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:

絕對不能馬虎......
剛開始一看到 最大字段和 同時輸出 字段的起始和終止位置 就不讀題目了
最後錯了兩次 ， 才發現： 是輸出 最大和 加 最大字段和 起始處的數字 和 結束位置的數字 ；
若是 整個數列 所有爲負數 ，則最大和爲 0 ， 字段和的起始位置 爲 0 ， 終止位置爲 n - 1 ; 

若是前綴和爲負數則不能給後面的 子段和 帶來增益， 因此置 零10 1 4

#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 ; 
}