[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("\n");
}
return 0;
}
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