hdu 4908 Bestcoder Problem Description Mr Potato is a coder.Mr Potato is the BestCoder. One night, an amazing sequence appeared in his dream. Length of this sequence is odd, the median number is M, and he named this sequence as Bestcoder Sequence.…
BestCoder Sequence Problem DescriptionMr Potato is a coder.Mr Potato is the BestCoder.One night, an amazing sequence appeared in his dream. Length of this sequence is odd, the median number is M, and he named this sequence as Bestcoder Sequence.As th…
BestCoder Sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 573 Accepted Submission(s): 201 Problem Description Mr Potato is a coder.Mr Potato is the BestCoder. One night, an amazing s…
BestCoder Sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 271 Accepted Submission(s): 112 Problem Description Mr Potato is a coder. Mr Potato is the BestCoder. One night, an amazing…
BestCoder Sequence Problem Description Mr Potato is a coder. Mr Potato is the BestCoder. One night, an amazing sequence appeared in his dream. Length of this sequence is odd, the median number is M, and he named this sequence as Bestcoder Sequence. A…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4908 BestCoder Sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 618 Accepted Submission(s): 214 Problem Description Mr Potato is a code…
# include <stdio.h> # include <algorithm> using namespace std; int main() { int n,m,i,sum,cot,flag,j; int map[80040]; int a[40010]; int ans1; int ans2; while(~scanf("%d%d",&n,&m)) { ans1=ans2=40010; memset(map,0,sizeof(map));…
A.预处理出来,0(1)输出. Task schedule Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 387 Accepted Submission(s): 193 Problem Description 有一台机器,而且给你这台机器的工作表.工作表上有n个任务,机器在ti时间运行第i个任务,1秒就可以完毕1个任务. 有m个…
BestCoder Sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1614 Accepted Submission(s): 566 Problem Description Mr Potato is a coder.Mr Potato is the BestCoder. One night, an amazing…
Sequence II Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 652 Accepted Submission(s): 164 Problem Description Long long ago, there is a sequence A with length n. All numbers in this sequenc…
Sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 712 Accepted Submission(s): 439 Problem Description Today we have a number sequence A includes n elements.Nero thinks a number sequenc…
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…
分析:大于等于m的变成1,否则变成0,预处理前缀和,枚举起点,找到第一个点前缀和大于m即可 找第一个点可以二分可以尺取 #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; typedef long long LL; ; int T,n,m,k,a[N],sum[N]; int main(){ scanf("%d…
分析:维护空隙的差,然后预处理前缀最大,后缀最大,扫一遍 #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; typedef long long LL; ; int a[N],T,n,b[N],l[N],r[N]; int main(){ scanf("%d",&T); while(T--){…
Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description Argestes has a lot of hobbies and likes solving query problems especially. One day Argestes came up with such a problem. Y…
Problem Description Today, Soda has learned a sequence whose n-th (n≥) item )+. Now he wants to know if an integer m can be represented as the sum of some items of that sequence. If possible, what are the minimum items needed? For example, =+++=+++.…
Argestes and Sequence Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 511 Accepted Submission(s): 127 Problem Description Argestes has a lot of hobbies and likes solving query problems espec…
树阵: 每个号码的前面维修比其数数少,和大量的这后一种数比他的数字 再枚举每一个位置组合一下 Sequence II Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 121 Accepted Submission(s): 58 Problem Description Long long ago, there is a sequen…
HDOJ5054 Alice and Bob Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 302 Accepted Submission(s): 229 Problem Description Bob and Alice got separated in the Square, they agreed that if they…
注意到查询次数不超过50次,那么能够从查询位置逆回去操作,就能够发现它在最初序列的位置,再逆回去就可以求得当前查询的值,对于一组数据复杂度约为O(50*n). Operation the Sequence Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 463 Accepted Submission(s): 187 Problem…
Today, Soda has learned a sequence whose n-th (n≥1) item is 3n(n−1)+1. Now he wants to know if an integer m can be represented as the sum of some items of that sequence. If possible, what are the minimum items needed? For example, 22=19+1+1+1=7+7+7+1…
Sequence Accepts: 59 Submissions: 650 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Problem Description \ \ \ \ Holion August will eat every thing he has found. \ \ \ \ Now there are many foods,but he does…
