Find Sequence

Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 279 Accepted Submission(s): 80

Problem Description

Give you an positive integer sequence a1,a2,…,ai,…,an, and they satisfy a1+a2+…+ai+…+an=M(0<M≤222).
We can find new sequence b1=aid1,b2=aid2,…,bx=aidx,…,by=aidy,…,bt=aidt, where if x != y then idx!=idy. and this sequence satisfy:
(1) b1≤b2≤…≤bt
(2) b2−b1≤b3−b2≤⋯≤bt−bt−1
We can find many sequences b1,b2,b3,…,bt. But we only want to know maximum t.

Input

The first line in the input file is an Integer T(1≤T≤30).
The first line of each test case contains two integer n,M(0<M≤222).
Then a line have n integer, they represent a1,a2,…,ai,…,an.

Output

For each test case, output the maximum t.

Sample Input

2 6 19 3 2 1 3 4 6 1 4194304 4194304

Sample Output

5 1

Hint

For the first testcase, The Sequence is 1 2 3 4 6

Source

BestCoder Round #13

题意: 在一个数列里面取出一些数，满足一些条件

官方题解

首先考虑解的结构一定是C1,C1,…,C1,C2,C3,…,Cm这种形式，其中满足C1<C2<C3<…<Cm
所以对a1,a2,a3,…,an去重后从小到大排序得到c1,c2,c3,…,cx其中x是sqrt(M)级别的，用DP[i][j]表示以ci和cj结尾的满足条件的最长序列 首先初值化 DP[i][i]=count(ci) 即ci在原序列中的个数。 而dp[i][j]=max(dp[k][i] 其中k≤i还满足ci−ck≤cj−ci)+1 这样的复杂度是 O(x^3),在题中x最大为1000级别所以会超时，要使用下面优化 因为 dp[i][j]=max(dp[k][i] 其中k≤i还满足ci−ck≤cj−ci)+1 dp[i][j+1]=max(dp[k][i] 其中k≤i还满足ci−ck≤cj+1−ci)+1 注意到cj+1>cj 所以满足ci−ck≤cj−ci的dp[k][i]必然满足ci−ck≤cj+1−ci因而不必重复计算 即最后复杂度可以为O(x^2).

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<map>
#include<ctime>
#include<bitset>
#define LL long long
#define maxn (1<<22)+10
using namespace std;int cnt[maxn],c[2010],dp[2010][2010];
int main()
{int i,n,m,j,k;int tmp,T,x,ans;cin >>T ;while(T--){scanf("%d%d",&n,&m) ;memset(cnt,0,sizeof(cnt));for( i = 1 ; i <= n ;i++){scanf("%d",&x) ;cnt[x]++;}n = 0 ;memset(dp,0,sizeof(dp));ans=0;for( i =0 ; i <= m ;i++)if(cnt[i]){c[++n]=i;dp[n][n]=cnt[i];ans=max(cnt[i],ans);}for( i = 1 ; i <= n ;i++){tmp=dp[i][i];k = i ;for( j = i+1 ; j <= n ;j++){for( ; k >= 1 ;k--){if(c[j]-c[i] >= c[i]-c[k])tmp = max(tmp,dp[k][i]+1) ;else break ;}dp[i][j]=tmp;ans=max(ans,dp[i][j]);}}printf("%d\n",ans);}return 0 ;
}

View Code

转载于:https://www.cnblogs.com/20120125llcai/p/4032336.html

hdu 5064 Find Sequence相关推荐

HDU 1005 Number Sequence
[题目] Number Sequence Time Limit: 2000/1000 MS (Jav ...
HDU 5400 Arithmetic Sequence
HDU 5400 Arithmetic Sequence /** HDU 5400 Arithmetic Sequence 直接预处理求解就好了预处理找出以a[i]结尾最长的subArr长度(满足条 ...
HDU.1005 Number Sequence
原题 HDU.1005 Number Sequence 分类杂题题意给定一个数列{an}\left\{ a_n \right\}{an}的前两项a1a_1a1.a2a_2a2,以及其递推公 ...
HDU 1560 DNA sequence（DNA序列）
HDU 1560 DNA sequence(DNA序列) Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K ...
HDU 1711 Number Sequence(KMP算法)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1711 Number Sequence Time Limit: 10000/5000 MS (Java/ ...
HDU 6304 Chiaki Sequence Revisited
题目链接 : http://acm.hdu.edu.cn/showproblem.php?pid=6304 多校contest1 Problem Description Chiaki is inter ...
hdu 2454 Degree Sequence of Graph G
点击打开链接 Degree Sequence of Graph G Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/3276 ...
HDU 5306 Gorgeous Sequence
5306 ( Gorgeous Sequence ) 思路: 吉司机线段树维护最大值和次大值,大于最大值不改,在最大值和次大值之间的直接修改,小于次大值递归修改. 代码: #pragma GCC o ...
HDU 1711 -Number Sequence(KMP)
题目 Number Sequence Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Othe ...
hdu 1560 DNA sequence(迭代加深搜索)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1560 题意:从n个串中找出一个最短的公共串,,该公共串对于n个字符串不要求连续,即只要保持相对顺序就好 ...

hdu 5064 Find Sequence

Find Sequence

hdu 5064 Find Sequence相关推荐

最新文章

热门文章