hdu 5586 Sum【dp最大子段和】】的更多相关文章

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5586 Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 677    Accepted Submission(s): 358 Problem Description There is a number sequence A1,A2...…
Sum Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5586 Description There is a number sequence A1,A2....An,you can select a interval [l,r] or not,all the numbers Ai(l≤i≤r) will become f(Ai).f(x)=(1890x+143)mod1…
Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description There is a number sequence A1,A2....An,you can select a interval [l,r] or not,all the numbers Ai(l≤i≤r) will become f(Ai).f(x)=(1890x+143)mod1…
Problem Description There )mod10007.After that,the sum of n numbers should be as much as possible.What is the maximum sum? Input There are multiple test cases. First line of each ≤n≤) Next line contains n integers A1,A2....An.(≤Ai≤) It's guaranteed t…
Maximum sum Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 39599   Accepted: 12370 Description Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below: Your task is to calculate d(A). Input The input consists o…
Problem Description There is a number sequence A1,A2....An,you can select a interval [l,r] or not,all the numbers Ai(l≤i≤r) will become f(Ai).f(x)=(1890x+143)mod10007.After that,the sum of n numbers should be as much as possible.What is the maximum s…
题意:将数组A的部分区间值按照函数f(Ai)=(1890*Ai+143)mod10007修改值,区间长度可以为0,问该操作后数组A的最大值. 分析:先求出每个元素的增量,进而求出增量和.通过b[r]-b[l-1]求区间增量和,枚举r,而b[l-1]则是b[r]前所有元素的最小值,注意mi初始化为0,因为当前有可能的最优值为区间0~r. #pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio>…
HDOJ(HDU).1003 Max Sum (DP) 点我挑战题目 算法学习-–动态规划初探 题意分析 给出一段数字序列,求出最大连续子段和.典型的动态规划问题. 用数组a表示存储的数字序列,sum表示当前子段和,maxsum表示最大子段和.不妨设想:当sum为负数的时候: 1.当下一个数字a[i]为正数的时候,sum+a[i] < a[i],不如将sum归零重新计算 2.当下一个数字为负数的时候,sum+a[i]< 0 ,若再下一个数字还为负数,依旧可以得出和小于零--直到遇到一个正数,此…
Reference:http://www.cnblogs.com/wuyiqi/archive/2012/03/28/2420916.html 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2836 题目大意:计算序列有多少种组合,每个组合至少两个数,使得组合中相邻两个数之差不超过H,序列有重复的数.MOD 9901. 解题思路: 朴素O(n^2) 以样例1 3 7 5为例,如果没有重复的数. 设dp[i]前n个数的方案数,初始化为1. 那么fo…
不好理解,先多做几个再看 此题是很基础的斜率DP的入门题. 题意很清楚,就是输出序列a[n],每连续输出的费用是连续输出的数字和的平方加上常数M 让我们求这个费用的最小值. 设dp[i]表示输出前i个的最小费用,那么有如下的DP方程: dp[i]= min{ dp[j]+(sum[i]-sum[j])^2 +M }  0<j<i 其中 sum[i]表示数字的前i项和. 相信都能理解上面的方程. 直接求解上面的方程的话复杂度是O(n^2) 对于500000的规模显然是超时的.下面讲解下如何用斜率…