BestCoder Round#8 1003

 1 #include <algorithm>
 2 using namespace std;
 3 typedef long long LL;
 4 const int N = 10000 + 10;
 5 
 6 LL a[N],b[N],c[N];
 7 int n;
 8 LL dp[N][N];
 9 int lowbit(int t)
10 {
11     return t & (-t);
12 }
13 void update(int pos, LL val)
14 {
15     while(pos <= n)
16     {
17         c[pos] += val;
18         pos += lowbit(pos);
19     }
20 }
21 LL query(int pos)
22 {
23     LL ans = 0;
24     while(pos >= 1)
25     {
26         ans += c[pos] % 123456789;
27         pos -= lowbit(pos);
28     }
29     return ans;
30 }
31 int main()
32 {
33     int  m, i, j;
34     while(scanf("%d%d",&n, &m) != EOF)
35     {
36         memset(dp, 0, sizeof(dp));
37 
38         for(i=1; i<=n; ++i)
39         {
40             dp[i][1] = 1;
41             scanf("%I64d",&a[i]);
42             b[i] = a[i];
43         }
44         sort(b+1, b+n+1);
45         for(j=2; j<=m; ++j)
46         {
47             memset(c, 0, sizeof(c));
48             for(i=1; i<=n; ++i)
49             {
50                 int index = lower_bound(b+1, b+n+1,a[i]) - b ;//离散化，index 表示a[i]在数列中第几大
51                 dp[i][j] = query(index-1);//找出小于index的长度为j-1的递增子序列有多少个
52                 update(index,dp[i][j-1]);//更新以i结尾的，长度为j-1的递增子序列有多少个
53             }
54         }
55         LL ans = 0;
56         for(i=1; i<=n; ++i)
57             ans = (ans + dp[i][m]) % 123456789;
58         printf("%I64d\n",ans);
59     }
60 }