记\(dp(i)\)为以第i个结尾的上升子序列的个数，可以得到转移方程
\[dp(i)=\sum_{j=1,\,a[j]<a[i]}^{i-1}dp(j) + 1\]
然后最终的结果就是\(ans = \sum{dp(i)}\),所以求出每一个\(dp(i)\).
直接求复杂度\(O(n^2)\),自然是不行的。因为只和大小的比较有关系，跟数值本身是无关的，所以先将其离散化。变成1到pos的点。这样就可以开下一个树状数组。
树状数组可以很方便的求出前n项和，利用这个性质，可以做到\(O(nlogn)\).
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <queue>
#include <map>
#include <stack>
#include <string>
#include <math.h>
#include <bitset>
#include <ctype.h>
using namespace std;
typedef pair<int,int> P;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-9;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
int n;
LL c[N];
inline LL lowbit(int i)
{
return i&(-i);
}
void add(int x, LL val)
{
while(x <= n)
{
c[x] += val;
c[x] %= mod;
x += lowbit(x);
}
}
LL sum(int x)
{
LL sum = 0;
while(x)
{
sum += c[x];
sum %= mod;
x -= lowbit(x);
}
return sum;
}
int t,kase = 0;
struct Node
{
int id;
int val;
Node(){}
Node(int a, int b) { id = a; val = b;}
}node[N];
bool cmp1(Node a, Node b)
{
return a.val < b.val;
}
bool cmp2(Node a, Node b)
{
return a.id < b.id;
}
map<LL,int> mp;
int main()
{
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
for(int i = 1; i <= n; i++)
{
scanf("%d", &node[i].val);
node[i].id = i;
}
sort(node+1, node+1+n, cmp1);
mp.clear();
int pos = 1;
for(int i = 1; i <= n; i++)
{
if(!mp.count(node[i].val)) mp[node[i].val] = pos++;
node[i].val = mp[node[i].val];
}
sort(node+1, node+1+n, cmp2);
memset(c, 0, sizeof(c));
for(int i = 1; i <= n; i++)
{
int tmp = sum(node[i].val-1) + 1;
add(node[i].val, tmp);
}
printf("Case %d: %d\n", ++kase, sum(pos-1));
}
return 0;
}