HDU 1003:Max Sum】的更多相关文章

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): 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=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…
Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 36673    Accepted Submission(s): 13069 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" proble…
Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 21336    Accepted Submission(s): 7130 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…
题 题意 需要在o(n)时间内,求最大连续的子序列的和,及其起点和终点. 分析 一种方法是一边读,一边维护最小的前缀和 s[i] ,然后不断更新 ans = max(ans,s[j] - s[i]),以及起始位置. 另一种方法是尺取(算是吧),l 和 r 代表起点和终点,一开始l=0,r=1,如果s[r]-s[l]>=0那就往右扫 r++,不断更新  ans = max(ans,s[r] - s[l]),以及起始位置,如果小于0了,那就舍弃前面这段了,也就是后面必然不用考虑它更划算,l=r,r+…
题目链接:Max Sum Plus Plus Plus 题意:在n个数中取m段数使得这m段数之和最大,段与段之间不能重叠 分析:见代码 //dp[i][j]表示前i个数取了j段的最大值 //状态转移:dp[i][j]=max(dp[k][j-1]+(sum[k+l[j]-sum[k]或者sum[i]-sum[i-l[j]) (0<=k<=i-l[j]) // 这种做法是O(n^2)的 //如果改成O(n)的呢? //需要加一个数组max_dp[i][j]表示前i个数取j段的dp最大值 //如果…
Problem Description Given a circle sequence A[1],A[2],A[3]......A[n]. Circle sequence means the left neighbour of A[1] is A[n] , and the right neighbour of A[n] is A[1]. Now your job is to calculate the max sum of a Max-K-sub-sequence. Max-K-sub-sequ…
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…
Max Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 287192    Accepted Submission(s): 68202 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max su…