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 = 50001;
long long a[N];
int main(){
    int n, i;
    long long ma_ed, ans, sum = 0;
    scanf("%d", &n);
    for(i = 1; i <= n; ++i){
        scanf("%I64d", &a[i]);
        sum += a[i];
    }
    ma_ed = a[1];
    ans = max(0LL, a[1]);
    for(i = 2; i <= n; ++i){//1~n求最大子段和
        if(ma_ed > 0)
            ma_ed += a[i];
        else
            ma_ed = a[i];
        ans = max(ma_ed, ans);
    }
    long long mi_ed = a[1], s = a[1];
    //最大字段和首尾相接情况
    for(i = 2; i <= n; ++i){
        if(mi_ed < 0)
            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 0;
}