高精度模板整理_Rising_Flashlight的博客-程序员秘密

技术标签： 算法  

高精度模板整理

  1. 超级复杂各种高精度操作大合集（希望整个生涯都不会完整的敲一遍）

    #include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<iosfwd>
#define MAX_L 505 //最大长度，可以修改
using namespace std;
 
class bign
{
public:
    int len, s[MAX_L];//数的长度，记录数组
//构造函数
    bign();
    bign(const char*);
    bign(int);
    bool sign;//符号 1正数 0负数
    string toStr() const;//转化为字符串，主要是便于输出
    friend istream& operator>>(istream &,bign &);//重载输入流
    friend ostream& operator<<(ostream &,bign &);//重载输出流
//重载复制
    bign operator=(const char*);
    bign operator=(int);
    bign operator=(const string);
//重载各种比较
    bool operator>(const bign &) const;
    bool operator>=(const bign &) const;
    bool operator<(const bign &) const;
    bool operator<=(const bign &) const;
    bool operator==(const bign &) const;
    bool operator!=(const bign &) const;
//重载四则运算
    bign operator+(const bign &) const;
    bign operator++();
    bign operator++(int);
    bign operator+=(const bign&);
    bign operator-(const bign &) const;
    bign operator--();
    bign operator--(int);
    bign operator-=(const bign&);
    bign operator*(const bign &)const;
    bign operator*(const int num)const;
    bign operator*=(const bign&);
    bign operator/(const bign&)const;
    bign operator/=(const bign&);
//四则运算的衍生运算
    bign operator%(const bign&)const;//取模（余数）
    bign factorial()const;//阶乘
    bign Sqrt(const bign& num)const;//整数开根（向下取整）
    bign pow(const bign&)const;//次方
//一些乱乱的函数
    void clean();
    ~bign();
};
#define max(a,b) a>b ? a : b
#define min(a,b) a<b ? a : b
 
bign::bign()
{
    memset(s, 0, sizeof(s));
    len = 1;
    sign = 1;
}
 
bign::bign(const char *num)
{
    *this = num;
}
 
bign::bign(int num)
{
    *this = num;
}
 
string bign::toStr() const
{
    string res;
    res = "";
    int j;
    for(j=len-1;s[j]==0 && j>=0;j--);
    
    for (int i = 0; i <= j; i++)
        res = (char)(s[i] + '0') + res;
    if (res == "")
        res = "0";
    if (!sign&&res != "0")
        res = "-" + res;
    return res;
}
 
 
istream &operator>>(istream &in, bign &num)
{
    string str;
    in>>str;
    num=str;
    return in;
}
 
ostream &operator<<(ostream &out, bign &num)
{
    out<<num.toStr();
    return out;
}
 
bign bign::operator=(const char *num)
{
    memset(s, 0, sizeof(s));
    char a[MAX_L] = "";
    if (num[0] != '-')
        strcpy(a, num);
    else
    {
    	int len=strlen(num); 
		for (int i = 1; i < len; i++)
		        a[i - 1] = num[i];
	} 
    sign = !(num[0] == '-');
    len = strlen(a);
    for (int i = 0; i < len; i++)
        s[i] = a[len - i - 1] - 48;
    return *this;
}
 
bign bign::operator=(int num)
{
    if (num < 0)
        sign = 0, num = -num;
    else
        sign = 1;
    char temp[MAX_L];
    sprintf(temp, "%d", num);
    *this = temp;
    return *this;
}
 
bign bign::operator=(const string num)
{
    const char *tmp;
    tmp = num.c_str();
    *this = tmp;
    return *this;
}
 
bool bign::operator<(const bign &num) const
{
    if (sign^num.sign)
        return num.sign;
    if (len != num.len)
        return len < num.len;
    for (int i = len - 1; i >= 0; i--)
        if (s[i] != num.s[i])
            return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i]));
    return !sign;
}
 
bool bign::operator>(const bign&num)const
{
    return num < *this;
}
 
bool bign::operator<=(const bign&num)const
{
    return !(*this>num);
}
 
bool bign::operator>=(const bign&num)const
{
    return !(*this<num);
}
 
bool bign::operator!=(const bign&num)const
{
    return *this > num || *this < num;
}
 
