Project Euler：Problem 38 Pandigital multiples

最新推荐文章于 2019-05-07 16:11:49 发布

要个男盆友扭蛋

最新推荐文章于 2019-05-07 16:11:49 发布

阅读量619

点赞数

CC 4.0 BY-SA版权

分类专栏： C++ 欧拉计划文章标签： C++ Project Euler

本文链接：https://blog.youkuaiyun.com/youb11/article/details/46357637

C++ 同时被 2 个专栏收录

89 篇文章

订阅专栏

欧拉计划

85 篇文章

订阅专栏

本文探讨了如何通过编程找出能形成最大1到9全数字9位数的整数与其连续正整数倍数的拼接乘积。利用C++实现了一个算法，该算法能够有效地搜索出满足条件的最大数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Take the number 192 and multiply it by each of 1, 2, and 3:

192 × 1 = 192
192 × 2 = 384
192 × 3 = 576

By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)

The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).

What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?

#include <iostream>
#include <string>
#include <map>
using namespace std;

bool pan_mul(int a, int b,int &c)
{
	c = 0;
	map<int, int>mp;
	for (int i = 1; i <= b; i++)
	{
		int tmp = a*i;
		int num = tmp;
		int count = 0;
		while (tmp)
		{
			if (mp[tmp % 10] != 0)
				return false;
			mp[tmp % 10]++;
			tmp /= 10;
			count++;
		}
		if (mp[0] != 0)
			return false;
		if (i == 1)
			c += num;
		else
			c = c*pow(10, count) + num;
	}
	if (mp[0] != 0)
		return false;
	if (mp.size() == 10)
		return true;
	else
		return false;
}

int main()
{

	int maxn = 0;
	for (int i = 1; i <= 10000; i++)
	{
		for (int j = 2; j <= 15; j++)
		{
			int c = 0;
			if (pan_mul(i, j, c))
			{
				if (c > maxn)
				{
					//cout << i << " " << j << " " << c << endl;
					maxn = c;
				}
			}
		}
	}
	cout << maxn << endl;
	system("pause");
	return 0;
}