poj 1426 深搜dfs

Find The Multiple

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%lld & %llu

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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

这道题唯一的坑点就是任意值都可以，你输出来就行，我输出6的倍数与样例的不同，然后交了下试试然后就对了。要好好学英语了！！！

#include<stdio.h>
int state;///定义是否已找到那个倍数
long long ans;
void dfs(long long a, long long b, int k)
{
    if(state)
        return ;
    if(k==19)
        return ;
    if(b%a==0 && state==0)//这个加不加state==0都行 我害怕出意外就加上了
    {
        state=1;
        ans=b;
        return ;
    }
    dfs(a,10*b,k+1);
    dfs(a,10*b+1,k+1);
}
int main()
{
    long long n;
    while(scanf("%lld",&n),n)
    {
        state=0;
        dfs(n,1,0);
        printf("%lld\n",ans);
    }
    return 0;
}

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