1523. K-inversions URAL 求k逆序对，，，，DP加树状数组

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 andai1 > 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

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

 #include <iostream>
 #include <string.h>
 #include <stdio.h>
 using namespace std;
 #define mod 1000000000
 int a[],sum[][];
 int p[],n;
 int lowbit(int x)
 {
     return x&(-x);
 }
 void update(int x,int z)
 {
     while(x)
     {
         p[x]=(p[x]+z)%mod;
         x-=lowbit(x);
     }
 }
 int query(int x)
 {
     long long s=;
     while(x<=n)
     {
         s+=p[x];
         s%=mod;
         x+=lowbit(x);
     }
     return s;
 }
 int main()
 {
     //freopen("in.txt","r",stdin);
     int k,i,j;
     scanf("%d%d",&n,&k);
     memset(sum,,sizeof(sum));
     for(i=;i<=n;i++)
     scanf("%d",&a[i]),sum[a[i]][]=;
     for(i=;i<=k;i++)
     {
         memset(p,,sizeof(p));
         for(j=i-;j<=n;j++)
         {
             update(a[j],sum[a[j]][i-]);
             sum[a[j]][i]=query(a[j]+);
         }
     }
     long long s=;
     for(i=k-;i<=n;i++)
     {
         s=(s+sum[a[i]][k])%mod;
     }
     cout<<s<<endl;
 }