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最新推荐文章于 2019-03-16 17:05:09 发布

Dijkstra__

最新推荐文章于 2019-03-16 17:05:09 发布

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分类专栏： dfs 文章标签： poj 搜索

本文链接：https://blog.youkuaiyun.com/qq_32126633/article/details/69524695

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Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 30495		Accepted: 12684		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

这个题说是bfs做的，但是我是dfs做的，试过一次bfs，用的stl里面的queue，但是超时了，估计queue还是比较慢吧，后来看了人家的代码直接dfs，果然过了。

题目大意：给你一个数字，要你求一个只有0和1组成的数，并且这个数是给你的数字的倍数，答案不唯一，随便输出一个就行

在unsigned __int64范围内是可以做的

而且搜索的时候最多会搜索19层

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<set>
#include<map>
#include<queue>
#include<algorithm>
using namespace std;
bool t;
void dfs(unsigned __int64 m,int num,int k){
    if(t)
        return;
    if(m%num==0){
        t=1;
        printf("%I64u\n",m);
        return;
    }
    if(k==19)
        return;
    dfs(m*10,num,k+1);
    dfs(m*10+1,num,k+1);
}
int main(){
    int num;
    unsigned long long m,tmp;
    while(~scanf("%d",&num)&&num){
        t=0;
        dfs(1,num,0);
    }
    return 0;
}