Fence repair

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length Li (1 ≤ Li ≤ 50,000) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths Li). FJ is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made; you should ignore it, too.

FJ sadly realizes that he doesn't own a saw with which to cut the wood, so he mosies over to Farmer Don's Farm with this long board and politely asks if he may borrow a saw.

Farmer Don, a closet capitalist, doesn't lend FJ a saw but instead offers to charge Farmer John for each of the N-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.

Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the N planks. FJ knows that he can cut the board in various different orders which will result in different charges since the resulting intermediate planks are of different lengths.

Input

Line 1: One integer N, the number of planks Lines 2..N+1: Each line contains a single integer describing the length of a needed plank Output

Line 1: One integer: the minimum amount of money he must spend to make N-1 cuts Sample Input

3 8 5 8 Sample Output

34

也可以采用优先队列的方法，每次从队列取出最小的两个数，加和入队列，直到队列的元素只有一个。

typedef long long ll;

int N, L[20001];

int solve(){
  int ans = 0;

  while(N > 1){
    int m1 = 0, m2 = 1;

    if(L[m1] > L[m2]) swap(m1, m2);

    //找到当前数字中的第一小和第二小元素
    for(int i = 2; i < N; i++){
      if(L[i] < L[m1]){
        m2 = m1;
        m1 = i;
      }else if(L[i] < L[m2]){
        m2 = i;
      }
    }

    int tmp = L[m1] + L[m2];
    ans += tmp;

    if(m1 == N - 1)  swap(m1, m2);
    L[m1] = t;
    L[m2] = L[N - 1];
    N--;
  }
}