poj1220

maksyuki 发表于 oj 分类，标签: 基础算法-高精度

15 9月 2015

NUMBER BASE CONVERSION

Write a program to convert numbers in one base to numbers in a second base. There are 62 different digits:{ 0-9,A-Z,a-z }HINT: If you make a sequence of base conversions using the output of one conversion as the input to the next, when you get back to the original base, you should get the original number.

Input

The first line of input contains a single positive integer. This is the number of lines that follow. Each of the following lines will have a (decimal) input base followed by a (decimal) output base followed by a number expressed in the input base. Both the input base and the output base will be in the range from 2 to 62. That is (in decimal) A = 10, B = 11, ..., Z = 35, a = 36, b = 37, ..., z = 61 (0-9 have their usual meanings).

Output

The output of the program should consist of three lines of output for each base conversion performed. The first line should be the input base in decimal followed by a space then the input number (as given expressed in the input base). The second output line should be the output base followed by a space then the input number (as expressed in the output base). The third output line is blank.

Sample Input

62 2 abcdefghiz

10 16 1234567890123456789012345678901234567890

16 35 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2

35 23 333YMHOUE8JPLT7OX6K9FYCQ8A

23 49 946B9AA02MI37E3D3MMJ4G7BL2F05

49 61 1VbDkSIMJL3JjRgAdlUfcaWj

61 5 dl9MDSWqwHjDnToKcsWE1S

5 10

42104444441001414401221302402201233340311104212022133030

Sample Output

62 abcdefghiz

11011100000100010111110010010110011111001001100011010010001

1234567890123456789012345678901234567890

16 3A0C92075C0DBF3B8ACBC5F96CE3F0AD2

35 333YMHOUE8JPLT7OX6K9FYCQ8A

23 946B9AA02MI37E3D3MMJ4G7BL2F05

49 1VbDkSIMJL3JjRgAdlUfcaWj

61 dl9MDSWqwHjDnToKcsWE1S

42104444441001414401221302402201233340311104212022133030

10 1234567890123456789012345678901234567890

Source

Greater New York 2002

题目类型：高精度进制转换

算法分析：直接使用大整数除法模拟实现即可，详细可以参考高精度进制转换用法

#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

const int INF = 0x7FFFFFFF;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int MOD = 10000007;

const int maxn = 100000 + 66;

char str[maxn];//输入字符串
int start[maxn], ans[maxn], res[maxn];//被除数，商，余数
int oldBase, newBase;//转换前后的进制


int GetNum (char c)//这里进制字符是先数字，后大写字母，后小写字母的
{
    if(c >= '0' && c <= '9') return c - '0';//数字
    if(c >= 'A' && c <= 'Z') return c - 'A' + 10;//大写字母
    return c - 'a' + 36;//小写字母
}

//数字得到字符
char GetChar (int i)
{
    if (i >= 0 && i <= 9) return i + '0';
    if (i >= 10 && i <= 35) return i - 10 + 'A';
    return i - 36 + 'a';
}

void Change ()//把输入的字符串的各个数位还原为数字形式
{
    int i;
    start[0] = strlen (str);//数组的0位存的是数组长度
    for(i = 1; i <= start[0]; i++)
        start[i] = GetNum (str[i-1]);
}

void Solve ()
{
    memset (res, 0, sizeof (res));//余数初始化为空
    int y, i, j;
    //模n取余法，(总体规律是先余为低位，后余为高位)
    while (start[0] >= 1)
    {//只要被除数仍然大于等于1，那就继续“模2取余”
        y = 0;
        i = 1;
        ans[0] = start[0];
        //
        while (i <= start[0])
        {
            y = y * oldBase + start[i];
            ans[i++] = y / newBase;
            y %= newBase;
        }
        res[++res[0]] = y;//这一轮运算得到的余数
        i = 1;
        //找到下一轮商的起始处
        while ((i <= ans[0]) && (ans[i] == 0)) i++;
        //清除这一轮使用的被除数
        memset (start, 0, sizeof(start));
        //本轮得到的商变为下一轮的被除数
        for(j = i;j <= ans[0]; j++)
            start[++start[0]] = ans[j];
        memset (ans, 0, sizeof(ans)); //清除这一轮的商，为下一轮运算做准备
    }
}

void Output ()//从高位到低位逆序输出
{
    int i;
    for (i = res[0]; i >= 1; i--)
        printf ("%c", GetChar (res[i]));
    printf ("\n");
}

int main()
{
//    ifstream cin ("aaa.txt");

    int t;
    cin >> t;

    while (t--)
    {
        cin >> oldBase >> newBase >> str;
        cout << oldBase << " " << str << endl;

        Change ();
        Solve ();
        cout << newBase << " ";
        Output ();
        cout << endl;
    }

return 0;
}

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

121

122

123

124

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

const int INF = 0x7FFFFFFF;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int MOD = 10000007;

const int maxn = 100000 + 66;

char str[maxn];//输入字符串

int start[maxn], ans[maxn], res[maxn];//被除数，商，余数

int oldBase, newBase;//转换前后的进制

int GetNum (char c)//这里进制字符是先数字，后大写字母，后小写字母的

{

if(c >= '0' && c <= '9') return c - '0';//数字

if(c >= 'A' && c <= 'Z') return c - 'A' + 10;//大写字母

return c - 'a' + 36;//小写字母

}

//数字得到字符

char GetChar (int i)

{

if (i >= 0 && i <= 9) return i + '0';

if (i >= 10 && i <= 35) return i - 10 + 'A';

return i - 36 + 'a';

}

void Change ()//把输入的字符串的各个数位还原为数字形式

{

int i;

start[0] = strlen (str);//数组的0位存的是数组长度

for(i = 1; i <= start[0]; i++)

start[i] = GetNum (str[i-1]);

}

void Solve ()

{

memset (res, 0, sizeof (res));//余数初始化为空

int y, i, j;

//模n取余法，(总体规律是先余为低位，后余为高位)

while (start[0] >= 1)

{//只要被除数仍然大于等于1，那就继续“模2取余”

y = 0;

i = 1;

ans[0] = start[0];

while (i <= start[0])

{

y = y * oldBase + start[i];

ans[i++] = y / newBase;

y %= newBase;

}

res[++res[0]] = y;//这一轮运算得到的余数

i = 1;

//找到下一轮商的起始处

while ((i <= ans[0]) && (ans[i] == 0)) i++;

//清除这一轮使用的被除数

memset (start, 0, sizeof(start));

//本轮得到的商变为下一轮的被除数

for(j = i;j <= ans[0]; j++)

start[++start[0]] = ans[j];

memset (ans, 0, sizeof(ans)); //清除这一轮的商，为下一轮运算做准备

}

void Output ()//从高位到低位逆序输出

{

int i;

for (i = res[0]; i >= 1; i--)

printf ("%c", GetChar (res[i]));

printf ("\n");

}

int main()

{

// ifstream cin ("aaa.txt");

int t;

cin >> t;

while (t--)

{

cin >> oldBase >> newBase >> str;

cout << oldBase << " " << str << endl;

Change ();

Solve ();

cout << newBase << " ";

Output ();

cout << endl;

}

return 0;

}

poj1218

poj1251

超客的水墨烟雨

poj1220

分类目录

标签云

书签

近期评论

2024年4月
一	二	三	四	五	六	日
« 6月
1	2	3	4	5	6	7
8	9	10	11	12	13	14
15	16	17	18	19	20	21
22	23	24	25	26	27	28
29	30