POJ1426 Find The Multiple (宽搜思想)
Find The Multiple
Time Limit: 1000MS | Memory Limit: 10000K | |||
Total Submissions: 24768 | Accepted: 10201 | 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
题意:求n个一个倍数,但这个数必须是01串
其实就是利用宽搜的思想枚举: 第一个数是1 , 然后下一个就是 10 (1 * 10) , 11( 1 * 10 + 1),再下一个就是 100(10 * 10), 101(10 * 10 + 1),110 (11 * 10),111(11 * 10 + 1) ...然后比较坑的就是只要是n个一个倍数就行,结果不用跟这个样例一样也可以,然后用long long也能过=_=
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <queue> 5 using namespace std; 6 typedef long long LL; 7 void bfs(int n) 8 { 9 queue<LL> q; 10 q.push(1); 11 while (!q.empty()) 12 { 13 LL temp = q.front(); 14 q.pop(); 15 if (temp % n == 0) 16 { 17 printf("%lld\n", temp); 18 return; 19 } 20 q.push(temp * 10); 21 q.push(temp * 10 + 1); 22 } 23 } 24 int main() 25 { 26 int n; 27 while (scanf("%d", &n) != EOF && n) 28 { 29 bfs(n); 30 } 31 return 0; 32 }