#1 #2 #9 #10 TLE 其余AC

用的快速幂

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int p;
int a[100005];
int b[100005];
int c[100005];
string ans;
string operator *(string sa, string sb) {
	memset(a, 0, sizeof a);
	memset(b, 0, sizeof b);
	memset(c, 0, sizeof c);
	reverse(sa.begin(), sa.end());
	reverse(sb.begin(), sb.end());
	for (int i = 0; i < sa.size(); i++) {
		a[i] = sa[i] - '0';
	}
	for (int i = 0; i < sb.size(); i++) {
		b[i] = sb[i] - '0';
	}
	for (int i = 0; i < sa.size(); i++) {
		for (int j = 0; j < sb.size(); j++) {
			c[i + j] += a[i] * b[j];
		}
	}
	for (int i = 0; i < sa.size() + sb.size(); i++) {
		if (c[i] >= 10) {
			c[i + 1] += c[i] / 10;
			c[i] %= 10;
		}
	}
	int l = sa.size() + sb.size();
	while (c[l] == 0) {
		l--;
	}
	string ret;
	for (int i = l; i >= 0; i--) {
		ret += c[i] + '0';
	}
	return ret;
}
string operator -(string sa, int b) {
	memset(a, 0, sizeof a);
	reverse(sa.begin(), sa.end());
	for (int i = 0; i < sa.size(); i++) {
		a[i] = sa[i] - '0';
	}
	a[0] -= b;
	for (int i = 0; i < sa.size(); i++) {
		if (a[i] < 0) {
			a[i + 1]--;
			a[i] = 9;
		}
	}
	int l = sa.size() + 7;
	while (a[l] == 0) {
		l--;
	}
	string ret;
	for (int i = l; i >= 0; i--) {
		ret += a[i] + '0';
	}
	return ret;
}
string power(string a, int b) {
	string ret = "1";
	while (b != 0) {
		if (b % 2 == 1) {
			ret = ret * a;
		}
		a = a * a;
		b /= 2;
	}
	return ret;
}
int main() {
	cin >> p;
	ans = power("2", p) - 1;
	int len = 500 - ans.size();
	cout << ans.size() << endl;
	for (int i = 1; i <= len; i++) {
		ans = "0" + ans;
	}
	if (ans.size() > 500) {
		ans = ans.substr(ans.size() - 500);
	}
	for (int i = 0; i < ans.size(); i++) {
		cout << ans[i];
		if ((i + 1) % 50 == 0) {
			cout << endl;
		}
	}
	return 0;
}