[编程题]进制转换
//M%N就是N进制的第一位，然后M/N后再取余数就是N进制第二位，以此类推，用栈可以解决。注意负数
#include <iostream>
#include <stack>
using namespace std;
int main(int argc, const char * argv[]) {
int M,N;
cin>>M>>N;
stack<char> answer;
if(M<0)
{
cout<<"-";
M=-M;
}
while(M!=0)
{
if(M%N<10)
{
answer.push(char(M%N+48));
M/=N;
}
else
{
answer.push(char(M%N+55));
M/=N;
}
}
while(answer.empty()!=1)
{
cout<<answer.top();
answer.pop();
}
return 0;
}
发表于 2018-01-30 22:14:00
回复(1)
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
String m=in.next();
int n=in.nextInt();
BigInteger bi=new BigInteger(m,10);
System.out.println(bi.toString(n).toUpperCase());
}
}
发表于 2017-09-02 22:30:31
回复(2)
可以用StringBuilder的insert不断在前面插入新的数字，也可以用append最后再reverse。
记得对负数先求绝对值再处理，最后输出对应符号即可。
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int num = in.nextInt();
int radix = in.nextInt();
char[] chars = new char[]{'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};
boolean negative = false;
if (num < 0) negative = true;
StringBuilder sb = new StringBuilder();
num = Math.abs(num);
while (num >= radix) {
sb.insert(0, chars[num % radix]);
num /= radix;
}
sb.insert(0, chars[num]);
System.out.println((negative ? "-" : "") + sb.toString());
}
}
发表于 2017-04-04 09:42:06
回复(1)
python解法
本來想用递归解的，结果超出最大深度，只能用循环做了。
def baseN(num, b):
res = ""
if num > 0:
while num:
res = "0123456789ABCDEFGHIGKLMNOPQRSTUVWXYZ"[num % b] + res
num = num // b
return res
else:
num = -num
while num:
res = "0123456789ABCDEFGHIGKLMNOPQRSTUVWXYZ"[num % b] + res
num = num // b
return "-" + res
a, b = map(int, input().split())
print(baseN(a, b))
发表于 2018-04-13 10:21:05
回复(2)
// 这个只需要按照进制转换直接计算就行了，要注意负数的问题
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
HashMap<Integer,Character> hm = new HashMap<Integer,Character>();
hm.put(10,'A');hm.put(11,'B');
hm.put(12,'C');hm.put(13,'D');
hm.put(14,'E');hm.put(15,'F');
while(sc.hasNext()){
int m = sc.nextInt();
int n = sc.nextInt();
StringBuilder sb = new StringBuilder();
boolean flag = false;
if(m<0){
flag = true;
m = 0-m;
}
while(m>0){
int t = m%n;
sb.append(hm.get(t)==null?t:hm.get(t)+"");
m = m/n;
}
if(flag)
System.out.println("-" + sb.reverse().toString());
else
System.out.println(sb.reverse().toString());
}
}
}
发表于 2017-09-05 10:09:34
回复(0)
import java.util.*;
import java.util.ArrayList;
public class Main {
public static void main(String[] args) {
// //N进制转换器
Scanner scanner = new Scanner(System.in);
int M = scanner.nextInt();
int N = scanner.nextInt();
//查询表,
String table = "0123456789ABCDEF";
int flag = 1;
if (M < 0) {
M = -M;
flag = 0;
}
Stack S = new Stack();
decimalBaseToNBaseConvertor(M, N, S, table);
//output
if (flag == 0) {
S.push('-');
}
while (!S.empty()) {
System.out.print(S.pop());
}
}
//十进制转二进制,占用空间较多;递归算法;
/*改进:
1. 使用堆栈来存储余数,并且利用堆栈的特点,倒序输出余数的值,N进制.堆栈是wrapper Character Object
2. 输入用Scanner类型到存储System.in的值.（Java语言不熟悉）
3. Object 有String 和Character.A Java String is an object of the class java.lang.
对象有方法.比C++的String更加方面(单纯char的数组）.
This Character class also offers a number of useful class (i.e., static) methods
for manipulating characters.
