1007 Maximum Subsequence Sum (25)（25 分）

1007 Maximum Subsequence Sum (25)（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


绝对不能马虎......
刚开始一看到 最大字段和 同时输出 字段的起始和终止位置 就不读题目了
最后错了两次 ， 才发现： 是输出 最大和 加 最大字段和 起始处的数字 和 结束位置的数字 ；
如果 整个数列 全部为负数 ，则最大和为 0 ， 字段和的起始位置 为 0 ， 终止位置为 n - 1 ; 

如果前缀和为负数则不能给后面的 子段和 带来增益， 所以置 零

#include <iostream>
#include <algorithm>

using namespace std ; 

#define maxn 20000
#define LL long long
int n ; 
LL num[maxn] ; 
LL maxSum ; 
int startPos , endPos ; 
int tempStartPos ; 

int main(){

    while(cin >> n){
        maxSum = -1 ; 
        startPos = tempStartPos = 0 ; 
        endPos = n - 1 ; 

        LL sum = 0 ; 
　　　　 
        for(int i=0 ; i<n ; i++){
            cin >> num[i] ; 
            
            sum += num[i] ;
            if(sum < 0){
                sum = 0 ; 
                // start pos 
                tempStartPos = i + 1 ; 
            }
            else if(sum > maxSum){
                maxSum = sum ; 
                startPos = tempStartPos ; 
                endPos = i ; 
            }     
        }
        if(maxSum < 0) maxSum = 0 ; 

        cout << maxSum << " " << num[startPos] << " " << num[endPos] << endl ; 
    }
    return 0 ; 
}