Find The Multiple

Time Limit: 1000MS	Memory Limit: 10000K
Total Submissions: 33882	Accepted: 14173	Special Judge

Description

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

題意：給你一個int範圍內的正整數n，輸出能夠整除這個正整數n且由01組成的任意一個數x。
思路：同餘定理的應用，所求x是能整除n的數，但是現在x我們未知，只能一步一步從高位到低位遞推出x。
　　　由於x只能由01序列組成，所以我們每次從x的最高位到低位進行查詢的過程中有兩個選擇，取0或者取1，最高位只能為1。
　　　如果我們用bfs模擬除法運算這個過程，用一個數組mod記錄每次的餘數且每次將mod[i]%10+str[i]-'0'對n取餘的餘數入隊，下一次進行運算時，就能通過mod[i-1]得到
　　　mod[i],不過我自己用bfs超時到爆炸，不知道咋回事，就換成了一位大佬的方法，用陣列模擬bfs。
 

借鑑部落格

#include<stdio.h>
#define N 600000
long long  mod[N],num[N],n;
int j,i;
int main()
{
    while(scanf("%lld",&n),n!=0)
    {
        i = j = 1;
        mod[i++] = 1%n;//初始化x的最高位數 
        while(mod[i-1]!= 0)
        {//mod[i/2]是上一步取餘的結果 
            mod[i] = (mod[i/2]*10+i%2)%n;//模擬bfs雙入口搜尋 
            i++;//解釋：i%2,i為偶數時,模擬+0，i為奇數時,模擬+1 
        }
        i--;
        while(i)
        {
            num[j++] = i%2;
            i/=2;
        }
        while(j > 1)
        {//逆序輸出 
            printf("%d",num[--j]);
        }
        printf("\n");
    }
    return 0;
}