* 时间复杂度:
* 空间复杂度:*/
public static void decimalBaseToNBaseConvertor(int M, int N, Stack<Character> S, String table) {
if (M == 0) {
return;
}
//入栈
S.push(table.charAt(M % N));
//下一次
decimalBaseToNBaseConvertor(M / N, N, S, table);
}
}
编辑于 2018-04-03 13:27:38
回复(0)
被测试用例带着做的节奏
import java.util.Scanner;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int M = sc.nextInt();
int N = sc.nextInt();
boolean isMinByZero = (M < 0);
M = Math.abs(M);
String str = "";
str = M%N >= 10 ? (char)('A'+(M%N-10)) + str : (M % N) + str;
while (M / N > 0){
M /= N;
str = M%N >= 10 ? (char)('A'+(M%N-10)) + str : (M % N) + str;
}
if (isMinByZero) {
System.out.println("-" + str);
}else {
System.out.println(str);
}
}
}
发表于 2017-09-25 19:21:29
回复(0)
importjava.util.Stack;importjava.util.Scanner;publicclassMain{publicstaticvoidmain(String[] args){intn,base;Scanner scanner = newScanner(System.in);Stack S=newStack();while(scanner.hasNextInt()){n=scanner.nextInt();base=scanner.nextInt();chardigit[]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};while(n>0){S.push(digit[n%base]);n/=base;}while(!S.empty()){System.out.print(S.pop());}System.out.println();}}}
发表于 2017-01-06 20:21:20
回复(3)
#include <iostream>
using namespace std;
char buf[50];
int main()
{
int m, n;
while (scanf("%d%d", &m, &n) != EOF)
{
int index = 0;
int mm = m;
if(m<0) m = -m;//注意m为负的情况，直接转换为正就行，因为负的结果实际上就是正的结果的负
do{
buf[index++] = (m%n < 10)?m%n+'0':m%n-10+'A';
m/=n;
}while(m);
if(mm<0)//在数组的最后一位加上一个负号，一会儿直接打印出来
buf[index++] = '-';
for(int i = index-1; i >= 0; i--)
printf("%c", buf[i]);
printf("\n");
}
return 0;
}
发表于 2020-07-06 14:15:48
回复(0)
#include<iostream>
#include<string>
using namespace std;
int main()
{
int m, n;
string s, table = "0123456789ABCDEF";
cin >> m >> n;
if (m == 0) cout << "0" << endl;
while (m)
{
if (m < 0)
{
m = -m;
cout << "-";
}
s = table[m%n] + s;
m /= n;
}
cout << s << endl;
return 0;
}
发表于 2019-10-11 14:27:10
回复(0)
import java.util.Scanner;
public class Main{
public static String convert(int num,int radix){
if (num == 0){
return "0";
}
char[] table = new char[]{'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'};
StringBuilder builder = new StringBuilder();
//默认是非负数
boolean flag = false;
if(num < 0){
flag = true;
num = - num;
}
while(num > 0){
builder.append(table[num % radix]);
num /= radix;
}
if(flag){
builder.append('-');
}
return builder.reverse().toString();
}
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
int num = scanner.nextInt();
int radix = scanner.nextInt();
System.out.println(convert(num,radix));
}
}
发表于 2019-07-20 13:34:44
回复(0)
#include<iostream>
using namespace std;
char hashT[16]={//查表省去判断条件
'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'
};
void change(int m,int n){//递归处理
if(m==0) return;
change(m/n,n);
cout<<hashT[m%n];
}
int main(){
int m,n;
while(cin>>m>>n){
if(m<0||m>0){
if(m<0){//考虑负数的情况
cout<<"-";
m=-m;
}
change(m,n);
cout<<endl;
}else{
cout<<'0'<<endl;//考虑为0时的情况
}
}
return 0;
}
发表于 2019-01-01 13:42:08
回复(0)
package train;
import java.util.Scanner;
/**
* 将十进制数M转化为N进制数 输入描述: 输入为一行，M(32位整数)、N(2 ≤ N ≤ 16)，以空格隔开。
* 输出描述: 为每个测试实例输出转换后的数，每个输出占一行。如果N大于9，则对应的数字规则参考16进制（比如，10用A表示，等等）
*
* @author jun
*
*/
public class JavaTest_87 {
public static void getValue(int m, int n) {
int a[] = new int[100];
int i = 0;
while (m != 0) {
a[i] = m % n;
m = m / n;
i++;
}
for (int j = i; j > 0; j--) {
if (a[j - 1] < 0) {// j-1是因为m=0时，i++算了一次。所以j=i,j需要减去多的1。
for (int k = 0; k < i - 1; k++) {
a[k] = Math.abs(a[k]);
}
}
if (a[j - 1] > 9 || a[j - 1] < -9) {
System.out.print((char) (a[j - 1] + 55));//ABCEF字符转换
} else
System.out.print(a[j - 1] + "");
}
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int m = in.nextInt();
int n = in.nextInt();
getValue(m, n);
}
}
import java.util.Scanner;
/**
* 将十进制数M转化为N进制数 输入描述: 输入为一行，M(32位整数)、N(2 ≤ N ≤ 16)，以空格隔开。
