E - Find The Multiple

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给定正整数n，编写程序找到一个非零的n的倍数，其十进制表示仅包含0和1。可以假设n不大于200，并且存在一个不超过100位十进制数的对应m。输入文件可能包含多个测试用例，每行包含一个n的值（1 <= n <= 200）。以零结尾的行结束输入。对于输入中的每个n值，打印相应的m值。m的十进制表示不应超过100位。如果给定的n有多个解，接受任何一个。

https://vjudge.net/contest/402204#problem/E

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

Sample Output

10
100100100100100100
111111111111111111

Sponsor

代码：

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include<time.h>
using namespace std;
#define ll long long
const int mod=1e9+7;
const int maxn=2e2+10;
ll n,m,s;
queue<ll>q;
void bfs()
{
    while(!q.empty()) q.pop();
    q.push(1);
    while(1)
    {
        ll x=q.front();
        q.pop();
        if(x%n==0)
        {
            cout<<x<<endl;
            break;
        }
        q.push(x*10);
        q.push(x*10+1);
    }
}
int main()
{
    while(scanf("%lld",&n))
    {
        if(n==0)
            break;
        bfs();
    }
    return 0;
}