POJ 1426 Find The Multiple

D - Find The Multiple

Time Limit:1000MS    Memory Limit:10000KB    64bit IO Format:%I64d & %I64u

SubmitStatusPracticePOJ 1426

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

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BFS搜索。

题目大概意思是给一个数n，让你找出一个只有1，0，组成的十进制数，要求是找到的数可以被n整除。不能超过100个数字。输出一个满足的数组即可。

假如n=6；

1余n=1；

10余n=1*10%6=4；

11余n=（1*10+1）%6=5；

100余n=4*10%6=4；

101余n=（4*10+1）%6=5；

110余n=5*10%6=2;

111余n=（5*10+1）%6=3；

可以推知10........01余n=（10.........0余n*10+1）余n；

当然这个搜索的效率貌似不高。438ms.

#include<stdio.h>#include<queue>using namespace std;void bfs(int n){        queue<long long>q;        q.push(1);        while(!q.empty())        {            long long x;            x=q.front();            q.pop();            if(x%n==0)            {                printf("%lld\n",x);                return ;            }            q.push(x*10);            q.push(x*10+1);        }}int main(){        int n;        while(scanf("%d",&n)>0&&n)        {            bfs(n);        }        return 0;}