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  • poj 1426 Find The Multiple

    poj 1426 的传送门

    Language:
    Find The Multiple
    Time Limit: 1000MS Memory Limit: 10000K
    Total Submissions: 21735 Accepted: 8939 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

    2
    6
    19
    0
    Sample Output

    10
    100100100100100100
    111111111111111111

    题目大意:给你一个数,让你找可以整除这个数的并且仅仅含有0和1 的数,比方说
    input :3;
    output : 111;
    可能有多个答案。仅仅须要输出一个就可以;
    解题思路:用dfs搜索。仅仅搜关于0和1 的数,详情见代码。,,,

    上代码:

    #include <iostream>
    
    using namespace std;
    bool fo;
    void dfs(unsigned long long t, int k, int m)//unsigned一定要有,这是一个无符号字符类型,
    {
        if(fo)//一定要有这句话,要不然会有非常多个答案的。
            return ;
        if(t % m == 0)
        {
            cout<<t<<endl;
            fo=1;//标记
            return;
        }
        if(k == 19)//long long 最多有19位,
            return;
        dfs(t*10, k+1, m);//搜*10的
        dfs(t*10+1, k+1, m);//搜*10+1的
    }
    int main()
    {
        int m;
        while(cin>>m,m)
        {
            fo=0;
            dfs(1, 0, m);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/yutingliuyl/p/7008545.html
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