只用输入用cin 输出  cout  每个数学符号都可以用   超级强大

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <stack>
using namespace std;
#define maxn 120
class DividedByZeroException {}; class BigInteger
{
private:
vector<char> digits;
bool sign; // true for positive, false for negitive
void trim(); // remove zeros in tail, but if the value is 0, keep only one:)
public:
BigInteger(int); // construct with a int integer
BigInteger(string&) ;
BigInteger();
BigInteger (const BigInteger&);
BigInteger operator=(const BigInteger& op2); BigInteger abs() const;
BigInteger pow(int a); //binary operators friend BigInteger operator+=(BigInteger&,const BigInteger&);
friend BigInteger operator-=(BigInteger&,const BigInteger&);
friend BigInteger operator*=(BigInteger&,const BigInteger&);
friend BigInteger operator/=(BigInteger&,const BigInteger&) throw(DividedByZeroException);
friend BigInteger operator%=(BigInteger&,const BigInteger&) throw(DividedByZeroException); friend BigInteger operator+(const BigInteger&,const BigInteger&);
friend BigInteger operator-(const BigInteger&,const BigInteger&);
friend BigInteger operator*(const BigInteger&,const BigInteger&);
friend BigInteger operator/(const BigInteger&,const BigInteger&) throw(DividedByZeroException);
friend BigInteger operator%(const BigInteger&,const BigInteger&) throw(DividedByZeroException); //uniary operators
friend BigInteger operator-(const BigInteger&); //negative friend BigInteger operator++(BigInteger&); //++v
friend BigInteger operator++(BigInteger&,int); //v++
friend BigInteger operator--(BigInteger&); //--v
friend BigInteger operator--(BigInteger&,int); //v-- friend bool operator>(const BigInteger&,const BigInteger&);
friend bool operator<(const BigInteger&,const BigInteger&);
friend bool operator==(const BigInteger&,const BigInteger&);
friend bool operator!=(const BigInteger&,const BigInteger&);
friend bool operator>=(const BigInteger&,const BigInteger&);
friend bool operator<=(const BigInteger&,const BigInteger&); friend ostream& operator<<(ostream&,const BigInteger&); //print the BigInteger
friend istream& operator>>(istream&, BigInteger&); // input the BigInteger public:
static const BigInteger ZERO;
static const BigInteger ONE;
static const BigInteger TEN;
};
// BigInteger.cpp const BigInteger BigInteger::ZERO=BigInteger();
const BigInteger BigInteger::ONE =BigInteger();
const BigInteger BigInteger::TEN =BigInteger(); BigInteger::BigInteger()
{
sign=true;
} BigInteger::BigInteger(int val) // construct with a int integer
{
if (val >= )
sign = true;
else
{
sign = false;
val *= (-);
}
do
{
digits.push_back( (char)(val%) );
val /= ;
}
while ( val != );
} BigInteger::BigInteger(string& def)
{
sign=true;
for ( string::reverse_iterator iter = def.rbegin() ; iter < def.rend(); iter++)
{
char ch = (*iter);
if (iter == def.rend()-)
{
if ( ch == '+' )
break;
if(ch == '-' )
{
sign = false;
break;
}
}
digits.push_back( (char)((*iter) - '' ) );
}
trim();
} void BigInteger::trim()
{
vector<char>::reverse_iterator iter = digits.rbegin();
while(!digits.empty() && (*iter) == )
{
digits.pop_back();
iter=digits.rbegin();
}
if( digits.size()== )
{
sign = true;
digits.push_back();
}
} BigInteger::BigInteger(const BigInteger& op2)
{
sign = op2.sign;
digits=op2.digits;
} BigInteger BigInteger::operator=(const BigInteger& op2)
{
digits = op2.digits;
sign = op2.sign;
return (*this);
} BigInteger BigInteger::abs() const
{
if(sign) return *this;
else return -(*this);
} BigInteger BigInteger::pow(int a)
