POJ1426 Find The Multiple —

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 34218		Accepted: 14337		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

Sample Output

10

100100100100100100

111111111111111111

Source

Dhaka 2002

题解:

题目说答案的位数不会超过100，所以以为是：高精度 + 模运算。这样也太丧心病狂了吧！

后来实在想不到其他方法，就看了题解。结果是一道普通的搜索题，答案用long long存就够了（题目不是说位数<=100），感觉被骗了。

再一次受到了启发：面对一些题目（人生也如此）时，如果想到的方法似乎不能解决问题，但除此之外又无其他办法，那就就要放开手脚试一试，不要畏手畏脚。如果想到的唯一方法都搁置不试，那就相当于交白卷了；试一试，或许能出现意想不到的结果。

代码如下：

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <cmath>

 #include <algorithm>

 #include <vector>

 #include <queue>

 #include <stack>

 #include <map>

 #include <string>

 #include <set>

 #define ms(a,b) memset((a),(b),sizeof((a)))

 using namespace std;

 typedef long long LL;

 const int INF = 2e9;

 const LL LNF = 9e18;

 const int MOD = 1e9+;

 const int MAXN = +;

 queue<LL>q;

 LL bfs(int n)

 {

     while(!q.empty()) q.pop();

     q.push();

     while(!q.empty())

     {

         LL x = q.front();

         q.pop();

         if(x%n==)

             return x;

         q.push(1LL*x*);

         q.push(1LL*x*+);

     }

 }

 int main()

 {

     int n;

     while(scanf("%d",&n) && n)

     {

         cout<< bfs(n) <<endl;

     }

 }

巴特西

POJ1426 Find The Multiple —— BFS

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