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…

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…

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…

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…

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…

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…

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…

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…

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…

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: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…

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…

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…

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…

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…

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…

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…

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…

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…

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…

完全想不到啊,同余模定理没学过啊,想起上学期期末考试我问好多同学'≡'这个符号什么意思,都说不知道,你们不是上了离散可的吗?不过看了别人的解法我现在会了,同余模定理介绍及运用点这里点击打开链接 简单说一下同余模定理:如果(a - b) / m = 0,说明a%m等于b%m,那么对于本题应该如何运用呢?  已知a % n = m,那么(a * 10 + x) % n = a * 10 % n + x % n = (a % n * 10 + x ) % n = (m *10 + x ) % n,有了…

题意:给出一大数K(4 <= K <= 10^100)与一整数L(2 <= L <= 106),K为两个素数的乘积(The cryptographic keys are created from the product of two primes) 问构成K的最小素数是否绝对小于L,若是,则输出BAD p,p为最小素数,否则输出GOOD; 分析:从小到大枚举1~10^6内的素数p,while(p<L)时,判断K是否能被p整除,若能则证明构成K的最小素数绝对小于L,反之则大于L…

题目链接:http://poj.org/problem?id=1426 Find The Multiple Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 34218   Accepted: 14337   Special Judge Description Given a positive integer n, write a program to find out a nonzero multiple m of n w…

POJ 2251 题目大意: 给出一三维空间的地牢,要求求出由字符'S'到字符'E'的最短路径,移动方向可以是上,下,左,右,前,后,六个方向,每移动一次就耗费一分钟,要求输出最快的走出时间.不同L层的地图,相同RC坐标处是相连通的.(.可走,#为墙) 解题思路:从起点开始分别往6个方向进行BFS(即入队),并记录步数,直至队为空.若一直找不到,则困住. /* POJ 2251 Dungeon Master --- 三维BFS(用BFS求最短路) */ #include <cstdio> #i…

Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 11435   Accepted: 3040 Description The young and very promising cryptographer Odd Even has implemented the security module of a large system with thousands of users, which is now in use in…

题目链接:https://www.cnblogs.com/kuangbin/archive/2012/04/01/2429463.html 题意:给出大数s (s<=10100) ,L (<=106),s是两个素数的乘积,求其最小因子即这两个素数中较小者是否小于L. 思路:先通过欧筛法打表计算出106以内的素数,大概有8e4个.然后用千进制表示s,如将12345678表示成[012][345][678],用整型数组Kt表示,前面补0,这样做之后利用同余模定理计算Kt对x的模.如计算[012][…