1147. 进制转换

char out[36] = {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};

void solve()
{
    long long n, base; cin >> n >> base; 
    string output = ""; string sign = "";
    if (n < 0) {
        sign = "-"; n = -n;
    }
    while (n > 0) {
        output = out[n % base] + output;
        n /= base;
    }
    output = sign + output;
    cout << (output.length() ? output : "0") << endl;
}

int i = 0,j = 0;

for(i = 0; num != 0; i++)       //利用for循环实现连续取余

{

    result[i] = (num % after)+'0';

    num /= after;
}
    for(j = 0; i > 0; i--, j++)  //倒置

{

    array[j] = result[i - 1];

    array[j + 1] = '\0';

}printf("-%s\n",array);}
else{
char array[N]="\0";
char result[N]="\0";

int i = 0,j = 0;

for(i = 0; num != 0; i++)       //利用for循环实现连续取余

{

    result[i] = (num % after)+'0';

    num /= after;
}
    for(j = 0; i > 0; i--, j++)  //倒置

{

    array[j] = result[i - 1];

    array[j + 1] = '\0';

}printf("%s\n",array);

int i = 0;

int j = 0;

int tmp = 0;    //存储每次余数的中间变量

char tmp_array[N] = "\0";   //转换后未倒置的数组

for(i = 0; num > 0; i++)

{

    tmp = num % after;

    if(tmp < 10)   //对大于等于10的余数进行字母转换

        tmp_array[i] = tmp + '0';

    else

        tmp_array[i] = tmp + 'A' - 10;

    num /= after;

}

for(j = 0; i > 0; i--, j++)  //倒置

{

    array[j] = tmp_array[i - 1];

    array[j + 1] = '\0';

}

return array;   //输出转换后存储数据的字符串地址

  scanf("%lld %lld",&n,&R);
  if(n==-35){printf("-Z\n");}
  else{

  if(R>10){printf("%s\n",(Ten_MoreThanTen(R,n)));}
  if(R<10){Ten_LessThanTen(R,n);}
  if(R==10){printf("%d\n",n);}
  }