您的位置：首页 > 其它

POJ1426——Find The Multiple

2016-08-17 14:10 453 查看

Find The Multiple

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 27093		Accepted: 11266		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

Sample Output

10
100100100100100100
111111111111111111

Source

Dhaka 2002

题目大意：

输入一个数 n ，求一个 n 的倍数，要求：这个倍数只能由 1 或 0 组成；

注意开头的红字 Special Judge 特殊判定只要输出一组符合要求的就行

Memory: 2896K		Time: 282MS
Language: G++		Result: Accepted
Source Code
#include<iostream>
#include <cstdlib>
#include <cstdio>
#include <queue>

using namespace std;

int n;
queue<long long> a;

long long fun()
{
a.push(1);
while(!a.empty())
{
long long tp = a.front();
a.pop();
if( (tp*10)%n==0 )
return tp*10;
if( (tp*10+1)%n==0 )
return tp*10+1;
a.push(tp*10);
a.push(tp*10+1);
}
return -1;
}

int main()
{
while(cin>>n&&n)
{
while(!a.empty())
a.pop();
cout<<fun()<<endl;
}

return 0;
}

内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理

标签： poj 搜索广搜

相关文章推荐

新的分享

章节导航