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

題目大意：給出n個數，要將n個數相加，每次相加所得的值為當次的計算量，完成所有的求和運算後，要求總的計算量最小。

解題思路：每次都找出最小的兩個數相加，記錄這個和，然後把和放回原來的數組替換掉之前的兩個數，重復這個操作直到剩下一個和，然後把之前記錄的和剩下的和全部相加，就是答案。

#include
#include
#include
#include
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;
}