高精度模板
/* 大整数Bignum类 开始写于2016.06.29 #----- Version 1.0.0 支持有符号运算 已实现+,-,*(位数较大(位数之和>800)时FFT，较小时模拟), abs(绝对值) 代码供学习交流使用，版权没有，欢迎盗版。 -- 作者:sxysxy #----- Version 1.0.1 更新于2016.07.02 实现 /(二分法), sqrt(二分法), pow(快速幂)，log(暴力，因为考虑到答案一般都比较小) lg(即log(10)), factorial(阶乘) 添加printB函数便于调试 -- sxysxy LICENSE: This file is under GPL. You can redistribute it and/or modify it under the terms of the GNU General Public License as published by the Free Software Foundation, either version 3 of the License, or (at your option) any later version. */ #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include <vector> #include <list> #include <map> #include <cstring> #include <string> #include <iostream> #include <fstream> #include <cmath> using namespace std; #define BIGNUM_DEBUG 2333 class Bignum { void mknum(const char *s, int len = -1) { sign = 0; if(*s == '-') { mknum(s+1); sign = 1; return; } int l; if(len == -1) l = strlen(s); else l = len; l = strlen(s); bits.clear(); bits.resize(l); for(int i = l-1; i >= 0; i--) bits[l-i-1] = s[i] - '0'; maintain(); } void mknum(string &s) { mknum(s.c_str(), s.length()); } // ------------- void us_addto(Bignum &b) // unsigned add to { int mlen = max(b.bits.size(), bits.size()); int slen = bits.size(); int olen = b.bits.size(); bits.resize(mlen); for(int i = 0; i < mlen; i++) { int s = 0; if(i < slen)s += bits[i]; if(i < olen)s += b.bits[i]; bits[i] = s; } maintain(); } class FFTer { class Complex { public: double real, image; Complex(double a = 0, double b = 0) { real = a; image = b; } Complex operator + (const Complex &o){return Complex(real+o.real, image+o.image);} Complex operator - (const Complex &o){return Complex(real-o.real, image-o.image);} Complex operator * (const Complex &o){return Complex(real*o.real-image*o.image, real*o.image+o.real*image);} Complex operator * (double k){return Complex(real*k, image*k);} Complex operator / (double k){return Complex(real/k, image/k);} }; public: vector<Complex> a; //系数向量 int n; //多项式次数上界 FFTer(vector<int> &vec) { a.resize(vec.size()); for(int i = 0; i < vec.size(); i++) a[i].real = vec[i]; n = vec.size(); } void transform() { int j = 0; int k; for(int i = 0; i < n; i++) { if(j > i)swap(a[i], a[j]); k = n; while(j & (k >>= 1))j &= ~k; j |= k; } } void FFT(bool IDFT = false) { const double Pi = IDFT?-acos(-1.0):acos(-1.0); //IDFT与DFT选择方向相反（符号相反） transform(); //交换元素（翻转二进制位，具体看下面注释，再具体看算导 for(int s = 1; s < n; s <<= 1) { //算法导论上是fot s = 1 to lgn,考虑到精度问题改为上面那个... for(int t = 0; t < n; t += s<<1) { //合并[t, t+s-1]与 [t+s, t+2*s-1] (算导上以指数形式给出，注意到他的s....) //合并为[t, t+2*s-1] (看起来像是废话) (有示例图在算导上，画得很形象的) /* 一个更简单的示例图 (翻转过程) 翻转 合并 00 -> 00 0-|--|--------------------------- | | 01 -> 10 1-|--|---\ /--------------------- | | X 10 -> 01 2-|--|---/ \--------------------- | | 11 -> 11 3-|--|--------------------------- */ double x = Pi/s; Complex omgn(cos(x), sin(x)); Complex omg(1.0, 0.0); //单位向量 for(int m = 0; m < s; m++) { //旋转 int a1 = m+t; int a2 = m+t+s; //取两边系数向量的系数 //算导上管这个叫公共子表达式消除 //(其实就是一个变量计算一次然后保存下来用多次...