{
BigInteger res();
for(int i=; i<a; i++)
res*=(*this);
return res;
} //binary operators
BigInteger operator+=(BigInteger& op1,const BigInteger& op2)
{
if( op1.sign == op2.sign )
{
//只处理相同的符号的情况,异号的情况给-处理
vector<char>::iterator iter1;
vector<char>::const_iterator iter2;
iter1 = op1.digits.begin();
iter2 = op2.digits.begin();
char to_add = ; //进位
while ( iter1 != op1.digits.end() && iter2 != op2.digits.end())
{
(*iter1) = (*iter1) + (*iter2) + to_add;
to_add = ((*iter1) > ); // 大于9进一位
(*iter1) = (*iter1) % ;
iter1++;
iter2++;
}
while ( iter1 != op1.digits.end() ) //
{
(*iter1) = (*iter1) + to_add;
to_add = ( (*iter1) > );
(*iter1) %= ;
iter1++;
}
while ( iter2 != op2.digits.end() )
{
char val = (*iter2) + to_add;
to_add = (val > ) ;
val %= ;
op1.digits.push_back(val);
iter2++;
}
if( to_add != )
op1.digits.push_back(to_add);
return op1;
}
else
{
if (op1.sign)
return op1 -= (-op2);
else
return op1= op2 - (-op1);
} } BigInteger operator-=(BigInteger& op1,const BigInteger& op2)
{
if( op1.sign == op2.sign )
{
//只处理相同的符号的情况,异号的情况给+处理
if(op1.sign)
{
if(op1 < op2) // 2 - 3
return op1=-(op2 - op1);
}
else
{
if(-op1 > -op2) // (-3)-(-2) = -(3 - 2)
return op1=-((-op1)-(-op2));
else // (-2)-(-3) = 3 - 2
return op1= (-op2) - (-op1);
}
vector<char>::iterator iter1;
vector<char>::const_iterator iter2;
iter1 = op1.digits.begin();
iter2 = op2.digits.begin(); char to_substract = ; //借位 while ( iter1 != op1.digits.end() && iter2 != op2.digits.end())
{
(*iter1) = (*iter1) - (*iter2) - to_substract;
to_substract = ;
if( (*iter1) < )
{
to_substract=;
(*iter1) += ;
}
iter1++;
iter2++;
}
while ( iter1 != op1.digits.end() )
{
(*iter1) = (*iter1) - to_substract;
to_substract = ;
if( (*iter1) < )
{
to_substract=;
(*iter1) += ;
}
else break;
iter1++;
}
op1.trim();
return op1;
}
else
{
if (op1 > BigInteger::ZERO)
return op1 += (-op2);
else
return op1 = -(op2 + (-op1));
}
}
BigInteger operator*=(BigInteger& op1,const BigInteger& op2)
{
BigInteger result();
if (op1 == BigInteger::ZERO || op2==BigInteger::ZERO)
result = BigInteger::ZERO;
else
{
vector<char>::const_iterator iter2 = op2.digits.begin();
while( iter2 != op2.digits.end() )
{
if(*iter2 != )
{
deque<char> temp(op1.digits.begin() , op1.digits.end());
char to_add = ;
deque<char>::iterator iter1 = temp.begin();
while( iter1 != temp.end() )
{
(*iter1) *= (*iter2);
(*iter1) += to_add;
to_add = (*iter1) / ;
(*iter1) %= ;
iter1++;
}
if( to_add != )
temp.push_back( to_add );
int num_of_zeros = iter2 - op2.digits.begin();
while( num_of_zeros--)
temp.push_front();
BigInteger temp2;
temp2.digits.insert( temp2.digits.end() , temp.begin() , temp.end() );
temp2.trim();
result = result + temp2;
}
iter2++;
}
result.sign = ( (op1.sign && op2.sign) || (!op1.sign && !op2.sign) );
}
op1 = result;
return op1;
} BigInteger operator/=(BigInteger& op1 , const BigInteger& op2 ) throw(DividedByZeroException)
{
if( op2 == BigInteger::ZERO )
throw DividedByZeroException();
BigInteger t1 = op1.abs(), t2 = op2.abs();
if ( t1 < t2 )
{
op1 = BigInteger::ZERO;
return op1;
}
//现在 t1 > t2 > 0
//只需将 t1/t2的结果交给result就可以了
deque<char> temp;
vector<char>::reverse_iterator iter = t1.digits.rbegin(); BigInteger temp2();
while( iter != t1.digits.rend() )
{
temp2 = temp2 * BigInteger::TEN + BigInteger( (int)(*iter) );
char s = ;
while( temp2 >= t2 )
{
temp2 = temp2 - t2;
s = s + ;
}
temp.push_front( s );
iter++;
}
op1.digits.clear();
op1.digits.insert( op1.digits.end() , temp.begin() , temp.end() );
op1.trim();
