POJ 1426 Find The Multiple (dfs)-优快云博客

本文链接：https://blog.youkuaiyun.com/CriminalCode/article/details/49850337

针对给定整数n，寻找其仅由0和1组成的倍数m，m不超过100位。通过深度优先搜索策略，在longlong范围内解决此问题。

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Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22773		Accepted: 9387		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

题目链接：http://poj.org/problem?id=1426

题目大意：给一个数字n，求出它的每位只由0和1组成的倍数。

解题思路：因为n的大小不超过200，所以满足条件的倍数在long long 范围内，直接深搜。

代码如下：

#include <cstdio>
#include <cstring>
#define ll long long 
ll n;
bool p;
int m;
void dfs(ll x,int cnt)
{
    if(x%n==0)   //找到答案，返回
    {
        p=true;
        printf("%I64d\n",x);
        return ;
    }
    if(cnt>=18)         //在long long 的范围内，超过这个范围直接返回 
        return ;
    for(int i=0;i<=1;i++)   //枚举下一位数字是0或者1
    {
        dfs(x*10+i,cnt+1);
        if(p)
            return ;
    }
}
int main(void)
{
    while(scanf("%I64d",&n)!=EOF&&n)
    {
        m=0,p=false;
        dfs(1,0);
    }
}