嗯算导总是这么有逼格) Complex comm = omg * a[a2]; a[a2] = a[a1] - comm; a[a1] = a[a1] + comm; //这两个顺序不要反了 omg = omg * omgn; } } } if(IDFT) for(int i = 0; i < n; i++) a[i] = a[i] / n; } void mul(FFTer &o) { int s = 1; while(s < n + o.n)s <<= 1; n = o.n = s; a.resize(s); o.a.resize(s); FFT(false); o.FFT(false); for(int i = 0; i < n; i++) a[i] = a[i] * o.a[i]; FFT(true); } }; void us_multo(Bignum &b) { FFTer x(bits); FFTer y(b.bits); x.mul(y); bits.clear(); bits.resize(x.a.size()); for(int i = 0; i < x.n; i++) bits[i] = (int)(x.a[i].real+0.5); maintain(); } void us_multo_simu(Bignum &b) { vector<int> r; r.resize(max(length(),b.length())<<1); for(int i = 0; i < length(); i++) for(int j = 0; j < b.length(); j++) r[i+j] += bits[i] * b.bits[j]; *(&(this -> bits)) = r; maintain(); } void us_subto(Bignum &b) // abs(self) >= abs(other) { int mlen = length(); int olen = b.length(); for(int i = 0; i < mlen; i++) { int s = bits[i]; if(i < olen)s -= b.bits[i]; bits[i] = s; if(bits[i] < 0) { bits[i] += 10; bits[i+1] -= 1; } } for(int i = bits.size() - 1; !bits[i] && i >= 1; i--)bits.pop_back(); if(bits.size() == 1 && bits[0] == 0)sign = 0; } void us_divto(Bignum &b) { if(length() == 1 && bits[0] == 0)return; Bignum L("0"); L.sign = 0; Bignum R(*this); R.sign = 0; Bignum two("2"); R *= two; Bignum one("1"); one.sign = 0; while(L + one != R) { Bignum M = L+R; M.divto2(); Bignum t = M*b; if(t > *this) { R = M; }else if(t < *this) { L = M; }else { *this = M; L = M; break; } } *this = L; } public: int sign; vector<int> bits; int length() { return bits.size(); } void maintain() { for(int i = 0; i < bits.size(); i++) { if(i + 1 < bits.size()) bits[i+1] += bits[i]/10; else if(bits[i] > 9) bits.push_back(bits[i]/10); bits[i] %= 10; } if(bits.size() == 0) { bits.push_back(0); sign = 0; } for(int i = bits.size() - 1; !bits[i] && i >= 1; i--)bits.pop_back(); } Bignum(string &s) { Bignum(); mknum(s); } Bignum(const char *s) { Bignum(); mknum(s); } Bignum(int n) { Bignum(); char buf[15]; sprintf(buf, "%d", n); mknum(buf); } Bignum() { sign = 0; bits.push_back(0); } Bignum(const Bignum& b) { copy(b); } void copy(const Bignum& b) { sign = b.sign; bits = b.bits; } // ------------------------------------------ bool us_cmp(Bignum &b) //无符号的比较 { if(length() != b.length())return false; int l = length(); for(int i = 0; i < l; i++) if(bits[i] != b.bits[i]) return false; return true; } bool us_larger(Bignum &b) { if(length() > b.length())return true; else if(length() < b.length())return false; int l = length(); for(int i = l-1; i >= 0; i--) if(bits[i] > b.bits[i]) return true; else if(bits[i] < b.bits[i]) return false; return false; } bool operator== (Bignum &o) { if(sign != o.sign) return false; return us_cmp(o); } bool operator!= (Bignum &o) { return !(*this == o); } bool operator> (Bignum &o) { if(sign == 0 && o.sign == 1)return true; if(sign == 1 && o.sign == 0)return false; if(sign == o.sign && sign)return !us_larger(o); return us_larger(o); } bool operator< (Bignum &o) { return !