Crazy Thairs POJ - 3378（10000进制加法 树状数组）

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 
#include 
#include 
#include 
#include 
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;
}