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;
}
};
