1523. K-inversions

1523. K-inversions

Time limit: 1.0 second Memory limit: 64 MB

Consider a permutation a₁, a₂, …, a_n (all a_i are different integers in range from 1 to n). Let us call k-inversion a sequence of numbers i₁, i₂, …, i_k such that 1 ≤ i₁ < i₂ < … < i_k ≤ n and a_i1 > a_i2 > … > a_ik. Your task is to evaluate the number of different k-inversions in a given permutation.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 20000, 2 ≤ k ≤ 10). The second line is filled with n numbers a_i.

Output

Output a single number — the number of k-inversions in a given permutation. The number must be taken modulo 10⁹.

Samples

input	output
3 2 3 1 2	2
5 3 5 4 3 2 1	10

Problem Author: Dmitry Gozman Problem Source: Dmitry Gozman Contest 1, Petrozavodsk training camp, January

************************************************************************************************

树状数组（加快运算时间）

************************************************************************************************

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int   N=20005;
const  int MOD=1000000000;
int a[N],dp[2][N];
int c[N];
int i,j,k,n;
int sum;
int lowbit(int x)//低位技术
{
    return x&(-x);
}
void insert(int x,int v)//修改
 {
     for(;x<=n;x+=lowbit(x))
      c[x]=(c[x]+v)%MOD;
 }
 int query(int x)//查询
  {
      int suma=0;
      for(;x>0;x-=lowbit(x))
       suma=(suma+c[x])%MOD;
      return suma;
  }
  int main()
  {
      cin>>n>>k;
      for(i=1;i<=n;i++)
      {
          cin>>a[i];
          dp[1][i]=1;
      }
      int now=0;
      for(i=1;i<k;i++,now^=1)
       {
           memset(c,0,sizeof(c));
           sum=0;
           for(j=i;j<=n;j++)
            {
                sum=(sum+dp[now^1][j])%MOD;//滚动数组
                insert(a[j],dp[now^1][j]);//满足dp[i][j]=dp[i-1][i……n]
                int  temp=query(a[j]);
                if(sum<=temp)sum+=MOD;
                dp[now][j]=sum-temp;
                if(sum>MOD)sum-=MOD;

            }
       }
       sum=0;
       for(i=k;i<=n;i++)
        sum=(sum+dp[now^1][i])%MOD;
       if(sum>MOD)sum-=MOD;
       cout<<sum<<endl;
       return 0;


  }