重点是无解时输出'Impossible!'，感叹号很重要，很容易错
var
n,i,len,k,j,x,y:longint;
s:string;
f:boolean;
num,num2,num3:array[1..100] of integer;
begin
readln(n);
readln(s);
len:=length(s);
for i:=1 to len do
begin
j:=len-i+1;
case s[i] of
'0'..'9' : val(s[i],num[j]);
'A' : num[j]:=10;
'B' : num[j]:=11;
'C' : num[j]:=12;
'D' : num[j]:=13;
'E' : num[j]:=14;
'F' : num[j]:=15;
end;
end;
k:=0;
repeat
inc(k);
for i:=1 to len do num2[len-i+1]:=num[i];
for i:=1 to len do num3[i]:=num[i]+num2[i];
for i:=1 to len do
if num3[i]>=n then
begin
inc(num3[i+1],num3[i] div n);
num3[i]:=num3[i] mod n;
end;
if num3[len+1]<>0 then inc(len);
num:=num3;f:=true;x:=1;y:=len;
while x<=y do
begin
if num3[x]<>num3[y] then f:=false;
inc(x);dec(y);
end;
until f or (k>30);
if f then writeln('STEP=',k)
else writeln('Impossible!');
end.

一种思路：查表
def change(N,M):
list=['0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F']
num=len(M)
flag=0
value=[]
for i in range(num):
temp=list.index(M[i])+list.index(M[num-i-1])+flag
if (temp>=N):
flag=1
temp=temp%N
else:
flag=0
value.append(list[temp])
if flag:
value.append(list[1])
return value

def judge(value):
flag=1
num=len(value)
for i in range(num//2+1):
if (value[i]!=value[num-i-1]):
flag=0
break

return flag

N=eval(input())
M=list(input())
fl=1

for i in range(1,31,1):
M=change(N,M)
if judge(M):
print('STEP={:d}'.format(i))
fl=0
break

if fl:
print('Impossible!')

STL大法好，STL+高精解法

#include<bits/stdc++.h>
using namespace std;
char cnums[17] = "0123456789ABCDEF";
inline int ctoi(char c,int cyy) {
    string s;
    s.push_back(c);
    return stoi(s, 0, cyy);
}
string add(string a,string b,int cyy) {
    string res = "";
    int pa, pb, sum, cy = 0;
    pa = a.length();
    pb = b.length();
    do
    {
        pa == 0 ? a[pa] = '0' : pa--;
        pb == 0 ? b[pb] = '0' : pb--;
        sum = ctoi(a[pa], cyy) + ctoi(b[pb], cyy) + cy;
        res = cnums[sum % cyy] + res;
        cy = sum / cyy;
    } while (pa!=0||pb!=0||cy!=0);
    return res;
}

int main() {
    int cyy, time = 0;
    string snum, revstr;
    cin >> cyy >> snum;
    while (true)
    {
        if (time == 30) {
            cout << "Impossible!";
            return 0;
        }
        revstr.resize(snum.size());
        reverse_copy(snum.begin(), snum.end(), revstr.begin());
        if (revstr == snum) break;
        time++;
        snum = add(snum, revstr, cyy);
    }
    cout << "STEP=" << time;
    return 0;
}

#include <cstring>
#include <iostream>
bool check(char s[])
{
    bool YorN = 1;
    for (int i = 0; i < strlen(s); i++)
    {
        if (s[i] != s[strlen(s)-1-i])
        {
            YorN = 0; 
        }
    }
    return YorN;
}

void change(int n, char s[])
{
    int x = 0, lenc = 0, i, num_tou, num_wei;
    int c[62510];
    for (i = strlen(s)-1; i >= 0; i--)
    {
        if (s[i] <= '9') {num_tou = s[i] - '0'; }
        else {num_tou = s[i] - 55;}
        if (s[strlen(s) - 1 - i] <= '9') {num_wei = s[strlen(s) - 1 - i] - '0'; }
        else {num_wei = s[strlen(s) - 1 - i] - 55;}
        c[lenc] = num_tou + num_wei + x;
        x = c[lenc] / n;
        c[lenc] %= n;
        lenc ++;
    }
    c[lenc] = x;
    lenc = lenc - (c[lenc]== 0);    
    for (i = 0; i <= lenc; i++)
    {
        if (c[lenc - i] < 10) {s[i] = c[lenc - i] + '0';}
        else {s[i] = c[lenc - i] + 55;}
    }
}
int main()
{
    char s[62510];
    int count = 0, n;
    scanf("%d\n%s", &n, s);
    for (int i = 0; i < 30; i++)
    {
        if(!check(s))
        {
            change(n, s);
            count += 1;
        }
        else
        {
            printf("STEP=%d", count);
            return 0;
        }
    }
    printf("Impossible!");
    return 0;
} 