bool bign::operator==(const bign&num)const
{
    return !(num != *this);
}
//luogu openjudge 验证
bign bign::operator+(const bign &num) const
{
    if (sign^num.sign)
    {
        bign tmp = sign ? num : *this;
        tmp.sign = 1;
        return sign ? *this - tmp : num - tmp;
    }
    bign result;
    result.len = 0;
    int temp = 0;
    for (int i = 0; temp || i < (max(len, num.len)); i++)
    {
        int t = s[i] + num.s[i] + temp;
        result.s[result.len++] = t % 10;
        temp = t / 10;
    }
    result.sign = sign;
    return result;
}
 
bign bign::operator++()
{
    *this = *this + 1;
    return *this;
}
 
bign bign::operator++(int)
{
    bign old = *this;
    ++(*this);
    return old;
}
 
bign bign::operator+=(const bign &num)
{
    *this = *this + num;
    return *this;
}
 
//luogu openjudge 验证
bign bign::operator-(const bign &num) const
{
    bign b=num,a=*this;
    if (!num.sign && !sign)
    {
        b.sign=1;
        a.sign=1;
        return b-a;
    }
    if (!b.sign)
    {
        b.sign=1;
        return a+b;
    }
    if (!a.sign)
    {
        a.sign=1;
        b=bign(0)-(a+b);
        return b;
    }
    if (a<b)
    {
        bign c=(b-a);
        c.sign=false;
        return c;
    }
    bign result;
    result.len = 0;
    for (int i = 0, g = 0; i < a.len; i++)
    {
        int x = a.s[i] - g;
        if (i < b.len) x -= b.s[i];
        if (x >= 0) g = 0;
        else
        {
            g = 1;
            x += 10;
        }
        result.s[result.len++] = x;
    }
    result.clean();
    return result;
}
 
bign bign::operator * (const bign &num)const
{
    bign result;
    result.len = len + num.len;
 
    for (int i = 0; i < len; i++)
        for (int j = 0; j < num.len; j++)
            result.s[i + j] += s[i] * num.s[j];
 
    for (int i = 0; i < result.len; i++)
    {
        result.s[i + 1] += result.s[i] / 10;
        result.s[i] %= 10;
    }
    result.clean();
    result.sign = !(sign^num.sign);
    return result;
}
 
bign bign::operator*(const int num)const
{
    bign x = num;
    bign z = *this;
    return x*z;
}
bign bign::operator*=(const bign&num)
{
    *this = *this * num;
    return *this;
}
 
bign bign::operator /(const bign&num)const
{
    bign ans;
    ans.len = len - num.len + 1;
    if (ans.len < 0)
    {
        ans.len = 1;
        return ans;
    }
 
    bign divisor = *this, divid = num;
    divisor.sign = divid.sign = 1;
    int k = ans.len - 1;
    int j = len - 1;
    while (k >= 0)
    {
        while (divisor.s[j] == 0) j--;
        if (k > j) k = j;
        char z[MAX_L];
        memset(z, 0, sizeof(z));
        for (int i = j; i >= k; i--)
            z[j - i] = divisor.s[i] + '0';
        bign dividend = z;
        if (dividend < divid) { k--; continue; }
        int key = 0;
        while (divid*key <= dividend) key++;
        key--;
        ans.s[k] = key;
        bign temp = divid*key;
        for (int i = 0; i < k; i++)
            temp = temp * 10;
        divisor = divisor - temp;
        k--;
    }
    ans.clean();
    ans.sign = !(sign^num.sign);
    return ans;
}
 
bign bign::operator/=(const bign&num)
{
    *this = *this / num;
    return *this;
}
 
bign bign::operator%(const bign& num)const
{
    bign a = *this, b = num;
    a.sign = b.sign = 1;
    bign result, temp = a / b*b;
    result = a - temp;
    result.sign = sign;
    return result;
}
 
bign bign::pow(const bign& num)const
{
    bign result = 1;
    for (bign i = 0; i < num; i++)
        result = result*(*this);
    return result;
}
 
bign bign::factorial()const
{
    bign result = 1;
    for (bign i = 1; i <= *this; i++)
        result *= i;
    return result;
}
 
void bign::clean()
{
    if (len == 0) len++;
    while (len > 1 && s[len - 1] == '\0')
        len--;
}
//luogu验证，应该差不多，只是乘法太慢t了。 
bign bign::Sqrt(const bign& num)const
{
    if(*this<0)return -1;
    if(*this<=1)return *this;
    bign l=0,r=*this,mid;
    while(r-l>1)
    {
        mid=(l+r)/2;
        
