高精度模板

不念不忘少年蓝@ 2021-11-09 07:34 416阅读 0赞

#include<bits/stdc++.h>
    #define MAXN 9999
    #define MAXSIZE 10
    #define DLEN 4
    typedef long long ll;
    using namespace std;
     
    class BigNum
    {
    private:
        ll a[50];    //可以控制大数的位数
        ll len;       //大数长度
    public:
        BigNum(){ len = 1;memset(a,0,sizeof(a)); }   //构造函数
        BigNum(const ll);       //将一个ll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 ll   &) const;    //重载除法运算符，大数对一个整数进行相除运算
     
        BigNum operator^(const ll  &) const;    //大数的n次方运算
        ll    operator%(const ll  &) const;    //大数对一个ll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 ll & t)const;      //大数和一个ll类型的变量的大小比较
    
    };
    BigNum::BigNum(const ll b)     //将一个ll类型的变量转化为大数
    {
        ll 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
    {
        ll 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(ll 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
    {
        ll 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
    {
        ll 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];
        ll i = -1;
        in>>ch;
        ll l=strlen(ch);
        ll count=0,sum=0;
        for(i=l-1;i>=0;)
        {
            sum = 0;
            ll t=1;
            for(ll 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
    {
        ll 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);
        ll 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
    {
        ll 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;
        ll i,j,up;
        ll 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 ll & b) const   //大数对一个整数进行相除运算
    {
        BigNum ret;
        ll 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;
    }
    ll BigNum::operator %(const ll & b) const    //大数对一个ll类型的变量进行取模运算
    {
        ll i,d=0;
        for (i = len-1; i>=0; i--)
        {
            d = ((d * (MAXN+1))% b + a[i])% b;
        }
        return d;
    }
    BigNum BigNum::operator^(const ll & n) const    //大数的n次方运算
    {
        BigNum t,ret(1);
        ll i;
        if(n<0)
            exit(-1);
        if(n==0)
            return 1;
        if(n==1)
            return *this;
        ll 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
    {
        ll 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 ll & t) const    //大数和一个ll类型的变量的大小比较
    {
        BigNum b(t);
        return *this>b;
    }
     
    ll n,a[105],m,ans=0;
    ll gcd(ll a,ll b){
         return b==0?a:gcd(b,a%b);}
    ll lcm(ll a,ll b){
         return a/gcd(a,b)*b;}
     
    void dfs(ll d,ll tot,ll cnt){
      if(d==n+1){
        if(cnt){
          if(cnt&1) ans+=m/tot;
          else ans-=m/tot;
        }
        return;
      }
     
      dfs(d+1,tot,cnt);
     
      BigNum tmp=BigNum(tot)/gcd(tot,a[d])*BigNum(a[d]);
      if(tmp>BigNum(m)) return;
      dfs(d+1,lcm(tot,a[d]),cnt+1);
    }
     
    int main()
    {
        scanf("%lld%lld",&n,&m);
        for(ll i=1;i<=n;i++) scanf("%lld",&a[i]);
     
        dfs(1,1,0);
     
        ans=max((ll)0,m-ans);
        printf("%lld\n",ans);
        return 0;
    }

转载于:https://www.cnblogs.com/nublity/p/11148833.html