URAL1495搜索_背包？

Posted 2020-10-31 小螺号打豆豆

1495. One-two, One-two 2

Time limit: 2.0 second

Memory limit: 64 MB

A year ago the famous gangster Vito Maretti woke up in the morning and realized that he was bored of robbing banks of round sums. And for the last year he has been taking from banks sums that have only digits 1 and 2 in their decimal notation. After each robbery, Vito divides the money between N members of his gang. Your task is to determine the minimal stolen sum which is a multiple of N.

Input
The input contains the number N (1 ≤ N ≤ 106).

Output
Output the minimal number which is a multiple of N and whose decimal notation contains only digits 1 and 2. If it contains more than 30 digits or if there are no such numbers, then output "Impossible".

Samples

input:5

output:
Impossible

Input:8

Output:112

题意

给定一个整数n，求n的倍数sum，要求sum的每一位都是1或者2,sum应是最小的，sum不存在或者位数超过30位时输出Impossible

解法

一个公式： (a*b)%c = (a%c)*(b%c)%c
bfs搜索，从低位开始搜索，范围限定在0到 n-1 ， 用string记录每个结果，每个结果对应着一个%n的值，如果这个值=0 那么这个就是最终结果
一开始竟然从先搜索低位！还调了半天+ + 迷
据说可以用背包的思想解，后续补坑。。。

#include <bits/stdc++.h>
using namespace std;
typedef pair<int,string>P;
const int maxn = 1e6+10;
int vis[maxn];
char x[]="012";
int n;
string bfs()
{
    queue<P>q;
    q.push(P(0,""));
    while(!q.empty())
    {
        string now = q.front().second;
        int num = q.front().first;
        q.pop();
        for(int i=1;i<=2;i++)
        {
            int mid = (num*10+i)%n;
            if(mid ==0) return now+x[i];
            if(!vis[mid]&&now.size()<30) q.push(P(mid,now+x[i]));
            vis[mid]=1;
        }
    }
    return " ";
}

int main()
{
    while(cin>>n){
        memset(dp,0,sizeof(dp));
        memset(vis,0,sizeof(vis));
        string fin=bfs();
        if(fin!=" ")cout<<fin<<endl;
        else cout<<"Impossible"<<endl;
    }
    return 0;
}