#include<stdio.h>
int MaxSubseqSum1(int a[],int n);
int MaxSubseqSum2(int a[],int n);
int MaxSubseqSum3(int a[],int n);
int MaxSubseqSum4(int a[],int n);
int main(void){
int n;scanf("%d",&n);
int a[n];
int i;for(i=0;i<n;i++){scanf("%d",&a[i]);}
int max;// max=MaxSubseqSum1(a,n);// max=MaxSubseqSum2(a,n);// max=MaxSubseqSum3(a,n);
max=MaxSubseqSum4(a,n);printf("%d\n",max);return0;}
int MaxSubseqSum1(int a[],int n){
int i,j,k;
int sum;
int max=0;for(i=0;i<n;i++){for(j=i;j<n;j++){
sum=0;for(k=i;k<=j;k++){
sum+=a[k];}if(sum>max){
max=sum;}}}return max;}
int MaxSubseqSum2(int a[],int n){
int i,j;
int sum;
int max=0;for(i=0;i<n;i++){
sum=0;for(j=i;j<n;j++){
sum+=a[j];if(sum>max){
max=sum;}}}return max;}
int Max3( int A, int B, int C){/* 返回3个整数中的最大值 */returnA>B?A>C?A:C:B>C?B:C;}
int DivideAndConquer( int List[], int left, int right ){/* 分治法求List[left]到List[right]的最大子列和 */
int MaxLeftSum, MaxRightSum;/* 存放左右子问题的解 */
int MaxLeftBorderSum, MaxRightBorderSum;/*存放跨分界线的结果*/
int LeftBorderSum, RightBorderSum;
int center, i;if( left == right ){/* 递归的终止条件,子列只有1个数字 */if( List[left]>0)return List[left];elsereturn0;}/* 下面是"分"的过程 */
center =( left + right )/2;/* 找到中分点 *//* 递归求得两边子列的最大和 */
MaxLeftSum =DivideAndConquer( List, left, center );
MaxRightSum =DivideAndConquer( List, center+1, right );/* 下面求跨分界线的最大子列和 */
MaxLeftBorderSum =0; LeftBorderSum =0;for( i=center; i>=left; i--){/* 从中线向左扫描 */
LeftBorderSum += List[i];if( LeftBorderSum > MaxLeftBorderSum )
MaxLeftBorderSum = LeftBorderSum;}/* 左边扫描结束 */
MaxRightBorderSum =0; RightBorderSum =0;for( i=center+1; i<=right; i++){/* 从中线向右扫描 */
RightBorderSum += List[i];if( RightBorderSum > MaxRightBorderSum )
MaxRightBorderSum = RightBorderSum;}/* 右边扫描结束 *//* 下面返回"治"的结果 */returnMax3( MaxLeftSum, MaxRightSum, MaxLeftBorderSum + MaxRightBorderSum );}
int MaxSubseqSum3( int a[], int n ){/* 保持与前2种算法相同的函数接口 */returnDivideAndConquer( a,0, n-1);}
int MaxSubseqSum4(int a[],int n){
int i;
int sum=0,max=0;for(i=0;i<n;i++){
sum+=a[i];if(sum>max){
max=sum;}if(sum<0){
sum=0;}}return max;}