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

求最大的连续子序列的和，并输出子序列的首尾元素，就是没注意这点，一直输出首尾索引号，一直WA
思路有两种：
1）子序列的和sum = sum[j] - sum[i] (j >= i)，所以只需在求每个sum[j]的时候，找到j前面最小的那个sum[i]，即可，这里的最小的sum[i]应该与求sum[j]的时候同时维护，否则就是暴力破解。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAXN 10005
struct node{
    int num;
    int l;
    int r;
    int l_sum;
} seq[MAXN];
int sum[MAXN];
int main()
{
    int n;
    int flag = 0;
    int min;
    int index;
    min = 0;
    index = -1;
    scanf("%d", &n);
    for (int i = 0; i < n; i++){
        scanf("%d", &seq[i].num);
        if(seq[i].num >= 0)
            flag = 1;
        sum[i] = sum[i - 1] + seq[i].num;
        seq[i].l_sum = sum[i] - min;
        seq[i].l = index + 1;
        seq[i].r = i;
        if(sum[i] < min){
            min = sum[i];
            index = i;
        }
    }
    if(!flag){
        printf("0 %d %d\n", seq[0].num, seq[n - 1].num);
    }
    else{
        int max;
        int l, r;
        max = -1;
        for (int i = 0; i < n; i++){
            if(max < seq[i].l_sum){
                max = seq[i].l_sum;
                l = seq[i].l;
                r = seq[i].r;
            }
        }
        printf("%d %d %d\n", max, seq[l].num, seq[r].num);
    }
    system("pause");
    return 0;
}

View Code

   2)见https://blog.csdn.net/liuchuo/article/details/52144554