poj 1426 Find The Multiple（DFS）

最新推荐文章于 2019-08-11 15:04:19 发布

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最新推荐文章于 2019-08-11 15:04:19 发布

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本文介绍了一个简单DFS算法，用于解决POJ 1426题目的问题：给定正整数n，寻找一个由0和1组成的n的倍数m，其中m不超过100位数。

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题目链接：http://poj.org/problem?id=1426

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 27481		Accepted: 11438		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

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题目大意：输入一个数字n，求n的倍数（由0和1组成的十进制）

解析：简单的DFS，搜它的倍数，如果搜了19次还没找到就return （因为当前搜素没答案，再往下搜就爆unsigned long long）

代码如下：

#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<string>
#include<cstdio>
#include<cstring>
#include<cctype>
#include<cmath>
#define N 19
using namespace std;
const int inf = 1e9;
const int mod = 1<<30;
const double eps = 1e-8;
const double pi = acos(-1.0);
typedef long long LL;
unsigned long long m, n;
int f;

void dfs(unsigned long long u, int k)
{
    if(f || k > 18) return ;
    if(u % n == 0)
    {
        m = u; f = 1; return ;
    }
    
    dfs(u * 10, k + 1);
    dfs(u * 10 + 1, k + 1);
}

int main()
{
    while(scanf("%lld", &n), n)
    {
        f = 0;
        dfs(1, 0);
        printf("%lld\n", m);
    }
    return 0;

}