/**
 * @author wpx
 * @version V1.0
 * @Package com.algorithm
 * @date 2021/5/28 17:02
 */
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

public class Main {


    public static void main(String[] args) {
        Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        Integer base = Integer.valueOf(sc.nextLine());
        String num = sc.nextLine();
        String tmpAddNum = num;
        int stepNum = 0;
        for(stepNum = 0; stepNum < 30; stepNum++){
            String athorNum = new StringBuilder(tmpAddNum).reverse().toString();
            final String resultR = addFun(tmpAddNum, athorNum, base);
            String result = new StringBuilder(resultR).reverse().toString();
            if(result.equals(resultR)) {
                break;
            } else {
                tmpAddNum = result;
            }
        }
        if(stepNum == 30) {
            System.out.println("Impossible!");
        } else {
            System.out.println("STEP=" + (stepNum + 1));
        }
    }

    private static String addFun(String one, String two, Integer base) {
        // 逐位相加
        // 取最大的位数进行遍历
        int maxLen = one.length() > two.length() ? one.length() : two.length();
        int oneLastPos = one.length() - 1;
        int twoLastPos = two.length() - 1;

        // 从最后一位往前
        StringBuilder result = new StringBuilder();
        boolean hasCarry = false;
        for(int i = 0; i < maxLen; i++) {
            int onePos = oneLastPos - i;
            int twoPos= twoLastPos - i;

            int oneChar = one.charAt(onePos);
            int twoChar = two.charAt(twoPos);
            if(oneChar > 57) {
                oneChar -= 55;
            } else {
                oneChar -= 48;
            }

            if(twoChar > 57) {
                twoChar -= 55;
            } else {
                oneChar -= 48;
            }


            Integer carryValue = hasCarry ? 1 : 0;
            int sumValue = oneChar + twoChar + carryValue;
            Integer appendInValue = sumValue % base;
            String appendValue;
            if(appendInValue > 9) {
                appendValue = String.valueOf((char) (65 + ( appendInValue - 10)));
            } else {
                appendValue = String.valueOf(appendInValue);
            }
            result.append(appendValue);
            // 处理进位
            if((sumValue) >= base) {
                hasCarry = true;
            } else {
                hasCarry = false;
            }
        }
        if(hasCarry){
            result.append("1");
        }
        return result.toString();
    }
}

#include<iostream>
#include<string>
#include<math.h>
#define MAX 2000
#define ll long long
using namespace std;
ll N_to_ten(int N,int a[],int n)//把一个n进制数转化为10进制数
{
ll sum=0;
for(int i=0;i<n;i++)
{
sum=sum*N+a[i];
}
return sum;
}
bool hw(int a[],int n)
{
for(int i=0;i<n;i++)
{
if(a[i]!=a[n-i-1])
{
return false;
}
}
return true;
}
ll Add(int N,int a[],int n)//实现进制的加减
{
int b[n];
for(int i=0;i<n;i++)
{
b[i]=a[n-i-1];
}
ll sum=0;
sum=N_to_ten(N,a,n)+N_to_ten(N,b,n);
return sum;
}
void panduan(int N,int a[],int n)
{
if(hw(a,n))
{
cout<<"STEP="<<0<<endl;
}
else
{
int setp=1;
ll sum=Add(N,a,n);
int k[MAX];
do
{
int i=0;
while(sum)
{
k[i]=sum%N;
sum/=N;
i++;
}
if(hw(k,i))
{
cout<<"STEP="<<setp<<endl;
return;
}
else
{
sum=Add(N,k,i);
setp++;
}
}while(setp<=30);
}
cout<<"Impossible!"<<endl;

