HDU5568/BestCoder Round #63 (div.2) B.sequence2 dp+高精度

sequence2

Problem Description

Given an integer array bi with a length of n, please tell me how many exactly different increasing subsequences.

P.S. A subsequence bai(1≤i≤k) is an increasing subsequence of sequence bi(1≤i≤n) if and only if 1≤a1<a2<...<ak≤n and ba1<ba2<...<bak.
Two sequences ai and bi is exactly different if and only if there exist at least one i and ai≠bi.

Input

Several test cases(about 5)

For each cases, first come 2 integers, n,k(1≤n≤100,1≤k≤n)

Then follows n integers ai(0≤ai≤109)

Output

For each cases, please output an integer in a line as the answer.

Sample Input

3 2
1 2 2
3 2
1 2 3

Sample Output

2
3

题解：我们可以通过dp[i][j]dp[i][j]表示第ii个数，当前这个数为序列中的第jj个数的方案总数。 转移为dp[i][j] = sum{dp[k][j-1]}(k < i, b_k < b_i)dp[i][j]=sumdp[k][j−1](k<i,b​k​​<b​i​​)。 本题需要高精度。

//
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

const int inf=~0u>>;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//**************************************************************************************

#define MAX_L 405 //最大长度，可以修改

class bign
{
public:
    int len, s[MAX_L];//数的长度，记录数组
//构造函数
    bign();
    bign(const char*);
    bign(int);
    bool sign;//符号 1正数 0负数
    string toStr() const;//转化为字符串，主要是便于输出
    friend istream& operator>>(istream &,bign &);//重载输入流
    friend ostream& operator<<(ostream &,bign &);//重载输出流
//重载复制
    bign operator=(const char*);
    bign operator=(int);
    bign operator=(const string);
//重载各种比较
    bool operator>(const bign &) const;
    bool operator>=(const bign &) const;
    bool operator<(const bign &) const;
    bool operator<=(const bign &) const;
    bool operator==(const bign &) const;
    bool operator!=(const bign &) const;
//重载四则运算
    bign operator+(const bign &) const;
    bign operator++();
    bign operator++(int);
    bign operator+=(const bign&);
    bign operator-(const bign &) const;
    bign operator--();
    bign operator--(int);
    bign operator-=(const bign&);
    bign operator*(const bign &)const;
    bign operator*(const int num)const;
    bign operator*=(const bign&);
    bign operator/(const bign&)const;
    bign operator/=(const bign&);
//四则运算的衍生运算
    bign operator%(const bign&)const;//取模（余数）
    bign factorial()const;//阶乘
    bign Sqrt()const;//整数开根（向下取整）
    bign pow(const bign&)const;//次方
//一些乱乱的函数
    void clean();
    ~bign();
};
#define max(a,b) a>b ? a : b
#define min(a,b) a<b ? a : b

bign::bign()
{
    memset(s, , sizeof(s));
    len = ;
    sign = ;
}

bign::bign(const char *num)
{
    *this = num;
}

bign::bign(int num)
{
    *this = num;
}

string bign::toStr() const
{
    string res;
    res = "";
    for (int i = ; i < len; i++)
        res = (char)(s[i] + '') + res;
    if (res == "")
        res = "";
    if (!sign&&res != "")
        res = "-" + res;
    return res;
}

istream &operator>>(istream &in, bign &num)
{
    string str;
    in>>str;
    num=str;
    return in;
}

ostream &operator<<(ostream &out, bign &num)
{
    out<<num.toStr();
    return out;
}

bign bign::operator=(const char *num)
{
    memset(s, , sizeof(s));
    char a[MAX_L] = "";
    if (num[] != '-')
        strcpy(a, num);
    else
        for (int i = ; i < strlen(num); i++)
            a[i - ] = num[i];
    sign = !(num[] == '-');
    len = strlen(a);
    for (int i = ; i < strlen(a); i++)
        s[i] = a[len - i - ] - ;
    return *this;
}

bign bign::operator=(int num)
{
    char temp[MAX_L];
    sprintf(temp, "%d", num);
    *this = temp;
    return *this;
}

bign bign::operator=(const string num)
{
    const char *tmp;
    tmp = num.c_str();
    *this = tmp;
    return *this;
}

bool bign::operator<(const bign &num) const
{
    if (sign^num.sign)
        return num.sign;
    if (len != num.len)
        return len < num.len;
    for (int i = len - ; i >= ; i--)
        if (s[i] != num.s[i])
            return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i]));
    return !sign;
}

bool bign::operator>(const bign&num)const
{
    return num < *this;
}

bool bign::operator<=(const bign&num)const
{
    return !(*this>num);
}

bool bign::operator>=(const bign&num)const
{
    return !(*this<num);
}

bool bign::operator!=(const bign&num)const
{
    return *this > num || *this < num;
}

