POJ 1426 Find The Multiple（数论——中国同余定理）

题目链接：

http://poj.org/problem?id=1426

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
题意描述：
输入n (1 <= n <= 200)
输出一个由0和1组成的数并且能够被n整除
解题思路：
看了题解有了一些思路，用mod[i]=mod[i/2]*10+i%2;这个式子计算二进制i在十进制的形式并存进mod[i]数组,所以使用long long 定义mod数组，很容易理解

#include<stdio.h>   
long long mod[600001];  
int main()  
{  
    int n;  
    while(scanf("%d",&n),n)  
    {  
        int i;  
        for(i=1; mod[i-1]%n != 0;i++)  
            mod[i]=mod[i/2]*10+i%2;// mod[i]表示十进制形式的二进制i  
        printf("%I64d
",mod[i]);  
    }  
    return 0;  
} 

但是延伸一下，如果超过200或者更大呢，请参考

#include<stdio.h>
#define N 600000
int mod[N];
int ans[200];//根据情况增大数组
int main()
{
    int i,k,n,j;
    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;
        //printf("i-1=%d
",i-1);
        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;
}