大数模板（线上赛专业模板） HDU5568 大数加DP

发布时间：2020-12-14 02:12:03 所属栏目：大数据来源：网络整理

导读：sequence2 Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 533????Accepted Submission(s): 192 Problem Description Given an integer array? b i ?with a length of? n ,please tell me how

sequence2

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 533????Accepted Submission(s): 192

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

Sample Output

2
3

Source

BestCoder Round #63 (div.2)

题意：在这n个数中存在多少个这样的递增序列（这些递增序列只有一个不同）

定义dp[i][j]表示以i结尾的长度为J的个数，则对于b[sb1]<b[sb2]&&sb1<sb2;

dp[sb2][x]+=dp[sb1][x-1](sb>=1&&sb<=k)

重点在于那个模板，以后大数就好搞了，

#pragma warning(disable:4996)
#include <iostream>
#include<stdio.h>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#include <map>
using namespace std;
typedef long long ll;
const int maxn = 105;
int n,k;
int b[105];

#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4

class BigNum
{
private:
    int a[40];    //可以控制大数的位数
    int len;       //大数长度
public:
    BigNum()
    {
        len = 1;    //构造函数
        memset(a,sizeof(a));
    }
    BigNum(const int);       //将一个int类型的变量转化为大数
    BigNum(const char*);     //将一个字符串类型的变量转化为大数
    BigNum(const BigNum &);  //拷贝构造函数
    BigNum &operator=(const BigNum &);   //重载赋值运算符，大数之间进行赋值运算

    friend istream& operator>>(istream&,BigNum&);   //重载输入运算符
    friend ostream& operator<<(ostream&,BigNum&);   //重载输出运算符

    BigNum operator+(const BigNum &) const;   //重载加法运算符，两个大数之间的相加运算
    BigNum operator-(const BigNum &) const;   //重载减法运算符，两个大数之间的相减运算
    BigNum operator*(const BigNum &) const;   //重载乘法运算符，两个大数之间的相乘运算
    BigNum operator/(const int   &) const;    //重载除法运算符，大数对一个整数进行相除运算

    BigNum operator^(const int  &) const;    //大数的n次方运算
    int    operator%(const int  &) const;    //大数对一个int类型的变量进行取模运算
    bool   operator>(const BigNum & T)const;   //大数和另一个大数的大小比较
    bool   operator>(const int & t)const;      //大数和一个int类型的变量的大小比较

    void print();       //输出大数
};
BigNum::BigNum(const int b)     //将一个int类型的变量转化为大数
{
    int c,d = b;
    len = 0;
    memset(a,sizeof(a));
    while (d > MAXN)
    {
        c = d - (d / (MAXN + 1)) * (MAXN + 1);
        d = d / (MAXN + 1);
        a[len++] = c;
    }
    a[len++] = d;
}
BigNum::BigNum(const char*s)     //将一个字符串类型的变量转化为大数
{
    int t,k,index,l,i;
    memset(a,sizeof(a));
    l = strlen(s);
    len = l / DLEN;
    if (l%DLEN)
        len++;
    index = 0;
    for (i = l - 1; i >= 0; i -= DLEN)
    {
        t = 0;
        k = i - DLEN + 1;
        if (k<0)
            k = 0;
        for (int j = k; j <= i; j++)
            t = t * 10 + s[j] - '0';
        a[index++] = t;
    }
}
BigNum::BigNum(const BigNum & T) : len(T.len)  //拷贝构造函数
{
    int i;
    memset(a,sizeof(a));
    for (i = 0; i < len; i++)
        a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n)   //重载赋值运算符，大数之间进行赋值运算
{
    int i;
    len = n.len;
    memset(a,sizeof(a));
    for (i = 0; i < len; i++)
        a[i] = n.a[i];
    return *this;
}
istream& operator>>(istream & in,BigNum & b)   //重载输入运算符
{
    char ch[MAXSIZE * 4];
    int i = -1;
    in >> ch;
    int l = strlen(ch);
    int count = 0,sum = 0;
    for (i = l - 1; i >= 0;)
    {
        sum = 0;
        int t = 1;
        for (int j = 0; j<4 && i >= 0; j++,i--,t *= 10)
        {
            sum += (ch[i] - '0')*t;
        }
        b.a[count] = sum;
        count++;
    }
    b.len = count++;
    return in;

}
ostream& operator<<(ostream& out,BigNum& b)   //重载输出运算符
{
    int i;
    cout << b.a[b.len - 1];
    for (i = b.len - 2; i >= 0; i--)
    {
        cout.width(DLEN);
        cout.fill('0');
        cout << b.a[i];
    }
    return out;
}

