POJ 1426 <Find The Multiple> <搜索>

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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

题解：首先，题目没有提示，也就是最后答案肯定不需要大数模板，也就是long long存的下，那就是18位的10进制数够了。直接暴力跑一遍就行。

#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
bool found;
 void DFS(unsigned long long t, int n, int k)
 {
     if (found)
         return;
   if (t%n == 0)
     {
         printf("%I64u\n", t);
         found = true;
         return;
     }
   if (k == 19)
         return;
     DFS(t * 10, n, k + 1);  
     DFS(t * 10 + 1, n, k + 1);   
 }
 int main()
 {
     int n;
     while (scanf("%d",&n)&&n)
     {
         found = false;
         DFS(1, n, 0);
     }
     return 0;
 }

POJ 1426 <Find The Multiple> <搜索>

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原文地址：http://www.cnblogs.com/914295860-jry/p/5822537.html