poj1426 Find The Multiple dfs搜索答案

最新推荐文章于 2019-08-11 15:04:19 发布

原创最新推荐文章于 2019-08-11 15:04:19 发布 · 288 阅读

0 ·

CC 4.0 BY-SA版权

------------解题报告------------- 同时被 3 个专栏收录

208 篇文章

订阅专栏

--------搜索----------

36 篇文章

订阅专栏

DFS

24 篇文章

订阅专栏

本文介绍了一道经典算法题目FindTheMultiple的解决方案。题目要求对于给定的正整数n，找到一个由0和1组成的数m，使得m能够被n整除。通过深度优先搜索的方法，在不超过100位的限制下寻找符合条件的数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 28704		Accepted: 11898		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

题意：给你一个n，求一个仅有1、0组成的数（十进制环境）能被n整除。

直接搜索搜索无脑搜搜搜搜

之所以能过是因为这题的数据很温柔。。。優しいです，所有的数不会超出unsigned long long

電車遅いな。。。。。。眠いです

/*
 ━━━━━┒
 ┓┏┓┏┓┃μ'sic foever!!
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 20;
int n,m;
unsigned long long res;
bool vis;
void dfs(unsigned long long now,int lenth){
    if(lenth==19||vis){
        return;
    }
    if(now%n==0){
        vis = true;
        res = now;
        return;
    }
    dfs(now*10,lenth+1);
    dfs(now*10+1,lenth+1);
}
int main(){
    while(scanf("%d",&n)&&n){
        vis = false;
        dfs(1,0);
        printf("%llu\n",res);
    }
    return 0;
}