题意很明确，求一个数组中单调递增子序列（长度为m）的数量，可惜比赛的时候很犯2地漏了一个细节

第一步是离散化数据，没什么好说的

对于数组的每个数num[i]，求出以num[i]结尾的长度为j的单调递增子序列的个数cnt [ num[i] ][j]，即求sum(cnt[1
~ num[i]-1][j-1])，对于区间求和可以用树状数组来优化

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef __int64 LL;
const int MAXN = 10002;
const LL mod = 123456789;
LL sum[102][MAXN];
int n, m, nm[MAXN];
int low_bit(int x)
{
    return x&(-x);
}
void add(int x, LL v, int c)
{
    while (x <= n)
    {
        sum[c][x] = (sum[c][x] + v)%mod;
        x += low_bit(x);
    }
}
LL get_sum(int x, int c)
{
    LL res = 0LL;
    while (x )
    {
        res += sum[c][x];
        res %= mod;
        x -= low_bit(x);
    }
    return res;
}
struct _node
{
    int d, idx, c;
    bool operator < (const _node & a) const
    {
        return d < a.d;
    }
}da[MAXN];
LL dp[105];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    while (scanf("%d%d", &n, &m)!=EOF)
    {
        for (int i = 1; i<= n; ++i)
            scanf("%d", &da[i].d), da[i].idx = i;;
        sort(da+1, da+n+1);
        da[0].d = -1; da[0].c = 1;
        for (int i = 1; i<= n; ++i)
        {
            if (da[i].d == da[i-1].d)
                da[i].c = da[i-1].c;
            else
                da[i].c = da[i-1].c+1;
        }
        for (int i = 1; i<=n; ++i)
        {
            nm[da[i].idx] = da[i].c;
        }
        LL res = 0;
        memset(sum, 0, sizeof sum);
        add(1, 1, 0);   // 初始化要注意
        for (int i = 1; i<= n; ++i)
        {
            for (int j = m; j>0; --j)
            {
                LL kk = get_sum(nm[i]-1, j-1);
                add(nm[i], kk, j);
            }
            res += get_sum(nm[i]-1, m-1);
            res %= mod;
        }
        printf("%I64d\n", res);
    }
    return 0;
}