topcoder srm 684 div1

problem1 link

首先由$P$中任意两元素的绝对值得到集合$Q$。然后枚举$Q$中的每个元素作为集合$D$中的最大值$Max$，这样就能确定最后集合$D$中的最小值要大于等于$Min=\frac{Max+k-1}{k}$。然后再枚举$S$中元素的最小值即可依次从小到大确定$S$中的所有值。因为假设$S$中最小的元素是$x$，且当前最大元素是$y$，那么对于一个元素$z，z>y$,要么不选它；如果要选进来就要满足$z-x\leq Max$且$z-y\geq Min$。这样可以用一个简单的dp来解决。

problem2 link

设$s_{i}$表示连续$i$个元素任意两个相邻元素$x,y$满足$x>y$且$x$%$y$=0的方案数。$s$数组可以通过简单的dp来计算得到。

$f_{x}$表示长度为$x$的数组$A$的方案数。答案为$f_{n}$。$f_{0}=1$,$f_{t}=\sum_{r=1}^{t}f_{t-r}s_{r}(-1)^{r+1}=f_{t-1}s_{1}-f_{t-2}s_{2}+f_{t-3}s_{3}...$.其中$s_{1}=k$。

这个容斥可以这样理解：首先不理会$A_{n-1}$和$A_{n}$的关系，那么$A_{n}$有$s_{1}=k$种选择，所以$f_{n}=f_{n-1}s_{1}$。这样会导致$f_{n-1},f_{n}$不满足第二个条件，那么$f_{n-2}s_{2}$包含了所有的这种情况，同时$f_{n-2}s_{2}$还包含了$f_{n-2}$和$f_{n-1}$不满足第二个条件的情况，所以减去$f_{n-2}s_{2}$然后继续向后计算。

由于$k$最多只有$50000$，那么最多只会出现连续 16个数字是倍数关系，也就是说当$r>16$时$s_{r}=0$。所以$f_{t}=\sum_{r=1}^{min(t,16)}f_{t-r}s_{r}(-1)^{r+1}$

problem3 link

枚举排列的第一个数字。所有的情况都要乘上$m!$，所以把它最后再乘。

当$n=2$时，设为$a_{1},a_{2}$，那么答案为$\frac{1}{a_{1}+a_{2}}=a_{2}^{-1}\left (\frac{1}{a_{1}+a_{2}}  \right )$

当$n=3$时，设为$a_{1},a_{2},a_{3}$，那么第一个固定时答案为

$\frac{1}{(a_{1}+a_{2})(a_{1}+a_{2}+a_{3})}+\frac{1}{(a_{1}+a_{3})(a_{1}+a_{3}+a_{2})}$

$=a_{3}^{-1}\left ( (a_{1}+a_{2})^{-1}-(a_{1}+a_{2}+a_{3})^{-1} \right )+a_{2}^{-1}\left ( (a_{1}+a_{3})^{-1}-(a_{1}+a_{3}+a_{2})^{-1} \right )$

$=(a_{2}a_{3})^{-1}\left ( \frac{a_{2}}{a_{1}+a_{2}}+\frac{a_{3}}{a_{1}+a_{3}}-\frac{a_{2}+a_{3}}{a_{1}+a_{2}+a_{3}} \right )$

当$n=4$时，前两项为$a_{1},a_{2}$时，后两项有两种组合，和为$(a_{3}a_{4})^{-1}\left ( \frac{a_{3}}{a_{1}+a_{2}+a_{3}}+\frac{a_{4}}{a_{1}+a_{2}+a_{4}}-\frac{a_{3}+a_{4}}{a_{1}+a_{2}+a_{3}+a_{4}} \right )$，这相当于$a_{1}+a_{2}$是$n=3$时的$a_{1}$。

所以对答案的贡献为$\frac{1}{a_{1}+a_{2}}(a_{3}a_{4})^{-1}\left ( \frac{a_{3}}{a_{1}+a_{2}+a_{3}}+\frac{a_{4}}{a_{1}+a_{2}+a_{4}}-\frac{a_{3}+a_{4}}{a_{1}+a_{2}+a_{3}+a_{4}} \right )$

$=(a_{3}a_{4})^{-1}\left ( \frac{1}{a_{1}+a_{2}}-\frac{1}{a_{1}+a_{2}+a_{3}}+\frac{1}{a_{1}+a_{2}}-\frac{a_{1}}{a_{1}+a_{2}+a_{4}}-(\frac{1}{a_{1}+a_{2}}-\frac{1}{a_{1}+a_{2}+a_{3}+a_{4}}) \right )$

$=(a_{3}a_{4})^{-1}\left ( \frac{1}{a_{1}+a_{2}}-\frac{1}{a_{1}+a_{2}+a_{3}}-\frac{1}{a_{1}+a_{2}+a_{4}}+\frac{1}{a_{1}+a_{2}+a_{3}+a_{4}} \right )$

$=(a_{2}a_{3}a_{4})^{-1}\left ( \frac{a_{2}}{a_{1}+a_{2}}-\frac{a_{2}}{a_{1}+a_{2}+a_{3}}-\frac{a_{2}}{a_{1}+a_{2}+a_{4}}+\frac{a_{2}}{a_{1}+a_{2}+a_{3}+a_{4}} \right )$

