POJ 1426-Find The Multiple(bfs)

Find The Multiple

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 19272 Accepted: 7813 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

题意：给一个数n，求其只含有数字0或1的倍数，输出其任意的可行解(答案不唯一

思路：从1开始判断是否为n的倍数，然后乘10或者乘10加1，找到一个退出即可

Ps:要不是这道题在bfs里我都不知道怎么搞。这个题我也不知道怎么回事，用g++交AC，用c++交MLE，saddd

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>#include <set>#include <queue>#include <stack>using namespace std;long long n;queue<long long >q;long long bfs(){    long long res;    while(!q.empty())        q.pop();    q.push(1);    while(!q.empty()){        res=q.front();        q.pop();        if(res%n==0)            return res;        q.push(res*10);        q.push(res*10+1);    }}int main(){    while(~scanf("%lld",&n)){        if(n==0) break;        printf("%lld\n",bfs());    }    return 0;}