楼天城之做男人就过八道题（第一题…
关于楼教主的第一题，感觉最费时间的还是高精度的实现
。实现一万位以上整数基本运算的
万进制高精度
模版就写了快一天（题目里用到大概也有几百位吧）
，只能说太容易出错了，得不断地调试修改。
题目源地址是北大poj1737
题，
题目本身的组合数学的算法是参考某位大牛的文章，在此附上地址 http://hi.baidu.com/accplaystation/item/0b3b4e1654a07548e75e06f9
下面分别是49和50个点时的结果以及源代码，供参考：
49：
下面分别是49和50个点时的结果以及源代码，供参考：
49：
10263013515700550779116289728670421776807355856352253452035361907379108631238572
44548313982876228994987864700400759811456244128889754306386459557887432298148719
59173497103061147469088590424739631395981885494059279529144993759879407051757016
75516079509792662379977972835636452421052447375208813714109600679021766298295142
56225641238164014573644333472284672
50：
57775629806264131981532128463353986108213291999872288565750767218860631769630192
41340682335187078778417692523562748348836783209222917852889522593249600859338855
72481476441044041666245632947630667669900623389069655523344952222114179660086674
25130052344927925607827177068266427605834927592260049347147617842015437801204857
1333436567365397136152469165480980158369042006016
#include "iostream"
#include "string"
#include "cstring"
#include "stdio.h"
using namespace std;
bool remember[55];
struct bigNumber
{
int num[100];
int ans;
bigNumber()
{
memset(num,0,sizeof(num));
ans=1;
}
bigNumber operator =(int k)
{
int i=0;
num[i]=k000;
while(k>=10000)
{
i++;
k/=10000;
num[i]=k000;
}
ans=i+1;
return *this;
}
};
bigNumber number[55];
bigNumber bigNumber1,bigNumber2,bigNumber3;
ostream& operator << (ostream& os, constbigNumber& big)//输出高精度数运算符重载
{
os<
for(inti=big.ans-2;i>=0;i--)
{
if(big.num[i]==0)
os<<"000";
else if(big.num[i]<10)
os<<"000";
else if(big.num[i]<100)
os<<"00";
else if(big.num[i]<1000)
os<<"0";
os<
}
return os;
}
istream& operator>>(istream& is, bigNumber &big)//输入高精度运算符的重载
{
string s;
is>>s;
intlength=s.size();
for(int i=0;i
{
char a=s[i];
s[i]=s[length-1-i];
s[length-1-i]=a;
}
int m=1;
for(int i=0;i
{
int a=s[i]-'0';
big.num[i/4]+=a*m;
m*=10;
if(m==10000)
m=1;
}
big.ans=(length-1)/4+1;
return is;
}
bigNumber operator*(bigNumber big1,bigNumberbig2)//两个高精度数相乘
{
bigNumber big3;
for(int i=0;i
{
for(int j=0;j
{
inta=big1.num[i]*big2.num[j];
big3.num[i+j]+=a000;
big3.num[i+j+1]+=a/10000;
}
for(int k=i;k<=i+big2.ans;k++)
{
if(big3.num[k]>=10000)
{
big3.num[k+1]+=big3.num[k]/10000;
big3.num[k]=big3.num[k]000;
}
}
}
big3.ans=big1.ans+big2.ans;
while(big3.num[big3.ans]==0)
big3.ans--;
big3.ans++;
if(big3.ans==0)
big3.ans++;
returnbig3;
}
bigNumber operator*(bigNumber big,intk)//高精度乘以普通常数（注意常数的大小范围不可太大）
{
bigNumber tool;
for(int i=0;i
{
int product=big.num[i]*k;
tool.num[i]+=product000;
int j=i;
while(product>=10000)
{
product/=10000;
tool.num[++j]+=product000;
}
}
tool.ans=big.ans+10;
while(tool.num[tool.ans]==0)
tool.ans--;
tool.ans++;
big=tool;
return big;
}
bigNumber operator+(bigNumber big1,bigNumberbig2)//两个高精度数相加，将和赋在第一个数上
{
bigNumber tool=big1;
for(int i=0;i
{
tool.num[i]+=big2.num[i];
int j=i;
while(tool.num[j]>=10000)
{
inta=tool.num[j];
tool.num[j]=a000;
tool.num[j+1]+=a/10000;
j++;
}
}
tool.ans=max(big1.ans+1,big2.ans+1);
while(tool.num[tool.ans]==0)
tool.ans--;
tool.ans++;
return tool;
}
bigNumber operator/(bigNumber &big1,int k)
{
int i=big1.ans-1;
int a;
while(i>=0)
{
a=big1.num[i];
big1.num[i]=0;
while(i>0&&a
{
a*=10000;
a+=big1.num[--i];
big1.num[i]=0;
}
big1.num[i]=a/k;
if(i==0)
break;
if(i>0)
big1.num[--i]+=(a%k)*10000;
}
while(big1.num[big1.ans]==0)
big1.ans--;
big1.ans++;
return big1;
}
bigNumber function0(int n)
{
if(n==1||n==2)
{
bigNumber big;
big.num[0]=1;
return big;
}
if(remember[n])
return number[n];
bigNumber tool;
for(int k=1;k
{
bigNumber big1=function0(k);
bigNumber big2=function0(n-k);
bigNumber big3;
big3.num[0]=1;
for(int i=n-2;i>n-2-k+1;i--)
{
big3=big3*i;
}
for(int i=2;i<=k-1;i++)
big3=big3/i;
bigNumber big4;
big4.num[0]=1;
for(int i=1;i<=k;i++)
big4=big4*2;
big4.num[0]-=1;
tool=tool+(big1*big2*big3*big4);
}
remember[n]=true;
number[n]=tool;
return tool;
}
int main()
{
for(inti=0;i<=50;i++)
remember[i]=false;
int n;
while(scanf("%d",&n)!=EOF&&n!=0)
cout<
return 0;
}