op1.sign = ( (op1.sign && op2.sign) || (!op1.sign && !op2.sign) );
return op1;
} BigInteger operator%=(BigInteger& op1,const BigInteger& op2) throw(DividedByZeroException)
{
return op1 -= ((op1 / op2)*op2);
} BigInteger operator+(const BigInteger& op1,const BigInteger& op2)
{
BigInteger temp(op1);
temp += op2;
return temp;
}
BigInteger operator-(const BigInteger& op1,const BigInteger& op2)
{
BigInteger temp(op1);
temp -= op2;
return temp;
} BigInteger operator*(const BigInteger& op1,const BigInteger& op2)
{
BigInteger temp(op1);
temp *= op2;
return temp; } BigInteger operator/(const BigInteger& op1,const BigInteger& op2) throw(DividedByZeroException)
{
BigInteger temp(op1);
temp /= op2;
return temp;
} BigInteger operator%(const BigInteger& op1,const BigInteger& op2) throw(DividedByZeroException)
{
BigInteger temp(op1);
temp %= op2;
return temp;
} //uniary operators
BigInteger operator-(const BigInteger& op) //negative
{
BigInteger temp = BigInteger(op);
temp.sign = !temp.sign;
return temp;
} BigInteger operator++(BigInteger& op) //++v
{
op += BigInteger::ONE;
return op;
} BigInteger operator++(BigInteger& op,int x) //v++
{
BigInteger temp(op);
++op;
return temp;
} BigInteger operator--(BigInteger& op) //--v
{
op -= BigInteger::ONE;
return op;
} BigInteger operator--(BigInteger& op,int x) //v--
{
BigInteger temp(op);
--op;
return temp;
} bool operator<(const BigInteger& op1,const BigInteger& op2)
{
if( op1.sign != op2.sign )
return !op1.sign;
else
{
if(op1.digits.size() != op2.digits.size())
return (op1.sign && op1.digits.size()<op2.digits.size())
|| (!op1.sign && op1.digits.size()>op2.digits.size());
vector<char>::const_reverse_iterator iter1,iter2;
iter1 = op1.digits.rbegin();
iter2 = op2.digits.rbegin();
while( iter1 != op1.digits.rend() )
{
if( op1.sign && *iter1 < *iter2 ) return true;
if( op1.sign && *iter1 > *iter2 ) return false;
if( !op1.sign && *iter1 > *iter2 ) return true;
if( !op1.sign && *iter1 < *iter2 ) return false;
iter1++;
iter2++;
}
return false;
}
}
bool operator==(const BigInteger& op1,const BigInteger& op2)
{
if( op1.sign != op2.sign || op1.digits.size() != op2.digits.size() )
return false;
vector<char>::const_iterator iter1,iter2;
iter1 = op1.digits.begin();
iter2 = op2.digits.begin();
while( iter1!= op1.digits.end() )
{
if( *iter1 != *iter2 ) return false;
iter1++;
iter2++;
}
return true;
} bool operator!=(const BigInteger& op1,const BigInteger& op2)
{
return !(op1==op2);
} bool operator>=(const BigInteger& op1,const BigInteger& op2)
{
return (op1>op2) || (op1==op2);
} bool operator<=(const BigInteger& op1,const BigInteger& op2)
{
return (op1<op2) || (op1==op2);
} bool operator>(const BigInteger& op1,const BigInteger& op2)
{
return !(op1<=op2);
} ostream& operator<<(ostream& stream,const BigInteger& val) //print the BigInteger
{
if (!val.sign)
stream << "-";
for ( vector<char>::const_reverse_iterator iter = val.digits.rbegin(); iter != val.digits.rend() ; iter++)
stream << (char)((*iter) + '');
return stream;
} istream& operator>>(istream& stream, BigInteger& val)
{
//Input the BigInteger
string str;
stream >> str;
val=BigInteger(str);
return stream;
} int T, n;
BigInteger a[]; bool solve()
{
int cnt = ;
for(int i=; i<n; i++)
{
if(a[i] == )
cnt ++;
} if(cnt && cnt!=n)
return false; for(int i=; i<n-; i++)
{
if( a[i-]*a[i+] != a[i]*a[i])
return false;
}
return true;
} int main()
{
int T;
BigInteger q, p, ans;
scanf("%d", &T);
while(T--)
{
cin>>q>>p;
q = q-;
ans = (q+)*q/%p;
cout<<ans<<endl;
}
return ;
}

