題目：https://pta.patest.cn/pta/test/15/exam/4/question/709

#include <iostream>
using namespace std;

int MaxSubseqSum2(int A[], int N);
int MaxSubseqSum3(int A[], int left, int right);		//(陣列名,元素左起0,元素終點n-1)
int MaxSubseqSum4(int A[], int N);			//(陣列名, 元素個數)


int main()
{
	int n;
	cin >> n;
	// int numSeq[n];				//C++是不支援用變數初始化陣列的
	int numSeq[100000];

	for(int i=0; i<n; i++){
		cin >> numSeq[i];
	}

	// int maxSum = MaxSubseqSum4(numSeq, n);
	int maxSum = MaxSubseqSum3(numSeq, 0, n-1);
	if(maxSum< 0) {
		maxSum = 0;
	}
	cout<<maxSum<<endl;

	return 0;
}


// 演算法2  窮舉法
int MaxSubseqSum2(int A[], int N){	
	int ThisSum, MaxSum = 0;
	int i, j, k;
	for(i=0; i<N; i++){						//i是子列左端位置
		ThisSum = 0;						//ThisSum是從A[i]到A[j]的子列和
		for(j=i; j<N; j++){					//j是子列右端位置
			ThisSum += A[j];
			//對於相同的i，不同的j，只要在j-1次迴圈的基礎上累加1項即可
			if(ThisSum > MaxSum)			//如果剛得到的這個子列和更大
				MaxSum = ThisSum;			//則更新結果
		}	//j迴圈結束
	}	//i迴圈結束
	return MaxSum;
}

int compareThreeNums(int a, int b, int c){
	return a>b ? (a>c?a:c) : (b>c?b:c)  ;		//if a>b: then if a>c, else if b>c
}
// 演算法3  分而治之
int MaxSubseqSum3(int A[], int left, int right){
	int leftMax, rightMax;

	//如果list中只剩一個元素了，則left==right
	if(left == right){
		return A[left];
	}
	//list不只一個元素，二分+治之 遞迴，獲得左右最大
	int middle = (left + right) / 2;
	leftMax = MaxSubseqSum3(A, left, middle);
	rightMax = MaxSubseqSum3(A, middle+1, right);
	
	//尋找跨越邊界的最大
	// int tempLeftMax, leftBorderMax = 0;						//未顯式賦值，造成錯誤假內容
	// int tempRightMax, rightBorderMax = 0;
	int tempLeftMax = 0, leftBorderMax = 0;				
	for (int i = middle; i >= left; i--){
		tempLeftMax += A[i];
		if(tempLeftMax > leftBorderMax){
			leftBorderMax = tempLeftMax;	
		}
	}
	int tempRightMax = 0, rightBorderMax = 0;
	for (int i = middle+1; i <= right; i++){
		tempRightMax += A[i];
		if(tempRightMax > rightBorderMax){
			rightBorderMax = tempRightMax;	
		}
	}
	int middleMax = rightBorderMax + leftBorderMax;

	int maxSum = compareThreeNums(middleMax, leftMax, rightMax);
	return maxSum;
}

// 演算法4  線上方法/動態規劃  
int MaxSubseqSum4(int A[], int N){
	int maxSum, thisSum = 0;
	int i = 0;
	for(i = 0; i<N; i++){
		thisSum += A[i];					/* 向右累加 */
		if(thisSum > maxSum){
			maxSum = thisSum;				/* 發現更大和則更新當前結果 */ 
		} else if (thisSum < 0){			/* 如果當前子列和為負 */
			thisSum = 0;					/* 則不可能使後面的部分和增大，拋棄之 */
		}									
	}
	//此題裡，不需要記錄最大和的具體項，發現新最大立刻更新，當前和已經小於0時，則加後面的大項必然可以大過加已有的負數
	return maxSum;
}