poj -- 1426 Find The Multiple (搜索)

 

题目连接：http://poj.org/problem?id=1426

 

 

 

Find The MultipleTime Limit: 1000MS Memory Limit: 10000K
Total Submissions: 8420 Accepted: 3443 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

 

题意：给出一个数字n，求一个不等于0并且能把n整除的数m,并且m只包含0和1

 

解法：1.模拟数字求余过程，进行搜索。

         2.学会打表

         3. 值得注意的是，如果n是偶数,则不用进行计算，直接=ans[i/2]+"0";

 

代码如下

//192K,16MS #include <iostream> #include <queue> #include <string> #include <CSTDIO> using namespace std; string ans[210]; bool mark[210]; struct node { string ans; int mod; }; string BFS(int n) { memset(mark,0,sizeof(mark)); queue<node>q; node s; s.ans="1"; s.mod=1; q.push(s); mark[1]=true; while(!q.empty()) { node now=q.front(),temp=now; q.pop(); int c=(now.mod*10+1)%n; int d=(now.mod*10)%n; if(d==0) { temp.ans+="0"; return temp.ans; } if(c==0) { temp.ans+="1"; return temp.ans; } if(!mark[d]) { mark[d]=1; temp.ans+="0"; temp.mod=d; q.push(temp); } if(!mark[c]) { mark[c]=1; now.ans+="1"; now.mod=c; q.push(now); } } } int main() { int n; for(int i=1;i<=200;i++) { if(i&1) ans[i]=BFS(i); else ans[i]=ans[(i>>1)]+"0"; } while (scanf("%d",&n)&&n) { cout<<ans[n]<<endl; } }