加基数n
在这个问题上，给出了两个数字。这些数字的底数为n。我们还必须在以n为底的加法中找到这些数字的结果。
首先，将数字转换为十进制数字。从十进制值中，我们可以简单地将它们相加。最后，数字再次转换为以n为底的数字。
n个基数以字符串形式给出，因为对于那些基数大于9的数字，它可能包含一些字母来表示数字，例如十六进制数字，则有6个字母(AF)。
输入输出
Input: The base of a number system: 16 First number 2C Second number 5F Output: The result of addition is: 8B
算法
baseNtoDec(number,base)
输入-以N为底的数字字符串，以N为底的值。
输出-以N为底的数字的十进制等效项。
Begin
len := length of number
power := 1
num := 0
for i := len -1 down to 0, do
if number[i] >= base, then
return invalid number
num := num + number[i] * power
power := power * base
done
return num
EnddecToBaseN(dec,base)
输入：十进制数，以N为基数的十进制数。
输出：以N为底的数字字符串。
Begin
while dec > 0, do
res := concatenate (dec mod base) with res
dec := dec / base
done
reverse the result
return res
EndaddBaseN(num1,num2,base)
输入： 以N为底的两个数字，即N的值。
输出： 加N后的数字。
Begin dec1 := baseNtoDec(num1, base) dec2 := baseNtoDec(num2, base) sum := decToBaseN(dec1 + dec2, base) return sum End
示例
#include<iostream>
#include<algorithm>
using namespace std;
int getVal(char c) {
if(c >= '0' && c<='9')
return int(c-'0'); //decimal value of given number
else
return int(c-'A'+10); //for Alphanumeric numbers
}
char revVal(int n) {
if(n >= 0 && n <=9)
return char(n+'0'); //character value of given number
else
return char(n+'A'-10); //for Alphanumeric numbers, get alphabet from decimal
}
int baseNtoDec(string number, int base) {
int len = number.size();
int power = 1;
int num = 0;
for(int i = len-1; i>= 0; i--) { //from last digit to first digit
if(getVal(number[i]) >= base)
return INT_MIN; //when a digit is >= base, return -ve infinity as error
num += getVal(number[i])*power;
power = power*base;
}
return num;
}
string decToBaseN(int dec, int base) {
string res = ""; //empty string
while(dec > 0) {
res += revVal(dec%base);
dec /= base;
}
reverse(res.begin(), res.end()); //reverse the string to get final answer
return res;
}
int main() {
int base;
string num1, num2, sum;
cout << "Enter Base: "; cin >> base;
cout << "Enter first number in base "<<base<<": ";cin >> num1;
cout << "Enter second number in base "<<base<<": ";cin >> num2;
sum = decToBaseN((baseNtoDec(num1, base) + baseNtoDec(num2, base)), base);
cout << "The result of addition is: " << sum;
}输出结果
Enter Base: 16 Enter first number in base 16: 2C Enter second number in base 16: 5F The result of addition is: 8B