1007. Maximum Subsequence Sum (25)

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本文介绍了一道经典的编程题目——最大子列和问题，并提供了一个C++实现示例。通过对输入序列的处理，该算法能够高效地找到具有最大和的连续子序列及其起始和结束元素。

1007. Maximum Subsequence Sum (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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：

给N个数，找出其中的最大子列和及其2个首尾。MOOC浙大数据结构入门第一题，一道在线处理的水题。。。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

struct Node{
	int sum;
	int start;
	int end;
};

int main(){
	
	int N;
	int a[10005];
	scanf("%d",&N);
	int flag = 0;
	for(int i=0; i<N; i++){
		scanf("%d",a+i);
		if(a[i]>=0)
			flag = 1;
	}
	if( flag==0 ){
		printf("0 %d %d\n",a[0],a[N-1]);
		return 0;
	}
	Node max_node,tmp_node;
	tmp_node.sum = tmp_node.start = tmp_node.end = 0;
	max_node.sum = max_node.start = max_node.end = -1;
	for(int i=0; i<N; i++){
		tmp_node.sum += a[i];
		tmp_node.end = i;
		if( tmp_node.sum<0 ){
			tmp_node.sum = 0;
			tmp_node.start = tmp_node.end = i+1;
		}
		if( tmp_node.sum>max_node.sum )
			max_node = tmp_node;
	}
	printf("%d %d %d\n",max_node.sum,a[max_node.start],a[max_node.end]);
	return 0;
}