这题只需要注意16进制的不同即可AC
10进制一下需要-‘0’
16进制则需要-55（-'A'+10）
加法进位时不是=10进位，而是=进制（看输入）才进位。若是最后一位进位则要将长度+1
最后判断是否回文。

评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 804 KiB, score = 25
测试数据 #1: Accepted, time = 0 ms, mem = 800 KiB, score = 25
测试数据 #2: Accepted, time = 0 ms, mem = 804 KiB, score = 25
测试数据 #3: Accepted, time = 15 ms, mem = 804 KiB, score = 25
Accepted, time = 15 ms, mem = 804 KiB, score = 100
代码（C++）

#include<iostream>  
#include<cstring>  
#include<cstdio>  
using namespace std;  
string x;  
int jz,n,a[10001],s=1;  
int pdhw(int n)  
{  
    for(int i=1;i<=n/2;i++)  
        if(a[i]!=a[n-i+1])  
            return 0;  
    return 1;  
}  
int jf(int n)  
{  
    int c[10001]={0},s=1;  
    for(int i=1;i<=n;i++)  
    {  
        c[i]=a[i]+a[n-i+1]+c[i];  
        c[i+1]+=c[i]/jz;  
        c[i]%=jz;  
    }  
    if (c[n+1]!=0)  
        n++;  
    for(int i=n;i>=1;i--)  
    {  
        a[s]=c[i];  
        s++;  
    }  
    return n;  
}  
int main()  
{  
    cin>>jz>>x;  
    n=x.size();  
    for(int i=1;i<=n;i++)  
    {  
        if(x[i-1]<65)  
            a[i]=x[i-1]-'0';  
        else  
            a[i]=x[i-1]-55;  
    }  
    while(s<=30)  
    {  
        if(pdhw(n)==1)  
        {  
            cout<<"STEP="<<s-1;  
            return 0;  
        }  
        s++;  
        n=jf(n);  
    }  
    cout<<"Impossible!";  
}

祝大家刷机愉快！

//fuck it
#include <stdio.h>
#include <strings.h>
int len=0;
int is[101];
int N;
void plus(){
    int rev[101];
    for(int k=0;k<len;k++){
        rev[k]=is[len-k-1];
    }
    for(int i=0;i<len;i++){
        is[i]+=rev[i];
    }
    for(int j=0;j<len;j++){
        if(is[j]>=N){
            is[j+1]+=is[j]/N;
            is[j]%=N;
            if(j==len-1){
                len++;
                return;
            }
        }
    }
}
char fuck(){
    for(int i=0;i<len;i++){
        if(is[i]!=is[len-i-1])
            return 0;
    }
    return 1;
}
int main()
{
    
    char strs[101];
    scanf("%d%s",&N,strs);
    len=strlen(strs);
    for(int i=0;i<len;i++){
        is[len-i-1]=strs[i];
        if(is[len-i-1]>='A'){
            is[len-i-1]-=('A'-10);
        }else{
            is[len-i-1]-='0';
        }
    }
    int times=0;
    for(;times<30;times++){
        plus();
        if(fuck()){
            printf("STEP=%d",times+1);
            return 0;
        }
    }
    printf("Impossible!");
}

这道题分为两个模块：
1.判断回文数
2.数据的正逆之和
☆重点讲下第二模块
①数据逆序
②对两个数组，N进制->十进制，再求和
重复进行这两个模块就能得到结果。
此处定义了反转函数，进制转换函数n->10,10->n
大家可以复制下来进行使用。

