题意：给一串数字，问长度为m的严格上升子序列有多少个
解法：首先可以离散化为10000以内，再进行dp，令dp[i][j]为以第i个元素结尾的长度为j的上升子序列的个数，
则有dp[i][j] = SUM(dp[k][j-1]) (a[k] < a[i] && k < i)
不可能直接遍历，所以考虑优化，可以看出dp方程相当于一个区间求和，所以可以用树状数组来优化。
代码：
#include <iostream> #include <cmath> #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> #define SMod 123456789 #define lll __int64 using namespace std; #define N 10007lll c[N],a[N],b[N]; int n,m; lll dp[N][104];int lowbit(int x) {return x & (-x); }void modify(int pos,lll val) {while(pos <= n){c[pos] += val;pos += lowbit(pos);} }lll getsum(int pos) {lll res = 0;while(pos > 0){res = (res+c[pos])%SMod;pos -= lowbit(pos);}return res; }int main() {int i,j;while(scanf("%d%d",&n,&m)!=EOF){for(i=1;i<=n;i++){scanf("%I64d",&a[i]);b[i] = a[i];}sort(b+1,b+n+1);memset(dp,0,sizeof(dp));for(i=1;i<=n;i++)dp[i][1] = 1;for(j=2;j<=m;j++){memset(c,0,sizeof(c));for(i=1;i<=n;i++){int ind = lower_bound(b+1,b+n+1,a[i])-b;dp[i][j] = getsum(ind-1);modify(ind,dp[i][j-1]);}}lll ans = 0;for(i=1;i<=n;i++)ans = (ans + dp[i][m])%SMod;printf("%I64d\n",ans);}return 0; }
