1 #include<stdio.h>
2 #include<algorithm>
3 using namespace std;
4 const int N = 50001;
5 long long a[N];
6 int main(){
7 int n, i;
8 long long ma_ed, ans, sum = 0;
9 scanf("%d", &n);
10 for(i = 1; i <= n; ++i){
11 scanf("%I64d", &a[i]);
12 sum += a[i];
13 }
14 ma_ed = a[1];
15 ans = max(0LL, a[1]);
16 for(i = 2; i <= n; ++i){//1~n求最大子段和
17 if(ma_ed > 0)
18 ma_ed += a[i];
19 else
20 ma_ed = a[i];
21 ans = max(ma_ed, ans);
22 }
23 long long mi_ed = a[1], s = a[1];
24 //最大字段和首尾相接情况
25 for(i = 2; i <= n; ++i){
26 if(mi_ed < 0)
27 mi_ed += a[i];
28 else
29 mi_ed = a[i];
30 s = min(mi_ed, s);
31 }
32 //首尾相接时，答案为序列数的总和 与 其中数相加最小的和（负值） 之差
33 printf("%I64d\n", max(ans, sum - s));
34 return 0;
35 }
