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

参考文章:题解 构造矩阵 矩阵快速幂

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;
}