1007 Maximum Subsequence Sum (25分)（一个字短！）

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本文详细解析了求解最大子序列和问题的算法，通过分析连续子序列中具有最大元素和的序列，并提供了一个具体的代码实现案例。该算法适用于处理一系列整数，找出其中连续子序列的最大和及其起始和结束元素。

1007 Maximum Subsequence Sum (25分)

题目

Given a sequence of K integers { N₁, N₂, …, 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

题意

求最大和子序列

分析

若从i到j的序列和为正，则以第j+1个元素结尾的子序列的最大和为从i到j+1的子序列的和；若从i到j的序列和为负，则以第j+1个元素结尾的子序列的最大和为从j+1到j+1的子序列的和

代码

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main() {
 	int n, i, j = 0, m;
 	scanf("%d", &n);
 	vector<int> v(n), vv(n);
 	for (i = 0; i < n; ++i)scanf("%d", &v[i]);
 	m = vv[0] = v[0];
 	for (i = 1; i < n; i++) {
  		vv[i] = max(v[i], v[i] + vv[i - 1]);
  		if (m < vv[i]) {
   			m = vv[i];
   			j = i;
  		}
 	}
 	if (m < 0) {
  		printf("0 %d %d", v[0], v[n - 1]);
  		return 0;
 	}
 	for (i = j; i >= 0; --i) {
  		if (vv[i] < 0)break;
 	}
 	printf("%d %d %d", i < 0 ? vv[j] : vv[j], v[i + 1], v[j]);
 	return 0;
}