Find The Multiple - POJ 1426

Find The Multiple - POJ 1426 - Virtual Judge (vjudge.net)

题意:

给一个n,求一个能被它整除的数只由01组成

思路:

宽搜,先放进去一个1,然后每次把所有的添进去,在后面加一个1或者0,然后有符合要求的就输出并退出。

#include <iostream>
#include <queue>
#define int long long
using namespace std;
const int N = 1e3 + 10;
int n;

void bfs()
{
    
    
        queue<int> q;
        q.push(1);
        while (!q.empty())
        {
    
    
                int t = q.front();
                q.pop();
                int a = t * 10;
                int b = a + 1;
                if (a % n ==0)
                {
    
    
                        cout << a << endl;
                        break;
                }
                if (b % n == 0)
                {
    
    
                        cout << b << endl;
                        break;
                }
                q.push(a);
                q.push(b);
        }
}

signed main()
{
    
    
        while (cin >> n)
        {
    
    
                if (n == 0) break;
                bfs();
        }
        return 0;
}

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Origin blog.csdn.net/qq_34682765/article/details/121892681