#include<iostream>
#include<string>
#include<string.h>
#include<cmath>
using namespace std;
bool Judge(const char* a) {//进制回文数的判断
    int size = strlen(a);
    for (int i = 0; i < size; i++)if (a[i] != a[size - i - 1]) {
        return false;
    }
    return true;
}
string turn_str(string M) {//字符串反转
    string M_copy = M;
    for (int i = 0; i < (int)strlen(M.c_str()); i++)M[i] = M_copy[(int)strlen(M.c_str()) - i - 1];
    return M;
}
long int demical(string M, int n) {//(n)进制->10进制
    long int dem=0;
    for (int i = 0; i < (int)strlen(M.c_str()); i++)
    {
        if ((long int)M[i]>=65)dem += ((long int)M[i] - 55) * (long int)pow(n, (int)strlen(M.c_str())-i-1);
        else dem += ((long int)M[i] - 48) * (long int)pow(n, (int)strlen(M.c_str())-i-1);
    }
    return dem;
}
string ochange(long int num, int n) {//10进制->(n)进制
    char* a=new char[100];
    int i;
    for (i = 0; num != 0; i++) {
        char b;
        if (num % n < 10) b = (char)(num % n + 48);
        else b = (char)(num % n + 55);
        num = (num - (num % n)) / n;
        a[i] = b;
    }
    a[i] = 0;
    string a2 = string(a);
    a2 = turn_str(a2);
    return a2;
}
string sum(string M, int n) {//字符串正逆之和
    long int num = demical(M, n) + demical(turn_str(M), n);
    string a = ochange(num, n);
    return a;
}
int main()
{
    int N, step=0;
    string M;
    cin >> N >> M;
    for (int i = 0; i < 30; i++)
    {
        if (Judge(M.c_str()))break;
        step++;
        M = sum(M, N);
    }
    if (step == 30)cout << "Impossible!";
    else cout << "STEP=" << step;
    return 0;
}

#include<iostream>
#include<cstdio>
using namespace std;

int main()
{
int n;
string m;
cin>>n>>m;

int v[20],vr[20]; //v是字符串转换为整数 下标小的放低位 下标大的放高位
int len=m.length();
for(int i=len-1,j=0;i>=0;i--,j++)
{
if(m[i]>='A')
v[j]=m[i]-'A'+10; //处理十六进制

else
v[j]=m[i]-'0';
}

//逆序放入vr
for(int i=0;i<len;i++)
{
vr[i]=v[len-i-1];
}

//相加
for(int i=1;i<=30;i++)
{
int jinwei=0;
for(int j=0;j<len;j++)
{
v[j]+=vr[j]+jinwei;
jinwei=v[j]/n; //进位等于加起来除以进制
v[j]%=n; //如果大于等于进制那么就会改变
}
if(jinwei>0)
{
v[len++]=jinwei; //如果最后有进位产生 那么就进一位
}

//判断是否是回文
int s=0,e=len-1;
while(s<=e)
{
if(v[s]!=v[e])
break;

s++;e--;
}

if(s>e)
{
printf("STEP=%d\n",i);
return 0;
}

else
{
//逆序放入vr
for(int i=0;i<len;i++)
{
vr[i]=v[len-i-1];
}
}

}

printf("Impossible!");

return 0;
}

U41君来了!
Pascal
var
n,step,i:longint;
a,b,c:array [0..101] of longint;
m:string;
y:boolean;
begin
readln(n);
readln(m);
step:=0;
y:=false;
fillchar(a,sizeof(a),0);
fillchar(b,sizeof(b),0);
a[0]:=length(m);
for i:=1 to a[0] do
case m[a[0]-i+1] of
'0'..'9':a[i]:=ord(m[a[0]-i+1])-48;
'A'..'F':a[i]:=ord(m[a[0]-i+1])-55;
end;
b[0]:=a[0];
for i:=1 to a[0] do b[i]:=a[a[0]-i+1];
repeat
inc(step);
y:=true;
fillchar(c,sizeof(c),0);
for i:=1 to a[0] do
begin
c[i]:=c[i]+a[i]+b[i];
c[i+1]:=c[i+1]+c[i] div n;
c[i]:=c[i] mod n;
end;
if c[a[0]+1]>0 then c[0]:=a[0]+1 else c[0]:=a[0];
for i:=1 to c[0] div 2 do
if c[i]<>c[c[0]-i+1] then begin y:=false;break;end;
if y then break;
a:=c;
for i:=1 to a[0] do b[i]:=a[a[0]-i+1];b[0]:=a[0];
until step=30;
if y then writeln('STEP=',step)
else writeln('Impossible!');

end.

