高精度乘法

#include <iostream>
#include <cmath>
using namespace std;
unsigned long long a[510] = {0, 1}, t;
int p;
int main()
{
    cin >> p;
    cout << (int)(p * log10(2)) + 1 << endl;
    for (; p >= 0; p -= 60)
    {
        t = 0;
        for (int j = 1; j <= 500; j++)
        {
            if (p >= 60) a[j] <<= 60;
            else a[j] <<= p;
            a[j] += t;
            t = a[j] / 10;
            a[j] %= 10;
        }
    }
    a[1] -= 1;
    for (int i = 500; i >= 1; i--) 
    {
        cout << (char)('0' + a[i]);
        if (i % 50 == 1) cout << endl;
    }
    return 0;
}

高精+快速幂即可。

位数是 $\log_{10} 2^n=n\times\log_{10} 2$

#include <bits/stdc++.h>
using namespace std;
const int N = 1005, k = 500;
int n;
struct fig {
	int a[N];
	void print() {
		a[1] -= 1;
		for (int i = 1; i <= k; i ++) {
			cout << a[k - i + 1];
			if (i % 50 == 0)
				cout << '\n';
		}
		cout << '\n';
	}
	fig operator *= (fig x) {
		fig s = *this;
		memset(a, 0, sizeof a);
		for (int i = 1; i <= k; i ++) {
			fig t = s;
			for (int j = k; j >= 1; j --) {
				t.a[j] *= x.a[i];
				t.a[j + 1] += t.a[j] / 10, t.a[j] %= 10;
			}
			for (int j = i; j <= k; j ++) {
				a[j] += t.a[j - i + 1];
				if (a[j] > 9) a[j + 1] += a[j] / 10, a[j] %= 10;
			}
		}
		return *this;
	}
	fig(int x) {
		a[1] = x;
	}
} bas(2), ans(1);
int real() {
	int res = 0;
	for (int i = k; !ans.a[i]; i --)
		res ++;
	return k - res;
}
int main() {
//	freopen("mason.in", "r", stdin);
//	freopen("mason.out", "w", stdout);
	cin >> n;
	cout << int(n * log10(2) + 1) << '\n';
	while(n) {
		if (n & 1) ans *= bas;
		bas *= bas;
		n >>= 1;
	}
	ans.print();
	return 0;
}