vjudge 贪心+分治专题-N题

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, lets 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 1 + 3 = 4, cost = 4 2 + 3 = 5, cost = 5
3 + 3 = 6, cost = 6 2 + 4 = 6, cost = 6 1 + 5 = 6, cost = 6
Total = 9 Total = 10 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
3
1 2 3
4
1 2 3 4
0
Sample Output
9

19

解决方案：一开始思路就有点错了，要先加最小的那两个，然后这两个数就替换成了两者之和，再从新的数组里寻找两个最小的。
如果每次都对新的数组排序的话，会很耗时，所以可用优先队列。

1.优先队列：#include <cstdio>  #include <cstring>  #include <queue>  #include <vector>  #include <algorithm>  using namespace std; struct cmp{    bool operator()(int x,int y)  {      return x>y;   //从小到大排序。即x小的优先级高。  }  };   int main() {      int n;      while(scanf("%d",&n) != EOF && n) {      priority_queue<int,vector<int>,cmp>q;         // priority_queue<int,vector<int>,greater<int> > q;          int num;              for(int i = 0; i < n; i++) {              scanf("%d",&num);              q.push(num);          }          int cost,sum = 0;          int tmp1,tmp2;          while(q.size() >= 2) {              tmp1 = q.top();              q.pop();              tmp2 = q.top();              q.pop();              cost = tmp1 + tmp2;              sum += cost;              q.push(cost);          }          printf("%d\n",sum);      }      return 0;  }  2.每次sort排序：#include<stdio.h>#include<algorithm>using namespace std;int main(){int n,sum1;int a[50010];int b[50010];while(scanf("%d",&n),n){sum1=0;for(int i=0;i<n;i++){scanf("%d",&a[i]);}int cnt=0,cnt2=0;while(cnt<n-1){sort(a+cnt,a+n);a[cnt+1]+=a[cnt];b[cnt2++]=a[cnt+1];cnt++;}for(int i=0;i<cnt2;i++){sum1+=b[i];}printf("%d\n",sum1);}return 0;}