高精度

#include 
#include
#include
#include 
#include
#include
using namespace std;

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

class BigNum
{
private:
	int a[500];    //可以控制大数的位数
	int len;       //大数长度
public:
	BigNum(){ len = 1;memset(a,0,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 = 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 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,0,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,0,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,0,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,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   //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(void) 
{
    int i, n;
    BigNum x[101]; //定义大数的对象数组
    x[0] = 0, x[1] = 1;
    for (i = 2; i < 101; i++)
        x[i] = x[i - 1]*(4 * i - 2) / (i + 1);
    
    
    while (scanf("%d", &n) == 1 && n != -1)
        x[n].print();
}


大数开方模板
输入x，求sqrt（2*n)
代码

#include
#include
#include
#include
using namespace std;
#define MAXN 2000
int big(char s1[],char s2[]){
    int len1,len2,i,q;
    q=0;
    while(s1[q]=='0') q++;
    strcpy(s1,s1+q);
    if(strlen(s1)==0){
        s1[0]='0';
        s1[1]=0;
    }
    q=0;
    while(s2[q]=='0') q++;
    strcpy(s2,s2+q);
    if(strlen(s2)==0){
        s2[0]='0';
        s2[1]=0;
    }
    len1=strlen(s1);
    len2=strlen(s2);
    if(len1>len2)
        return 1;
    else if(len1s2[i])
                return 1;
            else if(s1[i]=0;i--){
        k=t*(s[i]-'0')+left;
        re[j++]=(k%10)+'0';
        left=k/10;
    }
    while(left>0){
        re[j++]=(left%10)+'0';
        left/=10;
    }
    re[j]=0;
    len=strlen(re);
    for(i=0;i=0){
        temp=a[len1]-b[len2]+left;
        if(temp<0){
            temp+=10;
            left=-1;
        }
        else
            left=0;
        a[len1]=temp+'0';
        len1--;
        len2--;
    }
    while(len1>=0){
        temp=a[len1]-'0'+left;
        if(temp<0){
            temp+=10;
            left=-1;
        }
        else
        left=0;
        a[len1]=temp+'0';
        len1--;
    }
    j=0;
    while(a[j]=='0') j++;
    strcpy(a,a+j);
    if(strlen(a)==0){
        a[0]='0';
        a[1]=0;
    }
    return;
}
void sqr(char s[],char re[]){  //开方
    char temp[MAXN];
    char left[MAXN];
    char p[MAXN];
    int i,j,k,len1,len2,q;
    len1=strlen(s);
    if(len1%2==0){
        left[0]=s[0];
        left[1]=s[1];
        left[2]=0;
        j=2;
    }
    else{
        left[0]=s[0];
        left[1]=0;
        j=1;
    }
    re[0]='0';
    re[1]=0;
    q=0;
    while(j<=len1){
        mul(re,20,temp);
        len2=strlen(temp);
        for(i=9;i>=0;i--){
            temp[len2-1]=i+'0';
            mul(temp,i,p);
            if(!big(p,left))
            break;
        }
        re[q++]=i+'0';
        re[q]=0;
        sub(left,p);
        len2=strlen(left);
        left[len2]=s[j];
        left[len2+1]=s[j+1];
        left[len2+2]=0;
        j+=2;
    }
}

int main(){
    char s[MAXN],s2[MAXN],re[MAXN];
    int  an[MAXN];
    char ans[MAXN];
    int i;
    while(scanf("%s",s)!= EOF ){
        mul(s,2,s2);
        strcpy(s,s2);
        re[0]=0;
        sqr(s,re);
        i=0;
        while(re[i]=='0') i++;
        strcpy(re,re+i);
        printf("%s\n",re);
    }
    return 0;
}