poj-1426-Find The Multiple(dfs)
Find The Multiple
Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 40713 Accepted: 17088 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
0Sample Output
10
100100100100100100
111111111111111111给定一个正整数 n，请编写一个程序来寻找 n 的一个非零的倍数 m，这个 m 应当在十进制表示时每一位上只包含 0 或者 1。你可以假定 n 不大于 200 且 m 不多于 100 位。 提示：本题采用 Special Judge，你无需输出所有符合条件的 m，你只需要输出任一符合条件的 m 即可。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 | #include"iostream" using namespace std; typedef unsigned long long ll; int n; bool sign; void dfs(ll x,int count) { if(sign) return ; if(x%n==0){ sign=true; cout<<x<<endl; return ; } if(count==19)//m 最多 200 位 return ; dfs(x*10,count+1); dfs(x*10+1,count+1); //每两位数后两位有两种情况，10 或 11，深搜所有情况，找到一种就返回，找不到找另外一颗子树 } int main() { while(cin>>n&&n) { sign=false; dfs(1,0); } return 0; } |
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