1523. K-inversions

1523. K-inversions

Time limit: 1.0 second Memory limit: 64 MB
Consider a permutation a1, a2, …, an (all ai are different integers in range from 1 to n). Let us call k-inversion a sequence of numbers i1, i2, …, ik such that 1 ≤ i1 < i2 < … < ik ≤ n and ai1 > ai2 > … > aik. 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 ai.

Output

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

Samples

inputoutput
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）
************************************************************************************************
 1 #include<iostream>
 2 #include<string>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cmath>
 7 using namespace std;
 8 const int   N=20005;
 9 const  int MOD=1000000000;
10 int a[N],dp[2][N];
11 int c[N];
12 int i,j,k,n;
13 int sum;
14 int lowbit(int x)//低位技术
15 {
16     return x&(-x);
17 }
18 void insert(int x,int v)//修改
19  {
20      for(;x<=n;x+=lowbit(x))
21       c[x]=(c[x]+v)%MOD;
22  }
23  int query(int x)//查询
24   {
25       int suma=0;
26       for(;x>0;x-=lowbit(x))
27        suma=(suma+c[x])%MOD;
28       return suma;
29   }
30   int main()
31   {
32       cin>>n>>k;
33       for(i=1;i<=n;i++)
34       {
35           cin>>a[i];
36           dp[1][i]=1;
37       }
38       int now=0;
39       for(i=1;i<k;i++,now^=1)
40        {
41            memset(c,0,sizeof(c));
42            sum=0;
43            for(j=i;j<=n;j++)
44             {
45                 sum=(sum+dp[now^1][j])%MOD;//滚动数组
46                 insert(a[j],dp[now^1][j]);//满足dp[i][j]=dp[i-1][i……n]
47                 int  temp=query(a[j]);
48                 if(sum<=temp)sum+=MOD;
49                 dp[now][j]=sum-temp;
50                 if(sum>MOD)sum-=MOD;
51 
52             }
53        }
54        sum=0;
55        for(i=k;i<=n;i++)
56         sum=(sum+dp[now^1][i])%MOD;
57        if(sum>MOD)sum-=MOD;
58        cout<<sum<<endl;
59        return 0;
60 
61 
62   }
View Code

 

posted @ 2013-08-09 08:10  persistent codeants  阅读(205)  评论(0)  编辑  收藏  举报