POJ-1426-Find The Multiple (BFS +DFS)两种解法

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 35878 Accepted: 14986 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

题解：还是一道简单的dfs题，已知n 求m 是n的倍数 ，m（十进制中只有包含0和1 ） 不会超过100个数位。unsigned __int64 可存放结果，包含两条搜索路径，x*10 和 x*10+1 这两条.

当然也可以用BFS来做,见AC2.

AC1

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;bool vis;int n;void dfs(unsigned __int64 x,int n,int k){    if(vis) return;    if(x%n==0)    {        printf("%I64u\n",x);        vis=true;        return;     }    if(k==19) return;    dfs(x*10,n,k+1);    dfs(x*10+1,n,k+1);}int main(){    while(scanf("%d",&n))    {        if(n==0) break;        vis=false;        dfs(1,n,0);    }    return 0;}

AC2

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>using namespace std;int n;void bfs(){    queue<long long > q;    while(!q.empty())        q.pop();    q.push(1);    while(1)    {        long long tmp=q.front();        if(tmp%n==0)        {            cout<<tmp<<endl;            return;        }        q.pop();        q.push(tmp*10);        q.push(tmp*10+1);    }}int main(){     while(scanf("%d",&n),n)         bfs();    return 0;}