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:给出一个n,找n的倍数且数字都为0或1.
我刚开始也是这样想的,从1开始找,但不知道怎么找,发现是从1到10和11,10到100和101,11到110和111,,,;
也就是从低位开始两个入口(*10+0)和(*10+1),一直dfs下去,直到找到那个数。那么又一个问题来了,长度太长无法保存,那么就用到模运算了(其实我没想到),用6举例如下:
模运算
(a+b)%n=(a%n+b%n)%n
(a*b)%n=((a%n)*(b%n))%n
if(1%6=1)
{
if((1*10+0)%6=4)
{
if((4*10+0)%6=4)//((((1*10+0)%6)*10+0)%6=4)根据模运算下面同理
{
if((4*10+0)%6=4)
{
}
else if((4*10+1)%6=5)
{
}
}
else if((4*10+1)%6=5)
{
if((5*10+0)%6=2)
{
}
else if((5*10+1)%6=3)
{
}
}
}
else if((1*10+1)%6=5)
{
if((5*10+0)%6=2)
{
if((2*10+0)%6=2)
{
}
else if((2*10+1)%6=3)
{
}
}
else if((5*10+1)%6=3)
{
if((3*10+0)%6=0)
{
得到答案
}
else if((3*10+1)%6=1)
{
}
}
}
}
#include<stdio.h>
#include<string>
#include<string.h>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
int a[500000];
int b[500000];
int main()
{
int n;
while(~scanf("%d",&n)&&n)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
a[1]=1;
int i;
for(i=2;a[i-1]!=0;i++)
a[i]=(10*a[i/2]+i%2)%n;//实在想不到还有这种操作,佩服
i--;
int len=0;
while(i)
{
b[len++]=i%2;
i/=2;
}
for(int i=len-1;i>=0;i--)
{
printf("%d",b[i]);
}
printf("
");
}
return 0;
}