}
int main()
{
int N;
cin>>N;
string str;
cin>>str;
//把这个str类型的数转化为int型
int length =str.size();
int a[MAX];//存放改变的字符串
int a_length=0;
for(int i=0;i<length;i++)
{
if(str[i]<65)
{
a[i]=str[i]-'0';//其他进制0-9的转化
a_length++;
}
else
{
a[i]=str[i]-55;//16进制ABCD的转化
a_length++;
}
}
/*for(int i=0;i<a_length;i++)
{
cout<<a[i];
}*/
//Add(N,a,a_length);
//cout<<N_to_ten(N,a,a_length);
//cout<<hw(a,a_length);
panduan(N,a,a_length);
return 0;
}

利用string进行判断
```cpp
#include <bits/stdc++.h>
#define fu(a, b, c) for (int a = b; a <= c; a++)
#define fd(a, b, c) for (int a = b; a >= c; a--)
using namespace std;
int n;
string s;

int main() {
cin >> n >> s;
for (auto &c : s)
c=c>=65?c-55:c-48;
fu(i, 0, 30) {
string s1 = s;
reverse(s1.begin(), s1.end());
if (s1 == s) {
cout <<"STEP="<<i;
return 0;
}
int sz = s.size() - 1;
fd(i, sz, 1) {
s[i]+=s1[i];
s[i - 1]+=s[i]/n,s[i]%=n;
}
s[0] += s1[0];
if (s[0] >= n)
s.insert(0,1,char(s[0]/n)),s[1]%=n;
}
cout << "Impossible!";
}
```

/*
一开始无论怎样都过不了第二个，后来发现16进制的字母要用大写！！！
*/
#include <iostream> //[1999提高组-B]回文数
#include <algorithm>
#include <string>
using namespace std;

int H(char ch)
{
    if (ch >= '0' && ch <= '9')
        return ch - '0';
    else if (ch >= 'A' && ch <= 'Z')
        return ch - 'A' + 10;
}

string HEX = "0123456789ABCDEF";
int A[100], B[100], C[102];
string add(string a, string b, int n)
{
    string ans;
    int len1 = a.length(), len2 = b.length();
    int len = max(len1, len2), t = 0;
    for (int i = 0; i < len1; i++)
        A[i] = H(a[len1 - 1 - i]);
    for (int i = 0; i < len2; i++)
        B[i] = H(b[len2 - 1 - i]);

    for (int i = 0; i < len; i++)
    {
        int k = A[i] + B[i] + t;
        C[i] = k % n;
        t = k / n;
    }
    for (int i = 0; i < len; i++)
        ans += HEX[C[i]];
    if (t > 0)
        ans += '1';
    reverse(ans.begin(), ans.end());
    return ans;
}
int step(string num, int n)
{
    string re = num;
    int ans = 0;
    reverse(re.begin(), re.end());
    while (re != num)
    {
        re = num = add(num, re, n);
        reverse(re.begin(), re.end());
        ans++;
        if (ans == 31)
            break;
    }
    return ans;
}

int main()
{
    int n, ans;
    string num;
    cin >> n >> num;
    ans = step(num, n);
    //cout << ans << endl;
    if (ans == 31)
        cout << "Impossible!" << endl;
    else
        cout << "STEP=" << ans << endl;

    return 0;
}

#include<iostream>//wa的两次impossible拼错了，发现以后没有注释掉自己输出的东西又wa一发，也是醉了。 
#include<cstring>//编译错的两次发现评测机不允许用strrev？？？ 
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;

int to_num(char C){
    if(C >= '0' && C <= '9')
        return C - '0';
    else if(C >= 'A' && C <= 'Z')
        return C - 'A' + 10;
    else if(C >= 'a' && C <= 'z')
        return C - 'a' + 10;
}
char N[16] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'};

int main()
{
    char A[100000];//正序 
    char B[100000];//反序 
    char C[100001];
    int n;//进制 
    scanf("%d", &n);
    scanf("%s", &A);
    int time = 0;

while(1){
        for(int i = 0 ; i < strlen(A); i++)
            B[i] = A[strlen(A) - 1 - i];
        
