Find The Multiple
| Time Limit: 1000MS | Memory Limit: 10000K | |||
| Total Submissions: 64606 | Accepted: 26229 | 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
Source
一开始以为是一道数学题，做了以后发现似乎并没有什么特别的规律，所以这里其实还是用的二进制枚举，值得注意的是由于空间范围限制的非常小，所以结构体里面存数字用的是bool形式存的。
1 #include <cstdio> 2 #include <cmath> 3 #include <cstring> 4 #include <cstdlib> 5 #include <queue> 6 #include <stack> 7 #include <vector> 8 #include <iostream> 9 #include "algorithm" 10 using namespace std; 11 const int MAX=21; 12 int n; 13 struct Num{ 14 bool num[MAX]; 15 int len; 16 int mod; 17 }zt,tt; 18 queue <Num> q; 19 int main(){ 20 freopen ("multiple.in","r",stdin); 21 freopen ("multiple.out","w",stdout); 22 int i,j; 23 while (scanf("%d",&n),n!=0){ 24 while (!q.empty()) q.pop(); 25 zt.len=0; 26 zt.num[++zt.len]=1;zt.mod=1%n; 27 q.push(zt); 28 while (!q.empty()){ 29 zt=q.front(); 30 if (zt.mod==0) break; 31 q.pop(); 32 tt=zt; 33 tt.num[++tt.len]=1;tt.mod=(tt.mod*10+1)%n; 34 q.push(tt); 35 tt=zt; 36 tt.num[++tt.len]=0;tt.mod=(tt.mod*10)%n; 37 q.push(tt); 38 } 39 for (i=1;i<=zt.len;i++) 40 printf("%d",zt.num[i]); 41 printf(" "); 42 } 43 return 0; 44 }