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         else20             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         else29             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 }