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  • Find The Multiple 分类: 搜索 POJ 2015-08-09 15:19 3人阅读 评论(0) 收藏

    Find The Multiple
    Time Limit: 1000MS Memory Limit: 10000K
    Total Submissions: 21851 Accepted: 8984 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

    Source
    Dhaka 2002
    刚开始看到的时候以为是大数,要用JAVA,可是渣渣不会JAVA,搜了一下题解,原来long long 就可以过,bfs飞过

    #include <map>
    #include <list>
    #include <climits>
    #include <cmath>
    #include <queue>
    #include <stack>
    #include <string>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    #define eps 1e-9
    #define LL unsigned long long
    #define PI acos(-1.0)
    #define INF 0x3f3f3f3f
    #define CRR fclose(stdin)
    #define CWW fclose(stdout)
    #define RR freopen("input.txt","r",stdin)
    #define WW freopen("output.txt","w",stdout)
    
    int n;
    
    long long bfs()
    {
        queue<long long>Q;
        long long a;
        Q.push(1);
        while(!Q.empty())
        {
            a=Q.front();
            Q.pop();
            if(a%n==0)
            {
                return a;
            }
            Q.push(a*10);
            Q.push(a*10+1);
        }
        return 0;
    }
    
    int main()
    {
        while(scanf("%d",&n),n)
        {
            cout<<bfs()<<endl;
        }
        return 0;
    }
    

    版权声明:本文为博主原创文章,未经博主允许不得转载。

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  • 原文地址:https://www.cnblogs.com/juechen/p/4721907.html
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