1 solutions

  • 0
    @ 2024-12-10 23:27:11

    C++ :

    #include<iostream>
    #include<cstring>
    #include<cstdio>
    using namespace std;
    using std::istream;
    using std::ostream;
    using std::string;
    const int maxn=10000;/*精度位数*/
    /*(必选)类与基础功能定义,用法类似于unsigned(非负)*/
    class bign
    {
     	friend istream& operator>>(istream&,bign&);/*输入运算符友元*/
     	friend ostream& operator<<(ostream&,const bign&);/*输出运算符友元*/
     	friend bign operator+(const bign&,const bign&);/*加号运算符友元*/
     	friend bign operator*(const bign&,const bign&);/*乘号运算符友元*/
     	friend bign operator*(const bign&,int);/*高精度乘以低精度乘法友元*/
     	friend bign operator-(const bign&,const bign&);/*减号运算符友元*/
     	friend bign operator/(const bign&,const bign&);/*除法运算符友元*/
     	friend bign operator%(const bign&,const bign&);/*模运算符友元*/
     	friend bool operator<(const bign&,const bign&);/*逻辑小于符友元*/
     	friend bool operator>(const bign&,const bign&);/*逻辑大于符友元*/
     	friend bool operator<=(const bign&,const bign&);/*逻辑小于等于符友元*/
     	friend bool operator>=(const bign&,const bign&);/*逻辑大于等于符友元*/
     	friend bool operator==(const bign&,const bign&);/*逻辑等符友元*/
     	friend bool operator!=(const bign&,const bign&);/*逻辑不等符友元*/
    	private:
     	int len,s[maxn];
    	public:
     	bign(){memset(s,0,sizeof(s));len=1;}
     	bign operator=(const char* num)
     	{
      	   int i=0,ol;
      	   ol=len=strlen(num);
      	   while(num[i++]=='0'&&len>1)
       	   len--;
      	   memset(s,0,sizeof(s));
      	   for(i=0;i<len;i++)
       	   s[i]=num[ol-i-1]-'0';
      	   return *this;
     	}
     	bign operator=(int num)
     	{
      	  char s[maxn];
      	  sprintf(s,"%d",num);
      	  *this=s;
      	  return *this;
     	}
     	bign(int num){*this=num;}
     	bign(const char* num){*this=num;}
     	string str() const
     	{
      	  int i;
      	  string res="";
      	  for(i=0;i<len;i++)res=char(s[i]+'0')+res;
      	  if(res=="")res="0";
      	  return res;
     	}
    };
    /*(可选)基本逻辑运算符重载*/
    bool operator<(const bign& a,const bign& b)
    {
     	int i;
     	if(a.len!=b.len)return a.len<b.len;
     	for(i=a.len-1;i>=0;i--)
      	if(a.s[i]!=b.s[i])
       	return a.s[i]<b.s[i];
     	return false;
    }
    bool operator>(const bign& a,const bign& b){return b<a;}
    bool operator<=(const bign& a,const bign& b){return !(a>b);}
    bool operator>=(const bign& a,const bign& b){return !(a<b);}
    bool operator!=(const bign& a,const bign& b){return a<b||a>b;}
    bool operator==(const bign& a,const bign& b){return !(a<b||a>b);}
    /*(可选)加法运算符重载*/
    bign operator+(const bign& a,const bign& b)
    {
     	int i,max=(a.len>b.len?a.len:b.len),t,c;
     	bign sum;
     	sum.len=0;
     	for(i=0,c=0;c||i<max;i++)
     	{
      	  t=c;
      	  if(i<a.len)t+=a.s[i];
      	  if(i<b.len)t+=b.s[i];
      	  sum.s[sum.len++]=t%10;
      	  c=t/10;
     	}
     	return sum;
    }
    /*(可选)乘法运算符重载(高精度乘高精度)*/
    bign operator*(const bign& a,const bign& b)
    {
     	  int i,j;
     	  bign res;
     	  for(i=0;i<a.len;i++)
