5805 NanoApe Loves Sequence(想法题)】的更多相关文章

http://acm.hdu.edu.cn/showproblem.php?pid=5806 题意:给你一个n元素序列,求第k大的数大于等于m的子序列的个数. 题解:题目要求很奇怪,很多头绪但写不出,选择跳过的题,简称想法题. 首先考虑区间的更新方法:区间左端l不动,右端r滑动, 滑到有k个数>=m时,此区间符合条件,并且发现右端点再往右滑到底,此条件一直符合(因为若加入的数小于“第K大的数”,则毫无影响.若不然,加入该数会产生一个新的第k大数,保证>=“第K大的数”>=m) 所以一找…
题目:传送门. 题意:题目说的是求期望,其实翻译过来意思就是:一个长度为 n 的数列(n>=3),按顺序删除其中每一个数,每次删除都是建立在最原始数列的基础上进行的,算出每次操作后得到的新数列的相邻两数的差的绝对值的最大值,求这些n个最大值的总和. 题解:把n=3的情况单独拿出来直接算出来,就是abs(data[3]-data[2])+abs(data[2]-data[1])+abs(data[3]-data[1]),然后讨论n>=4的情况.首先遍历求出原始数列的相邻两数的差的绝对值的最大值m…
传送门 NanoApe Loves Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others) Total Submission(s): 1323    Accepted Submission(s): 521 Description NanoApe, the Retired Dog, has returned back to prepare for the Natio…
NanoApe Loves Sequence 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5805 Description NanoApe, the Retired Dog, has returned back to prepare for the National Higher Education Entrance Examination! In math class, NanoApe picked up sequences once aga…
先找相邻差值的最大,第二大,第三大 删去端点会减少一个值, 删去其余点会减少两个值,新增一个值,所以新增和现存的最大的值比较一下取最大即可 #include <iostream> #include <cstdio> #include <cmath> using namespace std; #define LL long long ; int t, n, p1, p2, p3; LL a[N]; LL s1[N], s2[N]; LL sum; int main() {…
处理出每个位置左边的最大值和右边的最大值.然后就可以o(1)计算去掉某位置的最大值了. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<map> #includ…
NanoApe Loves Sequence Accepts: 531 Submissions: 2481 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others) Problem Description NanoApe, the Retired Dog, has returned back to prepare for the National Higher Education Entr…
题目链接:NanoApe Loves Sequence Ⅱ Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)Total Submission(s): 339    Accepted Submission(s): 165 Problem Description NanoApe, the Retired Dog, has returned back to prepare for…
传送门 NanoApe Loves Sequence Ⅱ Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)Total Submission(s): 1585    Accepted Submission(s): 688 Description NanoApe, the Retired Dog, has returned back to prepare for for the…
NanoApe Loves Sequence Ⅱ 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5806 Description NanoApe, the Retired Dog, has returned back to prepare for for the National Higher Education Entrance Examination! In math class, NanoApe picked up sequences on…