s[k] - 'a' != j) flag = false;

      if (flag) {

        S[i][j] = k;

        break;

      }

    }

    // cout<<i<<' '<<j<<' '<<S[i][j]<<endl;

  }

  scanf("%s", t + 1); m = strlen(t + 1);

  rep(i, 0, m) rep(j, 0, 25) {

    int len = i + 1;

    rrep(k, min(len, m), 1) {

      bool flag = true;

      rep(_, 1, k - 1) if (t[i - k + _ + 1] != t[_]) {

        flag = false; break;

      }

      if (t[k] - 'a' != j) flag = false;

      if (flag) {

        T[i][j] = k;

        break;

      }

    }

    // cout<<i<<' '<<j<<' '<<T[i][j]<<endl;

  }

  rep(i, 0, len) rep(j, 0, n) rep(k, 0, m) f[i][j][k] = -1e9;

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

  rep(i, 0, len - 1) rep(j, 0, n) rep(k, 0, m) {

    if (f[i][j][k] == -1e9) continue;

    if (c[i + 1] != '*') {

      int nw = c[i + 1] - 'a', pl = 0;

      if (S[j][nw] == n) pl++;

      if (T[k][nw] == m) pl--;

      getmx(f[i + 1][S[j][nw]][T[k][nw]], f[i][j][k] + pl);

    } else {

      rep(nw, 0, 25) {

        int pl = 0;

        if (S[j][nw] == n) pl++;

        if (T[k][nw] == m) pl--;

        getmx(f[i + 1][S[j][nw]][T[k][nw]], f[i][j][k] + pl);

      }

    }

  }

  int ans = -1e9;

  rep(j, 0, n) rep(k, 0, m) {

    getmx(ans, f[len][j][k]);

  }

  printf("%d\n", ans);

}

int main() {

  #ifdef LZT

    freopen("in", "r", stdin);

  #endif

  work();

  #ifdef LZT

    Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);

  #endif

}

E Magical Permutation

// Copyright lzt

#include<stdio.h>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<vector>

#include<map>

#include<set>

#include<cmath>

#include<iostream>

#include<queue>

#include<string>

#include<ctime>

using namespace std;

typedef long long ll;

typedef std::pair<int, int> pii;

typedef long double ld;

typedef unsigned long long ull;

typedef std::pair<long long, long long> pll;

#define fi first

#define se second

#define pb push_back

#define mp make_pair

#define rep(i, j, k)  for (register int i = (int)(j); i <= (int)(k); i++)

#define rrep(i, j, k) for (register int i = (int)(j); i >= (int)(k); i--)

#define Debug(...) fprintf(stderr, __VA_ARGS__)

inline ll read() {

  ll x = 0, f = 1;

  char ch = getchar();

  while (ch < '0' || ch > '9') {

    if (ch == '-') f = -1;

    ch = getchar();

  }

  while (ch <= '9' && ch >= '0') {

    x = 10 * x + ch - '0';

    ch = getchar();

  }

  return x * f;

}

const int maxn = 200200;

int n, ans, pos;

int a[maxn], lb[20];

vector<int> vec;

bool vis[maxn << 2];

void add(int v) {

  int tmp = v;

  rrep(i, 20, 0) {

    if (v & (1 << i)) {

      if (lb[i]) v ^= lb[i];

      else {

        vec.pb(tmp);

        lb[i] = v;

        break;

      }

    }

  }

}

void dfs(int v, int num) {

  printf("%d ", v); vis[v] = 1;

  if (num == (1 << ans)) return;

  rep(i, 0, vec.size() - 1) if (!vis[v ^ vec[i]]) {

    dfs(v ^ vec[i], num + 1);

    break;

  }

}

void work() {

  n = read();

  rep(i, 1, n) a[i] = read();

  sort(a + 1, a + n + 1);

  ans = 0, pos = 1;

  rep(i, 1, 20) {

    while (pos <= n && a[pos] < (1 << i)) {

      add(a[pos]);

      pos++;

    }

    bool flag = true;

    rep(j, 0, i - 1) if (!lb[j]) flag = false;

    if (flag) ans = i;

  }

  memset(lb, 0, sizeof(lb));

  vec.clear();

  rep(i, 1, n) {

    if (a[i] < (1 << ans)) add(a[i]);

  }

  printf("%d\n", ans);

  dfs(0, 1);

}

int main() {

  #ifdef LZT

    freopen("in", "r", stdin);

  #endif

  work();

  #ifdef LZT

    Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);

  #endif

}

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