一. 题目描述

给出一个长度为\(n\)的序列\(a\)，选出其中连续且非空的一段使得这段和最大。

二. 题目解答

1. 分治法

算法思路：

给定一段数组\(A[low..high]\), 它的最大子数组所处的位置有三种情况:

• 完全位于左子数组中，即\(A[low..mid]\)
• 完全位于右子数组中，即\(A[mid+1..right]\)
• 跨越中点，即\(A[maxLeft..mid..maxRight]\),其中\(maxLeft\)与\(maxRight\)分别是跨越中点的最大子数组的左右下标
  

算法代码：

#pragma warning(disable:4996)
#include<iostream>
#include<cstdio>
#include<climits>
#include<cfloat>

using namespace std;
typedef long long ll;

const int maxn = 2e5 + 5;
int n;
ll a[maxn];

ll findCrossMax(int low, int mid, int high) {
	ll leftSum = LONG_MIN;
	ll sum = 0;
	for (int i = mid; i >= low; i--) {
		sum += a[i];
		if (sum > leftSum) {
			leftSum = sum;
		}
	}
	ll rightSum = LONG_MIN;
	sum = 0;
	for (int i = mid + 1; i <= high; i++) {
		sum += a[i];
		if (sum > rightSum) {
			rightSum = sum;
		}
	}
	return leftSum + rightSum;
}


ll findMax(int low, int high) {
	if (low == high) {
		return a[low];
	}
	else {
		int mid = (low + high) / 2;
		ll leftResult = findMax(low, mid);
		ll rightResult = findMax(mid + 1, high);
		ll crossResult = findCrossMax(low, mid, high);
		return max(leftResult,max( crossResult, rightResult));
	}
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%lld", &a[i]);
	}
	ll ans = findMax(1, n);
	printf("%lld", ans);
	return 0;
}