BigNum BigNum::operator+(const BigNum & T) const   //两个大数之间的相加运算
{
    BigNum t(*this);
    int i,big;      //位数
    big = T.len > len ? T.len : len;
    for (i = 0; i < big; i++)
    {
        t.a[i] += T.a[i];
        if (t.a[i] > MAXN)
        {
            t.a[i + 1]++;
            t.a[i] -= MAXN + 1;
        }
    }
    if (t.a[big] != 0)
        t.len = big + 1;
    else
        t.len = big;
    return t;
}
BigNum BigNum::operator-(const BigNum & T) const   //两个大数之间的相减运算
{
    int i,j,big;
    bool flag;
    BigNum t1,t2;
    if (*this>T)
    {
        t1 = *this;
        t2 = T;
        flag = 0;
    }
    else
    {
        t1 = T;
        t2 = *this;
        flag = 1;
    }
    big = t1.len;
    for (i = 0; i < big; i++)
    {
        if (t1.a[i] < t2.a[i])
        {
            j = i + 1;
            while (t1.a[j] == 0)
                j++;
            t1.a[j--]--;
            while (j > i)
                t1.a[j--] += MAXN;
            t1.a[i] += MAXN + 1 - t2.a[i];
        }
        else
            t1.a[i] -= t2.a[i];
    }
    t1.len = big;
    while (t1.a[len - 1] == 0 && t1.len > 1)
    {
        t1.len--;
        big--;
    }
    if (flag)
        t1.a[big - 1] = 0 - t1.a[big - 1];
    return t1;
}

BigNum BigNum::operator*(const BigNum & T) const   //两个大数之间的相乘运算
{
    BigNum ret;
    int i,up;
    int temp,temp1;
    for (i = 0; i < len; i++)
    {
        up = 0;
        for (j = 0; j < T.len; j++)
        {
            temp = a[i] * T.a[j] + ret.a[i + j] + up;
            if (temp > MAXN)
            {
                temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
                up = temp / (MAXN + 1);
                ret.a[i + j] = temp1;
            }
            else
            {
                up = 0;
                ret.a[i + j] = temp;
            }
        }
        if (up != 0)
            ret.a[i + j] = up;
    }
    ret.len = i + j;
    while (ret.a[ret.len - 1] == 0 && ret.len > 1)
        ret.len--;
    return ret;
}
BigNum BigNum::operator/(const int & b) const   //大数对一个整数进行相除运算
{
    BigNum ret;
    int i,down = 0;
    for (i = len - 1; i >= 0; i--)
    {
        ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
        down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
    }
    ret.len = len;
    while (ret.a[ret.len - 1] == 0 && ret.len > 1)
        ret.len--;
    return ret;
}
int BigNum::operator %(const int & b) const    //大数对一个int类型的变量进行取模运算
{
    int i,d = 0;
    for (i = len - 1; i >= 0; i--)
    {
        d = ((d * (MAXN + 1)) % b + a[i]) % b;
    }
    return d;
}
BigNum BigNum::operator^(const int & n) const    //大数的n次方运算
{
    BigNum t,ret(1);
    int i;
    if (n<0)
        exit(-1);
    if (n == 0)
        return 1;
    if (n == 1)
        return *this;
    int m = n;
    while (m>1)
    {
        t = *this;
        for (i = 1; i << 1 <= m; i <<= 1)
        {
            t = t*t;
        }
        m -= i;
        ret = ret*t;
        if (m == 1)
            ret = ret*(*this);
    }
    return ret;
}
bool BigNum::operator>(const BigNum & T) const   //大数和另一个大数的大小比较
{
    int ln;
    if (len > T.len)
        return true;
    else if (len == T.len)
    {
        ln = len - 1;
        while (a[ln] == T.a[ln] && ln >= 0)
            ln--;
        if (ln >= 0 && a[ln] > T.a[ln])
            return true;
        else
            return false;
    }
    else
        return false;
}
bool BigNum::operator >(const int & t) const    //大数和一个int类型的变量的大小比较
{
    BigNum b(t);
    return *this>b;
}

void BigNum::print()    //输出大数
{
    int i;
    cout << a[len - 1];
    for (i = len - 2; i >= 0; i--)
    {
        cout.width(DLEN);
        cout.fill('0');
        cout << a[i];
    }
    cout << endl;
}

int main()
{
    BigNum dp[105][105];
    int b[105];
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(dp,sizeof(dp));
        memset(b,sizeof(b));
        for(int i=1; i<=n; i++)
            scanf("%d",&b[i]);

        dp[0][0]=1;
        for(int i=1; i<=n; i++)
        {
            for(int j=0; j<i; j++)
            {
                if(b[j]<b[i])
                {
                    for(int sb=1; sb<=n; sb++)
                    {
                        dp[i][sb]=dp[i][sb]+dp[j][sb-1];
                    }
                }
            }

        }
        BigNum count=0;
        for(int i=1; i<=n; i++)
            count=count+dp[i][k];
        count.print();
    }
}

（编辑：李大同）

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