#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4
class BigNum
{
private:
int a[1500]; //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
void printline(); //输出大数并换行
};
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];
}
}
void BigNum::printline() {
print();
puts(“”);
}