所以加上前两项是$a_{1},a_{3}$以及$a_{1},a_{4}$，总的答案为

$(a_{2}a_{3}a_{4})^{-1}\left ( \frac{a_{2}}{a_{1}+a_{2}}+\frac{a_{3}}{a_{1}+a_{3}}+\frac{a_{4}}{a_{1}+a_{4}}-\frac{a_{2}+a_{3}}{a_{1}+a_{2}+a_{3}}-\frac{a_{2}+a_{4}}{a_{1}+a_{2}+a_{4}}-\frac{a_{3}+a_{4}}{a_{1}+a_{3}+a_{4}}+\frac{a_{2}+a_{3}+a_{4}}{a_{1}+a_{2}+a_{3}+a_{4}} \right )$

所以对于$n$项来说，第一项为$a_{1}$的答案为$\sum_{p\in P}(-1)^{1+|p|}\frac{\sum_{x\in p}x}{a_{1}+\sum_{x\in p}x}$。其中$P$表示$a_{2},a_{3},..,a_{n}$的任意非空子集的集合。理论上有$2^{n-1}-1$个。但是由于所有的数字之和最多为1000,所以可以直接统计和为$y$的有多少种。

code for problem1

#include <algorithm>
#include <vector>

class CliqueParty {
 public:
  int maxsize(std::vector<int> A, int k) {
    int n = static_cast<int>(A.size());
    std::sort(A.begin(), A.end());
    std::vector<int> a;
    for (int i = 0; i < n; ++i) {
      for (int j = i + 1; j < n; ++j) {
        a.push_back(A[j] - A[i]);
      }
    }
    std::sort(a.begin(), a.end());
    auto Cal = [&](const int Max) {
      const int Min = (Max + k - 1) / k;
      int ans = 0;
      for (int k = 0; k < n; ++k) {
        std::vector<std::vector<int>> f(n, std::vector<int>(n, -1));
        f[k][k] = 1;
        for (int i = k + 1; i < n; ++i) {
          for (int j = k; j <= i - 1; ++j) {
            f[i][j] = f[i - 1][j];
            if (f[i - 1][j] != -1 && A[i] - A[j] >= Min && A[i] - A[k] <= Max) {
              f[i][i] = std::max(f[i][i], f[i - 1][j] + 1);
            }
          }
        }
        for (int i = k; i < n; ++i) {
          ans = std::max(ans, f[n - 1][i]);
        }
      }
      return ans;
    };
    int result = 0;
    for (size_t i = 0; i < a.size(); ++i) {
      if (i == 0 || a[i] != a[i - 1]) {
        result = std::max(result, Cal(a[i]));
      }
    }
    return result;
  }
};

code for problem2

#include <vector>

class DivFree {
 public:
  int dfcount(int n, int k) {
    constexpr int kMaxT = 16;
    constexpr int kMod = 1000000007;
    std::vector<std::vector<int>> f(kMaxT + 1, std::vector<int>(k + 1));
    std::vector<int> s(kMaxT + 1);
    for (int i = 1; i <= k; ++i) {
      f[1][i] = 1;
    }
    s[1] = k;
    auto Add = [&](int &x, int y) {
      x += y;
      if (x >= kMod) {
        x -= kMod;
      }
    };
    for (int i = 2; i <= kMaxT; ++i) {
      for (int j = 1; j <= k; ++j) {
        for (int t = j + j; t <= k; t += j) {
          Add(f[i][t], f[i - 1][j]);
        }
      }
      for (int j = 1; j <= k; ++j) {
        Add(s[i], f[i][j]);
      }
    }
    std::vector<int> dp(n + 1);
    dp[0] = 1;
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= kMaxT && j <= i; ++j) {
        int k = static_cast<int>(1ll * dp[i - j] * s[j] % kMod);
        if (j % 2 == 1) {
          Add(dp[i], k);
        } else {
          Add(dp[i], kMod - k);
        }
      }
    }
    return dp[n];
  }
};

code for problem3

#include <algorithm>
#include <vector>

class Permutant {
 public:
  int counthis(const std::vector<int> &a) {
    constexpr int kMod = 1000000007;
    int n = static_cast<int>(a.size());
    int s = std::accumulate(a.begin(), a.end(), 0);
    std::vector<long long> inv(s + 1);
    inv[1] = 1;
    for (int i = 2; i <= s; ++i) {
      inv[i] = (kMod - kMod / i) * inv[kMod % i] % kMod;
    }
    auto Add = [&](int &x, long long y) {
      x += static_cast<int>(y % kMod);
      if (x >= kMod) {
        x -= kMod;
      }
    };
    int result = 0;
    for (int t = 0; t < n; ++t) {
      std::vector<std::vector<std::vector<int>>> dp(
          n + 1, std::vector<std::vector<int>>(2, std::vector<int>(s + 1)));
      dp[0][0][0] = 1;
      for (int i = 0; i < n; ++i) {
        for (int j = 0; j < 2; ++j) {
          for (int k = 0; k <= s; ++k) {
            Add(dp[i + 1][j ^ 1][k + a[i]], dp[i][j][k]);
            if (i != t) {
              Add(dp[i + 1][j][k], dp[i][j][k]);
            }
          }
        }
      }
      int num = 0;
      for (int i = 0; i < 2; ++i) {
        for (int j = 0; j <= s; ++j) {
          long long x = inv[j] * (j - a[t]) % kMod;
          if (i == 0) {
            Add(num, x * dp[n][i][j] % kMod);
          } else {
            Add(num, (kMod - x) * dp[n][i][j] % kMod);
          }
        }
      }
      for (int i = 0; i < n; ++i) {
        if (i != t) {
          num = static_cast<int>(inv[a[i]] * num % kMod);
        }
      }
      Add(result, num);
    }
    if (n == 1) {
      result = 1;
    }
    for (int i = 1; i <= s; ++i) {
      result = static_cast<int>(1ll * result * i % kMod);
    }
    return result;
  }
};