POJ_1426(搜索)

 

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal 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

26190

Sample Output

10100100100100100100111111111111111111

//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);    }}