     	  {
      	  for(j=0;j<b.len;j++)
      	  {
       	  res.s[i+j]+=(a.s[i]*b.s[j]);
       	  res.s[i+j+1]+=res.s[i+j]/10;
       	  res.s[i+j]%=10;
      	  }
     	  }
     	  res.len=a.len+b.len;
     	  while(res.s[res.len-1]==0&&res.len>1)res.len--;
     	  if(res.s[res.len])res.len++;
     	  return res;
    }
    /*高精度乘以低精度(注意:必须是bign*int顺序不能颠倒,要么会与高精度乘高精度发生冲突*/
    bign operator*(const bign& a,int b)
    {
     	int i,t,c=0;
     	bign res;
     	for(i=0;i<a.len;i++)
     	{
      	  t=a.s[i]*b+c;
      	  res.s[i]=t%10;
      	  c=t/10;
     	}
     	res.len=a.len;
     	while(c!=0)
     	{
      	  res.s[i++]=c%10;
      	  c/=10;
      	  res.len++;
     	}
     	return res;
    }
    /*(可选)减法运算符重载*/
    bign operator-(const bign& a,const bign& b)
    {
     	bign res;
     	int i,len=(a.len>b.len)?a.len:b.len;
     	for(i=0;i<len;i++)
     	{
      	  res.s[i]+=a.s[i]-b.s[i];
      	  if(res.s[i]<0)
      	{
       	  res.s[i]+=10;
       	  res.s[i+1]--;
      	}
     	}
     	while(res.s[len-1]==0&&len>1)len--;
     	res.len=len;
     	return res;
    }
    /*(可选)除法运算符重载(注意:减法和乘法运算和>=运算符必选)*/
    bign operator/(const bign& a,const bign& b)
    {
     	int i,len=a.len;
     	bign res,f;
     	for(i=len-1;i>=0;i--)
     	{
      	  f=f*10;
      	  f.s[0]=a.s[i];
      	  while(f>=b)
      	{
       	  f=f-b;
       	  res.s[i]++;
      	}
     	}
     	while(res.s[len-1]==0&&len>1)len--;
     	res.len=len;
     	return res;
    }
    /*(可选)模运算符重载(注意:减法和乘法运算和>=运算符必选)*/
    bign operator%(const bign& a,const bign& b)
    {
     	int i,len=a.len;
     	bign res,f;
     	for(i=len-1;i>=0;i--)
     	{
      	  f=f*10;
      	  f.s[0]=a.s[i];
      	  while(f>=b)
      	{
       	  f=f-b;
       	  res.s[i]++;
      	}
     	}
     	return f;
    }
    /*(可选)X=运算符重载(注意:X法必选)*/
    bign& operator+=(bign& a,const bign& b)
    {
     	a=a+b;
     	return a;
    }
    bign& operator-=(bign& a,const bign& b)
    {
     	a=a-b;
     	return a;
    }
    bign& operator*=(bign& a,const bign& b)
    {
     	a=a*b;
     	return a;
    }
    bign& operator/=(bign& a,const bign& b)
    {
     	a=a/b;
     	return a;
    }
    /*可选前缀++/--与后缀++/--(注意:加法必选)*/
    bign& operator++(bign& a)
    {
     	a=a+1;
     	return a;
    }
    /*bign& operator++(bign& a,int)
    {
     	bign t=a;
     	a=a+1;
     	return t;
    }*/
    bign& operator--(bign& a)
    {
     	a=a-1;
     	return a;
    }
    /*bign& operator--(bign& a,int)
    {
     	bign t=a;
     	a=a-1;
     	return t;
    }*/
    istream& operator>>(istream &in,bign& x)
    {
     	string s;
     	in>>s;
     	x=s.c_str();
     	return in;
    }
    ostream& operator<<(ostream &out,const bign& x)
    {
     	out<<x.str();
     	return out;
    }
    int main()
    {
    	bign a,sum=2,b,i;
    	cin>>a>>b;
    	if(a>=b)
    	cout<<"1";
    	else
    	cout<<"2";
    	
    	return 0;
    }
    
    • 1

    Information

    ID
    598
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    # Submissions
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