树状数组维护DP + 高精度

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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

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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("\n");

	}

	return 0;

}