[BZOJ 2729][HNOI2012]排队（排列组合+高精）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
typedef long long LL;
using namespace std;
struct Bign
{
    static const int base=100000;
    static const int width=5;
    LL len,s[10010];
    Bign()
    {memset(s,0,sizeof(s));len=1;}
    Bign(int num){*this=num;}
    void clear(){while(len>1&&!s[len-1])len--;} 
    Bign operator = (int num)
    {
        len=0;
        while(num)
        {
            s[len++]=num%base;
            num/=base;
        }
        return *this;
    }
    Bign operator + (const Bign& b) 
    {
        Bign c;c.len=0;
        for(int i=0,g=0;g||i<max(len,b.len);i++)
        {
            int x=g;
            if(i<b.len)x+=b.s[i];
            if(i<len)x+=s[i];
            c.s[c.len++]=x%base;
            g=x/base;
        }
        return c;
    }
    Bign operator * (const Bign& b)
    {
        Bign c;c.len=len+b.len;
        for(int i=0;i<len;i++)
        {
            for(int j=0;j<b.len;j++)
            {
                c.s[i+j]+=s[i]*b.s[j];
            }
        }
        for(int i=0;i<c.len;i++)
        {
            c.s[i+1]+=c.s[i]/base;
            c.s[i]%=base;
        }
        c.clear();
        return c;
    }
    Bign operator * (const int& b)
    {
        Bign c;
        c.len=0;
        for(int i=0,g=0;g||i<len;i++)
        {
            int x;
            if(i<len)x=s[i]*b+g;
            else x=g;
            c.s[c.len++]=x%base;
            g=x/base;
        }
        return c;
    }
    void print() 
    {
        string res;
        printf("%d",s[len-1]);
        for(int i=len-2;i>=0;i--)
        printf("%05d",s[i]);
    }
};
Bign A(int x,int y)
{
    if(x>y)return Bign();
    Bign c=1;
    for(int i=y-x+1;i<=y;i++)
    c=c*i;
    return c;
}
int n,m;
int main()
{
    scanf("%d%d",&n,&m);
    Bign ans;
    ans=A(n,n)*(A(2,n+1)*A(m,n+3)+A(1,n+1)*A(2,2)*A(1,m)*A(m-1,n+2)); 
    ans.print();
    return 0;
}