证明存在：http://blog.csdn.net/jcwKyl/article/details/3859155

自己的代码 M取余数能算的数比较大：

int _tmain(int argc, _TCHAR* argv[])
{
	int N;
	// 用栈存储似乎是个错误，费老劲了
	cout<<" input N "<<endl;
	cin>>N;
	stack<int> *bigInt = new stack<int>[N];
	for(int M=1,i=1;;M=M*10%N,i++)
	{
		int reminder = M%N;
		if(reminder==0)
		{
			bigInt[0].push(i);
			cout<<" in if "<<endl;
			break;//return 0;
		}
		else
		{
			cout<<" in else 1 "<<endl;
			if(bigInt[reminder].empty()==1)
				bigInt[reminder].push(i);
			for(int j=1;j<N;j++)// update reminder information
			{
				cout<<" in for 1 "<<endl;
				if(bigInt[j].empty()==0&&bigInt[(j+reminder)%N].empty()==1&&i>bigInt[j].top())
				{
					cout<<" in if in if  1 "<<endl;
					int jlength=bigInt[j].size();
					int *pt=new int[jlength];// pt用于存储从栈里拿出来的元素
					for(int k=0;k<jlength;k++)
					{
						int temp = bigInt[j].top();
						pt[jlength-1-k]=temp;
						bigInt[j].pop();//元素不能出栈 
						//bigInt[(j+reminder)%N].push(temp);
					}
					
					for(int k=0;k<jlength;k++)//保持原来的元素不变
					{
						bigInt[(j+reminder)%N].push(pt[k]);					
						bigInt[j].push(pt[k]);
					}
					bigInt[(j+reminder)%N].push(i);
				}
			}
			if(bigInt[0].empty()==0) break;//return 0;
		}

	}
	bool b[100] ={0};
	//for(int m=0;m<N;m++)
	//{
	int m=0;
		cout<<" in bool "<<endl;
		int len = bigInt[m].size();
		for(int i=0;i<len;i++)
		{
			
			int k = bigInt[m].top();
			cout<<k<<" ";
			b[k]=1;
			bigInt[m].pop();
		}
		cout<<" last "<<endl;
	//}
	for(int i=20;i>=0;i--)
	cout<<b[i]<<" ";
	cout<<endl;
	system("pause");
	//list<int> *li= new list<int>[N];
}

别人的算法： 用一个整数来存储，

#include <stdio.h>  
int iRes[1000];     
int OneAndZeroNum(int N)  
{  
    int i, X;       // X记录目前遍历的十进制位数  
    int num;        // num = 10^X   
    for(i = 0; i < N; i++)  
        iRes[i] = 0;  
    for(X = 0, num = 1; ; X++, num *= 10)  
    {  
        int tempRes = num % N;  
        if(tempRes != 0)    // 如果10^X不能整除N   
        {  
            if(iRes[tempRes] == 0)  
			{
				//cout<<"a X is "<<X<<" iRes[temp] is "<<iRes[tempRes]<<endl;
                iRes[tempRes] = 1 << X;   // 记录为tempRes时，十进制数的位数,先赋值运算 再移位运算
				//cout<<"b X is "<<X<<" iRes[temp] is "<<iRes[tempRes]<<endl;
			}
            for(i = 1; i < N; i++)       // 更新余数对应的十进制数 
                if(iRes[i] != 0 && iRes[(i+tempRes)%N] == 0 && 
                        ((iRes[i] & (1 << X)) == 0)) // 避免错误重复更新   
                    iRes[(i+tempRes)%N] |= (iRes[i] | (1 << X));  
            if(iRes[0] != 0)  
                return iRes[0];  
        }  
        else  
        {  
            iRes[0] = 0;  
            iRes[0] |= 1 << X;  
            return iRes[0];  
        }  
    }  
    return 0;  
}  

int _tmain(int argc, _TCHAR* argv[])
{
	int N;  
    int iBitMapNum, iFactor;  
    int Num;  
    while(1)  
    {  
        scanf("%d", &N);  
        iBitMapNum = OneAndZeroNum(N);  
        iFactor = 1;  
        Num = 0;  
        while(iBitMapNum)  
        {  
            Num += (iBitMapNum & 0x01) * iFactor;  
            iFactor *= 10;  
            iBitMapNum >>= 1;  
        }  
        printf("%d\n", Num);  
    }  
	system("pause");
    return 0;
}