题目描述
给你一个长度为(n)的排列(a)，每次要选择两个数，交换这两个数（这两个数可以相同）。总共要交换(k)次。
最后要统计数列中有多少位置(i)满足(max_{jleq i}a_i=a_i)。求前面这个东西的期望。
(nleq 100,kleq 80)
题解
我们枚举每个数(y)每在个位置(x)的贡献。把其他数中大于(y)的看成(1)，把其他数中小于(y)的看成(0)，然后DP。
设(f_{i,j,k})为交换了(i)次，(1)~(x-1)有(j)个(1)，（(k)在下面解释）的方案数
两条竖线中间是位置(x)
(k=0)：(|y|)
(k=1)：(y|0|)
(k=2)：(y|1|)
(k=3)：(|0|y)
(k=4)：(|1|y)
转移很多很繁琐，大家自己去推吧。。。
时间复杂度：总共有(O(n^2))次DP，每次(O(nk))，总的时间复杂度是(O(n^3k))
代码
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<ctime>
#include<cstdlib>
#include<utility>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
void open(const char *s)
{
#ifndef ONLINE_JUDGE
char str[100];
sprintf(str,"%s.in",s);
freopen(str,"r",stdin);
sprintf(str,"%s.out",s);
freopen(str,"w",stdout);
#endif
}
const ll p=1000000007;
ll fp(ll a,ll b)
{
ll s=1;
for(;b;b>>=1,a=a*a%p)
if(b&1)
s=s*a%p;
return s;
}
int x,y,i,j,k;
int n,m;
int a[100010];
ll f[90][110][5];
void add(ll &a,ll b)
{
a=(a+b)%p;
}
ll qian0(){return (k==1||k==2)?x-2-j:x-1-j;}
ll qian1(){return j;}
ll hou0(){return (k==2?y+j-x+1:(k==3?y+j-x-1:y+j-x));}
ll hou1(){return (k==2||k==4)?n-y-j-1:n-y-j;}
ll qian0(int k){return (k==1||k==2)?x-2-j:x-1-j;}
ll qian1(int k){return j;}
ll hou0(int k){return (k==2?y+j-x+1:(k==3?y+j-x-1:y+j-x));}
ll hou1(int k){return (k==2||k==4)?n-y-j-1:n-y-j;}
int now(){return (k==1||k==3)?0:((k==2||k==4)?1:-1);}
int where(){return (k==1||k==2)?0:1;}
ll sqr(ll x){return x*x%p;}
ll gao()
{
memset(f,0,sizeof f);
int h=0,hh;
for(i=1;i<x;i++)
if(a[i]>y)
h++;
for(i=1;i<=n;i++)
if(a[i]==y)
{
hh=i;
break;
}
if(hh==x)
f[0][h][0]=1;
else if(hh<x)
if(a[x]<y)
f[0][h][1]=1;
else
f[0][h][2]=1;
else
if(a[x]<y)
f[0][h][3]=1;
else
f[0][h][4]=1;
for(i=0;i<m;i++)
for(j=0;j<y;j++)
{
for(k=0;k<=4;k++)
if(f[i][j][k])
{
add(f[i+1][j][k],f[i][j][k]*sqr(qian0()+qian1()));
add(f[i+1][j][k],f[i][j][k]*sqr(hou0()+hou1()));
add(f[i+1][j][k],f[i][j][k]*2*qian0()%p*hou0());
add(f[i+1][j][k],f[i][j][k]*2*qian1()%p*hou1());
add(f[i+1][j][k],f[i][j][k]);
if(j)
add(f[i+1][j-1][k],f[i][j][k]*2*qian1()%p*hou0());
if(j<y-1)
add(f[i+1][j+1][k],f[i][j][k]*2*qian0()%p*hou1());
if(k)
add(f[i+1][j][k],f[i][j][k]);
}
if(f[i][j][0])
{
add(f[i+1][j][1],f[i][j][0]*2*qian0(0));
if(j)
add(f[i+1][j-1][2],f[i][j][0]*2%p*qian1(0));
add(f[i+1][j][3],f[i][j][0]*2*hou0(0));
add(f[i+1][j][4],f[i][j][0]*2*hou1(0));
}
if(f[i][j][1])
{
add(f[i+1][j][0],f[i][j][1]*2);
add(f[i+1][j][1],f[i][j][1]*2*(qian0(1)+qian1(1)));
add(f[i+1][j][3],f[i][j][1]*2*hou0(1));
if(j<y-1)
add(f[i+1][j+1][3],f[i][j][1]*2%p*hou1(1));
add(f[i+1][j][1],f[i][j][1]*2*(qian0(1)+hou0(1)));
add(f[i+1][j][2],f[i][j][1]*2*hou1(1));
if(j)
add(f[i+1][j-1][2],f[i][j][1]*2%p*qian1(1));
}
if(f[i][j][2])
{
if(j<y-1)
add(f[i+1][j+1][0],f[i][j][2]*2);
add(f[i+1][j][2],f[i][j][2]*2*(qian0(2)+qian1(2)));
add(f[i+1][j][4],f[i][j][2]*2*hou0(2));
if(j<y-1)
add(f[i+1][j+1][4],f[i][j][2]*2%p*hou1(2));
add(f[i+1][j][1],f[i][j][2]*2*hou0(2));
if(j<y-1)
add(f[i+1][j+1][1],f[i][j][2]*2%p*qian0(2));
add(f[i+1][j][2],f[i][j][2]*2*(qian1(2)+hou1(2)));
}
if(f[i][j][3])
{
add(f[i+1][j][0],f[i][j][3]*2);
add(f[i+1][j][3],f[i][j][3]*2*(hou0(3)+hou1(3)));
add(f[i+1][j][1],f[i][j][3]*2*qian0(3));
if(j)
add(f[i+1][j-1][1],f[i][j][3]*2*qian1(3));
add(f[i+1][j][3],f[i][j][3]*2*(qian0(3)+hou0(3)));
add(f[i+1][j][4],f[i][j][3]*2*hou1(3));
if(j)
add(f[i+1][j-1][4],f[i][j][3]*2*qian1(3));
}
if(f[i][j][4])
{
add(f[i+1][j][0],f[i][j][4]*2);
add(f[i+1][j][4],f[i][j][4]*2*(hou0(4)+hou1(4)));
add(f[i+1][j][2],f[i][j][4]*2*qian0(4));
if(j)
add(f[i+1][j-1][2],f[i][j][4]*2*qian1(4));
add(f[i+1][j][3],f[i][j][4]*2*hou0(4));
if(j<y-1)
add(f[i+1][j+1][3],f[i][j][4]*2*qian0(4));
add(f[i+1][j][4],f[i][j][4]*2*(qian1(4)+hou1(4)));
}
}
return f[m][0][0];
}
int main()
{
open("pltj");
scanf("%d%d",&n,&m);
int i;
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
ll ans=0;
for(x=1;x<=n;x++)
for(y=1;y<=n;y++)
if(y>=x)
ans=(ans+gao())%p;
ans=ans*fp(fp(n,2*m),p-2)%p;
printf("%lld
",ans);
return 0;
}