51nod 1050 循环数组最大子段和

题目链接：51nod 1050 循环数组最大子段和

 #include<stdio.h>
 #include<algorithm>
 using namespace std;
 const int N = ;
 long long a[N];
 int main(){
     int n, i;
     long long ma_ed, ans, sum = ;
     scanf("%d", &n);
     for(i = ; i <= n; ++i){
         scanf("%I64d", &a[i]);
         sum += a[i];
     }
     ma_ed = a[];
     ans = max(0LL, a[]);
     for(i = ; i <= n; ++i){//1~n求最大子段和
         if(ma_ed > )
             ma_ed += a[i];
         else
             ma_ed = a[i];
         ans = max(ma_ed, ans);
     }
     long long mi_ed = a[], s = a[];
     //最大字段和首尾相接情况
     for(i = ; i <= n; ++i){
         if(mi_ed < )
             mi_ed += a[i];
         else
             mi_ed = a[i];
         s = min(mi_ed, s);
     }
     //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
     printf("%I64d\n", max(ans, sum - s));
     return ;
 }