        if(time <= 30 && strcmp(A, B) == 0){
            cout<<"STEP="<<time<<endl;
            return 0;
        }
        else if(time > 30){
            cout<<"Impossible!"<<endl;
            return 0;
        }
        int carry = 0;
        for(int i = strlen(A) - 1 ; i >= 0 ; i--){
            C[strlen(A) - 1 - i] = N[(to_num(B[i]) + to_num(A[i]) + carry) % n];
            if(to_num(B[i]) + to_num(A[i]) + carry >= n)
                carry = 1;
            else
                carry = 0;
        }
        if(carry == 1)
            C[strlen(A)] = '1';
        strcpy(B, C);
        for(int i = 0 ; i < strlen(C) ; i++)
            A[i] = C[strlen(C) - 1 - i];
            
        time++;
    }
 } 

##python

把含有A-F的列表转换为数值列表

def str_to_int(list_str_x):
l = len(list_str_x)
list_int_return = []
for i in range(0,l):
if ord(list_str_x[i]) < 58:
list_int_return.append(ord(list_str_x[i]) - 48)
else:
list_int_return.append(ord(list_str_x[i]) - 55)
return list_int_return

##加法 返回两数之和
def jia_fa(list_int_x):
l1 = list_int_x
l2 = list_int_x[::-1]
len_x = len(list_int_x)
the_sum = []
for i in range(0, len_x+2):
the_sum.append(0)
for i in range(0, len_x):
each_plus = int(l1[i]) + int(l2[i]) + int(the_sum[i])
if each_plus >= n:
the_sum[i] = each_plus - n
the_sum[i+1] = 1
else:
the_sum[i] = each_plus
while True:
if the_sum[-1] == 0:
del the_sum[-1]
else:
break
return the_sum[::-1]

##判断是否回文
def hui_wen(x):
y = x[::-1]
if y == x :
return True
else:
return False

##输入及预处理
n = int(input())
in_put = input()
list_str_m = in_put
list_int_m = str_to_int(list_str_m)

##主体程序
out_put = 'Impossible!'
for i in range(0,30):
if hui_wen(list_int_m):
out_put = 'STEP='+str(i)
break
else:
list_int_m = jia_fa(list_int_m)

print(out_put)

#include <iostream>
#define max 300
using namespace std;
class cnum{
private:
int arr[max],r,t,times;
public:
cnum(char *c,int g,int sn);
int gettimes();
int check();
void puls();
};
cnum::cnum(char *c,int g,int sn){
t=g;times=0;r=sn;
for(int i=0;i<300;i++){
arr[i]=0;
}
for(int i=0;i<=sn;i++){
switch(c[sn-i]){
case 'A':arr[299-i]=10;break;
case 'B':arr[299-i]=11;break;
case 'C':arr[299-i]=12;break;
case 'D':arr[299-i]=13;break;
case 'E':arr[299-i]=14;break;
case 'F':arr[299-i]=15;break;
default:arr[299-i]=int(c[sn-i])-48;
}
}
}
int cnum::gettimes(){
for(int i=0;i<40;i++){
if(check()){
return times;
}else{
puls();
}
}
return times;
}
int cnum::check(){
for(int i=0;i<=r;i++){
if(arr[299-r+i]!=arr[299-i]){
return 0;
}
}
return 1;
}
void cnum::puls(){
int *fuck=new int[300];
for(int i=0;i<300;i++){
fuck[i]=arr[i];
}
for(int i=0;i<=r;i++){
arr[299-i]+=fuck[299-r+i];
if(arr[299-i]>=t){
arr[299-i-1]+=1;
arr[299-i]-=t;
}
}
if(arr[299-r-1]){
r++;
}
times++;
delete []fuck;
}
int main(){
int number,sb=0;char string[300]={' '};
cin>>number;
cin>>string;
for(int i=299;i>=0;i--){
if(string[i]){
sb=i;break;
}
}
cnum a(string,number,sb);
if(a.gettimes()>=0&&a.gettimes()<=30){
cout<<"STEP="<<a.gettimes();
}else{
cout<<"Impossible!";
}
return 0;
}
/////笨办法

