题目链接
题解
可以发现任意两个位置(A,B)最终位置关系的概率是相等的
如果数列是这样：
CCCCACCCCBCCCC
那么最终有(7)种位置关系
((A,B))
((A,C))
((B,A))
((B,C))
((C,A))
((C,B))
((C,C))
手玩出(7 imes 7)的转移矩阵，矩乘后即可得到各种位置关系的概率
然后枚举(B)，用树状数组维护与(A)有关的量组合计算即可
写得极烦
放开我我没疯
我只是计数时把大小弄反了
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
#define lbt(x) (x & -x)
using namespace std;
const int maxn = 500005,maxm = 100005,INF = 0x3f3f3f3f,P = 1000000007;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
return flag ? out : -out;
}
struct Matrix{
int s[7][7],n,m;
Matrix(){cls(s,0); n = m = 0;}
}M,F0,F;
inline Matrix operator *(const Matrix& a,const Matrix& b){
Matrix c;
if (a.m != b.n) return c;
c.n = a.n; c.m = b.m;
for (int i = 0; i < c.n; i++)
for (int j = 0; j < c.m; j++)
for (int k = 0; k < a.m; k++)
c.s[i][j] = (c.s[i][j] + 1ll * a.s[i][k] * b.s[k][j] % P) % P;
return c;
}
inline Matrix qpow(Matrix a,LL b){
Matrix re; re.n = re.m = a.n;
for (int i = 0; i < re.n; i++) re.s[i][i] = 1;
for (; b; b >>= 1,a = a * a)
if (b & 1) re = re * a;
return re;
}
inline int qpow(int a,int b){
int re = 1;
for (; b; b >>= 1,a = 1ll * a * a % P)
if (b & 1) re = 1ll * re * a % P;
return re;
}
inline void Add(int& x,int y){x += y; x >= P ? x -= P : 0;}
int n,K,A[maxn];
void init(int* a,int A,int B,int C,int D,int E,int F,int G){
a[0] = A; a[1] = B; a[2] = C; a[3] = D; a[4] = E; a[5] = F; a[6] = G;
}
void cal(){
M.n = M.m = 7;
int a = 1ll * (n - 2) * (n - 3) / 2 % P,b = ((1ll * n * (n - 1) / 2 % P - n) % P + P) % P;
int c = n - 2,d = n - 3,e = ((1ll * n * (n - 1) / 2 % P - 4) % P + P) % P;
init(M.s[0],a,1,1,0,0,1,0);
init(M.s[1],c,b,0,1,1,0,1);
init(M.s[2],1,0,a,1,1,0,0);
init(M.s[3],0,1,c,b,0,1,1);
init(M.s[4],0,1,c,0,b,1,1);
init(M.s[5],c,0,0,1,1,b,1);
init(M.s[6],0,d,0,d,d,d,e);
F0.n = 7; F0.m = 1; F0.s[0][0] = 1;
F = qpow(M,K) * F0;
}
int s[maxn],pos[maxn],vpos[maxn];
void add(int* s,int u,int v){while (u <= n) s[u] = (s[u] + v) % P,u += lbt(u);}
int query(int* s,int u){int re = 0; while (u) re = (re + s[u]) % P,u -= lbt(u); return re;}
void work(){
int ans = 0;
int p1 = F.s[0][0],p2 = F.s[1][0],p3 = F.s[2][0],p4 = F.s[3][0];
int p5 = F.s[4][0],p6 = F.s[5][0],p7 = F.s[6][0],inv = qpow(n - 2,P - 2);
int inv2 = qpow(2,P - 2);
LL sumf = 0,sumg = 0,a,b,fa,fb,ga,gb;
for (int i = 1; i <= n; i++){
a = query(s,A[i]); fa = query(pos,A[i]); ga = query(vpos,A[i]);
b = i - 1 - a; fb = sumf - fa,gb = sumg - ga;
Add(ans,1ll * b * p1 % P);
Add(ans,(1ll * a * (n - i) % P + 1ll * b * (i - 2) % P) * inv % P * p2 % P);
Add(ans,1ll * a * p3 % P);
Add(ans,1ll * (fb + ga) % P * inv % P * p4 % P);
Add(ans,(1ll * a * (i - 2) % P + 1ll * b * (n - i) % P % P) % P * inv % P * p5 % P);
Add(ans,1ll * (gb + fa) % P * inv % P * p6 % P);
add(s,A[i],1); add(pos,A[i],i - 1); add(vpos,A[i],n - i - 1);
sumf += i - 1; sumg += n - i - 1;
}
Add(ans,1ll * n * (n - 1) / 2 % P * inv2 % P * p7 % P);
printf("%d
",(ans % P + P) % P);
}
int main(){
n = read(); K = read();
REP(i,n) A[i] = read();
cal();
work();
return 0;
}