1523. K-inversions

http://acm.timus.ru/problem.aspx?space=1&num=1523

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

求k-1次逆序数，树状数组的应用
//关键在于理解求逆序数的含义
//其实求一次逆序就是求2个递减数对个数

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define N 20001
#define M 1000000000
int now,n,k;
int a[N],c[N],d[2][N];
int low(int x)
{
    return x&(-x);
}
void modify(int x,int v)
{
    for(;x<=n;x+=low(x))
      c[x]=(c[x]+v)%M;
}
int query(int x)
{
    int sum=0;
    for(;x>0;x-=low(x))
      sum=(sum+c[x])%M;
    return sum;
}
int main()
{
    int temp,sum;
    while(scanf("%d %d",&n,&k)!=EOF)
    {
        now=0;
        for(int i=1;i<=n;i++)
          d[1][i]=1,scanf("%d",a+i);
        for(int i=1;i<k;i++,now^=1)
        {
            memset(c,0,sizeof(c));
            sum=0;
            for(int j=i;j<=n;j++)
             {
                 sum=(sum+d[now^1][j])%M;
                 modify(a[j],d[now^1][j]);
                 temp=query(a[j]);
                 if(sum<=temp)  sum+=M;
                 d[now][j]=sum-temp;
                 if(sum>M) sum-=M;//开始这里忘记减回去了、WA了
             }
        }
        sum=0;
        for(int i=k;i<=n;i++) sum=(sum+d[now^1][i])%M;
        printf("%d\n",sum);
    }
    return 0;
}