P1015 [NOIP1999 普及组] 回文数

https://www.luogu.com.cn/problem/P1015

思路

将字符串m转换为10进制的值

翻转m，也转换为10进制值，

后运用10进制加这两个数

转换为对应进制n的字符串，

判断是否为回文

Code -- Partial AC

https://www.luogu.com.cn/record/101917196

四个测试过了三个，

检查发现，m可能有100位， 牵扯到大数加法运算，可参考：

https://www.cnblogs.com/LoginX/p/Login_X1.html

int n;
string m;

int strtoint(int n, string& m){
    int sum = 0;
    int pow = 1;
    int mlen = m.size();
    
    REP(i, mlen){
        char one = m[mlen-i-1];

        int ione = 0;

        if (one >= '0' && one <= '9'){
            ione = one - '0';
        } else if (one >= 'A' && one <= 'F'){
            ione = one - 'A' + 10;
        }

        sum += ione * pow;
        pow *= n;
    }
    
    return sum;
}

string inttostr(int n, int msum){
    string mstr;
    
//    cout << "n=" << n << "msum=" << msum << endl;
    
    while(msum>0){
        int r = msum % n;

//        cout << "r=" << r << endl;

        char one = '0';

        if (r >= 10 && r<=15){
            one = 'A' + r - 10;
        } else if (r >= 0 && r <= 9){
            one = '0' + r;
        }
        
        mstr.push_back(one);
        
        msum /= n;
    }
    
    reverse(mstr.begin(), mstr.end());
    
    return mstr;
}

bool ispalindrome(string str){
    int len = str.size();
    int half = len / 2;

    REP(i, half){
        char f = str[i];
        char s = str[len-i-1];
        
        if(f != s){
            return false;
        }
    }
    
    return true;
}

string reverse_add(int n, string m){
//    cout << "m=" << m << endl;

    int mint = strtoint(n, m);

//    cout << "mint=" << mint << endl;

    string mr = m;
    reverse(mr.begin(), mr.end());

//    cout << "mr=" << mr << endl;

    int mrint = strtoint(n, mr);

//    cout << "mrint=" << mrint << endl;

    int msum = mint + mrint;

//    cout << "msum=" << msum << endl;

    string msumstr = inttostr(n, msum);

//    cout << "msumstr=" << msumstr << endl;

    return msumstr;
}

int main()
{
    cin >> n >> m;
    
    int count = 0;
    
    while(true){
        bool ret = ispalindrome(m);
        //    cout << "ret=" << ret << endl;
        
        if (ret){
            break;
        }

        m = reverse_add(n, m);
        count++;
        
        if (count > 30){
            break;
        }
    }

    if (count <= 30){
        cout << "STEP=" << count << endl;
    } else {
        cout << "Impossible!" << endl;
    }

    return 0;
}