(*this == o || *this > o); //小于就是不等于也不大于 } bool operator<= (Bignum &o) { return *this < o || *this == o; } bool operator>= (Bignum &o) { return *this > o || *this == o; } // ------------------------------- Bignum& operator+= (Bignum &o) { if(!sign && !o.sign) { us_addto(o); sign = 0; } else if(sign && o.sign) { us_addto(o); sign = 1; } else if(sign && !o.sign) { if(o.us_larger(*this)) { Bignum t(o); t.us_subto(*this); *this = t; sign = 0; }else { us_subto(o); sign = 1; if(bits.size() == 1 && bits[0] == 0)sign = 0; } }else if(!sign && o.sign) { if(us_larger(o)) { us_subto(o); sign = 0; }else { Bignum t(o); t.us_subto(*this); *this = t; sign = 1; if(bits.size() == 1 && bits[0] == 0)sign = 0; } } return *this; } Bignum operator+ (Bignum &o) { Bignum t(*this); t += o; return t; } // ------------------------------ Bignum& operator*= (Bignum &o) { if(length() + o.length() > 800) us_multo(o); //FFT else us_multo_simu(o); //simulate if(sign == o.sign)sign = 0; else sign = 1; return *this; } Bignum operator* (Bignum &o) { Bignum t(*this); t *= o; return t; } // ------------------------------- Bignum& operator-= (Bignum &o) { if(!sign && !o.sign) { if(us_larger(o)) { us_subto(o); sign = 0; } else { Bignum t(o); t.us_subto(*this); *this = t; sign = 1; if(bits.size() == 1 && bits[0] == 0)sign = 0; } }else if(sign && o.sign) { if(us_larger(o)) { us_subto(o); sign = 1; if(bits.size() == 1 && bits[0] == 0)sign = 0; }else { Bignum t(o); t.us_subto(*this); *this = t; sign = 0; } }else if(!sign && o.sign) { us_addto(o); sign = 0; }else if(sign && !o.sign) { us_addto(o); sign = 1; } return *this; } Bignum operator- (Bignum &o) { Bignum t(*this); t -= o; return t; } // --------------------------------- Bignum& divto2() { if(!bits.size())return *this; bits[0] >>= 1; int i; for(i = 1; i < bits.size(); i++) { if(bits[i] & 1)bits[i-1] += 5; bits[i] >>= 1; } if(bits[i-1] == 0)bits.pop_back(); return *this; } Bignum& operator/= (Bignum &o) { us_divto(o); if(sign == o.sign)sign = 0; else sign = 1; return *this; } Bignum operator/ (Bignum &o) { Bignum t(*this); t /= o; return t; } // --------------------------------- Bignum abs() { Bignum t(*this); t.sign = 0; return t; } Bignum sqrt() { Bignum L("0"), R(*this); Bignum one("1"); Bignum m, t; while(L + one != R) { m = L+R; m.divto2(); Bignum t = m*m; if(t == *this)return m; else if(t > *this)R = m; else L = m; } return L; } //若e <= 0则会返回1 //底数（也就是this）是负数的话会根据次数决定符号 Bignum pow(Bignum &e) { if(e.sign)return 1; Bignum ans("1"); Bignum base(*this); Bignum zero("0"); Bignum exp(e); while(exp > zero) { if(exp.bits[0] & 1) { ans *= base; } base *= base; exp.divto2(); } if(sign && e.bits[0] & 1)ans.sign = 1; return ans; } //注意，如果本数小于底数返回1... Bignum log(Bignum &base) { if(sign)return 0; if(length() == 1 && bits[0] == 1)return 0; if(*this <= base)return 1; Bignum one("1"); Bignum r("1"); Bignum c("0"); while(r < *this) { r *= base; c += one; } if(r != *this)c -= one; return c; } Bignum lg() { Bignum ten("10"); return log(ten); } Bignum factorial() { Bignum r("1"); Bignum zero("0"); Bignum