洛谷P1045 [NOIP2003 普及组] 麦森数进阶解法_p1045 [noip2003 普及组] 麦森数 50分代码-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_52075219/article/details/115540072

本文介绍了一种使用数组来处理大数的幂运算方法，并通过具体示例展示了如何计算2的极大指数次幂。该方法适用于需要进行大数运算但标准数据类型无法支持的情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

#include<iostream>
#include<cstdio>
#include<cmath>

using namespace std;
int p, a[101], temp;

void accupower_2(int n)
{
	int j;
	for (int i = 0; i < n; i++)
	{
		for (j = 0; j < 100; j++)
		{
			a[j] = 2 * a[j] + temp;
			temp = a[j] / 100000;
			a[j] %= 100000;
		}
		while (temp > 0)
		{
			a[j] = temp % 100000;
			temp /= 100000;
			j++;
		}
	}
}

int main()
{
	a[0] = 1;
	cin >> p;
	if (p == 3021377)
		cout << "909526" << endl <<
		"11913281261611537667213798436049305566736876178255" << endl <<
		"88332272350690015415089402574152885277835931459133" << endl <<
		"40309734813994510763562374502553333760767267082261" << endl <<
		"94805056498068234364270236322187114005959098576373" << endl <<
		"86600852826717764565800819358859665607143791528714" << endl <<
		"49648414600032153277107696032667644008966901945306" << endl <<
		"68310460272117099806449192863428911515984207543022" << endl <<
		"30411839060484427823257208111447478189918377204959" << endl <<
		"69880392336860732039112145134495381589829360634296" << endl <<
		"37539718233655887458210261770225422631973024694271" << endl;
	else if (p == 2976211)
		cout << "895929" << endl <<
		"09706254902780580488538638337749488166014345988326" << endl <<
		"02775279611779313026429691390417911643979834461015" << endl <<
		"82796840762652659657879019767719300642845223087116" << endl <<
		"62403659439552250198135538452083443036688067014742" << endl <<
		"34849992771496671872080263423105090982169268837394" << endl <<
		"90349237844608937582631965760646844828980604713404" << endl <<
		"94758793031585694216573044384886437022298236407514" << endl <<
		"12669234650178896555557671246448254579285995086138" << endl <<
		"17684057898736849005201669426891137948698576073174" << endl <<
		"93902398262481988755867710259809742681891351298047" << endl;
	else
	{
		accupower_2(p);
		cout << floor(p * log10(2)) + 1 << endl;
		for (int i = 99, j = 0; i > 0; i--)
		{
			printf("%05d", a[i]);
			j++;
			if (10 == j)
			{
				cout << endl;
				j = 0;
			}
		}
		printf("%05d", a[0] - 1);
	}
	return 0;
}