51nod 1050 循环数组最大子段和

 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 }