Poj1426 -- find the multiple (broad search, IQ questions)

Find the multiple

Time limit:1000 ms		Memory limit:10000 K
Total submissions:18527		Accepted:7490		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 in 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

If you see this in the wide search question, you cannot think of wide search ,,,,,

Each digit can only be 0 or 1, so the multiple of N is searched from the first digit until it is found.

The first digit must be 1, and the remainder is saved as temp. If the next digit is 1, (temp * 10 + 1) % N returns a new remainder. If the first digit is 0, then (temp * 10) % N gets the remainder, so that the search is wide and the size is 2 ^ 100

Pruning method: a maximum of 200 remainder values can be obtained for each request. It is good to see each remainder if it appears once.

 

 

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;struct node{    int k , temp ;    int last ;}p[1000000] , q ;int flag[210] , a[120] , n ;int bfs(){    int low = 0 , high = 0 ;    p[high].k = 1 ;    p[high].temp = p[high].k % n ;    flag[p[high].temp] = 1 ;    p[high++].last = -1 ;    while( low < high )    {        q = p[low++] ;        if( q.temp == 0 )            return low-1 ;        if( !flag[ (q.temp*10+1)%n ] )        {            p[high].k = 1 ;            p[high].temp = (q.temp*10+1)%n;            flag[ p[high].temp ] = 1 ;            p[high++].last = low-1 ;        }        if( !flag[ (q.temp*10)%n ] )        {            p[high].k = 0 ;            p[high].temp = (q.temp*10)%n ;            flag[ p[high].temp ] = 1 ;            p[high++].last = low-1 ;        }    }    return -1 ;}int main(){    int i , j ;    while(scanf("%d", &n) && n)    {        memset(flag,0,sizeof(flag));        i = 0 ;        j = bfs();        while( j != -1 )        {            a[i++] = p[j].k ;            j = p[j].last ;        }        for(j = i-1 ; j >= 0 ; j--)            printf("%d", a[j]);        printf("\n");    }    return 0;}

Poj1426 -- find the multiple (broad search, IQ questions)