投机取巧了一波
转换来转换去速度大概是慢了许多

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdlib>
using namespace std;
char buffer[256];
char *stop;
long int M,A,sum;
int N,pos,cnt=0,flag=0;
void itoa(long int n, char*str, int base){
    if(base == 16){
        int i=0;
        while ( n > 0){
            str[i++] = n % base<10? (n%base)+'0':(n%base)-10+'A';
            n /= base;
        }
    }else{
        int i=0;
        while (n > 0){
            str[i++] = (n % base)+'0';
            n /= base;
        }
    }
}
void getReverse(){
    pos = 0;
    while((buffer[pos]<='9'&&buffer[pos]>='0')|| (buffer[pos]<='F'&&buffer[pos]>='A')){
        pos++;
    }
    reverse(buffer,buffer+pos);
}
int main(){
    scanf("%d%s",&N,buffer);
    while(cnt <= 30){
        M = strtol(buffer, &stop,N);
        getReverse();
        A = strtol(buffer, &stop,N);
        if(M == A){
            flag = 1;
            break;
        }
        sum = M + A;
        itoa(sum, buffer,N);
        cnt++;
    }
    if(flag) printf("STEP=%d",cnt);
    else printf("Impossible!");
}

/*
题目名称：回文数
题目难度：塑料
*/
#include<iostream>
#include<cstdio>
using namespace std;
int a[100];
int temp[100];
int l,n;
//每一次运行add，则将a的那啥相加算出来
//temp是辅助存储的数组
void add(){
int pi=0;//pi是记录进不进位的
int t=0;//t是辅助计算的
for(int i=1;i<=l;i++){
t=a[i]+a[l-i+1]+pi;
if(t>=n){
t-=n;
pi=1;
}else pi=0;
temp[i]=t;
}
if(pi){
temp[l+1]=1;
l++;
}
for(int i=1;i<=l;i++)a[i]=temp[i];
}
//check是用来检验是否为回文数的
bool check(){
for(int i=1;i<=l;i++){
if(a[i]!=a[l-i+1])
return false;
}
return true;
}
int main(){
scanf("%d\n",&n);
char s;
if(n==16){
for(l=1;;l++){
s=getchar();
if(s=='A')a[l]=10;
else if(s=='B')a[l]=11;
else if(s=='C')a[l]=12;
else if(s=='D')a[l]=13;
else if(s=='E')a[l]=14;
else if(s=='F')a[l]=15;
else{
if(s=='\n')break;
a[l]=s-'0';
}
}
}
else {
for(l=1;;l++){
s=getchar();
if(s=='\n')break;
a[l]=s-'0';
}
}

l--;
for(int i=1;i<=30;i++){
add();
if(check()){
printf("STEP=%d",i);
return 0;
}
}
printf("Impossible!");

return 0;
}

使用数组进行模拟，16进制的话只需要将字符替换成对应的数字就OK
我的数组记录顺序是从最低位到最高位，而且输入的时候不需要颠倒顺序
因为在STEP = 1的时候我已经将它倒序过来相加了
和颠倒顺序之后操作的结果一样
cpp
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int num = 505;
int A[num],B[num];
char array[num];
int k,l,ans = 0;
int getnum(char x) {
if(x >= '0' && x <= '9')
return x - '0';
else return x - 'A' + 10;
}
bool check() {
bool flag = true;
for(int i = 1;i <= l / 2;i++)
if(A[i] != A[l - i + 1]) {
flag = false;
break;
}
return flag;
}
void cnt() {
int push = 0;
for(int i = 1;i <= l;i++)
B[i] = A[l - i + 1];
for(int i = 1;i <= l;i++) {
int temp = A[i] + B[i];
temp += push;
A[i] = temp % k;
push = temp / k;
}
if(push != 0) A[++l] = push;
ans ++;
if(check() || ans > 30) return;
else cnt();
}
int main() {
cin >> k;
scanf("%s",array);
l = strlen(array);
for(int i = 1;i <= l;i++) {
A[i] = getnum(array[i - 1]);
}
cnt();
if(ans < 31) cout << "STEP=" << ans;
else cout << "Impossible!";
return 0;
}


