Problem
题目大意：
给一块长木板，现要将其锯成$n$段，共需锯$n-1$次，每次锯的代价为所锯木板的长度，求最小总代价。
Solution
弄个优先队列完事。。（STL大法好1
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using namespace std;
priority_queue<int, vector<int>, greater<int> > q;
int N;
int main()
{
long long ans = 0;
ios::sync_with_stdio(false);
cin >> N;
for (register int i = 1; i <= N; ++i)
{
int a;
cin >> a;
q.push(a);
}
while (q.size() >= 2)
{
int l1, l2;
l1 = q.top(), q.pop();
l2 = q.top(), q.pop();
ans += l1 + l2;
q.push(l1 + l2);
}
cout << ans << endl;
return 0;
}