ural 1523 K-inversions（dp+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

题意：求长度为k的不连续的严格递减子序列的总个数

题解：可以设dp【i】【j】表示以i结尾长度为j的子序列的个数，那么更新就是dp【i】【j】=∑dp【k】【j-1】，其中k<i，而且a【k】>a【i】。而要更新dp值，可以用树状数组维护，按顺序插入序列值，那么树状数组的值就可以表示比它小的长度为j-1的所有子序列的和，这样就可以在logn的时间更新dp值了，所以总复杂度是O（n*k*logn）

#include<stdio.h>
#include<string.h>
#define MAXN 20008
#define mod 1000000000
int a[MAXN],tree[MAXN],dp[MAXN][13];
int low_bit(int x){ return x&(-x); }
void add(int x,int y,int n)
{
while(x<=n)
{
tree[x]=(tree[x]+y)%mod;
x+=low_bit(x);
}
}
int query(int x)
{
int res=0;
while(x)
{
res=(res+tree[x])%mod;
x-=low_bit(x);
}
return res;
}
int main()
{
int i,j,n,k,res;

while(scanf("%d%d",&n,&k)>0)
{
for(i=1;i<=n;i++)
{
scanf("%d",a+i);
dp[i][1]=1;
}
for(i=1,j=n;i<j;i++,j--) res=a[i],a[i]=a[j],a[j]=res;
for(i=2;i<=k;i++)
{
memset(tree,0,sizeof(tree));
for(j=1;j<=n;j++)
{
dp[j][i]=query(a[j]);
add(a[j],dp[j][i-1],n);
}
}
for(res=0,i=1;i<=n;i++) res=(res+dp[i][k])%mod;
printf("%d\n",res);
}

return 0;
}