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. 系统设计摘录CAP

    系统架构设计理论与原则 这里主要介绍几种常见的架构设计理论和原则,常见于大中型互联系统架构设计. (一).CAP理论 1.什么是CAP 所谓CAP,即一致性(Consistency).可用性(Avai ...

  2. VBA 连接sql server的用法

    cnnstr = "Provider=sqloledb;Data Source=192.211.21.8;Initial Catalog=pub;UID=账号;PWD=密码" VB ...

  3. 洛谷 P2153 [SDOI2009]晨跑

    题目描述 Elaxia最近迷恋上了空手道,他为自己设定了一套健身计划,比如俯卧撑.仰卧起坐等 等,不过到目前为止,他坚持下来的只有晨跑. 现在给出一张学校附近的地图,这张地图中包含N个十字路口和M条街 ...

  4. 基于SAE的Python+Django部署

    本文主要参考:http://www.cnblogs.com/qtsharp/archive/2012/01/12/2320774.html,另外包括自己的实际操作. 一.申请SAE帐号以及创建应用ya ...

  5. Python3简明教程(四)—— 流程控制之分支

    我们通过 if-else 语句来做决定,来改变程序运行的流程. if语句 语法如下: if expression: do this 如果表达式 expression 的值为真(不为零的任何值都为真), ...

  6. 旅行商问题——状态压缩DP

    问题简介 有n个城市,每个城市间均有道路,一个推销员要从某个城市出发,到其余的n-1个城市一次且仅且一次,然后回到再回到出发点.问销售员应如何经过这些城市是他所走的路线最短? 用图论的语言描述就是:给 ...

  7. run_debug和run_demo的区别

    run_demo:给一张图,直接生成测试出来的框,输入不用给gt框 run_debug:生成ap值,生成的图片既有gt框也有测试得到的结果框 run_demo的源码demo_test放在example ...

  8. 使用snapshot继续训练网络

    注意:snapshots和weights不能同时使用 用预训练模型进行finetune是以下命令: ./build/tools/caffe train --solver=examples/XXX/le ...

  9. SQL比较两表字段和字段类型

    一.问题 业务需要把TB_Delete_KYSubProject表数据恢复到TB_KYSubProject,但提示错误,错误原因是两表字段类型存在不一致 insert into [TB_KYSubPr ...

  10. C指针类型转换问题

    先看下面的代码: #include<stdio.h> int main () { int a; char *x; x = (char *) &a; a = 512; x[0] = ...