bool bign::operator==(const bign&num)const
{
    return !(num != *this);
}

bign bign::operator+(const bign &num) const
{
    if (sign^num.sign)
    {
        bign tmp = sign ? num : *this;
        tmp.sign = ;
        return sign ? *this - tmp : num - tmp;
    }
    bign result;
    result.len = ;
    int temp = ;
    for (int i = ; temp || i < (max(len, num.len)); i++)
    {
        int t = s[i] + num.s[i] + temp;
        result.s[result.len++] = t % ;
        temp = t / ;
    }
    result.sign = sign;
    return result;
}

bign bign::operator++()
{
    *this = *this + ;
    return *this;
}

bign bign::operator++(int)
{
    bign old = *this;
    ++(*this);
    return old;
}

bign bign::operator+=(const bign &num)
{
    *this = *this + num;
    return *this;
}

bign bign::operator-(const bign &num) const
{
    bign b=num,a=*this;
    if (!num.sign && !sign)
    {
        b.sign=;
        a.sign=;
        return b-a;
    }
    if (!b.sign)
    {
        b.sign=;
        return a+b;
    }
    if (!a.sign)
    {
        a.sign=;
        b=bign()-(a+b);
        return b;
    }
    if (a<b)
    {
        bign c=(b-a);
        c.sign=false;
        return c;
    }
    bign result;
    result.len = ;
    for (int i = , g = ; i < a.len; i++)
    {
        int x = a.s[i] - g;
        if (i < b.len) x -= b.s[i];
        if (x >= ) g = ;
        else
        {
            g = ;
            x += ;
        }
        result.s[result.len++] = x;
    }
    result.clean();
    return result;
}

bign bign::operator * (const bign &num)const
{
    bign result;
    result.len = len + num.len;

    for (int i = ; i < len; i++)
        for (int j = ; j < num.len; j++)
            result.s[i + j] += s[i] * num.s[j];

    for (int i = ; i < result.len; i++)
    {
        result.s[i + ] += result.s[i] / ;
        result.s[i] %= ;
    }
    result.clean();
    result.sign = !(sign^num.sign);
    return result;
}

bign bign::operator*(const int num)const
{
    bign x = num;
    bign z = *this;
    return x*z;
}
bign bign::operator*=(const bign&num)
{
    *this = *this * num;
    return *this;
}

bign bign::operator /(const bign&num)const
{
    bign ans;
    ans.len = len - num.len + ;
    if (ans.len < )
    {
        ans.len = ;
        return ans;
    }

    bign divisor = *this, divid = num;
    divisor.sign = divid.sign = ;
    int k = ans.len - ;
    int j = len - ;
    while (k >= )
    {
        while (divisor.s[j] == ) j--;
        if (k > j) k = j;
        char z[MAX_L];
        memset(z, , sizeof(z));
        for (int i = j; i >= k; i--)
            z[j - i] = divisor.s[i] + '';
        bign dividend = z;
        if (dividend < divid) { k--; continue; }
        int key = ;
        while (divid*key <= dividend) key++;
        key--;
        ans.s[k] = key;
        bign temp = divid*key;
        for (int i = ; i < k; i++)
            temp = temp * ;
        divisor = divisor - temp;
        k--;
    }
    ans.clean();
    ans.sign = !(sign^num.sign);
    return ans;
}

bign bign::operator/=(const bign&num)
{
    *this = *this / num;
    return *this;
}

bign bign::operator%(const bign& num)const
{
    bign a = *this, b = num;
    a.sign = b.sign = ;
    bign result, temp = a / b*b;
    result = a - temp;
    result.sign = sign;
    return result;
}

bign bign::pow(const bign& num)const
{
    bign result = ;
    for (bign i = ; i < num; i++)
        result = result*(*this);
    return result;
}

bign bign::factorial()const
{
    bign result = ;
    for (bign i = ; i <= *this; i++)
        result *= i;
    return result;
}

void bign::clean()
{
    if (len == ) len++;
    while (len >  && s[len - ] == '\0')
        len--;
}

bign bign::Sqrt()const
{
    if(*this<)return -;
    if(*this<=)return *this;
    bign l=,r=*this,mid;
    while(r-l>)
    {
        mid=(l+r)/;
        if(mid*mid>*this)
            r=mid;
        else
            l=mid;
    }
    return l;
}

bign::~bign()
{
}

int  a[];
bign dp[][];
int n,k;
int main()
{
    while(cin>>n>>k)
    {

        for(int i=; i<=n; i++)
        {
            cin>>a[i];
        }
        bign aa=,bb=;
         for(int i=;i<=n;i++) {
            for(int j=;j<=k;j++) dp[i][j]=aa;
         }
         for(int i=;i<=n;i++)dp[i][]=bb;
         for(int i=;i<=n;i++) {
            for(int j=;j<i;j++) {
                if(a[i]>a[j]) {
                    for(int kk=;kk<=k;kk++) {
                        dp[i][kk]+=dp[j][kk-];
                    }
                }
            }
         }
         bign ans;
         ans=dp[][k];
         for(int i=;i<=n;i++)ans+=dp[i][k];
         cout<<ans<<endl;
    }

    return ;
}

代码