大数c++模板 超级好用的更多相关文章

  1. hdu_1041(Computer Transformation) 大数加法模板+找规律

    Computer Transformation Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/ ...

  2. JAVA 大数开方模板

    JAVA 大数开方模板 import java.math.BigInteger; import java.math.*; import java.math.BigInteger; import jav ...

  3. C++大数类模板

    友情提示:使用该模板的注意了,在大数减法里有一个小错误,导致减法可能会出错 // 原来的写法,将t1.len错写成了len ] == && t1.len > ) { t1.len ...

  4. Pollard_Rho大数分解模板题 pku-2191

    题意:给你一个数n,  定义m=2k-1,   {k|1<=k<=n},并且 k为素数;  当m为合数时,求分解为质因数,输出格式如下:47 * 178481 = 8388607 = ( ...

  5. java的大数运算模板

    import java.math.BigInteger; import java.util.Scanner; public class Main { public static void main(S ...

  6. icpc 江苏 D Persona5 组合数学 大数阶乘(分段阶乘) 大数阶乘模板

    Persona5 is a famous video game. In the game, you are going to build relationship with your friends. ...

  7. C++大数运算模板

    #include<iostream> #include<cstring> #include<cstdio> #include<iomanip> #inc ...

  8. PAT A1024题解——高精度大数相加模板

    PAT:A1024 Palindromic Number A number that will be the same when it is written forwards or backwards ...

  9. nyoj 28-大数阶乘 (大数模板)

    28-大数阶乘 内存限制:64MB 时间限制:3000ms Special Judge: No accepted:19 submit:39 题目描述: 我们都知道如何计算一个数的阶乘,可是,如果这个数 ...

随机推荐

  1. HttpURLConnection读取http信息

    废话不多说,直接上code. package mytest; import java.io.BufferedReader; import java.io.IOException; import jav ...

  2. UVA 11346 Probability 概率 (连续概率)

    题意:给出a和b,表示在直角坐标系上的x=[-a,a] 和 y=[-b,b]的这样一块矩形区域.给出一个数s,问在矩形内随机选择一个点p=(x,y),则(0.0)和p点组成的矩形面积大于s的概率是多少 ...

  3. SAP成都研究院安德鲁:自己动手开发一个Chrome Extension

    各位好,我叫何金鑫(He Andrew), 团队同事亲切地称呼在下为安德鲁.如果你在附近找到wifi热点名为 「安德鲁森面包房5g」,可能是我就在附近,我们可以去喝杯咖啡,聊聊最近有趣的东西. 鄙人现 ...

  4. 3.12 在运算和比较时使用NULL值

    问题:NULL值永远不会等于或不等于任何值,也包括NULL值自己,但是需要像计算真实值一样计算可为空列的返回值.例如,需要在表emp中查出所有比“WARD”提成(COMM)低的员工,提成为NULL(空 ...

  5. (转)编码剖析Spring依赖注入的原理

    http://blog.csdn.net/yerenyuan_pku/article/details/52834561 Spring的依赖注入 前面我们就已经讲过所谓依赖注入就是指:在运行期,由外部容 ...

  6. git 本地与远程仓库出现代码冲突解决方法

    提交过程中报错: [python@heaven-00 Selesystem]$ git push -u origin masterUsername for 'https://github.com': ...

  7. JavaEE-07 过滤器和监听器

    学习要点 过滤器 监听器 过滤器Filter 过滤器的概念 过滤器位于客户端和web应用程序之间,用于检查和修改两者之间流过的请求和响应. 在请求到达Servlet/JSP之前,过滤器截获请求. 在响 ...

  8. SqlSugar直接执行Sql

    参考:http://www.codeisbug.com/Doc/8/1132 我的思路: 1.数据库中写好sql 2.用SqlSugar直接执行sql,获取DataTable的数据 3.DataTab ...

  9. c++类流操作运算符的重定义

    对于流操作运算符我们需要注意的是函数的返回类型应该是流输入类型的引用或者流输出类型的引用,因为如果代码是 cout<<a<<b; 我们对a执行完cout函数之后,我们应该再次将 ...

  10. POJ-1163 递推

    代码很容易看明白,就不详解了. 这个是空间优化的代码. #include <iostream> #include <algorithm> #define MAX 101 usi ...