POJ-Find The Multiple

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 33514 Accepted: 14053 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

26190

Sample Output

10100100100100100100111111111111111111

Source

Dhaka 2002

Hint:

即找到一个是n(0<n<200)的倍数且由‘0’和‘1’组成的数，注意此数不可超过unsigned long long的范围，用DFS,BFS均可求解

///BFS

#include <iostream>
#include <queue>
using namespace std;
long long BFS(int n) ///注意返回的一定是long long类型的,否则WA
{
    long long m, m1,m2;
    queue<long long>q;
    q.push(1);
    while(!q.empty())
    {
        m=q.front();
        q.pop();
        if(m%n==0)
            return m;
        m1=m*10;
        m2=m*10+1;
        q.push(m1);
        q.push(m2);
    }
    return 0;
}
int main()
{
    int n;
    while(cin>>n&&n)
    {
        cout<<BFS(n)<<endl;
    }
    return 0;
}

DFS:

#include <iostream>
using namespace std;
int n,flat;
unsigned long long b;
void DFS(unsigned long long a,int step)
{
    if(flat||step==19)///unsigned long long整数有二十位，unsigned即signed的两倍
                      ///例如:unsigned long long:0-18446744073709551615(20位),signed long long:(-+)9223372036854775807(19位)
    {
        return ;
    }
    if(a%n==0)
    {
        printf("%I64u\n",a);///I64U:64位无符号整数
        flat=1;///提前终止回溯,防止输出多余的a.
    }
    else
    {
        DFS(a*10,step+1);
        DFS(a*10+1,step+1);
    }
    return ;
}
int main()
{
    while(scanf("%d",&n),n)
    {
        flat=0;
        DFS(1,0);
    }
    return 0;
}

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