【PAT甲级】1007 Maximum Subsequence Sum （25 分）

Given a sequence of K integers { N​1​​, N​2​​, ..., N​K​​ }. A continuous subsequence is defined to be { N​i​​, N​i+1​​, ..., N​j​​ } 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

【翻译】 ：输入K，表示接下来有K个数，在这有K个数的序列中，找出一个子序列，这个子序列的和最大，长度尽量长。

输出序列和，起点的数，终点的数（不是起点终点的下标），行尾没有空格，如果最大和不唯一，输出更小的下标时的数，如果最大和小于0，输出0.

【思路】：temp是当前序列和，sum是最大序列和，left,right记载序列起点终点下标，tempindex记载当前序列和的起点，temp从序列头头开始加，如果temp<0那么前面的都丢弃，因为负数只会减少序列和，此时更新tempindex下标，如果temp>0了，这时候再看他是否大于sum，如果比之前得出的最大和大，那么就更新最大和，和最大和情况下的左右下标。

#include 
using namespace std;
int a[10002];
 
int main(){
	int k;
	cin>>k;
	
	int left = 0,right = k - 1,temp = 0,tempindex = 0,sum = -1;
	
	for(int i = 0; i < k; i++){
		cin>>a[i];
		temp += a[i];
		if(temp < 0){
			temp = 0;//
			tempindex = i + 1;
		}else if(temp > sum){
			left = tempindex;
			right = i;
			sum = temp;
		}
	}
	if(sum < 0) sum = 0;
	cout<