POJ_1426(搜索)

 

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6690		Accepted: 2809		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

//BFS

//BFS
#include <cstdlib>
#include <iostream>
#include <cstdio>

using namespace std;

const int MAX = 1100000;

__int64 BFS(__int64 queue[], int value)
{
    int head = 0, tail = 0;
    __int64 temp = 0;
    
    queue[tail++] = 1;
    
    while(head != tail)
    {
        __int64 temp = queue[head++];
        
        if(temp % value == 0)
            return temp;
        
        queue[tail++] = temp * 10;   //push left child and right child into queue
        queue[tail++] = temp * 10 + 1;
    }
}

int main(int argc, char *argv[])
{
    int value;
    while(scanf("%d", &value) && value != 0)
    {
        __int64 queue[MAX];
        
        printf("%I64d/n", BFS(queue, value));      
    }

    return EXIT_SUCCESS;
}
 

对于那个大数组queue, 如果在声明的时候初始化， 就TLE, 去掉后125ms...

//DFS

#include <cstdio>
#include <cstdlib>

using namespace std;

const int MAX = 201;
int queue[MAX];
bool isFind;

void display(int count)
{
    for(int i=0; i<count; i++)
        printf("%d", queue[i]);
    printf("/n");
}

void DFS(int value, int deep, int rem)
{
    if(!isFind && deep < 100)
    {
        if(rem == 0)
        {
            isFind = true;
            display(deep);
        }
        else
        {
            queue[deep] = 0;  //handle left child
            DFS(value, deep + 1, (10 * rem) % value);
            
            queue[deep] = 1;  //handle right child
            DFS(value, deep + 1, (10 * rem + 1) % value);
        }
    }
}

int main(int argc, char *argv[])
{
    int value;
    while(scanf("%d", &value) && value != 0)
    {
        isFind = false;
        queue[0] = 1;
        
        DFS(value, 1, 1);
    }
}