Codeforces-461A Appleman and Toastman_acm-461a-优快云博客

本文链接：https://blog.youkuaiyun.com/j_sure/article/details/38868259

本文深入探讨游戏开发过程中的关键策略与优化技巧，包括如何高效利用资源、提高游戏性能和用户体验，以及采用先进编程技术和工具提升开发效率。通过实例分析，揭示游戏引擎和框架的选择与应用策略，为开发者提供实用的指导。

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Appleman and Toastman play a game. Initially Appleman gives one group of n numbers to the Toastman, then they start to complete the following tasks:

Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman.
Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.

After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get?

Input

The first line contains a single integer n (1 ≤ n ≤ 3·105). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106) — the initial group that is given to Toastman.

Output

Print a single integer — the largest possible score.

Sample test(s)

input
3
3 1 5

output
26

input
1
10

output
10

Note

Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions.

————————————————————集训27.2的分割线————————————————————

前言：一开始以为每次都扔掉最小的数不是正解，后来证明了每次扔最小的数就是正解。

思路：如果倒过来看。假设三个数字。把最后一次扔掉当做第一次添加，假设扔掉之前不会单独添加该数。

S = A1

S = A1 + A2

S = A1 + A2 + A3

由此可见，A从大到小排序是最优的。

最后把单独的数字加回来。

代码如下：

/*
ID: j.sure.1
PROG:
LANG: C++
*/
/****************************************/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <climits>
#include <iostream>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
/****************************************/
const int N = 3e5+5;
int a[N];

bool cmp(int a, int b)
{
	return a > b;
}

int main()
{
#ifdef J_Sure
//	freopen("000.in", "r", stdin);
//	freopen(".out", "w", stdout);
#endif
	int n;
	scanf("%d", &n);
	LL ans = 0;
	for(int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
		ans += a[i];
	}
	sort(a, a+n, cmp);
	LL tmp = a[0];
	for(int i = 1; i < n; i++) {
		tmp += a[i];
		ans += tmp;
	}
	printf("%I64d\n", ans);
	return 0;
}