AC了来这献个丑吧

#include <cstdio>
using namespace std;
long long ten(int x)
{
    long long y = 1;
    for(int i = 0; i < x; i++) y *= 10;
    return y;
}
long long flip(long long a, int x)
{
    int w = 10;
    if(x == 16) w = 16;
    long long c = a, b = 0;
    while( c ){
        b *= w;
        b += c%w;
        c /= w;
    }
    return b;
}
long long add(long long a, int x)
{
    if(x == 16 || x == 10) return a+flip(a, x);
    long long b = flip(a,x);
    long long c = 0;
    int i = 0;
    char s[100001] = {0};
    while( a || b ){
        s[i] += a%10 + b%10;
        if(s[i] >= x){
            s[i+1] += s[i]/x;
            s[i] %= x;
        }
        a /= 10;
        b /= 10;
        i++;
    }
    for(int j = 0; j <= i; j++){
        c += s[j]*ten(j);
    }
    return c;
}
int main()
{

    int N;
    scanf("%d", &N);
    long long M;
    if(N == 16) scanf("%llX", &M);
    else scanf("%lld", &M);
    int i;
    for(i = 0; i <= 30; i++){
///////////
//      if(N == 16) printf("%llX\n", M);
//      else printf("%lld\n", M);
// ///////////
        if(M == flip(M, N)){
            break;
        }
        M = add(M, N);
    }
    if(i > 30){
        printf("Impossible!\n");
    }else{
        printf("STEP=%d\n", i);
    }
    return 0;
}

import java.util.Scanner;

public class Main {
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
String num = scanner.next();
scanner.close();
int i = 0;
while(i<30){
if(reverse(num).equals(num)){
System.out.println("STEP="+i);
return;
}
num = add(num, n);
i ++;
}
System.out.println("Impossible!");
}
//针对16进制数字和字母的转换
private static String change(int m){
switch (m) {
case 10:
return "A";
case 11:
return "B";
case 12:
return "C";
case 13:
return "D";
case 14:
return "E";
case 15:
return "F";
default:
return String.valueOf(m);
}
}
//字符串反转
private static String reverse(String str) {
if(str == null || str.length() == 0) {
return str;
}
int len = str.length();
if(len == 1) {
return str;
} else {
return reverse(str.substring(1))+ str.charAt(0);
}
}
//防止溢出，直接转换为string的相加
private static String add(String num, int n){
int len = num.length();
String num_reverse = reverse(num);
char[] ch1 = num.toCharArray();
char[] ch2 = num_reverse.toCharArray();
int jw = 0;
StringBuilder sb = new StringBuilder();;
for(int i=len-1;i>=0;i--){
int add1 = getAdd(ch1[i]);
int add2 = getAdd(ch2[i]);
int sum = add1 + add2 + jw;
jw = sum / n;
sb.insert(0, change(sum % n));
}
if(jw == 1){
sb.insert(0, "1");
}
return sb.toString();
}
//针对16进制字母和数字的转换
private static int getAdd(char ch){
if(Character.isDigit(ch)){
return ch - '0';
}else{
return ch - 'A' + 10;
}
}
}

#include <cstdio>
#include <cctype>
#include <vector>
#include <cstring>
#include <algorithm>

#define maxn 2333

using namespace std;
char buf[maxn];
int n;
vector<int> a, b;

int main(){
    scanf("%d%s", &n, buf);
    for(int i=0,len=strlen(buf);i<len;i++){ // 简单的按位读入
        if(buf[i]>='0'&&buf[i] <= '9') a.push_back(buf[i]-'0');
        else a.push_back(islower(buf[i])?buf[i]-'a'+ 10:buf[i]-'A'+10);
    }
    b=a; // 将a copy到b里。
    reverse(a.begin(),a.end()); // 反转a。
    if(a==b) return puts("STEP=0"), 0; //反转后相等即为回文
    for(int ans=1;ans<=30;ans++) {
        for(int i=0,siz=a.size();i<siz;i++) {
            a[i]+=b[i]; // 加法
            if(i!=siz-1) a[i+1]+=a[i]/n; // 处理进位
            else if(a[i]>=n) a.push_back(a[i]/n);
            a[i]%=n;
        }
        b=a;
        reverse(b.begin(), b.end());
        if(a==b) return printf("STEP=%d",ans),0; // 判回文
    }
    return puts("Impossible!"), 0;
}

