1261 字串数.cpp

威士忌的HDU题解.大概有260多题的源码。对于学习非常有好处。

/*
1261 字串数
Time Limit : 1000 ms  Memory Limit : 32768 K  Output Limit : 256 K

GUN C++
*/
//组合公式很简单，组合运算量不大，只涉及大数乘除
#include <iostream>
#include <string>
#include <iomanip>
#include <algorithm>
using namespace std;

#define MAXN 9999
#define DLEN 4

class BigNum{
private:
   int a[300];//DLEN digs for a position
   int len;
public:
   BigNum(){len = 1;memset(a,0,sizeof(a));}
   BigNum(const int b);
   BigNum(const char* b);
   BigNum(const BigNum & T);

   bool     Bigger(const BigNum &) const;
   BigNum & operator=(const BigNum &);
   BigNum & Add(const BigNum &);
   BigNum & Sub(const BigNum &);
   BigNum operator+(const BigNum &) const;
   BigNum operator-(const BigNum &) const;
   BigNum operator*(const BigNum &) const;
   BigNum operator/(const int   &) const;
   void Print();

   BigNum operator+=(const BigNum &) ;
   BigNum operator-=(const BigNum &) ;
   BigNum operator*=(const BigNum &) ;
   BigNum operator/=(const int   &) ;
};
BigNum::BigNum(const int b)
{
   int c,d = b;

   len = 0;
   memset(a,0,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 BigNum & T) : len(T.len)
{
   int i;
   memset(a,0,sizeof(a));
   for(i = 0 ; i < len ; i++)
      a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n)
{
   len = n.len;
   memset(a,0,sizeof(a));
   for(int i = 0 ; i < len ; i++)
      a[i] = n.a[i];
   return *this;
}
BigNum & BigNum::Add(const BigNum & T)
{
   int i,big;

   big = T.len > len ? T.len : len;
   for(i = 0 ; i < big ; i++)
   {
      a[i] = a[i] + T.a[i];
      if(a[i] > MAXN)
      {
         a[i + 1]++;
         a[i] = a[i] - MAXN - 1;
      }
   }
   if(a[big] != 0) len = big + 1;
   else len = big;

   return *this;
}
BigNum & BigNum::Sub(const BigNum & T)
{
   int i,j,big;

   big = T.len > len ? T.len : len;
   for(i = 0 ; i < big ; i++){
      if(a[i] < T.a[i]){
         j = i + 1;
         while(a[j] == 0) j++;
         a[j--]--;
         while(j > i) a[j--] += MAXN;
         a[i] = a[i] + MAXN + 1 - T.a[i];
      }
      else a[i] -= T.a[i];
   }
   len = big;
   while(a[len - 1] == 0 && len > 1) len--;
   return *this;
}
BigNum BigNum::operator+(const BigNum & n) const
{
   BigNum a = *this;

   a.Add(n);
   return a;
}
BigNum BigNum::operator-(const BigNum & T) const
{
   BigNum b = *this;

   b.Sub(T);
   return b;
}
BigNum BigNum::operator*(const BigNum & T) const
{
   BigNum ret;
   int i,j,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
{
   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--;
   return ret;
}
void BigNum::Print()
{
   int i;

   cout << a[len - 1];
   for(i = len - 2 ; i >= 0 ; i--){
      cout.width(DLEN);
      cout.fill('0');
      cout << a[i];
   }
   cout << endl;
}
//////////////
const int mMax=12;

BigNum C(int n,int r)
{
    int temp,i;
    BigNum btemp;
    
    temp=n-r;
    if(temp<r)
        r=temp;
    if(r==0)
        return 1;
    else
    {
        for(i=n,btemp=1;i>n-r;i--)
            btemp=btemp*i;
        for(i=2;i<=r;i++)
            btemp=btemp/i;
        return btemp;
    }
}

int main()
{
    int m[26],n,i,all;
    BigNum ans;

    while(cin>>n && n!=0)
    {
        all=0;ans=1;
        for(i=0;i<n;i++)
        {
            cin>>m[i];
            all+=m[i];
        }
        for(i=0;i<n;i++)
        {
            ans=ans*C(all,m[i]);
            all-=m[i];
        }
        ans.Print();
    }
    return 0;
}