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 ≤ i < j < k < l < m ≤ N
- 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
题意：
寻找上升五元组的个数
思路：
本质上是树状数组优化dp。
状态就是dp[i][j]，到了第j个数字的i元上升组。
从比j小的i - 1元上升组转移过来，那么这个求和的过程可以用树状数组维护。
比较坑的就是得用大整数，而且复杂度卡的很死，得用10000进制加法，于是找了个10000进制加法的代码改了一下。
#pragma GCC optimize(2)
#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;const int maxn = 10;
const int BASE = 10000;struct bign{int d[maxn], len;void clean() { while(len > 1 && !d[len-1]) len--; }
// bign() { memset(d, 0, sizeof(d)); len = 1; }
// bign(int num) { *this = num; }
// bign(char* num) { *this = num; }bign():len(0) {}bign(int n):len(0) {for( ; n > 0; n /= BASE) d[len++] = n%BASE;}bign operator = (const char* num){memset(d, 0, sizeof(d)); len = strlen(num);for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0';clean();return *this;}bign operator = (int num){char s[20]; sprintf(s, "%d", num);*this = s;return *this;}// bign operator + (const bign& b){// bign c = *this; int i;// for (i = 0; i < b.len; i++){// c.d[i] += b.d[i];// if (c.d[i] >= BASE) c.d[i]%=BASE, c.d[i+1]++;// }// while (c.d[i] >= BASE) c.d[i++]%=BASE, c.d[i]++;// c.len = max(len, b.len);// if (c.d[i] && c.len <= i) c.len = i+1;// return c;// }bign operator + (const bign& b) { //不使用 = 运算bign c;int i, carry = 0;for(i = 0; i < this->len || i < b.len || carry > 0; ++i) {if(i < this->len) carry += this->d[i];if(i < b.len) carry += b.d[i];c.d[i] = carry%BASE;carry /= BASE;}c.len = i;return c;}bign operator - (const bign& b){bign c = *this; int i;for (i = 0; i < b.len; i++){c.d[i] -= b.d[i];if (c.d[i] < 0) c.d[i]+=10, c.d[i+1]--;}while (c.d[i] < 0) c.d[i++]+=10, c.d[i]--;c.clean();return c;}bign operator * (const bign& b)const{int i, j; bign c; c.len = len + b.len;for(j = 0; j < b.len; j++) for(i = 0; i < len; i++)c.d[i+j] += d[i] * b.d[j];for(i = 0; i < c.len-1; i++)c.d[i+1] += c.d[i]/10, c.d[i] %= 10;c.clean();return c;}bign operator / (const bign& b){int i, j;bign c = *this, a = 0;for (i = len - 1; i >= 0; i--){a = a*10 + d[i];for (j = 0; j < 10; j++) if (a < b*(j+1)) break;c.d[i] = j;a = a - b*j;}c.clean();return c;}bign operator % (const bign& b){int i, j;bign a = 0;for (i = len - 1; i >= 0; i--){a = a*10 + d[i];for (j = 0; j < 10; j++) if (a < b*(j+1)) break;a = a - b*j;}return a;}bign operator += (const bign& b){*this = *this + b;return *this;}bool operator <(const bign& b) const{if(len != b.len) return len < b.len;for(int i = len-1; i >= 0; i--)if(d[i] != b.d[i]) return d[i] < b.d[i];return false;}bool operator >(const bign& b) const{return b < *this;}bool operator<=(const bign& b) const{return !(b < *this);}bool operator>=(const bign& b) const{return !(*this < b);}bool operator!=(const bign& b) const{return b < *this || *this < b;}bool operator==(const bign& b) const{return !(b < *this) && !(b > *this);}string str() const{char s[maxn]={};for(int i = 0; i < len; i++) s[len-1-i] = d[i]+'0';return s;}void Print() {if(len == 0) {cout << "0" << endl; return ;}cout << d[len - 1];for(int i = len - 2; i >= 0; --i) {for(int j = BASE/10; j > 0; j /= 10) {cout << d[i] / j % 10;}}cout << endl;}
};istream& operator >> (istream& in, bign& x)
{string s;in >> s;x = s.c_str();return in;
}ostream& operator << (ostream& out, const bign& x)
{out << x.str();return out;
}const int maxm = 5e4 + 7;int a[maxm];
bign c[6][maxm];
int n;struct Node
{int x,id;
}nodes[maxm];bool cmp(Node a,Node b)
{return a.x < b.x;
}void add(int num,int x,bign v)
{while(x <= n){c[num][x] += v;x += x & (-x);}
}bign query(int num,int x)
{bign res = 0;while(x){res += c[num][x];x -= x & (-x);}return res;
}int main()
{ios::sync_with_stdio(false);while(cin >> n && n){for(int i = 1;i <= n;i++){cin >> nodes[i].x;nodes[i].id = i;}sort(nodes + 1,nodes + 1 + n,cmp);int cnt = 0;a[nodes[1].id] = ++cnt;for(int i = 2;i <= n;i++){if(nodes[i].x == nodes[i - 1].x)a[nodes[i].id] = cnt;else a[nodes[i].id] = ++cnt;}for(int i = 1;i <= 5;i++){for(int j = 1;j <= n;j++){c[i][j] = 0;}}for(int i = 1;i <= n;i++){int x = a[i];add(1,x,1);for(int j = 2;j <= 5;j++){add(j,x,query(j - 1,x - 1));}}bign ans = query(5,n);ans.Print();}return 0;
}