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

const int S=303;
int n,a[S],l;
char c[S],d[S];
inline void add()
{
    for (int i=0;i<l;++i)
        d[l-i-1]=c[i];
    l+=2;
    for (int i=0;i<l;++i){
        c[i]+=d[i];
        if(c[i]>=n)
            c[i+1]++,c[i]-=n;
    }
    while(!c[l-1])
        --l;
}

inline bool pd()
{
    for(int i=0;i<l;++i)
        if(c[i]!=c[l-1-i])
            return false;
    return true;
}

int main()
{
    cin>>n>>c;
    l=strlen(c);
    for (int i=0; i<l; ++i){
        if(c[i]>='0' && c[i]<='9')
            c[i]-='0';
        else
            c[i]=c[i]-'A'+10;
    }
    int step=0;
    while(!pd()){
        ++step;
        if(step>30)
            break;
        add();
    }
    if(step<=30)
        cout<<"STEP="<<step;
    else
        cout<<"Impossible!";
    return 0;
}

```trans = {'0': 0, '1': 1, '2': 2, '3': 3, '4': 4, '5': 5, '6': 6, '7': 7, '8': 8,
         '9': 9, 'A': 10, 'B': 11, 'C': 12, 'D': 13, 'E': 14, 'F': 15}
transform = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F']


def calculator(jinzhi, num_1, num_2):
    result = ""
    carry = 0
    for i in range(len(num_1)):
        _sum = trans[num_1[-1 - i]] + trans[num_2[-1 - i]] + carry
        carry = _sum // jinzhi
        result += transform[_sum % jinzhi]
    if carry != 0:
        result += transform[carry]
    return result[::-1]


if __name__ == '__main__':
    flag = False
    num = 0
    jinzhi = int(input())
    number_1 = input()
    for i in range(30):
        number_2 = number_1[::-1]
        number_1 = calculator(jinzhi, number_1, number_2)
        if number_1 == number_1[::-1]:
            flag = True
            num = i + 1
            break
    if flag:
        print('STEP=' + str(num))
    else:
        print('Impossible!')

请大佬指点
#1 Accepted 1ms 216.0 KiB
#2 Accepted 0ms 228.0 KiB
#3 Accepted 0ms 216.0 KiB
#4 Accepted 0ms 232.0 KiB
C语言

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int Cal(char A[],int time,int N){
    int l=strlen(A),num=0,out=0,in=0,i=0;
    char B[10001]={'\n'};
    while(i<l||out!=0){
        if(i<l){        
            if(A[i]>64){
                num=A[i]-'0'-7;
            }
            else{
                num=A[i]-'0';
            }
            if(A[l-1-i]>64){
                num=num+A[l-1-i]-'0'-7;
            }
            else{
                num=num+A[l-1-i]-'0';
            }
        }
        else{
            num=0;
        }
        in=(num+out)%N;
        if(in<10){
            B[i]=in+'0';
        }
        else{
            B[i]=in+'0'+7;
        }
        out=(num+out)/N;
        i++;
    }
    strcpy(A,B);
    time++;
    return time;
}

int main(){
    int N;
    char M[10001]={'\n'};
    int time=0,l=0,yes=0;
    scanf("%d",&N);
    getchar();
    scanf("%s",M);
    while(time<30){
        time=Cal(M,time,N);
        l=strlen(M);
        for(int i=0;i<(l/2);i++){
            if(M[i]!=M[l-1-i]){
                yes=1;
                break;
            }
        }
        if(yes==0){
            printf("STEP=%d",time);
            break;
        }
        if(time==30&&yes==1){
            printf("Impossible!");
            break;
        }
        yes=0;
    }
}

#include <iostream>
#include <algorithm>
using namespace std;
int n;
char hexx[6]={'A','B','C','D','E','F'};
bool huiWen(string t){
  for(int i=0;i<t.size()/2;++i)
    if(t[i]!=t[t.size()-i-1]) return false;
  return true;
}
string add(string t1,string t2){
  string res(t1.size(),0); //初始化长度
  int jw=0,t=0;
  for(int i=t1.size()-1;i>=0;--i){
    if(n<=10){
      t=t1[i]-'0'+t2[i]-'0'+jw;
      res[i]=t%n+'0';
      jw=t/n;
    }else{
      t=0;
      if(t1[i]>='0'&&t1[i]<='9') t+=t1[i]-'0';
      else t+=t1[i]-'A'+10;
      if(t2[i]>='0'&&t2[i]<='9') t+=t2[i]-'0';
      else t+=t2[i]-'A'+10;
      t+=jw;
      if(t%n>=10) res[i]=hexx[t%n-10];
      else res[i]=t%n+'0';
      jw=t/n;
    }
  }
  if(jw>0) res=(char)(jw+'0')+res; //jw<=3
  return res;
}
int main()
{
  cin>>n;
  string s,s0; cin>>s; s0=s;
  int cnt=0;
  if(huiWen(s)) {cout<<"0"; return 0;}
  while(1){
    reverse(s0.begin(),s0.end());
    s0=s=add(s0,s);
    ++cnt;
    if(huiWen(s)) break;
    if(cnt>30) break;
  }
  if(cnt>30) cout<<"Impossible!";
  else cout<<"STEP="<<cnt;
    return 0;
}

不要用字符串存数字
开个数组存当前数字
至于加法，新开一个数组倒着存
由于是加法，所以只可能进1，那么一个if就好啦（虽然在这里体现不出优越性，但%的确要慢一点），特判一下，如果当前数位大于等于进制，就把下一位++，然后当前位-=n
```cpp
#include<bits/stdc++.h>
using namespace std;

