石子合并

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...

#include<bits/stdc++.h>
using namespace std;

const int MAXN = 1e6 + 5 ;

int n , N , min_ans = MAXN , max_ans;
int a[401] ;
int f_max[401][401] , f_min[401][401];

int read(){
    char c ;
    int sign = 1 ;
    while((c = getchar()) < '0' || c > '9') 
        if(c == '-') sign = -1 ;
    int Ans = c - '0' ;
    while((c = getchar()) >= '0' && c <= '9')
        Ans = Ans * 10 + c - '0' ;
    return Ans * sign ;
}

int main(){
    n = read() ;
    N = (n << 1) ;
    for(int i = 1 ; i <= n ; ++ i){
        a[i] = a[i + n] = read() ;
    }
    for(int i = 2 ; i <= N ; ++ i){
        a[i] += a[i - 1] ;
    }
    for(int i = 2 ; i <= n ; ++ i){
        for(int j = 1 ; j + i - 1 <= N ; ++ j){
            int end = j + i - 1 ;
            f_min[j][end] = MAXN , f_max[j][end] = 0 ;
            for(int k = j ; k <= end - 1 ; ++ k){
                f_min[j][end] = min(f_min[j][end] , f_min[j][k] + f_min[k + 1][end] + a[end] - a[j - 1]) ;
                f_max[j][end] = max(f_max[j][end] , f_max[j][k] + f_max[k + 1][end] + a[end] - a[j - 1]) ;
            }
        }
    }
    for(int i = 1 ; i <= n ; ++ i){
        min_ans = min(min_ans , f_min[i][i + n - 1]) ;
        max_ans = max(max_ans , f_max[i][i + n - 1]) ;
    }
    cout << min_ans << endl << max_ans ;
    return 0 ;
}