题意大致就是在给出的数列中不断取出 i，j 两个数，ai>aj , 然后把 ai-aj重新放入，直到不能做为止
我用了个优先队列模拟。 因为可能同时有几个最大值。。所以不能单纯的比较最大值和后面一个最大值相等就停止
题目：
Fox Ciel is playing a game with numbers now.
Ciel has n positive integers: x1, x2, ..., xn. She can do the following operation as many times as needed: select two different indexes i and j such that xi > xj hold, and then apply assignment xi = xi - xj. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
The first line contains an integer n (2?≤?n?≤?100). Then the second line contains n integers: x1, x2, ..., xn (1?≤?xi?≤?100).
Output a single integer — the required minimal sum.
2
1 2
2
3
2 4 6
6
2
12 18
12
5
45 12 27 30 18
15
In the first example the optimal way is to do the assignment: x2 = x2 - x1.
In the second example the optimal sequence of operations is: x3 = x3 - x2, x2 = x2 - x1.
代码：
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <vector> 5 #include <queue> 6 #include <stack> 7 #include <cmath> 8 #include <cstring> 9 #include <string> 10 #include <cstdlib> 11 using namespace std; 12 #define LL lolng long 13 const double pi = acos(0.) * 2; 14 priority_queue<int >q; 15 vector<int> ans; 16 int n; 17 int num[200]; 18 int main() 19 { 20 scanf("%d", &n); 21 for(int i=0;i<n;i++) 22 { 23 int d; 24 scanf("%d",&d); 25 q.push(d); 26 } 27 long long ret =0; 28 while(1) 29 { 30 int f,s; 31 f = q.top();q.pop(); 32 if( f==q.top()) 33 { 34 ans.push_back(f); 35 while(!q.empty()&&f == q.top()) 36 { 37 f= q.top(); 38 ans.push_back(q.top());q.pop(); 39 } 40 if(!q.empty()) 41 { 42 int p = ans.front(); 43 ans.pop_back(); 44 q.push( p-q.top()); 45 for(int i=0;i<ans.size();i++) 46 q.push(p); 47 ans.clear(); 48 } 49 else 50 break; 51 } 52 else 53 q.push(f-q.top()); 54 } 55 56 while(!ans.empty()) 57 { 58 ret+= ans.back(); 59 ans.pop_back(); 60 } 61 printf("%d\n",ret); 62 return 0; 63 }