Find The Multiple（POJ 1426）

最新推荐文章于 2021-08-08 22:14:40 发布

luckyrass

最新推荐文章于 2021-08-08 22:14:40 发布

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分类专栏：宽度优先遍历

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本文介绍了一种算法，用于在不超过200的整数范围内，找出能被指定整数整除且仅包含0和1的数。通过输入整数作为参数，算法会输出满足条件的最小的数。示例输入为2、6、19，对应输出分别为10、100100100100100100、111111111111111111。

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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

参考网站：http://blog.youkuaiyun.com/lyy289065406/article/details/6647917

#include<iostream>
using namespace std;
int mod[1048578];
int main(){
    int n,m,pm;
    while(cin>>n,n){
        mod[1]=1%n;
        m=10%n;
        int i;
        for(i=2;mod[i-1];i++){
            mod[i]=(mod[i/2]*m+i%2)%n;
        }
        i--;
        pm=1;
        while(i){
            mod[pm++]=i%2;
            i/=2;
        }
        while(--pm) cout<<mod[pm];
        cout<<endl;
    }
}