poj 1426 Find The Multiple ( BFS+同余模定理)

poj 1426 Find The Multiple ( BFS+同余模定理)

Find The Multiple Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 18390 Accepted: 7445 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

同余模定理：

（a*b）%n = （a%n *b%n）%n；

（a+b）%n = （a%n +b%n）%n；

枚舉前14個數字為：1 、10 、11、100、101、110、111、1000、1001、1010、1011、1100、1101、1110。。。

1110%6=（110*10+0）%6=((110%6)*10)%6+(0%6);

mod[x]=(mod[x/2]*10+x%2)%n;

#include
#define N 600000
int mod[N];
int ans[200];
int main()
{
    int i,k,n;
    while(scanf("%d",&n),n)
    {
        mod[1]=1%n; 
        for(i=2;mod[i-1]!=0;i++)
        {                    
            mod[i]=(mod[i/2]*10+i%2)%n;
        }
        i--;
        k=0;
        while(i)
        {
            ans[k++]=i%2;
            i/=2;
        }
        for(i=k-1;i>=0;i--)
            printf("%d",ans[i]);
        puts("");
    }
    return 0;
}