HDU1223 Order Count 动态规划 组合数

Order Count

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 954    Accepted Submission(s): 371

Problem Description

If we connect 3 numbers with "<" and "=", there are 13 cases:
1) A=B=C
2) A=B<C
3) A<B=C
4) A<B<C
5) A<C<B
6) A=C<B
7) B<A=C
8) B<A<C
9) B<C<A
10) B=C<A
11) C<A=B
12) C<A<B
13) C<B<A

If we connect n numbers with "<" and "=", how many cases then?

 

Input

The input starts with a positive integer P(0<P<1000) which indicates the number of test cases. Then on the following P lines, each line consists of a positive integer n(1<=n<=50) which indicates the amount of numbers to be connected.

 

Output

For each input n, you should output the amount of cases in a single line.

 

Sample Input

2
1
3

 

Sample Output

1
13

动态规划+组合数+大数：

从左向右，n的数，第i个数是h[i]，和h[i]==h[i-1]+1或者h[i]==h[i-1]

DP把每种情况表示出来，组合数来排列（第i个数代表谁）

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<memory.h>
#include<cstring>
#include<string.h>
using namespace std;
long long f[][],ans[];
long long c[][];
int i,j;
void _getc()
{
    for(i=;i<=;i++) c[i][]=;
    for(i=;i<=;i++)
     for(j=;j<=i;j++)
      c[i][j]=c[i-][j]+c[i-][j-];
}
void _getf()
{
    for(i=;i<=;i++) f[i][]=;
        for(i=;i<=;i++)
    {
          for(j=;j<=i;j++)
           for(k=;k<j;k++)
             f[i][j]+=f[i][k]*c[i-k][j-k];
           ans[i]=f[i][i];
    }
}
int main()
{
    int T,k;
    _getc();
    _getf();
    scanf("%d",&T);
    while(T--)
    {
       scanf("%d",&k);
       printf("%lld\n",ans[k]);
    }
    return ;
}

（前面代码不要拿来用。没有写高精度。后面用的模板，所以有点长）

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<memory.h>
#include<cstring>
#include<string>
using namespace std;
#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4

class BigNum
{
private:
    int a[];    //可以控制大数的位数
    int len;       //大数长度
public:
    BigNum(){ len = ;memset(a,,sizeof(a)); }   //构造函数
    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 = ;
    memset(a,,sizeof(a));
    while(d > MAXN)
    {
        c = d - (d / (MAXN + )) * (MAXN + );
        d = d / (MAXN + );
        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=;
    for(i=l-;i>=;i-=DLEN)
    {
        t=;
        k=i-DLEN+;
        if(k<)
            k=;
        for(int j=k;j<=i;j++)
            t=t*+s[j]-'';
        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 =  ; 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 =  ; 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*];
    int i = -;
    in>>ch;
    int l=strlen(ch);
    int count=,sum=;
    for(i=l-;i>=;)
    {
        sum = ;
        int t=;
        for(int j=;j<&&i>=;j++,i--,t*=)
        {
            sum+=(ch[i]-'')*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 - ];
    for(i = b.len -  ; i >=  ; i--)
    {
        cout.width(DLEN);
        cout.fill('');
        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 =  ; i < big ; i++)
    {
        t.a[i] +=T.a[i];
        if(t.a[i] > MAXN)
        {
            t.a[i + ]++;
            t.a[i] -=MAXN+;
        }
    }
    if(t.a[big] != )
        t.len = big + ;
    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=;
    }
    else
    {
        t1=T;
        t2=*this;
        flag=;
    }
    big=t1.len;
    for(i =  ; i < big ; i++)
    {
        if(t1.a[i] < t2.a[i])
        {
            j = i + ;
            while(t1.a[j] == )
                j++;
            t1.a[j--]--;
            while(j > i)
                t1.a[j--] += MAXN;
            t1.a[i] += MAXN +  - t2.a[i];
        }
        else
            t1.a[i] -= t2.a[i];
    }
    t1.len = big;
    while(t1.a[t1.len - ] ==  && t1.len > )
    {
        t1.len--;
        big--;
    }
    if(flag)
        t1.a[big-]=-t1.a[big-];
    return t1;
} 

BigNum BigNum::operator*(const BigNum & T) const   //两个大数之间的相乘运算
{
    BigNum ret;
    int i,j,up;
    int temp,temp1;
    for(i =  ; i < len ; i++)
    {
        up = ;
        for(j =  ; j < T.len ; j++)
        {
            temp = a[i] * T.a[j] + ret.a[i + j] + up;
            if(temp > MAXN)
            {
                temp1 = temp - temp / (MAXN + ) * (MAXN + );
                up = temp / (MAXN + );
                ret.a[i + j] = temp1;
            }
            else
            {
                up = ;
                ret.a[i + j] = temp;
            }
        }
        if(up != )
            ret.a[i + j] = up;
    }
    ret.len = i + j;
    while(ret.a[ret.len - ] ==  && ret.len > )
        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 = ;
    for(i = len -  ; i >=  ; i--)
    {
        ret.a[i] = (a[i] + down * (MAXN + )) / b;
        down = a[i] + down * (MAXN + ) - ret.a[i] * b;
    }
    ret.len = len;
    while(ret.a[ret.len - ] ==  && ret.len > )
        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=;
    for (i = len-; i>=; i--)
    {
        d = ((d * (MAXN+))% b + a[i])% b;
    }
    return d;
}
BigNum BigNum::operator^(const int & n) const    //大数的n次方运算
{
    BigNum t,ret();
    int i;
    if(n<)
        exit(-);
    if(n==)
        return ;
    if(n==)
        return *this;
    int m=n;
    while(m>)
    {
        t=*this;
        for( i=;i<<<=m;i<<=)
        {
            t=t*t;
        }
        m-=i;
        ret=ret*t;
        if(m==)
            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 - ;
        while(a[ln] == T.a[ln] && ln >= )
            ln--;
        if(ln >=  && 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 - ];
    for(i = len -  ; i >=  ; i--)
    {
        cout.width(DLEN);
        cout.fill('');
        cout << a[i];
    }
    cout << endl;
}
BigNum f[][],ans[];
BigNum c[][];
int i,j,k;
void _getc()
{
    for(i=;i<=;i++) c[i][]=;
    for(i=;i<=;i++)
     for(j=;j<=i;j++)
      c[i][j]=c[i-][j]+c[i-][j-];
}
void _getf()
{
    for(i=;i<=;i++) f[i][]=;
    for(i=;i<=;i++)
    {
      for(j=;j<=i;j++)
       for(k=;k<j;k++)
        f[i][j]=f[i][j]+f[i][k]*c[i-k][j-k];
       ans[i]=f[i][i];
    }
}
int main()
{
    int T,Tmp;
    _getc();
    _getf();
    scanf("%d",&T);
    while(T--)
    {
       scanf("%d",&Tmp);
       ans[Tmp].print();
    }
    return ;
}