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；
枚举前14个数字为：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;
}