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  • POJ 1426 Find The Multiple

    题目链接:POJ 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

    2
    6
    19
    0

    Sample Output

    10
    100100100100100100
    111111111111111111

    题意

    多组输入,输入一个数n,然后寻找到一个只包含0和1的十进制数,并且这个数可以被n整除,然后输出它

    题解:

    刚开始想复杂了,以为有什么巧妙的公式,后来发现这是完全的暴力题好吧??枚举从0,1,10,11开始依次类推的所有数,因为题目的长度限制,不用担心超时的问题,而且在找到的数不唯一时,输出任意一个就可以了。用bfs,从0,1开始,每次*10入队,*10+1入队,然后一直找下去,符合条件就输出。

    代码

    #include<iostream>
    #include<queue>
    using namespace std;
    queue<long long>P;
    int main() {
    	int n;
    	long long ans;
    	while(cin>>n,n){
    		while (!P.empty())
    			P.pop();
    		P.push(1);
    		long long t;
    		while (!P.empty()) {
    			t = P.front();
    			P.pop();
    			if (t%n == 0)
    			{
    				ans = t;
    				break;
    			}
    			P.push(t * 10);
    			P.push(t * 10 + 1);
    		}
    		cout << ans << endl;
    	}
    	return  0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Titordong/p/9588539.html
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