Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14. Input The first line of the input contains an inte…

题目链接http://acm.hdu.edu.cn/showproblem.php?pid=3415 大意是给出一个有n个数字的环状序列,让你求一个和最大的连续子序列.这个连续子序列的长度小于等于k. 由于是环状的数列,一般采取的措施是在数列的后面再复制一部分前面的数列;a[n]=a[0],a[n+1]=a[1]---- 用sum数组存储前i个数的和,那么对于j属于范围(i-k,i-1)来说,最大连续子序列就等于sum[i]减去在该范围内最小的sum[j] 那么问题就变成了求在每k个范围内最小的…

测试样例之间输出空行,if(t>0) cout<<endl; 这样出最后一组测试样例之外,其它么每组测试样例之后都会输出一个空行. dp[i]表示以a[i]结尾的最大值,则:dp[i]=max(dp[i]+a[i],a[i]) 解释: 以a[i]结尾的最大值,要么是以a[i-1]为结尾的最大值+a[i],要么是a[i]自己本身,就是说,要么是连同之前的 构成一个多项的字串,要么自己单独作为一个字串,不会有其他的可能了. 状态规划的对状态的要求是:当前状态只与之前的状态有关,而且不影响下一…

HDU 1003    相关链接   HDU 1231题解 题目大意:给定序列个数n及n个数,求该序列的最大连续子序列的和,要求输出最大连续子序列的和以及子序列的首位位置 解题思路:经典DP,可以定义dp[i]表示以a[i]为结尾的子序列的和的最大值,因而最大连续子序列及为dp数组中的最大值.   状态转移方程:dp[1] = a[1]; //以a[1]为结尾的子序列只有a[1]:  i >= 2时, dp[i] = max( dp[i-1]+a[i],  a[i] ); dp[i-1]+a[i…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 250714    Accepted Submission(s): 59365 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=1003 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 180817    Accepted Submission(s): 42261 Problem Description Given a sequence a[1],a[2],a[3]....…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003 Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 158421    Accepted Submission(s): 37055 Problem Description Given a sequence a[1],a[2]…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 141547    Accepted Submission(s): 32929 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max s…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the m…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 211310    Accepted Submission(s): 49611 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

题目大意:求一串数字中,几个连续数字加起来最大值,并确定起始和最末的位置. 思路:这是一题DP题,但是可以用尺取法来做.我一开始不会,也是看了某大神的代码,然后有人告诉我这是尺取法,现在会了. //尺取法 #include<stdio.h> #include<string.h> ]; main() { int t,flag; scanf("%d",&t); flag=t; while(t--) { memset(que,,sizeof(que)); ;…

#include <stdio.h> int main(){ int i,t,j,n,x; int start,end,temp,max,sum; scanf("%d",&t); ;i<t;i++){ temp=; max=-; sum=; scanf("%d",&n); ;j<n;j++){ scanf("%d",&x); sum+=x; if(sum>=max){ max=sum; sta…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 154155    Accepted Submission(s): 35958 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.   Input The first line of the input contai…

点我看题目 题意 : 就是让你从一个数列中找连续的数字要求他们的和最大. 思路 : 往前加然后再判断一下就行. #include <iostream> #include<stdio.h> using namespace std; int main() { int n,start,end; cin>>n; int m ; ; k <= n ; k++) { cin>>m; ,sum = ,flag = ; ; i <= m- ; i++) { in…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 135262    Accepted Submission(s): 31311 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max s…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 161294    Accepted Submission(s): 37775 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max s…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 294096    Accepted Submission(s): 69830 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max s…

转载于acm之家http://www.acmerblog.com/hdu-1003-Max-Sum-1258.html Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 242353    Accepted Submission(s): 57218 Problem Description Given a sequence…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 292444    Accepted Submission(s): 69379 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max s…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 72615    Accepted Submission(s): 16626 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum…

 #include <iostream> #include<stdio.h> #include<stdlib.h> using namespace std; int main() {     int t;     scanf("%d",&t);     for(int i=0;i<t;i++)     {         int n;         scanf("%d",&n);         int…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 237978    Accepted Submission(s): 56166 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 330535    Accepted Submission(s): 78678 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…

Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 174588    Accepted Submission(s): 40639 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max s…

http://acm.hdu.edu.cn/showproblem.php?pid=1003 给你一串数列,让你求出其中 一段连续的子数列 并且 该子数列所有数字之和最大,输出该子数列的和.起点与终点序号. 具体细节不再赘述,看原题就好了. 说实话我觉得这个题有点难...想了好久,智商不够... 但实际的代码还算是相当简单的,主要的逻辑就在中间那两个 if 里了,仔细看看就能理解了. #include<stdio.h> #include<string.h> int main() {…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1003 题目意思: 即给出一串数据,求连续的子序列的最大和 解题思路: 因为我们很容易想到用一个max来存放找到的子序列的和中的最大值,通过不断比较,对max的值进行更新,最后我们就能够得到最大子序列的和,于是很容易想到用暴力搜索,见上一篇博客,这样的时间复杂度为O(n^3),是超时的. 又因为想到只要一个数不是负数,不管它再小,加上去也是会使和变大的,所以我们需要用另一个变量来判断即将要加上的一个…

A - Max Sum Plus Plus Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1024 Appoint description: Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a bra…

http://acm.hdu.edu.cn/showproblem.php?pid=1024 Max Sum Plus Plus Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are fac…