AOJ 907.进制转换

126


题目



点击显/隐题目

给定一个十进制正整数a和一个数字d,你需要将a转换成d进制并输出。
数字a转换成d进制后,每一个数位都是0~d-1之间的十进制整数。
对于样例2:
196=350^1 + 4650^0


第一行一个十进制正整数a(a<=10^3000)。
第二行一个正整数d(2<=d<=100)


将数字a转换成d进制从高位到低位依次输出(不要前导0),中间用空格隔开。


样例一:
53
2
样例二:
196
50


样例一:
1
1
0
1
0
1
样例二:
3
46




题解


高精度下的进制转换

代码


点击显/隐代码
#include <cstdio>
#include <cstring>
#include <iostream>
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。
* 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;
}
发布评论
  • 点击查看/关闭被识别为广告的评论