poj1426(dfs)

最新推荐文章于 2022-04-11 18:37:41 发布

原创最新推荐文章于 2022-04-11 18:37:41 发布 · 286 阅读

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本文介绍了一个编程挑战，目标是找出一个正整数n的非零倍数m，且m的十进制表示仅由0和1组成。文章提供了一种深度优先搜索（DFS）的方法来解决这个问题，并附上了AC代码实现。

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Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits.
Input
The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.
Output
For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.
Sample Input
2
6
19
0
Sample Output
10
100100100100100100
111111111111111111
这题第一反应就是用深搜，每增加一位数就就有两中选择，“1和0”，用dfs遍历访问，找到符合的就结束遍历。
但题目说可以假设不超过100位数，但去网上查了题解，才知道答案是不会超过19位数的，所以用longlong不会超。
ac代码

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <math.h>
#include <cstdlib>
#include <queue>
#include<iomanip>
using namespace std;
int dis, m, n;
void dfs(long long m, int step)
{
	int ms = m % n;
	if (dis == 0 || step > 19)return;
	if ( ms == 0 )
	{
		dis = 0;
		cout << m << endl;
		return;
	}
	else if (ms)
	{
		dfs(m * 10, step + 1);
		dfs(m * 10+1, step + 1);
	}
}
int main()
{
	
	while (cin >> n, n)
	{
		dis = 1;
		dfs(1, 1);
	}
}

#include
#include
#include
#include
#include <math.h>
#include
#include
#include
using namespace std;
int dis, m, n;
void dfs(long long m, int step)
{
int ms = m % n;
if (dis == 0 || step > 19)return;
if ( ms == 0 )
{
dis = 0;
cout << m << endl;
return;
}
else if (ms)
{
dfs(m * 10, step + 1);
dfs(m * 10+1, step + 1);
}
}
int main()
{

while (cin >> n, n)
{
	dis = 1;
	dfs(1, 1);
}

}