高精度模板

Posted 2022-03-24 nublity

#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;