C 解法, 执行用时: 1ms, 内存消耗: 380KB, 提交时间: 2020-08-08

/*
 将十进制数转换为2~9进制，十进制范围（0~1e18）
 输入： 33 2
 输出：100001
*/
#include <stdio.h>

/* 编译时链接 -lm */
#include <math.h>

void DecimalTurn (unsigned long long DecimalNum, unsigned int n)
{
    int i = 0;
    unsigned long long k = 0;
    unsigned int TurnResult[64] = {};

    do
    {
        TurnResult[i] = DecimalNum % n;
        DecimalNum = DecimalNum / n;
        i++;
    }while (0 != DecimalNum);

 for(i--; i >= 0; i--)
 {
    printf("%d", TurnResult[i]);
 }
 
    printf("\n");
    return;
}
  
 int  main ()
 {
    unsigned long long DecimalNum = 0;
    unsigned int n = 0;
    
    scanf("%lld %d", &DecimalNum, &n);
    DecimalTurn(DecimalNum, n);
     
    return 0;
 }

C 解法, 执行用时: 1ms, 内存消耗: 432KB, 提交时间: 2020-08-21

#include<stdio.h>
int main()
{
    long long int data;
    int format;
    scanf("%lld %d",&data,&format);
    int temp;
    char output[20000]={0};
	int count = 0;
    do
    {
        temp = data%format;
        data = data/format;
		output[count++] = temp;
    }while(data != 0);
	count--;
	for(;count>=0;count--)
	{
		printf("%d",output[count]);
	}
	return 0;
}

C++14 解法, 执行用时: 1ms, 内存消耗: 492KB, 提交时间: 2020-05-20

#include <stdio.h>
#include <stdlib.h>
using namespace std;
//思路和步骤是，找到运算核心和合适的数据结构。运算核心：相除取余。数据结构：双向链表。要注意不要让技术问题成为思路实现的障碍
struct yu_store
{
    yu_store* front;
    int a;
    yu_store* next;
};
  
int main()
{
    long int num;
    long int yu = 1;
    int n;
    int count;
    yu_store* f;
    yu_store* p;
    yu_store* s;
      
    scanf("%ld",&num);
    getchar();
    scanf("%d",&n);
      
    p = (yu_store *)malloc(sizeof(yu_store));
    f = p;
    f->front = NULL;
    f->next = NULL;
      
    if(num == 0)
    {
        printf("%ld",num);
        return 0;
    }
      
    while(num != 0)
    {
        yu = num%n;
        num = num/n;
        s = (yu_store *)malloc(sizeof(yu_store));
        p->a = yu;
        p->next = s;
        s->front = p;
        s->next = NULL;
        p = s;
    }
      
    while(f != p)
    {
        p = p->front;
        printf("%d",p->a);
        free(s);
        s = p;
    }
    free(f);
    return 0;
}

C++14 解法, 执行用时: 2ms, 内存消耗: 368KB, 提交时间: 2020-05-12

#include <stdio.h>
#include <stdlib.h>
using namespace std;
//思路和步骤是，找到运算核心和合适的数据结构。运算核心：相除取余。数据结构：双向链表。要注意不要让技术问题成为思路实现的障碍
struct yu_store
{
    yu_store* front;
    int a;
    yu_store* next;
};
 
int main()
{
    long int num;
    long int yu = 1;
    int n;
    int count;
    yu_store* f;
    yu_store* p;
    yu_store* s;
     
    scanf("%ld",&num);
    getchar();
    scanf("%d",&n);
     
    p = (yu_store *)malloc(sizeof(yu_store));
    f = p;
    f->front = NULL;
    f->next = NULL;
     
    if(num == 0)
    {
        printf("%ld",num);
        return 0;
    }
     
    while(num != 0)
    {
        yu = num%n;
        num = num/n;
        s = (yu_store *)malloc(sizeof(yu_store));
        p->a = yu;
        p->next = s;
        s->front = p;
        s->next = NULL;
        p = s;
    }
     
    while(f != p)
    {
        p = p->front;
        printf("%d",p->a);
        free(s);
        s = p;
    }
    free(f);
    return 0;
}

C 解法, 执行用时: 2ms, 内存消耗: 372KB, 提交时间: 2020-08-11

#include <stdio.h>

int main()
{
    unsigned long long x; 
    int k;
    int j = 0;
    char delta = '0';
    char a[100] = {'0'};

    scanf("%lld %d", &x, &k);
    
    if(k <= 1 || k > 10)
        return -1;
    
    while(x > 0)
    {
        a[j++] = x%k + delta;
        x /= k;
    }
    if(j == 0)
    {
        printf("0");
        return 0;     
    }
    for(k = j -1; k >= 0; k--)
    {
        printf("%c", a[k]);
    }
    return 0;
}