int n, tot;
int num[100000];

inline void Reverse(){
int temp[100000];
memset(temp,0,sizeof(temp));
for(register int i=1; i<=tot; i++){
temp[i] += num[i]+num[tot-i+1];
if(temp[i]>=n){
temp[i+1] ++;
temp[i] -= n;
}
}
if(temp[tot+1]) tot++;
for(register int i=1; i<=tot; i++)
num[i] = temp[tot-i+1];
}

inline bool Check(){
for(register int i=1; i<=tot/2; i++)
if(num[i]!=num[tot-i+1])
return false;
return true;
}

int main(){
int step=0;
char ch;
scanf("%d\n", &n);
while((ch=getchar())!=-1){
if(isdigit(ch))
num[++tot]=ch-'0';
else if(ch>='A'&&ch<='F')
num[++tot]=ch-'A'+10;
}
//for(register int i=1; i<=tot; i++) printf("%d", num[i]);
while(step<=30){
step++;
Reverse();
//for(register int i=1; i<=tot; i++)
//printf("%d", num[i]);
//putchar('\n');
if(Check()){
printf("STEP=%d\n", step);
return 0;
}
}
printf("Impossible!\n");
return 0;
}
```

解释在代码里了，用函数做模拟真的爽

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int n;//n进制

int judge(string a)  //判断是不是回文串
{
    string t=a;
    reverse(a.begin(),a.end());
    for(int i=0;i<a.size();i++)
    {
        if(a[i]!=t[i])
            return 0;
    }
    return 1;
}

string sum(string a)//n在[2，,10]之间的
{
    string t=a;
    reverse(a.begin(),a.end());
    string ans;
    int add=0;
    for(int i=a.size()-1;i>=0;i--)
    {
        char temp= (  ( (a[i]-'0') + (t[i]-'0') + add )%n  )+'0';
        ans.insert(ans.end(),temp);
        add=( (a[i]-'0') + (t[i]-'0') + add )  / n ;
    }
    if(add!=0)
        ans.insert(ans.end(),(char)(add+'0'));
    return ans;
}

int turn(char c)
{
    if(c>='A'&&c<='F')
        return c-'A'+10;
    else
        return c-'0';
}

char temp_turn(int key)
{
    if(key>=10)
    {
        char c=(char)(key-10+'A');
        return c;
    }
    else
        return (char)(key+'0');

}

string sum_sixteen(string a)//16进制的加法
{
    string t=a;
    reverse(a.begin(),a.end());
    string ans;
    int add=0;
    for(int i=a.size()-1;i>=0;i--)
    {
        char temp= temp_turn( (turn(a[i])+turn(t[i])+add)%16 ) ;
        ans.insert(ans.end(),temp);
        add=( turn(a[i])+turn(t[i])+add ) /16 ;
    }
    if(add!=0)
        ans.insert(ans.end(),temp_turn(add));
    return ans;
}

int main()
{
    string a;
    int ans_sum=1;
    cin>>n;
    cin>>a;
    if(n==16)
    {
        int tag=0;
        while(tag!=1&&ans_sum<=30)
        {
            a=sum_sixteen(a);
            if(judge(a)==0)
                ans_sum++;
            else
            {
                tag=1;
                break;
            }
        }
        if(tag==0)
            cout<<"Impossible!"<<endl;
        else
            cout<<"STEP="<<ans_sum<<endl;
    }
    else
    {
        int tag=0;
        while(tag!=1&&ans_sum<=30)
        {
            a=sum(a);
            if(judge(a)==0)
                ans_sum++;
            else
            {
                tag=1;
                break;
            }
        }
        if(tag==0)
            cout<<"Impossible!"<<endl;
        else
            cout<<"STEP="<<ans_sum<<endl;
    }

    return 0;
}

def palindrome(m, n):
    for i in range(30):
        m = to_dec(m, n) + to_dec(m[::-1], n)
        m = to_n(str(m), n)
        if m == m[::-1]:
            return 'STEP=%d' % (i+1)
    return 'Impossible!'
def to_dec(m, n):
    return int(m.upper(), n)
def to_n(m, n):
    num = int(m)
    if num == 0:
        return '0'
    result = ''
    while True:
        if num == 0:
            break
        remainder = num%n