#include <iostream>
#include<string>
#include<stdlib.h>
using namespace std;
int main()
{
string str;
int N;
int step = 0;
int v0[1000], v1[1000], v2[1000] = { 0 };
cin >> N;
cin >> str;
int p = str.size();
if (N == 16)
{
for (int i = 0; i < p; i++)
{
if (str[i] == 'A' || str[i] == 'B' || str[i] == 'C' || str[i] == 'D' || str[i] == 'E' || str[i] == 'F')
v0[i] =str[i] - 55;
else
v0[i] = str[i] - 48;
}
}
else
{
for (int i = 0; i < p; i++)
{
v0[i] = str[i] - 48;
}
}
for (int j = p - 1, i = 0; i < p, j >= 0; i++, j--)
v1[i] = v0[j];
int pos = 0;
for (int i = 0; i < p; i++)
{
if (v0[i] != v1[i])
{
pos = 1;
break;
}
}
if (pos == 0)
{
cout << "step=" << step;
}
else
{
while (step <= 30)
{
pos = 0;
int ksp = 0;//进位
int j = 999;
for (int i = p - 1; i >= 0; i--)
{
int o = v0[i] + v1[i] + ksp;
if (o >= N)
{
v2[j] = o - N;
ksp = 1;
}
else
{
v2[j] = o;
ksp = 0;
}
j--;
}
if (ksp == 1)
{
v2[j] = ksp;
ksp = 0;
}
else
j++;
int g = 0;
while (j < 1000)
{
v0[g] = v2[j];
j++;
g++;
}
for (int j = g - 1, i = 0; i <= g, j >= 0; i++, j--)
v1[i] = v0[j];
for (int j = 0; j <= g; j++)
{
if (v0[j] != v1[j])
{
pos = 1;
break;
}
}
p = g;
step++;
if (pos == 0)
break;
}
if (pos == 0)
cout << "STEP=" << step;
else
cout << "Impossible!";
}
system("pause");
return 0;
}

#include<bits/stdc++.h>
using namespace std;
int N;
long long N2dec(string s)//其他进制转10进制
{

    int len = s.size();
    long long num = 0;
    for(int i = 0; i < len; i++)
    {
        num *= N;
        if(isdigit(s[i]))
        {
            num += s[i] - '0';
        }
        else
            num += s[i] - 'A' + 10;
    }
    return num;
}
string dec2N(long long num){//10进制转其他进制
    string str = "";
    while(num != 0)
    {
        str += num % N + '0';
        if(str[str.size() - 1] > '9')
            str[str.size() - 1] += 'A'-'0' - 10 ;
        num /= N;
    }
    reverse(str.begin(), str.end());
    return str;
}
bool judge(string s){
    int len = s.size();
    for(int i = 0; i < s.size(); i++){
        if(s[i] != s[len-i-1]) return 0;
    }
    return 1;
}
int main()
{
    while(cin >> N){
        string input;
        cin >> input;

        string str;
        int ok = 0, times;
        for(times = 0; times <= 30; times++)
        {
            if(judge(input))
            {
                ok = 1;
                break;
            }
            long long a = N2dec(input);
            reverse(input.begin(), input.end());
            long long b = N2dec(input);
            input = dec2N(a + b);

        }
        if(!ok)
            printf("Impossible!\n");
        else
            printf("STEP=%d\n", times);

    }
}

/*Tokgo*/
/*
简单字符处理
用高精度加法
不必为进制发愁
*/
#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
vector <int> a;
string in;
int n;
int ans;
void change(){
    for(int i=in.size()-1;i>-1;--i){
        if(in[i]>='0'&&in[i]<='9')
            a.push_back(in[i]-'0');
        else
            a.push_back(in[i]-'A'+10);
    }
}
void jia(){
    for(int i=0;i<(a.size()>>1);++i){
        a[i]=a[a.size()-1-i]=a[i]+a[a.size()-1-i];
    }
    if((a.size()&1)==1)
        a[a.size()/2]*=2;
    for(int i=0;i<a.size();++i){
        if(a[i]>(n-1)){
            if(i!=a.size()-1)
                a[i+1]+=(a[i]/n);
            else
                a.push_back(a[i]/n);
            a[i]%=n;
        }
    }
}
bool re(){
    bool jud=1;
    for(int i=0;i<=(a.size()-1)/2;++i){
        if(a[i]!=a[a.size()-1-i]){
            jud=0;
            break;
        }
    }
    return jud;
}
int main(){
    cin>>n;
    cin>>in;
    change();
    for(int i=0;i<=31;++i){
        if(ans=re()){
            ans=i;
            break;
        }
        jia();
    }
    if(ans)
        cout<<"STEP="<<ans<<endl;
    else
        cout<<"Impossible!"<<endl;
    return 0;
}