负进制转换

//输入一个10进制数，一个负进制，输出转化后的数字

#include <bits/stdc++.h>
using namespace std;

char a[50] = {
    '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'
};

int main(){
    int T,n,m,k,t;
    int base;
    string ans = "";
    while(scanf("%d%d",&n,&base) != EOF){
        printf("%d=",n);
        while(n != 0){
            k = n % base;
            t = n / base;
            if(k < 0){
                k -= base;
                t ++;
            }
            n = t;
            ans = a[k] + ans;
            //printf("%c",a[k]);
        }
        cout << ans;
        printf("(base%d)\n",base);
        ans = "";
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/love-fromAtoZ/p/9077697.html