Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28550   Accepted: 11828   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains…

POJ 1426 Find The Multiple 题意:给定一个整数n,求n的一个倍数,要求这个倍数只含0和1 参考博客:点我 解法一:普通的BFS(用G++能过但C++会超时) 从小到大搜索直至找到满足条件的数,注意最高位一定为1 假设 n=6  k即为当前所求的目标数,不满足条件则进一步递推 (i 为层数(深度),在解法二的优化中体现,此时可以不管) 1%6=1 (k=1) i=1 { (1*10+0)%6=4 (k=10) i=2 { (10*10+0)%6=4 (k=100) i=4…

POJ 1426   Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 25734   Accepted: 10613   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representati…

POJ 1426 Find The Multiple(寻找倍数) Time Limit: 1000MS    Memory Limit: 65536K Description - 题目描述 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 as…

POJ.1426 Find The Multiple (BFS) 题意分析 给出一个数字n,求出一个由01组成的十进制数,并且是n的倍数. 思路就是从1开始,枚举下一位,因为下一位只能是0或1,故这个数字只能是1 * 10或者1 * 10 + 1.就按照这种方式枚举,依次放入队列,如果是其的倍数,就输出. 一开始没理解题意,以为是找一个能整除的二进制数,错了半天. 代码总览 #include <cstdio> #include <cstring> #include <algo…

题目传送门 /* 题意:找出一个0和1组成的数字能整除n DFS:200的范围内不会爆long long,DFS水过~ */ /************************************************ Author :Running_Time Created Time :2015-8-2 14:21:51 File Name :POJ_1426.cpp *************************************************/ #include…

题目链接:id=1426">Find The Multiple 解析:直接从前往后搜.设当前数为k用long long保存,则下一个数不是k*10就是k*10+1 AC代码: /* DFS */ #include <cstdio> #include <iostream> #include <algorithm> #include <queue> using namespace std; long long n; int DEEP; bool…

转载自:優YoU  http://user.qzone.qq.com/289065406/blog/1303946967 以下内容属于以上这位dalao http://poj.org/problem?id=1426 题意 给出一个整数n,(1 <= n <= 200).求出任意一个它的倍数m,要求m必须只由十进制的'0'或'1'组成. 分析 首先暴力枚举肯定是不可能的 1000ms 想不超时都难,而且枚举还要解决大数问题.. 要不是人家把这题放到搜索,怎么也想不到用BFS... 解题方法: B…

http://poj.org/problem?id=1426 测试了一番,从1-200的所有值都有long long下的解,所以可以直接用long long 存储 从1出发,每次向10*s和10*s+1转移,只存储余数即可, 对于余数i,肯定只有第一个余数为i的最有用,只记录这个值即可 #include <cstdio> #include <cstring> #include <queue> using namespace std; const int maxn=222…

Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 18012   Accepted: 7297   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains…

E - Find The Multiple Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u Submit Status Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only…

题目: 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…

题目链接:http://poj.org/problem?id=1426 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 correspo…

题目链接: 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…

题目:http://poj.org/problem?id=1426 题意:输入一个数,输出这个数的整数 倍,且只有0和1组成 程序里写错了一个数,结果一直MLE.…… #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #include<iomanip> #inclu…

题目链接 http://poj.org/problem?id=1426 题意 给出一个数 要求找出 只有 0 和 1 组成的 十进制数字 能够整除 n n 不超过 200 十进制数字位数 不超过100 思路 其实 十进制数字位数 不超过 20 下就有可以满足的答案 所以直接用 unsinged long long 就可以过了 .. AC代码 #include <cstdio> #include <cstring> #include <ctype.h> #include…

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 mo…

没什么好说的 从1开始进行广搜,因为只能包涵0和1,所以下一次需要搜索的值为next=now*10 和 next=now*10+1,每次判断一下就可以了,但是我一直不太明白我的代码为什么C++提交会错,G++则正确. #include<cstdio> #include<stdio.h> #include<cstdlib> #include<cmath> #include<iostream> #include<algorithm> #i…

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 mo…

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 mo…

Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 21436   Accepted: 8775   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: 18390   Accepted: 7445   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains…

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 mo…

注:本人英语很渣,题目大意大多来自百度~=0=   这个题有点坑,答案不唯一   题目大意:给你一个数n, 你需要输出的是一个由1和0组成的数,此数能被n整除   解题思路:用s = 1做数的起点, s*10则相当于在后面加上0, s*10+1代表在后面加1, 用long long 来保存s足够了, 每次判断一下s % n  符合条件则输出 当然用dfs不能无限寻找下去  比如你第一次10000...0 假如n是奇数不可能有结果 所以每次递归记录一下层数: long long 的范围是-9223…

Avoid The Lakes Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8173   Accepted: 4270 Description Farmer John's farm was flooded in the most recent storm, a fact only aggravated by the information that his cows are deathly afraid of wate…

问题:打冰球.冰球可以往上下左右4个方向走,只有当冰球撞到墙时才会停下来,而墙会消失.当冰球紧贴墙时,不能将冰球往那个方向打.冰球出界就当输,超过10次还没将冰球打到目标位置也当输.求用最小次数将冰球打到目标位置,或输出-1表示输了. 分析:一般来说,求最小步数之类的迷宫问题都是用BFS解决的,但这题涉及到迷宫状态的变化(墙),BFS要不断记录状态的变化很复杂,不过网上好像也有人用BFS做的.DFS更加适合这种状态一直变化的,只不过要保存最优值而已,其实最优值也方便剪枝(当前步数已经是当前最优值…

DFS入门的一道经典题目:LakeCounting 用栈或队列来实现: #include<cstdio> #include<stdlib.h> #include<iostream> #include<stack> using namespace std; int n,m; int pla[10][3]={{1,0},{1,1},{1,-1},{-1,-1},{-1,0},{-1,1},{0,-1},{0,1}};//对坐标进行移动的向量 struct pla…

Pet Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 535 Accepted Submission(s): 258 Problem Description One day, Lin Ji wake up in the morning and found that his pethamster escaped. He searched in…

[算法入门] 郭志伟@SYSU:raphealguo(at)qq.com 2012/05/12 1.前言 深度优先搜索(缩写DFS)有点类似广度优先搜索,也是对一个连通图进行遍历的算法.它的思想是从一个顶点V0开始,沿着一条路一直走到底,如果发现不能到达目标解,那就返回到上一个节点,然后从另一条路开始走到底,这种尽量往深处走的概念即是深度优先的概念. 你可以跳过第二节先看第三节,:) 2.深度优先搜索VS广度优先搜索 2.1演示深度优先搜索的过程 还是引用上篇文章的样例图,起点仍然是V0,我们修…

版权声明: 本文由Faye_Zuo发布于http://www.cnblogs.com/zuofeiyi/, 本文可以被全部的转载或者部分使用,但请注明出处. 上周学习了数组和链表,有点基础了解以后,这周初步探索了一下深度优先搜索.对于文科生的我来说,这个名词听起来有点可怕.于是我通过leetcode上的一个难度为medium的题目来逐渐认识这个概念的.这道题目是Validate Binary Search Tree(题号为98).下面我将通过这道题作为引子,整理一下上周学习到的东西. 一.树 这…