https://codeforces.com/contest/1151/problem/F
题意:给定一个01串,每次取出一组(l,r)1≤l<r≤n,交换s[l],s[r],求经过k次之后变成升序的概率是多大?
题解:构造矩阵+矩阵快速幂
现在我们知道有sn snsn个1和n−sn n-snn−sn个0,我们还知道多少个1在它的位置了,
共有四种0和1,
1 在规定位置,不在规定位置
0 在规定位置 不在规定位置
每一次交换可能引起的变化有三种:不变,+1,-1
目的是1全在规定位置,按照取出l,r列出转移矩阵就ok了
构造矩阵我看不懂,感觉还有一点DP的思想???
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,p,l,r,u,v;
int cnt,flag,temp,sum;
int a[N];
struct Matrix{
int n;
Matrix(int nn = 1):n(nn){ memset(a,0,sizeof(a));};
long long a[N][N];
void print(){
for(int i = 0;i <= n; ++i)
for(int j= 0;j <= n; ++j)
printf("%I64d%c",a[i][j]," \n"[j==n]);
}
Matrix operator*(const Matrix &b)const{
Matrix c(n);
for(int i = 0;i <= n; ++i){
for(int j = 0;j <= n; ++j){
for(int k = 0;k <= n; ++k){
c.a[i][j] += a[i][k] * b.a[k][j];
c.a[i][j] %= MOD;
}
}
}
//c.print();
return c;
}
};
Matrix ans,fac;
ll POW(ll a,ll b,ll c){
ll res=1;
ll base=a%c;
while(b){
if(b&1)res=(res*base)%c;
base=(base*base)%c;
b>>=1;
}
return res;
}
void MatrixPOW(ll k){
while(k){
if(k&1)ans=ans*fac;
fac=fac*fac;
k>>=1;
}
}
void init(){
cnt=0;
for(int i=0;i<n;i++) cnt += a[i]==1;
int now = 0;
for(int i = n-cnt;i < n;++i)
now += a[i]==1;
ans.n = fac.n = cnt;
ans.a[0][now] = 1;
for(int i = 0;i <= cnt; ++i){
// 四种情况 1 :i,sn-i
// 0 :sn-i,n-i-sn+i-sn+i = n-2*sn+i
fac.a[i][i] = n*(n-1)/2-(cnt-i)*(cnt-i)-i*(n-2*cnt+i);// 保持原状
if(i+1<=cnt)
fac.a[i][i+1] = (cnt-i)*(cnt-i);// 加1
if(i)
fac.a[i][i-1] = i*(n-2*cnt+i); // 减1
}
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
scanf("%I64d%I64d",&n,&k);
for(int i=0;i<n;i++)scanf("%d",&a[i]);
init();
//ans.print();
//fac.print();
MatrixPOW(k);
//ans.print();
cout<<ans.a[0][cnt]*POW(POW(n*(n-1)/2,k,MOD),MOD-2,MOD)%MOD<<endl;
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}