Fence Repair

Fence Repair
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 64801 Accepted: 21314
De ion

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
Hint

He wants to cut a board of length 21 into pieces of lengths 8, 5, and 8.
The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into 16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).

显然，越长的木板加的次数越少，则开销最少。
第一次做的时候，每次先对数组进行排序，找出a[0]+a[1]，然后sum+=a[0]+a[1]，再更新数组，木板数减一，结果TLE了；后来再尝试TOP2，还是TLE（毕竟我菜）;后来尝试了去寻找数组中最小的元素，AC；

#include<stdio.h>
int a[20000];
int main()
{
int n,i;
long long int sum=0;
scanf(\"%d\",&n);
for(i=0;i<n;i++)
{
scanf(\"%d\",&a[i]);
}
if(n==1)
{
printf(“0”);
}
else
{
while(n>1)
{
int m1=0,m2=1;
if(a[m1]>=a[m2])
{
int temp;
temp=a[m1];
a[m1]=a[m2];
a[m2]=temp;
}//保证a[m1]<=a[m2]；
for(i=2;i<n;i++)
{
if(a[i]<a[m1])
{
m2=m1;
m1=i;
}
else if(a[i]<a[m2])
{
m2=i;
}
}//找到数组中最小的两个数；
sum+=(a[m1]+a[m2]);
a[m1]=a[m1]+a[m2];
a[m2]=a[n-1];//更新数组；
n–;
}
}
printf(\"%lld\",sum);
return 0;
}