        if(num*mid.len-num+1>(*this).len || mid.pow(num)>*this)
            r=mid;
        else 
            l=mid;
    }
    return l;
}
 
bign::~bign()
{
}
 
int main()
{
}

  2. 只可以进行正整数操作的高精度模板，实测速度一般

    #include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<iosfwd>
#define ll short
using namespace std;

const ll N=221;
struct bignum
{
	ll len,num[N];
	bignum()
	{
		len=0;
		memset(num,0,sizeof(num));
	}
} one,zero,ans;

inline bignum read()
{
	bignum ans;
	string s;
	cin>>s;
	int len=s.size();
	for(ll i=0; i<len; i++) ans.num[s.size()-i]=s[i]-48;
	ans.len=s.size();
	return ans;
}

inline void write(bignum s)
{
	for(ll i=s.len; i>=1; i--) putchar(s.num[i]+48);
}

inline void operator ==(bignum &a,ll b)
{
	for(; b; b/=10) a.num[++a.len]=b%10;
}

inline bool operator <=(bignum a,bignum b)
{
	if(a.len>b.len) return 0;
	if(a.len<b.len) return 1;
	for(ll i=a.len; i>=1; i--)
	{
		if(a.num[i]>b.num[i]) return 0;
		if(a.num[i]<b.num[i]) return 1;
	}
	return 1;
}

inline bool operator >(bignum a,bignum b)
{
	if(a.len>b.len) return 1;
	if(a.len<b.len) return 0;
	for(ll i=a.len; i>=1; i--)
	{
		if(a.num[i]>b.num[i]) return 1;
		if(a.num[i]<b.num[i]) return 0;
	}
	return 0;
}

inline bignum operator +(bignum a,bignum b)
{
	bignum c;
	c.len=max(a.len,b.len);
	for(ll i=1; i<=c.len; i++)
	{
		c.num[i]+=a.num[i]+b.num[i];
		c.num[i+1]+=c.num[i]/10;
		c.num[i]%=10;
	}
	if(c.num[c.len+1]) c.len++;
	return c;
}

inline bignum operator -(bignum a,bignum b)
{
	bignum c;
	c.len=a.len;
	for(ll i=1; i<=c.len; i++)
	{
		c.num[i]=a.num[i]-b.num[i];
		if(c.num[i]<0)
		{
			c.num[i]+=10;
			a.num[i+1]--;
		}
	}
	while(c.len>1&&c.num[c.len]==0) c.len--;
	return c;
}

inline bignum operator *(bignum a,bignum b)
{
	bignum c;
	for(ll i=1; i<=a.len; i++)
	{
		for(ll j=1; j<=b.len; j++)
		{
			c.num[i+j-1]+=a.num[i]*b.num[j];
			c.num[i+j]+=c.num[i+j-1]/10;
			c.num[i+j-1]%=10;
		}
	}
	c.len=a.len+b.len;
	while(c.len>1&&c.num[c.len]==0) c.len--;
	return c;
}

inline bignum operator /(bignum a,ll b)
{
	ll sum=0;
	bignum c;
	for(ll i=a.len; i>=1; i--)
	{
		sum=sum*10+a.num[i];
		c.num[i]=sum/b;
		sum%=b;
	}
	c.len=a.len;
	while(c.len>1&&c.num[c.len]==0) c.len--;
	return c;
}

int main()
{
}

  3. 只可以进行正整数操作的高精度模板， f f t fft fft使用高精度乘法，实测速度巨慢，比2还慢。

    //测试之后发现跑得巨慢... 
#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<iosfwd>
#define ll long long
using namespace std;

const ll N=20010;
const double pi=acos(-1);
ll m;
struct bignum
{
	ll len,num[N];
	bignum()
	{
		len=0;
		memset(num,0,sizeof(num));
	}
} n,one,zero,ans,c;

inline void clear(bignum &a)
{
	memset(a.num,0,sizeof(a.num));
	a.len=0;
}

inline bignum read()
{
	bignum ans;
	string s;
	cin>>s;
	for(ll i=0; i<s.size(); i++) ans.num[s.size()-i]=s[i]-48;
	ans.len=s.size();
	return ans;
}