one("1"); Bignum t(*this); while(t > zero) { r *= t; t -= one; } return r; } // ----------------------------------- friend istream& operator>>(istream &is, Bignum &b) { string s; is >> s; b.mknum(s); return is; } friend ostream& operator<<(ostream &os, Bignum b) { if(b.sign)os << '-'; for(int i = b.bits.size()-1; i >= 0; i--)os << b.bits[i]; return os; } string to_string() { int sz = length(); string s; if(sign) s.resize(sz+1); else s.resize(sz); int i = 0; if(sign)s[i++] = '-'; for(int j = sz-1; i < sz+sign; i++, j--) s[i] = bits[j] + '0'; return s; } }; // -- #ifdef BIGNUM_DEBUG #ifdef __GNUC__ __attribute__((noinline)) //禁止内联 #endif #ifdef __MINGW32__ __attribute__((noinline)) #endif char* printB(Bignum &b) { //仅仅是用于能在gdb中使用它来输出自己 string s = b.to_string(); char *buf = (char *)malloc(sizeof(char) * s.length()); //这个函数仅用于调试，这里申请的内存不会释放 //因此非调试时不要使用这个函数 strcpy(buf, s.c_str()); return buf; //然后gdb中就可以 print printB(一个Bignum ) } #endif int main() { return 0; }
2更简短的模板
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <algorithm> using namespace std; const int MAXN = 410; struct bign { int len, s[MAXN]; bign () { memset(s, 0, sizeof(s)); len = 1; } bign (int num) { *this = num; } bign (const char *num) { *this = num; } bign operator = (const int num) { char s[MAXN]; sprintf(s, "%d", num); *this = s; return *this; } bign operator = (const char *num) { for(int i = 0; num[i] == '0'; num++) ; //去前导0 len = strlen(num); for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0'; return *this; } bign operator + (const bign &b) const //+ { bign c; c.len = 0; for(int i = 0, g = 0; g || i < max(len, b.len); i++) { int x = g; if(i < len) x += s[i]; if(i < b.len) x += b.s[i]; c.s[c.len++] = x % 10; g = x / 10; } return c; } bign operator += (const bign &b) { *this = *this + b; return *this; } void clean() { while(len > 1 && !s[len-1]) len--; } bign operator * (const bign &b) //* { bign c; c.len = len + b.len; for(int i = 0; i < len; i++) { for(int j = 0; j < b.len; j++) { c.s[i+j] += s[i] * b.s[j]; } } for(int i = 0; i < c.len; i++) { c.s[i+1] += c.s[i]/10; c.s[i] %= 10; } c.clean(); return c; } bign operator *= (const bign &b) { *this = *this * b; return *this; } bign operator - (const bign &b) { bign c; c.len = 0; for(int i = 0, g = 0; i < len; i++) { int x = s[i] - g; if(i < b.len) x -= b.s[i]; if(x >= 0) g = 0; else { g = 1; x += 10; } c.s[c.len++] = x; } c.clean(); return c; } bign operator -= (const bign &b) { *this = *this - b; return *this; } bign operator / (const bign &b) { bign c, f = 0; for(int i = len-1; i >= 0; i--) { f = f*10; f.s[0] = s[i]; while(f >= b) { f -= b; c.s[i]++; } } c.len = len; c.clean(); return c; } bign operator /= (const bign &b) { *this = *this / b; return *this; } bign operator % (const bign &b) { bign r = *this / b; r = *this - r*b; return r; } bign operator %= (const bign &b) { *this = *this % b; return *this; } bool operator < (const bign &b) { if(len != b.len) return len < b.len; for(int i = len-1; i >= 0; i--) { if(s[i] != b.s[i]) return s[i] < b.s[i]; } return false; } bool operator > (const bign &b) { if(len != b.len) return len > b.len; for(int i = len-1; i >= 0; i--) { if(s[i] != b.s[i]) return s[i] > b.s[i]; } return false; } bool operator == (const bign &b) { return !