01-复杂度2 Maximum Subsequence Sum (25分)

Given a sequence of KK integers{ N_1N​1​​, N_2N​2​​,..., N_KN​K​​ }.A continuous subsequence is defined to be { N_iN​i​​, N_{i+1}N​i+1​​,..., N_jN​j​​ }where 1 \le i \le j \le K1≤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 KK (\le10000≤10000).The second line contains KK 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 subsequenceis not unique, output the one with the smallest indices ii and jj (asshown by the sample case). If all the KK numbersare 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

思路：其实就是最大子列和问题的延伸，同样的套路，只是需要记录下来最大子列的首尾项并输出

但要注意的是，这道题在不同平台上的测试数据不尽相同，尤其当有零项的时候较为让人困惑

例如当序列为0 -1 1 0 1 1 1 -1 时，最大和为3，按照下面的算法，可得首项为0（下标为3），而实际上的首项是1（下标为4）？还是首项为0，（下标为0）？下标从0开始。

这个问题我也不知道该怎么回答，根据pta上的测试用例来看，下面的代码是正确的

代码：

#include<cstdio>#define maxsize 100010void Maxsum(int A[],int N){int thissum=0,maxsum=0;int first=0,last=N;           //定义首尾项的下标int flag=0;                   //flag来判断是否全为非零项for(int i=0;i<N;i++){thissum+=A[i];if(A[i]>=0) flag=1;if(thissum>maxsum){maxsum=thissum;last=i;               //先记录末项的下标，并由末项下标来推断首项下标}else if(thissum<0){thissum=0;}}int cnt=0;for(int i=last;i>=0;i--){cnt+=A[i];                           if(cnt==maxsum){first=i;                  //找到首项  这一步可能有较大的歧义}}if(flag==0) printf("0 %d %d",A[0],A[N-1]);else{if(maxsum==0) printf("0 0 0");else printf("%d %d %d\n",maxsum,A[first],A[last]);}} int main(){int K;scanf("%d",&K);int a[maxsize]={0};for(int i=0;i<K;i++){scanf("%d",&a[i]);}int max;Maxsum(a,K);return 0;}