poj 1426 Find The Multiple

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 49001 Accepted: 20448 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

这题题目可能有些同学看不懂吧，这题说的意思是：找一个只由1,0组成的数，对于给定的n，能够整除这个数。
这题思路很简单，深搜和广搜代码都简短。由于由1,0组成，故只需在原来基础上10或10+1就可使数只由1,0组成。
直接上代码：

BFS

#include <iostream>
#include <queue>

using namespace std;

typedef __int64 ll;
void bfs(int n)
{
    ll s = 1;
    queue<ll> q;
    q.push(s);
    while(!q.empty())
    {
        ll sum = q.front();
        q.pop();
        if(sum % n == 0)
        {
            cout << sum << endl;
            return;
        }
        q.push(sum*10);
        q.push(sum*10+1);
    }
}
int main()
{
    int n;
    while(cin >> n && n)
    {
        bfs(n);
    }
    return 0;
}

DFS

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int n;
int solve;//标记量，在递归中，如果solve为1了，说明之前已经找到了符合条件的不用继续递归下去了
void DFS(long long num,int k){
    if(solve==1)return;
    if(num%n==0){
        printf("%ll\n",num);
        solve = 1;
        return;
    }
    if(k==19)return;//题目要求不超100位，如果没有这句话可能会一直递归下去，直到溢出   
    DFS(num*10,k+1);
    DFS(num*10+1,k+1);
}
int main(){
    while(~scanf("%d",&n)&&n){
        solve = 0;
        DFS(1,0);//深搜从1开始
    }
    return 0;
}

还有一种思路是通过数组来写，因为每次都对一个数操作两次，故后来的数的数组下标是之前数的下标的两倍，故可根据此来写

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
long long mod[1000000];
int main(){
    int n;
    while(~scanf("%d",&n)&&n){
        int i;
        for(i = 1;; i++){//这里要从一开始，为什么看下面
            mod[i] = mod[i/2]*10+i%2;
            //这里就巧妙用到下标奇偶性质，奇数模2为1，就相当与加一，而偶数相当于不加，所以下标从1开始是方便的，
            if(mod[i]%n==0){
                printf("%lld\n",mod[i]);
                break;
            }
        }
    }
    return 0;
}