问题描述
如何在不使用字符串或数组的情况下从base n转换为base 10（其中n是2到9之间的输入）？转换的数字是一个长整数类型。
我试过的代码返回输入到函数中的相同数字（num）。
我尝试过：
int calcBase10（long long num，int base） ）
{
int base10;
int商;
int余数;
if（base == 10）
{
base10 = num;
return（base10）;
}
商= num / base;
余数= num％base;
if（商= = 0）
{
base10 =余数;
}
else
{
base10 =（base * calcBase10（商，基数）+余数）;
}
return（ base10）;
}
How do I convert from base n to base 10 without using strings or arrays(where n is an input between 2 and 9)? The number being converted is a long long integer type.
The code that I have tried returns the same number (num) that was input into the function.
What I have tried:
int calcBase10(long long num, int base)
{
int base10;
int quotient;
int remainder;
if(base == 10)
{
base10 = num;
return(base10);
}
quotient = num / base;
remainder = num % base;
if(quotient == 0)
{
base10 = remainder;
}
else
{
base10 = (base * calcBase10(quotient, base) + remainder);
}
return(base10);
}
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