1007. Maximum Subsequence Sum (25)

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最新推荐文章于 2024-09-23 14:10:40 发布

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本文探讨了最大子序列和问题的解决方案，采用了一种高效算法来找出具有最大和的连续子序列及其边界元素。文章提供了详细的输入输出规格，并附带了一个示例案例。

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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

#include<iostream>

using namespace std;
int main() {
	int N;
	cin >> N;
	long long *number = new long long[N];
	long long sque=0, max = -100000000;
	int i, sbegin, send,j,k;
	for (i = 0; i < N; i++) {
		cin >> number[i];
		if (number[i] > max) {
			max = number[i];        //判断是否都是负数或者最大是否为0
			 sbegin=i;
		}
	}
	if (max == 0)
		cout << max << " " << 0 << " " << 0 << endl;
	else if (max < 0)				//注意题设说满足这个条件是都是负数的时候，有0是不包括的
		cout << 0 << " " << number[0] << " " << number[N - 1] << endl;
	else {
		sque = 0; sbegin = 0; send = 0; max = 0;//利用 陈越老师ppt里的那种方法求解
		for (i = 0; i < N; i++) {
			sque += number[i];
			send = i;
			if (sque < 0) {
				sque = 0;
				sbegin = i + 1;
			}
			if (sque > max) {
				max = sque;
				j = sbegin;
				k = send;
			}
		}
		cout << max << " " << number[j] << " " << number[k];//输出的不是下标，是值
	}
}

感想：

1.注意要用数据结构课里介绍的方法

2.所有坑过我的点都标注出来了