1523. K-inversions

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

Consider a permutation a ₁, a ₂, …, a_n (all a_i are different integers in range from 1 to n). Let us call k-inversion a sequence of numbers i ₁, i ₂, …, i_k such that 1 ≤  i ₁ <  i ₂ < … <  i_k ≤  n and a_{i 1} >  a_{i 2} > … >  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 10 ⁹.

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;

}