        if remainder >= 10:
            remainder = chr(remainder-10+ord('A'))
        else:
            remainder = str(remainder)
        result += remainder
        num /= n
    return result[::-1].lstrip('0')

n = int(raw_input())
m = raw_input()
print palindrome(m, n)

新人第一道题, 因为内存给的不够和语文不好搞了一个Runtime Error和WA......
看到题解大佬们各种高端的解法, 感觉我这个有点too young too simple了

#include <iostream>

using namespace std;

long toDec(char s[], long d){
    long ret = 0;
    for(long i = 0; s[i] != '\0'; i++){
        if(s[i] <= '9' && s[i] >= '0')
            ret = ret * d + s[i] - '0';
        else
            ret = ret * d + s[i] - 'A' + 10;
    }
    return ret;
}

void Decto(long o, char s[], long d){
    long t = 0, i = 0;
    for(;o != 0; i++){
        t = o % d;
        if(t <= 9)
            s[i] = t + '0';
        else
            s[i] = t - 10 + 'A';
        o /= d;
    }
    s[i + 1] = '\0';
}

bool Check(char s[]){
    long i = 0;
    for(;s[i]!='\0';i++);
    for(long j = 0; j < i/2; j++)
        if(s[j] != s[i - j - 1])
            return false;
    return true;
}

void Flip(char s[]){
    long i = 0;
    for(;s[i]!='\0';i++);
    for(long j = 0; j < i/2; j++){
        char c = s[j];
        s[j] = s[i - j - 1];
        s[i - j - 1] = c;
    }
}

int main(int argc, char **argv) {
    char input[32] = { '\0' };
    char buff[32] = { '\0' };
    long b = 0;
    cin >> b >> input;
    long i = 0;
    long now = 0;
    if(Check(input)){
        cout << "STEP=0" << endl;
        return 0;
    }
    
    for(long i = 0;input[i]!='\0';i++)
        buff[i] = input[i];
    
    for(; i<=30;){
        i++;
        now = toDec(buff, b);
        Flip(buff);
        now += toDec(buff, b);
        Decto(now, buff, b);
        if(Check(buff)){
            cout << "STEP=" << i << endl;
            return 0;
        }
        Flip(buff);
    }
    cout << "Impossible!" << endl;     //一开始输出是STEP=Impossible!, 智障如我
    return 0;
}

初次接触oi, 萌新烦请(女装)大佬赐教.

#include<stdio.h>
#include<string.h>
#include<math.h>
void add_num(char result[],char num[], char addto[], int system)
{
    int numlen = strlen(num);
    int addtolen = strlen(addto);
    int  i = 0;
    int j = 0;
    int t = 0;
    int num1 = 0, num2 = 0;
    int re = 0;
    char temp;
    int carry = 0;
    if (system == 16)
    {
        for (i = numlen - 1; i >= 0; i--)
        {
            if (num[i] >= 'A' && num[i] <= 'F') num1 = (num[i] - 'A') + 10;
            else num1 = num[i] - '0';
            if (addto[i] >= 'A' && addto[i] <= 'F') num2 = (addto[i] - 'A') + 10;
            else num2 = addto[i] - '0';
            re = (carry + num1 + num2) % system ;
            if (re >= 10)
                result[t++] = 'A' + re - 10;
            else
                result[t++] =  re+'0';
            carry = (carry + num1 + num2) / system;
        }
        if (carry != 0)
            result[t] += result[t] + carry + '0';

    }
    else
    {

        for (i = numlen - 1; i >= 0; i--)
        {
            result[t++] = (carry + (num[i] - '0') + (addto[i] - '0')) % system + '0';
            carry = (carry + (num[i] - '0') + (addto[i] - '0')) / system;

        }
        if (carry != 0)
            result[t] += result[t] + carry + '0';
    }
    

    i = 0; j = strlen(result)-1;
    while (i < j)
    {
        t = result[i];
        result[i] = result[j];
        result[j] = t;
        i++;
        j--;
    }
    
    //printf("%s\n", result);

}
void change(char addto[], char num[])
