POJ 3378 (大整数加法+树状数组)

　　题意：给一个长度为N(N < 50000)的序列，求这个序列中长度为5的递增子序列的个数。

思路：对于长度为5的子序列，先考虑长度为1的子序列，第i个位置以i结尾长度为1的子序列个数为1。第i个位置以i结尾长度为2的子序列个数为i之前并且小于a[i]的长度为1的子序列的个数。同理，第i位置长度为3的子序列个数为i之前小于a[i]的长度为2的子序列的个数。。。依次往后推。

比如原序列为：(1, 2, 4, 6, 3) 推： -> (1, 1, 1, 1, 1)₁ -> (0, 1, 2, 3, 2)₂ -> (0, 0, 1, 3, 1)₃ -> (0, 0, 0, 1, 0)₄ -> (0, 0, 0, 0, 0)₅

 每次统计用树状数组就可以，中间过程和结果都有可能出现超long long的情况。所以用大整数加法。

ps：加法最后一次进位忘了。。。wa了无数次。。。

//#pragma comment(linker,"/STACK:327680000,327680000")
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>

#define CL(arr, val)    memset(arr, val, sizeof(arr))
#define REP(i, n)       for((i) = 0; (i) < (n); ++(i))
#define FOR(i, l, h)    for((i) = (l); (i) <= (h); ++(i))
#define FORD(i, h, l)   for((i) = (h); (i) >= (l); --(i))
#define L(x)    (x) << 1
#define R(x)    (x) << 1 | 1
#define MID(l, r)   (l + r) >> 1
#define Min(x, y)   (x) < (y) ? (x) : (y)
#define Max(x, y)   (x) < (y) ? (y) : (x)
#define E(x)        (1 << (x))
#define iabs(x)     (x) < 0 ? -(x) : (x)
#define OUT(x)  printf("%I64d\n", x)
#define Read()  freopen("data.in", "r", stdin)
#define Write() freopen("data.out", "w", stdout);

typedef long long LL;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int inf = ~0u>>2;

using namespace std;

const int N = 50010;
const int Base = 10000;

int a[N], b[N];

class BigNum {
public:
    int num[7], len;
    BigNum():len(0) {}
    BigNum(int n):len(0) {
        for( ; n > 0; n /= Base)    num[len++] = n%Base;
    }

    BigNum Bigvalueof(LL n) {
        len = 0;
        while(n) {
            num[len++] = n%Base;
            n /= Base;
        }
        return *this;
    }

    BigNum operator + (const BigNum& b) {   //++
        BigNum c;
        int i, carry = 0;
        for(i = 0; i < this->len || i < b.len || carry > 0; ++i) {
            if(i < this->len)   carry += this->num[i];
            if(i < b.len)   carry += b.num[i];
            c.num[i] = carry%Base;
            carry /= Base;
        }
        c.len = i;
        return c;
    }

    BigNum operator += (const BigNum& b) {  //+=
        *this = *this + b;
        return *this;
    }

    void Print() {
        if(len == 0)    {puts("0"); return ;}
        printf("%d", num[len - 1]);
        for(int i = len - 2; i >= 0; --i) {
            for(int j = Base/10; j > 0; j /= 10) {
                printf("%d", num[i]/j%10);
            }
        }
        puts("");
    }
};



class TreeArray : public BigNum{
public:
    BigNum c[N];
    int n;
    TreeArray() {}
    ~TreeArray() {}
    void init(int x) {
        n = x; CL(c, 0);
    }

    int lowbit(int i) {return i&(-i);}

    void add(int p, const BigNum& val) {
        while(p <= n) {
            c[p] += val;
            p += lowbit(p);
        }
    }

    BigNum sum(int p) {
        BigNum res;
        res.Bigvalueof(0);
        while(p > 0) {
            res += c[p];
            p -= lowbit(p);
        }
        return res;
    }
};

int bsearch(int b[], int m, int key) {
    int l = 0, r = m, mid;
    while(l <= r) {
        mid = (r + l) >> 1;
        if(b[mid] == key)   return mid + 1;
        else if(b[mid] < key)   l = mid + 1;
        else    r = mid - 1;
    }
    return -1;
}

int main() {
    //freopen("data.in", "r", stdin);

    int n, m, p, i, j;
    BigNum ans, tmp;
    TreeArray ta[6];

    while(~scanf("%d", &n)) {
        REP(i, n) {
            scanf("%d", a + i);
            b[i] = a[i];
        }
        sort(b, b + n);
        m = 0;
        for(i = 1; i < n; ++i) {
            if(b[m] != b[i])    b[++m] = b[i];
        }
        ++m;
        REP(i, 6)   ta[i].init(m + 5);
        for(i = 0; i < n; ++i) {
            //printf("%d ", a[i]);
            a[i] = bsearch(b, m, a[i]);
            //printf("%d\n", a[i]);
        }
        ans = ans.Bigvalueof(0);
        for(i = 0; i < n; ++i) {
            p = a[i];
            ta[1].add(p, tmp.Bigvalueof(1));
            for(j = 2; j <= 5; ++j) {
                tmp = ta[j-1].sum(p - 1);
                if(tmp.len) ta[j].add(p, tmp);
                if(j == 5)  ans += tmp;
            }
        }
        ans.Print();
    }
    return 0;
}