(*this > b) && !(*this < b); } bool operator != (const bign &b) { return !(*this == b); } bool operator <= (const bign &b) { return *this < b || *this == b; } bool operator >= (const bign &b) { return *this > b || *this == b; } string str() const { string res = ""; for(int i = 0; i < len; i++) res = char(s[i]+'0') + res; return res; } }; 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, b, c, d, e, f, g; while(cin>>a>>b) { a.clean(), b.clean(); c = a+b; d = a-b; e = a*b; f = a/b; g = a%b; cout<<"a+b"<<"="<<c<<endl; // a += b cout<<"a-b"<<"="<<d<<endl; // a -= b; cout<<"a*b"<<"="<<e<<endl; // a *= b; cout<<"a/b"<<"="<<f<<endl; // a /= b; cout<<"a%b"<<"="<<g<<endl; // a %= b; if(a != b) printf("YES\n"); else printf("NO\n"); } return 0; }
3
#include<iostream> #include<string> #include<iomanip> #include<algorithm> using namespace std; #define MAXN 9999 #define MAXSIZE 10 #define DLEN 4 class BigNum { private: int a[500]; //可以控制大数的位数 int len; //大数长度 public: BigNum(){ len = 1;memset(a,0,sizeof(a)); } //构造函数 BigNum(const int); //将一个int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 int &) const; //重载除法运算符，大数对一个整数进行相除运算 BigNum operator^(const int &) const; //大数的n次方运算 int operator%(const int &) const; //大数对一个int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 int & t)const; //大数和一个int类型的变量的大小比较 void print(); //输出大数 }; BigNum::BigNum(const int b) //将一个int类型的变量转化为大数 { int 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 { int 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(int 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 { int 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 { int 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]; int i = -1; in>>ch; int l=strlen(ch); int count=0,sum=0; for(i=l-1;i>=0;) { sum = 0; int t=1; for(int 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 { int 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); int 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 { int 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; int i,j,up; int 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 int & b) const //大数对一个整数进行相除运算 { BigNum ret; int 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; } int BigNum::operator %(const int & b) const //大数对一个int类型的变量进行取模运算 { int i,d=0; for (i = len-1; i>=0; i--) { d = ((d * (MAXN+1))% b + a[i])% b; } return d; } BigNum BigNum::operator^(const int & n) const //大数的n次方运算 { BigNum t,ret(1); int i; if(n<0) exit(-1); if(n==0) return 1; if(n==1) return *this; int 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 { int 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 int & t) const //大数和一个int类型的变量的大小比较 { BigNum b(t); return *this>b; } void BigNum::print() //输出大数 { int i; cout << a[len - 1]; for(i = len - 2 ; i >= 0 ; i--) { cout.width(DLEN); cout.fill('0'); cout << a[i]; } cout << endl; } int main() { int i,n; BigNum x[101]; //定义大数的对象数组 x[0]=1; for(i=1;i<101;i++) x[i]=x[i-1]*(4*i-2)/(i+1); while(scanf("%d",&n)==1 && n!=-1) { x[n].print(); } }