{
    int i = 0, j = 0;
    for (i = 0, j = strlen(num) - 1; i < strlen(num); i++, j--)
        addto[i] = num[j];
}
int ispalindrome(char result[])
{
    int i = 0, j = 0;
    i = 0; j = strlen(result) - 1;
    while (i < j)
    {
        if (result[i] != result[j])
            return 0;       
        i++;
        j--;
    }
    return 1;
}

int main()
{
    //freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
    int N = 0;
    int i = 0;
    char num[500];
    char addto[500];
    char result[500];
    scanf("%d", &N);
    scanf("%s", num);
    

    memset(result, 0, sizeof(result));
    memset(addto, 0, sizeof(addto));

    
    for (i = 0; i < 30; i++)
    {
        change(addto, num);
        add_num(result, num, addto, N);
        //printf("%s\n", result);
        if (ispalindrome(result) == 1)
        {
            printf("STEP=%d\n", i+1);
            return 0;
        }
        else
        {
            strcpy(num, result);
            memset(result, 0, sizeof(result));
            memset(addto, 0, sizeof(addto));        
        }
    }
    printf("Impossible!\n");
    return 0;
}

//新手强答一波，模拟高精
#include <bits/stdc++.h>
const int MAXN = 100;

using std::cin;
using std::cout;
using std::endl;
using std::string;
using std::ostream;

class BigInteger
{
private:
    int data[MAXN];
    int length;
public:
    BigInteger(string str);
    BigInteger(const BigInteger & s);
    BigInteger(int arr[], int _length);

    friend BigInteger add(const BigInteger & x, const BigInteger & y, const int N);
    friend BigInteger reverse(const BigInteger & x);
    friend bool check(const BigInteger & x);
    friend ostream & operator<<(ostream & os,const BigInteger x);
    
};

BigInteger::BigInteger(string str)
{
    length = str.size();
    for(int i = length - 1; i >= 0; i--)
        if(str[i] >= '0' && str[i] <= '9')
          data[length - i - 1] = str[i] - '0';
        else if(str[i] >= 'A' && str[i] <= 'Z')
          data[length - i - 1] = str[i] - 55;
}

BigInteger::BigInteger(const BigInteger & s)
    {
    length = s.length;
    for(int i = 0; i < length; i++)
        data[i] = s.data[i];
    }

BigInteger::BigInteger(int arr[], int _length)
    {
     length = _length;
     for(int i = 0; i < length; i++)
        data[i] = arr[i];
    }

BigInteger add(const BigInteger & x, const BigInteger & y, const int N)
   {
        int arr[MAXN] = {0}, _length = (x.length > y.length) ? x.length : y.length;
    
        for(int i = 0; i < _length; i++)
            arr[i] = x.data[i] + y.data[i];
        for(int i = 0; i < _length; i++)
        { 
           arr[i + 1] += arr[i] / N;
           arr[i] %= N;
        }
       if(arr[_length] != 0) _length++; 

       BigInteger tmp(arr, _length);
       return tmp; 
    }

BigInteger reverse(const BigInteger & x)
    {
        int arr[MAXN],_length;
        _length = x.length;
        for(int i = 0; i < _length; i++)
            arr[i] = x.data[x.length - i - 1];
        BigInteger tmp(arr, _length);
        return tmp; 
    }

bool check(const BigInteger & x)
    {
        for(int i = 0; i < x.length/2; i++)
            if(x.data[i] != x.data[x.length - i - 1])
               return false;
        return true;
    } 

ostream & operator<<(ostream & os,const BigInteger x)
    {
        for(int i = x.length - 1; i >= 0; i--)
            cout << x.data[i]; 
        return os;   
    }  


    int main()
    { 
        int N, count = 1;
