题解 | #数制转换#

//a-f|A-F用数组下标解决，这个想法很好
#include "stdio.h"
#include "string"
#include "math.h"
#include "algorithm"
using namespace std;

int main(){
    int num1,num3;//num1为原进制，num3为转换后的进制
    char num2[30];
    string change = "0123456789ABCDEF";//eg：A的下标为10，所以A代表10
    scanf("%d %s %d",&num1,num2,&num3);
    string str = num2;
    for (int i = 0; i < str.size(); ++i) {
        if (str[i] >= 'a' && str[i] <= 'z')
            str[i] -= 32;
    }
    int sum = 0;
    for (int i = str.size()-1; i >= 0; --i) {//计算出对应的十进制
        for (int j = 0; j < change.size(); ++j) {
            if (change[j] == str[i]){//change的下标j即为当前位的数字
                sum += j*pow(num1,str.size()-i-1);
                break;
            }
        }
    }
    string End = "";
    int remainder;
    while (sum != 0){
        remainder = sum%num3;
        sum = sum/num3;
        End += change[remainder];
    }
    reverse(End.begin(),End.end());
    printf("%s",End.c_str());
}