POJ 1426

B - Find The Multiple

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

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

这个题又是用暴力做的，感觉自己做题就是在拼智力，真正的算法基本不会，好苦恼，不过一定会加强学习哒~做题的过程和做会一道题的感觉都是很好的。

把这个代码记录下来主要是想记录一下找只有10组成的数的过程，是模仿二进制进位的过程。

#include<stdio.h>
#include<string.h>
int main()
{
    int a[30];
    int n,i,m = 1;
    long long int s = 1;
    scanf("%d",&n);
    while(n > 0)
    {
        while(s % n != 0)
        {
            for(i = 1;a[i] == 1;i++)
                a[i] = 0;
            a[i] = 1;
            if(i > m)m = i;
            for(s = 0,i = m;i >= 1;i--)
            {
                s =s * 10 + a[i]; 
            }
        }
        printf("%lld
",s);
        scanf("%d",&n);
        memset(a,0,30*sizeof(int));
        s = 1;
        m = 1;
    }
    return 0;
}