高精度重载运算符模板

2017年3月21日 | ljfcnyali

高精度重载运算符一直是很多人入门的基础，我也不例外。

这么久的努力奋斗以来，突然发现自己连一个高精度重载运算符的模板都没有打，立刻开始敲键盘。

这个高精度重载运算符其实很简单，会使用”operator”的就会，不会也没有任何办法。

所以，我决定不写方法，也就仅仅贴一个代码，给理解了”operator”的各位一个可以仿照的模板。

保证正确（但是除法最后不想谢了，250行的代码那么好些么，所以没有写除法，敬请谅解），经过实验（代码中就有）。

代码如下：

/*************************************************************************
> File Name: 模板\高精度模板.cpp
> Author: ljf-cnyali
> Mail: ljfcnyali@gmail.com
> Created Time: 2017/3/21 19:28:59
************************************************************************/

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>

using namespace std;

#define REP(i, a, b) for(int i = (a), _end_ = (b);i <= _end_; ++ i)
#define mem(a) memset((a), 0, sizeof(a))
#define str(a) strlen(a)

const int maxn = 9999;
const int Dlen = 5;
const int MaxSize = 10;

typedef long long Huge;

struct BigNum {
private :
int len, a[500];
public :
BigNum() {
len = 1;
mem(a);
}
BigNum(const int);
BigNum(const char*);
BigNum(const BigNum &);
BigNum &operator = (const BigNum &);

friend istream& operator >> (istream &, BigNum &);
friend ostream& operator << (ostream &, BigNum &);

BigNum operator + (const BigNum &) const;
BigNum operator * (const BigNum &) const;
BigNum operator - (const BigNum &) const;

bool operator > (const BigNum & T) const;
bool operator > (const int & t) const;

void print();
};

BigNum::BigNum(const int x) {
int u, v = x;
len = 0;
mem(a);
while(v > maxn) {
u = v - (v / (maxn + 1) * (maxn + 1));
v = v / (maxn + 1);
a[len ++] = u;
}
a[len ++] = v;
}

BigNum::BigNum(const char*s) {
int t, k, index, l;
mem(a);
l = str(s);
len = l / Dlen;
if(l % Dlen)
++ len;
index = 0;
for(int i = l - 1; i >= 0; i -= Dlen) {
t = 0;
k = i - Dlen + 1;
if(k < 0)
k = 0;
REP(j, k, i)
t = t * 10 + s[j] - '0';
a[index ++] = t;
}
}

BigNum::BigNum(const BigNum &T) : len(T.len) {
memcpy(a, T.a, len * sizeof * a);
}

BigNum & BigNum:: operator = (const BigNum & n) {
len = n.len;
memcpy(a, n.a, len * sizeof * a);
return *this;
}

istream& operator >> (istream & in, BigNum & b) {
char c[MaxSize * 4];
int i = -1;
in >> c;
int l = str(c);
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 += (c[i] - '0') * t;
b.a[count ++] = sum;
}
b.len = count ++;
return in;
}

ostream& operator << (ostream& out, BigNum& b) {
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 {
BigNum t(*this);
REP(i, 0, T.len > len ? T.len : len) {
t.a[i] += T.a[i];
if(t.a[i] > maxn) {
t.a[i + 1] ++;
t.a[i] -= maxn + 1;
}
}
int num = T.len > len ? T.len : len;
if(t.a[num] != 0)
t.len = num + 1;
else
t.len = num;
return t;
}

BigNum BigNum::operator - (const BigNum  & T) const {
int num;
bool flag;
BigNum t1, t2;
if(*this > T) {
t1 = *this;
t2 = T;
flag = 0;
}
else {
t1 = T;
t2 = *this;
flag = 1;
}
num = t1.len;
int j;
REP(i, 0, num - 1)
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 = num;
while(t1.a[len - 1] == 0 && t1.len > 1) {
t1.len --;
num --;
}
if(flag)
t1.a[num - 1] = 0 - t1.a[num - 1];
return t1;
}

BigNum BigNum::operator * (const BigNum & T) const {
BigNum ret;
int up, temp, temp1, j, i;
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;
}

bool BigNum::operator > (const BigNum & T) const {
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 {
BigNum b(t);
return *this > b;
}

void BigNum::print() {
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() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
BigNum x[101];
x[0] = 1;
int n;
REP(i, 1, 100)
x[i] = x[i - 1] * (i + 2 - 2);
REP(i, 1, 100)
x[i].print();
return 0;
}

本文转自：http://ljf-cnyali.cn/index.php/archives/96