inline void write(bignum s)
{
	for(ll i=s.len; i>=1; i--) cout<<s.num[i];
}

inline void operator ==(bignum &a,ll b)
{
	for(; b; b/=10) a.num[++a.len]=b%10;
}

inline bool operator <=(bignum a,bignum b)
{
	if(a.len>b.len) return 0;
	if(a.len<b.len) return 1;
	for(ll i=a.len; i>=1; i--)
	{
		if(a.num[i]>b.num[i]) return 0;
		if(a.num[i]<b.num[i]) return 1;
	}
	return 1;
}

inline bool operator >(bignum a,bignum b)
{
	if(a.len>b.len) return 1;
	if(a.len<b.len) return 0;
	for(ll i=a.len; i>=1; i--)
	{
		if(a.num[i]>b.num[i]) return 1;
		if(a.num[i]<b.num[i]) return 0;
	}
	return 0;
}

inline bignum operator +(bignum a,bignum b)
{
	clear(c);
	c.len=max(a.len,b.len);
	for(ll i=1; i<=c.len; i++)
	{
		c.num[i]+=a.num[i]+b.num[i];
		c.num[i+1]+=c.num[i]/10;
		c.num[i]%=10;
	}
	if(c.num[c.len+1]) c.len++;
	return c;
}

inline bignum operator -(bignum a,bignum b)
{
	clear(c);
	c.len=a.len;
	for(ll i=1; i<=c.len; i++)
	{
		c.num[i]=a.num[i]-b.num[i];
		if(c.num[i]<0)
		{
			c.num[i]+=10;
			a.num[i+1]--;
		}
	}
	while(c.len>1&&c.num[c.len]==0) c.len--;
	return c;
}

void FFT(complex<double> *a,ll n,ll op)
{
	if(!n) return;
	complex<double> a0[n],a1[n];
	for(ll i=0; i<n; i++)
	{
		a0[i]=a[i<<1];
		a1[i]=a[i<<1|1];
	}
	FFT(a0,n>>1,op);
	FFT(a1,n>>1,op);
	complex<double> W(cos(pi/n),sin(pi/n)*op),w(1,0);
	for(ll i=0; i<n; i++,w*=W)
	{
		a[i]=a0[i]+w*a1[i];
		a[i+n]=a0[i]-w*a1[i];
	}
}

complex<double> x[N],y[N];
inline bignum operator *(bignum a,bignum b)
{
	memset(x,0,sizeof(x));
	memset(y,0,sizeof(y));
	ll n=a.len-1,m=b.len-1;
	for(ll i=0; i<=n; i++) x[i]=a.num[i+1];
	for(ll i=0; i<=m; i++) y[i]=b.num[i+1];
	for(m+=n,n=1; n<=m; n<<=1);
	FFT(x,n>>1,1);
	FFT(y,n>>1,1);
	for(ll i=0; i<n; i++) x[i]*=y[i];
	FFT(x,n>>1,-1);
	bignum c;
	for(ll i=0; i<=m; i++) c.num[i+1]=(ll)(fabs(x[i].real()/n+0.5));
	c.len=m+1;
	while(c.len>1&&c.num[c.len]==0) c.len--;
	for(ll i=1; i<=c.len; i++)
	{
		c.num[i+1]+=c.num[i]/10;
		c.num[i]%=10;
	}
	while(c.num[c.len+1])
	{
		c.len++;
		c.num[c.len+1]+=c.num[c.len]/10;
		c.num[c.len]%=10;
	}
	return c;
}

inline bignum operator /(bignum a,ll b)
{
	ll sum=0;
	clear(c);
	for(ll i=a.len; i>=1; i--)
	{
		sum=sum*10+a.num[i];
		c.num[i]=sum/b;
		sum%=b;
	}
	c.len=a.len;
	while(c.len>1&&c.num[c.len]==0) c.len--;
	return c;
}

inline bool check(bignum a,ll b)
{
	bignum ans=one;
	for(ll i=1; i<=b; i++)
	{
		ans=ans*a;
		if(ans>n) return 0;
	}
	return 1;
}

int main()
{
	cin>>m;
	n=read();
	one==1;
	zero==0;
	bignum l=one,r=n;
	while(l<=r)
	{
		bignum mid=(l+r)/2;
		if(check(mid,m))
		{
			ans=mid;
			l=mid+one;
		}
		else r=mid-one;
	}
	write(ans);
	return 0;
}

