本蒟蒻写了快速幂 , 带高精乘 , TLE了5个点 , 测试信息

代码:

#include<algorithm>
#include<iostream>
#include<vector>
#include<cmath>
using namespace std;
int A[100010],B[100010],C[100010];
struct vli  
{
	vector <char> number;
	bool minis_zero = false;
	void Swap(vli b)
	{
		vector <char> n = b.number;
		b.number = number;
		number = n;
		bool x = b.minis_zero;
		b.minis_zero = minis_zero;
		minis_zero = x;
	}
	void get_string(string s)
	{
		if(s[0] == '-')
		{
			minis_zero = true;
			s.erase(s.begin() + 0);
		}
		number.clear();
		for(int i = 0;i < s.size();i++)
			number.push_back(s[i]);
	}
	void show()
	{
		for(int i = 0;i < number.size();i++)
			cout << number[i];
		cout << '\n';
	}
};
vli operator-(vli a,int b)
{
	vli ans;
	int lena = a.number.size();
	for(int i = lena - 1;i >= 0;i--)
		A[lena - i] = a.number[i] - '0';
	A[1] -= b;
	for(int i = 1;i <= lena;i++)
	    if(A[i] < 0)
	        A[i] += 10,A[i + 1]--;
	for(;A[lena] == 0;)
	    lena--;
	for(int i = lena;i >= 1;i--)
	    ans.number.push_back(A[i] + '0');
	for(int i = 0;i < 1010;i++)
		A[i] = 0,C[i] = 0;
	return ans;
}
vli operator*(vli a,vli b)
{
	vli ans;
	int lena = a.number.size(),lenb = b.number.size(),len = lena + lenb;
	for(int i = 0;i < lena;i++)
		A[lena - i] = a.number[i] - '0';
	for(int i = 0;i < lenb;i++)
		B[lenb - i] = b.number[i] - '0';
	for(int i = 1;i <= lena;i++)
		for(int j = 1;j <= lenb;j++)
			C[i + j - 1] += A[i] * B[j];
	for(int i = 1;i <= len;i++)
	{
		C[i + 1] += (C[i] / 10);
		C[i] %= 10;
	}
	for(;C[len] == 0;)
		len--;
	for(int i = len;i >= 1;i--)
		ans.number.push_back(C[i] + '0');
	for(int i = 0;i < 10010;i++)
		A[i] = B[i] = C[i] = 0;
	return ans;
}
int main()
{
	vli ans,two,s;
	int p,c = 0;
	ans.get_string("1");
	two.get_string("2");
	s = two;
	cin >> p;
	while(p > 0)
	{
		if(p & 1)
			ans = ans * s;
		s = s * s;
		p >>= 1;
	}
	ans = ans - 1;
	cout << ans.number.size() << '\n';
	if(ans.number.size() >= 500)
		for(int i = ans.number.size() - 500;i < ans.number.size();i++)
		{
			c++;
			cout << ans.number[i];
			if(c % 50 == 0)
				cout << '\n';
		}
	else
	{
		for(int i = 1;i <= 500 - ans.number.size();i++)
		{
			c++;
			cout << 0;
			if(c % 50 == 0)
				cout << '\n';
		}
		for(int i = 0;i < ans.number.size();i++)
		{
			c++;
			cout << ans.number[i];
			if(c % 50 == 0)
				cout << '\n';
		}
	}
	return 0;
}

请求各位$dalao$帮个忙QQL