[poj3378] Crazy Thairs (DP + 树状数组维护 + 高精度)

树状数组维护DP + 高精度

---

Description

These days, Sempr is crazed on one problem named Crazy Thair. Given N (1 ≤ N ≤ 50000) numbers, which are no more than 109, Crazy Thair is a group of 5 numbers {i, j, k, l, m} satisfying:

1. 1 ≤ i < j < k < l < m ≤ N
2. Ai < Aj < Ak < Al < Am

For example, in the sequence {2, 1, 3, 4, 5, 7, 6},there are four Crazy Thair groups: {1, 3, 4, 5, 6}, {2, 3, 4, 5, 6}, {1, 3, 4, 5, 7} and {2, 3, 4, 5, 7}.

Could you help Sempr to count how many Crazy Thairs in the sequence?

Input

Input contains several test cases. Each test case begins with a line containing a number N, followed by a line containing N numbers.

Output

Output the amount of Crazy Thairs in each sequence.

Sample Input

5
1 2 3 4 5
7
2 1 3 4 5 7 6
7
1 2 3 4 5 6 7

Sample Output

1
4
21

---

题目大意

给你一个长度为 n 的序列，输出长度为5
的上升子序列的个数。

题解

拿到这道题，可以很快的想出dp定义，
dp[i][j] 表示以 j 这个数结尾能组成长度为 i 的上升子序列的个数。可以推出转移方程为

(dp[i][j] = sum_{k=1}^{j-1}dp[i-1][k])

i 最大为5 ，j可以达到(10^9),但个数只有(10^5),所以我们可以考虑将 j 离散化,开 5 个树状数组来维护。
最后，这道题最坑的是，答案可能非常大，需要高精度。。。

代码

#include <iostream>
#include <vector>
#include <algorithm> 
#include <cstring>
#include <cstdio>
using namespace std;
#define lowbit(x) (x & -x)
#define base 10000

const int maxn = 5e4 + 5;
int n;
int a[maxn];
vector <int> v;


struct Bign {
	int c[20],l;
	Bign() {memset(c,0,sizeof(c));l = 1;}
	void reset() {memset(c,0,sizeof(c));l = 1;}
	
	void Print() {
		printf("%d",c[l]);
		for(int i = l - 1;i > 0;i--)printf("%04d",c[i]);
	}
	
	void operator = (const int &a) {
		int x = a;
		reset();
		c[1] = x % base;x /= base;
		while(x) {
			c[++l] = x % base;x /= base;
		}
	}
	
	Bign operator + (const Bign &a) {
		Bign res;
		res.l = max(l,a.l);
		for(int i = 1;i <= res.l;i++) {
			res.c[i] += c[i] + a.c[i];
			res.c[i+1] = res.c[i] / base;
			res.c[i] %= base;
		}
		if(res.c[res.l+1])res.l++;
		return res;
	}
	
	Bign operator += (const Bign &a) {
		*this = *this + a;
		return *this;
	}
}t[5][maxn];
	

inline int getid(int x) {return lower_bound(v.begin(),v.end(),x) - v.begin() + 1;}

inline void update(int pos,int x,Bign a) {
	while (x <= n) {
		t[pos][x] += a;
		x += lowbit(x);
	}
}

inline Bign query(int pos,int x) {
	Bign res;
	while(x) {
		res += t[pos][x];
		x -= lowbit(x);
	}
	return res;
}

int main() {
	ios::sync_with_stdio(false);cin.tie(0);
	while(cin >> n) {
		v.clear();
		for(int i = 0;i < n;i++) {
			cin >> a[i];v.push_back(a[i]);
		}
		Bign ans;
		sort(v.begin(),v.end());v.erase(unique(v.begin(),v.end()),v.end());
		memset(t,0,sizeof(t));
		for(int i = 0;i < n;i++) {
			int x = getid(a[i]);
			Bign xx;xx = 1;
			update(0,x,xx);
			for(int i = 1;i < 5;i++) {
				update(i,x,query(i-1,x-1));
			}
		}
		ans = query(4,n);
		ans.Print();printf("
");
	}
	return 0; 
}