poj-1426 Find The Multiple-蒲公英云

poj-1426 Find The Multiple

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

题意：找一个大于等于N的数，使这个数每一位只有0和1，且能被N整除

同余模定理：找一个大于等于N的数，使这个数每一位只有0和1，且能被N整除

代码：1

#include<stdio.h>
#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;
}
代码：2
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,flag;
void f( unsigned __int64 x,int n,int k )
{
    if(flag)
        return ;
    if(x%n==0)
    {
        printf("%I64u\n",x);
        flag=1;
        return ;
    }
    if(k==19)
        return ;
    f(x*10,n,k+1);
    f(x*10+1,n,k+1);
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            break;
        flag=0;
        f(1,n,0);
    }
}