#题解#洛谷P1045 麦森数#快速幂#高精度乘法#

P1045 [NOIP 2003 普及组] 麦森数 - 洛谷

分析

1. 2^{p-1的位数，易知：跟2}p的位数相同，为[log10(2)*p]+1

2. 求末尾500位数字：高精度乘法+快速幂

代码

#include<bits/stdc++.h>
#define int  long long
#define endl '\n'
using namespace std;
const int N = 1500;

int ans[N], a[N];
int l = 1;
int len = 1;
int c[N];
void pw(int b)
{
	ans[1] = 1;
	a[1] = 2;

	for (; b; b >>= 1)
	{
		if (b & 1)
		{
			memset(c, 0, sizeof c);
			for (int i = 1; i <= len; i++)
				for (int j = 1; j <= l; j++)
					c[i + j - 1] += ans[i] * a[j];

			len = min(len + l - 1,500ll);

			for (int i = 1; i <= len ; i++)
				c[i + 1] += c[i] / 10, c[i] %= 10;

			while (c[len + 1] and len < 500)
			{
				len = min(500ll, len + 1);
				c[len + 1] += c[len] / 10;
				c[len] %= 10;
			}

			for (int i = 1; i <= len; i++)
				ans[i] = c[i];
		}
		memset(c, 0, sizeof c);
		for (int i = 1; i <= l; i++)
			for (int j = 1; j <= l; j++)
				c[i + j - 1] += a[j] * a[i];

		l = min(2 * l - 1,500ll);

		for (int i = 1; i <= l; i++)
			c[i + 1] += c[i] / 10, c[i] %= 10;

		while (c[l + 1] and l < 500)
		{
			l = min(500ll, l + 1);
			c[l + 1] += c[l] / 10;
			c[l] %= 10;
		}

		for (int i = 1; i <= l; i++)
			a[i] = c[i];
	}
}
signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int p;
	cin >> p;
	pw(p);
	int xx = p * log10(2) +1;
	cout << xx  << endl;
	for (int i = 500; i >= 1; i--)
	{
		if (i == 1)
			cout << ans[i] - 1;
		else
			cout << ans[i];
		if ((500 - i + 1) % 50 == 0)
			cout << endl;
	}
	return 0;
}

错误/Trick总结

1. N开太大memset导致TLE

2. 只需对500位以内进行高精度乘法