[做题记录-计数相关] [AGC023E] Inversions
题意
一个长度为\(n\)的排列数组， 每个位置有上限的限制， 求所有合法的排列的逆序数的和。
\(n \leq 10^5\)
题解
经典套路是先计数一下序列的个数然后考虑每一对数对答案的贡献。
考虑从大往小填数， 记\(b_i\)表示\(a_i \ge i\)的位置个数， 那么合法的排列个数为：
\[cnt = \prod_i b_i - (n - i)
\]
然后现在对于序列上一对位置考虑。不妨设\(i < j\), \(p_i\)表示填好以后\(i\)位置上的数是什么。
当\(a_i = a_j\)时， 显然\(p_i >p_j\)和\(p_i < p_j\)的情况数相同， 贡献是\(\frac{cnt}{2}\)。
当\(a_i < a_j\)的时候， 讨论\(a_j\)的取值。当\(a_j \in [a_i + 1, a_j]\)的时候这里肯定是没有贡献的， 那么考虑强行让\(a_j = a_i\)， 那么这样的话会发现\(b_i : which \ i \in [a_i + 1, a_j]\)会减少\(1\)。那么这里的贡献是:
\[\frac{cnt}{2} \prod _{k = a_i + 1}^{a_j}\frac{b_k - (n - k) - 1}{b_k - (n - k)}
\]
当\(a_i > a_j\)的时候， 不妨把\(i, j\)反过来， 讨论就变成了\(a_j < a_i\)， 那么这里的顺序对就和上面情况中逆序对的情况是一样的， 用总方案减去出现顺序对的情况即可。
\[cnt - \frac{cnt}{2} \prod _{k = a_j + 1}^{a_i}\frac{b_k - (n - k) - 1}{b_k - (n - k)}
\]
然后考虑快速计算， 以\(a_i < a_j\)为例， 考虑从小往大枚举\(a_j\)， 维护一个全局的数据结构， 每次\(a_j\)增大的时候全局乘， 查询的时候查位置小于\(j\)的所有位置的权值和， 然后在\(j\)位置加入一个\(\frac{cnt}{2}\)。
/*
QiuQiu /qq
____ _ _ __
/ __ \ (_) | | / /
| | | | _ _ _ | | _ _ / / __ _ __ _
| | | | | | | | | | | | | | | | / / / _` | / _` |
| |__| | | | | |_| | | | | |_| | / / | (_| | | (_| |
\___\_\ |_| \__,_| |_| \__, | /_/ \__, | \__, |
__/ | | | | |
|___/ |_| |_|
*/
#include <bits/stdc++.h>
using namespace std;
class Input {
#define MX 1000000
private :
char buf[MX], *p1 = buf, *p2 = buf;
inline char gc() {
if(p1 == p2) p2 = (p1 = buf) + fread(buf, 1, MX, stdin);
return p1 == p2 ? EOF : *(p1 ++);
}
public :
Input() {
#ifdef Open_File
freopen("a.in", "r", stdin);
freopen("a.out", "w", stdout);
#endif
}
template <typename T>
inline Input& operator >>(T &x) {
x = 0; int f = 1; char a = gc();
for(; ! isdigit(a); a = gc()) if(a == '-') f = -1;
for(; isdigit(a); a = gc())
x = x * 10 + a - '0';
x *= f;
return *this;
}
inline Input& operator >>(char &ch) {
while(1) {
ch = gc();
if(ch != '\n' && ch != ' ') return *this;
}
}
inline Input& operator >>(char *s) {
int p = 0;
while(1) {
s[p] = gc();
if(s[p] == '\n' || s[p] == ' ' || s[p] == EOF) break;
p ++;
}
s[p] = '\0';
return *this;
}
#undef MX
} Fin;
class Output {
#define MX 1000000
private :
char ouf[MX], *p1 = ouf, *p2 = ouf;
char Of[105], *o1 = Of, *o2 = Of;
void flush() { fwrite(ouf, 1, p2 - p1, stdout); p2 = p1; }
inline void pc(char ch) {
* (p2 ++) = ch;
if(p2 == p1 + MX) flush();
}
public :
template <typename T>
inline Output& operator << (T n) {
if(n < 0) pc('-'), n = -n;
if(n == 0) pc('0');
while(n) *(o1 ++) = (n % 10) ^ 48, n /= 10;
while(o1 != o2) pc(* (--o1));
return *this;
}
inline Output & operator << (char ch) {
pc(ch); return *this;
}
inline Output & operator <<(const char *ch) {
const char *p = ch;
while( *p != '\0' ) pc(* p ++);
return * this;
}
~Output() { flush(); }
#undef MX
} Fout;
#define cin Fin
#define cout Fout
#define endl '\n'
using LL = long long;
inline int log2(unsigned int x);
inline int popcount(unsigned x);
inline int popcount(unsigned long long x);
template <int mod>
class Int {
private :
inline int Mod(int x) { return x + ((x >> 31) & mod); }
inline int power(int x, int k) {
int res = 1;
while(k) {
if(k & 1) res = 1LL * x * res % mod;
x = 1LL * x * x % mod; k >>= 1;
}
return res;
}
public :
int v;
Int(int _v = 0) : v(_v) {}
operator int() { return v; }
inline Int operator =(Int x) { return Int(v = x.v); }
inline Int operator =(int x) { return Int(v = x); }
inline Int operator *(Int x) { return Int(1LL * v * x.v % mod); }
inline Int operator *(int x) { return Int(1LL * v * x % mod); }