4
#include<stdio.h> #include<iostream> #include<string> #include<string.h> #include<algorithm> const int ten[4]={1,10,100,1000}; const int maxl=1050; using namespace std; char tmp[maxl*4]; struct BigNumber{//只进行整数操作，如果有负数需要特判 ,除法太麻烦我暂时没看懂.. int d[maxl]; BigNumber(string s){//读入一个字符串(待求数) int len=s.size();//深度 d[0]=(len-1)/4+1; int i,j,k; for(i=1;i<maxl;++i)d[i]=0;//初始化 for(i=len-1;i>=0;--i){ j=(len-i-1)/4+1; k=(len-i-1)%4;//每四位算一个数 d[j]+=ten[k]*(s[i]-'0'); } while(d[0]>1 && d[d[0]]==0)--d[0];//得到缩位后的位数 } BigNumber(){//不知道用来初始化this指针干什么.. *this=BigNumber(string("0")); } string toString(){//转换为string类型(可以直接输出) string s(""); int i,j,temp; for(i=3;i>=1;--i)if(d[d[0]]>=ten[i])break;//找到最后位置如果大于10就对最后位置进行操作 temp=d[d[0]]; for(j=i;j>=0;--j){ s=s+char(temp/ten[j]+'0');//其实是相当于把最高位放在第一位 temp%=ten[j]; } for(i=d[0]-1;i>0;--i){ temp=d[i]; for(j=3;j>=0;--j){ s=s+(char)(temp/ten[j]+'0'); temp%=ten[j]; } } return s; } bool operator <(const BigNumber &b){//比较 if(d[0]!=b.d[0])return d[0]<b.d[0];//如果返回位数小的为小的 int i; for(i=d[0];i>0;--i){ if(d[i]!=b.d[i]){//逐四位比较 return d[i]<b.d[i]; } } return false;//如果等于，那就是不小于 } BigNumber operator +(const BigNumber &b){//加法 BigNumber c; c.d[0]=max(d[0],b.d[0]); int i,x=0; for(i=1;i<=c.d[0];++i){//取最大位数 x=d[i]+b.d[i]+x;//四位相加 c.d[i]=x%10000;//相加超过10000了就减去10000，为当前四位的值 x/=10000;//多出来的x继承到下一次循环 } while(x!=0){//如果超过两个数的位数 c.d[++c.d[0]]=x%10000; x/=10000; } return c; } BigNumber operator -(const BigNumber &b){//减法 BigNumber c; c.d[0]=d[0]; int i,x=0; for(i=1;i<=c.d[0];++i){ x=10000+d[i]-b.d[i]+x;//先加上10000防止出现负数，相当于借位 c.d[i]=x%10000;//得到当前位实际值 x=x/10000-1;//因为一开始加了10000所以要减去1，即下一部减去借的那一位 } while((c.d[0]>1)&&(c.d[c.d[0]]==0))--c.d[0];//减去后可能有多位为空，最大位-- return c; } BigNumber operator *(const BigNumber &b){//乘法 BigNumber c; c.d[0]=d[0]+b.d[0]; int i,j,x; for(i=1;i<=d[0];++i){//逐四位相乘 x=0; for(j=1;j<=b.d[0];++j){//这两个的乘积假定全部在i+j-1的位置 x=d[i]*b.d[j]+x+c.d[i+j-1];//对这一位操作时除了本身之外还要加上原来这一位的值 c.d[i+j-1]=x%10000;//对位进行再运算 x/=10000; //计算乘法超额部分 } c.d[i+b.d[0]]=x;//超额部分实际上为b的最大位的下一位(x每次除以10000之后得到的) } while((c.d[0]>1)&&(c.d[c.d[0]]==0))--c.d[0];//前面的c.d[0]开的可能会比较大 return c; } BigNumber operator *(const int &k){//乘常数 BigNumber c; c.d[0]=d[0]; int i,x=0; for(i=1;i<=d[0];++i){ x=d[i]*k+x; c.d[i]=x%10000; x/=10000; } while(x>0){ c.d[++c.d[0]]=x%10000; x/=10000; } while((c.d[0]>1)&&(c.d[c.d[0]]==0))--c.d[0]; return c; } bool operator ==(const BigNumber &b){//判断是否相等 if(d[0]!=b.d[0])return false; for(int i=1;i<=d[0];++i)if(d[i]!=b.d[i])return false; return true; } }; string getnum(){ string s; char tmp=getchar(); while(tmp>='0'&&tmp<='9'){ s=s+tmp; tmp=getchar(); } return s; } int main(){ BigNumber a,b,c; a=BigNumber(getnum()); b=BigNumber(getnum()); c=a*b; cout<<(string)c.toString(); return 0; }
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