Catch That Cow Time Limit: 2000 MS Memory Limit: 65536 KB 64-bit integer IO format: %I64d , %I64u Java class name: Main [Submit] [Status] [Discuss] Description Farmer John has been informed of the location of a fugitive cow and wants to catch her imm…

题目链接:https://vjudge.net/problem/POJ-1426 题意:给出n(1<=n<=200),求出全部由01组成的能整除n的正整数. 思路:此题在unsigned long long以内就可以找到满足条件的数,因此限制递归深度为20,然后枚举每一位两种可能即可. AC代码: #include<cstdio> #include<algorithm> using namespace std; ]; void dfs(int pos,int pre){…

POJ1426--Find The Multiple 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 corre…

Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 24768   Accepted: 10201   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains…

Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14622   Accepted: 5938   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains…

图的遍历也称为搜索,就是从图中某个顶点出发,沿着一些边遍历图中所有的顶点,且每个顶点仅被访问一次,遍历可采取两种不同的方式:深度优先搜索(DFS)和广度优先搜索(BFS). 1.DFS算法思想` 从顶点v出发深度遍历图G的算法 ① 访问v0顶点,置vis[v0]=1,搜索v0未被访问的邻接点w,若存在邻接点w,则dfs(w),直到到达所有邻接点都被访问过的顶点u为止,接着退回一步,看是否还有其他没有被访问的邻接点.如果有,则访问此顶点,进行前述类似的访问,如果没有,就在退回一步进行搜索,重复上述…

Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16505   Accepted: 6732   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains…

Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 36335   Accepted: 15194   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains…

一开始模拟了一波大数取余结果超时了,最后改成long long过了emmm... 本题大意:给出一个200以内的数n,让你找出一个m使得m % n == 0,要求m只有1和0组成. 本题思路:BFS模拟即可. 参考代码: #include <cstdio> #include <queue> using namespace std; int mod; typedef long long LL; LL bfs() { LL now = ; queue <LL> s; s.p…

http://poj.org/problem?id=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 c…