#include <bits/stdc++.h>
#define int long long
using namespace std;
int n,cnt;
string add(string x,string y){
	int a[10010] = {},b[10010] = {},lenx = x.size(),leny = y.size();
	for(int i = lenx - 1 ; i >= 0 ; i --){
		if(x[lenx-i-1]<='9')a[i]=x[lenx-i-1]-'0';
		else a[i]=x[lenx-i-1]-'A'+10;
	}
	for(int i = leny - 1 ; i >= 0 ; i --){
		if(y[leny-i-1]<='9')b[i]=y[leny-i-1]-'0';
		else b[i]=y[leny-i-1]-'A'+10;
	}
	int len = max(lenx,leny);
	for(int i = 0 ; i < len ; i ++){
		a[i] += b[i];
		a[i+1]+=a[i]/n;
		a[i] %= n;
	} 
	if(a[len])len ++;
	string res = "";
	for(int i = len - 1; i >= 0 ; i --){
		if(a[i]<=9)res+=a[i]+'0';
		else res+=a[i]-10+'A';
	}
	if(res.size() == 0)return "0";
	return res;
}
bool check(string s){
	string x=s;
	reverse(s.begin(),s.end());
	return x==s;
}

signed main() {
	string m,x;
	cin >> n>>m;
	while(cnt<=30){
		if(check(m)){
			cout <<cnt;
			return 0;
		}
		x=m;
		reverse(m.begin(),m.end());
		m=add(x,m);
		cnt++;
	}
	cout <<"Impossible";
    return 0;
}

高精度加法（模板）

string add(string x,string y){
	int a[1010] = {},b[1010] = {},lenx = x.size(),leny = y.size();
	for(int i = lenx - 1 ; i >= 0 ; i --)a[i] = x[lenx - i - 1] - '0';
	for(int i = leny - 1 ; i >= 0 ; i --)b[i] = y[leny - i - 1] - '0';
	int len = max(lenx,leny);
	for(int i = 0 ; i < len ; i ++){
		a[i] += b[i];
		a[i + 1] += a[i] / 10;
		a[i] %= 10;
	} 
	if(a[len])len ++;
	string res = "";
	for(int i = len - 1; i >= 0 ; i --)res += a[i] + '0';
	if(res.size() == 0)return "0";
	return res;
}

高精度乘法（模板）

string mul(string x,string y){
	int a[2010] = {},b[2010] = {},c[4010] = {},lenx = x.size(),leny = y.size();
	for(int i = 0 ; i < lenx ; i ++)a[i] = x[lenx - i - 1] - '0';
	for(int j = 0 ; j < leny ; j ++)b[j] = y[leny - j - 1] - '0';
	for(int i = 0 ; i < lenx ; i ++){
		for(int j = 0 ; j < leny ;j ++){
			c[i + j] += a[i] * b[j];
			c[i + j + 1] += c[i + j] / 10;
			c[i + j] %= 10;
		}
	}
	int len = lenx + leny;
	while(len > 1 && c[len - 1] == 0)len -- ;
	string res = "";
	for(int i = len - 1; i >= 0 ;i --)res += c[i] + '0';
	return res;
}

高精度减法（模板）

string sub(string x,string y){//计算两个高精度数字加法的结果 
	string ans = "";
	if(x.size() < y.size() || x.size() == y.size() && x < y){
		swap(x,y);
		ans += "-";
	} 
	int a[101000] = {},b[101000] = {},lenx = x.size(),leny = y.size();
	//x[lenx - 1 - i] - '0'是表示把字符转换成数字 
	for(int i = 0 ; i < lenx ; i ++)a[i] = x[lenx - 1 - i] - '0';//倒序存储
	for(int i = 0 ; i < leny ; i ++)b[i] = y[leny - 1 - i] - '0';//倒序存储
	int len = max(lenx,leny);//取长度更长的数字长度
	for(int i = 0 ;i < len ; i ++){
		a[i] -= b[i];
		if(a[i] < 0){
			a[i] += 10;
			a[i + 1] --;
		}
	} 
	if(a[len] == 1)len ++;//判断最高位有没有进位，有的话那么数字长度变长
	for(int i = len - 1 ; i >= 0; i -- )ans += a[i] + '0';
	return ans; 
}

高精度除法（高精除以低精度）（模板）

struct node{
	string s;
	int res;
};

node divd(string a,int b){
	node ans = {"",0};
	for(int i = 0 ; i < a.size() ;i ++){
		ans.res = ans.res * 10 + a[i] - '0';//余数和下一位结合 
		ans.s += ans.res / b + '0';//计算每一位商 
		ans.res %= b;//重新计算余数 
	}
	while(ans.s[0] == '0')ans.s = ans.s.substr(1);
	if(ans.s == "")ans.s = "0";
	return ans;
}

二分函数：

upper_bound();
lower_bound();

L2单词的长度

#include <iostream>
#include <string>
using namespace std; 
int main() {
    string text;
    getline(cin,text);
    int start = 0;
    int end;
    while ((end = text.find(' ', start)) != string::npos) {
        string word = text.substr(start, end - start);
        if(word.length() != 0)
           cout << word.length() << "," ;
        start = end + 1;
    }
    if (start < text.length()) {
        string word = text.substr(start);
        if(word.length() != 0)
           cout << word.length() << endl;
    }
 
    return 0;
}```cpp

Gesp大纲

https://gesp.ccf.org.cn/101/attach/1579675000242208.pdf

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