这个要用动态规划写...

然后合并果子又要用贪心...

石子和果子...

大概编题目的老师也有苦衷的吧...

1 #include
     
       2 using namespace std; 3  4 const int MAXN = 1e6 + 5 ; 5  6 int n , N , min_ans = MAXN , max_ans; 7 int a[401] ; 8 int f_max[401][401] , f_min[401][401]; 9 10 int read(){11     char c ;12     int sign = 1 ;13     while((c = getchar()) < '0' || c > '9') 14         if(c == '-') sign = -1 ;15     int Ans = c - '0' ;16     while((c = getchar()) >= '0' && c <= '9')17         Ans = Ans * 10 + c - '0' ;18     return Ans * sign ;19 }20 21 int main(){22     n = read() ;23     N = (n << 1) ;24     for(int i = 1 ; i <= n ; ++ i){25         a[i] = a[i + n] = read() ;26     }27     for(int i = 2 ; i <= N ; ++ i){28         a[i] += a[i - 1] ;29     }30     for(int i = 2 ; i <= n ; ++ i){31         for(int j = 1 ; j + i - 1 <= N ; ++ j){32             int end = j + i - 1 ;33             f_min[j][end] = MAXN , f_max[j][end] = 0 ;34             for(int k = j ; k <= end - 1 ; ++ k){35                 f_min[j][end] = min(f_min[j][end] , f_min[j][k] + f_min[k + 1][end] + a[end] - a[j - 1]) ;36                 f_max[j][end] = max(f_max[j][end] , f_max[j][k] + f_max[k + 1][end] + a[end] - a[j - 1]) ;37             }38         }39     }40     for(int i = 1 ; i <= n ; ++ i){41         min_ans = min(min_ans , f_min[i][i + n - 1]) ;42         max_ans = max(max_ans , f_max[i][i + n - 1]) ;43     }44     cout << min_ans << endl << max_ans ;45     return 0 ;46 }