  4. 跑得超级快的压位高精

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<complex>
#include<string.h>
#include<math.h>
#include<string>
#include<iosfwd>
#define ll long long
using namespace std;

const ll N=30;
const double pi=acos(-1);
const ll mod=100000000,length=log(mod)/log(10);
struct bignum
{
	ll len,num[N];
	bignum()
	{
		len=0;
		memset(num,0,sizeof(num));
	}
} one,zero,ans,QAQ;

inline bignum read()
{
	bignum ans;
	string s;
	cin>>s;
	for(ll r=s.size()-1; r>=0; r-=length)
	{
		ans.len++;
		ll l;
		if(r>=length-1) l=r-length+1;
		else l=0;
		for(ll i=l; i<=r; i++) ans.num[ans.len]=ans.num[ans.len]*10+s[i]-48;
	}
	return ans;
}

inline void write(bignum s)
{
	printf("%lld",s.num[s.len]);
	for(ll i=s.len-1; i>=1; i--)
	{
		for(ll j=mod/10; j>s.num[i]; j/=10) putchar('0');
		if(s.num[i]) printf("%lld",s.num[i]);
	}
}

inline void operator ==(bignum &a,ll b)
{
	for(; b; b/=mod) a.num[++a.len]=b%mod;
}

inline bool operator <=(bignum a,bignum b)
{
	if(a.len>b.len) return 0;
	if(a.len<b.len) return 1;
	for(ll i=a.len; i>=1; i--)
	{
		if(a.num[i]>b.num[i]) return 0;
		if(a.num[i]<b.num[i]) return 1;
	}
	return 1;
}

inline bool operator >(bignum a,bignum b)
{
	if(a.len>b.len) return 1;
	if(a.len<b.len) return 0;
	for(ll i=a.len; i>=1; i--)
	{
		if(a.num[i]>b.num[i]) return 1;
		if(a.num[i]<b.num[i]) return 0;
	}
	return 0;
}

inline bignum operator +(bignum a,bignum b)
{
	bignum c;
	c.len=max(a.len,b.len);
	for(ll i=1; i<=c.len; i++)
	{
		c.num[i]+=a.num[i]+b.num[i];
		c.num[i+1]+=c.num[i]/mod;
		c.num[i]%=mod;
	}
	if(c.num[c.len+1]) c.len++;
	return c;
}

inline bignum operator -(bignum a,bignum b)
{
	bignum c;
	c.len=a.len;
	for(ll i=1; i<=c.len; i++)
	{
		c.num[i]=a.num[i]-b.num[i];
		if(c.num[i]<0)
		{
			c.num[i]+=mod;
			a.num[i+1]--;
		}
	}
	while(c.len>1&&c.num[c.len]==0) c.len--;
	return c;
}

inline bignum operator *(bignum a,bignum b)
{
	bignum c;
	for(ll i=1; i<=a.len; i++)
	{
		for(ll j=1; j<=b.len; j++)
		{
			c.num[i+j-1]+=a.num[i]*b.num[j];
			c.num[i+j]+=c.num[i+j-1]/mod;
			c.num[i+j-1]%=mod;
		}
	}
	c.len=a.len+b.len;
	while(c.len>1&&c.num[c.len]==0) c.len--;
	return c;
}

inline bignum operator /(bignum a,ll b)
{
	ll sum=0;
	bignum c;
	for(ll i=a.len; i>=1; i--)
	{
		sum=sum*mod+a.num[i];
		c.num[i]=sum/b;
		sum%=b;
	}
	c.len=a.len;
	while(c.len>1&&c.num[c.len]==0) c.len--;
	return c;
}

int main()
{
}

模板均来自网络，并略作修改。

模板1 来自 c s d n csdn csdn 用户 H a r r y G u o 2012 HarryGuo2012 HarryGuo2012的博文 https://blog.csdn.net/harryguo2012/article/details/42036829

模板2、3、4来自 l u o g u luogu luogu 用户 x u k u a n xukuan xukuan的题解 https://www.luogu.com.cn/problemnew/solution/P2293?page=2
版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。
本文链接：https://blog.csdn.net/Rising_shit/article/details/105394142

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