CSUOJ 1131（Nim-B* Sum）

这道题要我们先将10进制数x和y在B进制下进行加法取模运算，然后再将结果转换成10进制。

#include<iostream>
#include<string.h>
using namespace std;

const int MAX = 32;
int a[MAX], b[MAX];

int main()
{
    int n, cas, B, x, y;
    cin >> n;
    while(n --)
    {

        cin >> cas >> B >> x >> y;
        int j = 0;
        while(x > 0)
        {
            a[j++] = x % B;
            x /= B;
        }
        int len1 = j;
        j = 0;
        while(y > 0)
        {
            b[j++] = y % B;
            y /= B;
        }
        int len2 = j;
        int val = 0, k = 0;
        int c = 1;

        if( len1 >= len2 )
        {
            for( int i = 0; i < len2; i ++)
            {
                a[i] = ( a[i] + b[i] ) % B;
            }
            for( int i = 0; i < len1; i ++)
            {
                val += a[i] * c;
                c *= B;
            }
        }
        else {
            for( int i = 0; i < len1; i ++)
            {
                b[i] = ( b[i] + a[i] ) % B;
            }

            for( int i = 0; i < len2; i ++)
            {
                val += b[i] * c;
                c *= B;
            }
        }
        cout << cas << "" << val << endl;
    }
    return 0;
}

不过代码写得罗嗦了 980k 28ms