目录
一、题目
1、题目描述
2、输入输出
2.1输入
2.2输出
3、原题链接
二、解题报告
1、思路分析
2、复杂度
3、代码详解
一、题目
1、题目描述
2、输入输出
2.1输入
2.2输出
3、原题链接
G - Another Shuffle Window
二、解题报告
1、思路分析
不难用树状数组计算全局逆序对tot
我们可以在滑窗过程中维护当前 k-子数组的逆序对数目cur(不计入滑窗内的数和滑窗外的数字构成的逆序对)
对于滑窗的每个时刻,滑窗外元素对逆序对的贡献为 tot - cur
也就是说,每个滑窗,不带滑窗内会有tot - cur个逆序对
每个滑窗这一部分的贡献为 (tot - cur) * (n - k + 1),因为 n - k + 1个滑窗等概率,所以贡献比例相同,这一部分的答案我们滑窗过程中直接累加
关键在于如何理解滑窗内元素之间的逆序对数目?
我们发现我们只需计算子数组内产生的逆序对,而对于一个数组而言,数组内任意两个数之间构成逆序对的数目为 1 / 2,而 k-数组 pair 的数目为 k * (k - 1) / 2,每个pair 贡献1个逆序对的概率为1/2,根据二项分布的数学期望计算公式:E = np可得,k-数组的逆序对贡献期望为 k * (k - 1) / 4
那么答案就是把两部分加起来,详细看代码
2、复杂度
时间复杂度: O(NlogN)空间复杂度:O(N)
3、代码详解
#include <bits/stdc++.h>
// #define DEBUG
using u32 = unsigned;
using i64 = long long;
using u64 = unsigned long long;
constexpr int inf32 = 1E9 + 7;
constexpr i64 inf64 = 1E18 + 7;
template<class T>
constexpr T power(T a, i64 b) {
T res = 1;
for (; b; b /= 2, a *= a) {
if (b % 2) {
res *= a;
}
}
return res;
}
constexpr i64 mul(i64 a, i64 b, i64 p) {
i64 res = a * b - i64(1.L * a * b / p) * p;
res %= p;
if (res < 0) {
res += p;
}
return res;
}
template<i64 P>
struct MLong {
i64 x;
constexpr MLong() : x{} {}
constexpr MLong(i64 x) : x{norm(x % getMod())} {}
static i64 Mod;
constexpr static i64 getMod() {
if (P > 0) {
return P;
} else {
return Mod;
}
}
constexpr static void setMod(i64 Mod_) {
Mod = Mod_;
}
constexpr i64 norm(i64 x) const {
if (x < 0) {
x += getMod();
}
if (x >= getMod()) {
x -= getMod();
}
return x;
}
constexpr i64 val() const {
return x;
}
explicit constexpr operator i64() const {
return x;
}
constexpr MLong operator-() const {
MLong res;
res.x = norm(getMod() - x);
return res;
}
constexpr MLong inv() const {
assert(x != 0);
return power(*this, getMod() - 2);
}
constexpr MLong &operator*=(MLong rhs) & {
x = mul(x, rhs.x, getMod());
return *this;
}
constexpr MLong &operator+=(MLong rhs) & {
x = norm(x + rhs.x);
return *this;
}
constexpr MLong &operator-=(MLong rhs) & {
x = norm(x - rhs.x);
return *this;
}
constexpr MLong &operator/=(MLong rhs) & {
return *this *= rhs.inv();
}
friend constexpr MLong operator*(MLong lhs, MLong rhs) {
MLong res = lhs;
res *= rhs;
return res;
}
friend constexpr MLong operator+(MLong lhs, MLong rhs) {
MLong res = lhs;
res += rhs;
return res;
}
friend constexpr MLong operator-(MLong lhs, MLong rhs) {
MLong res = lhs;
res -= rhs;
return res;
}
friend constexpr MLong operator/(MLong lhs, MLong rhs) {
MLong res = lhs;
res /= rhs;
return res;
}
friend constexpr std::istream &operator>>(std::istream &is, MLong &a) {
i64 v;
is >> v;
a = MLong(v);
return is;
}
friend constexpr std::ostream &operator<<(std::ostream &os, const MLong &a) {
return os << a.val();
}
friend constexpr bool operator==(MLong lhs, MLong rhs) {
return lhs.val() == rhs.val();
}
friend constexpr bool operator!=(MLong lhs, MLong rhs) {
return lhs.val() != rhs.val();
}
};
template<>
i64 MLong<0LL>::Mod = i64(1E18) + 9;
template<int P>
struct MInt {
int x;
constexpr MInt() : x{} {}
constexpr MInt(i64 x) : x{norm(x % getMod())} {}
static int Mod;
constexpr static int getMod() {
if (P > 0) {
return P;
} else {
return Mod;
}
}
constexpr static void setMod(int Mod_) {
Mod = Mod_;
}
constexpr int norm(int x) const {
if (x < 0) {
x += getMod();
}
if (x >= getMod()) {
x -= getMod();
}
return x;
