小乐乐在课上学习了二进制八进制与十六进制后,对进制转换产生了浓厚的兴趣。因为他的幸运数字是6,所以他想知道一个数表示为六进制后的结果。请你帮助他解决这个问题。
import java.util.Scanner ;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
StringBuilder r = new StringBuilder();
while(n>0){
r.append( n%6 );
n /= 6 ;
}
System.out.println( r.reverse());
}
} void modSix(int n)
{
if(n){
modSix(n / 6);
printf("%d", n % 6);
}
}
int main()
{
int n = 0;
scanf("%d",&n);
modSix(n);
return 0;
} #include<stdio.h>
int main()
{
int n,i=0,j;
scanf("%d",&n);
int arr[100];
while(n/6!=0)
{
arr[i]=n%6;
n=n/6;
i++;
}
arr[i]=n;
for(j=i;j>=0;j--)
printf("%d",arr[j]);
return 0;
} int main() {
int a = 0;
int i = 0;
int arr[10] = {0};
scanf("%d", &a);
while (a != 0) //,循环的条件是这个数被除尽为止,即本数为零
{
arr[i] = a % 6; //短除法将一个数除以六取余,将余数放在数组中
a = (a / 6);
i++;
}
for (int j = i-1; j >= 0; j--) //因为短除法倒着取余,所以循环倒着打印每一位
{
printf("%d", arr[j]);
}
return 0;
} #include <stdio.h>
#include<math.h>
#include<time.h>
#include<stdlib.h>
#define JZ 6
int o(int x) //算最大次方位
{
while (1)
{
int n = rand() % 20;
if (pow(JZ, n + 1) > x && pow(JZ, n) <= x)
{
return n;
}
}
}
int p(int o, int x)//算最大次方位,填几
{
while (1)
{
int n = rand() % 10;
if (pow(JZ, o) * (n + 1) > x && pow(JZ, o) * (n) <= x && n!=0)
{
return n;
}
}
}
int main()
{
srand((unsigned int)time(NULL));
int n = 0;
int a, b;
int i = 0;
scanf("%d", &n);
int c = o(n);
int arr[] = { 0 };
for (i = 0; i <= c; i++)
{
arr[i] = 0;
}
while (n)
{
a = o(n);
//printf("%d ", a);
b = p(a, n);
// printf("%d\n", b);
arr[a] = b; //数组 存放a这个次方位的值
n = n - b * pow(JZ, a);
}
for (i = c; i >= 0; i--)
{
printf("%d", arr[i]);
}
return 0;
} #include <stdio.h>
int main() {
char a[15];
int i = 0;
int n = 0;
scanf("%d", &n);
int Hexadecimal = 6;
while (n) {
//余数存放到字符串中,在从后向前打印字符串
*(a + i) = n % Hexadecimal;
n /= Hexadecimal;
i++;
}
//打印余数组成的字符串,但字符串是反过来的,因为要打印为0的位置i可以为0
while (i-- && i >= 0) {
printf("%d", *(a+i));
}
return 0;
} #include <stdio.h>
int main() {
long n, arr[10], i = 0, sum = 0;
while (scanf("%ld", &n) != EOF) {
while (n) {
arr[i] = n % 6;
n /= 6;
i++;
}
for (int j = i - 1; j >= 0; j--) {
printf("%ld", arr[j]);
}
}
return 0;
}