PAT 甲级1007 Maximum Subsequence Sum

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本文介绍了一种求解最大子序和问题的算法实现，通过遍历整数序列找到具有最大元素和的连续子序列，并输出该子序列的首个与末尾数值。讨论了特殊情况的处理方式，如所有数均为负数时的解决方案。

1007 Maximum Subsequence Sum（25 分）

Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } 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

分析：求给定序列的最大子序和，并输出所求序列的第一个和最后一个数，如果最大子序和小于0，则令它为0，并且输出整个序列的第一个数和最后一个数。（不知道为什么下标从1开始就错了一个点0.0）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=10005;
int main()
{
    int a[maxn];
    int k;
    scanf("%d", &k);
    int sx = 0, ex = k-1, t = 0, maxx = -1, b = 0;
    for(int i = 0;i < k; i++){
        scanf("%d", &a[i]);
        b += a[i];
        if(b < 0){
            b = 0;
            t = i + 1;
        }else if(b > maxx){
            maxx = b;
            sx = t;
            ex = i;
        }
    }
    if(maxx < 0) maxx = 0;
    printf("%d %d %d\n", maxx, a[sx], a[ex]);
    return 0;
}