        cin >> N;
        string a;
        cin >> a;
        BigInteger A(a);
        while(count <= 30)
        {
            BigInteger tmp = add(A, reverse(A), N);
            if(check(tmp))
                {
                    cout << "STEP=" << count << endl;
                    return 0;
                }

            count++;
            A = tmp;    
        }
        cout << "Impossible!" << endl;
        return 0;
    }

#include <iostream>
#include <cstdio>
#include <string.h>
#include <cmath>
using namespace std;
const int maxn = 10100;
char a[maxn];
int res[maxn];
int t[maxn];
bool fun(int *t,int n){ //判断是否是回文
for(int i=0;i<n/2;i++){
if(t[i]!=t[n-i-1])
return false;
}
return true;
}
int main(){
int n,i;
scanf("%d",&n);
scanf("%s",a);
for(i=0; i<strlen(a); ++i){ //char类型转换成int
if(a[i]<='9'&&a[i]>='0'){
res[i] = a[i] - '0';
}else{
res[i] = a[i] - 'A' + 10;
}
}
bool flag = false;
for(int j=0;j<=30;j++){
memset(t,0,sizeof(t));
if(fun(res,i)){
printf("STEP=%d\n",j);
flag = true;break;
}else{
for(int k=0;k<i;k++){
t[k] += (res[k] + res[i-k-1]);
t[k+1] = t[k] / n;
t[k] %= n;
}
if(t[i]!=0)
i++;
for(int k=0;k<i;k++){
res[k] = t[k];
}
}
}
if(flag == false)
printf("Impossible!\n");
return 0;
}

#include <stdio.h>

int slong(char *a)
{
int i=99;
while(1){
if(a[i]=='\0') i--;
if(a[i]!='\0') break;
}return (i+1);
}
int huiwen(char *a)
{
int i,k;
k=slong(a);///printf("k=%dk",k);
for(i=0;i<k;i++)
if(a[i]!=a[k-i-1]) return 0;
return 1;
}
int main()
{
int N;
scanf("%d",&N);
if(N!=16){
int i,k,step=0;
char a[100],b[100];
for (i=0;i<100;i++) a[i]='\0';
scanf("%s",a);k=slong(a);for (i=0;i<k;i++) a[i]=a[i]-'0';
//printf("%d",huiwen(a));
while(1)
{
if (huiwen(a)==1) break;
if (step>=30) break;
k=slong(a);
for(i=0;i<k;i++) b[i]=a[i]+a[k-i-1];
for(i=0;i<k;i++) {a[i]=b[i];b[i]=0;}
for(i=0;i<k;i++)
{
a[i+1]+=a[i]/N;
a[i]=a[i]%N;
}
step++;
}
if(step<30) printf("STEP=%d",step);
else printf("Impossible!");
}
if(N==16){
int i,k,step=0;
char a[100],b[100];
for (i=0;i<100;i++) a[i]='\0';
scanf("%s",a);k=slong(a);
for (i=0;i<k;i++) {
if(a[i]>='0'&&a[i]<='9')
a[i]=a[i]-'0';
else a[i]=a[i]-'A'+10;
}
//printf("%d",huiwen(a));
while(1)
{
if (huiwen(a)==1) break;
if (step>=30) break;
k=slong(a);
for(i=0;i<k;i++) b[i]=a[i]+a[k-i-1];
for(i=0;i<k;i++) {a[i]=b[i];b[i]=0;}
for(i=0;i<k;i++)
{
a[i+1]+=a[i]/N;
a[i]=a[i]%N;
}
step++;
}
if(step<30) printf("STEP=%d",step);
else printf("Impossible!");
}
return 0;
}

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
using namespace std;
int a[1000000]={0},c[1000000],p,tot=0,ta=1,t1=1;
string n;
int power(int m[],int k)//请无视这个函数名..做别的题忘了改
{
int l=0;
memset(c,0,sizeof(c));
for(int j=1;j<=ta;++j)
{
c[j]=m[ta-j+1]+m[j];
}
for(int t=1;t<=ta+1;t++)
if(c[t]>=k)
{
c[t+1]+=c[t]/k;
c[t]%=k;
}
ta=ta+1;
while(1)
{
if(c[ta]==0)ta--;
else break;
}
for(int t=ta;t>=1;--t)
{
if(c[t]==c[ta-t+1])l++;
m[t]=c[t];
}
if(l==ta)return 1;
else return 0;
}
void get(string z,int m[])
{
int l1=z.length();
for(int i=1;i<=l1;++i)
{
switch(z[l1-i])//要注意16进制时读入高精度要进行判断
{
case 'A':{m[i]=10;break;}
case 'B':{m[i]=11;break;}
case 'C':{m[i]=12;break;}
case 'D':{m[i]=13;break;}
case 'E':{m[i]=14;break;}
case 'F':{m[i]=15;break;}
default:m[i]=z[l1-i]-'0';
}
}
}
int main()
{
cin>>p>>n;
get(n,a);
ta=n.length();
do
{
tot++;
if(tot==30){cout<<"Impossible!"<<endl;return 0;}
}while(power(a,p)==0);
cout<<"STEP="<<tot;
return 0;
}

忘了注释掉调试用语句了，错了两次poi~
~~~C++
#include <iostream>
#include <string>
using namespace std;