inline Int operator +(Int x) { return Int( Mod(v + x.v - mod) ); }
inline Int operator +(int x) { return Int( Mod(v + x - mod) ); }
inline Int operator -(Int x) { return Int( Mod(v - x.v) ); }
inline Int operator -(int x) { return Int( Mod(v - x) ); }
inline Int operator ~() { return Int(power(v, mod - 2)); }
inline Int operator +=(Int x) { return Int(v = Mod(v + x.v - mod)); }
inline Int operator +=(int x) { return Int(v = Mod(v + x - mod)); }
inline Int operator -=(Int x) { return Int(v = Mod(v - x.v)); }
inline Int operator -=(int x) { return Int(v = Mod(v - x)); }
inline Int operator *=(Int x) { return Int(v = 1LL * v * x.v % mod); }
inline Int operator *=(int x) { return Int(v = 1LL * v * x % mod); }
inline Int operator /=(Int x) { return Int(v = v / x.v); }
inline Int operator /=(int x) { return Int(v = v / x); }
inline Int operator ^(int k) { return Int(power(v, k)); }
} ;
using mint = Int<(int) (1e9 + 7)>;
const int N = 2e5 + 10;
const mint inv2 = ~ mint(2);
int n;
int a[N], b[N];
mint cnt;
struct Node {
Node *ls, *rs;
mint cj, tg;
int l, r;
Node() {}
Node(int _l, int _r) : l(_l), r(_r), cj(0), tg(1), ls(NULL), rs(NULL) {}
void upd() {
cj = ls -> cj + rs -> cj;
}
void downcj(mint v) { cj *= v; tg *= v; }
void pushdown() {
if(tg != 1) {
ls -> downcj(tg);
rs -> downcj(tg);
tg = 1;
}
}
void modify(int pos, mint v) {
if(l == r) { cj = v; return ; }
pushdown();
int mid = (l + r) >> 1;
if(pos <= mid) ls -> modify(pos, v);
else rs -> modify(pos, v);
upd();
}
mint qry(int L, int R) {
if(L <= l && r <= R) return cj;
pushdown();
int mid = (l + r) >> 1;
mint res = 0;
if(L <= mid) res += ls -> qry(L, R);
if(R > mid) res += rs -> qry(L, R);
return res;
}
void mul(mint v) { downcj(v); return ; }
} ;
Node *root;
Node *build(int l, int r) {
Node * x = new Node(l, r);
if(l == r) return x;
int mid = (l + r) >> 1;
x -> ls = build(l, mid);
x -> rs = build(mid + 1, r);
return x;
}
using pii = pair<int, int>;
int c[N];
#define lowbit(x) (x & -x)
void upd(int x, int y) {
for(; x <= n; x += lowbit(x)) c[x] += y;
}
int qry(int x) {
int ans = 0;
for(; x; x -= lowbit(x)) ans += c[x];
return ans;
}
int main() {
cin >> n;
for(int i = 1; i <= n; i ++) cin >> a[i];
for(int i = 1; i <= n; i ++) b[a[i]] ++;
for(int i = n; i >= 1; i --) b[i] += b[i + 1];
cnt = 1;
for(int i = 1; i <= n; i ++) cnt = cnt * (b[i] - (n - i));
root = build(1, n);
mint ans = 0;
static vector<int> lim[N];
for(int i = 1; i <= n; i ++) lim[a[i]].push_back(i);
for(int i = 1; i <= n; i ++) {
mint value = b[i] - (n - i) - 1;
value = value * (~ (value + 1));
root -> mul(value);
for(int j : lim[i]) ans += root -> qry(1, j);
for(int j : lim[i]) root -> modify(j, cnt * inv2);
mint t = lim[i].size();
ans += t * (t - 1) * inv2 * cnt * inv2;
}
//cout << ans << endl;
for(int i = n; i >= 1; i --) {
ans += cnt * qry(a[i] - 1);
upd(a[i], 1);
}
// cerr << ans << endl;
root = build(1, n);
for(int i = 1; i <= n; i ++) {
mint value = b[i] - (n - i) - 1;
value = value * (~ (value + 1));
root -> mul(value);
for(int j : lim[i]) ans -= root -> qry(j, n);
for(int j : lim[i]) root -> modify(j, cnt * inv2);
//mint t = lim[i].size();
//ans -= t * (t - 1) * inv2 * cnt * inv2;
}
cout << ans << endl;
return 0;
}
inline int log2(unsigned int x) { return __builtin_ffs(x); }
inline int popcount(unsigned int x) { return __builtin_popcount(x); }
inline int popcount(unsigned long long x) { return __builtin_popcountl(x); }
