zoj 1530 Find The Multiple

 

Find The Multiple

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Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge

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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 (0) 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

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Source: Asia 2002, Dhaka (Bengal)

 

题意：给你一个数n，让你找n的倍数m，m的十进制中只能有1和0，问最小的m是几。

思路：若没有是n的倍数的条件，如何构造01序列？开一个队列，先把1放进去。然后取出1，把1*10,1*10+1放进去，以此类推是n的倍数只需要判断一下即可，但答案可能爆long logn，所以用同余定理 若a%b=c，那么ax%b=cx%b （x！=0），用字符数组记录01序列，同时记录这个01序列%n的余数。还有一个问题，空间消耗巨大，还是用同余定理：如果a%n=c，b%n=c，a<b，那么这个时候b就不用入队了，因为答案只看余数

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct nond{
    string s;
    int mod;
}cur,net;
int n;
bool vis[1000001];
int main(){
    while(1){
        queue<nond>que;
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        if(n==0)    break;
        cur.s="1";cur.mod=1;
        que.push(cur);
        while(!que.empty()){
            cur=que.front();
            que.pop();
            net.mod=(cur.mod*10)%n;
            net.s=cur.s+'0';
            if(net.mod==0){
                cout<<net.s<<endl;
                break;
            }
            if(!vis[net.mod]){
                que.push(net);
                vis[net.mod]=1;
            }
            net.mod=(cur.mod*10+1)%n;
            net.s=cur.s+'1';
            if(net.mod==0){
                cout<<net.s<<endl;
                break;
            }
            if(!vis[net.mod]){
                que.push(net);
                vis[net.mod]=1;
            }
        }
    }
    
}