POJ1426 Find The Multiple(BFS)
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
題意:
給出一個數n,找出任意一個數m,m是n的倍數且只有0與1組成。
解題思路:
BFS。從1開始,依次添0和1,即
1
10 11
100 101 110 111
......
只要找到一個滿足條件的m即可結束搜尋。
AC程式碼:
#include<cstdio> #include<cstring> #include<queue> using namespace std; queue<long long> q; void bfs(int n) { while(!q.empty()) { q.pop(); } q.push(1); while(!q.empty()) { if(q.front()%n==0) { printf("%lld\n",q.front()); return; } q.push(q.front()*10); q.push(q.front()*10+1); q.pop(); } } int main() { int n; while(~scanf("%d",&n)&&n) { bfs(n); } return 0; }