题目
{% fold 点击显/隐题目 %}
题解
高精度下的进制转换
代码
{% fold 点击显/隐代码 %}```cpp 进制转换 https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份
#include
#include
#include
using namespace std;
/*
-
完全大数模板 修改版
-
输出cin>>a
-
输出a.print();
-
注意这个输入不能自动去掉前导0的,可以先读入到char数组,去掉前导0,再用构造函数。
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by kuangbin GG.
*/
#define MAXN 9999
#define MAXSIZE 1010
#define DLEN 4
class BigNum {
private:
int a[5000]; //可以控制大数的位数
int len;public:
BigNum() {
len = 1;
memset(a, 0, sizeof(a));
} //构造函数
BigNum(const long long); //将一个int类型的变量转化成大数
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 read(const char *s); //从字符串读入
void pre0(char *s); //取出字符串的前导0;
};
BigNum::BigNum(const long long b) //将一个int类型的变量转化为大数
{
long long 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) {
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--) {
printf("%04d", 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;
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
{
BigNum ret;
int i, j=0, 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
{
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)
return -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;
printf("%d", a[len - 1]);
for (i = len - 2; i >= 0; i--)
printf("%04d", a[i]);
printf("\n");
}
void BigNum::read(const char *s) {
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;
}
}
void pre0(char *s) {
int pos = 0;
int len = strlen(s);
for (int i = 0; i < len; i++) {
if (s[i] != '0') {
pos = i;
break;
}
}
for (int i = pos; i < len; i++)
s[i - pos] = s[i];
s[len - pos] = '\0';
}
char s[3005];
BigNum n, zero;
int ans[150000];
int main() {
int d;
while (~scanf("%s%d", s, &d)) {
pre0(s);
n = BigNum(s);
zero = BigNum(0LL);
int pos = 0;
while (n > zero) {
ans[pos++] = n % d;
n = n / d;
}
for (int i = pos - 1; i >= 0; --i) {
printf("%d\n", ans[i]);
}
printf("\n");
}
return 0;
}
{% endfold %}