max Sum（简单动态规划）

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HDU 1003 Max Sum【动态规划求最大子序列和详解】

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…

hdu 1003 MAX SUM 简单的dp，测试样例之间输出空行

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 Max sum(简单DP)

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…

hdu 1003 Max Sum （动态规划）

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…

杭电 1003 Max Sum （动态规划）

参考:https://www.cnblogs.com/yexiaozi/p/5749338.html #include <iostream> #include <cstdio> #include <cstring> using namespace std; ; int a[N],dp[N]; int n; void test() { cout<<"\n---------\n"; ;i<=n;i++) { cout<<'[…

HDU 1024 Max Sum Plus Plus （动态规划）

HDU 1024 Max Sum Plus Plus (动态规划) 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 faced with a more difficult problem. Given…

HDU-1003 Max Sum(动态规划，最长字段和问题)

Max Sum Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 193355 Accepted Submission(s): 45045 Problem Description Given a sequence a[1],a[2],a[3]--a[n], your job is to calculate the max sum of a su…