int string_into_int(char a)
{
if ((48 <= a + 0) && (a + 0 <= 57)) {
return (a - 48);
}
if ((65 <= a + 0) && (a + 0 <= 90)) {
return (a - 55);
}
}

bool hw(string str)
{
string rst;
for (auto f : str) {
rst = f + rst;
}
if (rst == str){
return true;
}
else { return false; }
}

int main()
{
string s = "0123456789ABCDEF";
string s1, s2, result;
int n, count=0;
while (cin >> n >> s1) {
count = 0;
result = s1;
while (hw(result) == false){
if (count == 30) { cout << "Impossible!" << endl; break; }
count += 1;
if (result == "result") { result = ""; }
else { s1 = result; result = ""; }
for (auto f : s1) {
s2 = f + s2;
}
int plusone = 0;
for (int i = s1.size() - 1; i != -1; --i) {
int res = string_into_int(s1[i]) + string_into_int(s2[i]) + plusone;
plusone = res / n;
result = s[res - plusone * n] + result;
}
if (plusone == 1) {
result = '1' + result;
plusone = 0;
}
//cout << result << endl;
}
if (count != 30) {cout << "STEP=" << count << endl; }
}
return 0;
}
~~~

其实在判断16进制时并不用转换成'A'....而用数组的特性就可以完成,需要注意输入时字母的情况（本人就错在这）
代码：
var
m,a:array[0..1000] of longint;
n,i,j,jw:longint;
ch:char;
function check:boolean;
var
i:longint;
begin
for i:=1 to m[0] div 2 do
if m[i]<>m[1+m[0]-i] then
exit(false);
exit(true);
end;
begin
readln(n);
while not eoln do
begin
inc(m[0]);
read(ch);
if ch in ['A'..'Z'] then m[m[0]]:=ord(ch)-55
else m[m[0]]:=ord(ch)-48;
end;
for i:=1 to 30 do
begin
jw:=0;
for j:=1 to m[0] do
begin
a[j]:=m[j]+m[1+m[0]-j]+jw;
jw:=a[j] div n;
a[j]:=a[j] mod n;
end;
a[0]:=m[0];
if jw>0 then
begin
inc(a[0]);
a[a[0]]:=jw;
end;
m:=a;
if check then
begin
writeln('STEP=',i);
halt;
end;
end;
writeln('Impossible!');
end.