负进制转换模板

参考文章 传送门

思路和正进制差不多,只不过当余数小于0的时候向商借数,类似减法

#include <bits/stdc++.h>
#define maxn 200005
typedef long long ll;
using namespace std;
char b[maxn];
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    ll n,m;
    cin >> n >> m;
    cout << n << "=";
    ll k = 0;
    ll t = 0;
    while(n != 0)
    {
        t = n % m;
        n /= m;
        if(t < 0){t -= m; n += 1;}
        if(t < 10)b[k++] = t + '0';
        else b[k++] = t - 10 + 'A';
    }
    for(int i = k-1; i >= 0; i --)
    {
        cout << b[i];
    }
    cout << "(base" << m << ")" << endl;
    return 0;
}

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转载自blog.csdn.net/zzzanj/article/details/81251802