}
constexpr int val() const {
return x;
}
explicit constexpr operator int() const {
return x;
}
constexpr MInt operator-() const {
MInt res;
res.x = norm(getMod() - x);
return res;
}
constexpr MInt inv() const {
assert(x != 0);
return power(*this, getMod() - 2);
}
constexpr MInt &operator*=(MInt rhs) & {
x = 1LL * x * rhs.x % getMod();
return *this;
}
constexpr MInt &operator+=(MInt rhs) & {
x = norm(x + rhs.x);
return *this;
}
constexpr MInt &operator-=(MInt rhs) & {
x = norm(x - rhs.x);
return *this;
}
constexpr MInt &operator/=(MInt rhs) & {
return *this *= rhs.inv();
}
friend constexpr MInt operator*(MInt lhs, MInt rhs) {
MInt res = lhs;
res *= rhs;
return res;
}
friend constexpr MInt operator+(MInt lhs, MInt rhs) {
MInt res = lhs;
res += rhs;
return res;
}
friend constexpr MInt operator-(MInt lhs, MInt rhs) {
MInt res = lhs;
res -= rhs;
return res;
}
friend constexpr MInt operator/(MInt lhs, MInt rhs) {
MInt res = lhs;
res /= rhs;
return res;
}
friend constexpr std::istream &operator>>(std::istream &is, MInt &a) {
i64 v;
is >> v;
a = MInt(v);
return is;
}
friend constexpr std::ostream &operator<<(std::ostream &os, const MInt &a) {
return os << a.val();
}
friend constexpr bool operator==(MInt lhs, MInt rhs) {
return lhs.val() == rhs.val();
}
friend constexpr bool operator!=(MInt lhs, MInt rhs) {
return lhs.val() != rhs.val();
}
};
template<>
int MInt<0>::Mod = 998244353;
template<int V, int P>
constexpr MInt<P> CInv = MInt<P>(V).inv();
constexpr int P = 998244353;
using Z = MInt<P>;
template<typename T>
class FenWick {
private:
int n;
std::vector<T> tr;
public:
FenWick(int _n) : n(_n), tr(_n + 1)
{}
FenWick(const std::vector<T> &_init) : FenWick(_init.size()) {
init(_init);
}
void init(const std::vector<T> &_init) {
for (int i = 1; i <= n; ++ i) {
tr[i] += _init[i - 1];
int j = i + (i & -i);
if (j <= n)
tr[j] += tr[i];
}
}
void add(int x, int k) {
for (; x <= n; x += x & -x) tr[x] += k;
}
void add(int l, int r, T k) {
add(l, k);
if (r + 1 <= n)
add(r + 1, -k);
}
T query(int x) const {
T res = T{};
for (; x; x &= x - 1) {
res += tr[x];
}
return res;
}
T query(int l, int r) const {
if (l > r) return T{};
return query(r) - query(l - 1);
}
int select(int k) {
int x = 0;
T cur{};
for (int i = 1 << std::__lg(n); i; i /= 2) {
if (x + i <= n && cur + tr[x + i] < k) {
x += i;
cur = cur + tr[x];
}
}
return x + 1;
}
void clear() {
tr.assign(n + 1, T{});
}
};
void solve() {
int n, k;
std::cin >> n >> k;
std::vector<int> p(n);
for (int i = 0; i < n; ++ i) {
std::cin >> p[i];
}
FenWick<Z> fen(n);
Z tot = 0;
for (int i = n - 1; ~i; -- i) {
tot += fen.query(p[i]);
fen.add(p[i], 1);
}
fen.clear();
Z cur = 0;
for (int i = k - 1; ~i; -- i) {
cur += fen.query(p[i]);
fen.add(p[i], 1);
}
Z ans = tot - cur;
for (int i = 0; i + k < n; ++ i) {
fen.add(p[i], -1);
cur -= fen.query(p[i]);
cur += fen.query(p[i + k], n);
fen.add(p[i + k], 1);
ans += tot - cur;
}
std::cout << (ans * Z(n - k + 1).inv() + Z(4).inv() * Z(k) * Z(k - 1)) << '\n';
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
#ifdef DEBUG
int START = clock();
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
int t = 1;
// std::cin >> t;
while (t --) {
solve();
}
#ifdef DEBUG
std::cerr << "run-time: " << clock() - START << '\n';
#endif
return 0;
}