POJ 1426-Find The Multiple_poj 1426 c语言版-优快云博客

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 19305		Accepted: 7831		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

小结：

    接触BFS也已经有一段时间了，说实话，与其说这是博客，还不如说是我的个人日志比较恰当。

    前面应该发表过同一道题目的另外一种解法，参考一位前辈的很有趣的解法，这里我想写写一般传统的做法，虽然代码长度较长，但是占用内存少，时间快，这些都是很明显优越的地方，看着这些前辈的神一般的方法，深深感觉到，“路漫漫其修远兮，吾将上下而求索”。

    不说废话了，再说下去可能就没有地方来放代码了。（我语文功底还是不错的，呵呵！）

以下是AC代码，一如既往还是用C语言实现的：

#include <stdio.h>
#include <math.h>
#include <string.h>
#define maxn 999999

int n;
int flag[200];
int ans[200];
struct node
{
    int k,temp,last;
}num[maxn],cur;

int bfs()
{
    int top,end;
    top=end=0;
    cur.k=1;
    cur.temp=1%n;
    cur.last=-1;
    flag[cur.temp]=1;
    num[top++]=cur;
    while(top>=end)
    {
        cur=num[end];
        end++;
        if(!cur.temp)
            return end-1;
        if(!flag[(cur.temp*10+1)%n])
        {
            num[top].k=1;
            num[top].temp=(cur.temp*10+1)%n;
            flag[(cur.temp*10+1)%n]=1;
            num[top++].last=end-1;
        }
        if(!flag[cur.temp*10%n])
        {
            num[top].k=0;
            num[top].temp=cur.temp*10%n;
            flag[cur.temp*10%n]=1;
            num[top++].last=end-1;
        }
    }
    return -1;
}
int main()
{
    while(~scanf("%d",&n),n)
    {
        memset(flag,0,sizeof(flag));
        int j=bfs();
        int i=0;
        while(j!=-1)
        {
            //printf("%d",num[j].k);
            ans[i++]=num[j].k;
            j=num[j].last;
        }
        for(int t=i-1;t>=0;t--)
        printf("%d",ans[t]);
        printf("\n");
    }
    return 0;
}