01-复杂度2 Maximum Subsequence Sum (25 分)

01-复杂度2 Maximum Subsequence Sum (25 分)

Given a sequence of K integers A continuous subsequence is defined to be 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

Note

1. 逻辑题， 思路有些混乱， 分类情况
   start的更新： sum < 0 && 接下来的子序列比max大 , start更新为i + 1
   end的更新： sum要比max来的大，更新为i， 是大于不是大于等于保证max多次时start，end取小的
   特殊情况： 更新到最后发现sum从来就没大于零 则 输出 0 头 尾
   注：max 初始化为-1 保证能区分全负序列 和 除零以外其他全负序列

Code

#include<iostream>
using namespace std;
const int MAX = 1e4 + 10;
int a[MAX];
int main() {
    int num, sum = 0, max = -1, temp = 0;
    int start = 0, end = 0;
    cin >> num;
    for(int i = 0; i < num; i++) {
        cin >> a[i];
        sum += a[i];
        if(sum > max) { max = sum;  end = i;    start = temp;}
        else if(sum < 0 && i < num - 1){   sum = 0; temp = i + 1;}
        else if(max < 0 && i == num - 1) {max = 0; start = 0; end = num - 1;}
    }
    cout << max << " " << a[start] << " " << a[end];
    return 0;
}