int solve(int left, int right)
{
	if(left == right)
		return num[left];

	mid = (left + right) / 2;
	lans = solve(left, mid);
	rans = solve(mid + 1, right);

	sum = 0, lmax = num[mid], rmax = num[mid + 1];
	for(i = mid; i >= left; i--) {
		sum += num[i];
		if(sum > lmax) lmax = sum;
	}
	sum = 0;
	for(i = mid + 1; i <= right; i++) {
		sum += num[i];
		if(sum > rmax) rmax = sum;
	}

	ans = lmax + rmax;
	if(lans > ans) ans = lans;
	if(rans > ans) ans = rans;
	return ans;
}

int main(void)
{
	scanf("%d", &n);
	for(i = 1; i <= n; i++)
		scanf("%d", &num[i]);

	printf("%d\n", solve(1, n));
	return 0;
}
 

以上二分法的時間複雜度分析：
solve函式採用了二分法的思想，先走一遍分析一下：開頭將陣列一分為二，時間複雜度為T(n) = 2T(n/2);接下來分別從中間向左向右搜尋，時間複雜度為n，因此T(n) = 2T(n/2) + n；繼續向下迭代，T(n) = 2*(2T(n/4)+n/2)+n=2^2T(n/22)+2n；假設迭代了k次，則T(n) = 2^k T(n/2^k) + kn，最後又n/2^k = 1,得到k = log2n,帶入得T(n) = nT(1) + nlog2n。即nlog2n。