poj2081

poj2081 Recaman's Sequence

C-Recaman'sSequenceCrawlinginprocess...CrawlingfailedTimeLimit:3000MS    MemoryLimit:60000KB    64bitIOFormat:%I64d&%I64uSubmitStatusPracticePOJ2081Appointdescription:SystemCrawler(2016-05-11)Descript
su20145104009·2016-05-11 10:00

POJ2081(Recaman's Sequence)

题目链接 简单动态规划题。 View Code 1 #include <stdio.h> 2 #include <string.h> 3 #define N 500002 4 #define MAX 10000000 5 char vis[MAX]; 6 int c[N]; 7 int main() 8 { 9 int n,i,tmp; 10
·2015-11-12 22:47

poj2081

简单题 View Code #include < iostream > #include < cstdio > #include < cstdlib > #include < cstring &
·2015-11-07 12:27

poj2081

Recaman's Sequence Time Limit: 3000MS Memory Limit: 60000K Total Submissions: 15860 Accepted: 6666 Description The Recaman's sequence is defined by a0 = 0 ; for m
·2015-10-31 10:15

POJ2081：Recaman's Sequence

DescriptionTheRecaman'ssequenceisdefinedbya0=0;form>0,am=am−1−mifthersultingamispositiveandnotalreadyinthesequence,otherwiseam=am−1+m.ThefirstfewnumbersintheRecaman'sSequenceis0,1,3,6,2,7,13,20,12,21,
libin56842·2014-03-01 21:00

打表递推poj2081

 题目连接：http://poj.org/problem?id=2081题目解析：要看清题目在开始做题目，还有就是开的数组要足够的大；开两个数组，一个是打表用的，一个是用判断是不是出现过了，从而判断用哪个公式。代码如下： #include #include #include #include #include usingnamespacestd; #definePI3.1415927
yujuan_Mao·2012-10-10 21:00

poj2081

又是简单的递推。#include usingnamespacestd; constintkMax=500001; intlist[kMax]; boolFind[3500000]; voidmakeList() { inti; list[0]=0; Find[0]=true; for(i=1;i0&&!Find[tmp]) list[i]=tmp; else list[i]=list[i-1
Non_Cease·2011-09-24 19:00
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