1007 Maximum Subsequence Sum（25分）

Given a sequence of K integers { N1​, N2​, ..., NK​ }. A continuous subsequence is defined to be { Ni​, Ni+1​, ..., Nj​ } where 1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (≤10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

10
-10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main(){
 4     int k;
 5     int n[10005],sum;
 6     int head,tail,mmax;
 7     cin>>k;
 8     for(int i=1;i<=k;++i){
 9         cin>>n[i];
10     }
11     n[0]=-1;
12     head=1;tail=1;
13     mmax=-1;
14     for (int i=1;i<=k;++i){ 
15         //每到一个n[i]为正数且n[i-1]为负数的地方就进行一轮计算
16         if (n[i]>=0&&n[i-1]<0){ 
17             sum=0;
18             for (int j=i;j<=k;++j){
19                 sum+=n[j];
20                 if (mmax<sum){
21                     mmax=sum;
22                     head=n[i];
23                     tail=n[j];
24                 }
25             }
26         }
27     }
28     if (mmax<0){
29         cout<<"0 "<<n[1]<<" "<<n[k];
30     }
31     else{
32         cout<<mmax<<" "<<head<<" "<<tail;
33     }
34 }

微量级dp，一开始开了个数组存取临时sum，ac后想了一下发现完全没必要，因为sum是线性变化，直接用一个变量就可以表示了，没有dp数组总觉得像暴力...

剪枝条件为最大子序列的两端必为正数。