* 输出描述: 为每个测试实例输出转换后的数，每个输出占一行。如果N大于9，则对应的数字规则参考16进制（比如，10用A表示，等等）
*
* @author jun
*
*/
public class JavaTest_87 {
public static void getValue(int m, int n) {
int a[] = new int[100];
int i = 0;
while (m != 0) {
a[i] = m % n;
m = m / n;
i++;
}
for (int j = i; j > 0; j--) {
if (a[j - 1] < 0) {// j-1是因为m=0时，i++算了一次。所以j=i,j需要减去多的1。
for (int k = 0; k < i - 1; k++) {
a[k] = Math.abs(a[k]);
}
}
if (a[j - 1] > 9 || a[j - 1] < -9) {
System.out.print((char) (a[j - 1] + 55));//ABCEF字符转换
} else
System.out.print(a[j - 1] + "");
}
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int m = in.nextInt();
int n = in.nextInt();
getValue(m, n);
}
}
发表于 2018-08-14 17:03:24
回复(0)
#include <stdio.h>intmain(){intnumber, nBase, rOfNumber, iIndex=0, nFlag=0;charresult[100];scanf("%d %d", &number, &nBase);if(number==0){printf("0");return 0;}else if(number<0){printf("-");number = -number;}while(number!=0){rOfNumber = number%nBase;if(rOfNumber<10)result[iIndex++] = '0'+rOfNumber;elseresult[iIndex++] = 'A'+rOfNumber-10;number /= nBase;}while(iIndex){printf("%c", result[--iIndex]);}return 0;}
发表于 2018-06-10 22:29:58
回复(0)
#include<iostream>
#include<vector>
using namespace std;
//进制转换,要考虑mode>9的情况，还要考虑负数
int main()
{
int num,mode;
while(cin>>num>>mode)
{
vector<int> ret;
int flag = 0;
if(num <0)
{
num = -num;
flag =1;
}
while(num>0)
{
ret.push_back(num % mode);
num /=mode;
}
vector<int>::reverse_iterator it;
for(it = ret.rbegin();it!=ret.rend();++it)
{
if(flag == 1)
{
cout<<"-";
flag = 0;
}
if(*it<9)
cout<<*it;
else
cout<<(char)((*it-10)+'A');
}
cout<<endl;
}
return 0;
}
发表于 2018-06-08 19:09:15
回复(0)
def solution(N,num):
ref = "0123456789ABCDEF"
result = [ref[num%N]]
while num//N !=0:
num = num//N
result.append(ref[num%N])
result.reverse()
return "".join(result)
num, N = map(int, input().split())
if num < 0:
print("-"+solution(N,-num))
else:
print(solution(N,num))
发表于 2018-04-10 00:16:41
回复(0)
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
String rul = "0123456789ABCDEF";
int val = sc.nextInt();
int raw = val;
int conv = sc.nextInt();
boolean neg = false;
StringBuilder sb = new StringBuilder();
while(val!=0){
sb.append(rul.charAt(Math.abs(val%conv)));
val/=conv;
}
if(raw<0)
sb.append("-");
System.out.println(sb.reverse().toString());
}
}
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
String rul = "0123456789ABCDEF";
int val = sc.nextInt();
int raw = val;
int conv = sc.nextInt();
boolean neg = false;
StringBuilder sb = new StringBuilder();
while(val!=0){
sb.append(rul.charAt(Math.abs(val%conv)));
val/=conv;
}
if(raw<0)
sb.append("-");
System.out.println(sb.reverse().toString());
}
}
发表于 2018-03-30 14:46:55
回复(0)
我是真的好菜啊，大佬的脑子里都是装的啥啊！
import java.util.Scanner;
import java.util.Stack;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int num = sc.nextInt();
int tar = sc.nextInt();
convert(num,tar);
}
public static void convert(int num, int tar){
boolean simple = true;
if (num < 0) {
num = -num;
simple = false;
}
String[] strs = {"0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"};
Stack stack = new Stack();
while((num)!=0){
stack.push(strs[(num%tar)]);
num = num/tar;
}
if (!simple){
System.out.print("-");
}
while (stack.size() != 0){
System.out.print(stack.pop());
}
}
}
发表于 2018-03-21 16:53:13
回复(0)
//记得处理符号
#include<stdio.h>
int main() {
int m,n;
while (scanf("%d%d", &n, &m) != EOF) {
int tag = 1;
if (n < 0) { n = -n; tag = 0; }
char A[40];
int p = 0;
int x;
do {
x= n % m;
A[p++] = x < 10 ? x + '0' : x + 'A' - 10;
n /= m;
} while (n);
if (tag == 0) printf("-");
for (p--; p >= 0; p--) {
printf("%c", A[p]);
}
printf("\n");
}
return 0;
}
发表于 2018-03-04 22:55:38
回复(0)
问题信息
通过挑战的用户
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2023-02-06 18:50:33
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