题解报告：poj 1426 Find The Multiple（bfs、dfs）

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
解题思路：题目的意思就是找一个不小于n的10进制数x，x只由1和0组成，并且满足x%n==0，输出这个x即可。由于x只由0和1组成，因此容易想到构建一棵平衡二叉树，从根节点值为1开始，每个左节点的值都为父节点值的10倍，每个右节点的值都为父节点值的10倍加1，这样只要找到x%n==0即可。亲测了一下x的值在long long范围内，于是bfs或dfs暴力即可。
AC代码之bfs：

#include<iostream>
#include<queue>
#include<cstdio>
using namespace std;
typedef long long LL;LL n;
queue<LL> que;
void bfs(LL n){//在long long的范围内查找，按层遍历
    while(!que.empty())que.pop();//清空
    que.push(1LL);//模拟一棵平衡二叉树，先将入根节点的值1入队
    while(!que.empty()){
        LL x=que.front();que.pop();
        if(x%n==0){cout<<x<<endl;return;}//只要满足条件，立即退出
        que.push(x*10);//左节点的值为父节点值的10倍
        que.push(x*10+1);//右节点的值为父节点值的10倍加1
    }
}
int main(){
    while(cin>>n&&n){bfs(n);}
    return 0;
}

AC代码之dfs：

#include<iostream>
#include<queue>
#include<cstdio>
using namespace std;
typedef long long LL;LL n;bool flag;
void dfs(int num,LL x,LL n){
    if(num>19||flag)return;//如果x的位数num超过19位即溢出了long long 范围，直接返回到上一层；如果已找到即flag为true则一直回退，直到栈空，表示不再深搜下去
    if(x%n==0){cout<<x<<endl;flag=true;return;}
    dfs(num+1,x*10,n);//左递归
    dfs(num+1,x*10+1,n);//右递归
}
int main(){
    while(cin>>n&&n){flag=false;dfs(1,1,n);}//从位数1，根节点值为1开始深搜
    return 0;
}