uva 10954 Add All（排序）

uva 10954 Add All

Yup!! The problem name reflects your task; just add a set of numbers. But you may feel yourselves condescended, to write a C/C++ program just to add a set of numbers. Such a problem will simply question your erudition. So, let’s add some flavor of ingenuity
to it.



Addition operation requires cost now, and the cost is the summation of those two to be added. So, to add
1 and 10, you need a cost of 11. If you want to add
1, 2 and 3. There are several ways –

1 + 2 = 3, cost = 3 3 + 3 = 6, cost = 6 Total = 9

1 + 3 = 4, cost = 4 2 + 4 = 6, cost = 6 Total = 10

2 + 3 = 5, cost = 5 1 + 5 = 6, cost = 6 Total = 11

I hope you have understood already your mission, to add a set of integers so that the cost is minimal.

Input

Each test case will start with a positive number, N (2 ≤ N ≤ 5000) followed by
N positive integers (all are less than 100000). Input is terminated by a case where the value of
N is zero. This case should not be processed.

Output

For each case print the minimum total cost of addition in a single line.

Sample Input Output for Sample Input

3 1 2 3 4 1 2 3 4 0

9 19

题目大意：给出n个数，要将n个数相加，每次相加所得的值为当次的计算量，完成所有的求和运算后，要求总的计算量最小。

解题思路：每次都找出最小的两个数相加，记录这个和，然后把和放回原来的数组替换掉之前的两个数，重复这个操作直到剩下一个和，然后把之前记录的和剩下的和全部相加，就是答案。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int num[5005], sum[5005];
int main() {
	int n;
	while (scanf("%d", &n), n) {
		for (int i = 0; i < n; i++) {
			scanf("%d", &num[i]);
		}
		int cnt = 0, cnt2 = 0;
		while (cnt < n - 1) {
			sort(num + cnt, num + n);
			num[cnt + 1] += num[cnt];
			sum[cnt2++] = num[cnt + 1];
			cnt++;
		}
		int Sum = 0;
		for (int i = 0; i < cnt2; i++) {
			Sum += sum[i];
		}
		printf("%d\n", Sum);
	}
	return 0;
}