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
这道题最后的答案可以用__int64或者long long储存，所以可以用队列直接进行宽搜，每次*10或者*10+1；第二种方法是用公式(a*b)%n=((a%n)*(b%n))%n;(a+b)%n=(a%n+b%n)%n;用mod[i]储存第i次的余数。那么mod[i]=(mod[i/2]*10+i%2)%n;
代码1：
#include<stdio.h>
#include<string.h>
int mod[911111],a[110];
int main()
{
int n,i,j,t;
while(scanf("%d",&n)!=EOF && n!=0)
{
if(n==1){
printf("1
");continue;
}
memset(mod,0,sizeof(mod));memset(a,0,sizeof(a));
mod[1]=1;
for(i=2;;i++){
mod[i]=(mod[i/2]*10+i%2)%n;
if(mod[i]==0)break;
}
t=0;
while(i){ //这里发现最后答案就是把i化为二进制的数
a[++t]=i%2;
i=i/2;
}
for(i=t;i>=1;i--){
printf("%d",a[i]);
}
printf("
");
}
return 0;
}
代码2：（直接用宽搜）
#include<stdio.h>
#include<string.h>
#define maxn 10000000
int n;
long long q[maxn];
long long bfs()
{
int front=1,rear=1;
//memset(q,0,sizeof(q));这里不能清空，否则会超时；
q[front]=1;
while(1)
{
long long x=q[front];
front++;
if((x*10)%n==0){
return x*10;
}
if((x*10+1)%n==0){
return x*10+1;
}
q[++rear]=x*10;
q[++rear]=x*10+1;
}
}
int main()
{
int i,j;
while(scanf("%d",&n)!=EOF && n!=0)
{
printf("%lld
",bfs());
}
return 0;
}